Part-Contents
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Thermodynamic concepts
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Prelude
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The Legendre transform
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Euler’s Theorem on Homogeneous Functions
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Postulates and Definitions
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Equations of state
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State changes at constant composition
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Closed control volumes
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Dynamic systems
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Open control volumes
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Thermodynamic methods TKP4175
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Tore Haug-Warberg
Department of Chemical Engineering
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Exercise 1
Table of contents
Exercise 156
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min(H )p ⇔ min (U )
S ,N
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List of symbols
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Tore Haug-Warberg (Chem Eng, NTNU)
S ,V ,N
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Part-Contents (2)
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Part-Contents (3)
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Departure Functions
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Simple vapour–liquid equilibrium
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Multicomponent phase equilibrium
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Chemical equilibrium
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Simultaneous reactions
.........................................
Heat engines
...............................................
Entropy production and available work
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Plug flow reactor
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Material Stability
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Thermofluids
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T , s and p , v -Diagrams
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Gas dynamics
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708
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Direct Substitution
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Linear Programming
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Nine Concepts of Mathematics
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Code Snippets
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SI units and Universal Constants
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Newton–Raphson iteration
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Part
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Thermodynamic concepts
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see also Part-Contents
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Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
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News Olds
ENGLISH text is missing.The language
New concepts
1
System
2
State
3
Process
2 Thermodynamic . . .
Thermodynamics
8
Since the rise of thermodynamics over the period 1840–1880, the 5 Some of the concepts are concrete, whilst others are abstract, and
original beliefs have been replaced by a more (ope)rational interpreta- it is by no means easy to understand them on a first read-through.
tion.
6
However, the aim is to convince readers that a good grasp of the
9
Meanwhile, the content has become so well-established and thor- basic concepts is within reach, and that such a grasp is a prerequisite
oughly tested that there is little room for innovation, although there is for mastering the remaining chapters of the book.
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still scope to inject fresh life into old ideas.
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[a] James A. Beattie and Irwin Oppenheim. Principles of Thermodynamics. Elsevier,
[a] As illustrated by Professor Gustav Lorentzen’s article: Bør Stortinget oppheve 1979.
2. hovedsetning? (Should the Norwegian parliament abolish the second law of thermodynamics?), Ingeniør-nytt, 23(72), 1987 — written as a contribution to the debate [a] Rubin Battino, Laurence E. Strong, and Scott E. Wood. J. Chem. Eng. Educ.,
on thermal power stations in Norway.
74(3):304–305, mar 1997.
2 Thermodynamic . . .
Thermodynamics describes natural phenomena in idealised terms.
1
The scope of the description can vary, and must be adapted to our
needs at any given time.
2
The abstract concept of a “system” is at the heart of our understanding.
4
4
Property
5
Extensive vs. intensive
6
Equilibrium
7
Heat and work
3
The more details we want to understand, the more information we
must include in the description.
A thermodynamic system simplifies physical reality into a mathematical model.
5
Note that the physical size and shape of the system is quite irrelevant — all simple thermodynamic systems have uniform properties and
can always be described by a single set of state variables regardless of
the actual geometry.
7
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Properties
2 Thermodynamic . . .
1
The state of a thermodynamic system is determined by the system’s properties — and vice versa.
3
For example, the energy of the system is given by the formula
U = (∂U /∂X1 ) X1 + (∂U /∂X2 ) X2 + · · · + (∂U /∂Xn ) Xn once the state
variables Xi and the derived properties (∂U /∂Xi ) are known.
2
The logic is circular, but experimentally the state is defined when all
of the thermodynamic properties of the system have been measured.
6
(∂U /∂Xi ) can then be calculated from the function
U (X1 , X2 , . . . , XN ) through partial differentiation with respect to
Xi .
4
This determines the value of U in one specific state.
5
If we need to know the value of U in a number of different states, it
is more efficient to base our calculations on a mathematical model.
7
The measurements are no longer directly visible to us, but are instead encoded in the shape of model parameters (that describe the observations to some degree at least).
8
Normally it is only the first and secord derivatives of the model that
are relevant for our calculations, but the study of so-called critical points
requires derivatives up to the third or maybe even the fifth order.
9
Such calculations put great demands on the physical foundation of
the model.
6
We can then use the model to perform thought experiments that
reveal certain characteristics of the system’s behaviour.
8
Tore Haug-Warberg (Chem Eng, NTNU)
1
Language shapes our ability to think independently and to communicate and collaborate with other people.
2
It has enabled us to go far beyond simply meeting our basic needs
associated with survival and social interaction.
3
Our ability to observe, describe and record our thoughts about
physical phenomena is vital to a subject like thermodynamics.
4
For us it is particularly important to reach a common physical understanding of abstract concepts such as energy and entropy[a], but as
languages are always evolving, the premises for a shared conception
ENGLISH text is missing.Notation
are constantly shifting.
1
Our understanding of thermodynamics depends on some vital, and
5
All authors have to contend with that dilemma, and few of us have
precisely defined, concepts that form part of a timeless vocabulary.
the good fortune to be writing for future generations.
2
Accurate use of language improves our insight into the subject,
6
The benefit of writing about a phenomenological subject such as
thereby reducing the risk of misunderstandings.
thermodynamics is that as long as the observed phenomena remain
3
However, an exaggerated emphasis on precision may overwhelm
unchanged, the subject will endure.
readers and thus be counterproductive.
7
The challenge you face with a classical subject is combining the
preservation of an intellectual heritage with the injection of new ideas, 4 In this chapter we will revise and explain the key concepts in therand it is rather utopian to believe that the latter is possible in our case. modynamic analysis[a][a] in accurate, but nevertheless informal, terms.
That is an absolute principle.
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§ 001
§ 156
§2
News Olds
2 Thermodynamic . . .
bo
a
und
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§ 001 System
2 Thermodynamic . . .
1
A system is a limited part of the universe with a boundary that
may be a mathematical surface of zero thickness, or a physical barrier
between it and the surroundings .
ry
m
Expanded footnote [a]:
st
e
[a] We must be able to either actively control the system’s mass and composition,
impulse (or volume) and energy, or passively observe and calculate the conserved
quantities at all times. All transport properties ρ must satisfy limx →0+ ρ = limx →0− ρ
so that ρ is the same for the system and the surroundings across the control surface.
In physics, this is called a continuum description. The exception to this is the abrupt
change in state variables observed across a shock front. Well-known examples of
this include high-explosive detonations and the instantaneous boiling of superheated
liquids. It is important to note that the concept of phase boundary in Theme 4 does not
satisfy the continuity requirement and is therefore unsuitable as a control surface.
sy
2
A reservoir is a system that can interact with other systems without
undergoing any change in its state variables.
system
State variables
2 Thermodynamic . . .
State variables only relate to the current state of the system.
1
4
In the simplest case, there are C + 2 such variables, where C is
the number of independent chemical components and the number 2
represents temperature and pressure.
2
In other words, they are independent of the path taken by the system to reach that state.
3
The number of state variables may vary, but for each (independent)
interaction that exists between the system and its surroundings there
must be one associated state variable.
5
The situation determines the number of chemical composition variables required.
6
For example, natural fresh water can be described using C = 1
component (gross chemical formula H2 O) in a normal steam boiler, or
using C = 5 components involving 9 chemical compounds in an isotope
enrichment plant[a].
[a] Naturally occurring water consists of a chemical equilibrium mixture of 1 H 1 H 16 O +
2
H 2 H 16 O = 2(1 H 2 H 16 O) and others, formed by the isotopes 1 H, 2 H, 16 O, 17 O and
18
O.
6
A real-world system will never be perfectly uniform, nor completely
unaffected by its surroundings, but it is nevertheless possible to make
some useful simplifying assumptions.
3
Our physical surroundings such as the air, sea, lakes and bedrock
represent unlimited reservoirs for individual human beings, but not for
the entire world population[a].
4
Similarly, a hydropower reservoir represents a thermodynamic
reservoir for the turbine, but not for the power company.
5
This may appear trivial, but it is still worth trying to justify the above
statements to yourself or a fellow student. These reservoirs, and other
simple systems, have properties that are spatially and directionally uniform (the system is homogeneous with isotropic properties).
[a] The first proof that the oceans were polluted by man came from Thor Heyerdahl’s
Ra II expedition in 1970.
environment
7
Thus, a closed system, in contrast to an open one, is unable to
exchange matter with its surroundings.
Explain in your own words what is meant by the concepts:
System, boundary and environment or surroundings.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
An isolated system can neither exchange matter nor energy.
8
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Processes
02 September 2013
9
Note that the term control volume is used synonymously with open
system.
10 The boundary is then called a control surface.
11 In this context an adiabatic boundary is equivalent to a perfect insulator, and a diabatic boundary is equivalent to a perfect conductor.
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2 Thermodynamic . . .
§ 002
§1
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News Olds
§3
2 Thermodynamic . . .
§ 002 The state concept
2 Thermodynamic . . .
1
The state of the system can be changed through what we call a
process.
1
A thermodynamic state is only fully defined once all of the relevant
thermodynamic properties are known.
2
Terms such as isothermal, isobaric, isochoric, isentropic, isenthalpic, isopiestic and isotonic are often used to describe simple physical processes.
3
In simple systems, the properties are (by definition) independent of
location and direction, but correctly identifying the system’s state variables is nevertheless one of the main challenges in thermodynamics.
2
In this context, a thermodynamic property is synonymous with a
state variable that is independent of the path taken by the system[a].
[a] In simple systems, viscosity, conductivity and diffusivity also depend on the current
thermodynamic state, in contrast to rheology, which describes the flow properties of a
system with memory. Two examples of the latter are the plastic deformation of metals
and the viscosity of paint and other thixotropic liquids.
Explain in your own words what is meant by the terms:
State, property, process and path.
3
Without considering how to bring about these changes in practice,
the terms refer to various state variables that are kept constant, such
that the state change takes place at constant temperature, pressure,
volume, entropy, enthalpy, (vapour) pressure or osmotic pressure respectively.
4
Generically the lines that describe all of these processes are sometimes called isopleths.
5
Another generic expression, used in situations where the properties of ideal gases are of prime importance (aerodynamics, combustion, detonation, and compression) is the polytropic equation of state
p2 /p1 = (V1 /V2 )γ .
6
Here γ ∈ R can take values such as 0, 1 and cp /cv , which respectively describe the isobaric, isothermal and isentropic transformation of
the gas.
9
A cycle is the same as a closed path. The cycle can either be
temporal (a periodic process) or spatial (a cyclic process).
4
One aid in revealing the properties of the system is the process,
which is used to describe the change that takes place along a given
path from one state to another.
5
In this context, a path is a complete description of the history of the
process, or of the sequence of state changes, if you like.
6
Thermodynamic changes always take time and the paths are therefore time-dependent, but for a steady (flow) process the time is unimportant and the path is reduced to a static state description of the input
and output states.
7
The same simplification applies to a process that has unlimited time
at its disposal.
8
The final state is then the equilibrium state, which is what is of
prime concern in thermodynamic analysis.
y
x
11 A state that is in thermodynamic equilibrium appears static at the
macroscopic level, because we only observe the average properties of
a large number of particles, but it is nontheless dynamic at the molecular level.
10 This choice greatly affects the system description. In a steady
state, the variables do not change with time, whereas in dynamic systems they change over time. Between these two extremes you have a
quasi-static state: The state changes as a function of time, but in such
a way that the system is at all times in thermodynamic equilibrium, as
described in greater detail in Theme 4.
12 This means that we must reformulate the equilibrium principle
when the system is small, i.e. when the number of particles n → 0.
13 At this extreme, intensive properties such as temperature, pressure
and chemical potential will become statistical variables with some degree of uncertainty.
14 In a theoretical reversible process, it
is possible to reverse any change of state
by making a small change to the system’s
interaction with its surroundings. Thus,
−
two equal weights connected by a string
1±ε kg
running over a frictionless pulley will un+
dergo a reversible change of state if the
1 kg
surroundings add an infinitely small mass
to one of the weights. This is an idealised
model that does not apply in practice to an irreversible system, where a
finite force will be required to reverse the process.
15 Let us assume, for example, that the
movement creates friction in the pulley.
You will then need a small, but measurable, change in mass in order to set the
weights in motion one way or the other.
16 The total energy of the system is conserved, but the mechanical energy is converted into internal (thermal)
energy in the process — so it is irreversible.
17 If the friction is suddenly eliminated the weights will accelerate.
This change is neither reversible nor irreversible. Instead, it is referred
to as lossless, which means that the mechanical energy is conserved
without it necessarily being possible to reverse the process. For that to
happen, the direction of the gravitational field would also need to be reversed.
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Termodynamic methods TKP4175
02 September 2013
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Termodynamic methods TKP4175
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News Olds
ENGLISH text is missing.Degrees of freedom
2 Thermodynamic . . .
[a] A mole is defined in terms of the prototype kilogramme in Paris and not vice-versa.
Atomic mass is measured in atomic mass units, where the mass of one atom of the
12
C isotope is defined as 12 amu. The 1 amu was originally defined as being equal
in mass to one hydrogen atom, which deviates a little from the current definition. The
deviation is due to a difference in nuclear energy between the two nuclei and is related
to Einstein’s relation E = ∆mc 2 .
Experimentally it has been shown that the size of a system is pro1
portional to some of its properties. These are called extensive properties, and include volume, mass, energy, entropy, etc.
1
Systems of variable mass contain a minimum number of independent components that together make up the chemical composition of
the system.
Intensive properties, meanwhile, are independent of the size of the
system, and include temperature, pressure and chemical potential.
Only the composition at equilibrium can be described in these simple terms, which is because we are forced to specify as few as possible
variables that depend on mass or the number of moles.
2
In thermodynamics there are usually many (N > 2) state variables,
as well as an infinite number of derived properties (partial derivatives).
3
4
Mathematically these properties are defined by Euler functions of
the 1st and 0th degree respectively, as described in the separate chapter on the topic.
5
There are also properties that behave differently.
6
For example, if you increase the radius of a sphere, its surface area
and volume increase by r 2 and r 3 respectively.
7
This contrasts with the circumference, which increases linearly with
the radius (it is extensive), and the ratio between the circumference and
the radius, which always remains 2π (it is intensive).
3
§ 004
1
Let us consider a thermodynamic system that does not change
with time, which means that it must be in equilibrium.
ENGLISH text is missing.Kinetics
2
This minimises the degrees of freedom we have to specify.
1
True equilibrium does not assert itself instantaneously, and from
3
At high temperature chemical equilibrium, for example, it is suffi- our perspective it is not possible to judge whether equilibrium can be
cient to state the quantity of each of the atoms present.
attained, or whether the system has kinetic limitations that prevent this.
§3
§5
4
The distribution of the atoms amongst the many substances in the 2 All that thermodynamics does is describe the state of the system
mixture occurs by chemical reaction which is determined by the princi- at equilibrium, and not how long the process takes.
ple of equilibrium (see below) and the system’s equation of state.
§ 3 Intensiveness
3
To describe in detail how the system changes over time we need
5
In the case of phase equilibria, the chemical substances are sim- an understanding of kinetics and transport theory.
1
Another group of properties is not affected by any change in the
ilarly distributed across the system’s phase boundaries and it is suffisize of the system.
4
Kinetics is the study of forces and the motion of bodies, whilst reaccient to specify the total composition of the entire system.
2
These properties are referred to as intensive [a] properties.
tion kinetics looks specifically at rates of chemical reactions and phase
6
As a rule of thumb these problems are easy to specifiy, but they transitions.
3
Well-known examples include temperature T , pressure p and
do nevertheless require numerical solution by iteration which in many
§ 3 Densities
chemical potential µ.
5
Transport theory is a concept taken from nonequilibrium thermody§ 3 Extensiveness
1
Dividing one extensive property by another gives you a new, inten- cases is a challenging task.
namics that can be used to describe the changes that take place in a
1
In thermodynamics, size is not only a measure of the volume of a 4 Certain pairs of extensive and intensive variables combine to form sive variable that represents the ratio between the two properties.
7
If the system has no internal degrees of freedom in the form of system until it reaches equilibrium.
a product with a common unit (most commonly energy), and feature in
system, but also of any properties related to its mass.
chemical reactions or phase transformations, the equilibrium state will
important relationships such as U = TS − pV + µ1 N1 + µ2 N2 + · · · + 2 Let e.g. f = ax and g = bx be two extensive properties expressed
6
As a general rule, all transport problems must be formulated as
2
This implies that two systems with identical state descriptions beas functions of x . Then ρ =
ˆ f /g = a /b is independent of x , in other be determined by an ordinary (but multivariate) function, generally with- partial differential equations, whereas thermodynamic equilibrium probµn Nn .
out iteration.
come a system of double the size when combined.
words ρ is intensive.
lems can always be expressed using algebraic equations[a].
5
These pairs of T and S , p and V , and µi and Ni are called conjugate
3
Properties that can be doubled in this way, such as entropy S ,
3
These kinds of variables express generic densities and are widely 8 The general principles of equilibrium mean that the energy of the 7 This difference in mathematical treatment reflects the gradients in
variables.
volume V and the number of moles N , are proportional to the size of
used to describe systems in a way that does not refer to their size, but system will be minimised with respect to all the degrees of freedom that the system.
6
A mechanical analogy is the flow rate ρ~
V (extensive) and the gravthe system, and are referred to as extensive variables.
many fields have different practices, and it is often unclear whether the form the basis for the system description.
itational potential g ∆z (intensive) at a hydropower station.
definition is being expressed on a mass or mole basis (which is the 9 Alternatively the entropy is maximised with respect to its system 8 If they are not significant, it is sufficient to determine one represen4
This means that all of the extensive variables must be increased by
tative value for each of the scalar fields of temperature, pressure and
7
The power generated by the turbine can be written g ∆z ρ~
V , which most common source of misunderstanding).
the same factor.
variables.
chemical potential.
is the product of an intensive and extensive variable.
5
You cannot simply double the volume while keeping the number of
4
The difference between a specific and a molar quantity is that one 10 The degrees of freedom are at all times controlled by the physical
§3
8
Equivalent analogies can be found in electrical and mechanical is expressed per kilogramme and the other per mole of the substance nature of the system, and this determines which extremal principle to 9 This distinguishes a thermodynamic problem from a transport
1
Explain in your own words what is meant by a property being moles constant — entropy, volume and the number of moles must all be
problem, where the scalar fields must be determined simultaneously
engineering.
(or mixture).
apply.
intensive or extensive. If you divide an extensive property by the number doubled together.
throughout the space.
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5
Common examples include specific and molar heat capacity, spe- 11 The theorectical foundations for this topic are discussed in a later
of moles in the system or its mass, you get a molar or specific property 6 At the same time, other derived extensive properties like energy,
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total heat capacity, etc. are scaled the same amount.
cific and molar volume, etc.
chapter on Legendre transformations.
respectively. Show that the property obtained is intensive.
36 / 1776 [a] The vague definitions of extensive and intensive used here will later be replaced by
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ones involving Euler’s homogeneous functions of the 1st and 0th degree respectively.
2 Thermodynamic . . .
[a] This is the difference between a distributed and lumped description of the system.
bo
und
a ry
m
e
Gibbs phase rule
These measures are not of fundamental importance to thermodynamics, but they are important concepts in physical chemistry, and are
sometimes used in thermodynamic modelling.
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5
γ
st
2 Thermodynamic . . .
2
An alternative way of measuring mass is by looking at the number
of moles of the various chemical compounds that make up the system.
3
A mol is defined as the number of atoms[a] which constitute exactly
0.012 kg of the carbon isotope 12 C, generally known as the Avogadro
constant NA = 6.022136(7)1023 .
sy
Extensive–intensive
ENGLISH text is missing.Mass and mole
1
The system’s mass is a fundamental quantity, which is closely related to inertia, acceleration and energy.
4
In this context, molality and molarity are two measures of concentration stated in moles per kilogramme of solvent and moles per litre of
solution respectively.
β
2
A component should here be considered a degree of freedom in
Gibbs’ phase rule F = C + 2 − P , where C is the number of components
and P is the number of phases at equilibrium (see Theme 4).
p h a s e b’n
dar
y
4
This restriction influences our choice of components, and in practice the question is whether chemical reactions are taking place in the
system, although the conventions of the applied discipline are equally
important.
5
Generally components are selected from the system’s chemical
constituents or species, or from its reactants, whereas for electrolytes
and salt mixtures it is natural to use the ions that make up the system’s
chemical composition as components[a].
[a] Components do not necessarily represent physical constituents. One example is
ternary recipocral systems (salt mixtures) of the type NaCl + KBr = NaBr + KCl, which
have 4 possible substances (salts), but only 3 independent components. For example,
the salt NaCl can be described by the vector ( 1 0 0 0 ), or ( 0 −1 1 1 ),
or any linear combination of these two vectors. This illustrates the use of a seemingly
non-physical (in this case negative) number of moles for one of the components. The
specification does make sense, however, because the true amounts of each of the ions
in the mixture are non-negative, and hence real physical quantitites.
α
environment
Explain in your own words the following terms associated
with equilibrium and equilibrium states: Phase, phase
boundary, aggregate state, equilibrium and stability.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Termodynamic methods TKP4175
02 September 2013
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News Olds
1
2
A phase is defined as being a homogeneous, macroscopic subsystem separated from the rest of the system by a phase boundary.
3
It must be possible to separate a phase from other phases through
mechanical means alone.
4
This is an important prerequisite.
5
The system’s equilibrium phases are often designated by the Greek
letters α, β, γ, etc.
6
The entropy density, energy density and mass density are constant within the phases, but they vary discontinuously across the phase
boundaries.
7
Temperature, pressure and chemical potential, meanwhile, are
constant throughout the entire system (assuming equilibrium).
8
Spatially a phase can be discretely distributed across the available
volume, cf. fog particles in air and drops of fat in milk.
9
If the whole system only consists of one phase, it is said to be
homogeneous.
10 Otherwise, it is heterogenous.
11 A phase is referred to as incompressible if the volume is dependent
on the pressure, but in reality all phases are compressible to a greater
or lesser extent (particularly gases).
12 In principle a phase has two possible states of matter: Crystalline
and non-crystalline (glass and fluids).
13 For practical purposes a fluid may sometimes be referred to as
a gas without a specific volume, a liquid with surface tension, or an
electrically conductive plasma, but in thermodynamics there is a gradual
transition between these terms, and there are no strict criteria as to what
falls within which category.
stable
metastable
unstable
2 Thermodynamic . . .
Equilibrium is the state attained by the system when t → ∞.
(In)exact differential
2 Thermodynamic . . .
3
These two properties control the system’s path as a function of its
physical relationship to its surroundings.
4
On the one hand you have the abstract analysis of the system,
which deals with state functions and mathematical formulae, and on the
other you have an equally idealised interpretation of the surroundings.
5
Within this context, heat and work are defined as two different
modes of exchanging energy.
6
It is important to note that heat and work are not state variables
of the system: They simply describe two different mechanisms for exchanging energy between the system and its surroundings.
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State variables are variables that form part of a state function.
1
2
A state function always produces an exact differential, but not all
differentials in physics are exact .
If the stability limit is exceeded, the system will split into two or
more equilibrium phases (from the mechanical analogy in the figure it
is equally likely that the ball will fall to the right as to the left).
One example from thermodynamics is (dU )n = δQ − δW , which
describes the energy balance for a closed system. Here the energy U
is a state function with the total differential (dU )n .
19
16 An unstable equilibrium is a mathematical limit case.
17 It describes a material state that breaks down if exposed to infinitely
small perturbations.
18 It is impossible to physically create this kind of equilibrium, but it is
nevertheless very important from a theoretical point of view, as it sets a
limit on what can physically exist based on simple laws of physics and
mathematics.
2
Transporting energy affects the state of the system, but the heat
and work are not themselves accumulated in the system.
3
Work involves moving a macroscopic mass, or elementary particles, against an external force.
4
A moving piston, a rotating axle, electrons in an electric circuit and
water flowing through a turbine are all examples of this.
2 Thermodynamic . . .
5
Heat results from large numbers of random, microscopic movements that do not result in any net movement of mass.
6
For heat to be converted into work, the microscopic movements
must first be coordinated.
ENGLISH text is missing.The purpose of the anal-
7
The 2nd law of thermodynamics then dictates that some of the ysis
heat will be lost to a thermal reservoir of the same temperature as the 1 Having defined the fundamental concepts, what we now must do is
turn theory into practice.
surroundings.
8
The term energy dissipation is used to emphasise the fact that 2 That will require a good understanding of the physics involved in the
spontaneous processes always result in a reduction in available energy. problems that we will be looking at, some mathematics and in particular
9
The convention is for work done and heat supplied to be expressed differential equations and linear algebra, and a selection of relevant
descriptions of substances, known as equations of state.
as positive values.
δQ
δQ
− δW
δW
3
Last but not least, we need to know the purpose of our analysis.
10 Individually, neither heat nor work are state variables, but the difference between them gives us the change in the system’s energy, which 4 It is often based in a wish to find a simple model to explain the
is a state function.
changes in the system’s state.
1
11 In other words, any given change in energy can be produced in an 5
Thermodynamics is, in short, a subject that combines most of the
Explain in your own words the meaning of the terms: Heat, work and infinite number of ways by varying the contributions made by heat and things taught in undergraduate chemistry, physics and mathematics at
energy. Are all three state variables?
work.
the university.
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1
Mathematical descriptions allow us to understand some characteristic events that surround us, but they do not give us the whole picture.
2
However, it is somewhat optimistic to believe that mechanical engineering, thermodynamics, electromagnetism and other calculationheavy subjects can fully describe the world we live in, and that we are
in a position to decide how detailed an answer we want.
3
Mathematics provides us with a useful tool, but that does not mean
that models and reality are two sides of the same coin, and hence that
we, by carefully eliminating “all” assumptions, can reach an absolutely
true answer.
15 A metastable equilibrium is stable when exposed to minor perturbations, but it becomes unstable in the event of major displacements.
14 If the system returns to the same equilibrium state after exposure
to a large random perturbation (disturbance), the equilibrium is said to
be stable.
Das Ding an sich
§ 5 Heat and work
1
Heat and work are two closely related mechanisms for transporting
energy between the system and its surroundings.
§5
=
§ 004 The phase concept
News Olds
ENGLISH text is missing.Surroundings
1
The assumptions and limitations set out in this chapter, and in the
book as a whole, in relation to thermodynamic system analysis, are
sufficient for the purposes of analysing the effects of a large number of
changes of state, but they do not tell us anything about how the system
relates to its surroundings.
2
In order to analyse that, we must introduce two further terms,
namely work W and heat Q .
dU
News Olds
Expanded footnote [a]:
[a] Let e.g. f (x , y ) = xy + c be a state function with x and y as its state variables. Here
df = y dx + x dy is the total differential of the function. The right side of the equation
is then called exact. If as a pure thought experiment we change the plus operator on
the right side to minus, the differential becomes non-exact. It is possible to transform
the left-hand side into y 2 dg = y δx − x δy , where y 2 is an integrating factor for the
differential and dg is the total differential of g (x , y ) = xy -1 + c , but the new differential
y δx − x δy remains non-exact, since it cannot be expressed as the total differential of
any known function.
4
Immanuel Kant unified the most important strands of rationalism and empiricism with his interpretation of das Ding
an sich. Science is based on observable
events taking place in a world where time
and space are assumed to be independent of us, but from a philosophical point
of view Kant argues that we may be unable to see all the aspects of whatever we
are observing.
Expanded footnote [a]:
[a] Immanuel Kant, 1724–1804. German philosopher and logician.
3
5
He stressed that we can never be certain about the intrinsic, true
nature of anything, and that our understanding of the world is limited by
our subjective experiences in time and space.
For any change δQ − δW there is a unique value of (dU )n .
4
6
For any change (dU )n there are in principle an infinite number of
combinations of δQ and δW , since only the difference between heat
and work is observable in the system.
5
However, the reverse is not true.
Duck or rabbit ?
6
For example, does the picture on the left represent a duck or a
rabbit?
7
With a bit of imagination we can spot both animals, but we can only
see one of them at any given time.
8
The full picture is not available to us, and at first glance we don’t
even realise that there are two possibilities[a].
[a] At first glance, around half of the population will see the duck, whilst the other half
will see the rabbit.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Olds
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News Olds
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
ENGLISH text is missing.Isotopes
1
Based on our understanding of nature, isotopes are chemically
very similar, but in this case the mixing enthalpy measured is equivalent
to a fall in temperature of 0.43 K for an equimolar mixture of the two
isotopes, which is much greater than expected.
Das Ding an sich II
2 Thermodynamic . . .
1
The solid circles show calometric readings for the mixing enthalpy
∆mıx h of the system H2 O–D2 O at varying compositions (mole fractions x ) of the two compounds.
2
Thermodynamics is really a subject that describes das Ding für
uns [a] as opposed to das Ding an sich. The figure on the right illustrates
this distinction in a rather subtle manner.
[a] Referred to as Erscheinung in Kant’s thesis.
Expanded footnote [a]:
[a] D. V. Fenby and A. Chand. Aust. J. Chem., 31(2):241–245, 1978.
The line fits the data points very well,
and we can at least temporarily conclude
that this simple model adequately represents the readings taken.
4
∆rx h or ∆mıx h ?
3
The continuous line has been fitted using the semi-empiric model
ax (1 − x ), which gives a first-order approximation of ∆mıx h .
h
This posture is, in simple terms, how
the experiment look to us and what is
meant by das Ding für uns. The question is if this is the only rational explanation that fits the observations.
b b
b
Termodynamic methods TKP4175
b
b
b
b
⊗ ⊗⊗
⊗
⊗
bC
⊗⊗
bC
bC
bC
⊗
bC
bC
bC
bC
⊗
bC
x
02 September 2013
New concepts
49 / 1776
2
Manifolds
3
Canonical forms
4
Conjugated variables
5
Maxwell relations
6
Gibbs–Helmholtz equation
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
The last statement appears to be self-contradictory, as the compo-
ENGLISH text is missing.Thermodynamic proofs
2 Thermodynamic . . .
circles in the figure. Subsituting OD to OH changes the chemistry to CD3 OD–CH3 OH patent application, except requirements; 3.3.6 Insufficient clarity excludes inventions
which again allows for proton–deuteron exchange and the equimolar change in en- that “. . . are inherently impossible to produce, as doing so would require accepted
thalpy increases to an intermediate value of 8.4 J mol-1 or 0.1 K.
[a] Systems with many components are obviously even more complex.
physical laws to be broken — this applies e.g. to perpetual motion machines.”
System: A part of the universe with uniform T , p , µ
2
State: U = (∂U /∂X1 ) X1 + (∂U /∂X2 ) X2 + · · · as a function of Xi
3
Process: The change from one state A to another state B
4
Property: (∂U /∂Xi ), (∂2 U /∂Xi ∂Xj ), . . . in N + 2 variables
5
Extensive: Mass proportional property U , V , S , H , . . .
6
Intensive: Mass independent property T , p , µ, ρ, . . .
7
Equilibrium: Time invariant state Ueq = min U
8
Heat is not work: Q 9 W
9
Work means heat: W → Q
Part
4
The Legendre transform
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
see also Part-Contents
02 September 2013
53 / 1776
Energy
The formal definition of the canonical thermodynamic potentials is
one of the three fundamental pillars of thermodynamic theory, together
with the principle of equilibrium, and the somewhat opaque distinction
between heat and work .
Expanded footnote [a]:
[a] I.e. the functions U (S , V , N ), H (S , p , N ), A (T , V , N ), G (T , p , N ), etc.
Expanded footnote [a]:
Expanded footnote [a]:
[a] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,
2nd edition, 1985.
[a] Michael Modell and Robert C. Reid. Thermodynamics and Its Applications. Prentice
Hall, 2nd edition, 1983.
From a pragmatic point of view it is sufficient to remember that U is
useful for dynamic simulation, H for steady state calculations, A for the
simulation of storage tanks, etc.
4
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
1
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
Legendre transformation
5
nents react chemically at the same time as they take part in an ideal 1 The foundations of phenomenological thermodynamics are too
equilibrium.
weak to state hard facts about the true nature of the systems it de6
This apparent contradiction arises because of the unclear bound- scribes, but strangely enough thermodynamic theory can still be used
to falsify claims that break the laws upon which it is founded[a].
ary between a chemical reaction and a physical interaction.
7
For a thin gas there is an explanation for this, but for a liquid there 2 The theoretical basis is also sufficient for deriving important relationships between the state variables, but it does not constitute indeis no entirely satisfactory definition.
ENGLISH text is missing.The simpler the better
pendent evidence in the mathematical sense.
1
It is also worth remembering that even a small one-component
8
The reaction product HDO is not stable either since it decomposes
instantaneously to H2 O and D2 O, which prevents us from observing 3 Thermodynamic analysis is capable of confirming prior assump- system has maybe 1015 –1020 microscopic degrees of freedom that are
tions, or of demonstrating new relationships between existing results, modelled with only three thermodynamic state variables[a].
the substance in its pure state.
but the calculations are not necessarily correct even if the model ap2
This means that a significant amount of information about the sys9
Our understanding of nature is, in other words, based on evidence
pears to correspond with reality.
tem is lost along the way.
derived from indirect observations and mathematical models of the
4
Firstly, the result is limited to the sample space for the analysis
physical properties in the water phase.
3
It is therefore necessary to develop our ability to recognise what is
and to the underlying assumptions and secondly, there may be several
important for the modelling, so that we can perform the right calcula10 In reality all scientific knowledge is based on theoretical models of
equally good explanations for a single phenomenon (as shown in the
tions, rather than trying to look (in vain) for the very accurate descripone kind or the other, and therefore does not imply that we have any
enthalpy example above).
tion.
exact understanding of das Ding an sich.
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4
In keeping with that principle, we will mainly be looking at simple
models such as ideal gas and the Van der Waals equation, but that
[a] The perpetuum mobile is the most famous example of this. In Norway it is in fact
[a] In the system CD3 OD–CH3 OD, covalent bonds form between C–D and C–H, pre- impossible to apply for patent protection for a perpetual motion machine, cf. the Nor- does mean that we will be very precise in our analysis of the ones that
venting any isotopic reaction, meaning that the change in enthalpy is only 0.79 J mol-1 wegian Industrial Property Office’s Guidelines for processing patent applications. The we do look at.
52 / 1776
or 0.01 K; cf. T. Kimura et al. J. Therm. Anal. Cal., 64:231–241, 2001, shown as open online (2010) version of Part C: Preliminary examination; Chapter II Contents of the
1
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
1
3
The net reaction involves the protonation of H2 O ⇔ H+ + OH−
and the deuteronation of D2 O ⇔ D+ + OD− which are fast reactions.
b
32 J mol-1
5
Tore Haug-Warberg (Chem Eng, NTNU)
b
Looking back
2
The reason for this discrepancy is that the two components react
endothermically to form HDO.
4
The result is that H2 O + D2 O ⇔ 2 HDO reaches equilibrium[a] in
a virtually ideal mixture.
Energy (2)
4 The . . .
Here, the Legendre transformation is the key to new insight, as
6
it provides a clever yet simple formula that allows us replace some or
all of the free variables of a function with their corresponding partial
derivatives. For example, the variable V in internal energy U (S , V , N )
can be replaced by (∂U /∂V )S ,N , see Figure 4.1.
5
If a feasible solution to the problem already exists then this rulebased understanding is perfectly adequate, but when seeking a new
solution we need a deeper theoretical understanding to see all the alternatives that exist.
Expanded footnote [a]:
[a] Adrien-Marie Legendre, 1752–1833. French mathematician.
2
The theoretical background of the energy potentials and their mutual transformation properties form the background to what we will now
discuss, but for the moment we will refrain from examining to what extent mathematical formulae are of practical relevance to the applications
of the theory.
3
Complications arise from the fact that there are several energy
functions to choose from, and it can be difficult to know exactly which
function is best suited for a particular problem.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Energy (3)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
Energy (4)
4 The . . .
vmode (bug): hmode expected
H1
H2
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Transformation rules
1
Mathematically, the Legendre transformation φi of the function f is
defined by:
7
The new variable can be interpreted as the negative pressure π
and the resulting transformed function, called enthalpy H (S , π, N ), is in
many cases more versatile than U itself. In fact, as we will see later, H
has the same information content as U .
8
This is one of the key reasons why the Legendre transformation is
central to both the thermodynamic and mechanical theories.
(∂∂UVii ) Vi
Hi = Ui −
= Ui + pi Vi
φi (ξi , xj , xk , . . . , xn ) =
ˆ f (xi , xj , xk , . . . , xn ) − ξi xi ,
ξi =
ˆ ∂∂xfi
.
U1 U2 U3 U4
xj ,xk ,...,xn
H4
V1V2 V3
H=
ˆ UV (S , π, N ) = U (S , V , N ) − πV
π=
ˆ ∂∂VU S ,N
=
−p
V4
Note that the volume derivative of U is the negative pressure π, because
U diminishes when the system performs work on the surroundings —
not vice versa.
Figure 4.1: Legendre transformation of internal energy to enthalpy.
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Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Transformation rules (2)
4 The . . .
and then substitute in the total differential of the initial function f exP
pressed as:df = ξi dxi + jn,i (∂f /∂xj )xi ,xk ,...,xn dxj .
The simplification is obvious:
dφi = −xi dξi +
j ,i
∂f ∂xj x ,x ,...,x
n
i k
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Transformation rules (3)
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Tore Haug-Warberg (Chem Eng, NTNU)
dxj .
dφi =
∂φi ∂ξi xj ,xk ,...,xn
dξi +
4 The . . .
n
X
j ,i
∂φi ∂xj ξ ,x ,...,x
n
i k
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
5
If we now consider φi as a function of the derivative ξi rather than
of the original variable xi , the total differential of φi can be written
dφi = df − xi dξi − ξi dxi ,
n
X
Termodynamic methods TKP4175
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
3
To begin to get an understanding of the Legendre transform we
shall first write out the differential
4
(4.1)
2
As mentioned above, one example of this is the transformation of
internal energy to enthalpy:
H3
Tore Haug-Warberg (Chem Eng, NTNU)
4 The . . .
Repeated transformation
1
The latter property shows that further transformation is straightforward:
dxj .
6
Comparing the last two equations term-by-term gives us the
important transformation properties (∂φi /∂ξi )xj ,xk ,...,xn = −xi and
ˆ ξj .
(∂φi /∂xj )ξi ,xk ,...,xn = (∂f /∂xj )xi ,xk ,...,xn =
4 The . . .
φij (ξi , ξj , xk , . . . , xn ) =
ˆ φi (ξi , xj , xk , . . . , xn ) − ξj xj ,
∂φ ˆ ∂xji
ξj =
ξi ,xk ,...,xn
= ∂∂xfj
.
(4.2)
φij (ξi , ξj , xk , . . . , xn ) = f (xi , xj , xk , . . . , xn ) − ξi xi − ξj xj ,
(4.3)
xi ,xk ,...,xn
Combining Eqs. 4.1 and 4.2 gives the alternative, and conceptually
simpler, expression
2
where the sequential structure of the Legendre transforms in 4.1
and 4.2 has been replaced by a simultaneous transformation of two
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Repeated transformation (2)
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
(or more) variables. Moreover, as the Legendre transform is independent of the order of differentiation, we know that φij = φji .
3
The two alternative approaches are equivalent because ξj is the
same regardless of whether it is calculated directly from the transform
as (∂φi /∂xj )ξi ,xk ,...,xn or from the original function as (∂f /∂xj )xi ,xk ,...,xn This
is clear from the differential equation above, and from Theme 20 on
page 132. Knowing the initial function f and its derivatives is therefore
sufficient to define any Legendre transform.
4
Mathematically we say that the Legendre operator [a] commutes.
5
The three sets of variables (xi , xj , . . . , xn ), (ξi , xj , . . . , xn ) and (ξi ,
ξj , . . . , xn ) are particularly important, and are often referred to as the
canonical variables of the functions f , φi and φij = φji .
6
Canonical means simple and in this context it puts emphasis on
state descriptions that contain all the information about the system.
[a] Mathematical operators are often allocated their own symbols, but in thermodynamics it is more usual to give the transformed property a new function symbol.
Canonical potentials
4 The . . .
This is not trivial, as we will see that U (T , V , n) and H (T , −p , n), for
instance, do not have this property.
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§ 019
§ 18
§ 20
4 The . . .
2
The legal system of the Roman Catholic Church, also called canon
law, has medieval roots, and is based on a collection of texts that the
Church considers authoritative (the canon).
3
Today the word canonical is used to refer to something that is orthodox or stated in a standard form.
4
The latter definition is particularly used in mathematics, where e.g.
a polynomial written with the terms in order of descending powers is
said to be written in canonical form.
7
02 September 2013
02 September 2013
1
In thermodynamics, we refer to canonical potentials, meaning
those that contain all of the thermodynamic information about the system.
5
Here we will show that the Legendre transforms of internal energy
give us a canonical description of the thermodynamic state of a system.
Termodynamic methods TKP4175
Termodynamic methods TKP4175
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
6
Essentially, what we need to show is that U (S , V , n), A (T , V , n),
H (S , −p , n), etc. have the unique property that we can recreate all of
the available information from any single one of them.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Derive all of the possible Legendre transforms of
internal energy. State carefully the canonical variables
ˆ (∂U /∂S )V ,N ,
in each case. Use the definitions τ =
ˆ (∂U /∂N )S ,V to help you.
π=
ˆ (∂U /∂V )S ,N and µ =
Expanded footnote [a]:
[a] Let τ and π denote temperature and negative pressure respectively. This is to
emphasize that they are transformed quantities, like the chemical potential µ. In this
notation, all intensive derivatives of internal energy are denoted by lower case Greek
letters.
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Termodynamic methods TKP4175
02 September 2013
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 019 Energy functions
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
For a single-component system this means that there are 23 − 1 =
7 possible transformations.
1
2
For any given thermodynamic function with m = dim(n) + 2 variables there are 2m − 1 Legendre transforms.
3
By using Eq. 4.1 on each of the variables in turn we get three of
the transforms:
A (τ, V , N ) = U (S , V , N ) − ∂∂U
ˆ U − τS ,
(4.4)
S V ,N S =
H (S , π, N ) = U (S , V , N ) − ∂∂VU S ,N V =
ˆ U − πV ,
(4.5)
∂U X (S , V , µ) = U (S , V , N ) − ∂N S ,V N =
ˆ U − µN .
(4.6)
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§ 019 Energy functions (2)
4 The . . .
4
By using Eq. 4.3 on pairs of variables we can obtain three more
transforms:
G (τ, π, N ) = U (S , V , N ) − ∂∂VU S ,N V − ∂∂U
S V ,N S
=
ˆ U − πV − τS ,
Y (S , π, µ) = U (S , V , N ) −
∂U ∂V S ,N
Ω(τ, V , µ) = U (S , V , N ) −
∂U ∂S V ,N
=
ˆ U − πV − µ N ,
=
ˆ U − τS − µN .
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
HAUG
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Termodynamic methods TKP4175
V−
∂U ∂N S ,V
S−
∂U ∂N S ,V
(4.7)
§ 019 Energy functions (3)
5
Finally, by using Eq. 4.2 on all three variables successivly we can
obtain the null potential, which is also discussed on page 192 in Chapter 5:
∂U O (τ, π, µ) = U (S , V , N ) − ∂∂VU S ,N V − ∂∂U
S V ,N S − ∂N S ,V N
=
ˆ U − πV − τS − µN
N
≡0
(4.8)
§ 020
§ 19
(4.10)
N
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(4.9)
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
4 The . . .
§ 21
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
Internal energy U (S , V , N ) is used when looking at changes to
1
closed systems, and is in many respects the fundamental relationship
of thermodynamics.
§ 020 Differentiation I
4 The . . .
1
It is natural to start with Eq. 4.1, which we differentiate with respect
to xj :
∂φi ∂(f −ξi xi )
=
,
(4.11)
∂ xj
∂ xj
2
Note that the variables xk , . . . , xn are common to both φi and f and
may hence be omitted for the sake of clarity. We will therefore limit our
current analysis to the functions f (xi , xj ) and φi (ξi , xj ).
2
Several of the Legendre transforms of energy have their own
names:
Helmholtz energy A (τ, V , N ) is central to describing the state properties of multicomponent fluids.
ξi
The variable ξj in Eq. 4.2 can be defined in two ways.
Either as (∂φi /∂xj )ξi ,xk ,...,xn or as (∂f /∂xj )xi ,xk ,...,xn . Use
implicit differentiation to prove that the two definitions
are equivalent.
3
4
Gibbs energy G (τ, π, N ) has traditionally been the transform that is
of most interest in chemical thermodynamics and physical metallurgy.
ξi
where (∂ξi /∂xj )ξi is by definition zero.
Hence, at constant ξi :
3
d(f − ξi xi ) ξi =
Enthalpy H (S , π, N ) is important for describing thermodynamic processes in chemical engineering and fluid mechanics.
∂f ∂ xi xj
dxi +
=
ˆ ξi dxi +
= ∂∂xfj dxj
5
6
The grand canonical potential Ω(τ, V , µ) is used when describing
open systems in statistical mechanics.
7
Meanwhile, the null potential O (τ, π, µ) has received less attention
than it deserves in the literature, and has no internationally accepted
name, even though the function has several interesting properties, as
we shall later see.
8
This just leaves the two functions X (S , V , µ) and Y (S , π, µ), which
are of no practical importance.
9
However, it is worth mentioning that X is the only energy function
that has two, and always two, extensive variables no matter how many
chemical components there are in the mixture.
10 This function is therefore extremely well suited to doing stability
analyses of multicomponent phase equilibria.
11 In this case the basic geometry is simple, because the analysis is
done along a 2-dimensional manifold (a folded x , y plane) where the
chemical potentials are kept constant.
12 This topic is discussed further in a separate chapter on phase stability.
∂f ∂ xj x
∂f ∂ xj x
i
i
dxj − ξi dxi
dxj − ξi dxi
xi
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
131 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 020 Differentiation I (2)
02 September 2013
132 / 1776
4 The . . .
§ 021
§ 20
§ 22
1
It is a matter of personal preference, therefore, whether we want to
calculate ξj as (∂f /∂xj )xi before the transformation is performed, or to
calculate (∂φi /∂xj )ξi after the transformation has been defined.
2
Normally it is easiest to obtain all of the derivatives of the original
function first — especially when dealing with an explicitly defined analytic function — but in thermodynamic analyses it is nevertheless useful
to bear both approaches in mind.
3
The next paragraph shows the alternative definitions we have for
temperature, pressure and chemical potential.
135 / 1776
i
=
∂f ∂xj x
i
.
xi
leading to the conclusion that differentiation of φi with respect to the
untransformed variable xj gives the same derivative as for the original
function f .
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Termodynamic methods TKP4175
02 September 2013
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02 September 2013
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§ 021 Identities I
4 The . . .
1
From the Eqs. 4.4 and 4.5 we have φ1 = A (τ, V , N ) and φ2 =
H (S , π, N ), which on substitution into Eq. 4.13 give:
U
∂A = ∂∂N
,
(4.14)
∂N τ,V
S ,V
∂U ∂H
= ∂N S ,V .
(4.15)
∂N S ,π
(4.12)
ξi
Termodynamic methods TKP4175
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
ENGLISH text is missing.Before or after?
or, alternatively:
d(f −ξi xi )
dxj
ξ
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
5
The full derivative takes the same value as the corresponding partial derivative (only one degree of freedom). Substitution into Eq. 4.11
yields
∂φi = ∂∂xfj =
ˆ ξj
(4.13)
∂xj
4
Termodynamic methods TKP4175
2
Let f = U (S , V , N ) be the function to be transformed. The question
asks for the derivatives with respect to the mole number N and it is
tacitly implied that only S and V are to be transformed.
Use the result from Theme 20 on page 132 to show that
the chemical potential has four equivalent definitions:
µ = (∂U /∂N )S ,V =
ˆ (∂A /∂N )τ,V =
ˆ (∂H /∂N )S ,π =
ˆ
(∂G /∂N )τ,π . Specify the equivalent alternative definitions
for temperature τ and negative pressure π.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
3
The transform φ12 = φ21 = G (τ, π, N ) in Eq. 4.7 can be reached
either via A or H .
4
136 / 1776
Inserted into Eq. 4.13 the two alternatives become:
∂G = ∂∂AN τ,V ,
∂N τ,π
∂G
= ∂∂H
∂N τ,π
N S ,π .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
(4.16)
(4.17)
137 / 1776
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 021 Identities I (2)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
§ 022
6
So, in conclusion, the following is true for any single-component
system:
U
ˆ ∂∂H
.
(4.18)
µ = ∂∂AN τ,V =
ˆ ∂∂N
ˆ ∂∂G
N S ,π =
N τ,π =
S ,V
§ 21
§ 23
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
ENGLISH text is missing.Freedom to choose
This shows very clearly that we are free to choose whatever coordinates we find most convenient for the description of the physical
problem we want to solve.
1
2
For example, in chemical equilibrium theory it is important to know
the chemical potentials of each component in the mixture.
3
If the external conditions are such that temperature and pressure
ˆ (∂G /∂Ni )T ,p ,Nj,i , but if the entropy
are fixed, it is natural to use µi =
and pressure are fixed, as is the case for reversible and adiabatic state
changes, then µi =
ˆ (∂H /∂Ni )S ,p ,Nj,i is a more appropriate choice.
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2
As in Theme 20, the variables xk , . . . , xn are common to both f and
φi , and have therefore been omitted for the sake of clarity.
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
4 The . . .
xj
which leads to the following conclusion: The derivative of φi with respect
to a transformed variable ξi is the original variable xi , but with the sign
reversed.
4
In other words, there is a special relationship between the variables
xi in f (xi , xj ) and ξi in φi (ξi , xj )[a].
5
Due to the simple relationship set out in Eq. 4.22, the variables (ξi ,
xj ) are said to be the canonical variables of φi (ξi , xj ).
[a] The variables x and ξ are said to be conjugate variables.
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Non-canonical differentiation
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1
Systematically applying Eq. 4.22 to all of the energy functions in
Theme 19 on page 127 yields :
H
G
Y
,
(4.23)
=
ˆ ∂∂π
=
ˆ ∂∂π
−V = ∂∂π
S ,N
S ,µ
τ,N
∂A
∂G
∂Ω
−S = ∂τ V ,N =
ˆ ∂τ π,N =
ˆ ∂τ V ,µ ,
(4.24)
X
Y
=
ˆ ∂∂µ
.
(4.25)
=
ˆ ∂Ω
−N = ∂∂µ
∂µ
4 The . . .
4
Differentiating φi = f − ξi xi with respect to the original variable xi
gives:
2 ∂φi i
= ∂∂xfi
− ∂ξ
xi − ξi = − ∂x∂i ∂fxi
xi ,
∂ xi
∂xi
xj ,i
xj ,i
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xj ,i
Termodynamic methods TKP4175
xj ,i
02 September 2013
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§ 024
§ 23
§ 25
4 The . . .
ENGLISH text is missing.Systematics
1
Knowing the properties of the Legendre transform, as expressed
by Eqs. 4.13 and 4.22, allows us to express all of the (energy) functions
U , H , A , . . . O that we have considered so far in terms of their canonical
variables.
2
The pattern that lies hidden in these equations can easily be condensed into a table of total differentials for each of the functions (see
below).
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02 September 2013
[a] Sharp students will note the absence of −S = (∂O /∂τ)π,µ , V = (∂O /∂π)τ,µ and
−N = (∂O /∂µ)τ,p . These relations have no clear thermodynamic interpretation, however, because experimentally τ, π, µ are dependent variables, see also Theme 26 on
page 153.
§ 023
§ 24
1
The expression on the right-hand side is not particularly complex,
but unfortunately it is not canonically related to f . It is nevertheless an
important result, because φi must by definition be a function with the
same variables as f itself.
4 The . . .
2
The change in variable from xi to ξi is an abstract concept that will
only rarely produce an explicit expression stated in terms of ξi .
3
Hence, if we want to calculate the derivative of φi by means of an
explicit function, we have to go via f and its derivatives— there is simply
no way around it. A simple example shows how and why this is the
case:
C
∂2 A ∂S ∂U ˆ TV T = CV (T , V , N ) .
∂T V ,N = − ∂T ∂T V ,N T = ∂T V ,N T =
4
Here the internal energy is, at least formally speaking, a function of
the canonical variables S , V and N , but there are virtually no equations
of state that are based on these variables.
5
T , V and N are much more common in that context, so to lure U
into revealing its secrets we must use T rather S .
6
Note, however, that in this case S = − (∂A /∂T )V ,N is a function to
be differentiated and not a free variable.
7
This illustrates in a nutshell the challenges we face in thermodynamics: Variables and functions are fragile entities with changing interpretations depending on whether we want to perform mathematical
analyses or numerical calculations.
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Use the results obtained in Theme 22 to prove that the
derivatives (∂H /∂π)S ,N , (∂G /∂π)τ,N and (∂Y /∂π)S ,µ
are three equivalent ways of expressing the volume V
of the system. Specify the corresponding expressions
for the entropy S and the mole number N .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
§ 024 Canonical differentials
dU (S , V , N ) =
02 September 2013
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4 The . . .
Use the results from Themes 21 and 23 on pages
135 and 142 to find the total differentials of all the
energy functions mentioned in Theme 19.
τ dS + π dV + µ dN ,
(4.26)
dA ( τ , V , N ) = −S d τ + π dV + µ dN ,
(4.27)
dH (S , π, N ) =
τ dS − V dπ + µ dN ,
(4.28)
dG ( τ , π, N ) = −S d τ − V dπ + µ dN ,
(4.30)
dX (S , V , µ) =
dY (S , π, µ) =
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ENGLISH text is missing.Non-canonical differentiation
§ 22
τ dS + π dV − N dµ ,
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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(4.29)
τ dS − V dπ − N dµ ,
(4.31)
dO ( τ , π, µ) = −S d τ − V dπ − N dµ .
(4.33)
dΩ ( τ , V , µ) = −S d τ + π dV − N dµ ,
Tore Haug-Warberg (Chem Eng, NTNU)
(4.21)
1
The total differentials of the energy functions can be stated by taking the results from Eqs. 4.18–4.20 and 4.23–4.25 as a starting point:
Expanded footnote [a]:
τ,V
.
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
[a] The symbol π =
ˆ −p is used here as the pressure variable. This is quite deliberate,
in order to avoid the eternal debate about the sign convention for p . As used here,
τ, π and µ are subject to the same transformation rules. This means that the same
rules apply to e.g. (∂H /∂π)S ,N = −V and (∂A /∂τ)V ,N = −S , whereas the traditional
approach using (∂H /∂p )S ,N = V and (∂A /∂T )V ,N = −S involves different rules for p
and T derivatives. Note, however, that it makes no difference whether −p or p is kept
constant during the differentiation.
S ,π
Termodynamic methods TKP4175
∂ xi ∂ξi xj
2
The Eq. 4.22 is strikingly simple, and leads to a number of simplifications in thermodynamics.
3
In particular the properties explained in the previous paragraph are
of prime importance.
2
Let us start out once more from f = U (S , V , N ) and define[a] the
transform φ2 = H (S , π, N ) = U − πV . Inserted into Eq. 4.22, this gives
us (∂H /∂π)S ,N = −V .
S ,V
Tore Haug-Warberg (Chem Eng, NTNU)
xj
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
xj
Before we move on, however, we should investigate what happens
1
if we do not describe the Legendre transform in terms of canonical
variables.
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 023 Identities II
02 September 2013
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
3
From Eq. 4.3 we know that (∂f /∂xi )xj = ξi , which can easily be
substituted into Eq. 4.21 to produce
∂φi = −xi ,
(4.22)
∂ξi
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
xj
has been used to obtain the last line below:
∂φi ∂(ξ x )
∂f = ∂ξ
− ∂ξi i i
∂ξi xj
i xj
xj
∂ xi ∂f = ∂ξ
−
x
−
ξ
i ∂ξi x
i
i xj
j
∂ xi = ∂∂xfi
− xi − ξi
∂ξi
xj
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 022 Differentiation II (2)
xj
The Legendre transform was differentiated with respect
to one of the orginal variables xj in Theme 20. However,
the derivative with respect to the transformed variable ξi
remains to be determined. Show that (∂φi /∂ξi )xj ,xk ,...,xn
= −xi .
By performing the same operations on temperature and negative
pressure we obtain:
τ = ∂∂H
ˆ ∂∂U
ˆ ∂∂YS π,µ =
ˆ ∂∂XS V ,µ =
(4.19)
S π,N =
S V ,N ,
∂U ˆ ∂∂Ω
=
ˆ
π = ∂∂AV τ,N =
ˆ ∂∂XV S ,µ =
(4.20)
V τ,µ
∂V S ,N
7
Termodynamic methods TKP4175
4 The . . .
1
Let us start with φi (ξi , xj ) = f (xi , xj ) − ξi xi from Eq. 4.1 and differentiate it with respect to ξi . Note that the chain rule of differentiation
∂ xi ∂f = ∂∂xfi
∂ξi
∂ξi
5
Note that all the Eqs. 4.14–4.17 have one variable in common on
the left and right hand sides (V , S , τ and π respectively). For multicomponent systems this variable will be a vector; cf. xk , . . . , xn in Theme 20.
Tore Haug-Warberg (Chem Eng, NTNU)
§ 022 Differentiation II
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
(4.32)
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Manifolds
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
In what has been written so far we note that all of the state variables
τ, S , π, V , µ and N appear in conjugate pairs such as τ dS , −S dτ,
π dV , −V dπ, µ dN or −N dµ.
1
3
The obvious symmetry reflects Eq. 4.22, which also implies that
the Legendre transform is “its own inverse”.
2
This is an important property of the energy functions.
4
However, this is only true if the function is either strictly convex
or strictly concave, as the relationship breaks down when the second
derivative f (xi , xj , xk , . . . , xn ) is zero somewhere within the domain of
definition of the free variables (see also Section 4.4).
Manifolds (2)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
5
This is illustrated in Figure 4.2 based on the transformation of the
third order polynomial
f (x ) = x (1 − x 2 )
ˆ ∂∂xf = 1 − 3x 2 .
⇒
ξ(x ) =
ˆ f − ξx can be expressed in two
The Legendre transformation of f to φ =
different ways:
3/2
φ(x ) = 2x 3
.
(4.34)
⇒
φ(ξ) = ±2 1−ξ
3
6
Moreover, both x and f can be expressed as functions of the transformation variable ξ:
1/2
1/2 2+ξ
⇒
,
x = ± 1−ξ
f (ξ) = ± 1−ξ
3
3
3
7
This means that in total we have to consider three functions involving x , and three involving ξ: f (x ), φ(x ), ξ(x ), f (ξ), φ(ξ) and x (ξ).
8
In order to retain the information contained in f (x ), we need to know
either φ(x ) and ξ(x ), or f (ξ) and x (ξ), or simply φ(ξ). In principle, the
latter is undoubtedly the best option, and this is the immediate reason
why φ(ξ) is said to be in canonical form.
9
Nevertheless, there is an inversion problem when φ is interpreted
as a function of ξ rather than x . This is clearly demonstrated by the
above example, where x is an ambiguous function of ξ, which means
that φ(ξ), and similarly f (ξ), are not functions in the normal sense.
10 Instead, they are examples of what we call manifolds (loosely
speaking folded surfaces defined by a function) as illustrated in Figures 4.2c–4.2d[a].
[a] A function is a point-to-point rule that connects a point in the domain of definition
with a corresponding point in the function range.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
150 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Termodynamic methods TKP4175
02 September 2013
151 / 1776
Figure a–b). The function f = x (1 −
x 2 ) and its Legendre transform φ =
2x 3 defined as the intersection of the
tangent bundle of f and the ordinate
axis. Note that f shows a maximum
and a minimum while φ has no extrema.
f (ξ)
φ(ξ)
ξ
ξ
d)
c)
f (x )
φ(x )
x
x
a)
b)
Figure c–d). The same two functions
shown as parametric curves with ξ =
ˆ
(∂f /∂x ) = 1 − 3x 2 along the abscissa.
The fold in the plane(s) is located at
the inflection point (∂2 f /∂x ∂x ) = 0.
Figure 4.2: The Legendre transformation of f = x (1 − x 2 ) to φ = f − ξx where
√
√
ξ=
ˆ (∂f /∂x ) = 1 − 3x 2 . The two domains of definition are x ∈ [− 2/3, 2/3]
and ξ ∈ [−1, 1]. Together, the four graphs demonstrate how an explicit function
of x , i.e. f (x ) or φ(x ), is turned into an implicit manifold when expressed in
terms of ξ, i.e. f (ξ) or φ(ξ).
Tore Haug-Warberg (Chem Eng, NTNU)
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Termodynamic methods TKP4175
02 September 2013
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
ENGLISH text is missing.Solution criteria
§ 025
1
The most general solution can be expressed as y = c ξ =
ˆ
c (∂f /∂x ), where c is an arbitrary factor.
ENGLISH text is missing.From φ to f ?
ENGLISH text is missing.Vapour–liquid
§ 24
1
In this section we will explain how, and under what conditions, it
2
It is natural to choose c = 1, which means that y ≡ ξ, as we know
from our earlier discussion of the Legendre transform, but any value of
c ∈ R+ will give the same result, as the inverse transform in Eq. 4.36 is
insensitive to scaling.
1
The above example also has practical relevance. Many equations is possible to convert the Legendre transform φ back to the original
3
We say that x and ξ are the natural (or canonical) variables of the
of state, and particularly those for the fluid state, are explicit equations function f .
functions f (x ) and φ(ξ), as they allow transformation in both directions
of the form p = p (T , V , N ). These equations of state can be used to 2 This operation will be based on the transformation rule set out in
without the loss of any information.
calculate the Helmholtz energy of the fluid.
Eq. 4.1
4
As an aside, it should be mentioned that the function φ(x ) does not
1
Finally, let us look at what differentiating φ with respect to ξ implies. 2 The problem in relation to manifolds becomes clear when this enhave this property, because y = x is not a solution to Eq. 4.36.
φ(y ) =
ˆ f − ∂∂xf x ,
(4.35)
ergy is transformed into Gibbs energy, which requires us to know the
From Theme 22, we know that the answer is −x , but
5
In the context of thermodynamics, this means that e.g.
ENGLISH text is missing.Integration
ˆ A (V (p )) + pV (p ). Along the where the properties of the variable y are, as yet, unknown.
term V = V (p ) in the definition G (p ) =
1/2
∂φ
2
A (T , V , n) can be inverse transformed to U (S , V , n), provided that
= ∓ 1−ξ
3
sub-critical isotherms the relation V = V (p ) is equivalent to a manifold 3 What we need to know is whether the same transformation rule 1 For the inverse transform to exist, it must be true that (∂ f /∂x ∂x ) , 2
∂ξ!
(∂ U /∂S ∂S )V ,n , 0.
0. This is a necessary condition.
with 3 roots as illustrated in Figure 14.3 on page 759.
applies to the inverse transform of φ(y ), such that
initially gives us a manifold in ξ with two solutions, as shown in Fig2
It is also worth adding that in general, partial differential equations 6 However, the transformation rule is symmetrical, and the same re∂φ
?
3
This results in a characteristic folding of the energy surface as exure 4.2.
f = φ − ∂y y ,
(4.36)
do not have an analytical solution. However, thermodynamic problems quirement must also apply in the other direction: U (S , V , n) can only be
amplified by the p , µ diagram in Figure 14.4 on page 767; where the
2
It is only when we introduce x (ξ) from Eq. 4.34 that the full picture
transformed into A (T , V , n) if (∂2 A /∂T ∂T )V ,n , 0.
and, if so, what y is (we have an inkling that y = (∂f /∂x ), but we need are often unusually simple, and that is also the case here.
molar volume v is used as a parameter for p (v ) along the abscissa and
emerges: (∂φ/∂ξ) = −x .
to prove it).
7
The two requirements may appear contradictory, but in actual fact
3
Let us define ξ =
ˆ (∂f /∂x ), which gives us
for µ(v ) along the ordinate axis.
3
This does not mean that we need x (ξ) to calculate the values of
they are overlapping, because:
4
Substituting Eq. 4.36 into Eq. 4.35, rearranging the expression
-1
?
∂x 4
In view of these practical considerations, we cannot claim that Legy = ξ ∂∂ξx .
(∂φ/∂ξ).
∂y
-1
-1
2
slightly, and applying the chain rule, gives us:
∂2 U = ∂∂ST V ,n = ∂∂ST V ,n = − ∂∂T ∂AT V ,n
endre transforms are globally invertible, but it is true to say that they
∂S ∂S V ,n
?
∂φ
∂φ
4
However, the relationship between x , ξ and (∂φ/∂ξ) is needed for are locally invertible on curve segments where the sign of (∂2 f /∂x ∂x )
ˆ dy /y and dln ξ =
ˆ
− ∂y y = − ∂x ∂∂xy y = ∂∂xf x .
(4.37) 4 Introducing the logarithmic variables dln y =
system identification.
dξ /ξ produces an elegant solution to the problem (note how the chain 8 In states in which one of the derivatives tends to zero, and the other
remains unchanged. We shall therefore take the precaution of treating
5
Next we substitute Eq. 4.35 into Eq. 4.37:
one tends to infinity, the system is on the verge of being unstable.
rule is used in reverse in the second-last line):
5
If the three values have been independently measured, then the each curve segment as a separate function.
?
2 -1
− ∂∂xf − ∂∂x ∂fx x − ∂∂xf ∂∂yx y = ∂∂xf x
relation (∂φ/∂ξ) = −x can be used to test the experimental values for 5 For instance, in vapor–liquid equilibrium calculations we shall des9
Depending on the variables involved, this relates to either thermal
∂x = ∂∂lnx ξ
∂ ln y
consistency.
(T , S ), mechanical (p , V ) or chemical (µi , Ni ) stability.
? ∂f ∂2 f ∂x ∂ ln ξ ∂x ignate one of the curve segments as being vapour and the other as be=1
∂x ∂x x ∂y y = ∂x x
∂x
∂ ln y
6
The existence of these kinds of tests, which can involve a variety of ing liquid, although there is no thermodynamic justification for this dis10 These kinds of states are always buried inside a region with two or
2 -1
?
∂ ln ξ
∂x y = ∂∂xf ∂∂x ∂fx .
=1
(4.39)
(4.38)
∂y
∂ ln y
physical measurements, is one of the great strengths of thermodynam- tinction, other than the fact that (∂2 A /∂V ∂V )T ,N = 0 somewhere along
more phases, being found in what is referred to as the spinodal region
the isotherm.
of the system (from the gr. spinode, meaning “cusp”).
ics.
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ENGLISH text is missing.Consistency requirements
§ 26
4 The . . .
§ 025 Inverse transform
4 The . . .
To understand the nature of the problem, let us consider the following example. From Eq. 4.4 it follows that:
∂U (S ,V ,N )
S = U (S , V , N ) − τS =
U (S , V , N ) −
ˆ A (τ, V , N ) . (4.41)
∂S
V ,N
1
Use the result from Theme 22 on page 138 to show
that performing two Legendre transformations on f ,
first with respect to xi and then with respect to ξi ,
returns the original function.
2
Given Definition 4.1 and Eq. 4.22, the inverse Legendre transformation can be written
∂φ (4.40)
φi − ∂ξ i ξi = φi − (−xi ) ξi ≡ f ,
i
xj
but it is not entirely clear what variable set should be used to define f .
To approach it from the opposite direction, we have to calculate the
Legendre transform of A (τ, V , N ) with respect to the variable τ:
∂A (τ,V ,N )
A (τ, V , N ) −
τ = A (τ, V , N ) − (−S )τ ≡ U (−S , V , N ) . (4.42)
∂τ
V ,N
§ 025 Inverse transform (2)
4 The . . .
5
In order to get back to where we started, we must perform two
further Legendre transformations:
∂U (−S ,V ,N )
U (−S , V , N ) −
(−S ) = U (−S , V , N ) − (−τ)(−S )
∂(−S )
4
This clearly returns the original function U , but the set of canonical
variables has changed from S , V , N to −S , V , N .
V ,N
=
ˆ A (−τ, V , N ) ,
3
A (−τ, V , N ) −
∂A (−τ,V ,N )
∂(−τ)
V ,N
(4.43)
(−τ) = A (−τ, V , N ) − S (−τ)
≡ U (S , V , N ) .
(4.44)
6
In other words, performing repeated Legendre transformations reveals a closed cycle, where the original information contained in U is
retained, as shown in the figure below:
Expanded footnote [a]:
[a] In practice it may be easier to use the canonical variables (x , y ) = (− (∂φ/∂z )y , y )
for the inverse transformation, rather than (x , y ) = ((∂φ/∂z )y , y ) as has been done
here.
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 025 Inverse transform (3)
U (S , V , N )
↑ −τ
A (−τ, V , N )
S
−→
−S
←−
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
§ 026
§ 25
§ 27
1
This example shows that the Legendre transform is (locally) invertible, which means that we can choose whatever energy function is most
suitable for our purposes.
2
Moreover, Eq. 4.10 tells us that the Legendre transform of U with
respect to all of the state variables S , V and N is a function with very
special properties. The function, which is known as the null potential, is
identical to zero over the entire definition domain, and as such it has no
inverse.
3
Our line of reasoning breaks down for this function, and we must
expect O (T , p , µ) to have properties that differ from those of U and the
other energy functions.
4
Note that this is not a general mathematical property of the Legendre transform; rather it follows inevitably from the fact that U and all
of the other extensive properties are described by first-order homogeneous functions to which Euler’s theorem can be applied. These functions display a kind of linearity that results in their total Legendre transforms being identical to zero. Hence, the differentials are also zero
over the entire definition domain. In thermodynamics, this is called the
Gibbs–Duhem equation.
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§ 026 Null potential
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4 The . . .
The differential of the null potential in Eq. 4.10 is of course 0.
1
A (τ, V , N )
↓τ
This becomes clear if we differentiate O (τ, π, µ) in Eq. 4.10: dU −
τ dS − S dτ − V dπ − π dV − µ dN − N dµ = 0.
4
U (−S , V , N )
02 September 2013
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
ENGLISH text is missing.Invertibility
Termodynamic methods TKP4175
Show that performing a Legendre transformation
of internal energy U (S , V , N ) with respect to all the
canonical variables S , V and N gives the null potential
O (τ, π, µ) = 0. Next, show that the differential of O
is identical to the Gibbs–Duhem equation; see also
Theme 38 in Chapter 5.
2
However, since this result is valid for the entire domain of definition
it must have a bearing on the degrees of freedom of the system.
3
Mathematically, the O -function forms a hypersurface in dim(n) + 2
dimensions.
5
If we spot that τ dS + π dV + µ dN is the total differential of internal
energy, we can simplify the expression to:
S dτ + V dπ + N dµ = 0 .
(4.45)
6
This result is identical to the Gibbs–Duhem Eq. 5.28, which plays
an important role in checking the thermodynamic consistency of experimental data.
7
This is because Eq. 4.45 tells us about the relationship between τ,
π, µ.
8
If a series of measurements is taken in order to obtain experimental
values for all of the variables τ, π, µ, then the Gibbs–Duhem equation
allows us to check the quality of the measurements, given that Eq. 4.45,
or Eq. 4.10 for that matter, must be fulfilled.
9
In practice, this is possible because the chemical potential[a], can
be expressed as a function of temperature and negative pressure, or
alternatively as f (τ, π, µ) = 0.
[a] For single component or multicomponent systems at fixed concentration.
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 027
§ 26
§ 28
News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
4 The . . .
ENGLISH text is missing.Seven samurai
1
We have in this chapter introduced seven energy functions: U , H ,
A , G , X , Y and O , and six state variables: S , V , N , t , p and µ.
2
On reflection it is clear that only four of these variables are independent, because the discussion has relied entirely on the fact that internal energy U is a function of S , V and N .
3
All of the other variables have been defined at a later stage either
as partial derivatives or as transformed functions of U .
4
There is therefore a substantial degree of redundancy in the variables that we are dealing with.
5
This redundancy becomes even clearer when the energy functions
are differentiated twice with respect to their canonical variables.
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 027 Maxwell relations
4 The . . .
Let us begin by illustrating what is meant by a Maxwell relation
using U (S , V , N ) as our starting point. From the definitions of τ and π it
follows that:
i
h
h
i
2
∂ ∂U ∂2 U = ∂∂V ∂US N
⇒
= ∂∂V ∂∂U
∂S ∂V N
∂S ∂V S ,N V ,N
S V ,N S ,N
∂π ⇒
= ∂∂τV S ,N .
∂S V ,N
1
Set out all of the Maxwell relations that can be derived by
applying the Schwarz theorem ∂2 f /∂xi ∂xj = ∂2 f /∂xj ∂xi
to the functions: U , A , H , Y , G , Ω, X and O , restricting
yourself to single-component systems.
Expanded footnote [a]:
[a] Hermann Amandus Schwarz, 1843–1921. German mathematician.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Expanded footnote [a]:
[a] James Clerk Maxwell, 1831–1879. Scottish physicist.
Similarly, cyclic permutation of the variables S , V and N yields:
2
∂µ ∂2 U = ∂∂N ∂US V
⇒
= ∂∂τN S ,V ,
∂S ∂N V
∂S V ,N
2
∂µ ∂2 U = ∂∂N ∂UV S
⇒
= ∂∂π
∂V ∂N S
∂V S ,N
N S ,V .
2
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
−
∂V τ,µ
=
∂π
∂S ∂τ V ,µ
∂µ τ,V
=
4 The . . .
∂N ∂τ V ,µ
−
∂π
∂µ τ,V
=
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∂N 4 The . . .
3
By systematically comparing the second derivatives of all of the
Legendre transforms mentioned in Theme 19 on page 127, we get the
results shown below.
7
4
Also see Theme 24 on page 144 for a complementary table of total
differentials and first derivatives.
5
Note that the Maxwell relations which are derived from the null potential O (τ, π, µ) have no physical interpretation, as experimentally τ, π
and µ are dependent variables.
6
This means that e.g. (∂τ/∂π)µ is unpredictable because Eq. 4.45
on integrated form is equivalent to a relation g such that g(τ, π, µ) = 0.
These relations have therefore been omitted from the table.
−
−
∂π = ∂∂τV S ,N
∂S V ,N
∂S ∂V τ,N
∂τ =
=
∂π S ,N
∂τ =
∂V S ,µ
∂S =
∂π τ,N
∂τ − ∂π
=
S ,µ
∂π
∂τ V ,N
−
∂π ∂S V ,µ
−
∂V ∂S π,N
∂V ∂τ π,N
∂V ∂S π,µ
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 027 Maxwell relations (3)
∂S Termodynamic methods TKP4175
§ 027 Maxwell relations (2)
−
−
∂µ = ∂∂τN S ,V
∂S V ,N
∂µ
∂S = ∂τ V ,N
∂N τ,V
∂µ ∂τ
= ∂S π,N
∂N S ,π
∂τ
= ∂∂N
∂µ S ,V
S V ,µ
∂µ
∂S = ∂τ π,N
∂N τ,π
∂τ = ∂∂N
∂µ S ,π
S π,µ
Termodynamic methods TKP4175
−
−
−
∂µ = ∂∂π
∂V S ,N
N S ,V
∂µ ∂π = ∂V τ,N
∂N τ,V
∂µ
∂V
= ∂π S ,N
∂N S ,π
∂π
= ∂∂VN S ,µ
∂µ S ,V
∂µ
∂V = ∂π τ,N
∂N τ,π
N
∂V = ∂∂π
∂µ S ,π
S ,µ
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
Alternative procedure
4 The . . .
§ 028
§ 27
§ 29
4 The . . .
1
We are familiar with the key characteristics of Legendre transforms, having looked at topics like transformation geometry, differentiation rules, inverse transforms and Maxwell relations.
∂V τ,µ
The subject becomes even more interesting now, as we are going
3
to prove another rule, which is even simpler, although not so easy to
derive.
2
In the derivations we have focused primarly on the simple rule used
in Figure 4.1, which shows that the Legendre transform is equal to the
intersection of the tangent bundle of the original function and the ordinate axis.
Verify the Gibbs–Helmholtz equation (∂(G /τ)/∂(1/τ))π,N = H .
See if you can generalise this result.
4
The key to this is a simple change in variables from y and x in
(∂y /∂x ) to y /x and 1/x in the expression (∂(y /x )/∂(1/x )).
5
This relationship is generally known as the Gibbs–Helmholtz equa-
tion.
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
§ 028 Gibbs–Helmholtz
1
2
By differentiating and then substituting (∂G /∂τ)π,N = −S and G +
TS = H , we can confirm that the equation is correct, but there is more
to it than that.
4 The . . .
≡y+
1 ∂y
x ∂(1/x )
∂y
∂x
2
we can conclude that
∂(f /xi ) ∂(1/xi ) xj ,xk ,...,xn
Looking back
1
≡y −x
=
ˆ f − ξi xi =
ˆ φi ,
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News Canonical Manifolds Inversion Maxwell Gibbs–Helmholtz Olds
From the mathematical identity
∂(y /x )
∂(1/x )
Termodynamic methods TKP4175
π,N
π,N
Note that the enthalpy has been differentiated with respect to negative pressure, which is a canonical variable; the result follows an easily recognisable pattern. We will come back to this form of the Gibbs–
Helmholtz equation in Chapter 8, but there it will be used to integrate
equations of state at a fixed composition.
2
Finally, it is worth adding that every time we derive an expression of
the form y − x (∂y /∂x ) we can choose to replace it with (∂(y /x )/∂(1/x )).
These two alternatives look different, but they are nevertheless mathematically identical.
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E.g. H =
ˆ U − (−p )V = U + pV
3
Differentiation I: (∂φi /∂xj )ξi ,xk ,...,xn = (∂f /∂xj )xi ,xk ,...,xn
E.g. (∂H /∂S )−p ,N = (∂U /∂S )V ,N =
ˆ T
Part
(4.46)
5
One of the classic applications of the Gibbs–Helmholtz equation
is in calculating the temperature derivative of the equilibrium constant,
as shown in Chapter 16, where ln K =
ˆ −∆rx G◦ /RT is shown to be an
almost linear function of 1/T with a slope of ∆rx h◦ /R .
7
Canonical forms: U (S , V , N ), H (S , π, N ), A (T , V , N ), G (τ, π, N ), . . .
8
Conjugated variables: (τ, S ), (π, V ), (µ, N ), . . .
9
Maxwell relations: (∂2 U /∂S ∂V )N = (∂2 U /∂V ∂S )N , . . .
10
Gibbs–Helmholtz: (∂(G /τ)/∂(1/τ))π,N = H
6
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Transformation rule: φi (ξi , xj , xk , . . . , xn ) =
ˆ f (xi , xj , xk , . . . , xn ) − ξi xi
cf. the Legendre transform in Eq. 4.1. The identity applies generally,
including to φij and to any other derived transforms.
3
This means that Eq. 4.46 provides an alternative way of calculating
the value of a Legendre transform, without this changing the original
definition.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
4
4
Termodynamic methods TKP4175
4 The . . .
ENGLISH text is missing.Continued differentiation
1
It remains to be shown that it is straightforward to differentiate the
Gibbs–Helmholtz equation. To illustrate this, let us use the derivative of
the equation given in the worked example above:
∂(V /τ)
∂(G /τ)
∂H ∂ ∂(G /τ)
∂
= ∂π
= ∂(1/τ)
= ∂(1/τ)
.
∂π τ,N
∂(1/τ)
∂π τ,N
π,N τ,N
Tore Haug-Warberg (Chem Eng, NTNU)
5
Euler’s Theorem on Homogeneous Functions
Differentiation II: (∂φi /∂ξi )xj ,xk ,...,xn = −xi
E.g. (∂H /∂(−p ))S ,N = −V
see also Part-Contents
5
An equally useful equation is (∂(A /τ)/∂(1/τ))V ,N = U , and in fact
there are a whole series of similar equations that are based on the same
principle.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
New concepts
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
State description
5 Euler’s . . .
1
Homogeneity
2
Scaling laws
3
Extensive (order k = 1)
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
1
The thermodynamic state of a so-called simple system is described
mathematically by the mapping f (x) : x ∈ Rn+2 → R where n ∈ N
stands for the number of indentifiable chemical components in the
system.
Expanded footnote [a]:
[a] An empty chamber is without internal mass and has therefore zero chemical components, but it nevertheless constitutes a thermodynamic system of electromagnetic
radiation with two degrees of freedom.
Homogeneity
1
More precisely, the function f (x1 , . . . , xn , ξn+1 , . . . , ξm ) is homogeneous of order k in the variables x1 , . . . , xn if the following criteria are
satisfied:
2
The practical consequences of this linearity[a] will be clarified when
we start investigating the mathematical properties of f , but let us first
formalise the said linearity by stating that the function f (x1 , . . . , xn ) is
homogeneous of order k ∈ Z provided that the parametrised function
f (λx) is proportional to λk in the direction of x = (x1 , . . . , xn ).
3
Here, the parameter λ ∈ R+ is a dimensionless measure of the
distance from the origin to the coordinate tuple λx ∈ Rn . We shall later
learn that λ is also a measure for the physical size of the system.
[a] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,
2nd edition, 1985.
2
For the system to be physically realisable, it must be macroscopically uniform and the properties of the system must not explicitly depend on time — which is to say the system is without memory of any
kind.
3
The domain of definition is mathematically very flexible despite this
limitation and the only formal requirement we ask is that n has a finite
value.
4
This general description fits all geometric surfaces in n + 3 dimensions but the understanding of the thermodynamic systems is considerably simplified by the fact that its energy function f is linear along all
state vectors x starting at the origin.
4
Intensive (order k = 0)
5
Hessian matrix
6
Euler’s 1st theorem (Euler integration)
7
Euler’s 2nd theorem
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F (X ) vs. f (x )
k =−1
F (X ) vs. f (x ) (2)
5 Euler’s . . .
Note that F has exactly the same function
definition as f . Hence, it is only by convention that we distinguish the two forms. We
may therefore write:
k =1
k =0
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5 Euler’s . . .
2
The distinction between F and f is a little
contrived and can use a more thorough explanation:
3
In thermodynamics it is customary to use the same symbol for
closely related functions like e.g. S (T , V , N ), S (T , p , N ) and S (H , p , N ).
4
The state functions are supposedly all the same but their free variables differ so the function mappings must necessarily be different.
5
The current situation represents the opposite: Two different symbols F and f are used to map two different states while referring to the
same function definition.
6
The difference is only a matter of scaling of the system size[a].
7
We can easily explain this with an example.
[a] Let X = λx indicate a scaling of the system by the dimensionless factor λ. Here,
k = 1 is an important special case, under which the function F is considered extensive. In thermodynamics it is common to write F = X hf i in this case, where
X is an extensive physical quantity with an associated unit of measurement and
where hf i is the properly normalized version of f . How this can be understood in
terms of Euler’s theorem becomes clearer if we consider a system with one single
group of homogeneous variables. We could look, for example, at the total number
of moles for all components in the system while it is kept at a constant composition
(constant mass or mole fractions). In this case F (X ) = λf (x ) where X = λx and
X = {X } [X]. Here, {} denotes the magnitude of X and [] denotes its unit of measurement. Let us now choose x = X /{X } i.e. λ = {X }. It is then possible to write
F (X ) = {X } · f (X /{X }) = {X } [X] · f (X /{X })/ [X] = X · f (X /{X })/ [X]. The most common
examples in thermodynamics are X = N [mol], X = M [kg] and X = V [m3 ]. Specifically, when X = N [mol] we also get F (N ) = Nf (N = 1 mol)/ [mol] = N hf i. The expression is the same as F = λf except that N is a physical quantity whilst λ is a dimensionless number. This means that f has the unit of [F] while hf i is the associated
molar quantity [F mol-1 ]. Subtle, yes, but nevertheless essential because in our dayto-day work we use hf i rather than f . It is for instance quite common to write the total Gibbs energy for the system as G (τ, π, N ) = Ng(τ, π, x) where x is a vector of mole
ˆ hf i defractions, or, alternatively, as G = Ng(τ, π) if x = ( 1 ). The function g =
notes molar Gibbs energy in both cases. Correspondingly, we can write U = Nu(s , v ),
H = Nh (s , π) etc. for the other molar energies. Our Euler notation does not have oneto-one equivalents in other writings on the subject, but it is hard to avoid conflicts when
the established standard does not fully reflect the physico-mathematical reality.
F (X1 , . . . , Xn , ξn+1 , . . . , ξm ) = λk f (x1 , . . . , xn , ξn+1 , . . . , ξm ) ,
(5.1)
Xi =
ˆ λxi .
(5.2)
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
Let f (x ) = 3x 2 . Hence
Homogeneity
5 Euler’s . . .
ENGLISH text is missing.Differentiation of F (X )
1
Just by curiosity we can go on and see what happens when f (x ) is
differentiated to (∂f /∂x ) = 6x . Simultaneously, for F (X ):
∂3(λx )2 ∂F = ∂(λx ) = 6(λx ) = λ · 6x = λ ∂∂xf
∂X
2
There are two things to be noted here:
3
Firstly (∂F /∂X ) has exactly the same function definition as (∂f /∂x )
and secondly the derivative is no longer homogeneous of 2nd order but
of 1st order.
4
The first observation is what should be expected because the differentiation operator is depending on the function definition only and not
the value of x .
5
The second observation is the prelude to a prolonged discussion
that will occupy the rest of this chapter.
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1
It has tacitly been assumed that ξn+1 , . . . , ξm do not take part in the
homogeneity of f . That means the scaling laws in Eq. 5.1–5.2 is valid
for all choices of ξn+1 , . . . , ξm .
2
They are of course used in the function descriptions of F , and in all
the partial derivatives of F , but they do not take part in the scaling of xi .
Taking this a step further, one can for instance say that the volume
5
v of a cube with dimensions x , y and z is a homogeneous function of
order 1 in the height z given the base area xy , and a homogeneous
function of order 3 if all the dimensions x , y og z are scaled the same
amount.
3
Strange at first sight, maybe, but the grouping of variables into disjoint subsets is quite natural in physics.
4
For example the kinetic energy of an n-particle ensemble EK (m,
P
υ) =
ˆ 12 in mi υ2i is homogeneous of order 1 with respect to the
masses mi and homogeneous of order 2 with respect to the velocities υi . The same function is homogenous of order 3 if mass and velocity are considered simultaneously as independent variables.
F (X ) =
ˆ f (λx ) = 3(λx )2 = λ2 · 3x 2 = λ2 f (x )
x
which proves that f (x ) is homogeneous of 2nd order.
X
8
However, what is more important right now is that F (X ) is just another way of writing f (λx ).
F ⇄f
k =2
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
1
F
Termodynamic methods TKP4175
5 Euler’s . . .
9
The two function mappings are exactly the same although the function values (for any given x ) are differerent.
X
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From our daily lives
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
Differential in X and ξ
5 Euler’s . . .
Of particular interest to us are the energy functions U , A , . . . , O
with state variables belonging to S , V , and N or τ, π, and µ.
1
The energy functions, and entropy, volume and mole number, are
homogeneous functions of order 1, while temperature, pressure and
chemical potential are homogeneous functions of order 0.
2
Expanded footnote [a]:
[a] Here, τ and π are used for temperature and negative pressure respectively, as
already introduced in Chapter 4. In this context we want to stress that the properties
(in common with the chemical potential µ) are intensive quantities.
In the context of thermodynamics these quantities are referred to
as extensive (k = 1) and intensive (k = 0) state variables respectively
.
3
Expanded footnote [a]:
[a] A physical quantity is extensive if it is proportional to the system size and intensive
if it is insensitive to the system size.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
We are not initially aiming to embark on a general discussion of
the multicomponent functions in Eq. 5.1. Instead, we will start with a
detailed analysis of the simpler two-variable function F (X , ξ).
1
3
In order to exploit the function properties, it makes sense to use
the total differential of F expressed in X and ξ-coordinates:
dF = ∂∂XF ξ dX + ∂∂ξF dξ .
Differential in λ and f
5 Euler’s . . .
An alternative would be to make use of F = λk f from Eq. 5.1 as
the starting point for the derivation:
1
dF = k λk −1 f dλ + λk df .
2
The results will later be generalised in Chapter General Theory,
and we lose nothing by taking an easy approach here.
X
The variable X is defined as a function of x and λ in Eq. 5.2, and
by substituting the total differential of X , or more precisely dX = λ dx +
x dλ, we obtain
(5.3)
dF = ∂∂XF ξ x dλ + ∂∂XF ξ λ dx + ∂∂ξF dξ .
4
X
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
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02 September 2013
184 / 1776
2
Substitution of the total differential of f expressed in x and ξ-coordinates gives:
∂f dξ .
(5.4)
dF = k λk −1 f dλ + λk ∂∂xf ξ dx + λk ∂ξ
x
Note that Eq. 5.3 and Eq. 5.4 are two alternative expressions for
the same differential dF (λ, x , ξ).
3
4
Comparing the equations term-by-term reveals three relations of
great importance to thermodynamic methodology:
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
Case dλ
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
Comparing the dλ terms reveals that (∂F /∂X )ξ x = k λk −1 f . Multiplying both sides by λ gives
∂F λx = k λk f ,
∂X ξ
1
which if we substitute in X =
ˆ λx from Definition 5.2 and F = λk f
from Eq. 5.1 can be transformed into:
∂F X = kF .
(5.5)
∂X ξ
2
Case dx
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
Comparing the dx terms reveals that (∂F /∂X )ξ λ = λk (∂f /∂x )ξ .
Dividing each side by λ leads to:
∂F = λk −1 ∂∂xf ξ .
(5.6)
∂X ξ
1
3
2
Because F has the same function definition as f (it is only the
names of the free variables that differ) it follows that ∂F /∂X and ∂f /∂x
are equivalent functions.
Hence, (∂F /∂X )ξ is a homogeneous function of order k − 1.
Case dξ
1
Comparing the dξ-terms reveals that the homogeneity of the
derivative with respect to ξ is unchanged:
∂F ∂f = λk ∂ξ
,
(5.7)
∂ξ
X
x
Differentiation with respect to ξ therefore conserves the homogeneity in X .
3
4
Differentiation with respect to X therefore reduces the order of homogeneity of F by one.
5 Euler’s . . .
2
Using the same arguments as in the case above this implies that
of F with respect to ξ is a new homogeneous function of
the derivative
degree k .
4
The result is known as Euler s first theorem for homogeneous
functions, or simply as the Euler integration of F .
Expanded footnote [a]:
3
The closed form of Eq. 5.5 indicates that there is a general solution
R
R
to the indefinite integral F (X , ξ) = (dF )ξ = (∂F /∂X )ξ dX .
[a] Leonhard Euler, 1707–1783. Swiss mathematician.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
Extended notation I
2
In fact, any scaled variable x = λ-1 X satisfies the equation:
∂f ∂x ξ x = kf .
5 Euler’s . . .
(5.8)
3
The validity of this statement is confirmed by first combining
Eqs. 5.5 and 5.6, and then substituting in Eq. 5.1 and Definition 5.2.
4
This result emphasises the practical importance of Euler’s theorem
as outlined in Eq. 5.5.
5
Euler homogeneous functions define a small yet important function
class in mechanics, and in thermodynamics in particular.
One of the reasons of their success is that the Legendre transform
φi =
ˆ f (x, ξ) − (∂f /∂xi ) xi of f conserves the homogeneity of f . From the
differentiation rules layed out above we already know that (∂f /∂xi ) is
homogeneous of order k − 1. Hence, the multiplication with xi into the
product (∂f /∂xi ) xi regains the homogeneity of order k . Subtracting two
homogeneous functions with the same homogeneity order k does not
change the value of k and we can conclude that the Legendre transform
of f with respect to xi conserves the Euler homogeneous properties of
f (x, ξ).
189 / 1776
6
§ 029
1
The general properties of homogeneous functions will be explained further in Chapter General Theory, but to get a sense of the
overall picture we shall briefly mention what changes are required in
Eqs. 5.5–5.7 to make them valid for multivariate functions:
n
P
i =1
Xi ∂∂XFi
Xj ,i ,ξl
∂F ∂Xi Xj ,i ,ξl
∂F ∂ξk Xj ,ξl ,k
= kF
=
k
=λ
∂f ∂ξk xj ,ξl ,k
187 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
§ 30
5 Euler’s . . .
§ 029 Intensive functions
188 / 1776
2
The functions τ, π and µ mentioned above must first be defined.
5 Euler’s . . .
Mathematically, the total differential of U (S , V , N ) is:
∂U ∂U dU = ∂∂U
S V ,N dS + ∂V S ,N dV + ∂N S ,V dN .
Expanded footnote [a]:
3
Comparing this with the differential given in the text implies that τ =
ˆ
ˆ (∂U /∂V )S ,N and µ =
ˆ (∂U /∂N )S ,V , see also Theme 19
(∂U /∂S )V ,N , π =
on page 127 in Chapter 4.
[a] Actually an empty set. Here it is used to denote a missing variable or an empty
vector.
5
The function τ(S , V , N ) is obviously homogeneous of order 0 in the
variables S , V , N because it is independent of the scaling factor λ.
(5.11)
.
02 September 2013
Substitution of k = 1, Xi = S , Xj ∈ {V , N } and ξl = ∅ in Eq. 5.10
4
yields:
∂U (S ,V ,N )
∂u(s ,v ,n)
τ(S , V , N ) =
ˆ
= λ0
= τ(s , v , n) .
(5.12)
∂S
∂s
v ,n
V ,N
Internal energy U = U (S , V , N ) is an extensive function
in the variables S , V , and N . The total differential of U
is dU = τ dS + π dV + µ dN . Explore the homogeneity
associated with the functions for τ, π and µ.
(5.10)
,
§ 28
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
1
(5.9)
,
λk −1 ∂∂xfi
xj ,i ,ξl
02 September 2013
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
ENGLISH text is missing.Note
1
It should be stressed that the Euler integration in Eq. 5.5 is not
limited to one particular interpretation of X .
Termodynamic methods TKP4175
2
In other words, increasing the dimensionality of x does not add to
the complexity of the homogeneous properties of f very much.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
ENGLISH text is missing.Partial pressure
1
A prolonged experience with thermodynamic issues will slowly
strengthen our intuition with regard to the role of the intensive and extensive state variables but even at this early stage we can give a striking example on the importance of the Euler homogeneity in the measurement of physico-chemical quantities.
2
As all chemists know there is something called partial pressure, but
there are few of us, I believe, who have ever been given the opportunity
to reflect on the exact nature of this quantity.
3
Let us by chance look at Eq. 5.13 which tells us that pressure is an
intensive variable.
4
We can exploit this property further by substituting k = 0 into
Eq. 5.9, see below:
V
5
∂p ∂V T ,n
+
n
P
i =1
V
−NRT
V2
+
7
6
The temperature is commonly referred to as an intensive variable
and is taken to be independent of the system size[a].
[a] Valid only when the system is sufficiently big i.e. when the number of particles is
large enough to fix the temperature.
n
P
i =1
Ni
RT
V
=0·p =0
=0
Back-substituting p ıg = NRT /V :
After reorganisation and introduction of the mole fraction yi we get:
−p ıg +
§ 029 Intensive functions (2)
∂p ∂Ni T ,V ,Nj ,i
Ni
Substituting ideal gas behaviour:
6
7
p ıg =
n
P
i =1
n N
P
i ıg
p =0
N
i =1
yi · p ıg =
ˆ
n
P
i =1
pi
8
To our surprise the partial pressure is not due to an independent
definition of any kind but is actually an inevitable implication of the homogeneity properties of p ıg representing an ideal gas mixture.
5 Euler’s . . .
Similarly, differentiation of U with respect to V and N yields
∂U (S ,V ,N )
∂u(s ,v ,n)
ˆ
= λ0
= π(s , v , n) ,
(5.13)
π(S , V , N ) =
∂V
∂v
S ,N
s ,n
∂U (S ,V ,N )
0 ∂u(s ,v ,n)
ˆ
=
λ
=
µ(
s
,
v
,
n
)
,
(5.14)
µ(S , V , N ) =
∂N
∂n
S ,V
s ,v
where the (negative) pressure and the chemical potential are intensive
variables as well.
§ 030
9
In practise, pi also finds use in the description of non-ideal gases,
which we to a certain extent may find useful, but only if we at the same
time make clear that the property then is not thermodynamically measureable.
10 Vi cannot, therefore, define the partial pressure as e.g. “the fraction
of the pressure that is due to a single component i in the mixture”.
11 That would in fact be quite meaningless because the pressure in
a gas mixture is a communal property that is caused by all the components acting together.
12 Due to the mutual physical interactions between the molecules it is
not possible to sum up the contributions from each of the components
in an independent manner.
§ 29
§ 31
5 Euler’s . . .
13 The question is then: What is the correct interpretation of the partial
pressure?
Well, by winding backwards through the list of derivations pi can
alternatively be expressed on this form:
∂p ıg ∂p ıg Ni NRT
pi =
ˆ ∂ log N
= Ni ∂N
= Ni RT
ˆ yi p ıg
V = N · V =
14
i
T ,V ,Nj ,i
i
T ,V ,Nj ,i
15 The conclusion is that the partial pressure pi can be understood as
the partial derivative of the pressure of an ideal gas mixture with respect
to the logarithm of the mole number of component i .
16 On this ground the quantity is secured as a thermodynamic state
variable of the same kind as pressure itself.[a]
194 / 1776
[a] The discussion of the physical foundation of the partial pressure has a lot in common with the discussion of pH which is defined as “the negative logarithm of the concentration of H+ ions”. They do both refer to quantities that are not measureable by
thermodynamic means alone. I mean: What would be the pH in a sample of methanol?
Maybe 20 or even 40 since there are scarcely any H+ ions present in that medium.
We instinctively feel there is a lack of stringency which leads us into making wrong assumptions about the medium’s true nature. The true nature of the problem becomes a
little clearer if we incorporate the ionization constant of the solvent into the discussion
but even then the definition is not proof because are only able to measure the concentration of the H+ ion indirectly (using conductivity for instance).
§ 030 Euler integration
5 Euler’s . . .
1
Substituting k = 1, Xi ∈ {S , V , N } and ξl = ∅ into Eq. 5.9, together
with the definitions from Theme 29 on page 177 i.e. τ =
ˆ (∂U /∂S )V ,N ,
ˆ (∂U /∂N )S ,V yields Euler’s equation applied to
π=
ˆ (∂U /∂V )S ,N and µ =
internal energy:
Internal energy U = U (S , V , N ) is an extensive function
in the variables S , V and N . The total differential of
U is dU = τ dS + π dV + µ dN . What is the correct
integral form of U ? Explain the physical significance
of this integral.
U = τS + πV + µN .
(5.15)
2
We are not being asked about multicomponent mixtures, but
Eq. 5.15 is quite general and can (by analogy with Eq. 5.9) be extended
to:
n
P
µi Ni =
U = τS + πV +
ˆ τS + πV + µT n
(5.16)
i =1
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 031
§ 30
§ 32
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
ENGLISH text is missing.Remark
1
The total differential of internal energy for a single component
system is dU = τ dS + π dV + µ dN .
2
The differential can be integrated without actually solving any
partial differential equations because τ, π, µ are intensive (sizeindependent) variables.
3
Physically, this means that the system can be built up from zero
size[a] in a manner that keeps τ, π, µ constant during the process.
4
This is achieved by combining a large number of small systems at
fixed temperature, (negative) pressure and chemical potential.
5
When these subsystems are merged into one big system there will
be no changes in the intensive properties, because the requirements for
thermodynamic equilibrium are automatically fulfilled, see also Chapter 21 on page 1312.
197 / 1776
[a] By definition, a system of zero size has zero internal energy, i.e. U = 0. In this
context, “zero size” refers to zero volume, zero mass and zero entropy. Note, however,
that it is not sufficient to assume zero mass, because even an evacuated volume will
have radiation energy proportional to T 4 !
§ 031 Extensive functions
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 033
5 Euler’s . . .
1
First of all the functions S , V , and µ must be defined. Mathematically, the total differential of G (τ, π, N ) is:
G
dπ + ∂∂G
dG = ∂∂τG π,N dτ + ∂∂π
N τ,π dN .
τ,N
ENGLISH text is missing.Non-canonical variables
§ 32
1
Any further discussion of homogeneity should ideally be based on
the properly identified canonical state variables of the function.
§ 34
2
However, this is not an absolute condition, as the thermodynamic
state of a system is uniquely determined when any of its independent
variable sets is fully specified.
3
For example U is extensive in S , V , N while G is extensive in N at a
given τ and π, but we can alternatively state the opposite and say that
U is extensive in N at a given τ and π and that G is extensive in S , V ,
N.
4
There are plenty of alternative ways to describe the system, but a
combination of three arbitrary state variables is not always sufficient.
5
To demonstrate this, let us look at the specification H , τ, N .
6
For example, the ideal gas enthalpy can be written H ıg = H (τ, N ).
At a given τ, N the system is underspecified, because there is no way
we can determine the pressure.
5 Euler’s . . .
§ 32 Gibbs energy II
1
From the definition G (τ, π, N ) =
ˆ U − τS − πV , where U =
U (S , V , N ), τ =
ˆ (∂U /∂S )V ,N and π =
ˆ (∂U /∂V )S ,N , it follows that we
can express Gibbs energy as a function of S , V and N , because the
side of the equation only includes U and functions derived
right-hand
from U :
G (S , V , N ) = U (S , V , N ) − τ(S , V , N )S − π(S , V , N )V .
2
We already know that τ and π are intensive variables, whereas U ,
S and V are extensive variables.
3
From this it follows that
G (S , V , N ) = λu(s , v , n) − τ(s , v , n)λs − π(s , v , n)λv
=
ˆ λg (s , v , n) ,
which clearly demonstrates that G is homogeneous, in accordance with
7
If we try to specify H in addition to τ and N , this will give us a § 32
redundant (or even contradictory) statement.
1
Show that G (S , V , N ) is really an extensive function of the vari- the conjecture.
8
To avoid problems of this kind we shall therefore make use of ables S , V and N , thus verifying the conjecture set out in the introduc- 4 Note that the last equation tacitly assumes the homogeneity relacanonical variables unless otherwise stated.
tion to this chapter (on page 184).
tions S = λs , V = λv and N = λn.
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2
We will apply the same procedure as was used in Theme 29 on
page 177.
Gibbs energy G (τ, π, N ) is an extensive function of
the mole number N at a given temperature τ and
(negative) pressure π. The total differential of G is
dG = −S dτ − V dπ + µ dN . Explore the homogeneity
associated with the functions S and V .
Termodynamic methods TKP4175
3
Comparing this with the differential in the problem formulation
yields −S (τ, π, N ) = (∂G /∂τ)π,N , −V (τ, π, N ) = (∂G /∂π)τ,N and µ =
(∂G /∂N )τ,π .
5
From Eq. 5.10 it can be seen that the chemical potential is (still) a
homogeneous function of order 0, see also Eq. 5.14.
4
From Eq. 5.11 we can deduce that both volume and entropy are
homogeneous functions of order 1 in the mole number at specified temperature and (negative) pressure, i.e. they are extensive variables.
02 September 2013
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 033 Gibbs energy I
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
The function is nonlinear in N1 and N2 , and the isopleths must be
calculated iteratively using e.g. Newton–Raphson’s method: N2,k +1 =
N2,k − (Gk − G )/µ2 , where µ2 = ax12 + ln x2 is the partial derivative of G
with respect to N2 .
1
§ 033 Gibbs energy I (2)
6
5 Euler’s . . .
6
This indicates that the phase equilibrium condition is fulfilled whenever two points on the same isopleth have common tangents (remember that G takes constant values along each isopleth such that the criterion is reduced to µ1 = µ2 ).
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G
µ2
5
5
It can be proved (do it!) that the tangent intersects with the y and x
axes at G /µ2 and G /µ1 respectively.
4
G
µ1
3
2
Note that each isopleth defines a non-convex region, which can be
interpreted as a fundamental thermodynamic instability.
3
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
N2
Tore Haug-Warberg (Chem Eng, NTNU)
The Gibbs energy of a binary mixture is given as a
homogeneous function of order 1 in the mole numbers
N1 and N2 (at fixed T and p ). Make a contour diagram
illustrating the function G = aN1 x2 + N1 ln x1 + N2 ln x2
where N2 is plotted along the ordinate axis and N1 along the
ˆ N1 /(N1 + N2 ) and x2 =
ˆ N2 /(N1 + N2 ).
abscissa. Let x1 =
Show that the isopleths corresponding to constant
G (the contour lines) are equidistant for a series of
evenly distributed Gibbs energy values. Use a = 2.4
in your calculations.
2
A fixed value is selected for N1 , and N2 is iterated until Gk →∞ has
converged to G , see Program 1.4 in Appendix 31. The result obtained
is shown in Figure 5.1 on page 187.
1
The corresponding two-phase region (the symmetry of the model
reduces the phase equilibrium criterion to µ1 = µ2 ) can be calculated
from the total differential of G , rewritten here into the tangent of the
isopleth:
µ
dN2 = − µ12 .
dN1
4
0
T ,p ,G
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§ 034
§ 33
§ 35
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5 Euler’s . . .
§ 034 Euler’s 2nd theorem
X
For k = 1 the expression reduces to X d(∂F /∂X )ξ = (∂F /∂ξ)X dξ.
To proceed we need the differential of ∂F /∂X , and because F is a function of X and ξ it can be written as the total differential
2
2F dξ ,
d ∂∂XF ξ = ∂∂X ∂FX ξ dX + ∂∂X ∂ξ
2
For k = 1 this implies that (∂2 F /∂X ∂X )ξ = 0 and
(∂F /∂ξ)X = (∂2 F /∂X ∂ξ) X . Verify these results and
give a physical explanation for them.
4
6
Termodynamic methods TKP4175
§ 034 Euler’s 2nd theorem (2)
5 Euler’s . . .
The left-hand side of Eq. 5.5 is differentiated and the right-hand
side is replaced by the total differential of F :
X d ∂∂XF ξ + ∂∂XF ξ dX = k ∂∂XF ξ dX + k ∂∂ξF dξ .
X
2
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
1
Homogeneity causes a whole range of remarkable
results. One formula obtained by differentiating Eq. 5.5
is:
∂F ∂F X d ∂X ξ − k ∂ξ dξ = (k − 1) ∂∂XF ξ dX ,
0
N1
Figure 5.1: Contour diagram of Gibbs energy
(solid lines). Equidistant values
(open circles) are made visible along 3 rays (dotted lines) from the origin.
The two-phase region is spanned by the two outermost rays. One of the
isopleths indicates the phase equilibrium condition in the shape of a convex
hull construction.
3
Note that X and dξ are arbitrary.
5 Euler’s . . .
This implies that two non-trivial relations follow from Eq. 5.17 (one
7
equation in three variables gives two non-trivial relations), irrespective
of the actual values of X , dX , ξ and dξ:
∂2 F = 0,
(5.18)
∂ X ∂X ξ
2F X ∂∂X ∂ξ
(5.19)
= ∂∂ξF
6
The trick is to recognise that X , dX and dξ are independent variables.
X
5
Substituted into the equation above (letting k = 1), this yields the
intermediate result:
2
2F dξ = ∂∂ξF dξ .
X ∂∂X ∂FX ξ dX + X ∂∂X ∂ξ
(5.17)
4
where dX is also arbitrary.
X
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
Example
1
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
ENGLISH text is missing.Extended notation II
1
5 Euler’s . . .
In the general case, x and ξ would be vector quantities.
2
If F (x, ξ) is extensive in x it can be shown that the differential in
Theme 34 on page 188 takes the form:
T
(5.20)
xT d ∂∂xF ξ − ∂∂ξF dξ = 0 ,
x
where the differential of ∂F /∂x is written:
∂2 F d ∂∂xF ξ = ∂x∂x
dx +
∂2 F ∂x∂ξ dξ
.
(5.21)
3
The two quantities dx and dξ are independent, and by substituting
the differential 5.21 into Eq. 5.20 it follows that:
function F (X , ξ) can
§ 35 Linear potentials
∂2 F x = 0,
(5.22)
∂x∂x ξ
1
The three functions G , Y and Ω are extensive in N , S , and V
∂2 F (5.23)
x = ∂∂ξF .
respectively, i.e. in one variable each.
This stems from the fact that F (0, ξ) = 0 in addition to ∂2 F /∂X ∂X =
∂ξ∂x
x
0, see Eq. 5.18[a] The derivative of F with respect to ξ is therefore an
2
This makes Eq. 5.18 valid and the substitution of the definitions for
4
From a logical point of view Eq. 5.23 is a true generalisation of 5.19,
extensive function β′ (ξ)X where the second derivative of F with respect
τ, π and µ yields:
whereas Eq. 5.18 represents a specialisation of 5.22. This means that
to both X and ξ is equal to the intensive parameter β′ (ξ).
∂µ ∂2 G the second derivative of F (X , ξ) with respect to X is zero for all single= ∂N τ,π = 0 ,
(5.24)
∂N ∂N τ,π
3
The numerical value of ∂F /∂ξ can easily be obtained by integrating variable functions, while the corresponding Hessian F (x, ξ) for multi
∂2 Y § 35
= ∂∂τS π,µ = 0 ,
(5.25) § 36
∂S ∂S π,µ
∂2 F /∂X ∂ξ using the Euler method as shown in Eq. 5.19.
variate systems is singular in the direction of x, where x is proportional 1
Use the result from Theme 34 on page 188 to determine the
1
Show that the Hessian of A = A (τ, V , N ) has only one indepen∂2 Ω ∂π 210 / 1776 to one of the eigenvectors of the Hessian.
(5.26)
∂V ∂V τ,µ = ∂V τ,µ = 0 .
second derivatives (∂2 G /∂N ∂N )τ,π , (∂2 Y /∂S ∂S )π,µ and (∂2 Ω/∂V ∂V )τ,µ ,
dent element when the temperature τ is taken to be a constant paramwhere G = G (τ, π, N ) and Y = Y (S , π, µ) and Ω = Ω(τ, V , µ).212
eter. Use Eq. 5.22 as your starting point.
5
This is Euler’s second theorem for homogeneous functions.
/ 1776
214 / 1776
213 / 1776
211 / 1776
[a] It should be noted that it is the second derivative function which is zero. It is not
ENGLISH text is missing.Parametric form
From a physical point of view any extensive
be expressed in the parametric form F = β(ξ)X .
1
2
sufficient to say that the second derivative is zero at a given point.
§ 038
§ 37
§ 39
1
The solution of the homogeneous (!) system of equations can be
formulated in an infinite number of ways, one of which is:


  
 N −1   V   0 
N ∂µ (5.27)
 V
 
 =   .
V ∂N τ,V 
−1 V   N   0 
N
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
§ 36 Hessians
2
Note that the second derivative of Helmholtz energy with respect
to the mole number i.e. the matrix element H2,2 = V /N is non-zero in
Eq. 5.27, while the corresponding second derivative of Gibbs energy is
zero in Eq. 5.24. In the form of the chemical potential derivative we can
write:
∂µ from 5.24:
= 0,
∂N τ,π
∂µ , 0.
and from 5.27:
∂N τ,V
§ 37 Partial molarity
1
Replace F by G (ξ, n) in Eq. 5.23, where ξT = ( τ π ) and nT =
( N1 N2 · · · Nn ). This gives, without much difficulty:



 




 ∂G   ∂2 G 
 ∂S  s̄T n 



 



 ∂τ π,N  =  ∂τ∂nT π  n = −  ∂nT τ,π  n =
 .
ˆ − 


 

 ∂V  T 
 ∂G   ∂2 G 
v̄ n
∂π τ,N
∂π∂nT τ
∂nT τ,π
2
The partial derivatives of S and V on the right hand side are called
the partial molar entropy and partial molar[a] volume respectively.
3
These and other partial molar quantities occur so frequently in the
thermodynamic theory of mixtures that they have been given their own
differential operator, see Theme 10 on page 81.
4
A more compact notation is therefore S = nT s̄ and V = nT v̄.
3
This emphasises the fact that it is important to have a proper un- § 37
218 / 1776
derstanding of the many peculiarities of the energy functions, and it also 1
Find analytical expressions for (∂G /∂τ)π,N and (∂G /∂π)τ,N by
highlights that it is essential to know which variables are held constant using the Euler method to integrate the partial molar entropy and partial
during differentiation.
molar volume.
[a] A partial molar quantity is defined as f̄ = (∂F /∂n)τ,π , irrespective of whether or not
216 / 1776
217 / 1776 F has τ, π, N1 , . . . , Nn as canonical variables (it is only for Gibbs energy that there is a
correspondence between the two sets of variables).
1
2
Bearing in mind Eq. 5.22 the Hessian of A can be expressed as a
function of only one independent variable because the matrix, in addition to being symmetric, must satisfy the relation:


  
 ∂π  
  
∂π  ∂V τ,N ∂N τ,V   V   0 
∂2 A x=
ˆ 
 
 =   .
∂x∂x τ
∂µ  ∂µ   N   0 
∂V τ,N
∂N τ,V
Substitute U = U (S , V , N ) into Theme 34 on page 188
and show that S dτ + V dπ + N dµ = 0. Do you know
the name of this equation in thermodynamics? Does
it make any difference if you plug in G = G (τ, π, N )
rather than U (S , V , N )?
S dτ + V dπ + N dµ = 0 ,
3
Alternatively, if F = G (x, ξ) where x = N and ξ = ( τ π ), then
Eq. 5.20 takes the form:
∂G ∂G N d ∂∂G
N τ,π − ∂τ π,N dτ − ∂π τ,N dπ = 0 .
2
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5
The expression can therefore be reformulated as N dµ + S dτ +
V dπ = 0 which is identical to the Gibbs–Duhem equation.
Tore Haug-Warberg (Chem Eng, NTNU)
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 038 Gibbs–Duhem (2)
which is better known as the Gibbs–Duhem equation.
4
The partial derivatives of G with respect to N , τ and π have been
identified as µ, −S and −V in Theme 31 on page 182.
Altogether there are 4 matrix elements, with 3 associated relations.
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Tore Haug-Warberg (Chem Eng, NTNU)
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
(5.28)
T
3
The fact that the matrix, like all Hessians, must be symmetric, in
this case leads to (∂π/∂N )τ,V = (∂µ/∂V )τ,N .
Tore Haug-Warberg (Chem Eng, NTNU)
5 Euler’s . . .
Substituting F = U (x, ξ) into Eq. 5.20 where xT = ( S V N )
and ξ = ∅, reduces the expression to xT d(∂U /∂x) = 0. From the
definitions of τ, π and µ in Theme 29 on page 177 we can write
Helmholtz energy is extensive in V and N at any given temperature
τ.
4
§ 038 Gibbs–Duhem
§ 039
§ 38
§ 40
1
We have not been asked for any extensions, but by analogy to
Eqs. 5.9 and 5.16 the Gibbs–Duhem equation may be extended to a
multicomponent form:
S dτ + V dπ +
n
P
i =1
Ni dµi =
ˆ S dτ + V dπ + nT dµ = 0 .
(5.29)
Itensive state
1
Understanding the intensive properties of a system is of fundamenENGLISH text is missing.Canonical forms
tal importance for thermodynamic methodology.
It is important to realise that the information contained in U is 2 This emerges clearly from Euler’s 1st theorem in Eqs. 5.8 and 5.5,
The homogeneity, which effectively removes one degree of free- 1
dom in the function expression, shows up as a mutual dependency in conserved during the Legendre transformation to H , A , . . . , O .
and also from the Gibbs–Duhem equation 5.28.
2
For example, knowing the Gibbs energy G (τ, π, N ) really implies 3 The first two equations show that the extensive state of a system
the partial derivatives (of U ).
5
As a result, only n − 1 of the derived intensive state variables are full knowledge of U (S , V , N ), and vice versa.
is given as an analytic integral over the corresponding intensive proper3
The Gibbs–Duhem equation can therefore be derived from any of ties.
independent.
6
In single component systems this means that any arbitrary inten- the energy functions.
4
The last equation shows that τ, π and µ in a real system are not
4
sive variable can be expressed as a function of (at most) two other in- 4 In particular this also applies to the differential of the null-potential entirely independent but that they collectively must satisfy a differential
O (τ, π, µ) which is identical to Eq. 5.28, see Theme 26 on page 153.
tensive variables, see also Theme 39.
equation called the Gibbs–Duhem equation.
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
ENGLISH text is missing.Comment
2
Note that the Gibbs–Duhem equation follows inevitably from the
properties of extensive functions when differentiated, which in turn reflect their homogeneity.
3
This is true for all functions of this kind, and not only thermodynamic ones.
Termodynamic methods TKP4175
6
In fact, all Legendre transforms of U end up giving the same Gibbs–
Duhem equation.
§ 039 The intensive state
5 Euler’s . . .
1
A thermodynamic system is normally described by n + 2 independent state variables.
2
However, the intensive state can be determined once n + 1 (intensive) variables are known.
Starting from Theme 35 on page 191, can you tell
how many thermodynamic variables are needed to
determine the intensive state of a system?
4
In fact, the (single) extensive variable determines the system size
but has no influence on τ, π and µ.
3
This fact is illustrated by Eqs. 5.24–5.26, where the derivative of
an intensive property with respect to an extensive variable is zero if the
other variables are intensive, and are held constant during differentiation.
5
For a single-component system we can describe the intensive state
in three different ways:
τ = τ(π, µ) ,
π = π(τ, µ) ,
µ = µ(τ, π) .
Tore Haug-Warberg (Chem Eng, NTNU)
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Tore Haug-Warberg (Chem Eng, NTNU)
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
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Tore Haug-Warberg (Chem Eng, NTNU)
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
ENGLISH text is missing.Generalized Euler integration
1
We shall now examine to which extent these results are of general
validity, or whether they only apply in the case of an extensive (energy)
function F (X ) =
ˆ f (λx ).
2
As a basis for the discussion we choose Euler’s theorem 5.8.
3
The functional dependencies are made clearer by explicitly spelling
out the argument (x ) where needed:
∂f (x )
x = kf (x ) .
(5.8)
∂x ξ
4
The main challenge with our analysis so far is that the homogeneity
of ∂f /∂x depends on the homogeneity of f .
5
It is only for k = 1 that the derivative is in intensive form.
6
Then what if this is not the case?
7
What is the significance of the intensive state for a function that is
homogeneous of order k , 1?
8
To answer this question we shall tentatively write the Euler equation
on another form so that the derivative is always intensive.
9
A step in the right direction might be to multiply the left side of
Eq. 5.8 with 1 ≡ x −(k −1) · x k −1 .
§ 039 The intensive state (2)
10 The first factor can then compensate for the fact that the derivative
is homogeneous of order k − 1, tacitly understood that the function f is
homogeneous of order k , but it leads us nowhere unless we are able to
transform the entire expression to a form that agrees with Eq. 5.8.
11 It is therefore of considerable interest to show that the variable
transformation
y (x ) =
ˆ xk
is doing exactly this.
12
5 Euler’s . . .
The chain rule of differentiation gives
∂f (x )
∂f (x ) ∂y (x )
= ∂y (x )
=
∂x ξ
∂x ξ
ξ
∂f (x ) k −1
,
∂y (x ) ξ kx
which combined with Eq. 5.8 leads to
∂f (x ) kx k −1 x = kf (x ) ,
∂y (x )
ξ
and thereafter to
∂f (x ) y (x )
∂y (x ) ξ
= f (x ) ,
given the transformation rule x k −1 x = y (x ).
13 Because y (x ) is homogeneous of order k just like f itself so must
the partial derivative ∂f /∂y be homogeneous of order 0.
14
§ 040
It is thus an intensive function!
15 That means we have managed to write Euler’s theorem in a general
form applicable to all homogeneous functions, mind except those which
basically are intensive.
16
This result has far reaching consequences.
17 It means for example that all non-intensive thermodynamic properties can be calculated with knowledge of an intensive property — in the
general case several such properties — multiplied by a corresponding
coordinate that says something about the size of the system.
§ 39
§ 41
18 It also means that each of these properties has its own, we could
almost say private, Gibbs–Duhem equation that occupy one degree of
among the intensive properties, namely the partial derivatives
freedom
∂f /∂y .
19
5 Euler’s . . .
The scaling law does also apply still so that we by defining
X=
ˆ λx
and Y =
ˆ λk x k ,
can write
∂f (λx ) ∂y (λx ) ξ y (λx )
= f (λx )
⇒
∂F ( X ) ∂ Y ( X ) ξ Y (X )
= F (X ) .
20 We note that the expression to the right of the implication sign is a
continuation of Eq. 5.5 from earlier in this chapter.
21
A generalization to several Y variables gives similarly
n
P
i =1
Yi
∂F (X1 ,X2 ,...,Xn )
∂Yi (Xi )
Yj ,i ,ξl
= F (X1 , X2 , . . . , Xn ) ,
(5.30)
which is a continuation of Eq. 5.9.
22 Both of these equations are on the same form as those we derived
earlier and then must the derived properties be similar too.
23 Theme 36 for example, applies to any non-intensive thermodynamic property f (x ) as defined in the text above.
24 We may even dare to say that the variable transformation from x to
x k forms the basis for a generalization of Euler’s 1st theorem.
228 / 1776
§ 040 The intensive state
5 Euler’s . . .
(n )
To simplify the discussion, we write first AV on a form that shows
the homogeneity k as an explicit parameter:
1
We shall apply the generalization of the Euler’s 1st theorem
onto the nth derivative of the Helmholtz energy for an
ideal gas — on the occasion written as:
∂n (A ıg ) (n )
1−n
AV =
ˆ ∂V ∂V ···∂V τ,N = (−1)n (n − 1)! · NRT
.
V ·V
(n )
AV = (−1)k +1 k ! ·
3
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Here k ! must be understood as |k |! if k < 0.
· Vk .
By introducing new state variables Y1 = V k and Y2 = N k we get:
(n )
AV = (−1)k +1 k ! · RT · Y21/k · Y11−1/k .
(n )
The function AV is homogeneous of order k = 1 − n
and by observing the Hessian matrix of this quantity
with respect to the variables V k and N k , we can verify
that this matrix is singular in exactly the same manner
as the Hessian matrix of A with respect to V and N in
Eq. 5.27.
Tore Haug-Warberg (Chem Eng, NTNU)
2
NRT
V
4
229 / 1776
(n )
The first order partial derivatives of AV are
(n)
1 −1/k
∂ AV = (−1)k +1 k ! · RT · Y21/k · 1 −
Y
,
∂Y1 τ,Y2
k 1
(n)
1
∂ AV = (−1)k +1 k ! · RT ·
Y 1/k −1 · Y11−1/k ,
∂Y2 τ,Y1
k 2
Tore Haug-Warberg (Chem Eng, NTNU)
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News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 040 The intensive state (2)
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
5 Euler’s . . .
6
We also realize that these derivatives are intensive because Y1
and Y2 appearantly have exponents of the same magnitude but with
opposite signs.
5
and we can easily check that these two equations satisfy the generalized form of the Euler’s 1st theorem in Eq. 5.30.
News 1st Theorem 2nd Theorem Gibbs–Duhem Intensive Olds
§ 040 The intensive state (3)
Looking back
5 Euler’s . . .
8
The matrices 5.31 and 5.27 have the exact same structure despite
the fact that Eq. 5.31 has free variables that are different from Eq. 5.27.
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3
G=
4
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
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Tore Haug-Warberg (Chem Eng, NTNU)
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Integration of Gibbs–Duhem’s equation: Ideal gas
2
Canonical potentials: Helmholtz, Gibbs, Grand canonical
3
Molecular degrees of freedom: Translation, vibration, rotation
4
Atomic gases: Sackur–Tetrode’s equation
5
Solids: Einstein’s equation
6
Quantum mechanics: Photons, phonons, electrons
7
Debye temperature
8
High temperature limit: Dulong–Petit’s rule
9
Statistical mechanics: Virial theorem
10
Empirism: Van der Waals equation of state
Tore Haug-Warberg (Chem Eng, NTNU)
8
If we know the fundamental state functions referred to as the equations of state of the system
T = T (S , V , n) ,
(7.2)
p = p (S , V , n) ,
(7.3)
µi = µi (S , V , n) ,
(7.4)
then Eq. 7.1 can be integrated using Euler’s 1st theorem to give U =
P
TS − pV + ni=1 µi Ni , as shown in Theme 30 in Chapter 5. However,
in practice this doesn’t work, because the relationships defined by 7.2–
7.4 only exist as implicit functions .
Expanded footnote [a]:
[a] Only very rarely can entropy be written as an explicit (independent) variable in the
state functions.
i =1
µi (τ, π, n)Ni
Pn
Intensive state: τ = τ(π, µ), π = π(τ, µ) or µ = µ(τ, π)
!
∂2 f
Singularity:
x=0
∂x∂xT ξ
∂µ ∂µ Note: ∂N τ,π = 0 while ∂N τ,V , 0
Tore Haug-Warberg (Chem Eng, NTNU)
i =1 Ni
Termodynamic methods TKP4175
dµi = 0
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State description
7 Equations . . .
1
In this chapter we learn about a variety of useful concepts, such as
ideal gases, virial development, the Einstein–Debye phonon model and
Van der Waals theory, but first of all we need to define the concept of
an equation of state.
2
Although thermodynamics builds on some basic, fundamental principles of physical science, which form part of a well-defined mathematical theory, it is essentially an empirical science.
3
This is particularly true of the many semi-empirical models that are
used in engineering research.
4
That is how it was in the past, and that is how it will continue to be
in the future, because with the exception of a handful of simplifications
used in statistical mechanics[a] there are no precise models, and there
will always be the need for essentially physical models with adjustable
parameters.
5
A thermodynamic state description needs one or more functions of
state that define the asymptotes of the system (mixture) in a consistent
manner.
6
Applying parameters to the functions gives engineers the flexibility
they need, while it also ensures thermodynamic consistency.
[a] Above all crystalline phases, ideal gases and dilute electrolytes.
Let us start with any canonical differential, e.g. of internal energy
7
d U = T dS − p dV +
n
P
i =1
µi dNi ,
(7.1)
defined using the canonical variables S , V og Ni .
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State description (3)
9
n
P
6
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Tore Haug-Warberg (Chem Eng, NTNU)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
=0
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
1
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
State description (2)
T ,N
Gibbs–Duhem equation: S dτ + V dπ +
7
New concepts
Equations of state
T ,N
5
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7
is calculated by differentiation with respect to V
To our first surprise the cross derivatives are different.
Hence, the entire expression above is erroneous.
However, in the heat of the battle it is easy to overlook such details!
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i =1
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Part
(n )
That is, not across the variable sets X and Y .
A simple example illustrates the problem:
∂2 p ıg ∂2 p ıg = −2NRT
, ∂(1/V )∂V
V
∂V ∂(1/V )
7
9
10
Scaling: Xi =
ˆ λxi for all i ∈ [1, n]
n
P
A = π(τ, V , n)V +
µi (τ, V , n)Ni
2
(5.32)
02 September 2013
5 Euler’s . . .
This is not normally the case, but here it is.
5
6
8
But, partial differentiation with respect to V at constant 1/V does
not really make any sense because constant 1/V also entails constant
V.
F (X1 , . . . , Xn , ξn+1 , . . . , ξm ) = λk f (x1 , . . . , xn , ξn+1 , . . . , ξm )
1
8
Tore Haug-Warberg (Chem Eng, NTNU)
2
3
The property AV
in the first place.
4
When we afterwards replace the free variables V and N with Y1 =
V k and Y2 = N k we can no longer use Schwarz’s theorem on the cross
terms.
9
Certainly, the coefficient on the outside of the matrix is changed
from (∂µ/∂N )τ,V to h22 because of the variable shift, but this is due to
the nature of the problem since we have chosen to use ideal gas as a
specific example in the latter case.
7
The second order derivatives are elements in the Hessian matrix
(n )
of AV which satisfies the homogeneity condition

  

 Y2 −1   Y   0 
(n)
∂ 2 (A V )  Y1
  1   
Y2
y
=
h
(5.31)

 =   ,


 
Y1 22 
∂y∂y τ

  
−1 YY21   Y2   0 
where the lower right matrix element h22 can be written:
1 1/k −2 1
·
h22 = (−1)k +1 k ! · RT ·
Y
− 1 Y11−1/k .
k 2
k
The theorem of Schwarz
1
Repeated differentiation with respect to several variable sets puts
us in the danger of committing a systematic error — if the variables
would prove to be interdependent.
i =1
Ideell (perfect) gas
7 Equations . . .
An ideal gas consists of individual particles (atoms, molecules or
1
radicals) that move freely and independently in space, without any kind
of interaction.
(7.6)
7
Ideal gases are only valid as a special case of the true thermodynamic state when pressure approaches zero, and can never be verified
directly in a real (physical) system.
2
The term ideal (perfect) gas refers to ideal, and from a physical
point of view hypothetical, descriptions of the conditions in a real gas.
i =1
n
P
271 / 1776
(7.5)
or alternatively
(dA )T = −p dV +
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7 Equations . . .
It is therefore more normal to start with
n
P
µi dNi
(dG )T ,p =
Termodynamic methods TKP4175
3
This is the ideal situation because it involves ignoring the intermolecular forces that exist between the particles in a real gas.
4
This includes all of the interactions caused by induced dipoles
(Van der Waals forces), dipole-dipole forces, dipole-quadrupole forces,
etc.
5
In spite of this massive simplification, the total energy of the gas is
not equal to zero, as the particles have their kinetic energy (translation),
moment of inertia (rotation), internal degrees if freedom (vibration, internal rotation, torsion, electronic excitation) and chemical binding energy.
6
Furthermore, electron and nuclear spin can also play a role.
µi dNi ,
8
Nevertheless, we must be allowed to consider the ideal gas law
one of the great scientific discoveries.
9
Perhaps it is just as important as Maxwell’s equations, which describe electromagnetism, although that is a very subjective judgement,
as Maxwell’s equations are a coupled set of four complex partial differential equations, whereas the ideal gas law is just a simple algebraic
equation.
10 What they have in common, then, is not their mathematical formulae, but rather their understanding of the underlying physics and of the
practical and theoretical implications.
but it is still difficult to accurately describe the chemical potential of the
compounds in a multicomponent mixture.
10 which can be integrated in the same way if the state function(s)
µi (T , p , n) or p (T , V , n) and µi (T , V , n) are known,[a]
[a] Many textbooks refer to p (T , V , n) as if there were no other state functions, but that
is wrong!
11 The exceptions to this are ideal (gas) mixtures for which we can
obtain exact solutions using statistical mechanics or, as we shall do in
a moment, from integrating the Gibbs–Duhem’s equation using just one
additional assumption.
12 By taking the ideal gas model as a reference, it is relatively easy
R
to obtain the residual
functions G r,p (T , p , n) =
ˆ (V (T , p , n) − V ıg ) dp
R
or A r,v (T , V , n) =
ˆ (p (T , V , n) − p ıg ) dV , which describe the difference
between a real mixture and an equivalent ideal gas under the same
physical conditions.
13 If we combine this with an expression for the energy of the reference gas, we get the Gibbs or Helmholtz energy of the mixture, and a
full set of state functions can then be derived using differentiation.
14 This leaves us with just V (T , p , n) or p (T , V , n) as the state functions.
15 The alternative to the above approach is to derive the thermodynamic functions in non-canonical coordinates:
U (T , V , n) rather than U (S , V , n)
H (T , V , n) rather than H (S , p , n)
..
.
S (T , p , n) rather than S (H , p , n)
This is particularly useful if you only need to obtain U , H or S , but not
A or G .
16 Typically, this is the case for physical state changes that occur at a
constant composition like for instance fluid flow in pipes, compressors
and valves.
17 For systems of variable composition, where the chemical potential
is important, it is more suitable to use A and G because the expressions
µi =
ˆ (∂A /∂Ni )T ,V ,Nj,i =
ˆ (∂G /∂Ni )T ,p ,Nj,i can then be obtained using
differentiation with respect to the canonical variables.
18 If we choose to disregard canonical theory for a moment, then
the chemical potential must be defined as µi = (∂H /∂Ni )T ,p,Nj,i −
T (∂S /∂Ni )T ,p ,Nj ,i =
ˆ h̄i − T s̄i or a similar expression based on U , but
this approach normally involves more work.
Tore Haug-Warberg (Chem Eng, NTNU)
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Tore Haug-Warberg (Chem Eng, NTNU)
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
ENGLISH text is missing.The ideal gas law
Gibbs–Duhem I
1
It is not a new idea that the world is composed of more or less
distinct material points, but the experiments that underpin the ideal gas
law as we know it today are more recent:
Name
Boyle
Charles
Dalton
Gay-Lussac
Avogadro
Observation
( PV )T ,N = c1
( V /T )p ,N = c2
( p /N )T ,V = c3
( P /T )V ,N = c4
( V /N )T ,p = c5
Year
1662
1787
1801
1809
1811
Given what we know today, the above table accurately represents the
scientific progress, but at the time the scientists lacked our historical
perspective, and we should be cautious about attributing entirely rational motivations for their observations.
7 Equations . . .
2
It was only in 1834 that Émile Clapeyron successfully formulated
the ideal gas law in the simple form:
p ıg =
NRT
V
.
(7.7)
3
The question we are going to look at is to what extent, and if so how,
this law can be used to derive expressions for the Helmholtz energy of
the gas.
4
We can see that the unit of the product p · V is the same as for
A , which means that there may be a deeper relationship between pressure–volume and the energy functions.
5
The argument that follows is a great example of how physical and
mathematical analysis can be successfully combined with experimental
results (thanks to the insight of the old scientists, not the author).
6
We will admittedly need to make some choices along the way,
which mean that the results are not entirely conclusive, but this must
be seen in the context of established practice for measuring and reporting basic thermodynamic functions.
275 / 1776
The quickest way to obtain a useful result is to apply the Gibbs–
Duhem Eq. 5.28 described in Chapter 5. For a thermodynamic system
with only one chemical component, the folowing applies:
1
Gibbs–Duhem I (2)
7 Equations . . .
or:
µıg (T , p ) − µ◦ (T ) = −RT
Rv
v◦
dν
ν
= RT ln
v◦
v
2
It is also important to remember that thermodynamic energy functions are first-order homogeneous functions to which Euler’s theorem
applies.
6
From Chapter 4 on Legendre transforms we know the general
equations S = − (∂A /∂T )V ,N and U = A + TS , which means that it
also follows that:
"
!#
∂µ◦
− NRT = Nh◦ − NRT .
U ıg = N µ◦ − T
∂T
3
7
In other words, the internal energy of the gas depends on the system’s size N and temperature T through the constant of integration
called µ◦ .
Expanded footnote [a]:
[a] The internal energy U ıg is inevitably independent of both pressure and volume:
Ideal gases are a collection of free material points without any form of interaction. It
is only the internal degrees of freedom of the particles that contribute to their energy,
and these are also what determine the integration constant µ◦ .
Expanded footnote [a]:
[a] The Gibbs–Duhem equation includes the term v dπ where π is understood as the
equation of state π(τ, v ). The term v dπ can, therefore, not be given the interpretation
v (τ, π) dπ. That would require a full variable transformation given by v dπ = dpv −π dv .
In the case above, d(pv ) = 0 because it is restricted to an ideal gas at constant temperature and mass — thus making v (τ, π) dπ and v dπ (τ, v ) seemingly equivalent —
but the condition of constant mass is not universal and will in fact break in the next
section when we start talking about multicomponent systems.
(dµ )T = −
Tore Haug-Warberg (Chem Eng, NTNU)
8
The latter constant must be determined experimentally in each individual case, or theoretically–experimentally if there are good simulation models for the chemical compound (based on spectrometric measurements).
9
Our provisional conclusion is therefore that we can calculate the
Helmholtz energy of an ideal gas with only one chemical component
and its internal energy. From this the other energy functions can be
obtained in a similar way using Legendre transforms, apart from one
temperature dependent constant.
dv ,
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Tore Haug-Warberg (Chem Eng, NTNU)
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Gibbs–Duhem II
7 Equations . . .
Our next challenge is to generalise this result into one that is similarly valid for a multicomponent system. In order to do that, we must
write the ideal gas law in its extended form
P
p ıg = NiVRT
(7.8)
so that we can identify the components that constitute the mixture.
The Gibbs–Duhem equation from Chapter 5 follows directly from
the fact that U is a homogeneous function to which Euler’s theorem
applies when expressed using the variables S , V and n.
2
This is a function of state in the form p (T , V , n), which gives us a
good way to approach the question of Helmholtz energy A (T , V , n), but
in order to determine A using the Euler method of integration as shown
ıg
in Chapter 5, we also need to know the equation of state µi (T , V , n).
3
This function cannot be obtained experimentally in the same way
as p ıg , and must therefore be derived theoretically.
4
One way to do this is to apply the Gibbs–Duhem equation again:
P
S dτ + V dπ + Ni dµi = 0 .
i
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The normal convention is to assume a pure-component standard
state where V◦ /Ni◦ = RT /p◦ and where p◦ = 1 bar, or, in older reference
works, where p◦ = 1 atm.
1
3
Bearing in mind this fundamental understanding of what
means,
the chemical potential of the component can be expressed as
ıg
(7.9)
µi (T , V , n) = µ◦i (T , p◦ ) + RT ln Npi◦RT
V .
µ◦i
n
ıg
Here we should particularly note that (dµi )T ,V ,n = 0 must be true for all
values of p◦ .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
7 Equations . . .
6
This means that there is an interdependence between the differentials of U which can be usefully exploited, and using the equation of
state π =
ˆ −p ıg and the Gibbs–Duhem equation we will obtain a possible solution for µi .
i
P
(RT
or
i
i
i
dV
V
− RT
dNi
Ni
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
+ (dµi )T ) = 0 .
Gibbs–Duhem II (3)
13
9
Thermodynamics is a phenomenological science, and we cannot
be sure that the above expression correctly describes the gas mixture,
but at least it is a possible solution.
10 However, the result agrees with experimental observations of dilute
gases, which in the limit case p → 0 behave like a chaotic collection of
particles without any interaction, and where the macroscopic properties
associated with each particle are independent of the other particles.
11 Interestingly, deriving the chemical potiential of a component of an
ideal gas using statistical mechanics gives us exactly the same answer[a].
12 This is a convincing demonstration of the great power of formal
thermodynamic analysis.
[a] Statistical mechanical theory is based on a microscopic description of the properties
of particles, and involves calculating their average values based on a relatively small
number of axioms. Whereas at the macroscopic level we have to accept that our
result is merely a possible result, the theoretician can assert that it is a correct result
for a gas consisting of hypothetical particles without any physical extension or mutual
interaction. In other words, a perfect gas.
7 Equations . . .
Integrating the differential gives us
1
RT
ıg
(µi (T , V , n) − µi◦ (T )) = − ln VV◦ + ln NN◦i .
i
Note that the constant of integration µ◦i , which will hereafter be called
the standard state of component i , is a function of temperature only.
14 This means that for any given temperature T we must calculate or
estimate a new value for µ◦i .
15 If the function of state were available in the form p (S , V , n), then µ◦i
would become a genuine constant of integration.
16 In practice there are very few systems that can be described in this
way, but radiation from black bodies, as described in Theme 46, is one
example.
8
The simplest solution for an expression with a sum equal to zero is
to assume that each individual term is equal to zero:
ıg
(dµi )T
RT
Tore Haug-Warberg (Chem Eng, NTNU)
= − dVV +
dNi
Ni
;
∀i ∈ [1, n] .
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7 Equations . . .
2
The standard state µi◦ should in that case be interpreted as repıg
resenting µ◦i (T , p◦ ), or more properly µi (T , p◦ , N1 = 0, . . . , Ni = 1 mol,
. . . , N = 0).
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ˆ T constant as
7
Let us substitute for π =
ˆ −p ıg while keeping τ =
shown below:
P
P
P
ıg
Ni (dµi )T = 0 ,
−V N−iVRT2 dV − V RT
V dNi +
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Standard potential
02 September 2013
Gibbs–Duhem II (2)
i
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
1
5
Substituting in p ıg and µıg from the above equations gives us:
A ıg = −NRT + N µ◦ (T ) + NRT ln vv◦ .
4
We have thus derived the Helmholtz energy from the ideal gas law,
using the fact that Euler’s theorem applies to the extensive properties of
the gas, and the Gibbs–Duhem equation, which is itself a direct result of
Euler’s theorem. The interpretation of the two parameters µ◦ and v◦ is
postponed till we start discussing multicomponent mixtures in the next
section.
5
Here, it is more important to stress the fact that knowlegde of A
means knowlegde of all the energy functions.
2
This differential has no general solution, as s and v are two independent co-ordinates of the material state, but for a change of state that
is assumed to be isothermal, it become possible to replace s , v by τ, v
such that dτ = 0 and since v =
ˆ V /N it follows that dπıg = R τ/v 2 dv ,
giving us
RT
v
7 Equations . . .
1
The total Helmholtz energy of a single-component system can
therefore be written as the integral A = −Npv + µN .
.
s dτ (s , v ) + v dπ (s , v ) + dµ (s , v ) = 0 .
ıg
Euler’s homogeneous function theorem
Standard potential (2)
10 Essentially, standard states are presented in four different ways in
the literature:
This is of course the case by definition, since T , V and n completely
determine the thermodynamic state, but at the same time p◦ acts as a
ıg
parameter in the expression for µi .
5
It is therefore slightly unclear what happens if p◦ changes.
6
The passive role of the standard pressure becomes more obvious
ıg
if we partially differentiate µi with respect to p◦ .
7
This gives us (∂µi◦ /∂p◦ )T − RT /p◦ = 0, or put another way: If we
choose to change p◦ then we must also correct µ◦i to ensure that the
ıg
total impact on µi is zero.
8
It is important for our own self-esteem to acknowledge that in order to accurately determine the standard state one needs information
at a high level including precise calorimetric measurements, advanced
spectroscopy and quantum physics, or a combination of these things,
and that these activities lie outside our area of responsibility.
9
We must therefore adapt our way of working to the available thermodynamic databases, and not vice-versa, which means that it is important to use the data properly.
4
(7.10)
where f is any suitable parameterized function like e.g. f (T ) = a + bT +
cT ln T + · · · ,
µ◦i (T , p◦ )
µ◦i (T , p◦ )
=
=
∆f hi◦ (T ) − Tsi◦ (T , p◦ ) ,
∆f hi◦ (T◦ ) + ( hi◦ (T ) − hi◦ (T◦ ) )
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
Termodynamic methods TKP4175
02 September 2013
Standard potential (3)
7 Equations . . .
11 which follow from the definition G =
ˆ U + pV − TS =
ˆ H − TS from
Chapter 4 applied to the standard state as defined in the above paragraph, and finally:
RT
µi◦ (T , p◦ ) = TT◦ µi◦ (T◦ , p◦ ) + ∆f hi◦ (T◦ ) 1 − TT◦ + cp◦,i (τ) 1 − Tτ dτ .
T◦
(7.13)
(7.11)
− Tsi◦ (T , p◦ ) ,
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7 Equations . . .
µi◦ (T , p◦ ) = f (T ) ,
Tore Haug-Warberg (Chem Eng, NTNU)
(7.12)
283 / 1776
12 The last relation is derived from the Gibbs–Helmholtz equation in
Chapter 4:4.6 which in this case says that (∂(µ/T )/∂(1/T ))p = h .
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Standard potential (4)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
15 All of the formation properties (enthalpy, volume, entropy, etc.) of
any given compound Aa Bb · · · Zz in an ideal gas state are defined analogously by
13 Moreover, it is an experimental fact that all molecules, with the exception of monoatomic ones, have internal energy due to the bonding
within the molecule.
14 Normally this energy is reported as the enthalpy of formation
∆f hi◦ (T◦ ) of the compound.
Standard potential (5)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
ıg
hH
2
lıq
hBr
2
=
20 Hence, in the above equation
= 0 and the enthalpy of
ıg
◦
formation ∆f hHBr
≡ hHBr is given relative to this arbitrary reference.
23 There are therefore many ways of reporting the standard state, but
for all of them it is true that
RT
hi◦ (T )
− hi◦ (T◦ )
(7.14)
= cp◦,i (τ) dτ ,
◦
, 0, as the enthalpy reference by
21 In particular, note that ∆f hBr
2
definition relates to bromine in the liquid phase and not for bromine in
the (ideal) gas phase.
22 This is the main principle that determines how the thermodynamic
standard state is reported, but there are, of course, exceptions to the
rule: At least one important database (JANAF) uses the more specific
definition hi◦ (T ) = 0 for the elements at all temperatures T , while another database perhaps uses T = 0 K as its reference and yet another
one may use gi◦ (T◦ , p◦ ) instead of si◦ (T◦ , p◦ ).
ıg
ˆ hA
∆f hA◦ a Bb ···Zz =
β
a Bb ···Zz
− ahAα − bhB − · · · − zhZζ ,
where α, β and ζ indicate the phases (configurations) of the various
elements.
T◦
For example, hydrogen bromide would have the formula
(H2 )0.5 (Br2 )0.5 , giving the enthalpy of formation
18
◦
=
∆f hHBr
19 The reason for this discrepancy, with respect to the established
practice for chemical equilibrium reactions, is that according to the IUPAC convention the enthalpy of a chemical element is defined as zero
for the stable configuration of the element[a]
[a] Such as H2 (gas), Ar(gas), S(rhombıc), Br2 (lıq), Al(sol), etc.
si◦ (T , p◦ ) − si◦ (T◦ , p◦ ) =
16 The phase can be liquid or solid, in which case the crystalline configuration must also be given, or a hypothetical ideal gas for gaseous
elements.
17 The chemical formula Aa Bb · · · Zz should here be seen as the elemental composition, i.e. not the atomic composition, of the molecule.
ıg
hHBr
ıg
0.5hH
2
−
at T◦ = 298.15 K and p◦ = 1 bar.
Tore Haug-Warberg (Chem Eng, NTNU)
−
lıq
0.5hBr
2
.
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02 September 2013
Euler integration
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Tore Haug-Warberg (Chem Eng, NTNU)
7 Equations . . .
It is easy to lose one’s way in the details of these calculations.
It is therefore worth reiterating that using Euler’s theorem to integrate
euqations 7.8 and 7.9 always works, and that the Helmholtz energy of
the gas can be expressed in the following general form:
P
ıg
A ıg (T , V , n) = πıg V + Ni µi
i
i
Ni RT +
P
i
|
Ni µ◦i (T , p◦ ) +
{z
G ◦ (T ,p◦ )
}
P
i
Ni RT ln
Ni RT p◦ V
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Helmholtz potential A (T , V , n)
1
2
The consequences become even clearer if we take A ıg as our
starting point for differentiating U ıg = (∂(A ıg /T )/∂(1/T ))V ,n or −S ıg =
(∂A ıg /∂T )V ,n .
3
The partial derivative of A ıg is in both cases defined at constant
volume, but how is it possible to implement this for µ◦i (T , p◦ )?
4
Let us use U ıg as an example. We start with
P
P µ◦ (T ,p◦ ) P
A ıg
+ Ni R ln( Npi◦RT
Ni R + Ni i T
T =−
V )
i
i
followed by
d
A ıg T V ,n
=
−
i


!
!
∂(µ◦i /T )

 ∂(µ◦i /T )
dp◦ 
Ni 
d 1 +
∂(1/T ) p◦ T
∂p◦ T
h
i
Ni R T d T1 + p1◦ dp◦ .
P
i
P
i
5
Finally we use the Gibbs–Helmholtz equations from Theme 28 on
page 160 together with (∂(µ◦i /T )/∂p◦ )T = R /p◦ from the equivalent
discussion on page 279.
6
The differential can now be expressed in the form
!
P
P
d(A ıg /T )
= Ni hi◦ (T ) − Ni RT = H ıg − p ıg V ,
U ıg (T , n) =
d(1/T ) V ,n
i
i
which shows that U ıg is only a function of T and n because p◦ and V
cancel out during the derivation.
7
The total derivative now has one degree of freedom, and therefore takes the same value as the partial derivative (∂(A ıg /T )/∂(1/T ))
throughout the domain of definition.
8
But why is it necessary to go via the total differential?
9
Well, because otherwise we risk writing (∂(µ◦i /T )/∂(1/T ))V ,n = ui◦ .
10 Firstly µi◦ is a function of V◦ and not of V , and secondly V◦ varies
with T when p◦ has been chosen as the standard state.
11 It is easy to make a mistake in those circumstances, and it is therefore safest to go via the total differential.
12 The derivation also illustrates how risky it is to introduce noncanonical variables for the thermodynamic potentials.
In summary, the following equations apply to ideal gas mixtures:
i
h
P
P
A ıg (T , V , n) = Ni µi◦ (T , p◦ ) + Ni RT ln Npi◦RT
(7.16)
V −1
i
−S ıg (T , V , n) =
ıg
P
i
i
Ni si◦ (T , p◦ ) +
P
i
ıg
U (T , n) = H (T , n) − NRT
P
Ni hi◦ (T ) .
H ıg (T , n) =
.
7 Equations . . .
Ni R ln
Ni RT p◦ V
(7.17)
(7.18)
(7.19)
i
cp◦,i
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=
cvtrans
,i
+
cvrot
,i
+ cvvıb
+ cvelec
+R
,i
,i
and can be calculated from spectrometric measurements.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
4
The JANAF tables[a] are a good example of how far this development has gone.
5
For relatively small molecules (fewer than 8–10 atoms) it is now
possible to estimate cp◦,i more precisely than using high-quality calorimetric measurements.
[a] JANAF thermochemical
Suppl., 14(1), 1985.
tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,
02 September 2013
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Gibbs potential G (T , p , n)
7 Equations . . .
1
In our case, it is possible to calculate Gibbs energy by using the
Euler method to integrate the chemical potential in Eq. 7.9, and substituting in the ideal gas law from Eq. 7.7
ıg
N p
µi (T , p , n) = µi◦ (T , p◦ ) + RT ln Npi ◦ ,
(7.20)
2
Historically speaking, T , p , n has long been the favoured set of variables for modelling all chemical systems, probably because temperature and pressure are particularly easy to control in a laboratory.
which gives us:
G ıg (T , p , n) =
n
P
i =1
ıg
µi Ni =
n
P
i =1
µi◦ Ni + RT
n
P
i =1
Ni ln
Ni N
+ NRT ln
p
p◦
.
(7.21)
3
Alternatively, we can calculate the Gibbs energy as a Legendre
transform of Helmholtz energy (or vice versa).
7
This significantly complicates the practical application of Legendre
transforms.
4
This is undoubtedly possible, but since G ıg and A ıg have different
canonical variables, we also need to invert p (T , V , n) to V (T , p , n).
5
If this is impossible, the result of the transformation will be an implicit function [a].
6
For most equations of state, with the exception of ideal gases, the
2nd virial equation and certain other models, an explicit transformation
from T , p , n to T , V , n is impossible.
[a] The specification V (p ) has 3 solutions in the steam/liquid phases, and is not a
function in the ordinary sense of the word. On the other hand, p (V ) is an unambiguous
specification, which means that there is no problem with doing a Legendre transform
of Helmholtz energy to Gibbs energy.
3
In some strange manner, something has gone wrong here, but
what?
4
It is clear that we no longer have a canonical description of
A (T , V , n), and that the unfortunate decision to combine T , V and n
from the original set of variables of p ıg , with p◦ from the standard state
ıg
of µi , is creating unintended confusion.
02 September 2013
3
The enthalpy of formation is generally measured using calorimetry,
whereas the heat capacity of an ideal gas component essentially adheres to the superposition principle
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Note that, in this expression, the standard state term is a conventional Gibbs energy function, because T and p◦ have been selected as
the state variables of µ◦i , rather than T and V◦ .
Termodynamic methods TKP4175
2
The sections 3.1–3.4 explain a little bit about the fundamental physical principles that, in their different ways, contribute to cv◦,i .
[a] The gas constant R in the above equation is a result of the defintiion of H =
ˆ U + pV ,
ˆ U ıg + (pV )ıg = U ıg + NRT . If we then differentiate
which for an ideal gas gives us H ıg =
with respect to temperature we get cp◦,i = cv◦,i + R .
(7.15)
2
Tore Haug-Warberg (Chem Eng, NTNU)
1
In practice, all thermodynamic reference works are based on a
large number of carefully recorded measurements supported by one
or more theoretical calculations, which together produce figures for
∆f hi◦ (T◦ ), si◦ (T◦ , p◦ ) and cp◦,i (T ).
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
1
P
T◦
7 Equations . . .
Expanded footnote [a]:
cp◦,i (τ) dln τ .
24 One of these equations (it doesn’t matter which) can be viewed as
a definition of cp i.e. either cp =
ˆ (∂h /∂T )p ,n or cp =
ˆ T (∂s /∂T )p ,n , while
the other one follows from the partial differential (dh )p ,n = T ds .
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
=−
RT
The superposition principle
Tore Haug-Warberg (Chem Eng, NTNU)
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Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
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Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
ENGLISH text is missing.Grand canonical potential Ω(T , V , µ)
1
The ideal gas law is a simple model, which allows us to choose any
system variables that we like.
2
For example, we can rewrite µ(T , p , n) or µ(T , V , n) as expressions
for n(T , V , µ).
3
Very few models allow us to do this, and so we will make use of the
opportunity now that we have it.
4
An alternative way of expressing Eq. 7.20 is:
µi −µ◦ ıg
p◦ V
Ni (T , V , µ) = RT
exp RT i .
(7.22)
5
Substituted into the ideal gas law, Eq. 7.22 gives us an adequate
equation of state for the grand canonical potential Ω(T , V , µ):
p ıg (T , µ) =
RT
V
n
P
i =1
ıg
N i = p◦
n
P
i =1
exp
µi −µ◦i RT
(7.23)
.
6
The total differential of Ω for a one-component system was derived
in Chapter 4, see Theme 24 on page 144.
7
For an n-component system, we can state that
(dΩ)T ,µ = −p dV .
(7.24)
8
The differential (dΩ)T ,µ can be integrated directly, as long as the
function of state p (T , V , µ) is expressed explicitly. Eq. 7.23 is explicit,
and if we integrate using Euler’s method at a given T and µ we obtain
Ωıg (T , V , µ) = −p ıg V = −p◦ V
Molecular contributions
n
P
i =1
exp
µi −µi◦ RT
.
(7.25)
9
It may at first sight appear that p◦ is a free variable in the above
equations, but if we change p◦ , then µi◦ (T , p◦ ) varies accordingly, always
keeping the value of Ωıg unchanged (show this).
10 Also note that p ıg (T , µi ) is a function of only two variables in
Eq. 7.23, whereas Ωıg (T , V , µi ) in Eq. 7.25 requires three variables.
11 The intensive state is evidently given when n + 1 intensive variables have been determined, while the extensive state requires n + 2
variables.
12
This is a general result, which is not limited to ideal gases[a].
13 From Theme 35 in Chapter 5 we know that the partial differentials
of functions with only one extensive variable fulfil the requirements for
homogeneity[a]:
∂p ∂µ1 ∂T ∂S p ,µ ,...,µ = ∂V T ,µ ,...,µ = ∂N1
1
n
1
n
= ... =
T ,p ,µ2 ,...,µn
∂µn ∂Nn T ,p ,µ1 ,...,µn−1
= 0.
14 The intensive variables are clearly independent of the size of the
system, and the state is therefore given when n + 1 intensive variables
have been determined, cf. Theme 39 on page 195.
15 However, in order to determine the size of the system, we also
need an extensive variable.
16 In other words, a thermodynamic system requires n + 2 state variables, in spite of the fact that the intensive state can be described with
only n + 1 variables.
§ 41
291 / 1776 1
Derive the partial derivatives of Ωıg (T , V , µj ) = −N ıg RT with respect to T , V and µi . Show that the results can be interpreted as −S ıg ,
ıg
−p ıg and −Nj , without knowing the differentiation rules of the Legen[a] In contrast to e.g. U ıg = U (T , n), which only applies to ideal gases.
[a] Zero is zero, most people probably think, but the partial differentials have different dre transform beforehand.
292 / 1776
units in this case, and the similarity only applies to the measure (the absolute value) of
§ 41 Partial differentiation of Ω(T , V , µ)
1
It’s a good idea to start by establishing some shadow values before
starting on the differentiation process itself.
2
Instead of differentiating p ıg V in Eq. 7.25 we will use the fact that
P ıg
ıg
−Ωıg = p ıg V = i Ni RT = N ıg RT . If we differentiate Ni (T , V , µ) in
Eq. 7.22 we get:
ıg
ıg
ıg
◦
◦
N
N ¯ ıg
∂Ni ıg RTsi −R (µi −µi )
− Ti = RTi si − R ,
= Ni
2
∂T V ,µ
(RT )
ıg
ıg
∂Ni ∂V T ,µ
ıg
∂Ni
∂µ
j
=
Ni
V
=
δij Ni
RT
7 Equations . . .
,
ıg
T ,V ,µk ,j
.
3
Here we have made use of the relationship −si◦ = − (∂µ◦i /∂T )p ,n ,
which also means that −si◦ = − (∂µi◦ /∂T )V ,µ .
4
Think about this result for a moment.
5
Apart from this fundamental observation, the simplification of
RTsi◦ − R (µi − µ◦i ) in the first partial differential equation relies on an
understanding of certain principles of thermodynamics.
6
In the general case, µi = h̄i − T s̄i . In the special case of ideal
ıg
ıg
gases, h̄i = h̄i◦ , and hence Tsi◦ − (µi − µ◦i ) = T s̄i .
7
We are now ready to derive the partial differentials of Ωıg
∂Ωıg ∂T V ,µ
= −RT
∂Ωıg ∂V T ,µ
= −RT
∂Ωıg ∂µj T ,V ,µ
k ,j
= −RT
n
P
i =1
ıg
∂Ni ∂T V ,µ
− NR = −
n ∂N P
i
∂V T ,µ
i =1
n ∂N ıg P
i
∂µj T ,V ,µ
i =1
ıg
k ,j
n
P
i =1
= −RT
=−
n
P
i =1
ıg
Ni (s̄i − R ) − NR = −S ıg ,
n
P
i =1
ıg
Ni
V
ıg
Ni δ
ij
= −p ıg ,
ıg
= − Nj
293 / 1776
these quantities.
Without further introduction, let us assume that the relationship
between the particles’ microstates and Helmholtz energy is
1
2
The ideal gas law was established in several stages over the course
of the 17th-19th centuries, as an experimental law for the so-called permanent gases (air, N2 , H2 , etc.).
3
We now know that there is a direct link between the macroscopic
characteristics of the gas and the microscopic properties of the individual molecules.
4
The related discipline is called statistical mechanics or statistical
thermodynamics.
5
It would be far too big a job to discuss this theory in detail, but
we will take the time to look at some of the details that underpin our
understanding of ideal gases.
A
NRT
For a system with n particles in defined locations, Q =
the particles are not in a fixed location (as in a gas), then Q =
6
Expanded footnote [a]:
Expanded footnote [a]:
[a] D. F. Lawden. Principles of Thermodynamics and Statistical Mechanics. John Wiley
& Sons, 1987.
[a] Robert P. H. Gasser and W. Graham Richards. An Introduction to Statistical Thermodynamics. World Scientific Publishing Co. Pte. Ltd., 2nd edition, 1995.
.
q n , but if
1 n
n! q .
We can only barely scratch the surface of quantum mechanics, and
the descriptions given in this section have therefore been kept as simple
as possible.
9
8
In principle, this is all we need to know about the gas, but in order
to calculate the necessary properties, we must first know the quantum
mechanical description of each individual input to the model.
10 However, before we start looking at microstates, we will first establish the fundamental macroscopic thermodynamic equations that apply
to ideal gases.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
A particle-in-a-box has three identical degrees of translational freedom.
1
2
Even though the particle can in principle move freely within the
control volume, its kinetic energy is quantified (only standing waves are
permitted).
The molecular partition function q approximates to the product of
q ≈ qtrans qvıb qrot · · · over the (assumed) number of independent terms.
7
7 Equations . . .
3
In a cubic box, each direction will have its own quantum number jx ,
jy and jz , which is given the same unit of energy ε = h 2 /8ml 2 .
= − ln Q ,
where Q is the total partition function of the system
Free translation
02 September 2013
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4
Free translation (2)
7 Equations . . .
6
The surface area of the octant of the sphere is πj 2 /2, which leads
us to the integral
R∞
−j 2 ε
lim qtrans = π2 j 2 exp kT dj
kT ≫ε
The associated partition function can be written
∞ P
∞ P
∞
P
−(j 2 +j 2 +j 2 )ε
qtrans =
exp x kTy z .
0
=
Expanded footnote [a]:
[a] Per Chr. Hemmer. Statistisk mekanikk. Tapir Forlag, 1970. In Norwegian.
π kT 3/2
2 ε
jx =1 jy =1 jz =1
→
5
For high quantum numbers j 2 = jx2 + jy2 + jz2 ≫ 0 is a virtually
continuous function over (an eighth of) the surface of a sphere with a
radius of j .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
295 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
(2πmkT )
h3
R∞
2
x 2 e -x dx ,
0
|
3/2
V
{z
√
.
π/4
ˆ
x2 =
j2ε
kT
,
}
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Free translation (3)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Sackur–Tetrode S
7 Equations . . .
7
When going from summation to integration, it has been assumed
that kT ≫ εx . Combining this with j 2 ≫ 0 gives us j 2 ε/kT < 1 over
much of the area of integration, which ensures that the integral is a
good approximation of the sum.
02 September 2013
297 / 1776
§ 42
§ 44
Tb
[K]
20.39
4.224
27.09
2
The length of the sides of the box is l = 1 nm (which is right at the
limit of the range of validity of classical thermodynamics).
§ 42
3
Except in the case of He, ε/k is reassuringly far below the boiling
point of the compounds, and in practice all gases have a virtually continuous spectrum of translational energies[a].
299 / 1776
1
Calculate the typical translational temperature ε/k = h 2 /8ml 2 k
for the light gases hydrogen, helium and neon. Assume that the lengths
of the sides of the box are ≥ 1 nm. Comment on the answer.
[a] Note that ε ∝ l -2 . If the length of the box is changed to e.g. 10 nm, then ε/k will be
298 / 1776 reduced by a factor of 100. This means that even He behaves in the classical manner
in a macroscopic control volume.
(qtrans )
n
n!
n→∞ e n (qtrans )
→
nn
n
ε≪kT e n V (2πmkT )3/2 −→ n
h3
n
5
Two of thermodynamic properties that can be derived from this are
cvtrans /R = 3/2 and standard entropy
s trans (1 atm)
R
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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7 Equations . . .
This topic will be further discussed in the following sections on
vibrational and rotational degrees of freedom.
302 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
The atoms in a molecule are held together by forces which, when
subjected to small perturbations, allow the atoms to vibrate as if their
centres were connected by elastic springs.
The partition function for a harmonic oscillator of this kind, with a
characteristic frequency ν, quantum number j and energy factor ε =
h ν = h ω/2π =
ˆ ~ω, is
∞
∞
P
P
−(j + 21 )ε
−j ε
= exp 2−ε
exp
exp kT .
qvıb =
kT
kT
2
3
4
Termodynamic methods TKP4175
02 September 2013
303 / 1776
ln
T
[K]
+
3
2
ln
Mw
[g/mol]
,
√
u
305 / 1776
vıb
=
√
u
1−u
→
√
e -x
1−e -x
=
√ 1√
e x − e -x
√
7
NNA4 h 3
.
which is also known as the Sackur –Tetrode equation.
Expanded footnote [a]:
Expanded footnote [a]:
[a] O. Sackur. Ann. Physik, 36:398, 1911.
[a] H. Tetrode. Ann. Physik, 38:434, 1912. Correction in 39 (1912) 255.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Figure 7.1: Comparison of experimental standard entropies with the Sackur–
Tetrode equation (the margin of error for R , NA , h and Mw is less than
0.01 cal/mol K).
He
Ne
Ar
Mw
[g/mol]
s
[cal/K mol] [†]
s trans
[cal/K mol]
4.0026
20.179
39.948
35.01 ± 0.1
36.95 ± 0.2
30.11
34.93
36.97
Kr
Xe
Rn
Mw
[g/mol]
s
[cal/K mol] [†]
s trans
[cal/K mol]
83.80
131.29
222.0
39.17 ± 0.1
40.70 ± 0.3
-
39.18
40.52
42.08
[†] Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill Book
Company, Inc., 2nd edition, 1961.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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§ 044
§ 43
§ 45
7 Equations . . .
e -x , we get
.
5
We can obtain a canonical equation of state if we substitute
√
√
2 sinh(x /2) = e x − e -x .
6
V (2πMw RT )3/2 News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
If we divide the numerator and denominator by
q
j =0
The summation can be written as a geometric series qvıb =
√
j
u/(1 − u) where u =
ˆ e -x =
ˆ exp(−ε/kT ).
j =0 u =
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
Free vibration (2)
1
P∞
Termodynamic methods TKP4175
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
j =0
5
2
= 1 − ln
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
3
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Free vibration
= −1.164855 · · · +
= 1,
.
2
These calculations are more precise than experimental results, but
the theory cannot be tested on multi-atomic molecules, because the
internal degrees of freedom of the molecules also contribute energy.
Use the Sackur–Tetrode equation to estimate
for all of the noble gases. Compare your answer with
experimental values.
02 September 2013
µtrans
RT
Let us substitute in n = NA N , m = NA-1 Mw , k = NA-1 R and ln e = 1.
§ 043 Sackur–Tetrode
s ◦ (298.15 K, 1 bar)
Termodynamic methods TKP4175
p trans V
NRT
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
7 Equations . . .
4
Differentiating gives us a set of canonical equations of state for
Helmholtz energy:
This results in a function for communal entropy that only depends
on the system’s volume and the mass and number of particles, while
the geometry and bonds are irrelevant:
1
A convincing comparison of measured and calculated values is
shown in Table 7.1.
Tore Haug-Warberg (Chem Eng, NTNU)
Sackur–Tetrode S (2)
2
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
§ 043
ε/k
[K]
1.19
0.60
0.12
For an ensemble of n particles-in-a-box without fixed locations, the
total partition function becomes Q trans = n1! (qtrans )n .
3
Termodynamic methods TKP4175
Mw
[g/mol]
2.0159
4.0026
20.179
1
Q trans =
Tore Haug-Warberg (Chem Eng, NTNU)
H2
He
Ne
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
§ 42 Particle-in-a-box
1
The table below gives the values for ε/k and the standard boilingpoint of these three low-boiling-point compounds.
When inserted into x = ε/kT this produces
h
i
µvıb
vıb
ν
= ln 2 sinh 2hkT
.
RT = − ln q
The heat capacity of a harmonic oscillator has an
upper temperature limit of limT →∞ cvvıb = R . Show this.
Differentiating twice with respect to temperature gives us
cvvıb
R
Tore Haug-Warberg (Chem Eng, NTNU)
=
( 2hkTν )
2
ν
sinh2 ( 2hkT
)
(7.26)
.
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
§ 044 Harmonic oscillator
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
1 3
The power series expansion of sinh(x ) = x + 3!
x + O(x 5 ) applied
to kT ≫ h ν gives:
1
cvvıb
lim
kT ≫h ν R
x2
2
1 3
x ≪1 (x + 3!
x ··· )
= lim
= 1.
Einstein cv
Termodynamic methods TKP4175
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2
In quantum mechanics, the moment of inertia of isolated atoms is
zero, whereas diatomic (linear) molecules have degenerate moments of
inertia about two of their axes of rotation and a zero moment of inertia
about the third axis.
1
3, 2, 1, 1
1
1, 1, 1
[cm-1 ]
ε/2k
[K]
≈ 3000
900–2200
≈ 2100
1100–2700
qEinstein →
1
−ε
1−exp( kT
)
=
1
1−e -x
.
The heat capacity can then be expressed as Einstein’s equation:
5
Expanded footnote [a]:
[a] Albert Einstein, 1879–1955. German physicist.
cvEinstein =
x 2ex
(e x −1)2
(7.27)
.
6
Our intuition tells us that the Eqs. 7.26 and 7.27 are equivalent, because the ground state only makes a constant (temperature-independent) energy contribution. The explanation for this is by no means obvious at first glance, but if we compare the equations, we can see that it
is the case (show this).
Tore Haug-Warberg (Chem Eng, NTNU)
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Free rotation (2)
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
4
The partition function for a linear molecule with a moment of inertia
I, quantum number j and energy factor ε = ~2 /2I =
ˆ h 2 /8π2 I can be
expressed
∞
P
−j (j +1)ε
qrot,2D = 1σ (2j + 1) exp kT
,
Free rotation (3)
5
There is no analytical limit to the above summation, but expanding
ε
ε 3
1 ε 2
4
the power series gives us qrot = kT
ε (1 + 3kT + 15 ( kT ) + 315 ( kT ) +
· · · ) provided that 0.3kT > ε.
Expanded footnote [a]:
[a] Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill Book
Company, Inc., 2nd edition, 1961.
7 Equations . . .
6
The upper temperature limit can be calculated using integration to
obtain an approximation of the partition sum, as shown below:
R∞
−j (j +1)ε
lim qrot,2D = 1σ (2j + 1) exp
dj
kT
kT ≫ε
j =0
ε ∞
-j(j+1) kT
= − kT
σε e
0
→
8π2 IkT
σh
2
.
Expanded footnote [a]:
[a] Note the close analogy between q rot,3D and q trans . Both of the terms include a kinetic
energy contribution, and so there are clear similarties.
lim q
rot,3D
kT ≫ε
=
(8π2 IkT )3/2
σh 3
3
In multi-atomic (non-linear) molecules, the moments of inertia
about each axis of rotation are different, unless the symmetry of the
molecule dictates otherwise.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
The rotational behaviour of multi-atomic (non-linear) molecules is
7
more complex than that of linear molecules, and we will therefore simply
state the partition function without further explanation :
j =0
where σ corrects for any rotational symmetry of the molecule (σH2 O = 2,
σNH3 = 3, σCH4 = 12).
7 Equations . . .
If we ignore the ground state, the partition function simplifies to
λ-1
1,2,3,...
4162
1306, 1534, 2916, 3019
2885
1595, 3657, 3756
g1,2,3,...
H2
CH4
HCl
H2 O
3
Tore Haug-Warberg (Chem Eng, NTNU)
Free rotation
1
A rotating system with a fixed distribution of mass is called a rigid
rotor. The moment of inertia of the rotor determines how much kinetic
energy it stores at any given frequency.
2
The table below shows the characteristic vibrational temperatures
ε/k = h ν/k = hc /λk for a few di- and multi-atomic molecules.
3
Unlike the contributions from translation and rotation, the vibrational state is only fully developed at high temperatures:
2
In a multi-atomic molecule with many vibration frequencies, the
contributions of each individual oscillator must be summed (linearly) in
accordance with the equipartition principle.
7 Equations . . .
1
One of the striking features of the oscillator equations is that the
ground state (with the quantum number j = 0) makes a positive contribution to the total energy.
4
A harmonic oscillator has two equally important energy contributions (kinetic and potential energy), which gives us cv → R /2+ R /2 = R
as the upper temperature limit.
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
.
Free rotation (4)
7 Equations . . .
8
Here yi = xi − b
x represents the molecular coordinates of each of
the atoms i measured in relation to the molecule’s centre of mass b
x=
P -1
i Mw Mi xi .
In this case, the moment of inertia is a much more complicated function
of the atomic configuration of the molecule, but it still turns out that the
√
√
3
end result can be put on the simple form I = 3 λ1 λ2 λ3 = det Λ where
λ1 , λ2 and λ3 are the eigenvalues of the inertia tensor (a symmetric
tensor → real eigenvalues):
P M = Mi yTi yi I − yi yTi ∼ QΛQT .
i
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Termodynamic methods TKP4175
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
ENGLISH text is missing.Debye A
Debye cv
1
The Debye[b] model of vibration is based on a periodic crystal
lattice.
ENGLISH text is missing.Photon gas
2
At each lattice point there is a particle (atom, group of atoms or
molecule) oscillating in the force field of the neighbouring particles.
1
Thermal radiation (in a closed control volume) can be treated as a
thermodynamic system, provided that we’re not interested in the spectral distribution[b] of the radiation.
3
The partition function adheres to the Einstein model on page 301,
but in such a way that the frequency can vary for each individual particle.
2
In this case it is actually possible to set up the canonical equations
of state[b] for internal energy
q
1 3 3 S 4
p rad =
4b V ,
4
q
3
S
T rad = 43bV
,
4
Together all of the particles vibrate within a wide spectrum of frequencies (called phonons), right up to the critical frequency ωmax , which
is an indirect measure of the total number of degrees of freedom in the
system:
ωRmax
hvor
g (ω) dω = 3n .
0
b=
8π5 k 4
15h 3 c 3
= 7.56591 · · · × 10-16
J
m 3 K4
5
The function g (ω) indicates the multiplicity for each frequency, and
the next important assumption in the Debye model is g(ω) ∝ ω2 .
.
3
Note that the number of photons present in the radiation space is
not included in the equations (determined by S and V ).
§ 45 Moment of inertia
ENGLISH text is missing.Non-linear molecules
4
In this respect eletromagnetic radiation deviates from the classical
ENGLISH text is missing.Internal degrees of free1
Let us number the atoms H(1), H(2) and O(3).
1
Finally, the canonical equation of state describing the rotation of
notion that the particle number of a system can be treated as a free
dom
2
Their respective atomic masses are M1 = 1.00794 g mol-1 , M2 = non-linear molecules can be written
variable because it is countable.
1
Vibration and rotation are the “classical” contributions to the intra1.00794 g mol-1 and M3 = 15.9994 g mol-1 .
(8π2 IRT )3/2
µrot,3D = −RT ln σN 3 h 3 ,
5
An unexpected consequence of this is that the chemical potential
molecular energy function.
A
3
The molecule can be oriented in an infinite number of ways, and
of the photon must be set at zero (this follows from a Lagrange optimi√λ
the coordinates given below are simply based on an arbitrary choice as long as we understand that I = 3 1 λ2 λ3 represents the molar mo- 2 For simple molecules, these contributions can be added together,
sation of the equilibrium state).
provided that the amplitude of vibration is small and doesn’t disturb the
(origin at O(3) and H(1) along the x axis):
ment of inertia.
ENGLISH text is missing.Excitation & quantum 6 Using Euler’s method to do a simplified integration with respect to


molecule’s centre of mass (harmonic oscillator).


 
 0.958 · sin( 14.5·2π

360 ) = 0.240 
only S and V gives us the internal energy of the radiation space:
spin
 −.958 
 0  2 Most molecules, with the exception of hydrogen (isotopes), there



 fore have a relatively high moment of inertia, which means that the up- 3 However, the general description of the molecule is far more com

x3
x1
q
 

=  0  , x2 =  0.958 · cos( 14.5·2π
3 3 3S 4
360 ) = 0.927 
 , [Å] =  0  .
[Å]
[Å]
plex, and must take into account anharmonic vibration or internal rota- 1 Electron excitation is a phenomenon that typically occurs at high



U rad (S , V ) = T rad S − p rad V =
.


0
0 per temperature limit is as low as T . 50 K:
temperatures, and the outer electrons in a stable molecule will (gradu4 4bV
tional and torsional fluctuations.
0
ally) be excited at temperatures & 3000 K.
Tb
ε/k
λi ·1047
I·1047
7
By eliminating S we can obtain the Stefan[b]–Boltzmann[b] law of
Ref. [1] σ
4
In hydrocarbons, e.g. CH3 , groups will rotate freely around the
[K]
[K]
[kg m2 ]
[kg m2 ]
4
From this information, the inertia tensor follows
rad = bVT 4 . Alternatively, we can eliminate T and keep p ,
bond axis, whereas double bonds such as CH2 = CH2 only allow tor- 2 In radicals, excitation can occur at . 1000 K, because their elec- radiation U
H2
2 0.460 0.460
0
0.460 87.6 20.39


0 
trons are more weakly bonded.
 0.818 −.262
which gives us U rad = 3pV .
CH4
12 5.313 5.313 5.313 5.313
7.6 111.66
sional fluctuations.


M
 −.262 0.954
0  .
319 / 1776
2 = 
1 2.641 2.641
0
2.641 15.3 188.15
HCl


[g mol-1 Å ]
5
Internal rotation can also be partially prevented by sufficiently low 3 The spin of elementary particles can also contribute to the energy
0
0
1.772
2 1.022 1.920 2.942 1.794 22.5 373.15
H2 O
state of a molecule.
temperatures (staggered rotation).
6
This is not, of course, entirely accurate, but it is nevertheless a
much better assumption than saying that all of the particles vibrate with
the same frequency, as is assumed in the Einstein model.
7 Equations . . .
7
The explanation for this[b] is complicated, so we will not dwell on it
here, but the result is incredibly simple
g (ω) =
9n
ω3max
ω2 .
8
The partition function can now be expressed in a modified Einstein
model
ωQ
max
1
Q Debye =
,
−~ω g (ω)
ω [1−exp( kT )]
§ 46 Atomic bomb
where the ω represent all of the discrete frequencies in the crystal.
1
The density of the 235 uranium isotope is 19.1 g cm-3 and its molar 9 At the limit value n → ∞ there will be a continuum of frequencies,
and the Helmholtz energy can be expressed as
mass is 235.04 g mol-1 .
ωRmax
h1
i
2
If 1% of the mass is involved in the reaction, the energy density will
lim A Debye = 9ωNRT
ω2 ln − exp −~ω
dω .
3
kT
n→∞
max
be u =
ˆ V -1 U = 0.01 · 6.0221 · 1023 · 3.5 · 10-11 · 19.1 · 106 /235.04 =
0
1.71 · 1016 J m-3 .
§ 46
322 / 1776
4
rad = 6.9 · 107 K and p rad =
5
Diagonalising M y QΛQT gives us the eigenvalues λ1 = 0.615, 3 This gives us the classical heat capacity values of R and 3/2R for 6 An important theoretical limit in this context is the equipartition prin- 4 In the case of hydrogen, nuclear spin (cf. parahydrogen and ortho- [b] An energy distribution function which has frequency as a free variable.
1
One of the most important fission reactions in a uranium bomb 3 From u = bT = 3p we can obtain T
λ2 = 1.156 and λ3 = 1.772, which have the same units as M itself.
5.7 · 1010 bar.
235
91
1
142
1
linear and non-linear molecules respectively.
ciple, which states that a fully developed energy degree of freedom con- hydrogen) plays a key role at temperatures . 400 K.
§ 45
[b] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley, can be written 0 n +92 U → 56 Ba +36 Kr +0 n. The energy released is
[b] Petrus Josephus Wilhelmus Debije, alias Peter Joseph William Debye, 1884–1966.
316 / 1776 tributes 1 R to cv (vibration results in two degrees of freedom, consist- 5 Similarly, electron spin is significant if the molecule has unpaired 2nd edition, 1985.
3.5 · 10-11 J per fission event. Calculate T rad and p rad if 1% of the atoms 4 These figures are representative for the uranium bomb Little Boy,
1
Determine the moment of inertia of an H2 O molecule if the bond 6 The moment of inertia can now be calculated as I =
2
√0.615
2
Dutch–American physicist.
3
take part in the fission reaction.
· 1.156 · 1.772 = 1.080 g mol-1 Å = 1.794 · 10-47 kg m2 .
angle is 104.5◦ and the O–H distance is 0.958 Å.
ing of potential and kinetic energy).
electrons (cf. NO, O2 and NO2 ).
which was dropped on Hiroshima on 6 August 1945.
[b] Joz̆ef Stefan alias Joseph Stefan, 1835–1993. Slovenian physicist and poet.
315 / 1776
314 / 1776
320 / 1776
317 / 1776
318 / 1776
321 / 1776 [b] Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley,
[a] G. M. Barrow. Physical Chemistry. McGraw-Hill, 3rd edition, 1979.
[b] Ludwig Boltzmann, 1844–1906. Austrian physicist.
2nd edition, 1985.
When expressed in the dimensionless form it makes sense to introduce a characteristic vibrational temperature θD =
ˆ ~ωmax /k also referred to as the Debye temperature. The chemical potential then becomes
θDR/T
µDebye
T 3
x 2 ln (1 − e -x ) dx ,
RT = 9 θD
1
2
The canonical equation of state can be obtained in the normal way
through differentiation.
0
Termodynamic methods TKP4175
7 Equations . . .
3
If we differentiate the partition function twice with respect to temperature, it gives us
Debye
cv
R
=9
T 3
θD
θDR/T
0
x 4ex
(e x −1)2
dx ,
Debye
This introduces a thermodynamic inconsistency where cv
,
−T ∂2 µDebye /∂T ∂T .
02 September 2013
323 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
The Dulong–Petit law
7 Equations . . .
1
There is no analytical solution to the above integral, but at the upper
temperature limit θD /T → 0 and expanding the power series e x = 1 +
x + O(x 2 ) gives us
Debye
but note that these expressions are two independent approximations of
the partition function.
4
where x = ~ω/kT =
ˆ θD /T .
Tore Haug-Warberg (Chem Eng, NTNU)
Debye cv (2)
lim
T →∞
cv
R
=9
T 3
θD
θDR/T
0
x 4 (1+x +··· )
(x +··· )2
dx ≃ 9
T 3
θD
θDR/T
x 2 dx = 3 .
0
2
This result shows that each atom in the lattice oscillates in 3 independent directions if the temperature is sufficiently high.
3
Many metals (Pb, Sn, Ag, . . . ) are essentially in this state at room
temperature. The limit value is also referred to as the Dulong –Petit
law.
Termodynamic methods TKP4175
02 September 2013
324 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
Expanded footnote [b]:
Expanded footnote [b]:
[b] Pierre Louis Dulong, 1785–1838. French chemist.
[b] Aléxis Thérèse Petit, 1791–1820. French physicist.
02 September 2013
325 / 1776
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
The Dulong–Petit law (2)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Figure 7.2: Experimental heat capacities of 12 metals plus sulphur measured
by Dulong and Petit . It is maybe a little weird that sulphur is on this list but
according to the classic thoughts of the time the element was closely related
to the metals.
Low temperature limit
7 Equations . . .
Expanded footnote [b]:
[b] Aléxis T. Petit and Pierre L. Dulong. Ann. Chim. Phys., 10:395, 1819.
4
The table below shows the original data measured by the two men
in 1819. Recalculating their data to our present values for the atomic
masses and heat capacity of water gives the limit value cp /R = 3.020 ±
0.004 while the theoretical estimate stated above is 3.
Expanded footnote [b]:
[b] The value of cp · Mw for each of the substances is multiplied with the conversion factor Mw ,O · cp ,H2 O /(R · Mw ,H2 O ) where Mw ,O = 15.9994 g mol-1 , cp ,H2 O = 75.327 J mol-1 K-1 ,
R = 8.3145 J mol-1 K-1 and Mw ,H2 O = 18.0153 g mol-1 before the average value
P
and q
the standard deviation is calculated in the usual way as b
x = N -1 xi and
P
N -1 (xi − b
x )2 .
Bi
Au
Sn
Zn
Cu
Fe
S
5
This must said to be an extraordinary good fit — not least so when
considering that the measurements were done almost 200 years ago.
σ=
cp [†]
Mw [‡]
0.0288
0.0298
0.0514
0.0927
0.0949
0.1100
0.1880
13.30
12.43
7.35
4.03
3.957
3.392
2.011
cp · Mw
0.3830
0.3704
0.3779
0.3736
0.3755
0.3731
0.3780
Pb
Pt
Ag
Te
Ni
Co
-
cp [†]
Mw [‡]
0.0293
0.0314
0.0557
0.0912
0.1035
0.1498
-
12.95
11.16
6.75
4.03
3.69
2.46
-
cp · Mw
1
For the lower temperature limit, the ratio θD /T → ∞ and the integral is a constant.
2
0.3794
0.3740
0.3759
0.3675
0.3819
0.3686
-
This result has a practical application in the proportional relation-
ship
Debye
lim
T →0
cv
R
∝ T3
when determining the entropy of a substance by calorimetry.
[†] Relative heat capacity. [‡] Relative atomic mass. The reference values are the
heat capacity of liquid water and the atomic mass of oxygen respectively.
3
Instead of the highly time-consuming and difficult process of measuring near 0 K, all you need to do is measure the heat capacity down
to 15–20 K.
4
Often this is sufficient to determine the factor of proportionality
in the above equation, allowing you to extrapolate the measurements
down to 0 K.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
326 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Second virial coefficient
1
The bonding electrons in a metal belong to a group of particles
called fermions, named after the physicist Enrico Fermi[b].
2
One of the characteristics of fermions is that they cannot share the
same one-particle state (the Pauli[b] exclusion principle).
3
In a metal, the electron energy levels will therefore be filled up from
the lowest level up to the Fermi level equivalent to the chemical potential
(µ◦ ) of the electrons in the conduction band.
4
7 Equations . . .
The internal energy of these electrons can be expressed
U elec
NRT
=
u◦
RT
+
π2
4
·
kT
µ◦
+ ···
5
The chemical potential of electrons is a relatively abstract concept,
µ
and an alternative to this value is the Fermi temperature TF =
ˆ k◦ , which
for normal metals is approximately 105 K.
cvelec
lim
T →0 R
=
π2 T
2TF
∝T.
This is in contrast to crystalline phases with covalent bonds where
limT →0 (cv ) ∝ T 3 due to the phonon contribution.
7
8
The practical implication of this is that the heat capacity of metals is dominated by the fermion contribution at sufficiently low temperatures (T . 5 K).
329 / 1776
[b] Enrico Fermi, 1901–1954. Italian–American physicist.
[b] Wolfgang Ernst Pauli, 1900–1958. Austrian physicist.
The virial equation relates to ideal gases. Expand the power series
p /RT in molar density ρ = N /V :
1
2
There is an equivalent power series expansion for osmotic pressure
in dilute solutions, but it is harder to derive.
p vır V
NRT
=
∞
P
k =1
Bk (T )
N k −1
V
= 1 + B2 (T )
N
V
(7.28)
+ ···
4
Here the series has been truncated after the second term, which
means that the index 2 can be left off the symbol B :
BN
V
+ ··· ≃ 1 +
Bp
RT
+ ···
Termodynamic methods TKP4175
Second virial coefficient (2)
02 September 2013
330 / 1776
In practice B < 0 for T < TB ≈ 6Tc , where TB is known as the
Boyle temperature.
7
Statistical mechanics explains the theoretical relationship between
B and the associated pair potential φ(r ) for the interaction between two
molecules in the gas phase:
R∞ −φ(r )
B (T ) = 2πNA
(7.29)
1 − e kT r 2 dr .
8
9
This equation holds very well for non-polar molecules with virtually perfect spherical symmetry, but for rod-shaped molecules and
molecules with permanent dipole moments integration must be done
with respect to the (two) inter-molecular angles of orientation as well as
the inter-molecular distance.
10 This results in an integral that looks like the one in 7.29, but where
φ(r ) represents the mean pair potential of the molecules.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
331 / 1776
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
1
Potential well (2)
3
= ln α3α−1
⇔
α3 =
r
Let us assume that φ = 0 for r > σ and φ = ∞
for r ≤ σ.
φ
σ
Tore Haug-Warberg (Chem Eng, NTNU)
r 2 dr =
2πNA σ3
3
Termodynamic methods TKP4175
02 September 2013
332 / 1776
100
0
ε
−100
TB
−200
B hs
B sw
−300
B LJ
−400
0
200
400
600
800
T [K] of methane calculated using the
Figure 7.3: The second virial coefficient
hard sphere, square-well and Lennard-Jones 6:12-potential where σ = 3.85 Å
and ε = 1.96 · 10-21 J. Both B sw and B LJ satisfy the lower temperature limit
limT →0 B = −∞, whilst the upper temperature limit limT →∞ B = 0+ is only fulfilled for B LJ (this can only be seen once you reach T & 104 K).
σ
333 / 1776
Rσ
4
The hard sphere potential produces a reasonable approximation
of a real system at high temperatures, but it doesn’t explain why B is
temperature-dependent.
ε
Substituting this simple potential into the integral for B (where
α > 1), demonstrates that B (T ) is a strongly temperature-dependent
function with the hard-sphere potential as its upper temperature limit:
ασ
R ε
1 − e kT r 2 dr
B sw = B hs + 2πNA
02 September 2013
Substituting this into the integral for B gives
0
e kTB
That means viewing the molecules as hard
spheres with pairwise interaction within the potential well.
Termodynamic methods TKP4175
2
That means viewing the molecules as hard spheres, with no pairwise interaction.
us
B hs = NA
e kTB −1
3
Tore Haug-Warberg (Chem Eng, NTNU)
3
r
7 Equations . . .
2
i
h
ε
= B hs 1 + (α3 − 1)(1 − e kT ) .
7 Equations . . .
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Let us assume that φ = 0 for r > ασ, φ = −ε for
σ < r ≤ ασ and φ = ∞ for r ≤ σ.
σ
328 / 1776
1
B LJ [cm3 mol-1 ]
ε
kTB
02 September 2013
Hard spheres
4
Another connection between ε, TB and α is revealed if you solve
the equation B sw (TB ) = 0:
φ
−ε
7 Equations . . .
6
Experimentally B = limp →0 (V − RT /p ) is used, but this limit tells us
nothing about how B varies with temperature.
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Potential well
Termodynamic methods TKP4175
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
0
5
Multiplying both sides by NRT /p gives the volume-explicit form
V 2.vır = NRT /p + BN , which reveals that B is a correction factor for
the molar volume of the gas.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
[b] Robert Boyle, 1627–1691. English chemist and philosopher.
This is the normal way to express the virial development of gases,
but it is also possible to invert the series from p (ρ) to ρ(p ) as shown
below.
=1+
327 / 1776
Expanded footnote [b]:
3
pV 2.vır
NRT
02 September 2013
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
ENGLISH text is missing.Fermions in metals
6
Differentiating U elec gives us the lower temperature limit for the
electronic contribution to the heat capacity of metals:
Termodynamic methods TKP4175
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Lennard-Jones[2]
1
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
i
h
Let us assume that φ = 4ε ( σr )12 − ( σr )6 for
r > 0.
Expanded footnote :
The Van der Waals equation
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
For a given vector x consisting of the mole fractions x1 , x2 , . . . , xn
the Van der Waals equation of state can be expressed as
1
Expanded footnote [c]:
[c] Johannes Diderik van der Waals, 1837–1923. Dutch physicist.
[a] John Edward Lennard-Jones, 1894–1954. English mathematician and physicist.
φ
σ
−ε
r
The exponent 6 has a theoretical basis in
the London theory of dispersion forces involving instantaneous dipoles, whilst the exponent
12 is empirical.
3
2
This version is often referred to as the 6:12 potential.
Expanded footnote [b]:
[b] Fritz London, 1900–1954. German–american physicist.
p VdW (T , v , x) =
P P
RT
v −b (x)
−
a (x)
v2
(7.30)
,
where a (x) = i j aij xi xj tells us something about the attractive forces
P
between the molecules and b (x) = i bi xi represents the hard-sphere
volume.
The Van der Waals equation (2)
7 Equations . . .
2
This means that there must be a relationship between the
Van der Waals equation and the virial equation in Section 7.7. We
can demonstrate this relationship by expanding the power series
p VdW V /NRT at the limit V → ∞ (ideal gas):
p VdW V
a N
1
NRT = 1−b ( N ) − RT V
V
2
n N n
a N
− RT
= 1 + b VN + b 2 N
V + ··· + b V
V .
3
If we compare the coefficients in this series with the ones in
Eq. 7.28, we find that B2vdw = b − a /RT , B3vdw = b 2 , B4vdw = b 3 etc.
B LJ cannot be determined analytically, and numerical integration is
therefore required.
4
7
A typical temperature progression is shown in Figure 7.3 along with
the experimental data for methane .
5
This is simple in theory, but not quite so straightforward in practice.
6
In order to prevent numerical problems, the lower limit is estimated
using a hard-sphere contribution and the upper limit using a power series expansion, see the Program 1.5 in Appendix 31.
Expanded footnote [b]:
[b] J. H. Dymond and E. B. Smith. The Virial Coefficients of Pure Gases and Mixtures.
Oxford University Press, 1980.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
336 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
The Van der Waals equation (3)
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
§ 047
It is beyond the scope of this chapter to discuss all of the impli6
cations of the Van der Waals equation here, but we will at the very
least note one important property: The parameters aii and bi can be
obtained by measuring the critical point of each individual compound
experimentally.
§ 46
§ 48
7 Equations . . .
Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
7 Equations . . .
The two Eqs. 7.31 and 7.32 have the same algebraic structure,
which allows us to compare the coefficients term by term.
§ 047 Critical points (2)
5
3
Alternatively we could have solved the equations with respect to
e.g. a , b and R , but traditionally vc is favoured because it is subject to
a relatively large degree of experimental uncertainty.
4
However, this means that the Van der Waals equation of state has
too few degrees of freedom to accurately reproduce all measurements
of pc , Tc and vc .
3vc2
=
a
pc
,
vc3 =
ab
pc
,
RTc
pc
3
8
=
b=
a=
,
2
The values given can obviously be verified by direct substitution,
but we will choose a different approach.
3
This shows that (at constant pressure) the Van der Waals equation
can be treated as a cubic equation with volume as a free variable. From
the problem statement, we know that the pV isotherm has a degenerate
extreme point[c] as its critical point.
[c] Both a maximum and a minimum (a horizontal point of inflection).
v 3 − v 2 3vc + v 3vc2 − vc3 = 0 .
340 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
(7.32)
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
vc pc
RTc ,
1
3 vc ,
3pc vc2 ,
§ 048
§ 47
§ 49
7 Equations . . .
(7.33)
(7.34)
(7.35)
which can in turn be rewritten in the form used in the problem statement.
6
We should note that the critical compressibility factor zcVdW =
ˆ
pc vc /RTc = 83 behaves as a universal constant because vc is a dependent variable in the solution.
which must be solved with respect to a , b and vc .
7 Equations . . .
The solution to the above set of equations can be expressed
This gives rise to the following set of equations:
3vc = b +
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4
This implies that the equation of state must have three coincident
roots (v − vc )3 = 0, or in its expanded form:
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
1
2
Termodynamic methods TKP4175
02 September 2013
If you multiply both sides of the equation by v 2 (v − b ) you can
rewrite the equation of state 7.30 in the form p (v − b )v 2 = RTv 2 −
a (v − b ), or as
a
ab
v 3 − v 2 b + RT
(7.31)
p + v p − p = 0.
Show that the parameters b = RTc /8pc , vc = 3b and
a = 27(RTc )2 /64pc meet the criteria for a mechanical–
critical point defined by (∂p /∂v )T ,x = 0, (∂2 p /∂v ∂v )T ,x =
0 and p (Tc , vc ) = pc .
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
§ 047 Critical points
Cubic equations of state
1
7
This does not mean that the equation is valid under near-critical
conditions (it isn’t), but these parametric values produce universal behaviour which is very valuable.
Termodynamic methods TKP4175
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
4
Only B2vdw bears any resemblance to real systems; the other virial
coefficients are completely unrelated to reality.
5
In spite of the massive simplification, the Van der Waals equation
of state is often used as a starting point for more complicated equations
of state, see Chapter 13.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Pressure is an intensive property (homogeneous function
of degree zero), which implies that p (T , v , x) = p (T , V , n).
Rewrite the Van der Waals Eq. 7.30 in this form.
7
Experimentally this is no more than a rough approximation as
0.25 < zc < 0.35 for most molecular compounds.
Tore Haug-Warberg (Chem Eng, NTNU)
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02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
§ 048 VDW in the extensive form
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
§ 049
§ 48
§ 50
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
7 Equations . . .
1
Dimensional analysis shows that if v → V and x → N x in Eq. 7.30,
then homogeneity will be preserved.
2
p
VdW
(T , V , n) =
NRT
V −Nb (x)
−
p VdW =
Show that, in its reduced form, the Van der Waals
equation can be written as pr = 8Tr / (3vr − 1) − 3/vr2 .
Draw the function for Tr ∈ {0.75, 1, 1.5}. Identify all of
the asymptotes to the graph. Specify the physical
range of the function.
(7.36)
V2
7 Equations . . .
1
Rewrite the Van der Waals equation in its reduced form by substituting Eqs. 7.34 and 7.35 into Eq. 7.30:
Consequently, the equation of state can be written in the form
N 2 a (x)
§ 049 Corresponding state
RT
1
v − 3 vc
−
3pc vc2
v2
(7.37)
.
ˆ p /pc and
2
Then divide by pc and define reduced pressure as pr =
ˆ v /vc :
reduced volume as vr =
prVdW =
RT
pc vc
1
vr − 3
−
3
vr2
(7.38)
.
3
Finally substitute in R /(pc vc ) = 1/(zc Tc ) = 8/(3Tc ) from Eq. 7.33
ˆ T /Tc .
and define reduced temperature as Tr =
Tore Haug-Warberg (Chem Eng, NTNU)
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02 September 2013
345 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
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Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
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News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
ENGLISH text is missing.Condensed phases
1
According to Murnaghan[c]’s equation, isobaric expansivity α and
isothermal compressibility β are two independent state functions of temperature and pressure respectively.
2
The equation is most useful for pure solids and stoichiometric compounds.
3
To derive the chemical potential, we start with the total differential
=
of molar volume, for our current purposes expressed as (dln v )mur
N
α(T ) dT − β(p ) dp , also see Section 8.7 on page 377.
4
Two commonly used functions for α and β are:
α(T ) = a1 + a2 T +
β(p ) =
§ 049 Corresponding state (2)
7 Equations . . .
4
This gives us the Van der Waals equation as formulated in the
problem statement:
prVdW =
8Tr
3vr −1
3
vr2
−
Looking back
1
(7.39)
.
b
6
We call this a universal equation, and here we will take the opportunity to mention the principle of corresponding states, which states that
related substances behave virtually alike at reduced states.
8
The fluid is thermodynamically unstable in the mid-range where
(∂pr /∂vr ) > 0.
vr
Tore Haug-Warberg (Chem Eng, NTNU)
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02 September 2013
348 / 1776
6
1+b2 p◦
b2 −b1
hβ
◦ v
β v◦
,
1+b2 p 1−b1 /b2
1+b2 p◦
§ 50 The Murnaghan model
1
It is easy to make apparently independent assumptions without
checking whether they are mutually coherent.
2
For the model to be consistent, the following third-order Maxwell
equation
∂2 v
∂T ∂p
=
§ 51 Murnaghan CP
∂2 v
∂p ∂T
§ 52 Thermal instability
1
The fact that cp is pressure-dependent can be derived as shown
below.
must hold. Substituting in (∂v /∂T )p = vf (T ) and (∂v /∂p )T = −vg(p )
gives us
?
− ∂∂Tv p g (p ) = ∂∂pv f (T ) ,
2
T
where f (T ) and g (p ) are functions of different state variables.
10 A trivial integration of µmur using the Euler method gives us the § 50
1
Murnaghan’s equation is derived by integrating the expansivity 3 This avoids circular references.
Gibbs energy function
with respect to T at a given p◦ , and then integrating the compressibility 4 If we rewrite the last equation slightly we get
G mur (T , p , N ) = N µmur (T , p ) .
§ 51
with respect to p at constant T , but this provides no guarantee that our
?
vf (T )g (p ) = vg(p )f (T ) .
1
Assume that thermal expansivity is constant, i.e. that a2 = 0 K-2
starting point was thermodynamically consistent. Use differentiation to
349 / 1776 show that α = f (T ) and β = g (p ) genuinely represent a consistent basis Here the right-hand side is identical to the left-hand side, so the test has and a3 = 0 K. Find an expression for (∂cp /∂p )
that is true for this
T ,N
for a model.
simplified Murnaghan model. What is the sign of the derivative?
been satisfied.
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Note the Maxwell relationship used at the start of the equation:
h
i
∂cpmur ∂s ∂
∂
∂s ∂p T ,N = ∂p T ∂T p ,N T ,N = T ∂T ∂p T ,N
p ,N
h i
= − T ∂∂T ∂∂Tv p ,N
p ,N
∂(v α)
=
ˆ − T ∂T p ,N
= − Tv α2
≤
0
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1
The first condition is satisfied for all T and p if b1 > 0 and b2 > 0.
2
With respect to the second condition, in Theme 51 we showed that
(∂cpmur /∂p )
< 0 for all V > 0.
T ,N
3
Since cp < ∞, there will always be a pressure p < ∞ at which
cpmur = 0, unless
§ 52
R∞
p◦
?
v mur dp < ∞ .
1
For a one-component system to be thermodynamically stable, it 4 It can be demonstrated that this is not the case, and that Murmust satisfy the conditions β > 0 and cp > 0. Are these conditions naghan’s model therefore does not satisfy the requirement for thermal
(always) met for Murnaghan’s equation?
stability at all T , p .
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[c] F. D. Murnaghan. Proc. Natl. Acad. Sci. U. S. A., 30:244–247, 1944.
ıg
P
i
i
Ni si◦ (T , p◦ ) +
ıg
U (T , n) = H (T , n) − NRT
P
H ıg (T , n) = Ni hi◦ (T )
P
i
Ni R ln
p◦ V
Ni RT p◦ V
−1
hi◦ (T ) − hi◦ (T◦ ) =
si◦ (T ) − si◦ (T◦ ) =
4
2)
3)
µ◦i (T , p◦ ) = µ◦i (T , p◦ )
µi◦ (T , p◦ )
µi◦ (T , p◦ )
=
=
Tore Haug-Warberg (Chem Eng, NTNU)
5
∆f hi◦ (T ) − Tsi◦ (T , p◦ )
∆f hi◦ (T◦ ) + (hi◦ (T ) − hi◦ (T◦ ))
0
Termodynamic methods TKP4175
Part
Corresponding state principle:
−
3pc vc2
v2
RT
cp◦,i (τ) dln τ
T◦
The superposition principle:
02 September 2013
The virial expansion theorem:
∞
P
k −1
p vır V
Bk (T ) N
= 1 + B2 (T ) VN + · · ·
NRT =
V
356 / 1776
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α and β Ideal gas Waals Olds
02 September 2013
8
State changes at constant composition
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α and β Ideal gas Waals Olds
New concepts
R∞ −φ(r )
1 − e kT r 2 dr
RT
1
v − 3 vc
cp◦,i (τ) dτ ,
T◦
k =1
− Tsi◦ (T , p◦ )
The 2nd virial coefficient:
p VdW =
RT
cp◦,i = cvtrans
+ cvrot
+ cvvıb
+ cvelec
+R
,i
,i
,i
,i
Standard state contribution:
1)
7 Equations . . .
Temperature dependency:
i
i
2
Looking back (2)
3
Ni RT 7 Equations . . .
B (T ) = 2πNA
7
7 Equations . . .
− 1
− 1i ,
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
News Ideal gas Molecules Photons Phonons Electrons Virial Waals Murnaghan Olds
Looking back (3)
b1 v◦
b2 −b1 β◦
−S ıg (T , V , n) =
For the lowest isotherm (Tr = 0.75)
there are up to three possible solutions for
pr = pr (vr ). This results in a phase transition for the system as described in Chapter 14 on vapour–liquid equilibria.
7
p◦
= µ◦ (T , p◦ ) + v (T , p◦ )
= µ◦ (T , p◦ ) +
9
For solid phases at high pressures and polymorphous compounds,
it is normal to express the standard molar Gibbs energy as a (kind of)
power series of temperature:
µ◦
2
a
b
T
RT = · · · + T 2 + T + c + d ln T◦ + eT + fT + · · ·
i
The
E physical range is defined by vr ∈
D
1
,
∞
. Note that the equation contains no
3
arbitrary model parameters.
c.p.
a3
T2
.
7
We can now integrate (dµ)T ,N = v dp betweenRthe two states T , p
p
and T , p◦ . This produces µmur (T , p ) = µ◦ (T , p◦ ) + p v (T , p ) dp which
◦
with a little bit of effort can be rewritten as
Rp 1+b p -b1 /b2
dp
µmur (T , p ) = µ◦ (T , p◦ ) + v (T , p◦ ) 1+b22p◦
ˆ v (T , p◦ ).
where β◦ =
ˆ β(p◦ ) and v◦ =
8
It is now easy to produce a complete equation of state µmur (T , p ),
provided that we have an expression for the compound’s standard
chemical potential µ◦ (T , p◦ ).
Helmholtz energy of an ideal gas:
h
P
P
A ıg (T , V , n) = Ni µi◦ (T , p◦ ) + Ni RT ln
5
pr
b1
1+b2 p
5
Let us start by integrating α(T ) at a given reference pressure p◦
between the reference temperature T◦ and an arbitrary system temperature T . This gives us: ln[v (T , p◦ )/v (T◦ , p◦ )] = a1 (T − T◦ ) + 12 a2 (T 2 −
T◦2 ) − a3 (T -1 − T◦-1 ) = (T − T◦ )[a1 + 21 a2 (T + T◦ ) + a3 (TT◦ )-1 ].
6
Next we integrate β(p ) at constant T between the reference pressure p◦ and an arbitrary system pressure p . This gives us: ln[v (T , p )/
v (T , p◦ )] = −b2-1 b1 ln[(1 + b2 p )/(1 + b2 p◦ )].
8 State . . .
1
Non-canonical variables
2
The Bridgman table
3
Heat capacity
4
Isothermal expansivity
5
Isobaric compressibility
6
Critical point divergence
see also Part-Contents
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
U , H and S
8 State . . .
The most important quantities in process engineering include energy, enthalpy and entropy, generally expressed as functions of temperature and pressure (or density), and composition.
1
In many cases it is sufficient to know U , H and S as functions of
just one (common) set of variables, but from a theoretical point of view
there is great value in being able to choose the variables that are most
suited to the particular application.
4
2
These are particularly important to the energy balances for media
flowing through valves, turbines, compressors and pipelines.
3
It is therefore of great practical relevance to have a fundamental
understanding of the thermodynamic properties of a fluid (of constant
composition).
5
In this context, T , p , n and T , V , n are particularly relevant, as are
the functions’ canonical variables.
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8 State . . .
Assume for a change that we do not remember the rules of differentiation leading to Eqs. 8.6–8.8.
What alternatives exist for the differentiation in that case?
By inserting Eq. 8.6 on the left side and writing the partial derivatives as differential quotients we get:
d(−pV +µN )
d(U −TS )
=
,
−S =
dT
dT
V ,N
V ,N
−p dV −V dp +µdN +N dµ
dU −T dS −S dT =
=
.
dT
dT
V ,N
V ,N
3
Let us apply Eq. 8.6 to a single component system and use that as
an example.
4
Substitution of respectively the Legendre transform A =
ˆ U − TS
and the Euler form A = −pV + µN leads to:
∂(−pV +µN )
∂(U −TS )
∂A =
=
.
∂T V ,N
∂T
∂T
V ,N
V ,N
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02 September 2013
1
T
dU +
p
T
dV .
8 State . . .
(8.1)
(8.2)
(8.3)
Canonical means conforming to a pattern or rule, and in thermodynamics this refers to the structure of the differentials.
2
For example, S , V , n are canonical variables for U , whereas T , V , n
and T , p , n are not.
Other important functions (stated with their associated canonical
4
variables) include: H (S , p , n), A (T , V , n), G (T , p , n) and S (U , V , n).
The relationship between these functions can be derived from U , which
in this case is a fundamental energy function:
3
This becomes clear if we compare the above equations with 8.13
and 8.24, which show that (dU )n becomes more complex if we stray
away from the canonical description.
U (S , V , n) = TS − pV +
Tore Haug-Warberg (Chem Eng, NTNU)
366 / 1776
To shed some light on this issue we shall nevertheless have a
closer look at the possibility of using f (ξ) as an alternative to f (x ) and
φ(ξ). The following definition can serve us as an example:
1
i =1
Termodynamic methods TKP4175
µi Ni .
02 September 2013
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Canonical variables I (2)
8 State . . .
7
Thanks to our study of the theory behind Legendre transforms in
Chapter 4, we know that the new function is particularly useful if we
replace S with T as a free variable:
−S = ∂∂AT V ,n ,
(8.6)
∂A (8.7)
−p = ∂V T ,n ,
∂A µi = ∂Ni
.
(8.8)
5
Alternatively we can view S as a new fundamental relationship
that gives us the same information as U , in which case it is called a
Massieu[a] function:
n
P
µi
p
S (U , V , n) = T1 U + T V −
(8.5)
T Ni .
i =1
6
The remaining functions in the list can be obtained by performing
Legendre transformations on the relevant fundamental relationship[a].
For instance, transforming U with respect to S gives rise to a new function, which is called Helmholtz energy, which is defined by the equation
A=
ˆ U − TS .
[a] François Jacques Dominique Massieu, 1832–1896. French mathematician and
physicist.
[a] If you start out from U (S , V , n), the coordinates will always be positive. If on the
other hand you use S (U , V , n), the coordinates make more sense in relation to the
physical system, in the sense that it is easier to accept energy as a free variable than
entropy.
T ,V ,Nj ,i
8
This result is the reason behind the term canonical variables.
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α and β Ideal gas Waals Olds
6
Replacing the differential dU = T dS − p dV + µ dN in the middle of
the expression, and the Gibbs–Duhem equation S dT − V dp + N dµ = 0
in the last part of the expression, yields:
−p dV +µdN −S dT −p dV +µdN −S dT =
,
−S =
dT
dT
V ,N
V ,N
= −S
α and β Ideal gas Waals Olds
9
Also note that the differential of A with respect to T is −S , whereas
the differential of U with respect to S is T .
10 This duality means that S and T are referred to as conjugate variables.
11 The differentials of A with respect to V and n are the same as those
of U , but remember that T has to replace S as one of the variables that
is being held constant.
(8.4)
8 State . . .
Non-canonical variables
1
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α and β Ideal gas Waals Olds
ENGLISH text is missing.Non-canonical variables I
1
The canonical variables are of prime concern in the general description of thermodynamic states.
8 State . . .
2
The question we are now going to ask ourselves is to what extent
we can choose among the co-ordinates (S , V , n), (T , V , n), (T , p , n),
(U , V , n), etc.
3
We know that these variables are the natural choices for describing
U , A , G and S respectively, but could e.g. (T , V , n) serve as a common
set of variables for all of the functions?
4
That would at be immensely practical, if nothing else.
5
The answer proves to be ambiguous.
6
On the one side this choice is entirely possible but on the other side
it does complicate our formulas.
7
In order to get to the bottom of the matter, you need a thorough understanding of function transformations in general and Legendre transforms in particular.
8
That discussion is not based so much on thermodynamics, and so
it may be helpful to leave the thermodynamic equations out of it.
9
We will therefore study an anonymous function f (x ) with an associated Legendre transform φ(ξ).
368 / 1776
2
But, anonymity is not synonymous with denial and we shall continue to keep our thermodynamic cutlery sharp.
3
In particular we must inculcate ourselves that a thermodynamics
state is inextricably linked to the information contained in the graph
(x , f (x )).
4
From Chapter 4 we know that the Legendre transform φ =
ˆ f − ξx
preserves all of the information about the thermodynamic state.
5
Hence the graph (ξ, φ(ξ)) is equivalent to (x , f (x )).
6
At the same time we argued that neither f (ξ) nor φ(x ) possesses
the same property.
In other words, the knowlegde of f (x ) is sufficient for calculating
both the function value and ξ whereas f (ξ) is less valuable because
x (ξ) cannot be derived from this function.
8
The alternative is to use φ(ξ) which holds the same information as
f (x ).
7
= −S .
7
In the end, all the three derivatives were proved to be equal, although it is was not easy to be in charge of the operation and maintain
a good overview all the time.
8
We shall therefore make use of the next subchapter to learn more
about the limitations that start to play when, or if, we want to deviate
from the canonical formulations.
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8 State . . .
n
P
Canonical variables II (2)
α and β Ideal gas Waals Olds
Non-canonical variables III
(dS )n =
1
Find the total differentials of U (S , V , n), H (S , p , n) and S (U , V , n) The structure of these differentials is clearly simple, but this makes them
in the specified sets of variables. Assume that the composition is in all the more difficult to integrate, as the equations of state are rarely or
each case constant. How easy do you think it would be to integrate never known for T (S , V , n) and p (S , V , n), or alternatively T (S , p , n)
p
and V (S , p , n), or T1 (U , V , n) and T (U , V , n).
these differentials?
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News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
1
5
(dU )n = T dS − p dV ,
(dH )n = T dS + V dp ,
§ 53
1
α and β Ideal gas Waals Olds
Canonical variables II
2
Canonical variables I
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 53 Revision
1
As a warm-up exercise, we will start by revising the canonical
expressions for dU , dH and dS :
Mappings
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
8 State . . .
After substitution, the two functions are different.
The function f (ξ) should therefore be given a symbol different from
f (x ), but in thermodynamic applications it is usual to use the argument
to distinguish between the functions.
3
4
This is not an ideal situation, but the same practice is used in e.g.
computer programming[a].
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[a] One example is the so-called multiple dispatch methods where e.g. sqrt(x) in most
FORTRAN-like languages is meant to return the square root of x — be it a real or a
complex number — independently of whether the input value of x is a natural number
or a real number.
Termodynamic methods TKP4175
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α and β Ideal gas Waals Olds
Function ranges
1
Note that the range of f (ξ) is exactly the same as the range of f (x ).
This assumes that ξ(x ) is an invertible function, but provided that this
assumption holds, the two functions are identical until x is substituted
by x (ξ).
2
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α and β Ideal gas Waals Olds
Mappings (2)
8 State . . .
f (ξ)
The link between the original two functions f (x ) and ξ(x ) on the
one side and f (ξ), x (ξ), and φ(ξ) on the other side is illustrated below:
1
ξ(x )
2
The possibility of using φ(x ) is also true but it is of less practical
interest.
f (x )
f (x ) =
ˆ c + ln(x ) .
3
4
The transformed variable is defined by
ˆ ∂∂xf = x -1 ,
ξ(x ) =
e -1
1
x (ξ)
e
c −1
c
c+1
φ(ξ)
giving us x = ξ-1 and
φ(ξ) =
ˆ f − ξx = c − ln(ξ) − 1
f (ξ) =
ˆ f (x (ξ)) = c − ln(ξ)
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
370 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
372 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
373 / 1776
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
α and β Ideal gas Waals Olds
ENGLISH text is missing.U (T , V , N )
1
Chemical process simulation software often uses p (T , V , n) equations of state to predict the behaviour of fluids in phase transitions.
2
Based on what we have learnt so far, it seems wise to describe
the system in terms of its Helmholtz energy, and to derive the internal
energy, enthalpy and entropy from A (T , V , n), but below we will look at
a more direct approach to deriving U , H and S — partly because it is
useful practice and partly because it is a key topic in most textbooks.
3
The pivot of the derivations is the assumption that all real fluids
eventually behave as ideals gases whence the pressure is sufficiently
low. In terms of Helmholtz energi we write:
ıg
lim A = A .
V →∞
Mappings (3)
8 State . . .
Legendre transform
8 State . . .
§ 054
1
In order to recreate the original function graph we must know both
(ξ, f (ξ)) and (ξ, x (ξ)). Alternatively we can rely on the Legendre transform φ =
ˆ f − ξx , which has the special property that [∂/∂] φξ = −x and
that f (ξ) = φ − (∂φ/∂ξ) ξ.
2
In the figure x ∈ [1, e ] has been chosen as a suitable domain of
definition for f .
3
The mappings have the corresponding function domains:
4
How innocent it looks, this equation has far-reaching consequences.
5
Beacuse, if A approaches A ıg assymptotically when all its partial
derivatives must do the same!
§ 53
§ 55
6
In particular, we would like to differentiate with respect to T and V
since the composition is assumed constant. However, it would also be
of interest to include the Legendre transform(s) of A in the study, and
for this reason we shall include the derivative with respect to 1/T (by
applying the Gibbs–Helmholtz equation):


∂(A /T )
 U =

→ U ıg


∂( 1/T ) V ,n



∂( A )

→ S ıg
S = −
lim 
(8.9)

∂( T ) V ,n

V →∞ 


∂( A )


→ p ıg
 p = −
V
∂(
7
8 State . . .
) T ,n
The discussion of the limits does not end here.
8
We can in principle continue to differentiate A for as long as we
wish.
9
That is, however, not for very long because we in practice only need
to know three of the second order derivatives:

∂( p /T )
∂2 (A /T )
 ∂U 
→ 0
= −
=


∂( 1/T ) ∂V
∂( 1/T ) V ,n

 ∂V T ,n
ıg


C
CV

∂2 ( A )

 ∂S
=
ˆ
→ TV
= −

T
∂ T V ,n
∂( T ) ∂T
lim 
(8.10)

∂( p )
 ∂S ∂2 ( A )
V →∞ 

→ NR
 ∂V T ,n = − ∂( T ) ∂V =
V

∂( T ) V ,n



| {z }


Definition & Maxwell rules
10 These results make it possible to derive closed expressions for
U and S of an arbitrary real fluid where the reference is an ideal gas
sharing the same macroscopic state variables T , V , and n.
376 / 1776
2
In the present case ξ(x ) is invertible which means we can also
determine the inverse function x (ξ) = ξ-1 (x ).
3
But, after substituting x (ξ) into f (ξ) =
ˆ f (x (ξ)), it is no longer possible to reverse the transformation.
4
Information about x (ξ) has been lost, and even if f (x ) = f (ξ)
throughout the entire value domain, it is impossible to recreate x from
f (ξ).
f (x ) : [1, e ] → [c , c + 1]
ξ(x ) : [1, e ] → [1, e -1 ]
f (ξ) : [1, e -1 ] → [c , c + 1]
Set out the total differentials of U (T , V ), H (T , V ) and
ıg
S (T , V ) in a form that can be integrated if CP (T ) and
p (T , V ) are known state functions.
In other words, in a thermodynamic context we can recreate
the information contained in U (S , V , N ) either from the two functions
U (T , V , N ) and S (T , V , N ), or from the Legendre transform A (T , V , N ).
9
5
An equivalent description of the fucntion graph (x , f (x )) can then
be stated in terms of (ξ, φ(ξ)).
6
The new situation is also captured by the figure above.
7
The Legendre transform is appearantly sufficient for recreating the
entire function graph because x is equal to − (∂φ/∂ξ) and f (x ) is equal
to φ − (∂φ/∂ξ) ξ.
8
In addition we learn a new function φ to know which can be of
practical interest in itself.
10 If T and p are chosen as free variables we must either know the
three functions: U (T , p , N ), S (T , p , N ) and V (T , p , N ), or the Legendre
transform G (T , −p , N ).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
2
One general approach to problems of this type is to start out from
one of the canonical differentials 8.1–8.3 and to transform it by systematically eliminating unwanted variables (in this case S ).
(8.11)
If we compare this differential with the equivalent differential in Eq. 8.1
we can see that S needs to be replaced with T in order for us to make
any progress.
One possible approach is to convert dS into a differential expressed in terms of T and V :
(dS )n = ∂∂ST V ,n dT + ∂∂VS T ,n dV
C
∂p =
ˆ TV dT + ∂T V ,n dV .
(8.12)
3
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
§ 54
378 / 1776
§ 56
8 State . . .
[a] For the derivation of Maxwell relations, see Theme 27 on page 155.
Let us combine Eqs. 8.1 and 8.12: :
h ∂p i
(dU )n = CV dT + T ∂T V ,n − p dV
∂(p /T )
= CV dT − ∂(1/T )
dV
Expanded footnote [a]:
[a] In order to derive the second line from the first line we need to use the mathematical
∂(y /x )
∂y
∂y
≡ y + 1x
≡ y−x
identity
, also known as the Gibbs–Helmholtz’
∂(1/x )
∂(1/x )
∂x
equation, from Theme 28.
V ,n
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
Expanded footnote [a]:
6
Tore Haug-Warberg (Chem Eng, NTNU)
α and β Ideal gas Waals Olds
5
Also note that the Maxwell relation (∂S /∂V )T ,n = (∂p /∂T )V ,n has
been used in the last line.
(8.13)
§ 054 Differential of H (T , p )
02 September 2013
377 / 1776
α and β Ideal gas Waals Olds
8 State . . .
ENGLISH text is missing.U (T , V )
1
Eq. 8.13 can be integrated if we know CV (T , V ) and p (T , V ), but
ıg
ıg
ıg
since CV = CP − NR and (∂(p /T )/∂(1/T ))V ,n = 0 it is possible to
ıg
replace CV (T , V ) with CP (T ) by going via an ideal gas reference state.
This makes it much easier to integrate the equation.
2
Note that the canonical differential of U in Eq. 8.1 requires knowledge of altogether different and more complex equations of state.
3
Incidentally, the differential in Eq. 8.13 is in the same form as the
differential in Eq. 8.11, from which it follows that
∂(p /T )
∂U = − ∂(1/T )
,
(8.14)
∂V T ,n
V ,n
∂U =
ˆ CV ,
(8.15)
∂T V ,n
where 8.15 represents an alternative definition of CV .
4
Our next task is to determine the total differential of H (T , V ). The
easiest way to do that is to start from H = U + pV , but on this occasion
we will derive dH from first principles in order to demonstrate how it is
done.
380 / 1776
1
With reference to Eq. 8.2 we will here express dp as a differential
with respect to T and V :
∂p ∂p (8.16)
(dp )n = ∂T V ,n dT + ∂V T ,n dV .
2
If we substitute the above equation, along with Eq. 8.12, into
Eq. 8.2, we get the following result:
i
i
h
h ∂p ∂p ∂p (dH )n = CV + V ∂T V ,n dT + V ∂V T ,n + T ∂T V ,n dV .
(8.17)
4
Eqs. 8.12, 8.13 and 8.17 summarise dS , dU and dH as differentials
expressed in T , V coordinates. All of them can be integrated, provided
ıg
that you know the functions CP (T ) and p (T , V ).
3
Relationships equivalent to the ones that we found in Eqs. 8.14–
8.15 can be derived for H , but they are of little practical value.
Expanded footnote [a]:
ıg
[a] Normally CP is included in reference works on thermodynamics — rather than
CVıg (T ).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
RV
8 State . . .
dU ıg
R∞
T◦
T◦
V◦
0
02 September 2013
379 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 055 Integration of U (T , V )
dU
∞
RT
Find the integral expressions for U , H and S by starting
out from the differentials that were derived in Theme 54.
Use ∆f H ıg (T◦ ) and S ıg (T◦ , p◦ ) as the constants of
integration.
375 / 1776
4
This eliminates the problem of U being a free variable, as it is in
ˆ CV /T does not bring us much
Eq. 8.3, but the definition (∂S /∂T )V ,n =
closer to our goal — we still have some work to do.
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
02 September 2013
§ 054 Differential of U (T , V ) (2)
T
§ 055
02 September 2013
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
The total differential that we are after is
∂U (dU )n = ∂∂U
T V ,n dT + ∂V T ,n dV .
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
α and β Ideal gas Waals Olds
§ 054 Differential of U (T , V )
1
374 / 1776
U (T , V )
02 September 2013
381 / 1776
α and β Ideal gas Waals Olds
8 State . . .
ENGLISH text is missing.Ideal gas U
1
A selection of useful differentials were derived in Theme 54 on
page 347.
2
Since U , H and S are state functions, we can largely choose to
integrate the differentials in whatever way suits our purpose.
3
In the case of internal energy, the most sensible way to integrate
is as shown in Figure 8.1 on the previous page: First you integrate dU
from T◦ to T assuming ideal gas behaviour, and then you integrate from
V = ∞ to V substituting in the relevant equation of state. It may appear
we are switching between an ideal gas and a real fluid as if by magic,
but remember that
ıg
lim U = U .
V →∞
4
In other words, all real fluids approach ideal behaviour at the limit
V → ∞ ⇔ p → 0.
5
This is precisely where the switch from ideal to real takes place, so
there is no discontinuity in the method.
383 / 1776
1
Assuming that there is an equation of state that can be expressed
in the form p = p (T , V ), and that the heat capacity is a known function
of temperature, it is possible to integrate dU (T , V ) in Eq. 8.13 to get
dU ıg = 0
U ıg (T◦ , V◦ )
V◦-1
U (T , V ) = U ıg (T◦ ) +
RT
T◦
ıg
CV dT −
RV
∞
∂(p /T )
dV
∂(1/T ) V ,n
.
(8.18)
Residual: U r,v
2
The internal energy of an ideal gas is independent of the volume
of the gas. The contribution of the integral of dU from V◦ to V = ∞ is
therefore zero in the equation above .
V -1
Expanded footnote [a]:
[a] This means that the internal energy of the fluid can be given without specifying its
pressure, if you assume ideal gas behaviour.
Figure 8.1: Calculation of the internal energy of a fluid at constant composition. The most common standard state for fluids is an ideal gas at T◦ =
ıg
298.15 K and V◦ = RT◦ /p◦ where p◦ = 1 bar, but because (dU )T ,n < f (V ) the
standard state pressure is not important in this context.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
382 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
384 / 1776
This is evident in Eq. 8.13, where if you substitute in p ıg /T =
ıg
ıg
ıg
NR /V , you get (∂(p /T )/∂(1/T ))V ,n = 0 and (dU )n = CV dT .
3
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
385 / 1776
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 055 Integration of U (T , V ) (2)
8 State . . .
4
The residual term does on the other hand explain the departure
in the internal energy of the real fluid from the internal energy of a
hypothetical ideal gas assuming the two fluids are compared at the
same temperature, (volume) and chemical composition.
5
§ 055 Integration of H (T , V )
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
8 State . . .
ENGLISH text is missing.Aggregate state
1
From the definition H =
ˆ U + pV , the enthalpy can be determined
using Eq. 8.18, but here it is worth commenting on the standard state.
2
Normally the standard state is chosen to ensure that the elements
have zero enthalpy in the stable aggregate state at 298.15 K and 1 bar
(ideal gas, liquid or crystal).
387 / 1776
The internal energy must therefore be expressed in terms of the
standard enthalpy, and not vice-versa:
1
U ıg (T◦ ) =
The residual concept is more thoroughly explained in Chapter 13.
H ıg (T◦ ) − (p◦ V◦ )ıg
=
ˆ ∆f H ıg (T◦ ) − NRT◦ .
(8.19)
2
The definition is trivial, but once the zero point has been chosen,
there are certain consequences if it is changed. One of the consequences is that
§ 055 Integration of S (T , V )
S (T , V ) = S ıg (T◦ , V◦ ) +
R∞
V◦
NR
V
dV +
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
386 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 055 Integration of S (T , V ) (2)
Termodynamic methods TKP4175
8 State . . .
All real gases approach ideal gas behaviour when p → 0, and
the volume integral in Eq. 8.22 is well-defined at this limit, since the
integrand tends to zero there.
3
7
We will therefore rewrite Eq. 8.22 in a slightly different form using
p◦ as the standard state: Let us first calculate the ratio
02 September 2013
388 / 1776
V →∞
V →∞
V ,n
389 / 1776
[a] Integrating dS in Eq. 8.12, assuming ideal gas behaviour, confirms that the expression diverges.
RT
ıg
CV
V
V◦
+
dT +
T
RT
T
T◦
dT +
∂p dV
∂T V ,n
RV h
∞
∂p ∂T V ,n
−
NR
V
i
dV .
(8.22)
RV
2
Note the term ∞ (NR /V ) dV , which we have explicitly added and
taken away from the right-hand side in order to avoid divergence at the
limit V → ∞.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
R
8 State . . .
Here, the residual term in the last equation explains the departure
in the entropy of the real fluid from the entropy of a hypothetical ideal
gas when the two fluids are compared at the same temperature, volume
and chemical composition.
RV
∞
ıg
CV
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman
CV
P and CV
V
α and β Ideal gas Waals Olds
§ 055 Integration of S (T , V ) (3)
5
Once again, switching between an ideal gas and a real fluid does
not imply discontinuity because
∂p ⇔
lim ∂T
= NR
lim S = S ıg
V .
∞
dS −
R
02 September 2013
390 / 1776
α and β Ideal gas Waals Olds
dS ıg
∞
S (T , V )
T
Tore Haug-Warberg (Chem Eng, NTNU)
Finally, dS −dS ıg is integrated between V = ∞ and V in accordance
with the selected equation of state.
4
T◦
= S ıg (T◦ , V◦ ) + NR ln
(8.20)
(8.21)
3
First dS is integrated between V◦ and V , and then it is integrated
between T◦ and T assuming ideal gas behaviour.
ıg
[a] With reference to Eq. 8.21, the molar heat of formation of a compound Aa Bb · · · Zz
ıg
ıg
ıg
ˆ uAaBb
in the ideal gas state is defined as being: ∆f uAaBb ···Zz (T◦ ) =
(T◦ ) − auA (T◦ ) −
···Zz
ıg
ˆ ∆f hAaBb
buBıg (T◦ ) − · · · − zuZıg (T◦ ) =
(T◦ ) − RT◦ + aRT◦ + bRT◦ + · · · + zRT◦ , where
···Zz
the coefficients a , b , . . . , z are the number of formula units of the elements A , B , . . . , Z
in the compound. For example, water has the chemical formula (H2 )1 (O2 )1/2 because
H and O naturally occur as diatomic molecules. Only if a + b + · · · + z = 1 does
∆f uıg = ∆f h ıg .
U ıg (T◦ ) , ∆f U ıg (T◦ )
8 State . . .
1
In principle, S is determined in the same way as U , but here we
have to bear in mind that dS diverges at the limit V → ∞[a].
2
In order to get around this problem, dS is integrated in three steps,
as shown in Figure 8.2 on page 359.
If we substitute (∂p /∂T )V ,n = NR /V into Eq. 8.12, the integral
of S is as shown below, but the limit V → ∞ remains undefined, and
requires closer analysis:
1
Expanded footnote [a]:
H ıg (T◦ ) =
ˆ ∆f H ıg (T◦ ) ,
α and β Ideal gas Waals Olds
ENGLISH text is missing.Ideell gass Ideal gas S
9
10
RT
dS ıg
dS ıg
cancels
V◦
T◦
The residual concept is more thoroughly explained in Chapter 13.
RV
V
V◦
8
The result is substituted into Eq. 8.22.
Vp◦
NRT◦
=
=
Vp◦
NRT
·
T
T◦
T◦
4
In thermodynamics, an integral that expresses the difference between a thermodynamic property of a real gas and an ideal gas in the
same state is often called a residual.
5
Chapter 13 covers this topic in greater detail.
6
Unlike in the case of U , the standard state of S is a function of
volume. However, the ideal gas entropy is generally tabulated at T◦ =
298.15 K and p◦ = 1 bar, and in spite of the fact that the volume is a free
variable in the function, the standard volume V◦ = NRT◦ /p◦ will have a
numerical value.
S ıg (T◦ , V◦ )
.
ıg
At the same time we may take advantage of the relationship CP =
ıg
CV
0
V -1
V -1
Figure 8.2: Calculating the entropy of ◦a fluid at constant composition. The
hatched lines illustrate the nominal steps involved in integration. The steps
used in practise differ from the nominal ones because the integral diverges
at the limit V → ∞. It is possible to switch between the two routes because
ıg
(dS )T ,n < f (T ) and (dS )V ,n < f (V ). The most common standard state for fluids
is an ideal gas at T◦ = 298.15 K and V◦ = RT◦ /p◦ where p◦ = 1 bar.
+ NR :
S (T , V ) = S ıg (T◦ , p◦ ) + NR ln
+
RT
T◦
Tore Haug-Warberg (Chem Eng, NTNU)
ıg
CP
dT +
T
RV h
∞
∂p ∂T V ,n
p◦ V NRT
−
NR
V
Residual: S r,v
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
i
dV
02 September 2013
(8.23)
391 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
02 September 2013
392 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
02 September 2013
393 / 1776
α and β Ideal gas Waals Olds
ENGLISH text is missing.T , p versus T , V
1
Temperature and pressure occupy a unique place amongst the
thermodynamic state variables.
The Bridgman table
2
This is because we generally live under constant temperature and
pressure conditions, and consequently a large number of experiments
have been performed at room temperature and atmospheric pressure.
3
It is therefore a useful exercise to derive the formulae discussed in
the previous section again, but this time using p as a free variable.
4
Without going into details, the total differentials of U (T , p ), H (T , p )
and S (T , p ) can be derived in an analogous way to the equivalent differentials of internal energy, enthalpy and entropy expressed in T , V coordinates:
h
i
+ T ∂∂VT p ,n dp ,
(dU )n = CP − p ∂∂VT p ,n dT − p ∂∂Vp
(8.24)
T ,n
V /T (8.25)
dp ,
(dH )n = CP dT + ∂∂1/
T
8 State . . .
p ,n
(dS )n =
ˆ
CP
T
dT −
∂V dp
∂T p ,n
(8.26)
.
5
It is entirely possible to integrate these expressions if we know the
ıg
heat capacity CP (T ) and the equation of state V (T , p ).
6
In that case, we get
U (T , p ) = H (T , p ) − pV ,
H (T , p ) = ∆f H ıg (T◦ ) +
ıg
RT
ıg
T◦
S (T , p ) = S (T◦ , p◦ ) − NR ln
CP dT +
p
p◦ +
RT
T◦
Rp
ıg
CP
T
0
(8.27)
∂ V /T dp
∂1/T p ,n
dT −
Rp h
0
(8.28)
,
∂V ∂T p ,n
−
NR
p
i
dp ,
(8.29)
but in practice only the ideal gas law, the 2nd virial equation expressed
as pV /NRT = 1 + Bp , and certain equations of state for solids are
explicit functions of V (T , p , n).
7
This means that Eqs. 8.28 and 8.29 are used less than the equivalent Eqs. 8.18 and 8.23.
394 / 1776
In an arbitrarily chosen system of constant composition, there are
two possible degrees of freedom.
1
The Bridgman table (2)
8 State . . .
5
In view of the requirement for thermodynamic consistency, there
must also be 9 − 3 − 2 = 4 independent differentials that tell us what we
need to know about the behaviour of the system.
6
It is possible to imagine several alternative sets of differentials.
7
Here we will select the set that describes the change in the system’s energy in the simplest manner.
8
This is a practical choice, in keeping with the philosophy that underlies the rest of this book:
For practical (and historic) reasons, these degrees of freedom are
expressed in terms of 9 experimentally determined properties that are
all related to physical chemistry: T , p , s , v , u, h , α, β and cp .
2
3
T (ds )n =
ˆ cp dT − T αv dp ,
The latter three quantities in this list are defined as
ˆ −v -1 ∂∂pv , cp =
ˆ T ∂∂Ts p ,
α=
ˆ v -1 ∂∂Tv p , β =
p (dv )n =
ˆ pv α dT − p βv dp .
T
4
(du)n = T ds − p dv ,
(dh )n = T ds + v dp ,
cf. Eq. 8.37 on page 377 and Eq. 8.26.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
395 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
396 / 1776
The Bridgman table (3)
8 State . . .
9
The differentials are linearised functions (here as elsewhere), and
the information that they contain can (here as elsewhere) be summarised in matrix form.


 du 
  
 


 −1
0
1 −1
0
0   dh   0 
 

  


 0 −1
1
0
0
1   T ds   0 

 

 =   .
(8.30)
 


 0
0 −1
0
1
−T α   p dv   0 

  
 


  

 0
0
0 −1 pv αcp-1 −p β   cp dT   0 




v dp 
10 Doing so will, in this case, make the next stages of our task significantly easier.
11 We will choose to express the coefficient matrix in dimensionless
form, with the right-hand side equal to 0 J mol-1 :
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
397 / 1776
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
The Bridgman table (4)
The Bridgman table (5)
8 State . . .
12 By performing a straightforward Gaussian elimination of the rows
in the coefficient matrix, taking the pivot elements from columns 1, 2, 3
and 6, it is possible to derive the following set of coefficients:
−1
0
0 −1
0
0
0
0
(cp β − Tv α2 )/(cp β)
T α/(p β)
−1/(p β)
v α/(cp β)
(∂p /∂T )v ,n = αβ-1 ,
v (∂p /∂v )T ,n = −β-1 ,
(dh )n = (cp β − Tv α2 + v α)β-1 dT − (1 − T α)β-1 dv ,
cv = cp − Tv α2 β-1 ,
(ds )n = (cp β − Tv α2 )T -1 β-1 dT + αβ-1 dv ,
then we see that the result of the matrix operation is identical to the
calculations performed in Eqs. 8.13, 8.17, 8.12 and 8.16.
(dp )n = αβ-1 dT − v -1 β-1 dv .
0
8 State . . .
If we now bear in mind that
14
(du)n = (cp β − Tv α2 )β-1 dT − (p β − T α)β-1 dv ,
0
α and β Ideal gas Waals Olds
The Bridgman table (6)
8 State . . .
13 Back substitution assuming p dv and cp dT to be the free variables
gives us:
0
(cp β − Tv α2 + v α)/(cp β)
(−1 + T α)/(p β)
0 −1
0
(cp β − Tv α2 )/(cp β)
0 (−p β + T α)/(p β)
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
16 By systematically selecting the relevant pivot elements, we can now
calculate all of the 6 × 5 × 4 = 120 differentials that describe the system.
−1
15 This means that we have uncovered a general method for calculating an arbitrarily chosen differential expressed in the six primary variables: T , p , s , v , u og h .
17 A Ruby program that calculates all of these differentials is described in Appendix 31:3.1
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
398 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
Table 8.1: A
6 × 5 × 4 = 12
phase at cons
substance (or
(δT )p = −(δ p )T = 1
02 September 2013
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(δ v )T
=−
(δ s )T
=−
(δ u )T
=−
(δ h )T
=−
=−
(δ s )v
(δ s )p
=−
=−
(δ u )v
(δ u )p
=−
(δ h )p
=−
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
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d h ( s , p ) = T ds + v dp
dh (v , T ) =
T α−1
β
dh (v , p ) =
cp
vα
dv +
cp β−Tv α2 +v α
β
cp β−Tv α2 +v α
α
dv +
dT
dp
d h ( T , p ) = cp dT + (1 − T α)v dp
dp (u , h ) =
cp
(p βcp −pTv α2 −cp +pv α)v
dp (u , s ) =
cp
p (cp β−Tv α2 )v
dp (u , v ) =
α
cp β−Tv α2
du −
dp (u , T ) =
1
(p β−T α)v
du −
dp (h , s ) =
1
v
dp (h , v ) =
α
cp β−Tv α2 +v α
dp (h , T ) =
dh −
1
(1−T α)v
T
v
du −
cp −pv α
(p βcp −pTv α2 −cp +pv α)v
(cp −pv α)T
p (cp β−Tv α2 )v
cp −pv α
(cp β−Tv α2 )v
cp −pv α
(p β−T α)v
dh
dh −
ds
dv
dT
cp
(cp β−Tv α2 +v α)v
cp
(T α−1)v
dv
du +
(δ h )v
cp β−Tv α2 +v α
cp β−Tv α2
=−
dh (u , T ) =
1−T α
p β−T α
du +
p βcp −pTv α2 −cp +pv α
p β−T α
(δ u )s
=−
dh (u , p ) =
(δ h )s
=−
cp
cp −pv α
du −
(p βcp −pTv α2 −cp +pv α)v
cp −pv α
dh ( s , v ) =
(cp β−Tv α2 +v α)T
cp β−Tv α2
(δ h )u
=−
dh ( s , T ) =
T α−1
α
02 September 2013
404 / 1776
cp
(cp β−Tv α2 )v
dp (s , v ) =
Tα
cp β−Tv α2
dp (s , T ) =
−1
vα
ds +
cp
Tv α
dp (v , T ) =
−1
vβ
dv +
α
β
ds (u, h ) =
−cp
(p βcp −pTv α2 −cp +pv α)T
1
T
ds −
p
T
du +
ds (u, T ) =
−α
p β−T α
ds (u, p ) =
cp
(cp −pv α)T
p (cp β−Tv α )
(p β−T α)T
d s (h , p ) =
1
T
dh +
dh +
cp
(1−T α)T
dp
cp
(cp β−Tv α2 +v α)T
dv
dT
dp
cp β−Tv α
βT
Tore Haug-Warberg (Chem Eng, NTNU)
dh
dT
p (cp β−Tv α2 )v
(cp −pv α)T
du −
d s (h , T ) =
dv +
du +
2
α
T α−1
α
β
dv
p (cp β−Tv α2 )
(p βcp −pTv α2 −cp +pv α)T
2
dT
Termodynamic methods TKP4175
02 September 2013
du +
cp
Tα
ds +
Tore Haug-Warberg (Chem Eng, NTNU)
p βcp −pTv α2 −cp +pv α
cp β−Tv α2
ds −
cp
cp β−Tv α2
dp
dv
dT
Termodynamic methods TKP4175
405 / 1776
cp β−Tv α2
Tα
ds (v , p ) =
cp
Tv α
d s (T , p ) =
cp
T
dT ( u , h ) =
T α−1
p βcp −pTv α2 −cp +pv α
dv +
dT ( u , s ) =
Tα
p (cp β−Tv α2 )
dT ( u , v ) =
β
cp β−Tv α2
dT ( u , p ) =
1
cp −pv α
du +
du +
du −
dT ( h , v ) =
β
cp β−Tv α2 +v α
dT ( h , p ) =
1
cp
dT ( s , v ) =
βT
cp β−Tv α2
ds −
dT ( s , p ) =
T
cp
Tv α
cp
ds +
Tore Haug-Warberg (Chem Eng, NTNU)
dh
ds
dv
dp
ds
dh +
(T α−1)v
cp
403 / 1776
α and β Ideal gas Waals Olds
p β−T α
p βcp −pTv α2 −cp +pv α
(p β−T α)T
p (cp β−Tv α2 )
p β−T α
cp β−Tv α2
(1−T α)T
cp
Tα
cp
dh +
du +
(p β−T α)v
cp −pv α
dT ( h , s ) =
02 September 2013
dp
dT − v α dp
dh +
ds
dv
dT
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
dv
cp β−Tv α2
(cp β−Tv α2 +v α)T
v
T
402 / 1776
dT
du +
dh −
02 September 2013
dT
d s (h , v ) =
ds (v , T ) =
dT
Termodynamic methods TKP4175
dh (u , v ) =
ds (u, v ) =
ds
dh +
Tore Haug-Warberg (Chem Eng, NTNU)
du −
=−
Termodynamic methods TKP4175
400 / 1776
α and β Ideal gas Waals Olds
(p βcp −pTv α2 −cp +pv α)T
p (cp β−Tv α2 )
cp
p (cp β−Tv α2 )
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
Table 8.2: A complete listing of all of the 6 × 5 × 4/2 = 60 total differentials
of u, h , s , v , T and p in a homogeneous phase at constant composition, i.e.
either one kilogram or one mol of the substance (or alternatively the sum of
all substances in the mixture). The differentials are calculated by performing
Gaussian elimination on the system of Eqs. 8.30 on page 362 followed by
R
algebraic reduction and simplification in Maple.
dh ( u , s ) =
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
(δ v )p
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
1−T α
cp β−Tv α2 +v α
dv
dp
Tα
cp β−Tv α2
dv
dp
Termodynamic methods TKP4175
02 September 2013
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News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
β
α
dT ( v , p ) =
1
vα
du (h , s ) =
p (cp β−Tv α2 )
cp
du (h , v ) =
cp β−Tv α2
cp β−Tv α2 +v α
du (h , T ) =
p β−T α
1−T α
dh +
p βcp −pTv α2 −cp +pv α
T α−1
du (h , p ) =
cp −pv α
cp
dh +
(p βcp −pTv α2 −cp +pv α)v
cp
dv +
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
dp
dh −
(p βcp −pTv α2 −cp +pv α)T
cp
dh −
d u ( s , v ) = T ds − p dv
du (s , T ) =
−(p β−T α)
α
du (s , p ) =
(cp −pv α)T
cp
ds +
du (v , T ) =
−(p β−T α)
β
dv +
du (v , p ) =
cp −pv α
vα
ds +
dv +
p βcp −pTv α2 −cp +pv α
cp β−Tv α2 +v α
p (cp β−Tv α2 )
Tα
cp β−Tv α2
α
dv
dT
dp
dT
p (cp β−Tv α2 )v
cp
cp β−Tv α2
β
ds
dp
dT
dp
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
dv (u, s ) =
−1
p
dv (u, T ) =
−β
p β−T α
dv (u, p ) =
vα
cp −pv α
407 / 1776
cp β−Tv α2
p β−T α
du −
(cp β−Tv α2 )v
cp −pv α
dT
dh +
dv (h , T ) =
β
T α−1
cp β−Tv α2 +v α
1−T α
dT
dv (h , p ) =
vα
cp
(cp β−Tv α2 +v α)v
cp
dp
dv (s , T ) =
β
α
dv (s , p ) =
Tv α
cp
dh +
ds −
cp β−Tv α2
Tα
ds −
A total differential contains the same information whether it is expressed in T , V or T , p coordinates. Based on Eqs. 8.12 and 8.26, for
(dS )n we can write
CV
CP
∂p ∂V (8.33)
T dT + ∂T V ,n dV = T dT − ∂T p ,n dp .
1
2
If we assume constant volume we get: (dp )V ,n = (∂p /∂T )V dT .
3
Similarly, at constant pressure: (dV )p ,n = (∂V /∂T )p dT .
4
In both cases, the elimination of dT from the equation above gives:
∂p (8.34)
CP − CV = T ∂T V ,n ∂∂VT p ,n .
Termodynamic methods TKP4175
02 September 2013
411 / 1776
(cp β−Tv α2 +v α)T
cp
ds
dT
(cp β−Tv α2 )v
cp
Tore Haug-Warberg (Chem Eng, NTNU)
dp
Termodynamic methods TKP4175
02 September 2013
408 / 1776
5
We can now choose to get rid of either p or V as a dependent
variable. If we substitute dp = 0 into the total differential of p , we can
show that
∂p ∂p -1
∂V = − ∂T V ,n ∂V T ,n .
(8.35)
∂T p ,n
Expanded footnote [a]:
[a] The total differential of p at constant composition is: (dp )n = (∂p /∂T )V ,n dT +
(∂p /∂V )T ,n dV . Assume constant pressure i.e. dp = 0 and divide by dT . Then isolate
the derivative: (dV / dT )p,n = (∂V /∂T )p ,n = . . .
6
Similarly, by substituting dV = 0 into the total differential of V , we
can show that
-1
∂p .
(8.36)
= − ∂∂VT p ,n ∂∂Vp
∂T V ,n
(8.41)
.
8 State . . .
9
A third option, which is widely used in solid and fluid mechanics, is
to assume adiabatic conditions and to look at how energy varies with
temperature along an isentrope.
10 The relationship between these heat capacities can be derived
from the volumetric properties of the fluid.
11 In the next section we will look at the relationship between the first
two of the abovementioned heat capacities.
409 / 1776
Termodynamic methods TKP4175
Termodynamic methods TKP4175
02 September 2013
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.
02 September 2013
(8.31)
(8.32)
410 / 1776
α and β Ideal gas Waals Olds
8 State . . .
Other historically important properties include isobaric expansivity
1
and isothermal compressibility:
ˆ −v -1 ∂∂pv .
(8.37)
α=
ˆ v -1 ∂∂Tv p , β =
2
Heat capacity at constant pressure was one of the first thermodynamic properties to be investigated systematically.
T
3
For some applications it is useful to use these quantities as they
give a more compact set of formulae.
In the case of molar volume, for instance, it gives us
(dv )n = v α dT − v β dp ,
(8.38)
while for molar entropy we get
(ds )n =
Substitution of the last two equations into Eq. 8.34 produces the
two expressions that we set out to find .
7
(ds )n =
cp
T
cv
T
dT − αv dp ,
dT +
α
β
(8.39)
dv .
(8.40)
Expanded footnote [a]:
[a] With a pressure-explicit equation of state, it may be sensible to work in reduced
variables. For your own benefit, show that (cp − cv )/R = zc Tr (∂pr /∂Tr )2vr ,x (∂pr /∂vr )-1
Tr ,x
for one mole of fluid.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
§ 057
§ 56
§ 58
02 September 2013
412 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
8 State . . .
ENGLISH text is missing.Applications
1
Ideal gases are an important concept in thermodynamics, and it is
reassuring to know that this simple model gives a good description of
many real systems, e.g. in process engineering and meteorology.
2
This section is therefore one of the most important ones in the
whole book, at least in terms of its practical applications.
415 / 1776
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α and β Ideal gas Waals Olds
§ 057 Ideal gas I
8 State . . .
If we combine the ideal gas law p ıg = NRT /V with the definitions
in Eq. 8.37, it follows that
1
αıg =
NR
PV
= T -1
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
and
βıg =
NRT
P2V
ıg
ıg
cp − cv
= p -1 ,
which gives us the molar relationship
= pv /T = R when
substituted into Eq. 8.41. For an arbitrary system size, we can say that
What are the expansivity α and compressibility β of an
ideal gas mixture, expressed in terms of the variables
ıg
ıg
T and p ? And what is CP − CV for the mixture?
ıg
ıg
CP − CV = NR ,
also known as Mayer’s relation.
Tore Haug-Warberg (Chem Eng, NTNU)
T ,n
∂p -1
∂p 2
∂T V ,n ∂V T ,n
Definition of α and β
T ,n
4
These expressions are obtained by substituting 8.37 into the total
differentials of volume and entropy; see Theme 56 on page 374 for the
necessary details.
5
Incidentally, Eqs. 8.31–8.32 in Theme 56 are expressed in terms of
extensive variables.
Tv α2
β
§ 57
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
In the literature on thermodynamics, intensive variables are gener6
ally used, which makes the size of the system irrelevant to the problem,
and one widely used alternative to Eq. 8.31 is the molar form
cp − cv =
§ 55
8
you measure how energy varies with temperature at either constant
pressure, or at constant volume.
Tore Haug-Warberg (Chem Eng, NTNU)
α and β Ideal gas Waals Olds
8 State . . .
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
Heat capacity is a measure of the thermal inertia of the system.
3
For instance, if you were to measure the heat capacity of water
at 1 atm without being aware of the vapour–liquid transition, you would
conclude that cp → ∞ at 100 ◦ C.
5
Analogously, substances with a low heat capacity, e.g. carbon, for
which cp ≈ 3.5 J mol-1 K-1 in the temperature range T ∈ h0, 300i K, require only a small amount of energy for a large temperature change.
7
For practical reasons it is normal to define the heat capacity of a
closed system in two different ways:
ii) CP − CV = −T
§ 056 Heat capacity (2)
α and β Ideal gas Waals Olds
Definition of α and β (2)
1
2
If this inertia is very high, a lot of energy is needed to raise the
temperature by one degree.
4
But the integral of cp dT over a small temperature range dT ∈
h100 − ǫ, 100 + ǫi will be finite, and equal to what is defined as the
heat of vaporization.
6
Hence heating one gramme of carbon from 0 → 300 K will require
about the same amount of energy as is needed to vaporize just 0.04 g
of water.
Entropy and heat capacity are closely linked, and by
comparing the differentials dS (T , V )n and dS (T , p )n
from Eqs. 8.12 on page 348 and 8.26 on page 360,
it is possible to obtain two alternative relationships
between CP and CV . Show that these relationships
can be expressed as:
2
-1
i) CP − CV = −T ∂∂VT p ,n ∂∂Vp
,
dp
−(cp β−Tv α2 )
cp
dh −
§ 056
dh
α and β Ideal gas Waals Olds
ENGLISH text is missing.Calorimetry
ds
du +
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
T
p
du +
cp β−Tv α2
p βcp −pTv α2 −cp +pv α
du +
dv (h , s ) =
α and β Ideal gas Waals Olds
§ 056 Heat capacity
Tore Haug-Warberg (Chem Eng, NTNU)
−(cp β−Tv α2 +v α)
p βcp −pTv α2 −cp +pv α
d v ( T , p ) = v α dT − v β d p
d u ( T , p ) = (cp − pv α) dT + (p β − T α)v dp
Tore Haug-Warberg (Chem Eng, NTNU)
dv (u, h ) =
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
416 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
2
Consistency: Use the Eqs. 8.31 and 8.32 to verify this result.
Termodynamic methods TKP4175
02 September 2013
417 / 1776
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
250
20
200
S [J/K mol]
H [kJ]
Entropy
30
10
0
150
100
500
0
1000
0
500
T [K]
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
424 / 1776
6
ıg
5
Here the heat capacity ratio is defined as being γ =
ˆ cp /cv .
line.
s ,n
γ−1
=
ıg
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
427 / 1776
dT + NR ln
ıg
T
p◦ V NRT◦
dT − NR ln
NRT p◦ V
dT − NR ln
p
p◦
ıg
T
,
8 State . . .
Then we differentiate the equation of state with respect to T , which
gives us
2
∂(p /T )VdW
= − aN
.
∂(1/T )
V2
V ,n
If we substitute this into the expression for U (T , V ) in Theme 55 on
page 351 (Eq. 8.18), we get
4
CV dT +
RV
∞
aN 2
V2
Termodynamic methods TKP4175
=
=
=
γ−1
p2 γ
p1 S ,n
V2 -γ
V1 S ,n
V2 1−γ
V1 S ,n
§ 061 Isentropic relations
8 State . . .
2
For a isentropic process we also know that ds = 0, so we can set
up the following integral:
(8.46)
,
(8.47)
.
(8.48)
02 September 2013
α and β Ideal gas Waals Olds
1
Use the differential in Eq. 8.39 as a starting point and then assume
ideal gas behaviour for α.
,
Termodynamic methods TKP4175
V ,n
p2 ıg
p1 T ,n
T2 ıg
T1 p ,n
425 / 1776
T
R2
T1
ıg
cp (dln T )s ,n =
p
R2
p1
αıg v dp = R
Tore Haug-Warberg (Chem Eng, NTNU)
p1
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
8 State . . .
p
R2
dln p .
3
Integrate between states 1 and 2, assuming that the heat capacity
is constant over this range.
§ 062
§ 61
§ 63
02 September 2013
426 / 1776
α and β Ideal gas Waals Olds
8 State . . .
ENGLISH text is missing.Van der Waals p (T , V )
The Van der Waals equation represents an important milestone in
our understanding of fluids.
1
2
By itself, the equation is too inaccurate to be of any practical use,
but it has laid the foundations for a whole series of cubic[a] equations
of state that do have practical applications.
3
A number of these equations (RK, SRK, PR) are very widely used
in relation to liquid–gas separation and gas treatment equipment.
429 / 1776
[a] Known as cubic because v (p ) can be expressed as the solution to a cubic equation.
V ,n
=
=
V2 -1
V1 T ,n
Substitute the Van der Waals equation of state (7.30)
into Theme 55 on page 351. Go on to show that the
internal energy can be expressed
,
V2 V1 p ,n
U VdW (T , V ) = U ıg (T◦ ) + ∆U ıg (T ) −
These are evidently related to 8.46–8.48, but there is no value of γ
capable of converting that system of equations into this one.
9
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
428 / 1776
Reference
Tore Haug-Warberg (Chem Eng, NTNU)
8 State . . .
RT ıg
= ∆f H ıg (T◦ ) − NRT◦ + CV dT −
|
{z
} T◦
aN 2
V
.
CV integral
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 062 Van der Waals U (2)
§ 063
§ 62
aN 2
.
V
Residual
02 September 2013
430 / 1776
α and β Ideal gas Waals Olds
§ 64
8 State . . .
(8.49)
ıg
U (T◦ )
[a] Mixing up total and molar values is a common mistake in thermodynamics!
Tore Haug-Warberg (Chem Eng, NTNU)
T
CP
CP
ıg
4
7
The other relations referred to in the question can be derived by
analogous methods starting from a combination of Eqs. 8.39 and 8.40.
2
The form of the equation of state is important, because subsequently it is integrated with respect to the total (rather than molar[a])
volume V .
T◦
ıg
CV
RT
RT
§ 60 The entropy function
1
The expression for entropy becomes ∆S (T , p◦ ) = cp ln(T /T◦ ),
where T◦ = 298.15 K. The predictive accuracy of the function can be
seen on the right-hand side of Figure 8.3 on the preceding page.
Argon is a monatomic gas without any rotational and vibrational
2
Unsurprisingly, the distance between the calculated values and the
energy, nitrogen has one vibrational and two rotational degrees of freeıg
experimental ones increases as T moves away from T◦ .
1
Use cp (298.15 K) to obtain an approximation for the enthalpy dom, while methane has nine vibrational and three rotational degrees
H ıg (T ) − H ıg (298.15 K) of argon, nitrogen and methane in the tempera- of freedom.
3
An explanation of the behaviour at high temperatures follows from
§ 60
ture range 0–1000 K. Make a graphical illustration of the functions and
Theme 59, but what happens at low temperatures?
ıg
5
At 298.15 K it is essentially only the rotational energy that is rele- 1
Use cp (298.15 K) to obtain an approximation for the entropy
(8.44) compare with the values from JANAF[a]. Comment on the discrepanvant, whereas the vibrational energy only comes into play at higher tem- S ıg (T , p ) − S ıg (298.15 K, p ) of argon, nitrogen and methane in the 4 You should particularly note that S (0, p◦ ) < 0 (see inset figure).
cies.
◦
◦
peratures.
temperature range 0–1000 K. Make a graphical illustration of the func- 5 Negative entropy is theoretically possible and errors of prediction of
420 / 1776
(8.45)
6
This is why the enthalpy curves turn upwards, and why the devia- tions and compare the curves with the values from JANAF[0]. Com- this kind can have major consequences in conjunction with simulating
tion increases with the size of the molecule.
ment on the discrepancies.
real processes. Program 1.3 in Appendix 31 gives all of the details.
421 / 1776
423 / 1776
419 / 1776 [a] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,
422 / 1776
§ 59
For an isothermal, isobaric or isochoric change of state, the ideal
8
gas law immediately tells us that:
p T2 ıg
= p21
,
T1
Let us first state the Van der Waals equation in the form p (T , V , n);
cf. Eq. 7.36.
ıg
RT
T◦
T◦
T◦
2
The discrepancy between the calculated and measured values is
barely noticeable for argon, slightly bigger for nitrogen and biggest for
methane (see Program 1.3 in Appendix 31).
3
This difference between the elements is due to the internal degrees
of freedom of the molecules.
(8.43)
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
RT
(8.42)
ıg
§ 061 Isentropic relations (3)
1
U VdW (T , V ) = U ıg (T◦ ) +
S ıg (T , p ) = S ıg (T◦ , p◦ ) +
ıg
CV dT + NRT
What will the exponent in Eq. 8.46 be for an isochoric
process? And what is it for an isothermal process in
Eq. 8.47? And for an isobaric process in Eq. 8.48?
α and β Ideal gas Waals Olds
§ 062 Van der Waals U
3
= S ıg (T◦ , p◦ ) +
1
Rewrite the equations 8.18–8.23 and 8.27–8.29 assuming ideal
gas behaviour.
418 / 1776
1
The expression for enthalpy becomes ∆H (T ) = cp (T − T◦ ), where
T◦ = 298.15 K. The predictive accuracy of the function can be seen on
the left-hand side of Figure 8.3 on the previous page.
ıg
CV dT
CP dT ,
T2 ıg
T1 S ,n
.
Note that R = cp − cv , which has been used to obtain the final
Tore Haug-Warberg (Chem Eng, NTNU)
S ıg (T , V ) = S ıg (T◦ , p◦ ) +
§ 58
§ 59 The enthalpy function
ıg
RT
RT
T◦
ıg
CP dT ,
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
p
p2 γ
p1 s ,n
RT
T◦
T◦
RT
T◦
=
ˆ ∆f H ıg (T◦ ) − NRT◦ +
p2 ıg
p1 S ,n
α and β Ideal gas Waals Olds
4
This assumption holds perfectly for monatomic gases, whereas for
other gases it is necessary to use a mean value:
ıg
p ln TT21
= cRıg ln p21
,
s ,n
= ∆f H ıg (T◦ ) − NRT +
= ∆f H ıg (T◦ ) +
Suppl., 14(1), 1985.
1000
§ 061 Isentropic relations (2)
T2 ıg
T1 s ,n
§ 62
U ıg (T ) = ∆f H ıg (T◦ ) − NRT◦ +
H ıg (T ) = U ıg (T , V ) + (pV )ıg
T2 ıg
T1 S ,n
Figure 8.3: Approximated (linearised) ideal gas enthalpy and entropy calculated for Ar, N2 and CH4 , compared to precise values taken from the JANAF
tables.
Tore Haug-Warberg (Chem Eng, NTNU)
§ 60
0
−10
T [K]
§ 061
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 58 Ideal gas II
1
In general the results speak for themselves, but is worth noting
that U ıg and H ıg are only functions of T (and not of p or V ):
Show that the following relations are true for an ideal
gas mixture if the heat capacity is constant over the
interval of integration:
50
0
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
Enthalpy
The physical model for the internal energy of a Van der Waals fluid
involves an arbitrarily chosen reference state, a non-linear temperature
term and a composition term that is proportional to the system’s molar
density.
5
Find an expression for cvVdW based on Theme 62.
Determine the function of cpVdW − cvVdW and draw it
on a suitable diagram. What happens as the state
approaches the mixture’s critical point? What is the
upper temperature limit and what happens at extremely
high pressure?
dV
02 September 2013
431 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
432 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
433 / 1776
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
§ 063 Van der Waals CP
8 State . . .
provided that a is a constant parameter in the equation. This shows,
quite unexpectedly, that heat capacity at constant volume is the same
as for an ideal gas.
Now substitute the differentials into Eq. 8.32 on page 374 and express the result in terms of molar variables:
2
−T v R
−b
VdW
= 2a
(cp − cv )
.
− RT 2
v3
4
§ 063 Van der Waals CP (2)
2
Experimentally this is not true.
3
In order to determine the heat capacity at constant pressure we
need to derive the Van der Waals Eq. 7.36 again:
∂p VdW = VNR
(8.50)
−Nb ,
∂T V ,n
2
∂p VdW = −NRT 2 + 2VaN3 .
(8.51)
∂V T ,n
(V −Nb )
(cp −cv )VdW
R
1
=
1−
(3vr −1)2
4Tr vr3
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
§ 064
§ 63
§ 65
02 September 2013
434 / 1776
3
Funnily enough, the same limit also applies at extremely high pressures i.e. vr → 1/3.
4
In spite of the fact that at constant volume the heat capacity varies
in the same way as that of an ideal gas, at constant pressure it deviates
greatly.
5
It is particularly worth noting that cpVdW → ∞ at the critical point,
which is in line with experimental values.
437 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
435 / 1776
0
Tore Haug-Warberg (Chem Eng, NTNU)
10
vr
§ 064 Joule–Thomson coefficient
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
8 State . . .
1
02 September 2013
α and β Ideal gas Waals Olds
§ 064 Joule–Thomson coefficient (2)
8
5
and substitute this result into the expression for κ:
κVdW =
−b +
(v −b )2 2a
RT v 2
(v −b )2 2a
RT v 3
R + cv − cv
.
6
In Theme 63 we showed that the heat capacity cv of a
Van der Waals fluid is the same as that of an ideal gas.
7
In addition we need the relationship between the parameters a , b
and Tc , pc as discussed in Theme 47 on page 322, and the relationship
ıg
ıg
between cp and cv from Theme 56 on page 374.
436 / 1776
8 State . . .
In summary, we can state that:
2
This exercise would become incredibly unwieldly if we were to derive everything from first principles.
3
For the sake of clarity we will therefore employ a schematic solution with references to other chapters (this is not generally an advisable
approach, but every rule is made to be broken).
4
Next we take the partial derivatives of the Van der Waals equation
from Theme 63 on page 389
∂p VdW
= VNR
−Nb ,
∂T V ,n
VdW
2
∂p
−NRT
,
= (V −Nb )2 + 2aN
∂V T ,n
V3
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
Inversion temperature
Termodynamic methods TKP4175
Our starting point is a relevant expression for the Joule–Thomson
coefficient which we derived in Theme 93 on page 514:
∂p ∂p T ∂T V ,n + V ∂V T ,n
κ=
ˆ ∂∂Tp
=
.
H ,n
∂p 2
∂p T ∂T V ,n − CV ∂V T ,n
438 / 1776
0
Figure 8.4: Dimensionless heat capacities calculated from the Van der Waals
equation of state. The numbers in the figure refer to Tr = {1.0, 1.2, 1.5, 2.0,
5.0}.
1
The inversion temperature of a fluid is defined as the
temperature where (∂T /∂p )H ,n = 0. What are the
practical implications of this? Determine the function
for the Van der Waals equation of state and draw the
graph in a pr , Tr diagram.
1.5
2
5
10
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
8 State . . .
1
A few selected isotherms have been drawn on Figure 8.4 on the
preceding page using the Program 31:1.6.
2
Note that cpVdW − cvVdW → R due to ideal gas behaviour at high
temperatures Tr >> 1 and low pressures vr >> 1.
6
2
6
According to the Van der Waals theory, the volume is defined in the
range hb , ∞i.
7
This means that vr ∈ h1/3, ∞i, and moreover that cpVdW − cvVdW → ∞
in Eq. 8.52 if both vr → 1 and Tr → 1.
α and β Ideal gas Waals Olds
ENGLISH text is missing.Van der Waals cp − cv
1
1.2
4
At temperatures Tr < 1, the gas and liquid states can exist in equilibrium, but our method is not designed to deal with that situation. We
will therefore stick to supercritical temperatures Tr > 1.
8
(v −b )
Tore Haug-Warberg (Chem Eng, NTNU)
8
(8.52)
.
α and β Ideal gas Waals Olds
10
8 State . . .
5
If we substitute in a = 9RTc vc /8 and b = vc /3, the equation can
be rewritten using reduced variables:
Differentiating U VdW with respect to T at constant V gives us
VdW
ıg
CVVdW =
ˆ ∂∂U
T V ,n = CV ,
1
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
cp −cv
R
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
02 September 2013
439 / 1776
ıg
cp
ıg
cp
9
=
=
ıg
cv +
ıg
γcv
R
=
3
8
b
=
1
3
a
= 3pc vc2 =
vc
9
8
RTc vc
Substituted into the expression for κVdW this gives us
κVdW R
vc
Tore Haug-Warberg (Chem Eng, NTNU)
=
− 13 +
γ
γ−1
−
(3vr −1)2
4Tr vr2
1 (3vr −1)2
γ−1 4Tr vr3
Termodynamic methods TKP4175
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
02 September 2013
440 / 1776
α and β Ideal gas Waals Olds
Looking back
7
8 State . . .
vc pc
RTc
ıg
cvVdW = cv
8 State . . .
6
1
Assuming that the denominator is finite throughout the interval of
solution, the criterion for inversion will be fulfilled when the numerator is
zero. If we solve with respect to vr we get
(vrVdW )κ=0 =
6
4
As a final trial the answer needs to be converted into reduced coordinates.
5
We know the Van der Waals equation in reduced coordinates from
Theme 49 on page 328
prVdW =
8Tr
3vr −1
−
3
vr2
,
and it is easy enough to substitute in (vrVdW )κ=0 from the equation
above.
3−
√1 4
3 Tr
Tr
4
Miller
3
.
1
Van der Waals
5
2
The inversion state is defined at κVdW = 0.
3
In this state the enthalpy does not change with pressure, and hence
the temperature remains unchanged even if there is a slight fall in pressure.
2
That gives us our final result:
1
p
(prVdW )κ=0 = 24 3Tr − 12Tr − 27 .
0
This expression can usefully be compared with the empirical correlation (pr )κ=0 = 24.21 − 18.54Tr-1 − 0.825Tr2 , which gives a relatively accurate approximation of the experimental values for N2 , CO2 and some
simple hydrocarbons .
0
5
pr
10
7
8
In spite of the fact that the two equations are entirely different in
form, the graphs that they produce are qualitatively similar; cf. Figure 8.5[a].
Expanded footnote [a]:
[a] Donald G. Miller. Ind. Eng. Chem. Fundam., 9(4):585–589, 1970.
Tore Haug-Warberg (Chem Eng, NTNU)
[a] Theoretical expressions and empirical correlations are in some ways mathematical
half-brothers — they have the same origins and areas of application, but in terms of
their content and form they are completely different from each other.
Termodynamic methods TKP4175
02 September 2013
441 / 1776
Figure 8.5: Inversion temperature calculated from the Van der Waals equation
of state compared with an empirical correlation obtained by D. G. Miller (1970).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
442 / 1776
Canonical: U (S , V , N ), H (S , −p , N ), A (T , V , N ), etc.
2
Non-canonical: U (T , p , N ) + S (T , p , N ) + V (T , p , N ), etc.
3
The Bridgman table:

 −1
0
1 −1
0
0


 0 −1
1
0
0
1



 0
0 −1
0
1
−T α


 0
0
0 −1 pv αcp-1 −p β
Tore Haug-Warberg (Chem Eng, NTNU)


 
 
 
 
 
 
 
 
 
 
 
 



Termodynamic methods TKP4175
du
dh
T ds
p dv
cp dT
v dp


 

 
 
 
 
 
 = 
 
 
 
 
 




0 


0 


0 


0 
02 September 2013
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News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
Looking back (2)
4
5
Constant volume heat capacity:
ˆ
(dS )n = ∂∂ST V ,n dT + ∂∂VS T ,n dV =
Constant pressure heat capacity:
dV =
ˆ
(dS )n = ∂∂ST p ,n dT + ∂∂Sp
Isobaric expansivity: α =
ˆ V -1
∂V
∂T
dT +
CP
T
dT −
!
p ,n
∂V
∂p
∂p dV
∂T V ,n
lim
Tr ,pr →1,1
CPVdW = ∞
Part
∂V dp
∂T p ,n
9
Closed control volumes
!
7
Isothermal compressibility: β =
ˆ −V -1
8
Heat capacity relation: CP − CV = NTv α2 β-1
9
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
8 State . . .
Critical point divergence:
10
CV
T
α and β Ideal gas Waals Olds
Looking back (3)
8 State . . .
T ,n
6
News f (ξ) vs f (x ) f (T , V ) f (T , p ) Bridgman CP and CV
α and β Ideal gas Waals Olds
T ,n
see also Part-Contents
ıg
Van der Waals heat capacity: CVVdW = CV
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
444 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
New concepts
Closed control volume
2
The extended 1st law of thermodynamics
3
Compression work of ideal gas
02 September 2013
445 / 1776
Closed system
Continuity of mass, energy and momentum
5
Thermodynamic speed of sound
Termodynamic methods TKP4175
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
A closed control volume is synonymous with a closed system, and
hence its energy balance reflects the 1st law of thermodynamics, and
can be expressed as dE = δQ − δW .
1
The inexact differential δQ and − δW represent the transfer of heat
4
and work from the surroundings to the control volume. The symbol δ
highlights the fact that Q and W are not state functions.
There is no transport term in this equation, but that does not mean
that the control volume must remain stationary.
3
For example, later in this chapter we will study the changes of state
of a centre of mass moving along a streamline.
2
4
Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
1
Termodynamic methods TKP4175
Closed system (2)
9 Closed . . .
10 At thermodynamic equilibrium, for each independent Ni there will
be a chemical potential µi that is constant both in time and space.
14 In order to avoid insuperable mathematical complications, we will
here limit ourselves to quasi-static systems, which can be described
using ordinary differential equations.
11 Hence, if suitable values are selected for the thermodynamic variables, all dNi = 0, but if there are any irreversible reactions, diffusion
processes or phase transitions in the control volume, this no longer applies.
12 In such cases, the composition may vary both in space and in time,
and the energy equation must be integrated over small partial volumes
with virtually uniform properties.
13 This quickly leads us to partial differential equations (see Chapter 22 on page 1368).
5
The extent to which the system reacts to external changes is given
by the exact differential of E = EK + EP + U .
It is possible to choose values for the thermodynamic variables
9
that make them invariant in response to chemical reactions, which
results in all dNi ∈{1,n} = 0 , although this requires the system to be in
equilibrium.
6
In the absence of external fields and internal stress–strain conditions, the 1st law can be written:
n
P
M υ dυ + Mg dz + T dS − p dV + µi dNi = δQ − δW ,
1
but it is rarely necessary to include all of the terms in order to obtain a
satisfactory description of the system.
7
Remember that any mathematical solution to a problem involving
physical phenomena always simplifies what actually happens.
8
In thermodynamics you have to be particularly careful in relation to
the assumption of equilibrium, and we should note that the composition
of the system can vary as a result of chemical reactions in the control
volume.
Expanded footnote [a]:
[a] Except in the case of nuclear reactions, the number of moles of each atom will be
preserved in the control volume. One option is therefore to let Ni be the mole number
of each atom i .
Expanded footnote [a]:
[a] Admittedly there will be small, natural fluctuations in composition, but this will not
affect the energy, provided that the system is thermodynamically stable. We will not
discuss the concept of stability in any greater detail here, but will instead refer to
Chapter 21 on page 1312.
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Thermostatic state change
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Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
Isochoric pressure change
9 Closed . . .
This chapter started by discussing the 1st law of thermodynamics,
but a general analysis based on the principle of conservation of energy
is not always required or practicable.
1
§ 65 Refrigerator
1
Let T◦ = 294 K and T1 = 280 K.
2
Assuming that the door seals tightly, the number of moles inside
the refrigerator will remain constant during cooling.
3
Applying the ideal gas law gives us N =
pV
RT◦
=
pV
RT1 ,
which implies
that the negative gauge pressure is ∆p = p◦ − p1 = p◦ (1 −
14
5
T1
T◦ )
=
§ 66
294 · 10 = 4762 Pa.
1
An apparatus for measuring diffusion at high pressures consists
4
The negative gauge pressure in the refrigerator means that an ex- of a test cell placed in a thermostatically controlled environment kept at
a temperature of T = T◦ . Initially the cell is filled with oil that can be
ternal force will act inwards on the door.
RB
compressed to high pressures using a piston. The piston is locked into
5
The moment about the hinge is in this case M = ∆pL 0 x dx =
§ 65
position once the working pressure has been reached. The compres1
1
2
2
2 ∆pLB = 2 4762 · 1.4 · (0.5) = 833.3 N m.
1
The temperature inside a refrigerator is 7 ◦ C. The door is opened
sion is rapid, giving a virtually isentropic change of state. The subseand air at a temperature of 21 ◦ C fills the whole refrigerator. What is 6 If we assume that the handle is right at the edge of the door, a force quent equalisation of the temperature occurs at constant volume. Asthe negative gauge pressure in the refrigerator after the door has been equivalent to |F| B = 833.3 N m, i.e. |F| = 1666.7 N, must be applied in sume that all of the volume of oil is within the walls of the thermostaticlosed and the air has cooled back down to 7 ◦ C? How much force is order to open the door.
cally controlled environment. Describe the fall in pressure as the oil apneeded to open the door again if the refrigerator measures 50 × 140 cm 7 The volume of air in the refrigerator is not relevant to the problem, proaches thermal equilibrium, i.e. TCn H2n+2 → T◦ . What compression is
on the inside? The atmospheric pressure can be assumed to be 1 bar. as it will not affect the result, provided that the air can flow freely when needed if you want the final pressure in the cell to be p = 300 bar? Use
data from Table 9.1 on page 411.
How do you solve the negative gauge pressure problem in practice?
the door is open.
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
Isochoric pressure change (2)
9 Closed . . .
4
Obviously, the key to the problem is working out what compression
is required to achieve a final pressure in the cell of p = 300 bar.
1
3
Instead of following the naive experimental procedure of repeatedly
compressing the oil to 300 bar, and then allowing the temperature to
stabilise, the pressure can be raised higher than that, before levelling
off to 300 bar at t → ∞.
∆TS
T
eplacem
In many cases it is simpler to start with an explicit temperature
equation.
Termodynamic methods TKP4175
∆T
V
T◦
2
3
This is particularly true for temperature regulated systems and systems that are in contact with thermal reservoirs.
t
2
The oil is practically incompressible, so the changes in both volume
and temperature are modest, in spite of the big change in pressure.
It is therefore possible to obtain an approximation of ∆p by integrating
the differential
(dp )tot = (dp )s ,n + (dp )v ,n
=
ˆ γ-1 (dp )s ,n ,
using the constant heat capacity ratio γ =
ˆ cp /cv .
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 066 Pressure cell
1
2
The size of the system is irrelevant, and the solution can be expressed in terms of moles.
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
(ds )n =
9 Closed . . .
5
Finally we substitute in (dT )s ,n = Tv α/cp (dp )s ,n as the variable of
integration for the change in pressure:
Take the total differentials of s and v from Chapter 8:
cp
T
§ 066 Pressure cell (2)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
d T − v α dp ,
§ 066 Pressure cell (3)
10 As we don’t have data for hydraulic oil, we will use the data for
benzene:
7
Before the differential equation can be integrated, we need to know
how γ varies with temperature and pressure.
8
Far away from the oil’s critical point, the heat capacity ratio will be
relatively constant, at least across a moderate range of temperatures
and pressures.
9
In this case it is sufficient to know that γ is constant ±10%.
∆p
[bar]
(dp )tot = (dp )s ,n + (dp )v ,n
(dv )n = v α dT − v β dp .
=
In this case, the energy balance is replaced by the temperature
equation (dT )s + (dT )v = 0, because the system is in a thermostatcontrolled environment; , also see the Figure on page 406.
3
=
= 1−
Expanded footnote [a]:
[a] In a thermostat-controlled environment we can control the temperature at all times,
but we cannot normally control heat flow across the system boundary.
4
Let us combine the differentials of s and v and eliminate (dT )s ,n
using the temperature equation.
cp
-1
Tv α (dT )s ,n + αβ (dT )v ,n
cp
-1
Tv α − αβ (dT )s ,n
Tv α2
cp β
9 Closed . . .
≈
91.6
134.3
(300 − 1) = 204 .
(9.1)
12 An analogous calculation to the one shown in 9.1 reveals that the
initial pressure must be 439 bar in order to ensure that pt →∞ = 300 bar.
11 The final pressure deviates significantly from the desired pressure,
which has practical implications when setting the pressure prior to starting an experiment.
(dp )s ,n .
6
This expression can be significantly simplified by introducing the relationship cp − cv = Tv α2 β-1 , which is again from Chapter 8 on page 378.
This gives us the final result
(dp )tot = γ-1 (dp )s ,n ,
where γ =
ˆ cp /cv .
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Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
cp
α
β
v
cp − cv
cv
Units
J
K mol
104
K
1011
Pa
106 m3
mol
J
K mol
J
K mol
Ideal gas
H2 O(lıq)
C(diamond)
SiO2 (quartz)
NaCl
H2 O(sol)
Cu
Ag
Pb
−
75.5
6.07
44.5
49.7
36.6
24.5
25.5
26.8
33.54016
2.35
0.035
0.35
1.21
2.55
0.48
0.57
0.86
106
46
0.68
2.65
4.2
12
0.72
0.99
2.37
24790
18
3.4
22.6
27
19.6
7.1
10.2
18.7
8.3145
0.63
0.0018
0.31
2.8
3.2
0.68
1.28
1.74
−
74.9
6.07
44.2
46.9
33.4
23.8
23.3
25.1
§ 067
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§ 66
§ 68
Tore Haug-Warberg (Chem Eng, NTNU)
1
Heat can be transported by diffusion, convection and radiation, so
an intelligent question to ask is: When can we no longer rely on thermodynamic analysis?
2
When do we instead need to embark on more esoteric approaches
based on solving partial differential equations?
3
The answer to the latter question is relatively obvious: Thermodynamic functions have n + 2 variables, none of which describe the geometry of the system.
4
Consequently, we can only solve problems where the geometry is
irrelevant (i.e. where the system has uniform properties).
5
If there are gradients in e.g. T or p within the control volume, then
we must use a distributed model, i.e. a partial differential equation.
6
In practice, convection and radiation can (often) be described in
terms of simple thermodynamic relationships, whereas diffusion requires real mathematics.
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We are not at this point going to attempt to mathematically derive
CV for a two-phase water/steam system. Instead we will say that if the
flask is reasonably full, then its heat capacity will be almost entirely
determined by the water phase.
1
The change in temperature from 100 ◦ C to 0 ◦ C will reduce the
water vapour pressure from 1 atm to approximately 0.005 atm, but this
has little influence on the heat capacity of the water, and for all practical
purposes we can assume that CV ≈ CP ,H2 O + CP ,SiO2 .
2
§ 067 Thermos (2)
Isochoric cooling
T◦ +100
=−
Rt
0
Tore Haug-Warberg (Chem Eng, NTNU)
kA
CP ,H2 O ∆l
dt
⇒
Termodynamic methods TKP4175
θ
[◦ C]
(dU )V ,n = CV dT = δQ ,
A simple heat transfer model gives us
3
CV dT = − ∆k l A (T − T◦ ) dt .
10 The variation in pressure will be hugely different when there is
steam present compared to what is observed for a single phase liquid and we cannot simply use the value of CV for pure liquid water.
4
This equation assumes that the temperature in the flask is uniform
(i.e. it is based on a lumped model).
5
This gives a good approximation, as the flask is designed to prevent heat loss, while the water is a relatively good heat conductor (approximately 20 times better than expanded polystyrene).
6
However, if the thermos flask is filled with air, the assumption no
longer holds.
7
Next we need to consider whether the flask is full with water, or
whether there is steam present.
8
If we assume that there is steam in the flask, we must be careful
when calculating the CV term.
9
Admittedly CV is defined as being the derivative of internal energy
with respect to temperature at constant volume, but in this context the
definition refers to a two-phase system consisting of water and steam.
11 On the other hand, if we assume that the flask is initially full, then
the contraction of the water as it cools will cause a significant reduction
in pressure.
12 If the negative gauge pressure is significant, then steam will inevitably be produced, which takes us back to a two-phase system.
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
100
9 Closed . . .
4
Some representative design parameters are k = 0.025 W m-1 K-1 ,
-1
A = 0.06 m2 , CP ,H2 O = 75.5 · 1000
18 ≈ 4194 J K and ∆l = 0.02 m, which
gives us a time constant of τ = 15.53 h.
The latter depends on the geometry of the system, in so far as the
contact area A affects the calculation, but the area isn’t very significant
unless there are big changes to the shape.
5
80
60
1/1
40
1/3
1/8
20
0
0
5
10
15
20
Time [h]
Figure 9.2: Time–temperature profile for a thermos flask buried in snow. Solid
lines show temperature change if the flask is ignored. Dotted lines take the
flask into account. The flask is 1/1, 1/3 and 1/8 full respectively.
= 100 exp − τt .
02 September 2013
9 Closed . . .
or, if we want to emphasise the role of time: dU / dt = CV dT / dt .
For the moment we will ignore the heat capacity of the flask itself,
and say that
dT
(T −T◦ )
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2
It will be assumed that the thermos is closed (in other words there
is no evaporation).
3
T◦R+θ
02 September 2013
For a closed system, the 1st law of thermodynamics can be written:
1
dU = δQ − δW . In this case, the system (the thermos flask) doesn’t
perform any work, and the energy equation can be expressed as
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
Termodynamic methods TKP4175
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
ENGLISH text is missing.Limits of thermodynamics
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 067 Thermos
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A thermos flask is filled with 1 l of boiling water. It is
then buried in snow which has a temperature of 0 ◦ C.
Calculate how the temperature in the flask changes as
a function of time. Does it make any difference whether
the flask maintains a constant volume or constant
pressure? Does the geometry of the system affect
the calculation? Assume that the flask is insulated
with 2 cm thick expanded polystyrene. Find the time
constant. Is the material used for the inner flask, either
2 mm thick glass or 0.5 mm thick stainless steel, of
any practical significance? What about the amount of
liquid in the flask?
[†] G. M. Barrow. Physical Chemistry. McGraw-Hill, 3rd edition, 1979. p. 151.
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02 September 2013
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Figure 9.1: Physical properties of some common compounds. All values are
given for T = 298.15 K and p = 1 bar.
Property[†]
Termodynamic methods TKP4175
Temperature [◦ C]
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
Time constant
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
Including the flask in the calculation affects the temperature change
over time, but it does not make much difference whether it is made of
0.5 mm thick steel or 2 mm thick glass.
1
Time constant (2)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
U2 ≃ 0
If the flask contains less water, both the initial temperature and the
time constant fall, and if the flask is an eighth full, the initial temperature
will be 69.0 ◦ C, while the time constant will be only 2.8 h.
9
6
This assumes that the temperature of the flask is 0 ◦ C when it is
filled.
7
So, the initial temperature falls, while the time constant rises.
8
At a given point in time, in this case after 16.0 h, the water temperature will be the same for both calculation methods.
10 This has a huge impact on the water temperature, as can be seen
from Figure 9.2 on page 416.
11 The three temperature progressions were drawn using the Program 31:1.12.
U1 =
The volumetric heat capacity of steel is about a quarter of the
equivalent value for glass.
2
−aN12
V1
A rough but nevertheless realistic estimate is CP ,SiO2 ≈ 44 ·
= 228 J K-1 . This contribution will change the time constant
from 15.5 h to 16.4 h.
4
3
Bearing in mind the thickness of the flask, the contribution of the
steel and glass to the overall heat capacity is the same.
0.2·600·2.6
60.1
Moreover, the initial temperature will fall from 100 ◦ C to 94.7 ◦ C, as
the flask is raised to the same temperature as the water.
5
Figure 9.3: Gas leak in a closed room. The reference state is an ideal gas at
the original temperature of the system.
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 68 Closed room
1
We can safely ignore the potential energy contribution.
2
The kinetic energy is significant in the release phase, but gradually
its contribution tends to zero.
3
A differential energy balance equation is unnecessary for the purposes of this problem, as the equations can be solved algebraically at
limt →∞ .
4
This is because the system does not interact with its surroundings.
5
It is therefore sufficient to solve the energy and mass balance equations without looking at the sequence of events.
6
The energy balance equation for the room, including the gas in the
flask, can be expressed as
Mechanical energy balance
(∆U )V ,n = 0 .
7
We will start by finding the internal energy of the pressurised gas.
8
Let us convert a and b to SI units: a = 1.35 · 101325 · 10-6 =
0.13679 Pa m6 mol-2 and b = 3.8 · 10-5 m3 mol-1 .
9
Then we can calculate the number of moles of gas in the cylinder
using the equation
300·105
[Pa]
=
N1 R ·293.15
0.04−N1 b
−
N12 a
(0.04)2
,
which gives us N1 = 443.6 mol, cf. Program 1.9 in Appendix 31.
9 Closed . . .
10 Incidentally, the pressure in the cylinder is ≈ 10% lower than it
would be if the gas were ideal (which appears realistic).
11
In the rest of the room there are N2 = 2049.7 mol of air.
12 We will use an ideal gas at 293.15 K as our reference state for internal energy — see explanation in Figure 9.3 on the preceding page —
and will ignore non-ideal contributions in the expanded gas (verify that
this is a reasonable assumption).
§ 68
13 By using the results that we derived in Chapter 8, we can therefore
say that[a]
ıg
14 (N1 + N2 )c ∆T = −aN12 /V1 , or:
v
A 40 l gas cylinder pressurised to 300 bar is placed in an insulated
room of volume 50 m3 . The room temperature is 20 ◦ C and the air
0.13679 (443.6)2
∆T
= −13.0 .
=
[K]
pressure is 1 bar. The behaviour of the gas can be described using
0.04 (443.6 + 2049.7) (29 − 8.3145)
the Van der Waals equation of state p = RT /(v − b ) − a /v 2 , where
15 The change in temperature is modest, but the positive gauge presa = 1.35 atm l2 mol-2 and b = 0.038 l mol-1 . The valve fails and the
sure in the room is 0.1607 bar. On a 2 m2 window, this is equivalent to
gas flows out of the cylinder and into the room. Calculate how the
31237 N, which is far more than the frame can withstand.
temperature and pressure of the gas have changed once equilibrium is
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reached. The room has a window of area 2 m2 . How big is the force
that acts on the window frame?
468 / 1776 [a] Assuming that the heat capacity is independent of temperature — which is a good
1
approximation when ∆T is small.
§ 069
1
The thermos flask in Theme 67 on page 412 represents an extreme
case, from a thermodynamic point of view, in that the energy balance
equation is reduced to simply a heat transfer problem.
§ 68
§ 70
9 Closed . . .
This is the same as saying that the entropy is conserved, which
6
means that the energy balance equation can be written
3
This kind of system is more common than you might expect.
4
If state changes are rapid, e.g. the air in a bicycle pump or the gas
in a gas turbine can be described in terms of adiabatic changes of state.
5
For an ideal gas, the internal friction caused by flow resistance and
turbulence can be ignored.
(dU )s ,n = − δW = −p dV .
2
The internal energy of an ideal gas is only dependent on its temıg
perature and composition, so we can say that CV dT = δQ − p ıg dV .
5
The simplest thing to do is to start by looking at isobaric
compression at a given system pressure p◦ :
3
Note that assuming that the gas is ideal means that all of the parts
of the question have the same structure.
4
The only question that remains is how to integrate p dV .
Expanded footnote [a]:
[a] Perhaps compression is a misleading term given that the gas pressure is constant.
An alternative way of looking at it is to think of a cooling process in which the gas
contracts as a result of the fall in the temperature of the system.
ıg
Wp =
Here we will compare adiabatic compression with a couple of other
types of compression.
8
9 Closed . . .
1
For all of the parts of the question, the energy equation can be exıg
pressed in the form (dU )n = δQ − δW , where the work term is δWrev =
p ıg dV .
Determine the compression work done during a variable
change in volume in a closed system. The answer
evidently depends on how the compression takes
place. Find comparable expressions for the isobaric,
isothermal and isentropic compression of an ideal
ıg
gas. Assume a constant heat capacity of CP . How
much heat must be removed/added in each case?
An adiabatic system that only does pressure–volume work on its
surroundings is another extreme case.
2
§ 069 Compression work
RV
V◦
p◦ dν =
ˆ p◦ V◦ (κ − 1) = NRT◦ (κ − 1) .
7
Systems that exchange both heat and work with their surroundings
are an intermediate case.
9
We will return to the connection between heat and work in Chapter 18 on page 938.
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 069 Compression work (2)
9 Closed . . .
6
In the above equation, and in the rest of the section, κ refers to the
relative reduction in volume V /V◦ .
7
The amount of heat that must be removed during compression is
ıg
ıg
ıg
δQ = CV dT + p ıg dV = CV dT + NR dT = CP dT F (Q ıg )p = (∆H ıg )p .
ıg
RV
V◦
p (T◦ , ν) dν =
RV
V◦
NRT◦
ν
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8
The next item on our agenda is to calculate the isothermal compression work at the system temperature T◦ :
WT =
Termodynamic methods TKP4175
§ 069 Compression work (3)
15
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9 Closed . . .
10
In this case δQrev = 0, which means that (Q ıg )S = 0.
Show that this statement is true.
18 The smaller κ is, the greater the difference, as shown in Figure 9.4
on the following page.
19 See Program 31:1.2 for further details.
20 Note that W is defined as being positive when the system performs
work on its surroundings.
21 The compression work shown in Figure 9.4 is therefore negative.
dν =
ˆ NRT◦ ln κ .
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
17 What, then, is the difference between the three kinds of compression work? Well, for a given volume reduction of κ, (W )p < (W )T <
(W )S .
16
Tore Haug-Warberg (Chem Eng, NTNU)
W [J/mol]
Tore Haug-Warberg (Chem Eng, NTNU)
10
4
3
14 In order to save space, we will not do so again here. Instead, refer
to Eq. 8.47 on page 382:
ıg
9
Here the amount of heat released is δQ = CV dT + p ıg dV =
p ıg dV = δW i.e. (Q ıg )T = (W ıg )T .
10 Now what remains to be done is to obtain an expression for the
isentropic compression work.
11 Normally we only know the entropy of the system indirectly through
T , p or T , V .
12 The integral is therefore expressed in terms of these variables (so,
for the initial state, T◦ is used, rather than S◦ ).
13 We have already derived the isentropic relations for an ideal gas;
see Chapter 8:8.8.
ıg
WS =
RV
V◦
p (S◦ , ν) dν =
=
RV
V◦
p◦
NRT◦
1−γ
ν –γ
dν
V◦
=
ˆ p◦ V◦
κ1−γ − 1 .
Rκ
10
κ–γ dκ
1
2
0
0.2
0.4
1−
(9.2)
0.6
V
V◦
0.8
1
Figure 9.4: Isentropic, isothermal and isobaric compression work for one mole
of an ideal gas (γ = 1.40).
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
1
What remains to be done is to substitute in the differentials of
internal energy at constant entropy (dU )S ,n = −p dV and of kinetic
energy dEK = M υ dυ.
3
which gives us the following energy balance equation:
4
The differentials of s and v that we need can be found in Chapter 8:
5
Eliminating dT at constant molar entropy and composition gives
p (dv )s ,n = Mw υ dυ .
Heat capacity ratio γ
(dv )n = v α dT − v β dp ,
(ds )n =
cp
T
dT − v α dp .
us:
(dv )s ,n = v α cTp v α dp − v β dp
2 2
= Tvcpα − v β dp .
(dv )s ,n = −
vβ
γ
dp .
x
(9.4)
9 Closed . . .
6
Combining the differential from Eq. 9.4 with the relation cp − cv =
Tv α2
β , which is taken from Section 8.7 on page 377 gives us the final
result
(9.5)
7
From the energy equation we know that p (dv )s ,n = Mw υ dυ. Combining this with Eq. 9.5 gives us γ-1 v βp (dp )s ,n = −Mw υ dυ.
1
Based on statistical mechanics it can be shown that γ = 5/3 for a
monoatomic ideal gas, and that γ = 7/5 for a diatomic ideal gas with all
of its vibrational and rotational levels occupied.
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
x + ∆x
§ 70 Water hammer
2
Note that the size of the system is irrelevant, so we can express
the equations on an intensive form,
ENGLISH text is missing.Kinetic energy conver-
8
For isentropic changes of state in liquids, the molar volume and the
sion
speed of sound are roughly constant, so the maximum pressure can be
1
Modern kitchen taps have ceramic valves that stop the water flow 1 In this case, the energy balance equation for the control volume
expressed as
within a fraction of a second. This sudden blocking of the flow causes has an unusual form:
q
q
M γ
γ
∆p ≈ υ vwβ = υρ ρβ =
ˆ υρυ⋆
water hammer, which can damage pipe connections and plumbing fixdEK + (dU )S ,n = 0 .
(9.3)
tures. The water hammer originates where the valve is shut, and moves
where υ⋆ is the speed of sound in the liquid; see Eq. 9.10 on page 443.
up the pipes at the speed of sound. The physics of the interface be2
It is tempting to include a pV term, because instinctively you think For water flowing at a speed of 1 m s-1 , the water hammer will have a
tween the water and the pipe is complex, and we will only look at what
that the control volume is compressed by the liquid upstream of it.
pressure of 14.5 bar.
happens in the water phase. The compression of this phase is extremely rapid, and it is a good approximation to assume that it is isen- 3 However, this is not the case, as the size of the control volume is 9 This is at the limit of what pipes and plumbing fixtures can withtropic. Derive an expression for the pressure build-up in the water when arbitrary, and in theory it could encompass all of the liquid that is in stand, so the moral of the story is that modern taps should be turned
off with some care.
the flow of water is blocked. Calculate ∆pmax for a couple of represen- motion.
tative flow velocities and temperatures. Use the physical data from the 4 Hence there is no external mass that can perform pV work on the 10 The best taps have slow-closing valves to prevent pipes from comsystem.
ing loose from their supports.
preceeding sections.
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§ 70
§ 071
ENGLISH text is missing.Uniform flow
1
Classic thermodynamics deals with equilibria and the properties of
equilibrium states.
2
In this chapter we are applying control volume theory to flowing
media, which raises several fundamental questions about the nature of
a state.
§ 70
§ 72
3
It is far from obvious that results from thermodynamic equilibrium
theory apply to media in motion, but we will nevertheless assume this
without any further justification[a].
9 Closed . . .
4
Next we must write Newton’s 2nd law in a form that relates velocity
and time to thermodynamic state variables.
5
Velocity is distance over time, and multiplying it by the local crosssectional flow area gives us the volume transported per time unit.
6
By considering a control volume that travels υ dt over a short space
of time dt , it is possible to relate the velocity to familiar thermodynamic
variables such as mass, volume, energy etc, but it is not possible to
rigorously analyse the flow.
7
The theory in this section therefore only applies to the adiabatic,
stationary and quasi-one-dimensional flow of a frictionless fluid without
chemical reactions.
8
This may appear to be a hopelessly unrealistic special case, but in
practice there are many situations that fall within this category.
9
A number of important applications are discussed in Chapter 12
(gas dynamics).
481 / 1776
[a] This topic is covered in detail in Chapter 22 on page 1368.
2
The heat capacity ratio γ appears in a wide variety of contexts, and
it is worth saying a bit about what values it can have.
υx d t
Figure 9.5 on the previous page shows a volume
element moving with the fluid flow. There are no chemical
reactions in the control volume, and the flow is assumed
to be stationary. Show that the mass balance of the
dA
element can be expressed as ( dvv )n = dυ
υ + A where
υ is the flow velocity, v is the molar volume and A is
the cross-sectional flow area.
υx +∆x dt
However, it is worth noting a few exceptions: For solids with a high
modulus of elasticity, such as quartz and diamond, γ ≈ 1.
4
3
But what about the value of γ for liquids and solids? In their cases
it is harder to make any generalisations, but for many substances the
value lies between 1.2 and 1.5, as shown in Table 9.1 on page 411.
5
That is also the value for water at 4 ◦ C: This is the temperature
at which its density reaches a maximum, which means that α = v -1
(∂v /∂T )p ,n = 0, and therefore that cp − cv = 0 in Eq. 8.41 on page 378.
From this it follows that γ = 1, which must also be considered the
minimum value for γ.
6
Expanded footnote [a]:
[a] cp ≥ cv is a fundamental prerequisite for stability in thermodynamics, cf. Chapter 21
on page 1312.
Tore Haug-Warberg (Chem Eng, NTNU)
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Figure 9.5: Quasi one-dimensional flow in the horizontal plane. The impact
of the y and z planes has been ignored. The arrows indicate the distance
travelled by the control volume during the time interval dt (the fluid accelerates
to the right). The fluid has been shown simultaneously expanding in order to
focus attention on compressible flow.
Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 071 Continuity of mass
§ 071 Continuity of mass (2)
Figure 9.5 shows the control volume at t = 0 and t = dt . move
from the shaded areas to the dotted lines in the drawing.
2
In the time period dt the system boundaries[a]
[a] In fluid dynamics, the coordinate system is either based on a fixed point in space
(Euler), or it moves freely with the flow (Lagrange). The latter system has been chosen
here.
3
The left-hand system boundary moves υx dt in parallel with the
right-hand boundary moving υx +∆x dt .
4
This creates two sub-volumes, one at the left and one at the right
of the figure, which both have the same mass Mx = Mx +∆x , given
stationary flow.
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Tore Haug-Warberg (Chem Eng, NTNU)
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
( Avυ )x
Termodynamic methods TKP4175
§ 072
§ 71
§ 73
9 Closed . . .
( Avυ )x +∆x
If we divide by dt , the equation simplifies to
8
=
which means that Avυ is conserved in the direction of flow. In other
words ∆( Avυ ) = 0.
6
Since there are no chemical reactions in the control volume, we
can alternatively view v as a molar volume, but in that case the mass
balance must be expressed in the form Nx = Nx +∆x .
7
Regardless of whether Eq. 9.6 is expressed in molar or mass
terms, the differential dt will be a common factor to both sides of the
equation.
9
Provided that A , v and υ are continuous functions of the path length
x , then d Avυ n = 0, or expressed in logarithmic form:
∆( Avυ ) = 0
continuum
⇒
(dln v )n = dln υ + dln A .
Show that the energy balance of the (adiabatic) control
volume in Figure 9.5 can be expressed as (dh )n = −Mw υ
dυ where Mw and h are respectively the molar mass and
molar enthalpy of the fluid. The remaining information
you need follows from Theme 71.
(9.7)
11 Even if the derivative doesn’t exist, meaning that it cannot be defined experimentally, it will still be possible to satisfy ∆(y ) = 0 without
dy = 0.
10
If we substitute in the flow variables, the mass balance can be
expressed as
A υdt A υdt (9.6)
v x =
v x +∆x .
5
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
1
Termodynamic methods TKP4175
Note that the implication sign in Eq. 9.7 is pointing one way only.
12 This kind of sudden change in the state variables is referred to as
a normal shock in Chapter 12:12.5.
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
ENGLISH text is missing.Comments
1
In deriving equations 9.7 and 9.8, the main emphasis has been on
describing the physical properties of a flowing fluid, rather than on the
underlying mathematics.
2
In continuum mechanics, the same equations are derived in a more
universal and elegant way from the divergence theorem, but the more
abstract the derivation, the further removed it is from thermodynamics.
§ 072 Continuity of energy
§ 072 Continuity of energy (2)
9 Closed . . .
9 Closed . . .
1
The energy balance can be expressed (Me + pA υ dt )x = (Me + pA
υ dt )x +∆x since the total energy content E = Me , plus the pV work required to displace the fluid, must be the same for the two sub-volumes
.
§ 073
3
This difference will become even clearer when we now go on to
look at the last of the balance equations — the momentum balance
equation.
4
Unlike the continuity and energy equations, the momentum balance equation is not related to thermodynamics.
5
A thermodynamic equilibrium describes a static state, and is unrelated to the dynamics of the system, whereas the momentum equation
is based on Newton[a]’s 2nd law.
§ 72
§ 74
6
One of the ways that it can be formulated involves the integral of
force with respect to time (impulses).
R
R
7
From Newton’s 2nd law it follows that F dt = M a dt = M ∆υ. In
other words, an impulse changes the momentum of the system.
9 Closed . . .
8
With reference to Figure 9.5, any pressure gradients in the control
volume result in contact forces, which in turn affect the velocity of the
fluid.
9
If we relate the pressure to the mechanical state, it tells us something about the movement of the fluid.
10 But it is important to note that the force vector F has both a spatial
direction and a length (a measuring number combined with a unit).
11 This means that the momentum balance equation is vectorial,
which is unhelpful to us, since the thermodynamic variables are assumed to vary only along the x axis.
12 If we are to use scalar equations, it is no longer sufficient for the
flow to be quasi-one-dimensional, it must be genuinely one-dimensional
and have a constant cross-sectional flow area.
489 / 1776
[a] Sir Isaac Newton, 1642–1727 by the Julian calendar. English physicist and mathematician.
Expanded footnote [a]:
[a] These are still defined by the displacement of the control volume in the time period
dt
Show that in a frictionless control volume with a constant
cross-sectional flow area, the momentum equation
can be expressed in the form v (dp )A ,n = −Mw υ dυ.
The flow is only in the horizontal plane, it is genuinely
one-dimensional and frictionless, and there are no
chemical reactions in the fluid.
2
If we divide by Mx = Mx +∆x from the mass balance, we get (e +
pv )x = (e + pv )x +∆x , where v = M -1 A υ dt is the specific volume of the
fluid.
3
This shows us that e + pv ≈ u + ek + pv =
ˆ h + ek is conserved in
the horizontal plane (ignoring the potential energy ep ).
5
If we substitute in molar kinetic energy ek =
Expanded footnote [a]:
4
As there are no chemical reactions in the fluid, we can alternatively
use molar values.
[a] It has been assumed that υ2 = b
υ2 = υb2 . Thtat allows us to use the same velocity
~ υb2 .
variable in the expressions for (mean) mass flow ρAb
υ and (mean) kinetic energy 21 m
This simplification only applies in the case of fully developed turbulence.
with
∆(h + ek ) = 0
Tore Haug-Warberg (Chem Eng, NTNU)
continuum
⇒
1
2
2 Mw υ
(dh )n = −Mw υ dυ
Termodynamic methods TKP4175
we end up
02 September 2013
(9.8)
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 073 Continuity of momentum
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
By expressing the momentum equation in scalar form, we are
assuming that the flow is one-dimensional.
1
2
In that case, the pressure is isotropic, and Newton’s 2nd law can
t
A υdt
be formulated as ( A υd
v · υ)x +∆x − ( v · υ)x = (pA dt )x +∆x − (pA dt )x .
If we divide the balance equation by (A dt )x = (A dt )x +∆x we find
that p + v -1 υ2 is conserved in the direction of flow.
3
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§ 073 Continuity of momentum (2)
1
9 Closed . . .
Combining the equations 9.9 and 9.8 gives us (dh )n = v (dp )A ,n .
Bearing in mind that the differential (dh )n = T ds + v dp , it follows
that T ds = 0.
4
Since the momentum Eq. 9.9 requires frictionless flow, whereas
the energy Eq. 9.8 does not, it is true to say that “frictionless” in this
context is the same as “isentropic”.
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∆(p +
=0
continuum
⇒
v (dp )A ,n = −Mw υ dυ ,
Let us first isolate the v dp term from the 1st law and then substitute in the energy balance, mass balance and 2nd law in that order:
1
dh − T d s
=
ˆ −Mw υ dυ − T ds
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Speed of sound
=
ˆ −Mw υ2 v -1 dv − T ds
9 Closed . . .
For the purposes of the momentum Eq. 9.9, it is important
to assume that the flow is frictionless. Show that this
assumption also implies that the flow is isentropic.
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Tore Haug-Warberg (Chem Eng, NTNU)
9 Closed . . .
§ 075
[a] Based on a historical perspective and with the benefit of hindsight.
Energy equation;
Tore Haug-Warberg (Chem Eng, NTNU)
(dh )n
=−Mw υ dυ .
(dln v )n = dln υ + dln A .
1st law;
(dh )n
2nd law;
T (ds )n
= T ds + v dp .
≥ 0.
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§ 075 Choked flow (2)
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§ 74
§ 76
9 Closed . . .
According to the 2nd law of thermodynamics, ds / dt ≥ 0.
This only applies to adiabatic systems. Combining
this law with the balance equations from Themes 71
and 72, as well as with the differential (dh )n = T ds +
v dp , reveals that the flow velocity cannot exceed
the speed of sound of the fluid, which is given by
υ⋆2 =
ˆ −Mw-1 v 2 (∂p /∂v )s ,n . Show this.
The equations for an adiabatic control volume can be summarised
as follows:
2
However, it should be stressed that the derivation is subtle, arguably even philosophical, and that in many ways it appears to be a
retrospective[a] interpretation.
3
Hence it doesn’t tell us a great deal about speed of sound as a
physical phenomenon, but it does still give us a useful perspective for
looking at the connection between thermodynamics and flow.
Continuity equation;
Termodynamic methods TKP4175
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
1
It is perhaps surprising that a concept like the “speed of sound” is
related to thermodynamics, but in fact it can be derived by looking at a
simple control volume.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
4
Next we divide by dv assuming that s is constant, and isolate the
velocity term:
∂p υ2 ≤ υ⋆2 =
ˆ −Mw-1 v 2 ∂v s ,n .
5
This result is an expression for the speed of sound in the fluid.
6
The same definition is used in wave mechanics, although it is derived in a completely different way.
≤ −Mw υ2 v -1 dv .
§ 75
(9.9)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
v (dp )n =
υ2
v )A
where v represents the specific volume in the left-hand equation and
the molar volume in the right-hand one.
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 075 Choked flow
§ 73
4
Since we know that d v A ,n = 0 from the continuity Eq. 9.7 on
page 430, this means that the momentum equation can be expressed
in the form
4
In other words, the flow is isentropic.
Tore Haug-Warberg (Chem Eng, NTNU)
§ 074
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
2
3
9 Closed . . .
υ
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 074 Frictionless vs isentropic
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
2
According to the above inequality, v (dp )n will reach a maximum
under isentropic conditions.
§ 076
§ 75
§ 77
9 Closed . . .
ENGLISH text is missing.Comments
1
Here we must accept that the speed of sound and the maximum
flow velocity are two sides of the same coin, but it is fascinating to see
how well-known results from wave mechanics and gas dynamics can
be related to thermodynamics even when you have so little to go on.
2
Incidentally, the speed of sound is relevant far beyond the confines
of pipe flow, and the above equation correctly defines the speed of
sound in all homogeneous substances (but this cannot be shown using
thermodynamic principles).
3
This classical interpretation of the speed of sound is interesting in
its own right, but the reverse application is equally important,
4
since the speed of sound of a fluid can be measured relatively
easily and precisely.
5
This has made υ⋆ an important parameter for experimentally determining the equations of state of gases and liquids.
499 / 1776
As a general rule, there are many ways of expressing
thermodynamic relationships, and the speed of sound is
no exception. Show that υ⋆ in Theme 75 on page 440
can alternatively be expressed
c v
γ
∂p υ⋆2 =
ˆ −Mw-1 v 2 ∂v s ,n = Mwpcv β =
ˆ ρβ .
(9.10)
3
Under other conditions, −T ds will make a negative contribution to
the right-hand side.
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Termodynamic methods TKP4175
02 September 2013
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News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 076 Speed of sound I
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
Compare Eq. 9.10 with the equivalent expression in Theme 75 on
page 440. The exercise boils down to demonstrating that (∂p /∂v )s ,n =
−cp /(cv v β) is true.
1
2
The following three differentials from Chapter 8 on page 377 provide a useful starting point:
c
∂p (ds )n = Tv dT + ∂T v ,n dv ,
(ds )n =
cp
T
§ 076 Speed of sound I (2)
4
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
§ 077
At constant v , the volume differential can be rearranged to
(∂p /∂T )v ,n = αβ-1 , which if inserted into the above equation gives us
−T cv -1 αβ-1 (dv )s ,n = T cp -1 v α(dp )s ,n .
d T − v α dp ,
After dividing by (dv )s ,n , the final result is
cp
∂p dp dv s ,n = ∂v s ,n = − cv v β
Tore Haug-Warberg (Chem Eng, NTNU)
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Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 077
§ 76
§ 78
(2)
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§ 077 Speed of sound II
you can either use Eq. 9.10 on page 443, or you can
consult Eq. 9.11 on page 451.
1
He
H2
N2
SF6
Hg
CS2
1015
1318
352
136
1599
1149
υ⋆ [ ms ]
CCl4
C 6 H6
CHCl3
H2 O(lıq)
Diamond
Quartz
-1/2
For gases, υ⋆ ∝ Mw
100–1000 m s-1 .
T [K]
υ⋆ [ ms ]
283.148
293.145
303.143
313.140
1447.290
1482.320
1509.040
1528.800
323.137
333.134
347.131
363.126
1542.470
1550.880
1555.010
1550.320
NaCl
H2 O(sol)
Cu
Ag
Pb
Expanded footnote [a]:
Expanded footnote [a]:
[a] Del A. J. Grosso. J. Acoust. Soc. Am., 45:1287, 1969.
[a] W. Kroebel and K. H. Mahrt. Acustica, 35:154, 1976.
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§ 077 Speed of sound II (2)
9 Closed . . .
3
If you calculate υ⋆ for liquid water you get 1480 m s-1 at 25 ◦ C, which
is close to the value in the table.
υ⋆ [ ms ]
931
1312
1050
1480
6455
3780
υ⋆ [ ms ]
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
υ⋆ [ ms ]
T [K]
Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
9 Closed . . .
Calculate the speed of sound of each of the fourteen
compounds in Table 9.1 on page 411. Is there a
general pattern to which substances have a high or
low speed of sound? For water, compare the result obtained by calculation with the experimental
values in the above table. All of the values relate
to 1 [bar]. When doing calculations for ideal gases,
(dv )n = v α dT − v β dp .
3
Here we can eliminate dT from the first two equations, provided
that s is constant:
∂p − cTv ∂T v ,n (dv )s ,n = cTp v α(dp )s ,n .
§ 78
We still need to get rid of (∂p /∂T )v ,n .
5
6
§ 76
4
The other liquids have similar values to water, albeit generally
slightly lower.
3412
3152
3998
3153
2016
, which produces values in the range
2
Solids with a high modulus of elasticity (particularly diamond) have
the highest speeds of sound, often in the range 3–4000 m s-1 .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
§ 078
§ 77
§ 79
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Tore Haug-Warberg (Chem Eng, NTNU)
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
§ 078 Ideal gas III
1
In gas dynamics, the speed of sound in an ideal gas is
p
ıg
a very important concept. Show that υ⋆ = γRT /Mw
based on Theme 76.
For an ideal gas βıg = −v -1 (∂v /∂p )T ,n = p -1 .
Substituted into Eq. 9.10 this gives us:
3
The equation below follows from the definition γ =
ˆ cp /cv :
q
γRT
ıg
υ⋆ =
Mw
=
Looking back
1
cp RT
Mw cv .
2
2
3
(9.11)
4
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The 1st law of thermodynamics: d(EK + EP + U ) = δQ − δW
RV
ıg
ˆ NRT◦ ln κ
Isothermal compression: WT = V p (T◦ , ν) dν =
◦
RV
ıg
1−γ − 1
◦
Isentropic compression: WS = V p (S◦ , ν) dν =
ˆ NRT
1−γ κ
Continuity of mass: ∆( Avυ ) = 0
◦
Continuity of energy: ∆(h + ek ) = 0
6
Continuity of momentum: ∆(p +
8
Termodynamic methods TKP4175
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9 Closed . . .
5
7
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News State change Thermodynamics Mechanics Uniform flow Speed of sound Olds
9 Closed . . .
υ⋆2
Termodynamic methods TKP4175
Speed of sound:
υ⋆2
=
ˆ
∂p −Mw-1 v 2 ∂v s ,n
ıg
Applied to ideal gas: υ⋆ =
Tore Haug-Warberg (Chem Eng, NTNU)
υ2
v )
q
=0
=
cp v
Mw cv β
=
ˆ
γ
ρβ
γRT
Mw
Termodynamic methods TKP4175
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News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
New concepts
Part 10
Dynamic systems
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10 Dynamic . . .
1
Inlet and outlet control volumes
2
Emptying an ideal gas flask
3
Filling an ideal gas flask
4
Joule’s gas expansion experiment
5
Kvasistatic state
6
Helmholtz resonance
7
Eigenfrequency
Dynamic systems
10 Dynamic . . .
1
This chapter will look at what happens when a closed system is
opened, allowing mass transfer through an inlet or outlet.
2
The result is a dynamic system, which can be filled or emptied over
a period of time.
3
Apart from the equilibrium state, which is reached at the limit t →
∞, the system has no stationary state.
6
Figure 10.1 on the following page shows three identical cylinders
with the flow limiter — for instance a valve or a crack in the cylinder
wall — in different locations.
4
It is hard to conceive of this kind of system as an abstract control
volume without physical limits, so it can be helpful to look at a specific
object such as a storage cylinder.
5
However, although most people find it easier to get to grips with
a physical entity than an abstract system, by using one we create an
unintended problem for ourselves: Where should the system boundary
go?
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
b)
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Tore Haug-Warberg (Chem Eng, NTNU)
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
§ 079 Outflow
a)
Termodynamic methods TKP4175
c)
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10 Dynamic . . .
§ 79
1
Where would you draw the system boundaries in Figure 10.1 if
you wanted to describe the state inside the cylinder? Then assume that
the direction of flow is reversed. Where would you draw the system
boundaries now?
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Termodynamic methods TKP4175
§ 079 Inflow
10 Dynamic . . .
1
The system boundary should be drawn inside the cylinder, as close
to the valve as possible.
1
The system boundary should once again be drawn inside the cylinder, but this time so as to include the valve.
2
It is important to know the state of the fluid as it leaves the system,
so it is assumed that the control volume has uniform properties.
2
This is because we need to know the state of the fluid flowing into
the system.
3
Heat exchange with the cylinder walls must be considered on a
case-by-case basis.
3
Heat exchange with the walls of the cylinder must once again be
considered on a case-by-case basis.
4
In the figure on the left, you must also consider whether there is any
heat exchange between the small section of pipe inside the cylinder and
the fluid in the cylinder.
4
In the figure on the right, you must also consider whether there is
any heat exchange between the section of pipe outside the cylinder and
the flowing fluid.
Figure 10.1: Outflow from cylinder. Three alternative locations for the flow
limiter.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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§ 080
§ 79
§ 81
Termodynamic methods TKP4175
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10 Dynamic . . .
§ 080 Cylinder emptying
4
If we are dealing with a leak, e.g. through a hole in the wall, the
control surface may need to be inside the cylinder.
5
If the size of the leak is large in relation to the surface area of the
walls of the cylinder, we will need to put the control surface so far into
the cylinder that it will be at the expense of the control volume itself.
6
Then we lose track of a significant proportion of the contents of the
cylinder, and the description breaks down[a].
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[a] If a pipe breaks in two, we cannot treat it as a lumped problem because the crosssectional area of flow of the gas has the same dimensions as the control surface.
Provided that the cross-sectional area of flow is small in relation to
the typical cross-sectional area of the control volume, the energy and
mass balances for the emptying process can be expressed as
Derive an expression for the temperature change that
takes place when a cylinder containing an ideal gas
is emptied. Assume that the gas has a constant heat
capacity in the relevant temperature range. Ignore any
heat gained from the surroundings, including from the
walls of the cylinder. What will the final temperature
be in a compressed air cylinder with an initial state of
10 bar and 20 ◦ C? Does the size of the cylinder affect
the numerical result?
dUc-v
dt
dNc-v
dt
4
2
We have tacitly assumed that there are no chemical reactions taking place in the control volume.
3
If that is not the case, then the mass M must replace the mole
number N as the free variable.
where the reference energy has been defined as U ıg (0) = 0.
pV ,
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10 Dynamic . . .
7
By combining the mass and energy balances, while simultaneously
substituting in molar values, we get
ıg
d(Ncv T )c-v
dt
ıg
~ out = c ıg Tout
= −cp Tout N
p
dNc-v
dt
.
This brings us to the crux of the derivation.
10 which is a direct consequence of our foresight when choosing the
location of the control surface.
For an ideal gas with constant heat capacity we can say that
ıg
Ncp T
11 Expanding out the total differential on the left-hand side of the
equation gives us
,
ıg
cv T
ıg
= Ncp T follows from the definition
The relationship
ıg
which can be restated as H ıg = U ıg + NRT = N (cv + R )T
H ıg
Tore Haug-Warberg (Chem Eng, NTNU)
§ 080 Cylinder emptying (2)
8
ıg
5
02 September 2013
9
The temperature inside the cylinder is the same as the temperature
in the flowing fluid, This allows us to eliminate the indices from the
temperature equation.
~ out .
= −N
U ıg = Ncv T ,
6
Termodynamic methods TKP4175
~ out =
~ out ,
~ out − pout V
ˆ −H
= −U
H ıg =
Termodynamic methods TKP4175
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10 Dynamic . . .
ENGLISH text is missing.System boundary
1
The system boundary must be drawn in such a way as to allow us
to describe the state of the fluid passing through the control surface.
2
In order to avoid having to digress into a discussion of adiabatic
flow and gas dynamics, we will therefore aim to place the system boundary somewhere that allows us to disregard the flow of kinetic energy
through the control surface.
3
This means that the pressure and temperature of the flowing fluid
will be the same as inside the cylinder itself.
1
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
dN
dt
ıg dT
dt
+ Ncv
ıg
= cp T
dN
dt
.
H =
ˆ U+
ıg
= Ncp T .
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Termodynamic methods TKP4175
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§ 080 Cylinder emptying (3)
Now we can substitute in γ =
in logarithmic form:


2/3



ıg

dln T
=γ−1≈
2/5
dln N



 0
12
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
10 Dynamic . . .
(cp /cv )ıg and rewrite the expression
for He, Ne, Ar, Kr
for H2 , N2 , O2 , etc.
for ∞ degrees of freedom
13 Note that the differential dt has disappeared.
14 Thermodynamics is quite literally a timeless[a] discipline, and there
are never any time variables in thermodynamic expressions unless they
are linked to a transport term[a].
[a] Thermodynamics does not involve time variables, and can also be considered a
timeless discipline, as most of its principles are based on work carried out over 100
years ago, by people like Ludwig Ferdinand von Helmholtz, Rudolf Julius Emanuel
Clausius, James Clerk Maxwell and Josiah Willard Gibbs.
[a] From fluid mechanics, heat transport, diffusion etc.
Alternatively, by substituting in the values for an ideal gas we get
ıg
T2 = T1
p2 V γ−1 RT1 γ−1
RT2
p1 V
Tore Haug-Warberg (Chem Eng, NTNU)
= T1
p2 T1 γ−1
p1 T2
ıg
T2 = T 1
⇒
Termodynamic methods TKP4175
p2 p1
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5
In practice this is not true if: i) the volume of gas is small, or ii) the
cylinder is emptied slowly.
γ−1
γ
Figure 10.2: Control volume for rocket engine
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
§ 81 Rocket engine
dUc-v
dt
§ 083
~
Q
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Tore Haug-Warberg (Chem Eng, NTNU)
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2
The basic principles set out in Section 80 therefore remain valid,
but we must add a term for the kinetic energy of the exhaust gases to
the energy balance equation:
~
W
6
The thermodynamic description of a similar system is discussed in
detail in Chapter 17.
§ 84
between these two extremes.
[a] Even at a pressure of 100 bar the pV term comes to less than 500 J.
dUc-v
dt
dNc-v
dt
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
~s,
= −~
Hout − W
= −~
Nout .
3
Let us take a formula for isentropic work of expansion from Section 91 on page 504, Eq. 11.7:
"1
#
γ−1
~ ıg = γ RT − p◦ γ N
~ out
W
γ−1
p
S
γ
Nout .
=
ˆ γ−1 RT (1 − α)~
4
5
6
The parameter α ≤ 1 has been introduced purely out of laziness.
T , p is the state inside the cylinder and p◦ is the outflow pressure.
ıg
ıg
~ , as well as U ıg = Nc T
Let us now add the work of expansion W
S
v
ıg
10 Dynamic . . .
p1 /10, γ = 1.4 gives you T2 = 0.2318T1 .
§ 083 Filling
ENGLISH text is missing.Kinetic energy
1
This time we must assume that the kinetic energy of the fluid flowing into the cylinder can be disregarded.
10 Dynamic . . .
2
For this to be true, the valve must be located close to the cylinder
and the cylinder must be filled slowly, to ensure that the velocity of the
gas in the fill hose is low.
3
By analogy to Section 80 on page 459, it follows that the mass and
energy balances can be expressed
dUc-v
dt
dNc-v
dt
~ ın ,
=H
~ ın .
=N
4
Quite unexpectedly, the two balance equations have an algebraic
solution if the cylinder is evacuated prior to the filling.
5
Straighforward integration gives the result Uc-v = Hın and Nc-v =
Nın showing that the enthalpy from the supply line is converted to internal energy when the gas enters the cylinder.
6
Because H ıg > U ıg we can conclude that the temperature of the
cylinder gas must increase.
7
We can also deduce that this temperature increase will happen
momentarily because the U /H ratio is independent of the degree of
filling.
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~ =0
∆H
b)
ıg
In the same way as in Section 80, we can substitute in U ıg = Ncv
1
T and
3
2
Let T be the temperature inside the cylinder and T◦ be the temperature in the fill hose (which is assumed to be constant).
H ıg
=
ıg
Ncp T ,
but this time Tc-v , Tın .
Eliminating the mass balance and then differentiating gives us
ıg
cv T
dN
dt
ıg dT
dt
+ Ncv
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dN
dt
ıg
ıg
ıg dT
dt
(cv T − cp T◦ ) =−Ncv
dln N
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§ 083 Filling (2)
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§ 083 Filling (3)
10 Dynamic . . .
ıg
ıg
=
ıg
(cp −cv )T1
ıg
ıg
cp T1 −cv T2
=
ˆ
ıg
=− dln(cp T◦
Termodynamic methods TKP4175
dT ,
ıg
− cv T ) .
02 September 2013
10 Dynamic . . .
We should particularly note that T2 = γT1 when α → ∞.
3
Also note that the pressure is not explicitly part of the problem.
4
This follows from our decision to assume ideal gas behaviour,
5
since for an ideal gas H ıg = H (T , N ), which means that the enthalpy of the gas flowing into the cylinder is only a function of temperature.
6
Having said that, it should be added that the pressure is implicitly
present in the expression through the quantity α.
(γ−1)T1
γT1 −T2
.
8
Some numerical results are shown in Figure 10.4 on the following
page.
8
If we use the definition α =
ˆ (N2 /N1 )ıg , we can write the final expression as
9
To do the calculations, the temperature equation first has to be
ıg
restated as T2 = T1 γβ/(β + γ − 1) where β = p2 /p1 .
10
ıg
T 2 = T1
Tore Haug-Warberg (Chem Eng, NTNU)
γ(α−1)+1
α
Termodynamic methods TKP4175
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In order to achieve α ≫ 1, i.e. to fill the cylinder with a large quantity
7
of gas, the pressure of the fill hose must be much higher than the initial
pressure of the cylinder.
By substituting accordingly and taking the exponential we get
N2 ıg
N1
ıg
cp T◦ −cv T
If the cylinder has been evacuated prior to filling, this limit value
will be reached instantaneously at t → 0, and the temperature of the
cylinder’s contents will remain constant throughout the filling process.
.
In order to obtain as simple an expression as possible, with the fewest
possible parameters, we will assume that T◦ = T1 .
7
ıg
,
ıg
cv
2
ıg
ıg
ıg
cp T◦ −cv T2
.
Limiting case
1
6
Note that T1 describes the temperature in the cylinder at t = 0,
while T◦ is the temperature in the fill hose.
cp T◦ −cv T1 dN
dt
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5
Now let us integrate between two states: N1 , T1 and N2 , T2 . The
result of the integration is
= ln
Tore Haug-Warberg (Chem Eng, NTNU)
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10 Dynamic . . .
ıg
= cp T◦
4
Next substitute in γ = (cp /cv )ıg and then rearrange the differential
equation in two steps:
Figure 10.3: Alternative methods of emptying a gas cylinder to produce a
temperature drop: a) is isentropic and b) is isenthalpic.
N2 ıg
N1
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§ 82 Isentropic emptying
N -1 dN =
ln
02 September 2013
2
Apart from that, the principles are the same as in Section 80 on
page 459 (and Section 83 below):
ıg
§ 82
Termodynamic methods TKP4175
1
This time the control volume does work that needs to be included
in the energy balance equation.
= −~
Hout − E~K ,out .
4
This simplification is possible because all of the quantities are timeinvariant, and in that sense the description of the thermodynamic state
of a rocket is simpler than the description of a pressure vessel.
5
From the energy equation we can see that the outflow velocity
would be zero if the enthalpy of the exhaust gases exactly matched
the internal energy of the fuel.
and H = Ncp T , to the energy balance:
7
The lower the enthalpy of the exhaust gases (and thus the temperıg
ature), the greater the velocity.
d(Ncv T )
γ
ıg ~
= −cp T N
Nout .
out − γ−1 RT (1 − α)~
dt
§ 82 Isenthalpic emptying
8
Even supersonic flow is possible, if the nozzle is designed in a
1
It is easy to fail to see the wood for the trees, but what goes on 7 If we expand the left-hand side (using the chain rule) and substitute
particular way (see Chapter 12 on page 644 for a discussion of that).
inside the control volume is actually irrelevant to the changes of state. in the mass balance we get
9
In this exercise we will assume that the temperature of the gas
γ
ıg
ıg
ıg
2
In practice, it is only the interaction between the control volume and
flowing out of the rocket is the same as that of the unreacted fuel.
cv T ddNt + Ncv ddTt = cp T ddNt + γ−1 RT (1 − α) ddNt .
its surroundings that differentiate this example from Section 80.
10 In that case, the outflow velocity will be
ıg
ıg
q
q
3
If the gas is ideal, even that difference no longer exists, as the 8 Let us substitute in cp − cv = R , which gives us
2(Uc-v −Hout )
rx H
≃ −2∆
= 2229 m s-1 ,
υ=
γ
ıg
Mw
Mw
temperature is then the same on both sides of the valve.
Ncv ddTt = RT ddNt + γ−1 RT (1 − α) ddNt ,
where ∆rx H◦ = −2 × 241826 − 110527 + 365560 = −228619 J[a][a] and
4
Consequently, in an ideal gas there is no net heat exchange be- and then multiply the whole expression by dt /(c ıg TN ).
v
Mw = 92.054 g (values given for the reaction equation, not per mole).
tween the contents of the cylinder and the gas flowing out of it.
1
An ideal rocket engine is assumed to have the following charac9
Rearranging the equation slightly gives us the following expression:
11 This calculation is based on the assumption that U = H − pV ≃ H
5
However, the temperature of a real gas will fall as it passes through
teristics: Chemical equilibrium[a] in the combustion chamber; exhaust
dln T
for low-molecular-weight solid compounds, e.g. ammonium nitrate and
dln N = (γ − 1) + γ(1 − α) .
the valve (or rise if its state is outside the inversion curve shown in
gases follow the ideal gas law; flow is frictionless and adiabatic; and
§ 81
§ 82
carbon[a].
Figure 397).
uniform mass flow rate with a quasi-one-dimensional velocity field.
10 Compared with Section 80, this equation has the extra term γ(1 −
1
Figure 10.2 is a schematic diagram of a solid-fuel rocket. Set out a
Figure 10.3 on the preceding page shows two identical gas cylin526 / 1776 1
differential energy balance equation for the control volume shown, and 2 In spite of all of these assumptions, in many cases the numerical
ders being emptied in two different ways. Assume perfect heat ex- 6 This results in heat exchange between the gas and the contents of α) > 0 on the right-hand side, which tells us that the temperature drop
is greater in this case[a].
integrate it over the period that the rocket operates. Assume that the results are within 10% of the observed behaviour.
change between the gas that leaves the cylinder and the gas that re- the cylinder, which in turn changes the outflow conditions.
530 / 1776
525 / 1776 [a] The NBS tables of chemical thermodynamic properties. Selected values for inor- mains, and that heat loss through the wall is negligible. Derive expres- 7 Here it is worth noting that if the outflow pressure is sufficiently low,
chemical reaction C + NH4 NO3 → N2 +2 H2 O + CO has a constant
ganic and C1 and C2 organic substances in SI units. J. Phys. Chem. Ref. Data, 11(2):1– sions for the temperatures in the two cylinders as functions of time. the behaviour of the gas will match that of an ideal gas as described in
rate of reaction and that the state of the combustion chamber is station392, 1982.
ary. Derive an expression for the outflow velocity if all of the internal enWhich of the two methods of emptying results in the lowest final tem- Section 80, because the impacts of the gas’s non-ideal properties will [a] It is possible to obtain an exact answer, but to do so we first need to integrate the
[a] It is normal to assume either that the composition of the gas remains unchanged
γ−1
cancel each other out during the heat exchange.
ergy of the reactants is converted into kinetic energy.
differential equation. Hint: From the definition of α it follows that dln α = − γ (dln N +
through the nozzle (frozen equilibrium), or that the equilibrium is constantly changing [a] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data, perature?
.
(2−α1 )
N
529 / 1776 dln T ). Use this to show that ln αα12 (2−α
523 / 1776 throughout the nozzle (dynamic equilibrium). In real life, the truth lies somewhere Suppl., 14(1), 1985.
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= −2(γ − 1) ln N2 Substituting in p2 = p◦ =
2)
1
ENGLISH text is missing.Rocket engines
Repeat the operation from Section 80 on page 459,
but this time changing the direction of flow, so that
the cylinder is filled instead of being emptied. What
will be the final temperature in a helium cylinder if it is
filled using a high-pressure hose at 300 bar and 20 ◦ C?
The problem becomes simpler if you assume that the
cylinder has been evacuated prior to being filled.
Tore Haug-Warberg (Chem Eng, NTNU)
υ
3
In other words, it is hardly surprising that condensation forms when
you empty a compressed air cylinder.
1
From a thermodynamic point of view, there is little difference between a rocket engine and a gas cylinder being emptied.
a)
υ⋆
4
The size of the cylinder doesn’t affect the calculation, but that is
only because we have questionably assumed that the emptying process
is adiabatic.
3
Under stationary conditions, the energy equation can be integrated
to obtain EK ,out = Uc-v − Hout .
~
Q
υ≈0
In other words, the temperature inside the cylinder changes in the
same way as it does when a closed system expands adiabatically.
1
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
~ =0
∆S
10 Dynamic . . .
2
If a compressed air cylinder at T1 = 293.15 K and p1 = 10 bar is
emptied quickly until it reaches a final pressure of p2 = 1 bar, the final
temperature will be T2 = 151.8 K (assuming γ = 1.40).
15 Given an initial state T1 , N1 , the differential equation can be integrated to obtain
γ−1
ıg
2
T2 = T 1 N
.
N1
16
Temperature change
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Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
The figure was drawn using the Program 1.13 in Appendix 31.
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News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
§ 084
250
200
§ 83
§ 85
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10 Dynamic . . .
Expanded footnote [a]:
50
0
100
200
300
p [bar]
Figure 10.4: Temperature increase when filling a helium cylinder with four
different initial pressures: p1 = 1, 5, 25, 125 bar. The upper line shows the
theoretical limit for p1 → 0.
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Termodynamic methods TKP4175
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Quasi-steady state (2)
§ 084 Joule[1]’s experiment
1
V2
ıg
d(N1 cv T1 )
Figure 10.5: An isolated system consisting of two gas cylinders.
ıg dT1
dt
N1 cv
ıg
= cp T1
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Note that T2,t =0 = γ · T1t =0 .
12
This condition is necessary if
13
Hence T2 = γT1 at t = 0.
14
stiff.
02 September 2013
T2 = γ · 300 K
150
Expanded footnote [a]:
[a] Also see Section 83 on page 466.
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If N2 → 0 then dT2 / dt → ∞, which makes the set of equations
15 In order to avoid numerical problems, an adjusted heat transfer coefficient has been used in the Program 31:1.14
p
p
UA′ = UA min p21 , p12 .
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0.5
=
RT1
V1
RT2
V2
dN1
dt
dN2
dt
+
+
N1 R1
V1
N2 R2
V2
= sign(p2 − p1 )K
= sign(p1 − p2 )K
p
p
,
,
|p1 − p2 | ,
|p1 − p2 | ,
dN1
dt
-1 dN2
T2 ) N2 dt
= (γT1 − T1 ) N1-1
= (γT1 −
dT1
dt
dT2
dt
+ UA (N1 cv )-1 (T2 − T1 ) ,
+ UA (N2 cv )-1 (T1 − T2 ) .
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Termodynamic methods TKP4175
02 September 2013
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30
400
350
0
=
[a] It’s a different matter if we have to solve them by hand.
450
200
is to have a finite solution .
Termodynamic methods TKP4175
dp1
dt
dp2
dt
dN1
dt
dN2
dt
dT1
dt
dT2
dt
8
It is possible to reduce the number of equations from six to four, by
dp
eliminating e.g. dt2 and ddNt2 , but we will keep all six in order to emphasise their symmetry[a].
25
20
15
497 Hz
480 Hz
10
UA = 0
dT2
dt
ıg
Big cylinder
UA > 0
10 Dynamic . . .
9
This set of equations constitutes a classic initial value problem that
can be solved by any numerical method of integration, but it isn’t all
plain sailing.
+ UA (T2 − T1 ) .
Termodynamic methods TKP4175
300
250
16 This adjustment prevents artificially high heat transfer values when
t → 0, but it also means that the time-axis is non-linear.
17 The numerical results are shown in Figure 10.6 on the next page.
Tore Haug-Warberg (Chem Eng, NTNU)
dN1
dt
UA ≫ 0
N2 = 0 mol
T1 = 300 K
11
10 Dynamic . . .
p2 = 0 bar
mol
ıg
+ UA (T2 − T1 ) − cv T1
ıg
cv )T1 ddNt1
Small cylinder
T [K]
p1 V1
RT1
−
dN1
dt
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6
Dividing by N1 cv results in a pure temperature equation for the
small cylinder.
term to the right-
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
The initial values are as follows:
N1 =
ıg
(cp
dN1
dt
02 September 2013
By analogy we can obtain a temperature equation for the large
7
cylinder, and the differential equations can be summarised as
+ UA (T2 − T1 ) .
Expanding the left-hand side and moving the
hand side gives us:
=
p1 = 10 bar
dN1
dt
5
From this it follows that ( ddUt )1 + ( ddUt )2 = 0, which means that the
total energy of the system is conserved.
10
ıg
= cp T1
Termodynamic methods TKP4175
§ 084 Joule[2]’s experiment (2)
This makes use of the fact that the enthalpy rate can be expressed
~ 1 = − dN1 , which we know from the mass
~ 1 and that N
~ 1 = cp T1 N
as H
dt
balance equation for the cylinder.
11
§ 084 Joule[3]’s experiment (3)
10 Dynamic . . .
4
Combining the three energy equations gives us
dU dU ~ ~
dt 1 = Q − H1 = − dt 2 .
Termodynamic methods TKP4175
Tore Haug-Warberg (Chem Eng, NTNU)
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
2
In order to solve this problem, we need to know what happens to
the state variables T , p and N in each of the cylinders, in other words
six variables in total.
Expanded
footnote
: ideal
3
The differential equations for p and N can
be derived
from the
gas law and the equation for the valve given above, but the tempera[] James Prescott Joule, 1818–1889. English experimentalist.
tures require a bit more thought.
dt
Tore Haug-Warberg (Chem Eng, NTNU)
538 / 1776
Based on the energy balance for the small cylinder, and the proposed heat transfer model, we can say that
~
Q
10
02 September 2013
2
Let us moreover assume that the gas is flowing slowly out of the
smaller cylinder, and that both cylinders are in a quasi-steady state at
all times.
3
In the case of the larger cylinder, this assumption is questionable,
as you would expect the gas to be flowing quickly downstream of the
valve.
4
On the other hand, the kinetic energy will quickly be transformed
into thermal energy due to internal friction, so the assumption produces
accurate results over an extended period of time.
5
Thus, if the small cylinder is emptied over a long time period in relation to the typical delay time constant of the turbulence, the assumption will be valid.
6
This also means that E2 = U2 can be used as an approximation for
the energy of the large cylinder.
7
Finally we have to say something sensible about the valve.
8
It is out of the question to describe it in detail, as the situation there
is far too disordered.
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
10 Dynamic . . .
V1
Termodynamic methods TKP4175
However, if the small cylinder is much bigger than the control vol9
ume in the valve, then the latter will be in a quasi-stationary state
(in = out):
dU ~ ~
dt 1 = Q − H1 ,
dE ~
~
~
dt JT = H1 − (H2 + EK ,2 ) = 0 ,
dU ~
~
~
dt 2 = H2 + EK ,2 − Q .
Signal strength [dB]
T [◦ C]
[a] The set-up is the same as in James Prescott Joule’s famous experiment of 1843.
100
10 Dynamic . . .
1
Let us start by putting system boundaries on each side of the valve,
giving us three control volumes: The smaller cylinder, the valve and the
larger cylinder.
Two cylinders are connected in series as shown in
Figure 10.5 on page 474 . Calculate the temperature
changes in the cylinders if the flow rate through the
√
~ = K ∆p and the heat transfer between the
valve is N
~ = UA ∆T , where
two cylinders is assumed to be Q
UA is the overall heat transfer coefficient [J/K s]. The
smaller cylinder has an initial pressure of 10 bar and
a temperature of 300 K, while the larger cylinder has
been evacuated. The heat capacity ratio of the gas is
γ = 1.40. Use a total energy balance to show that
dU
dt = 0. Assume ideal gas behaviour. What will the
final state be if UA = 0? And if UA > 0?
150
Quasi-steady state
1
300
0
0.5
1
5
300
400
500
600
700
Frequency [Hz]
Figure 10.6: Simulation of Joule’s famous experiment to demonstrate that
the internal energy of a gas is unrelated to its volume. The graphs show
temperature variations over time for three different heat transfer coefficients
(time axis is not scaled).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
544 / 1776
Figure 10.7: Helmholtz resonance in a flask with the following dimensions:
∆V = 25 ml, V = 120 ml and L = 50 mm. The theoretical eigenfrequency
(blue) is 497 Hz, while the graph peaks at 480 Hz. The discrepancy is due
to the fact that the effective mass of the oscillating gas plug is bigger than
V=
ˆ AL suggests.
By increasing
L to
L ′ =TKP4175
53.5 mm we02get
a more
∆Haug-Warberg
Tore
(Chem Eng, NTNU)
Termodynamic
methods
September
2013 realistic
548 / 1776
result (red).
News Emptying a gas cylinder Filling a gas cylinder Acoustic resonance Olds
News Incompressible flow Compressible flow Olds
News Incompressible flow Compressible flow Olds
ENGLISH text is missing.Helmholtz resonance
1
Leaving behind flow processes for a moment, let us look at a control volume that is disturbed by acoustic pressure waves in the surrounding gas (typically air).
2
The term Helmholtz resonator refers to a physical control volume
(container) that interacts with its surroundings through a small hole (or
neck) through which the gas can move freely and uniformly in both
directions.
3
ENGLISH text is missing.Eigenfrequency
1
Looking back
1
2
3
4
5
Substituting in z =
ˆ S-1 x gives us the system of equations Λz′ = z.
2
Without getting bogged down in the details, we can affirm that the
solutions are in the form zi ∝ exp(t /λi ) = exp(±i √t c ).
In practice the resonator can be a beer bottle, a guitar, a flute, etc.
3
The eigenvalues are imaginary (two complex conjugate eigenvalues with a real component of zero), so the oscillations of the wave are
not damped.
4
When surrounded by a spectrum of sound frequencies, the instrument will pick up a characteristic frequency that resonates with the “air
cushion” inside the control volume[d].
4
This agrees with our initial assumption of isentropic changes in
density.
5
The air cushion changes volume in response to the sound waves
coming from the outside, while the gas that oscillates back and forth in
the neck is assumed to act like a stiff plug (piston).
6
The resonant frequency is mainly determined by the mass ratio
between the air cushion and the gas plug.
ENGLISH text is missing.Eigenvalue analysis
these assumptions don’t hold, the theory becomes more complex. A didgeridoo is an
example of a control volume where the assumptions break down.
[d] Waves that oscillate in the same direction as they are travelling, as opposed to
transverse (surface) waves, where the centre of mass oscillates at right angles to the
wave’s direction of travel.
[d] The audio file was recorded in mono (2.5 s) using GarageBand 2 on an iBook Mac
with a built-in microphone. The file was then converted from AIFF to WAVE format
546 / 1776 using iTunes (again on a Mac), before being read and processed with Matlab running
in Windows XP; see program 31:1.15.
ıg
Filling:
=
10 Dynamic . . .
SΛS-1 x′ = x .
Depletion: T2 = T1
ıg
T2
New concepts
5
If we apply Euler’s formula e ix = cos x + i sin x we find that the
√
characteristic period of oscillation is τ = 2π c .
6
Alternatively, this solution can be stated in terms of the eigenfre1
A pragmatic interpretation of Newton’s 2nd law is that “force equals
quency
7
Sound propagates as longitudinal waves[d] consisting of alter- mass times acceleration”.
q
nately compressed and rarefied air, and for small amplitudes there is
∆V
1 υ⋆
ν=
ˆ τ-1 = 2π
2
In this case A (p − p◦ ) is a force that causes the mass ρAL of the
L
V .
very little dissipation of energy.
2
air plug in the neck to move in the opposite direction of the pressure
ˆ υL⋆ ∆VV .
In the above equation, we have substituted in c =
8
It is therefore important to assume that the density variations in the gradient:
7
Note that we do not need an exact solution to the wave equation
control volume are isentropic:
2
ρAL d 2x = −A (p − p◦ ) .
(besides which, the limits are unclear), and an eigenvalue analysis of
dt
∂p (dp )s ,n = ∂v s ,n dv ,
the problem is sufficient to give us the resonant frequency of the control
3
Combining the last two equations gives the following 2nd order
−N dv = A dx .
volume which dominates the audible sound.
linear differential equation:
2
8
Figure 10.7 shows the frequency diagram[d] for a homemade
2
υ⋆2 ρA
9
Here A is the cross-sectional area of the hole, N is the mole num− d 2x = vN
ˆ υL⋆ ∆VV x =
ˆ cx .
ρL x =
dt
Helmholtz resonator made of glass (specifically a 100 ml round-bottom
ber of the air cushion and dx is the displacement of the air plug in the
4
∆V and V represent the air plug and the control volume respec- flask from the organic chemistry laboratory).
neck (which is assumed to be incompressible).
tively.
9
The shape of the container is fairly irrelevant, but is important for
10 For small density variations, the state variables are roughly constant, so using Eq. 9.10 the integral of dp over the amplitude can be 5 The differential equation can be written in the standard form cx ′′ + there not to be space for standing sound waves inside the control volume.
x = 0. This solution to the equation involves restating it as
given as:

 


υ2 ρA
∂p 10 In air, a 480 Hz sound wave has a wavelength of approximately
A
 0 −c   x1′   x1 
x = ⋆vN x .
p − p◦ ≈ − ∂v s ,n N

 = 

 
 1
0.7 m, and the typical length of the resonator must be much less than
0   x2′   x2 
that.
ˆ x1′ .
545 / 1776 where x2 =
√ 11 If the length is increased to around a metre, the acoustics will
6
Show that the coefficient matrix has the eigenvalues λ = ±i c
change dramatically (becoming more like those of a didgeridoo).
and calculate the eigenvectors in S. The diagonal form can be formally
[d] The control volume must be small in relation to the wavelength of the sound, and
547 / 1776
expressed as:
the volume of the neck must not be much bigger than the control volume itself. If
N2 γ−1
N1
= T1
γ−1
p2 γ
p1
γ(α−1)+1
T1
α
Part 11
Limit value: lim (T2 ) = γT1
α→∞
Helmholtz resonator: cx ′′ + x = 0
√
ODE form (eigenvalues λ = ±i c ):



 
 0 −c   x1′   x1 

  ′  = 

1
0
x
x 
Open control volumes
2
2
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
549 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Incompressible flow Compressible flow Olds
Termodynamic methods TKP4175
02 September 2013
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1
Pipe flow control volume
2
Velocity profiles
3
Laminar versus turbulent flow
4
The steady state
5
Incompressible flow
6
Bernoulli’s equation
7
Ideal gas compression work
8
Joule–Thomson coefficient
9
Linde plant
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News Incompressible flow Compressible flow Olds
Open control volumes
11 Open . . .
11 Open . . .
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02 September 2013
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News Incompressible flow Compressible flow Olds
Open control volumes (2)
11 Open . . .
§ 085
§ 84
§ 86
11 Open . . .
2
For example, it is possible to fill and empty the volume at the same
time, which will later enable us to study stationary processes.
1
It will therefore be assumed that all inflows and outflows have uni4
form properties measured along (or at) their respective control surfaces.
3
In this chapter we will focus primarily on describing the control volume itself, and less on properly understanding the flow at the inlets and
outlets.
~
Q
~ out
E
dE
dt
~ ın
E
What is the ratio between the mean velocity b
υ and the
maximum velocity υ◦ of laminar flow? The velocity
profile is defined by υ(R ) = υ◦ (1 − (R /R◦ )2 ); see
Figure 11.2 on the next page. What is the equivalent
ratio between the square of the mean velocity b
υ2 and
the mean square velocity υb2 ?
5
For practical reasons we will also assume that the control surfaces
are at right angles to the direction of flow.
~
W
6
Incidentally, these assumptions are so fundamental, and above all
so hard to get around, that in problem formulations they are often only
alluded to.
7
Before going further, we will therefore spend a moment to reflect
on them and to think about their practical consequences.
8
In particular, we will see how, in the absence of the assumption of
uniformity, the rates of energy and mass transport would diverge.
Figure 11.1: A simplified open control volume.
With only one inlet and one outlet, the control volume shown in Figure 11.1 is of limited validity, but in spite of that the model is general
enough for many practical purposes.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
552 / 1776
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News Incompressible flow Compressible flow Olds
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02 September 2013
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News Incompressible flow Compressible flow Olds
υ(R )
11 Open . . .
The mean velocity can be obtained by calculating the first moment
of the axial velocity component about R = 0:
1
R
2
b
υ=
R R◦
0
R R◦
0
υ2πR dR
2πR dR
=
2πυ◦
πR◦2
R
R◦ 0
1−
R 2
R◦
R dR =
υ◦
R◦2
2
R
R◦ 0
1−
R 2
R◦
b
υ = υ◦
R◦4
2R◦2
R◦2
02 September 2013
§ 085 Velocity profile (2)
554 / 1776
11 Open . . .
4
The mean square velocity can be calculated using the same principles, which gives us the following result:
υb2 =
2
dR .
R R◦
υ2 2πR dR
R R◦
2πR dR
0
0
=
υ2◦
R◦2
2
R
R◦ 0
1−
2
R 2
dR 2
R◦
=
υ2◦
3
.
Thus, for the square velocities, b
υ2 = 34 υb2 = 14 υ2◦ .
Integrating between the limits gives us the ratio
R◦2 −
Termodynamic methods TKP4175
News Incompressible flow Compressible flow Olds
§ 085 Velocity profile
R◦
Tore Haug-Warberg (Chem Eng, NTNU)
While we use the mean velocity to calculate the transport of in6
ternal energy and potential energy, we cannot use it for kinetic energy
transport.
5
This result is relevant to energy balances that include kinetic energy transport.
=
υ◦
2
.
3
For laminar pipe flow, the mean velocity is half the maximum velocity component.
7
In fact, if we were to use it for laminar flows, the kinetic energy
would be underestimated by as much as 25%.
8
In the rest of this chapter we will therefore stick to uniform (i.e.
turbulent) velocity profiles where b
υ = υ◦ and b
υ2 = υ2◦ .
9
This means that we can leave out the line above the variable and
just use the symbol υ to mean velocity.
Figure 11.2: Parabolic velocity profile for laminar pipe flow
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
555 / 1776
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Termodynamic methods TKP4175
02 September 2013
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News Incompressible flow Compressible flow Olds
§ 086
§ 85
News Incompressible flow Compressible flow Olds
§ 87
11 Open . . .
News Incompressible flow Compressible flow Olds
§ 086 Energy balance
11 Open . . .
In process engineering, a control volume will normally have virtually constant kinetic and potential energy. The total energy of the volume will therefore be E = U .
1
Set out the energy balance for the control volume
shown in Figure 11.1 on page 483. The balance should
include internal energy, kinetic energy and potential
energy. Assume that the inflows and outflows have
uniform properties over the cross-sectional area of flow.
What happens if the latter assumption does not hold?
Termodynamic methods TKP4175
02 September 2013
~ −W
~ )ın − (E
~ )out + Q
~ + pV
~ + pV
~
= (E
§ 86
d(Mu)
dt
~ −W
~ ın ( 1 υ2 + gz + h )ın − M
~ out ( 1 υ2 + gz + h )out + Q
~ .
=
ˆ M
2
2
(11.2)
(11.1)
~ ,W
~,H
~,Q
~ are energy rates [J/s] associated with the transHere E
port of mass, heat and work.
10 In practice this assumption means that υ = υ◦ and υ2 = υ2◦ , where
υ◦ is the net flow velocity.
~ −W
~ )ın − (E~K + E~P + H
~ )out + Q
~ .
=
ˆ (E~K + E~P + H
558 / 1776
~ −W
~ ın (ek + ep + h )ın − M
~ out (ek + ep + h )out + Q
~
=M
~ =
9
The mass flows in and out of the system are by definition M
A ρυ, but remember the assumption of a uniform velocity profile from
Theme 85.
~ −W
~ + E~K + E~P + p V
~ + E~K + E~P + p V
~
~ )ın − (U
~ )out + Q
=
ˆ (U
Tore Haug-Warberg (Chem Eng, NTNU)
News Incompressible flow Compressible flow Olds
§ 087
6
If the intensive properties of the fluid vary over the cross-sectional
area of flow Ω, they must be considered average energy
rates,
R
~ = hT n dΩ, where n is
7
which can be determined by integrating H
Ω
the unit normal vector of the control surface and h [J/m2 s] is the fluid’s
enthalpy flux vector.
2
This requires the control volume to be stationary, without any kinetic energy bound up as eddies or turbulence in the flow medium.
3
Moreover, the centre of gravity of the control volume must not move
up or down.
dU
dt
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02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News Incompressible flow Compressible flow Olds
§ 88
11 Open . . .
11 Open . . .
8
If we introduce the specific variables e = M -1 E and h = M -1 H ,
Eq. 11.1 can be expressed in the form
For a control volume that satisfies these conditions, the energy
4
balance can be expressed as
5
Tore Haug-Warberg (Chem Eng, NTNU)
§ 086 Energy balance (2)
§ 087 Stationary state
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02 September 2013
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News Incompressible flow Compressible flow Olds
11 Open . . .
ENGLISH text is missing.Stationary processes
For stationary processes the state variables of the control volume
are constant over a time period ∆t ≫ τ, where τ is a characteristic time
constant for the process.
1
2
In other words, the change in the state variables shouldn’t reflect
any transient conditions in the control volume, cf. a quasi-static process.
3
For example, it is a good approximation to assume that a heat
exchanger is stationary, even if the temperature of the cooling water
varies with the seasons.
562 / 1776
§ 088
§ 87
§ 89
11 Open . . .
1
For a stationary process, dU / dt = 0 and dM / dt = 0, which
~ ın = M
~ out .
means that the mass balance can be expressed M
2
Set out the energy balance for a stationary control
volume. Can the control volume be said to be stationary
if the states of the inflows and outflows vary as functions
of time? If so, what conditions must be fulfilled?
~ ın and M
~ out in Eq. 11.2 gives us
Eliminating the flow rates M
( 12
2
υ + gz + h )ın + q =
( 12
Assume that the flow through the control volume 11.1
on page 483 is steady, incompressible and lossless .
Show that the energy balance is an expanded Bernoulli
equation. Use this equation to look at how h , ek and ep
vary along the pipeline in Figure 11.3 on the following
page. A qualitative analysis is sufficient. How do T , p
and µ vary along the same pipeline?
2
υ + gz + h )out + w .
Expanded footnote [a]:
[a] Lossless means that the flow is frictionless (inviscid), and that there are no temperature gradients in the fluid, although heat can be transferred between the control volume and its surroundings.
3
The values of e , h , q and w are specific, i.e. they are stated in
[J/kg].
4
Conventionally the final result is expressed in ∆ notation:
1
2
∆υ2 + g ∆z + ∆h = q − w .
(11.3)
5
If there are no chemical reactions in the volume, the energy balance can alternatively be expressed in molar ∆ notation:
1
2
Mw ∆υ2 + Mw g ∆z + ∆h = q − w .
(11.4)
6
It probably makes sense to stick to the symbols h , q and w , although here the unit is [J/mol] rather than [J/kg] as in the previous
equation.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
561 / 1776
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News Incompressible flow Compressible flow Olds
z
υ◦
1
4
02 September 2013
563 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Incompressible flow Compressible flow Olds
§ 088 Bernoulli[1]’s equation
υ◦
1
Expanded footnote :
[] Daniel Bernoulli, 1700–1782. Dutch mathematician (son of Johann Bernoulli).
11 Open . . .
For lossless flow, δqrev = T ds .
564 / 1776
§ 088 Bernoulli[2]’s equation (2)
11 Open . . .
8
Provided that we know the state of the fluid at a given point in the
streamline, it is possible to calculate the pressure along the rest of the
streamline as a function of the cross-sectional area of flow (a simple
mass balance gives the velocity) and the elevation.
5
Here the unit of w is [J/m3 ], but that shouldn’t cause any major
confusion.
6
In the absence of chemical reactions, Bernoulli’s equation can also
be expressed in molar form:
1
2
4
1
2
02 September 2013
Mw ∆υ2 + Mw g ∆z + v ∆p = −w .
(11.6)
7
Bernoulli’s equation expresses a direct relationship between the
flow profile of p and the equivalent profiles of υ and z .
3
For a compressible fluid, i.e. a fluid with a constant specific volume
that is independent of the pressure, the differential can be integrated to
give (∆h )n,rev = q + v ∆p . If we substitute this into Eq. 11.3 and divide
by v [m3 /kg], it gives us the expanded form of Bernoulli’s equation.
x
Termodynamic methods TKP4175
News Incompressible flow Compressible flow Olds
2
Substituting this into (dh )n = T ds + v dp gives us (dh )n,rev = δq +
v dp .
~
W
~
Q
Termodynamic methods TKP4175
ρ∆υ2 + ρg ∆z + ∆p = −w .
9
The associated temperature profile follows indirectly from the assumption δqrev = T ds .
11
10 For example, in the case of isobaric flow there is the simple relationship cp dT = δqrev .
If we know both T and p , we can work out the profile of µlıq (T , p ).
The expansion is trivial and only involves the work term w :
(11.5)
14 The calculation involves using the simplifications µlıq ≈ µsat (T ) +
v (p − p sat (T )) and µıg = µ◦ (T , p◦ ) + RT ln(p /p◦ ).
It is also possible to calculate the difference between this potential
and the associated potential for a hypothetical ideal gas (steam) at the
same temperature and pressure as the liquid.
13 The steam is not physically present, as that would contradict the
assumption of incompressibility, but the difference ∆µ =
ˆ µlıq − µıg has
an important function: It tells us where in the system there is a risk of
boiling, i.e. where ∆µ < 0; see Figure 11.4.
12
16
15 If we apply them to the equilibrium state µlıq = µıg , we get µsat =
RT ln(p sat /p◦ ).
Hence ∆µ =
ˆ µlıq − µıg = RT ln(p sat /p◦ ) + v (p − p sat ).
17
Chapter 15 on page 791 discusses this topic in greater detail.
Figure 11.3: Simplified diagram of Bernoulli flow.
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Termodynamic methods TKP4175
02 September 2013
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News Incompressible flow Compressible flow Olds
∆p
0.5
News Incompressible flow Compressible flow Olds
∆T
15
News Incompressible flow Compressible flow Olds
Cavitation
11 Open . . .
§ 089
§ 88
§ 90
11 Open . . .
0
[mK]
[bar]
a)
−0.5
b)
10
5
−1
You forget to shut the damper of the fireplace on the
ground floor. The temperature difference between
the house and the outside air produces a convection
current in the chimney which means that you have to
use more heating to maintain your desired indoor
temperature. Express the extra heating required as a
function of the outdoor temperature. Assume that the
temperature of the air passing through the chimney is
constant from the inlet to the outlet, and that Bernoulli’s
principle ∆p + 12 ρ∆υ2 + ρg ∆z = 0 gives a reasonable
estimate of the flow velocity. The indoor temperature
is 20 ◦ C. The chimney has a standard 9” flue and
runs through two full stories plus a cold loft.
The total elevation is z = 10 m, the heat added is 55 kJ m-3 and
the work done is 75 kJ m-3 (see Program 1.20 in Appendix 31 for further
details).
1
2
Using Bernoulli’s equation to interpret variation in chemical potential isn’t a common approach, but Figure 11.4 confirms the usefulness
of the exercise.
3
The first three sub-figures show the p , T and µ profiles for flowing
water with the inlet conditions Tın = 50 ◦ C, pın = 1 bar and υın = 8 m s-1 .
This means that ∆µ varies considerably along the streamline, and
6
we should particularly note the coloured area where the difference is
negative (this represents the top of the pipeline just before it gets wider).
0
4
This results in an interesting µlıq profile, but more importantly it
gives us the graph shown in Figure 11.4d , which shows the difference
in chemical potential between liquid water and a hypothetical steam
phase at the same temperature and pressure.
5
In the liquid phase it is weakly dependent on pressure and strongly
dependent on temperature, whereas in the steam phase it is strongly
dependent on both temperature and pressure.
µvap − µlıq
∆µlıq
0
10
d)
[kJ/mol]
[J/mol]
c)
−1
−2
−3
7
Here the steam phase is more stable
both phases will be present in this area.
5
Expanded footnote [c]:
[c] As an aside, it should be noted that there is an equivalent test for multicomponent
systems, but it involves a more advanced criterion called the minimum tangent plane
distance; see Chapter 21 on page 1312.
than the liquid phase, so
8
This undermines the foundations for Bernoulli’s equation, and in
practice the steam bubbles can cause cavitation problems.
0
−5
Figure 11.4: T , p , µ profiles for the Bernoulli flow shown in Figure 11.3 (a
detailed explanation is given in the body of the text).
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ENGLISH text is missing.Chimneys
1
Dz
1
11 Open . . .
4
The velocity at the inlet of the chimney can be safely assumed to
be zero. The velocity right by the inlet is of course non-zero, but the
assumption is valid because somewhere or other in space at the same
height as the inlet the velocity is (more than) close enough to zero for
our purposes.
5
Remember that Bernoulli’s equation applies along a streamline,
which means that this assumption is consistent with the theory that
underpins the equation.
6
Also assume that the density of the air depends on temperature,
but not on pressure.
7
This may appear to be a strange assumption, but the difference
in air density due to the height difference ∆z is small compared with
the difference in density between the cold air outside and the warm air
inside the chimney[c].
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[c] In fluid mechanics, assuming incompressible gas flow and constant thermal expansivity is called the Boussinesq approximation. Named after the French mathematician
Valentin Joseph Boussinesq (1842–1929).
We know that the chimney runs through 2 stories.
In addition there is the chimney stack and the section running
through the loft.
2
3
4
In total this gives us approximately 8 m of chimney.
The cross-sectional area of flow is 9 × 9” i.e. A =
§ 089 Boussinesq flow (2)
6
Bernoulli’s equation can therefore be expressed 0 = ρınsıde g ∆z −
pM
ρoutsıde g ∆z + 12 ρınsıde υ2out , or since ρıg = RTw :
q
υout = 2g ∆z TT◦ − 1 .
2
Kinetic:
ρ∆υ
Potential:
ρg ∆z
Mechanical:
∆p
Tore Haug-Warberg (Chem Eng, NTNU)
1
2 ρınsıde
−
⇒
− ρoutsıde g ∆z .
⇒
~
Vp
RT◦
(T◦ − T ) =
υ Ap
CP out
RT◦
( T◦ − T ) .
We can eliminate the velocity υout by combining the last two equations:
q
2g ∆z
3/2
~ = CP Ap
Q
RT◦
T ( T◦ − T )
ρınsıde g ∆z .
Termodynamic methods TKP4175
02 September 2013
4
2
~ = CP N
~ (T◦ − T ) = CP
Q
9
υ2out .
⇒
6
ing:
The various pressure components of Bernoulli’s equation are
therefore:
5
1
2
8
Now we can calculate the amount of extra energy needed for heat-
7
0.052 m2 .
10
11 Open . . .
Q [kW]
§ 089 Boussinesq flow
Figure 11.5: Natural convection through a chimney flue.
A cross-section of the chimney is shown in Figure 11.5.
2
The pressure differential between the air inside and outside the
house is taken to be equal to the hydrostatic pressure gradient in the
atmosphere multiplied by the height ∆z .
3
This assumes that the convection current has not resulted in significant negative gauge pressure in the building (this is no doubt debatable).
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0
−20 −10
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0
10
20
30
Outdoor temperature [◦ C]
8
The heat lost from the room due to the convection current in the
chimney is thus proportional to the inside/outside temperature differential, and to the flow velocity in the chimney.
Figure 11.6: Convection losses through an open chimney at two different
indoor temperatures (18 ◦ C and 25 ◦ C) and for two different chimney lengths:
4 m (dotted) and 8 m (solid).
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§ 90 Hydro-power
1
Let us start by defining an open control volume around the waterfall; see Figure 11.7.
2
The water does not do any mechanical work apart from the pV
work related to the inflows and outflows.
3
Provided that there is little evaporation, as is the case in Norway, it
doesn’t exchange any heat with its surroundings either.
4
Under these conditions, the energy balance 11.4 simplifies to
1
2
Mw ∆υ2 + Mw g ∆z + ∆h = 0 .
5
We haven’t said anything about the size of the control volume, and
the equation
Figure 11.7: Control volume over a water
is just as applicable to the whole waterfall as to a small section of it.
§ 091
6
Assume that the vertical drop is 300 m and that the river’s flow
velocity is the same above and below the waterfall:
∆h = −Mw g ∆z .
7
The process takes place at constant pressure and the heat capacity remains virtually constant, i.e. CP ∆T ≈ −Mw g ∆z .
8
If we insert numerical values we get ∆T ≈ 0.018 · 300 · 9.82/75 =
0.7 K.
§ 90
§ 92
9
The work done by the turbines can
R be calculated from the energy
balance 11.6, assuming that qrev = T ds and that the water can be
treated as an incompressible liquid.
10 There is little change in the pressure of the surroundings between
the reservoir and the turbine outlet, and the pressure drop in the pipeline
can be largely ignored (less than 5% loss of pressure):
w = −Mw g ∆z .
11 An ideal turbine captures all of the available kinetic energy in the
water, and the temperature of the water therefore remains unchanged
as it travels through the power plant.
ENGLISH text is missing.Fireplaces
The extra heating required has been drawn using the Program 31:1.21 and is shown in Figure 11.6 on the previous page.
1
12 As a result, the temperature is 0.7 K lower with the turbine installed
than without it.
13 However, this temperature reduction is so small that any thermal
§ 90
pollution is probably due to other factors.
2
As can be seen, the impact is frightening at low outdoor tempera- 1
In the mainstream press, it is occasionally claimed that hytures, and the moral of the story should be clear: Seal your house, but dropower stations cause thermal pollution of the surroundings. Is it 14 Probably a much bigger problem is the release of cold water from
not to the extent that you damage your health by having a poor indoor possible that the imbalance is caused by the work done by the turbines, the bottom of the reservoir in the summer because the water enters a
climate (this is a difficult balancing act).
or could there be other factors at play? Hint: First set out the energy bal- biotope that is adapted to a higher water temperature.
3
Note that the impact of the chimney height is also evident from ance for the waterfall. Then calculate the temperature change across 15 In winter, conversely, the water from the bottom of the reservoir will
the turbine relative to the temperature change in free-falling water.
Figure 11.6.
be too warm.
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11 Open . . .
Work of steady flow
11 Open . . .
Work of steady flow (2)
1
Let us hold onto the idea of a closed control volume, which in this
case is a hypothetical volume element moving with the fluid flow, but
now in addition to the work of compression, the control volume performs flow work by displacing the gas downstream of it, while the gas
upstream of it performs flow work on the control volume itself.
11 Open . . .
2
The basic equations describing the isentropic compression of an
ideal gas in a closed system were derived in Theme 69 on page 421.
Here we will extend our analysis to include the compression of a gas
under stationary flow conditions.
The behaviour of a steadily flowing gas when it is
compressed or allowed to expand is of particular interest
in process engineering. Replace the incompressible
liquid in Theme 88 on page 494 with an ideal gas. Once
again assume that the flow is frictionless, and ignore
any heat transfer between the control volume and its
surroundings (adiabatic flow). Derive an expression for
the shaft power needed to compress the gas from its
state at the inlet Tın , pın to its state at the outlet Tout , pout .
Assume that the composition and heat capacity of
~.
the gas are constant. The molar flow rate is N
These two extra contributions are shown in Figure 11.8 as −(pv )out
and +(pv )ın , which represent the flow work done by and on the system
respectively.
pout
−(pv )out
+(pv )ın
3
pın
vout
vın
11.8: Isentropic compression of a steadily
R out flowing gas: ws =
RFigure
out
p dv − (pv )out + (pv )ın or alternatively ws = − ın v dp .
ın
5
4
The net amount of work involved in displacing the contents of the
control volume followed by compressing and moving the gas— from the
inlet to the outlet — is equal to the sum of the three components.
Thus, ws =
R out
ın
p dv − (pv )out + (pv )ın . Partial integration also
R out
v dp .
ın
enables us to state that ws = −
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News Incompressible flow Compressible flow Olds
News Incompressible flow Compressible flow Olds
Work of steady flow (3)
11 Open . . .
7
The result is mathematically correct, but by assuming the existence
of a closed control volume moving with the fluid flow in the velocity
field we have gone slightly against the terms in which the problem was
originally formulated.
A graphical
figure[c].
6
interpretation of the three components is shown in the
[c] The explanation for this is as follows: Due to adiabatic compression work the
internal energy of the closed
R control volume increases
R with (∆u)s ,n = −R p (dv )s ,n .
Partial integration gives us p (dv )s ,n = ∆(Rpv )s ,n − v (dp )s ,n . Combining the two
equations produces (∆u
ˆ u + pv it
R )s ,n + ∆(pv )s ,n = v (dp )s ,n , and because h =
follows that (∆h )s ,n = v (dp )s ,n , as stated in the body of the text. The temperature increase that takes place during compression means that ∆(pv )s ,n is positive.
Hence (∆h )s ,n is greater than (∆u)s ,n . In other words, more energy is required
to compress a fluid flowing through an open system than a fluid in a closed system. As an aside it should be noted that the ratio between the two work terms is
(∆h )s ,n /(∆u)s ,n = −R v (dp )s ,n /R v (dp )s ,n , which for an ideal gas with constant heat capacity becomes (∆h )s ,n /(∆u)s ,n = cp ∆RT /cv ∆T =
ˆ γ. This means that the ratio between the two integrals −R p (dv )s ,n and v (dp )s ,n in Figure 11.8 is also γ.
News Incompressible flow Compressible flow Olds
§ 091 Isentropic compression
11 Open . . .
~ out − W
~.
~ ın − H
From the energy Eq. 11.1, it follows directly that 0 = H
Or, expressed in molar terms, ∆h = −w ; also see Eq. 11.4. Assuming
ıg
ideal gas behaviour, this becomes (dh )s ,n = (RT /p ) dp .
1
§ 091 Isentropic compression (2)
6
ıg
5
If we set the upper and lower limits of integration, −ws expresses
the work required to compress one mole of a gas from its inlet conditions
Tın , pın to its outlet pressure pout and outlet temperature Tout > Tın .
~ the total power required
At any given (stationary) molar flow rate N
is
2
The difference in enthalpy at a given composition is (dh )n = T ds +
v dp , which for frictionless, adiabatic flow can be simplified to (dh )s ,n =
v dp .
~ s-s,ıg = Nw
~ ıg =
W
s
S
8
We will therefore derive the stationary work term again based on
an independent analysis of an open system.
From Theme 61 on page 382 we are familiar with the adiabatic
relationship (T /Tın )s ,n = (p /pın )(γ−1)/γ , which gives us the following
equation:
3
ıg
(dh )s ,n =
RT
p
ıg
(dp )s ,n =
If we integrate this we get
4
ıg
ıg
−ws = (∆h )s ,n = RTın
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11 Open . . .
In the case of totally reversible flow, the heat transfer is reversible,
i.e. δQrev = T dS , whereas for adiabatic flow δQ = 0.
3
poutR/pın
1
p
pın
−1
γ
γ−1
γ
d
dp = RTın
p
pın
=
γ
γ−1
p
pın
RTın
Termodynamic methods TKP4175
−1
γ
d
p
pın
p
pın
γ−1
γ
.
2
In fluid engineering, the terms totally reversible (externally and internally reversible, frictionless) flow and adiabatic flow are used.
1
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12
10 The work of compression in Eq. 11.7 is therefore greater than the
equivalent work in Eq. 9.2.
11 We can determine the ratio between the two terms by substituting
an isentropic relationship from Theme 61 into Eq. 9.2 and then comparing our result with eqution 11.7.
What we find is that (wss-s /wsc-s )ıg = γ.
Tore Haug-Warberg (Chem Eng, NTNU)
11 Open . . .
§ 092
1st law of thermodynamics: (dU )n = δQ − δW
Fundamental energy equation: (dU )n = T dS − p dV
Combining these equations gives us (T dS )n = δQ + p dV − δW .
13 In other words, isentropic-adiabatic flow is a special case of totally
reversible flow.
11 Hence a control volume that moves with the fluid flow cannot do
any work against friction, and the process must be reversible.
12 On the other hand, if the process is totally reversible, p (dV )n = δW
even if the entropy of the fluid element doesn’t remain constant.
From this it follows that T dSırr > 0.
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§ 092 Isenthalpic flow
1
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§ 093
§ 92
§ 94
11 Open . . .
[a] James Prescott Joule, 1818–1889. English experimentalist.
4
It is important to express the equation in terms of mass, and not
in terms of moles. This is because the problem formulation does not
make any assumptions about the composition.
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11 Open . . .
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11 Open . . .
3
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Termodynamic methods TKP4175
02 September 2013
Expanded footnote :
[b] William Thomson, 1st Baron (Lord) Kelvin, 1824–1907. Irish–Scottish physicist and
engineer.
(dH )n = T dS + V dp ,
C
∂p (dS )n = TV dT + ∂T V ,n dV ,
∂p ∂p (dp )n = ∂T V ,n dT + ∂V T ,n dV .
We will come back to this question in Chapter 17:17.7 in conjuction
with the calculation of adiabatic flame temperatures.
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2
From Chapter 8, and more specifically equations 8.1, 8.12
and 8.16, we know that
Derive an expression for the temperature change that
takes place when a flowing fluid is forced through a
valve, if the fluid can be described by an equation of
state in the form p = (T , v ). What result does this
give you if the fluid is an ideal gas? What about for a
gas that obeys the Van der Waals equation?
5
Tore Haug-Warberg (Chem Eng, NTNU)
§ 93
§ 093 Joule[3]–Thomson[4]
Expanded footnote :
This equation holds for turbulent and adiabatic flow in any fluid,
provided that the acceleration and change in potential energy can be
ignored.
02 September 2013
1
The question requires us to find a suitable expression for
(∂T /∂p )H .
2
3
13 Incidentally, the same result is derived using a slightly different
method in footnote on page 507.
14 In the reverse case, i.e. when pout < pın , work of expansion will be
done by the open system.
15 We can calculate this work by swapping the integration limits in the
above equation.
16 After a bit of manipulation we can obtain the same function, but as
~ ıg > 0.
the limits have been changed W
S
Termodynamic methods TKP4175
News Incompressible flow Compressible flow Olds
~ ın = H
~ out .
From Eq. 11.1 it follows that H
If we restate this in terms of the specific enthalpy, ∆h = 0,; also
see Eq. 11.3.
§ 91
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11 Open . . .
(11.7)
.
In Themes 88 and 91, the flow was totally reversible.
We are now going to look at the opposite extreme,
which is completely turbulent flow. Here the entropy
of the fluid changes, regardless of whether there is
any heat exchange between the control volume and
its surroundings. This kind of flow model gives a
good description of e.g. junctions, valves and burners.
Provided that the flow is stationary and adiabatic, and
no mechanical work is done, the energy balance can
be expressed in very simple terms. What are they?
7
There is often a great deal of confusion over the distinction between
totally reversible, adiabatic and isentropic processes, but as usual in
thermodynamics, the terms may be vague but the maths isn’t.
8
Once we have these definitions in place, we can derive most of
what we need from the following two equations:
4
If the temperature of the fluid changes due to friction, the flow cannot be totally reversible, since the entropy changes without any simultaneous heat transfer between the fluid and its surroundings.
5
#
9
That requires less work, since the gas remains in the control volume, and doesn’t need to be moved in the direction of flow.
For an isentropic process, (δQ )n = δW − p dV , and if the process
is also adiabatic, the work done on the system must be exactly equal to
the pressure–volume work: (p dV )n = δW .
Adiabatic: δQ = 0
Isentropic: dS = 0
Loss-free: δQ = T dS
γ−1
γ
~ ıg < 0. We can see
but note carefully: Since pout > pın , then W
S
that the delivered work done is negative, which means that the supplied
work of compression must be positive, as you would expect.
6
However, the flow can still be adiabatic, i.e. not involve any exchange of heat with the surroundings.
9
pout pın
News Incompressible flow Compressible flow Olds
Reversible vs frictionless (2)
10
In isentropic flow, on the other hand, δQ = 0 and dS = 0:
"
~ ın 1 −
NRT
8
Eq. 9.2 was an equivalent expression for the work required to compress a closed system.
pout /pın
.
02 September 2013
γ
γ−1
7
News Incompressible flow Compressible flow Olds
Reversible vs frictionless
1
RTın p p
pın
11 Open . . .
588 / 1776
If we eliminate dS and dV from the equations we get
h
i
∂p -1
∂p ∂p (dH )n = CV dT + T ∂T V ,n ∂V T ,n dp − ∂T V ,n dT + V dp
∂p -1
∂p -1
∂p 2
∂p = CV − T ∂T V ,n ∂V T ,n dT + T ∂T V ,n ∂V T ,n + V dp .
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News Incompressible flow Compressible flow Olds
News Incompressible flow Compressible flow Olds
§ 093 Joule[5]–Thomson[6] (2)
11 Open . . .
4
For an isenthalpic process (dH )n = 0, so the Joule–Thomson coefficient is
∂p ∂p T ∂T V ,n + V ∂V T ,n
=
.
(11.8)
κ(T , V , n) =
ˆ ∂∂Tp
2
H ,n
∂p ∂p T ∂T V ,n − CV ∂V T ,n
5
From Eq. 8.32 we can see that the denominator can be expressed
as CP (T , V , n), but that is not of any relevance to us here and now.
6
For an ideal gas (∂p /∂V )T ,n =
which means that κıg = 0.
−NRT /V 2
and (∂p /∂T )V ,n = NR /V ,
News Incompressible flow Compressible flow Olds
Pressure–temperature drop
~
Q
11 Open . . .
1
Assume that the depressurised gas in Theme 93 is heat exchanged
with the gas entering the system as shown in Figure 11.9 on the following page.
~
Q
2
If the Joule–Thomson coefficient is positive, the exchange of heat
will cause the valve inlet temperature to fall.
3
Eventually the temperature will become so low that some of the
gas will condense and in order to achieve a steady state the liquid must
be tapped.
6
Note that in the case of light gases such as hydrogen, helium and
neon, the Joule–Thomson coefficient is negative at room temperature.
4
This is the principle that allows a Linde plant to produce liquid air.
5
To improve efficiency, it is common to use a turbine to precool the
air before forcing it through the valve.
In other words, the enthalpy of an ideal gas is independent of its
pressure.
7
7
Under those circumstances, depressurisation leads to a rise in
temperature, which means that it is essential to precool the gas before passing it through the valve.
Figure 11.9: Schematic diagram of a Linde plant for producing liquid air.
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News Incompressible flow Compressible flow Olds
§ 94 Linde[7] plant
§ 095
1
In the absence of heat loss, the process becomes an isenthalpic
expansion.
2
Perfect heat exchange has also been assumed, which means that
the inflow and outflow temperatures are the same.
3
Under steady state conditions, the inflow and outflow rates will also
be the same.
4
This allows us to set out the following four relationships:
~ out , pın > pout ,
~ ın = H
H
§ 94
§ 96
Tın = Tout ,
11 Open . . .
~ ın = N
~ out .
N
5
Overall the flow is isenthalpic, which means that
dh = ∂∂Th p dT + ∂∂ph dp = 0, where h is the molar enthalpy as
T
~ .
~ = Nh
seen in H
6
Since the fluid temperature is the same at the inlet and outlet, it
follows that dT = 0.
7
We also know that fluids always flow through valves down a pressure gradient, and hence that dp < 0.
8
That allows us to formulate the necessary condition for the process
to be stationary: ∂∂hp = 0.
T
9
This condition is only fulfilled in an ideal gas.
With a real gas, the system will get colder and colder if the Joule–
Thomson coefficient is positive, and conversely it will get hotter if ∂∂Tp <
h
§ 94
0.
1
Use an energy balance to show that the system shown in Figure 11.9 cannot reach a steady state unless: i) the fluid is an ideal 11 In a Linde plant, this cooling is counteracted by removing liquid
gas, or ii) the Joule–Thomson coefficient is positive and liquid is tapped straight after the valve.
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straight after the valve. Assume perfect heat exchange between the two
fluids and zero heat loss from the process as a whole.
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10
[g] Carl Paul Gottfried von Linde, 1842–1934. German engineer.
p [bar]
1.01325
100
300
77.347 K
−3405.3 (lıq)
+1547.6 (vap)
-
240 K
290 K
6967.1
8425.0
6070.4
5232.8
7837.9
7254.9
§ 095
Termodynamic methods TKP4175
gas
vap
lıq
(1 − α) + hout α ,
hın = hout
(11.9)
2
The liquid fraction α can be calculated from Eq. 11.9: 7837.9 =
8425.0(1 − α) − 3405.3α, and thus α = 0.0496.
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Tore Haug-Warberg (Chem Eng, NTNU)
News Incompressible flow Compressible flow Olds
11 Open . . .
For nitrogen this means that TJT,N2 ≈ 750 K.
In our process Tın = 290 K, which is comfortably below the inversion temperature.
6
The amount of liquid produced tends to zero when pın → pout ,
and at an inlet pressure of 2 bar the liquid fraction falls to a measly
α = 5.58e − 4.
If the inlet conditions are changed to 240 K and 300 bar, on the
other hand, then α = 0.1672.
8
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11 Open . . .
and υb2 =
υ2◦
3
υ◦
2
2
3
Turbulent: b
υ = υ◦ and υb2 = υ2◦
Steady state:
1
2
4
Bernoulli’s equation:
5
Ideal gas compression work:
6
Adiabatic: δQ = 0
7
Isentropic: dS = 0
8
Lossless: δQ = T dS
Tore Haug-Warberg (Chem Eng, NTNU)
γ
γ−1
11 Open . . .
ˆ
κ(T , V , n) =
ρ∆υ2 + ρg ∆z + ∆p = −w
~ ıg =
~ s-s,ıg = Nw
W
s
S
"
~ ın 1 −
NRT
Termodynamic methods TKP4175
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Joule–Thomson coefficient:
∆υ2 + g ∆z + ∆h = q − w
1
2
02 September 2013
Looking back (2)
9
Laminar: b
υ=
1
Termodynamic methods TKP4175
News Incompressible flow Compressible flow Olds
Looking back
7
Termodynamic methods TKP4175
11 Open . . .
where α is the molar liquid fraction under stationary conditions. For nigas
trogen at an inlet temperature of 290 K and pressure of 100 bar, hın =
7837.9 J mol-1 .
[h] Richard T. Jacobsen and Richard B. Steward. J. Phys. Chem. Ref. Data, 2(4):757–
922, 1973.
3
Theoretically liquid will be produced at the valve if the inlet temperature is lower than the inversion temperature TJT of the gas.
Tore Haug-Warberg (Chem Eng, NTNU)
§ 095 Production of liquid air
or
Expanded footnote [h]:
4
The inversion temperature, which is defined at κ(TJT ) = 0, is a
function of the pressure (see Theme 64 on page 393), but TJT ≈ 6Tc is
a reasonable approximation at low pressures.
5
11 Open . . .
inlet pressure is reduced to 1 atm? Use the enthalpy
values [J mol-1 ] in the above table.
News Incompressible flow Compressible flow Olds
§ 095 Production of liquid air (2)
(2)
§ 96
1
Based on the analysis that we did in Theme 94, the energy balance
can be expressed in the form
h
i
~ gas = H
~ vap + H
~ lıq = N
~ h vap (1 − α) + h lıq α ,
H
ın
out
out
out
out
Calculate how much liquid nitrogen is produced if the
inlet conditions are 290 K and 100 bar and the outlet
pressure is 1 atm. Assume perfect heat exchange
between the two fluids. In order to improve the efficiency of the system, it is important to cool the gas
before it reaches the inlet. How much liquid nitrogen would be produced if the inlet conditions were
changed to 300 bar and 240 K? What happens if the
Tore Haug-Warberg (Chem Eng, NTNU)
§ 94
pout pın
γ−1
γ
∂T ∂p H ,n
T
=
T
∂p ∂T V ,n
∂p 2
∂T V ,n
+V
− CV
∂p ∂V T ,n
∂p ∂V T ,n
#
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
New concepts
Part 13
13 Departure . . .
1
Gibbs residual
2
Helmholtz residual
3
Fugacity coefficient
4
Redlich–Kwong’s equation of state
Multicomponent thermodynamics
13 Departure . . .
1
From a thermodynamic point of view, it makes little difference
whether a system is composed of one, two or many components, provided that their relative proportions are constant in time and space.
3
The situation becomes much more complicated, however, if the
components that vary with mass are included in the system description.
2
Such systems will only have three degrees of freedom[a] and
will have relatively simple descriptions, as discussed in Chapter 8 on
page 335.
Departure Functions
[a] Chemical reactions change the mole numbers of the compounds in a system without affecting the components that only vary with mass, i.e. the atoms or atom groups
that are reaction invariant. The internal degrees of freedom are hidden from the outside world, so the composition of the system is perceived to be constant, although the
size of the system may well vary with e.g. temperature and pressure. Isotopic mixtures can be particularly intricate in this respect. Natural water contains e.g. nine equilibrated isotopic compounds without being paid any attention by us under normal circumstances.
4
The number of degrees of freedom increases, and the thermodynamic functions describe abstract hyperplanes in more than three spatial dimensions.
5
This makes it difficult to visualize the mathematical description, and
as we do not want to restrict the theory to any specific chemical system,
we shall hereafter use generic indices i ∈ [1, n] instead of component
names.
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Multicomponent thermodynamics (2)
13 Departure . . .
§ 104
§ 103
§ 105
13 Departure . . .
ıg
CV (T , n)
=
ıg
CP (T , n) =
ıg
S (T◦ , p◦ , n) =
n
P
i =1
n
P
i =1
n
P
i =1
Ni ∆f hi◦ (T◦ ) ,
(13.1)
Ni cv◦,i (T ) ,
(13.2)
Ni cp◦,i (T ) ,
Express the Gibbs energy of an ideal gas based on
Eqs. 13.1–13.4. Calculate the standard state contribution
of G ıg by using si◦ , ∆f hi◦ and cp◦,i (T ). Show by differentiation
Ni si◦ (T◦ , p◦ ) − R
n
P
i =1
Ni ln
Ni N
+
that the chemical potential can be expressed as
p , n) = µi◦ (T , p◦ ) + RT ln[Ni p /(Np◦ )]. Identify the
function µ◦i .
(13.4)
.
§ 104 Ideal gas Gibbs energy
ıg
µi ( T ,
(13.3)
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13 Departure . . .
1
The Gibbs energy G = H − TS of an ideal gas mixture can be
derived from Eqs. 8.43 and 8.44:
RT ıg
G ıg (T , p , n) = ∆f H ıg (T◦ , n) + CP (τ, n) dτ − T S ıg (T◦ , p◦ , n)
6
The thermodynamics of multicomponent mixtures therefore has a
more formal structure than that of its single-component counterpart.
7
The virial equation will be used repeatedly as an example in this
chapter, but let us start by looking at how we can extend the results
from Chapter 8 on page 335 to systems with variable compositions.
i =1
n
P
Termodynamic methods TKP4175
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
In ideal gases, the composition dependence is relatively simple,
8
and by referring to the Eqs. 8.42–8.44 on 382 we can write:
∆f H ıg (T◦ , n) =
Tore Haug-Warberg (Chem Eng, NTNU)
RT
T◦
ıg
CP (τ,n)
dτ − NR ln
τ
T◦
p
p◦
(13.5)
.
9
Note that ∆f hi◦ , cv◦,i , cp◦,i and si◦ are contributions from the standard
state of the same kind as µ◦i in Eq. 7.20 on page 287.
10 For systems with constant composition, U , H and S are the key
functions, but in mixtures the chemical potential becomes increasingly
important.
11 However, this quantity cannot be obtained by differentiating
U or H with respect to their canonical variables because µi =
ˆ
(∂U /∂Ni )S ,V ,Nj,i =
ˆ (∂H /∂Ni )S ,p ,Nj,i requires S to be a free variable,
which is only possible in very special cases.
12 The solution to the problem is to take either the Gibbs[a] or the
ˆ (∂G /∂Ni )T ,p ,Nj,i =
ˆ
Helmholtz[a] energy as our starting point, as µi =
(∂A /∂Ni )T ,V ,Nj ,i can then be determined through explicit differentiation[a].
13 Whether we use G or A depends on the equation of state being
used (more on this later).
[a] Josiah Willard Gibbs, 1839–1903. American physicist.
[a] Hermann Ludwig Ferdinand von Helmholtz, 1821–1894. German physician and
physicist.
[a] The great majority of thermodynamic models, including both equations of state and
activity coefficient models, are explicit in T but not in S .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 104 Ideal gas Gibbs energy (2)
13 Departure . . .
T ,p ,Nj ,i
= ∆f hi◦ (T◦ ) +
− R ln
+
Ni N
RT cp◦,i (τ)
T◦
τ
−
T◦
n
P
k =1
cp◦,i (τ) dτ − T si◦ (T◦ , p◦ )
Nk R
dτ − R ln
p
p◦
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4
13 Departure . . .
By simplifying Eq. 13.6 we can state the chemical potential as
ıg
N p
µi (T , p , n) = µi◦ (T , p◦ ) + RT ln Npi ◦ ,
(13.7)
µ◦i (T ) = ∆f hi◦ (T◦ ) +
RT
T◦
cp◦,i (τ) dτ − Tsi◦ (T◦ , p◦ ) − T
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RT
T◦
cp◦,i (τ)
τ
dτ
(13.8)
ENGLISH text is missing.The thermodynamic
residual
1
In the coming sections we are going to examine these ideas further
and use our knowledge about ideal gas mixtures to describe the deviation between them and the real fluid.
2
The interpretation of the word “real” is a little vague, however, since
we most of the time will be dealing with different state models and not
with experimental data.
13 Departure . . .
3
For us the Van der Waals equation will be as real as the virial
equation — except for yielding different function expressions but that is
also all.
4
The theory behind is the same and that is what matters in this
context.
The deviation
called a residual.
5
§ 105
1
Ni .
Use a specific example to show that U ıg − TS ıg + pV ıg =
Pn
ıg
i =1 µi
719 / 1776
between the ideal gas mixture and the real fluid is
6
Traditionally, the residual is calculated at given a temperature, pressure and composition but we will also be using the volume as a free vari§ 105 Ideal gas internal energy
able whenever that suits our needs better.
1
It provides valuable experience to know that the general relations 7 It is important to understand that the residual concept is connected
we meet in thermodynamics are valid under special circumstances as to a change in the model space and not the state space.
well.
8
The thermodynamic coordinates T , p , n (or T , V , n) are frozen while
2
This paragraph underpins such an experience.
the model description is changed from the real behaviour to the hypo3
First, Eq. 13.7 is substituted into the right-hand side of the equation thetically ideal gas.
given in the paragraph text.
9
By integrating from a reference point with known properties we can
4
The left-hand side then follows by inserting H ıg = U ıg + NRT .
reach all other thermodynamic states we would like to investigate.
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Using what we already know about G ıg allows us to define the
residual Gibbs energy as:
1
2
The canonical variables of Gibbs energy include temperature and
pressure.
3
This makes it possible to compare the chemical potentials of the
components of real fluids with those of the same components in the
ideal gas state, at the same temperature and pressure.
G r,p (T , p , n) =
ˆ G (T , p , n) − G ıg (T , p , n) .
(13.9)
4
As p → 0, all fluids have virtually ideal properties in the sense
of the pressure of the fluid approaching NRT /V when the volume increases indefinitely. This approximation means that
r,p n G r,p = B2′ , . . . , ∂∂p ···∂
= (n − 1)!Bn′ +1 (13.10)
lim G r,p = 0, ∂G
∂p
p
5
This does not mean that the volume of the fluid is the same as that
of an ideal gas!
p →0
(13.6)
.
Termodynamic methods TKP4175
Termodynamic methods TKP4175
Gibbs residual G r,p
where the standard chemical potential µi◦ is defined as:
02 September 2013
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T ,n
T ,n
is meaningful and also defined such that the parameters Bi′ are not
functions of the pressure (but they can still be strong functions of temperature and composition).
3
The above equation contains Kronecker’s delta, which has the
value 1 if i = k and 0 in all other cases.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 104 Ideal gas Gibbs energy (3)
N δik N −Nk
Nk
N2
02 September 2013
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
2
Let us substitute Eqs. 13.1, 13.3 and 13.4 into Eq. 13.5 and differentiate with respect to Ni :
ıg ıg
µi (T , p , n) = ∂∂GNi
RT
Termodynamic methods TKP4175
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Termodynamic methods TKP4175
02 September 2013
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Termodynamic methods TKP4175
02 September 2013
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Gibbs residual G
8
6
In particular, we shall notice that B2′ , 0. This quantity describes
the low-pressure difference in the volume of the fluid and the ideal gas
when kept at the same total pressure and is very important for our physical understanding of the matter.
7
The disposition given here is consistent with the virial theorem in
Chapter 7 on page 268, and with the Legendre transform in Chapter 4,
from which it follows that (∂G /∂p )T ,n is the system volume regardless
of whether the fluid is ideal or real.
r,p
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
(2)
13 Departure . . .
By taking the limits from Eq. 13.10 it is possible to rewrite Eq. 13.9
as :
Expanded footnote [a]:
[a] Note that π is used as the integration variable in order to avoid confusion with p ,
which appears in the upper limit of the integral.
Rp
Rp G r,p (T , p , n) = (V − V ıg ) dπ = V (π) −
0
0
NRT
π
dπ .
(13.11)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Residual chemical potential
13 Departure . . .
1
An equivalent expression for the chemical potential µi =
(∂G /∂Ni )T ,p ,Nj,i is true for all conceivable Gibbs energy functions, including G r,p .
r,p
2
This allows us to define the residual chemical potential as µi
ıg
µi − µi , or
r,p r,p
ˆ ∂∂GNi
µi (T , p , n) =
=
ˆ
T ,p ,Nj ,i
=
Rp 0
∂V − RT
π
∂Ni T ,π,Nj ,i
dπ =
ˆ
Rp 0
v̄i −
RT
π
dπ ,
(13.12)
The Schwarz theorem
13 Departure . . .
1
Since the system pressure is only present in the limit of the integral,
we need only worry about the derivative of the kernel, but there the
pressure does vary (locally) throughout the integration!
2
To obtain Eq. 13.12 from Eq. 13.11, G r,p must be differentiated at
constant pressure[a].
[a] Note that to calculate the derivative of G r,p at a given pressure p we must know the
derivative of V − V ıg for all pressures π ∈ [0, p ] in the integration domain. This goes
against usual practice when performing partial differentiation, but G r,p is a functional
that is defined for all pressures p > 0, and not merely for one arbitrarily chosen system
pressure p . The integrand, on the other hand, only depends on the local pressure π, so
it will not change in any way if p changes in the upper limit of the integral. The integral in
question is thus uniquely defined in terms of the area under a specific curve, integrated
from the lower limit p = 0 up to the system pressure p . Admittedly the curve depends
on the specified isotherm T and specified composition n, but it does not depend on the
system pressure p . Consequently, the curve must be defined by differentiating V − V ıg
at every value of π such that we can choose to stop the integration at any value of
p > 0.
However, there is a given local pressure π at each stage of the in5
tegration, so the only remaining degree of freedom comes from variation in the composition of component i .
3
This is an unusual situation, and it is important to realise that the
corresponding differential of V (π) is defined locally as
dNi
(dV )T ,π,Nj,i = ∂∂NV
i
T ,π,Nj ,i
for all pressures π ∈ [0, p ].
4
What is unusual in this respect is that there are two different physical interpretions of pressure — one being the system pressure in the
upper limit of the integral and the other being the integration variable π.
6
Hence, the chemical potential must be integrated over the volume
derivative as shown in Eq. 13.12.
ˆ (∂V /∂Ni )T ,π,Nj,i is defined as the partial molar volume of the
where v̄i =
component i .
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Termodynamic methods TKP4175
02 September 2013
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
The Schwarz theorem (2)
13 Departure . . .
§ 106
§ 105
§ 107
13 Departure . . .
§ 106 The fugacity coefficient
2
ıg
8
This is a measure of the difference between the thermodynamic
state of a component in a real fluid and of the same component in ideal
gas under the system conditions T , p , n.
9
The fugacity coefficient has traditionally been given a great deal of
attention in the American literature, but really it is a vestige of the past.
10 For gases at low pressures and phase equilibrium calculations using the K -value method it has certain benefits, but in general it causes
more problems than it solves — it neither increases our basic understanding nor improves the numerical performance of thermodynamic
algorithms.
1.
r,p
Show that µi = µi + µi = µi◦ + RT ln[Ni ϕi p /(Np◦ )].
Verify that the lower limit limp →0 ϕi = 1. What is ϕi for
a component in the ideal gas state?
Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 107
§ 106
§ 108
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13 Departure . . .
The residual Gibbs energy can be determined by directly substituting the 2nd virial equation into Eq. 13.11:
1
Derive G r,p for the 2nd virial equation if pV 2.vır = NRT + Bp .
Differentiate to find RT ln ϕk2.vır . Use the quadratic
PP
mixing rule for B = N
xi xj Bij where Bij = Bji .
i
Rp 0
NRT
π
+B −
NRT
π
dπ = Bp .
(13.14)
G r,p ,
and hence B ,
2
In order to find RT ln ϕk we must differentiate
with respect to Nk , but differentiating with respect to (mole) fractions
has an unfortunate tendency to cause what is known as “code bloat” in
computing.
j
r,p,ıg
ıg
= 0 regard-
ıg
Hence ϕi = 1.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 107 Virial Gibbs energy I
G r,p,2.vır =
r,p
For an ideal gas v̄i = RT /π, which means that µi
less of the pressure.
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
13 Departure . . .
13 Departure . . .
The latter equation gives us limp →0 µi = 0 and hence limp →0 ϕi =
3
4
Tore Haug-Warberg (Chem Eng, NTNU)
725 / 1776
1
The chemical potential can be derived by combining Eqs. 13.7
and 13.12.
(13.13)
where ϕi (T , p , n) is the fugacity coefficient of component i .
02 September 2013
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
7
For historical reasons it is customary to rewrite the residual potential as
r,p
RT ln ϕi =
ˆ µi (T , p , n) ,
Termodynamic methods TKP4175
§ 107 Virial Gibbs energy I (2)
13 Departure . . .
3
We can considerably reduce the amount of writing that we need
P P
to do by using NB = i j Ni Nj Bij , which, even though it is an implicit
expression, is much better suited for differentiation:
P P ∂Ni Nj ∂NB Bij .
=
∂Nk
∂Nk
T ,Nl ,k
i
Nl ,k
j
4
Differentiate both sides and substitute in Kronecker’s delta, where
δii = 1 if i = j and δij = 0 if i , j :
!
PP
P
P
δ Nj + Ni δ Bij = Nj Bkj + Ni Bik .
B + N ∂∂NBk
=
T ,Nl ,k
i
j
ik
jk
j
i
5
We can simplify the right-hand side considerably by substituting
Bkj = Bjk and simultaneously changing one of the summation indices
from j to i .
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Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Termodynamic methods TKP4175
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News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 107 Virial Gibbs energy I (3)
13 Departure . . .
6
This gives us two identical terms on the right-hand side, so the
partial derivative of B can now be written
P
B
b̄k =
ˆ ∂∂NBk
= 2 NNi Bik − N
.
T ,Nl ,k
i
7
The residual potential can be calculated from Eqs. 13.13 and 13.14
on page 724, as shown below:
r,p,2.vır RT ln ϕk2.vır = ∂G∂Nk
= p ∂∂NBk
=
ˆ p b̄k .
T ,p ,Nl ,k
Helmholtz residual A
Expanded footnote [a]:
[a] It is worth noting that b̄k is the (pressure-independent) departure function of the
partial molar volume from the 2nd virial equation: v̄k2.vır − RT /π = b̄k (T ).
I
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
2
We must therefore use an equation of state in the form V = V (T , p ,
n), which in practice means that the virial equation V 2.vır = NRT /p + B
is of limited use.
3
However, there has been rapid development of equations of state
in the form p = p (T , V , n), which changes the canonical variables to
temperature and volume rather than pressure.
4
This makes it attractive to take Helmholtz energy rather than Gibbs
energy as our starting point.
Traditionally, the definition of the fugacity
not changed in the transposition from G to A .
5
Helmholtz residual A r,p I (2)
13 Departure . . .
ENGLISH text is missing.Helmholtz residual
1
Residual Gibbs energy has the canonical variables temperature
and pressure.
coefficient in Eq. 13.13 is
6
This causes a problem with the chemical potential, because ϕi
describes the difference between the real fluid and ideal gas at a given
pressure — not a given volume.
7
As a result, you end up with a non-canonical description, which is
discussed here mainly because of its historical interest.
733 / 1776
Our goal is to describe the difference in Helmholtz energy between
a real fluid and an ideal gas at a given temperature and pressure.
1
7
This is true for any Helmholtz energy function, regardless of
whether it refers to an ideal gas or a real fluid.
Expanded footnote [a]:
ıg
n −A ıg
) = 0 in accor[a] This follows from lim A − A ıg = 0, (∂A∂−VA )T ,n = 0, . . . , (∂∂VA···∂
V T ,n
V →∞
dance with the virial theorem.
A r,p (T , V (p ), n) =
2
To do this, we have to find the difference between two state functions applied to two different states.
r,p
ıg
=
≡ A (T , V (p ), n) − A ıg (T , V (p ), n)
ıg
ıg
(13.15)
ıg
+ A (T , V (p ), n) − A (T , V (p ), n) .
VR(p )
∞
ıg
A (T , V (p ), n) =
ˆ A (T , V (p ), n) − A (T , V (p ), n)
13 Departure . . .
6
From Chapter 4 we know that (∂A /∂V )T ,n = −p . Analogously to
Eq. 13.11 on page 717 we can therefore write
We can simplify our task by splitting the expression into two terms:
The first term compares the two functions in the same state, while the
second term compares the ideal gas in two different states:
3
T ,Nl ,k
8
Alternatively, we can achieve the same result by substituting the
partial derivative of B into Eq. 13.12 on page 718 and integrating .
r,p
(πıg − π) dν −
VR(p )
∞
NRT
ν
VR(p )
πıg dν
V ıg (p )
− π dν − NRT ln z (p ) ,
(13.16)
where the ideal gas volume is given by V ıg = NRT /p , and the comˆ V /V ıg = pV /(NRT ).
pressibility factor z is defined as z =
4
It should be emphasised that volume is a function of pressure p .
It implies, however, that the two fluids have different volumes, since
V , V ıg at a given p .
5
The pressure is by definition the same in the ideal gas and the real
fluid, i.e. p ıg = p .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
732 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Termodynamic methods TKP4175
02 September 2013
734 / 1776
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 108
1
Eq. 13.16 may look trivial, but please note that the volume is an
implicit variable in the integral.
2
The implicit form of the function A r,p (T , V (p ), p , n) prevents us from
calculating the chemical potential by differentiation with respect to the
canonical variables T , V , n[a].
3
so we must instead use
µi = ∂∂NAi
T ,V ,Nj ,i
§ 107
§ 109
⇒
ıg
µi =
∂A ıg ∂Ni T ,V ıg ,Nj ,i
,
r,p
4
If we substitute this into the residual chemical potential µi
ıg
µi we get
ıg r,p
µi = ∂∂NAi
− ∂∂ANi
T ,V (p ),Nj ,i
T ,V ıg (p ),Nj ,i
r,p = ∂∂AN
.
ıg
i
=
ˆ µi −
13 Departure . . .
T ,V (p ),V (p ),Nj ,i
5
Note that V (p ) is constant in one of the partial derivatives, while
V ıg (p ) is constant in the other one.
7
The residual chemical potential can therefore be expressed
VR(p ) r,p
RT
∂π dν − RT ln z (T , V (p ), n) .
ν − ∂Ni
µi (T , V (p ), n) =
∞
T ,ν,Nj ,i
(13.17)
8
Although this derivation is mathematically correct, it is hard to explain logically, as it relies on there being two different volumes (both of
which remain constant, in line with the definition) during the same differentiation.
736 / 1776
[a] If we redefine the fugacity coefficient to mean the difference between an ideal gas
and a real fluid at a given volume rather than a given pressure, it becomes easier to
derive the chemical potential; see Section 13.4 on page 731.
§ 108 Virial Helmholtz energy
13 Departure . . .
1
The residual Helmholtz energy is calculated by inserting the
2nd virial equation into Eq. 13.16, which gives us this surprising result:
RV NRT
NRT
V A r,p,2.vır =
(13.18)
ν − ν−B dν − NRT ln V −B = 0 .
∞
Find A r,p for the 2nd virial equation if pV 2.vır = NRT + Bp .
2
So there is no difference between Helmholtz energy for the
2nd virial equation and that for an ideal gas at the same pressure.
3
In other words: The volume term in
RT ln z 2.vır .
8
4
This does not mean that all residual properties cancel.
5
For example, G r,p,2.vır was found to be Bp in Eq. 13.14 on page 724,
and you should verify that if the 2nd virial equation is substituted into
Eq. 13.17, it gives the same answer as in Eq. 13.12 on page 718.
6
Moreover, even though A r,p,2.vır is zero, this does not imply that
r,p,2.vır
is also zero.
µi
r,p,2.vır
7
The fact is that G r,p,2.vır , and hence µi
, are direct measures of
the non-ideality of the gas, whereas A r,p,2.vır is not.
It is important to remember that
A r,p,2.vır
A r,p
Termodynamic methods TKP4175
happens to be equal to
is a non-canonical function.
(T , p , n) vs (T , V , n)
Termodynamic methods TKP4175
02 September 2013
737 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 110
§ 109
§ 111
13 Departure . . .
738 / 1776
1
2
The problems we experienced in the previous section were caused
by the disastrous decision to use pressure as a free variable in the
Helmholtz function.
r,p
2
ENGLISH text is missing.Helmholtz residual A r,p III
Let us first differentiate Eq. 13.15 with respect to the mole number 2
Ni :
Substituting G = A + pV into G r,p =
ˆ G − G ıg gives us:
G r,p = A (T , V (p ), n) + pV (p ) − [A ıg (T , V ıg (p ), n) + pV ıg (p )] . (13.24)
∂A ıg ∂A r,p (13.19)
∂Ni T ,V ,Nj ,i = µi − ∂Ni T ,V ,Nj ,i .
3
Furthermore, pV = zNRT and pV ıg = NRT . When we substitute
3
This time we face the problem that constant V does not necessarily these into the above equation, and combine it with Eqs. 13.15 and 13.16
on page 727, we get:
imply constant V ıg .
4
In fact, in order to differentiate A ıg with respect to Ni , we need to
ıg
G r,p = A r,p + NRT (z − 1)
know how V varies with Ni when V is kept constant.
VR(p )
NRT
=
(13.25)
ν − π dν − NRT ln z (p ) − NRT [z (p ) − 1] .
5
At a given temperature and at fixed amounts of all components j ,
∞
except i , the differential of A ıg is
4
Our one remaining task is to differentiate this expression with reıg ıg ıg
dNi + ∂∂VA ıg T ,n dV ıg = µi dNi − p ıg dV ıg ,
(dA ıg )T ,Nj ,i = ∂∂ANi
spect to the mole number of component i .
T ,V ıg ,Nj ,i
(13.20)
5
It is evident that we will need to obtain the partial derivative of
where ıg is stated explicitly in order to make the next step a little bit
z=
ˆ pV /NRT , so let us start by doing that:
clearer.
p v̄
pV
p v̄
zi =
ˆ ∂∂Nzi
= NRTi − N(NRT ) = NRTi − Nz .
If we bear in mind the condition p ıg = p , we can simplify Eq. 13.20
∂A ıg ∂Ni T ,V ,Nj ,i
7
ıg
= µi − p
∂V ıg ∂Ni T ,V ,Nj ,i
Combined with Eq. 13.19 this gives us:
r,p ıg ıg
− p ∂∂VNi
µi − µi = ∂∂ANi
T ,V ,Nj ,i
8
6
This is where things start to get tricky: How can we differentiate the
(13.21) integral?
.
T ,V ,Nj ,i
.
(13.22)
∞
T ,ν,Nj ,i
7
The upper limit clearly depends on the composition at fixed pressure, while the differential of π at a given volume ν is:
dNi .
(dπ)T ,ν,Nj ,i = ∂∂π
Ni
T ,ν,Nj ,i
Finally, Eq. 13.16 is differentiated:
RV ∂p RT
dν +
=
ν − ∂Ni
∂A r,p ∂Ni T ,V ,Nj ,i
9
§ 110 Redlich–Kwong[1]
02 September 2013
13 Departure . . .
Expanded footnote :
[] Otto Redlich and J. N. S. Kwong. Chem. Rev. (Washington, D. C.), 44:233–244,
1949.
At the critical point this equation must fulfil (v − vc )3 = 0, or v 3 −
3vc v 2 + 3vc2 v − vc3 = 0.
2
If compared term-by-term (see Theme 47 in Chapter 7), the two
polynomials can be used to produce three equations, which can then
be solved for a , b and vc :
3
3 vc =
744 / 1776
RTc
pc
,
Tore Haug-Warberg (Chem Eng, NTNU)
3vc2 =
a
pc Tc1/2
− b2 −
RTc b
pc
Termodynamic methods TKP4175
,
13 Departure . . .
T ,p ,Nj ,i
6
to
8
It is important to note that ν is an integration variable in Eq. 13.25,
§ 109 Virial Gibbs energy II
and that it remains constant during the above differentiation.
NRT ∂V ıg V ıg ∂Ni T ,V ,Nj ,i − RT ln z .
1
From Eq. 13.25 we know the general expression G r,p = A r,p +
9
Partial differentiation therefore gives us:
NRT (z − 1).
(13.23)
VR(p ) r,p
RT
NRT
∂π 2
Combined with A r,p,2.vır = 0 from Eq. 13.18 on the previous page
µi =
ν − ∂Ni T ,ν,Nj ,i dν + V (p ) − p v̄i
∞
this gives us G r,p,2.vır = NRT (z − 1).
− RT ln z − zi NRT
+ RT (z − 1) + zi NRT .
z
3
Substituting in z 2.vır = 1 + Bp /NRT gives G r,p,2.vır = Bp in accorıg
dance with Eq. 13.14 on page 724.
=
ˆ µi − µi is thus the same as 10 If we substitute in zi , most of the terms cancel, and the final result § 109
is identical to Eq. 13.17 on page 729.
1
Show that G r,p,2.vır = NRT (z − 1) = Bp when pV 2.vır = NRT + Bp . 4 The results from the various sections are thus internally consistent.
741 / 1776
739 / 1776
740 / 1776
742 / 1776
You will recognise NRT /V ıg = p ıg = p in Eq. 13.23.
10 This means that the term (∂V ıg /∂Ni )T ,V ,N disappears when the
j ,i
Eqs. 13.22 and 13.23 are combined.
r,p
11 The residual chemical potential µi
in Eq. 13.17 on page 729.
If we redefine the residual Helmholtz energy as
A r,v (T , V , n) =
ˆ A (T , V , n) − A ıg (T , V , n)
we are back into the world of canonical variables, and we can use the
same approach as the one used for residual Gibbs energy on page 716.
3
The alternative departure function can be written
A r,v =
RV ∞
NRT
ν
− p (ν) dν ,
while the associated chemical potential is
RV r,v RT
µir,v =
ˆ ∂∂ANi
=
ν −
Tore Haug-Warberg (Chem Eng, NTNU)
∞
∂p ∂Ni T ,ν,Nj ,i
Termodynamic methods TKP4175
dν .
02 September 2013
743 / 1776
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Let us first estimate a and b for a fluid composed of a single
component where we supposingly have a measurement of the critical
temperature and pressure of this component. For this purpose we will
rewrite the equation of state as a cubic polynomial of volume:
2
2
a
RTb
v 3 − RT
v − pTab1/2 = 0 .
p v + pT 1/2 − b − p
Derive an expression for A r,v and the first derivatives
∂A r,v /∂(T , V , n) for a fluid that conforms to the Redlich–
√
Kwong equation of state p RK = RT /(v − b ) − a / T v (v + b ).
P
P P √
Use the mixing rules b (x) = i bi xi and a (x) = i j ai aj
xi xj where bi and ai are component-specific parameters.
Finally, state bi and ai as functions of Tc and pc ; also
see Theme 47 on page 322.
Termodynamic methods TKP4175
02 September 2013
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
1
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
735 / 1776
ENGLISH text is missing.Helmholtz residual A II 1 A third option is to use Gibbs energy as our starting point, which is
1
At this point it may be helpful to offer an alternative, more intuitive a natural thing to do, since the residual chemical potential has the same
derivation of A r,p .
canonical variables as Gibbs energy.
T ,V ,Nj ,i
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
ENGLISH text is missing.Problems
6
This means that both V (p ) and V ıg (p ) must be kept constant when
differentiating A r,p , which in turn implies that z (p ) = V (p )/V ıg (p ) is a
constant factor.
Tore Haug-Warberg (Chem Eng, NTNU)
vc3 =
ab
pc Tc1/2
.
02 September 2013
745 / 1776
§ 110 Redlich–Kwong[2] (2)
13 Departure . . .
4
Combining the middle equation with the other two yields 2vc3 −(vc +
b )3 = 0, which can be solved for the positive root b = (21/3 − 1)vc .
5
If expressed in terms of the critical temperature and pressure (for
any component i ), this is equivalent to:
bi = Ωb
Ωb =
ˆ
RTc,i
pc,i
21/3 −1
3
ai = Ωa
ˆ
Ωa =
R 2 Tc,5/2
i
pc,i
1
9(21/3 −1)
6
We can now choose whether to express the rest of the derivation
in terms of extensive properties (mole number and volume) or intensive
properties (mole fraction and molar volume).
7
Here, the extensive properties have been chosen, in order to emphasise the homogeneity of the thermodynamic functions.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
746 / 1776
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
§ 110 Redlich–Kwong[3] (3)
8
§ 110 Redlich–Kwong[4] (4)
13 Departure . . .
The total Helmholtz energy of the fluid is thus
A
r,v,RK
=
RV ∞
NRT
ν
−p
RK
V V −B
dν = NRT ln
+
and the associated first derivatives are:
V ∂A r,v,RK = NR ln V V−B + ABT ln V +
B ,
∂T
V ,n
−B
A
∂A r,v,RK NRT
=
+
,
V (V −B )
V (V + B )
∂V
T ,n
∂A r,v,RK = RT ln V V−B + NRT Vb−iB + B1 Ai −
∂Ni
T ,V ,Nj ,k
Tore Haug-Warberg (Chem Eng, NTNU)
A
B
ln
V V +B
Abi
B
Termodynamic methods TKP4175
ln
V
V +B
,
i
−
02 September 2013
Abi
B (V + B )
.
13 Departure . . .
Helmholtz residual at given pressure:
VR(p )
∞
4
−p (ν) +
NRT
ν
i
αi1/2 Ni
2
Proper Gibbs residual:
1
G r,p (T , p , n) =
ˆ G (T , p , n) − G ıg (T , p , n)
Rp = V (π) − NRT
dπ
π
,
0
Proper Helmholtz residual:
2
A r,v (T , V , n) =
ˆ A (T , V , n) − A ıg (T , V , n)
= 2(αi A )1/2 ,
=
Termodynamic methods TKP4175
02 September 2013
748 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Clausius–Clapeyron Waals min A
T ,ν,Nj ,i
A r,p,2.vır = 0
6
A r,v,RK = NRT ln
7
Chemical potential from Gibbs residual:
r,p
µi (T , p , n) =
ˆ
Tore Haug-Warberg (Chem Eng, NTNU)
A
B
ln
=
ˆ
Rp 0
Termodynamic methods TKP4175
v̄i −
RT
π
749 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Part 14
r,p
see also Part-Contents
dπ
02 September 2013
750 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
New concepts
Saturation state
2
Clausius–Clapeyron’s vapour pressure equation
3
Van der Waals equation of state
4
Critical point behaviour
5
Maxwell’s area rule
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
1
Phase boundary
02 September 2013
751 / 1776
2
In our everyday lives, the same physical phenomena cause the
evaporation of solvents in paints and varnishes, allow us to enjoy the
aromas of wines, cheeses and coffees, make it possible to use propane
and butane in camping gas stoves, etc.
3
Moreover, at a global level, all life on Earth is inextricably linked to
the transport of sea water (as vapour) from warmer to cooler parts of
the globe, where it then condenses as rain or snow before it eventually
finds its way back to the ocean.
4
Although this cycle, which is driven by the sun, occurs under nearequilibrium conditions, it is responsible for the vast majority of mass
transport at or near the Earth’s surface.
5
It is therefore not true to say that near-equilibrium processes are
“slower” or in any way less important than irreversible processes.
754 / 1776
p vap = p lıq
Termodynamic methods TKP4175
02 September 2013
753 / 1776
dT vap = dT lıq
⇒
Tore Haug-Warberg (Chem Eng, NTNU)
dp vap = dp lıq
02 September 2013
02 September 2013
752 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Clapeyron equation
14 Simple . . .
1
It is normal to combine these three differentials into a single equation, which is referred to as the Clapeyron equation:
dp ∆v s
s vap −s lıq
ˆ ∆
= T∆∆vvhv .
dT ∆µ=0 = v vap −v lıq =
vv
2
The criteria on the left-hand side are the classical equilibrium conditions being derived in Section 14.4.
3
They must be fulfilled at all thermodynamic equilibrium points, but
from Theme 39 on page 195 we know that the intensive state[a] of the
system can be determinied when two of the three state variables T , p , µ
are known.
4
Consequently, the equilibrium state can be parametrised by either
T , p or T , µ or p , µ, the first of which is the most common choice.
5
For a small change in the variables, the conditions on the righthand side then apply.
[a] We have not formulated a mass balance for the equilibrium, so the phase distribution is unknown, but that is irrelevant in this context because (∂µ/∂N )T ,p = 0 for all
one-component systems, i.e. the equilibrium is the same regardless of the quantity of
vapour and liquid.
Expanded footnote [a]:
[a] Émile Clapeyron, 1799–1864. French engineer and mathematician.
7
Note that the relationship between the entropy and enthalpy of vaporization follows from the equilibrium condition ∆v µ = 0, which substituted into µ = h − Ts gives us ∆v s = ∆v h /T .
6
This equation brings together four important quantities in phase
theory: Vapour pressure, temperature, enthalpy of vaporization and saturation volume.
−s vap dT vap + v vap dp vap = −s lıq dT lıq + v lıq dp lıq
Termodynamic methods TKP4175
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
1
Particularly in the chemical process industry, vapour–liquid equilibria are essential to the design and operation of all kinds of distillation
and absorption columns, oil and gas separators, evaporators and other
similar process equipment.
2
The physical explanation of most of the abovementioned processes
require complex descriptions of multicomponent systems involving various phases.
3
The theory is too involved to be covered in one single chapter and
we shall rather learn from studying the characteristics of vapour–liquid
equilibria based on the behaviour of a realistic, but much simpler, onecomponent system.
4
This makes the formulation part of the theory far simpler because
it does not need to take into account the variation in chemical composition, and to some extent you would expect there to be analytical solutions to some of the selected problems.
T vap = T lıq
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
ENGLISH text is missing.Perspective
1
How to go about that is precisely what this chapter is about. According to our working hypothesis the conditions for thermodynamic
equilibrium along a continuous phase boundary curve in the T , p , µ
space are:
µvap = µlıq
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
V V +B
∂G r,p ∂Ni T ,p ,Nj ,i
News Clausius–Clapeyron Waals min A
dν
Simple vapour–liquid equilibrium
5
+
Fugacity coefficient: RT ln ϕi =
ˆ µi (T , p , n)
9
G r,p,2.vır = NRT (z − 1) = Bp
V V −B
NRT
ν
13 Departure . . .
∞
dν − NRT ln z (p )
−p (ν) +
Termodynamic methods TKP4175
Gibbs inspired chemical potential from Helmholtz residual:
VR(p ) r,p
RT
∂π dν − RT ln z (T , V (p ), n)
µi (T , V (p ), n) =
ν − ∂Ni
ıg
RV ∞
Looking back (3)
A (T , V (p ), n) =
ˆ A (T , V (p ), n) − A (T , V (p ), n)
=
P
=
13 Departure . . .
ai
T 1/2
Tore Haug-Warberg (Chem Eng, NTNU)
8
ıg
j
j
αi =
ˆ
747 / 1776
i
ai aj 1/2
Ni Nj
T
Looking back
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
Looking back (2)
r,p
PP
AT =
ˆ − 2AT ,
P 1/2
Ai =
ˆ 2α1/2
αj Nj
i
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
3
13 Departure . . .
9
The coefficients of the equation are defined below (note that αi has
the same units as aiVdW ):
P
B =
ˆ bi Ni ,
A =
ˆ
News G r,p (T , p , n) A r,p (T , p , n) A r,v (T , V , n) Olds
755 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
756 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Clausius–Clapeyron equation
News Clausius–Clapeyron Waals min A
ENGLISH text is missing.Reduced Clapeyron
equation
1
We could have stopped the derivation after the last section but
traditionally the expression is written on a dimensionless form.
2
Finally, the equation will be integrated using a couple of physically
based assumptions. Reducing the right-hand side
dp dT ∆µ=0
=
∆v h
T ∆v v
=
3
and then the left-hand side:
4
The answer is:
p p
− T cT r2 ·
c r
RTc
Tc vc
·
dln pr d(1/Tr ) ∆µ=0
dln pr d(1/Tr ) ∆µ=0
∆v h r
Tr ∆v vr
=
c
= − pRT
·
c vc
R
vc
Tr
pr
=
·
·
R
vc
·
∆v h r
Tr ∆v vr
∆v h r
∆v vr
∆v h r
Tr ∆v vr
14 Simple . . .
.
.
(14.1)
5
To come any further it is required to use a mix of measurements,
assumptions and models.
6
The ideal gas is one obvious candidate but how should the critical
constants be interpreted in that case?
There exist no ideal critical state and we must formally eliminate all
Tc , pc and vc from the equation (reversing all the previous operations
you can say).
7
8
At low pressures the denominator is taken over by the vapour volume such that ∆v v ≈ v vap .
9
We then get the limit case:
dln p ıg
lim
p →0 d(1/T ) ∆µ=0
= − ∆Rv h .
(14.2)
757 / 1776
1
If we integrate we get the Clausius –Clapeyron vapour pressure
equation, where b = ∆v h /R is assumed to be constant over the relevant temperature range:
p C-C
(14.3)
ln p◦
= a − Tb .
Expanded footnote [a]:
[a] Rudolf Julius Emanuel Clausius, 1822–1888. German physicist.
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Measured vapour pressure
News Clausius–Clapeyron Waals min A
14 Simple . . .
1
If Tr → 1 then we cannot assume ideal gas behaviour and the
volume of liquid is no longer negligible either. This undermines the
theoretical basis for
The enthalpy of vaporization is an independent quantity in this case
and must be measured experimentally.
6
The following is then true (see also Theme 47 on page 322):
pc vc VdW
RTc
=
3
,
8
lim prVdW =
Tr ≪1
8Tr
vap
3 vr
,
vap
lim ∆v vr = vr
Tr ≪1
.
Substitution into Eq. 14.1 gives the final result:
dln p VdW
lim d(1/Trr )
= −∆v hr .
Tr ≪1
−2
1
1.2
1.4
ˆ Tc /T
1/Tr =
Figure 14.1: Vapour pressure of CO2 , where ◦ is the measured value and ×
is the percentage deviation.
3
In order to attain a consistent description of the vaporization process vi have to consider a model that can describe both phases with a
single formula.
5
0
−1
−3
Being omnipresent throughout the rest of this chapter, the
Van der Waals equation 14.5 is a natural choice.
14 Simple . . .
1
2
4
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Measured vapour pressure (2)
ˆ ln(p /pc )
ln(pr ) =
News Clausius–Clapeyron Waals min A
(14.4)
∆µ=0
7
The two Eqs. 14.2 and 14.4 are similar notwithstanding the latter
equation is written on reduced form.
8
The conclusion is that the Van der Waals equation is consistent
with the ideal gas assumption at low pressures.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
02 September 2013
758 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Measured vapour pressure (3)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
the Clausius–Clapeyron equation, but the non-linear contributions
largely cancel each other out, so Eq. 14.3 still gives a reasonable approximation of the vapour pressure.
14 Simple . . .
2
The Van der Waals equation has been presented before in Chapters 7:7.8 and 8:8.9 but only in the context of single phase gases.
3
The focus is now shifted towards two-phase vapour–liquid equilibria.
4
However, we do not want to undertake the full discussion of multicomponent mixtures and for the present the parameters b and a are
considered to be true constants:
p VdW =
NRT
V −Nb
−
aN 2
V2
(14.5)
5
Let us next calculate the chemical potential of the Van der Waals
fluid using the p (T , V )-relation above.
[a] W. Duschek, R. Kleinrahm, and W. Wagner. J. Chem. Thermodynamics, 22:841–
864, 1990.
µr,v =
ˆ
=
5
The state description is not nearly precise enough for practical uses
in process engineering studies but it is the simplest equation we can use
to illustrate the peculiarities we are after.
RV ∞
RV ∞
RT
ν
−
RT
ν
−
= RT ln
∂p ∂N T ,ν
RT
ν−Nb
V V −Nb
−
+
and the ideal gas contribution is
NRTb
(ν−Nb )2
NRTb
V −Nb
−
News Clausius–Clapeyron Waals min A
761 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Van der Waals A
14 Simple . . .
It is worth noting that these two equations represent a complete set
of equations of state for Helmholtz energy.
Integration using the Euler method gives us A = −pV + µN , which
µ
pV
A
expressed in dimensionless form becomes NRT
= − NRTc + RTc . Rewritc
ing this in reduced variables:
2
A VdW
NRTc
.
= − 83 prVdW vr + µVdW
r
(14.8)
3
Note that the critical compressibility factor zc drops out as the constant number pc vc /RTc = 83 when the Van der Waals equation of state
is used to solve for the critical state of the fluid. The universal behaviour
indicated is what we call a corresponding state. A more thorough explanation is given in Themes 47 and 49 on page 328.
02 September 2013
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NRT Vp◦
dν
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Van der Waals p and µ (2)
14 Simple . . .
6
Combining the two equations above gives us the following expression for the chemical potential:
NRTb 2aN
µVdW = µ◦ (T , p◦ ) + RT ln p◦ (NRT
V −Nb ) + V −Nb − V .
p VdW
pc
,
µVdW
RTc
=
ˆ prVdW =
8Tr
3vr −1
−
3
vr2
(14.6)
,
=
ˆ µVdW
= µ◦r − Tr ln pr◦ + Tr ln
r
8Tr
3vr −1
+
Tr
3vr −1
−
9
4vr
(14.7)
.
.
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
14 Simple . . .
Given the expression of Helmholtz energy above it is easy to derive
the entropy of the fluid by differentiation with respect to T . which in the
current case becomes:
r
+ 1 + 3v1r −1
−srVdW = − 38 · 3v8r −1 · vr − sr◦ − ln pr◦ + ln 3v8rT−1
8Tr ◦
◦
(14.9)
= −sr − ln pr + ln 3vr −1 .
2
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
1
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Van der Waals S and H (2)
14 Simple . . .
4
The standard state contribution cancels out when calculating the
enthalpy of vaporization:
On a dimensionless form we have the general expression
∂(A /NRT )
−S
−sr =
ˆ NR
= (NR )-1 ∂∂AT V ,N = ∂(T /Tc )c
=
ˆ ∂∂Tarr ,
V ,N
vr
∆v hrVdW =
Tr
vap
3vr
−1
−
9
vap
4vr
−
Tr
lıq
3vr −1
+
9
lıq
4vr
.
(14.11)
5
The Eqs. 14.6–14.11 contain everything we need to know in order
to discuss the fundamental issues of vapour–liquid equilibria.
6
Theoretically, the expressions are not particularly difficult but their
mathematical handling depend on quite a few physical details.
7
The knowledge necessary is gained only with practical training and
a one single example may contribute more than a thousand words.
3
We can thereafter do Euler integration of H = TS + µN to find the
enthalpy:
Since µ◦ =
Termodynamic methods TKP4175
2aN
V
Van der Waals S and H
hr =
ˆ
Tore Haug-Warberg (Chem Eng, NTNU)
2aN
ν2
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
1
ˆ
arVdW =
Tore Haug-Warberg (Chem Eng, NTNU)
+
Termodynamic methods TKP4175
7
If we use the generalised parameters 3b = vc , a = 3pc vc2 and
8pc vc = 3RTc from Theme 47 on page 322, we can express the pressure and chemical potential in dimensionless form:
dν
µıg = µ◦ (T , p◦ ) + RT ln
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
News Clausius–Clapeyron Waals min A
From the discussion of residuals in Chapter 13 the chemical potential can be expressed as µVdW = µıg + µr,v , where the residual term is
defined as being
1
2
This relationship is illustrated in Figure 14.1, which shows a vapour
pressure curve that is surprisingly rectilinear, even close to the critical
point (cf. Program 1.11 in Appendix 31).
3
The percentage difference between the estimated and measured
vapour pressure[a] is given in the figure by × and there are no clues
that the linear estimate is fundamentally wrong.
Termodynamic methods TKP4175
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Van der Waals p and µ
4
To begin to understand why the vapour pressure can be plotted
as a straight line even at high pressures, and to start learning about
vapour–liquid equilibria in general, let us look at the Van der Waals
equation of state in greater detail.
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
h◦
−
Ts ◦
H
NRTc
=
TS +µN
NRTc
= Tr sr + µr .
the latter expression becomes:
Tore Haug-Warberg (Chem Eng, NTNU)
hrVdW = hr◦ +
Tr
3vr −1
−
9
4vr
Termodynamic methods TKP4175
(14.10)
.
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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News Clausius–Clapeyron Waals min A
Phase equilibrium
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
News Clausius–Clapeyron Waals min A
14 Simple . . .
ENGLISH text is missing.Brief comment
1
Theme 112 in Section 14.4 takes you into that direction where
Figure 14.3 is a practical example used to illustrate the properties of
Van der Waals equation.
2
The equations that are necessary for calculating such phase diagrams in general are treated in the next two sections.
767 / 1776
§ 111
1
In this section we are going to define the thermodynamic equilibrium of a closed two-phase system as a state of minimum Helmholtz
energy.
§ 112
2
In principle this task sounds quite easy but in practise it is a little
more complex.
14 Simple . . .
News Clausius–Clapeyron Waals min A
ar · ρr =
ˆ Avc /VRTc
Tore Haug-Warberg (Chem Eng, NTNU)
News Clausius–Clapeyron Waals min A
02 September 2013
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1
2
In optimization theory there are necessary and sufficient conditions
for the minimum.
3
The necessary conditions, being of first order, lead to the equilibrium relations.
4
The sufficient conditions, being of second order, lead to the stability
criteria.
1
14 Simple . . .
Here, the first order conditions will do fine.
vap
1
2
lıq
= −p vap dV vap + µvap dN vap − p lıq dV lıq + µlıq dN lıq
3
ρr =
ˆ vc /v
Figure 14.2: Helmholtz energy for the isotherms in Figure 14.4 on page 767.
The equilibrium state (the tangent plane at minimum energy) is given for the
second lowest isotherm. The standard state has been chosen for the purposes of clarity, and varies for each isotherm. The bell shaped curve indicates the system’s phase boundaries.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 111 Minimum of Helmholtz energy (2)
14 Simple . . .
7
The limits on the volume and mole number of the closed system
mean that dV lıq = − dV vap and dN lıq = − dN vap . Substituted into the
differential of A that gives us
(dA )T = −(p vap − p lıq ) dV vap + (µvap − µlıq ) dN vap = 0 .
8
(dA )T = (dA )T + (dA )T
0
02 September 2013
Since dV vap and dN vap are independent variables, then it must be
the case that p vap = p lıq and µvap = µlıq , which is what we needed to
show.
Taking the total differential for both phases gives us
6
0.5
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
5
A necessary condition for thermodynamic equilibrium is that
(dA )T = 0 for variations in the extensive variables V vap , N vap and
V lıq , N lıq .
1.5
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 111 Minimum of Helmholtz energy
2
0
Termodynamic methods TKP4175
c.p.
2.5
14 Simple . . .
Here e =
ˆ ( 1 1 ), vT =
ˆ ( V vap V lıq ) and nT =
ˆ ( N vap N lıq ).
vap
lıq
vap
lıq
Show that p
= p and µ = µ are necessary
conditions for equilibrium. Illustrate the energy surface
using a graph drawn in (N /V , A /V ) or (V /N , A /N )
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
3
(2)
co-ordinates. Use reduced variables ρr , vr and ar
where ρr ∈ h0, 3i when you draw the diagram.
ev = V ,
7
It can be shown that the axis intercepts in the diagram are proportional to p and µ respectively, which means that the tangent requirement is also equivalent to p vap = p lıq and µvap = µlıq .
8
This can be considered the result of a differential condition for equilibrium, whereas the Maxwell equal area rule in Theme 113 on page 762
gives an integral formulation of the same condition.
768 / 1776
§ 112
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
en = N .
6
From a geometrical point of view, this condition will be met if the
two points have a common tangent.
02 September 2013
§ 110
v,n
The equilibrium state is therefore characterised by having two
points on a single isotherm that can be connected by a line that is entirely below the rest of the function surface (the graph).
Termodynamic methods TKP4175
§ 111
Aeq = min(A )T ,
4
Admittedly we are going to seek the minimum of the Helmholtz
energy function, but we will do that by distributing the total mass across
two different phases.
Tore Haug-Warberg (Chem Eng, NTNU)
News Clausius–Clapeyron Waals min A
The necessary conditions for thermodynamic equilibrium
in a closed system with only one chemical component,
at a given temperature T , total volume V = V vap + V lıq
and total composition N = N vap + N lıq , are
Helmholtz energy is a function of the standard state of the fluid,
3
and since standard state properties do not have absolute values, the
minimum of the function won’t be an absolute value either.
5
§ 110
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
9
The next question is how we best should illustrate the minimum.
10 A minimum in two variables would normally belong to a surface in
three dimensions but for Helmholtz energy it is sufficient to map the
function value along a straight line in the (V , N ) plane.
11 The line must actually be traversing the entire (V , N ) plane.
12 It is not sufficient that the line departs from the origin.
13 This would imply a simpel scaling of the extensive variables V and
N.
14 A scaling of this kind does not influence on the intensive properties
like p and µ and it has because of this no effect on the equilibrium.
15 In order to traverse the entire (V , N ) plane we can e.g. keep V
constant while changing N or we can do the opposite.
16 Here, the first option is chosen.
17 Figure 14.2 shows selected isotherms in the requested (N /V , A /V )
diagram, along with the saturation curve of the fluid (see Program 1.10
in Appendix 31).
= 0.
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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02 September 2013
773 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
ENGLISH text is missing.The tangent plane
1
In order to understand the details of the figure we need to know the
slope of the graph and the intersection of the tangent bundle and the
ordinate axis.
2
The differential of A /V in reduced co-ordinates is:
µv
A
c
d(ar · ρr )V =
· Nv
ˆ d NRT
= VRTc c dN
V
c
V
3
4
Maxwell equal area rule I
3
2.5
pr =
ˆ p /pc
V
After re-calculating the right-hand side we get:
−pV +µN
µ
µ
pV
A
ˆ −pr ·
NRTc − RTc · ρr =
NRTc − RTc · ρr = NRTc · ρr =
a⋆ =
pc vc
RTc
11 This implies that the two points must actually share the same tangent.
12 Figure 14.2 shows what the construction looks like for the second
lowest isotherm.
§ 112 Corresponding state
1
The results calculated from the Van der Waals equation have been
drawn in reduced coordinates in Figure 14.3 on the preceding page
using the program in Appendix 31:1.8.
2
Theoretically, the two experiments should fit the lowest isotherm
but they are in this example located outside the calculated two-phase
domain.
3
This means that the priciple of corresponding states is not an exact
science.
4
Note in particular that the Van der Waals equation overestimates
the volume of N2 in the liquid phase.
5
This is a weakness that is common to all cubic equations of state.
The minimum principle was illustrated using a (N /V , A /V ) diagram
6
In the vapour phase the discrepancy is relatively speaking smaller,
§ 112
above but we could equally well have been using a (V /N , A /N ) diabecause there the molar volume is greater than in the liquid phase.
1
At 101 K nitrogen has the following saturation state: p = 8.344
gram.
bar, ρlıq = 24.360 mol dm-3 and ρvap = 1.222 mol dm-3 . The critical 7 Also note that the areas above and below the hatched phase
14 This would, however, change the definition domain of the abscissa
point exists at pc = 34.000 bar, ρc = 11.210 mol dm-3 and Tc = 126.2 K. boundary line at Tr = 0.8 based on experimental values are roughly
from ρr ∈ h0, 3i to vr ∈ h 13 , ∞i which is a little impractical.
Calculate the saturation state expressed in reduced coordinates and equal.
15 At the same time the values of the slope and the intersection would draw your result onto the graph prVdW = f (Tr , vr ); also see Theme 49 8
This is not a coincidence, and for a correctly calculated phase equiswap (consider it a bonus task to show this) without having any negative on page 328. Comment on the discrepancy between the calculated librium the Maxwell equal area rule states that the two areas will be
identical.
volumes of vapour and liquid and the experimental values.
side effects.
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13
14 Simple . . .
2
Maxwell’s area rule gives an elegant depiction of the thermodynamic criterion for phase equilibrium.
Maxwell equal area rule I (2)
14 Simple . . .
7
This being the case proves that the net area of the v (p ) graph
will be equal to zero when integrating the volume from one saturation
pressure to another.
8
The construction is conceptually simple but not very practical[a].
In this chapter we have been dealing exclusively with Helmholtz energy
and it would be better if the representation is based on a p (v ) graph.
9
That variable change is easily undertaken using integration by
parts like it is discussed in the paragraph below.
[a] If you rotate Figure 14.3 ninety degrees in the mathematical sense, turns the page
and holds it up against the light, you will understand why. The v (p ) graph forms a
pretty manifold which unfortunately is hard to exploit mathematically.
c.p.
5
1
3
We are next expanding the volume of this liquid (at constant temperature) thereby gradually changing its character from liquid to vapour.
4
In practise the liquid would evaporate through boiling but we shall
do this experiment in our minds only and with a mathematical model to
our disposal.
The isothermal state change is then described by: (dµ)T ,N = v dp .
6
By integrating the state from the saturated liquid to a saturated
vapour we get
0.5
µRvap
0
10
0
ˆ v /vc
vr =
10
µlıq
1
Figure 14.3: The Van der Waals equation of state expressed in pr , vr coordinates for Tr = {0.8, 1.0, 1.2}. The horizontal line shows the experimentally established two-phase region for nitrogen at Tr = 0.8.
Tore Haug-Warberg (Chem Eng, NTNU)
r
7
9
Furthermore, the solution of the energy minimization problem we
posed earlier tells us that the pressure and the chemical potential must
balance in the two phases.
10 The only way this can happen is that two points along the same
isotherm has the same slope and the same intersection.
Assume there exist a liquid phase brought to its boiling point by
external means.
1
2
1.5
The corresponding differential of N /V is:
vc
c
ˆ d Nv
(dρr )V =
V V = V dN
The elimination of dN gives the slope of the graph:
!
d(ar · ρr )
µ
ˆ µr
= RTc =
dρr
V
5
The intersection of the tangent and the ordinate is now easily calculated:
!
d(ar · ρr )
a ⋆ = a r · ρr −
· ρr
dρ
6
As a small curiosity we note that a⋆ is identical to the Legendre
transform of ar · ρr with respect to ρr .
8
In other words, the slope of the graph is equal to µr while the intersection of the tangent with the ordinate is proportional to pr .
Termodynamic methods TKP4175
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(dµ)T ,N =
pRvap
v dp = 0 ,
p lıq
assuming that the isothermal equilibrium state is governed by p vap =
p lıq and µvap = µlıq .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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§ 112
§ 114
News Clausius–Clapeyron Waals min A
14 Simple . . .
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 113 Maxwell equal area rule II
News Clausius–Clapeyron Waals min A
14 Simple . . .
1
There are at least two alternative ways of deriving this relationship.
2
Integration by parts is one. Applied to the Maxwell area rule we
get:
Maxwell equal area rule can be expressed as
R The
V vap
(p − psat )(dV )T ,N = 0 and be plotted on a figure
V lıq
equivalent to 14.3, but with the ordinary volume along
the abscissa. Show that the proposed equation actually
holds for a one-component phase equilibrium.
0=
pRvap
p lıq
vRvap
v dp = psat
p lıq
v lıq
dv −
vRvap
v lıq
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
02 September 2013
780 / 1776
(p − psat ) dv =
vRvap
v lıq
vRvap
v lıq
p dv = −
vRvap
v lıq
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
If we substitute in the temperature from Eq. 14.14 and rewrite the
equation in terms of molar density, the chemical potential equation can
be expressed as
3(x −y )
x (3−y )
6
(14.15)
ln y (3−x ) + 1 − x +
y (3−x )(3−y ) = 0 .
7
02 September 2013
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3
The solution is shown graphically in Figure 14.4 on the next page
along with a corresponding (pr , µr )-diagram .
2
The latter was implemented in the Function 2.5 shown in Appendix 31, which calculates x ∈ [1, 3i for a given value of y ∈ h0, 1].
Expanded footnote [a]:
[a] Drawn as the parametrised graph pr (vr ), µr (vr ) where the volume increases from
the top right corner (liquid) down to the bottom left corner (gas) of the figure.
Note that the slope of the graph asymptotically approaches −1 as
it nears the critical point, where xc = yc = 1. The average density of
the two phases therefore has the nearly constant value 12 (x + y ) ≃ 1.
vap
VRvap
V lıq
(p − psat ) (dV )Tsat ,N = −
VRvap h
V lıq
∂A ∂V
T ,n
?
14 Simple . . .
i
+ psat dV
= A lıq − A vap − psat V vap − V lıq
= 0.
(µN )lıq = (µN )vap .
?
5
(14.12)
(14.13)
It then follows that
µlıq = µvap ,
=
8Tr
vap
3vr −1
−
x 2 −y 2
.
3
vap 2
)
(v r
,
2y
2x
3−x − 3−y
(14.14)
5
Here x ∈ [1, 3i and y ∈ h0, 1] are reduced (molar) densities defined
lıq
vap
by x =
ˆ (vr )-1 and y =
ˆ (vr )-1 .
3vr −1
Tore Haug-Warberg (Chem Eng, NTNU)
4vr
r
r
Termodynamic methods TKP4175
r
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Molar density
14 Simple . . .
In order to determine the saturation state of the fluid, we need to
solve Eq. 14.15 for y = y (x ) or x = x (y ).
lıq
3
lıq
( vr ) 2
−
News Clausius–Clapeyron Waals min A
1
6
8Tr
lıq
3vr −1
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Pressure–potential diagram
2
The integral along the isotherm can then be expressed with the
following:
which is identical to the phase equilibrium criterion for a one-component
system.
782 / 1776
The fact that the pressure is the same in both phases means that
3vr −1
Tore Haug-Warberg (Chem Eng, NTNU)
ENGLISH text is missing.Maxwell equal area rule
III
1
Alternatively we may recognise −p as the partial derivative
of Helmholtz energy with respect to the volume along the chosen
isotherm.
3
Since the system only has one chemical component, it must also
be true that A = −pV + µN . Moreover, when the vapour and liquid phases are in equilibrium, p lıq = p vap = psat . Hence the equality
in 14.12 can be simplified to
4
The integration has been performed at constant temperature Tsat
and total composition N . Hence, as in 14.13, the mole number can
be considered a constant (common) factor for the vapour and liquid
phases.
6
Similarly, the fact that the chemical potential is the same in both
phases means that
Tr
Tr Tr
− 9lıq = Tr ln 3v8vap
+ 3v vap
− 4v9vap .
Tr ln 8lıqTr + lıq
−1
−1
(p − psat ) dv = 0 .
(p − psat ) dV = 0 .
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Vapour–liquid equilibrium (2)
2
At thermodynamic equilibrium, T , p and µ have uniform values over
the entire phase ensemble, which in this case is made up of the vapour
and liquid phases.
3
Since the Van der Waals equation is expressed on a pressure explicit form in terms of pr (Tr , vr ), the equilibrium equations have to be
vap lıq
solved simultaneously in vr , vr coordinates.
4
We can’t expect there to be an analytical solution of this (nonlinear) 2 × 2 equation system, but there is no harm in setting out the
equations.
8T r = 6
4
At constant composition dV = N dv which means we can also write
(the negative sign is not important here):
N
1
v lıq
3
Introducing the equilibrium condition p lıq = p vap = psat and writing
the difference in the phase volumes as an integral:
pRvap
Vapour–liquid equilibrium
which can be solved for temperature:
vRvap
vap vRvap
v dp = (vp )lıq −
p dv = v vap p vap − v lıq p lıq −
p dv .
v lıq
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Pressure–chem.pot.
3
3
2.5
2
c.p.
µr =
ˆ µ/RTc
§ 113
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
x=
ˆ ρlıq
ˆ vc /v lıq
r =
News Clausius–Clapeyron Waals min A
2
1.5
1
0
lıq
4
In the last figure the simultaneous solution of pr = pr and µr =
vap
µr appears as the triple point of a manifold with three regions: Vapour,
liquid and an unphysical fluid phase for which (∂p /∂v )T > 0.
5
The figures were drawn using the Program 31:1.10.
This means that the liquid density decreases at the same rate as
the vapour density increases at the far right in the left sub-figure.
7
8
Experimentally it has been demonstrated that the value of x + y is
not quite constant, but nevertheless it is virtually a linear function (hence
it is also referred to as the rectilinear density) of Tr over a wide range of
temperatures.
1
0
0.5
y=
ˆ ρvap
=
ˆ vc /v vap
r
1
−1
−1
0
1
2
pr =
ˆ p /pc
lıq
Figure 14.4: ρvap
r , ρr saturation graph, with the associated isotherms, expressed in pr , µr coordinates (drawn with volume as the parameter).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
02 September 2013
784 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Lower temperature limit
2
Not because it is required right now but in order to make the
vapour–liquid formulae complete:
Tr
9
(14.16)
∆v hrVdW = (y − x ) (3−x3)(3−
y) − 4 .
This is as far as we can get by looking at the general case, so we
will now move on to a discussion of the asymptotes of the left graph in
Figure 14.4.
3
4
The discussion will go on for next two sections but there are many
details and in order to appease the impatient reader we refer immediately to the final results in Table 14.7.
5
The table summarises the main results from the coming discussion
of the lower and upper temperature limits Tr ≪ 1 and Tr → 1.786
/ 1776
2
At temperatures far below the normal boiling point of the liquid
phase, the molar density approaches a constant value of limTr ≪1 (x ) = 3
while the corresponding molar vapour density y tends to zero.
if only the dominant terms are included.
3
The asymptotic solution of β is
lim β =
Tr ≪1
9
−3
α·exp( α )
(14.17)
,
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
ENGLISH text is missing.Enthalpy of vaporization
1
The last piece of information needed is the enthalpy of vaporization
from Eq. 14.11 expressed on density form.
At sufficiently low temperatures the reduced densities can therefore
be expressed as x = 3 − α and y = β, where α and β are small positive
numbers. When substituted into the equilibrium Eq. 14.15 this gives us
3(3−α−β)
(3−α)(3−β)
9
6
≃ ln βα
+ 1 − 3−α+β
− α3
0 = ln
βα
α(3−β)
1
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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785 / 1776
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Lower temperature limit (2)
14 Simple . . .
4
which shows that the vapour density declines exponentially with
falling temperature.
5
In that limit case, the temperature Eq. 14.14 can be simplified as
lim (8TrVdW ) = 6
Tr ≪1
(3−α)2 −(β)2
2β
2(3−α)
− 3−β
α
≃ 9α ,
(14.18)
and if we substitute that into the Van der Waals equation 14.5 we get
lim prVdW =
Tr ≪1
8Tr
3
β −1
− 3β2 ≃
α
3
9α
exp( α3 )−1
≃
27
exp( α3 )
.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
6
Combining these latter two equations allows us to eliminate α, giving us the temperature-explicit form:
lim ln prVdW = ln(27) − 827
Tr
(14.19)
Tr ≪1
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Lower temperature limit (3)
14 Simple . . .
8
In practice, the low-temperature limit can never be reached, because most liquids freeze at a temperature of Tr ≈ 0.7. Table 14.5 on
page 774 therefore gives a comparison with the heat of vaporization for
a few simple substances measured at the normal boiling point.
Expanded footnote [a]:
9
As can be seen from the table, the correlation between the two is
no more than reasonable.
10 Confer also Table 14.7 to get an overview of the main results obtained in this section.
[a] Robert C. Reid, John M. Prausnitz, and Thomas K. Sherwood. The Properties of
Gases and Liquids. McGraw-Hill, 3rd edition, 1977.
which conforms to the Clausius–Clapeyron vapour pressure Eq. 14.3,
with a slope of −b = − 27
8 = −∆v hr .
7
This result is confirmed by explicitly calculating the heat of vaporization due to Eqs. 14.16 and 14.18:
3Tr
− 94
lim ∆v hrVdW = (β − 3 + α) α(3−β)
Tr ≪1
≈ −(3 − α) Tαr − 94
≈
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
27
8
.
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
790 / 1776
News Clausius–Clapeyron Waals min A
§ 114
§ 113
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 115
News Clausius–Clapeyron Waals min A
14 Simple . . .
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 114 Clapeyron’s equation I
News Clausius–Clapeyron Waals min A
14 Simple . . .
As previously mentioned, when the temperature approaches zero,
the specific volume of the liquid tends to a ratio of limTr ≪1 (vrVdW ) = 13 .
1
vap
ˆ vr
lim ∆v vr =
Tr ≪1
lim (pr · ∆v vr ) ≈
News Clausius–Clapeyron Waals min A
791 / 1776
N2
Ar
O2
CH4
Kr
H2 O
Tc
[K]
∆v h
[cal/mol]
∆v h
RTc
∆v h VdW
[cal/mol]
77.35
87.27
90.17
111.66
119.74
373.15
126.2
150.8
154.6
190.4
209.4
647.3
1333
1560
1630
1955
2309
9717
5.32
5.21
5.31
5.17
5.55
7.55
846
1011
1037
1277
1404
4341
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
794 / 1776
= 2(4 − α2 ) ,
(14.21)
which can in turn be substituted into the Van der Waals equation 14.5,
to give us
− 3x 2 =
2(4−α2 )
3
1+α −1
− 3(1 + α)2 = 1 − α2 .
Tore Haug-Warberg (Chem Eng, NTNU)
=
8Tr
3−β
− 3β ≈
∆v hr
∆ v vr
8Tr
3
≈ − 83 ·
= 3α .
Tr
3α
·
27
8
= − 27
8 .
02 September 2013
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14 Simple . . .
Termodynamic methods TKP4175
If α2 is eliminated from the last two equations, the saturation pressure
can instead be written as a function with temperature as its free variable:
lim (prVdW ) = 4Tr − 3 .
02 September 2013
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
795 / 1776
7
6
Now let us expand ln(1 + x ) = x + O(x 2 ) and multiply out the terms
in brackets.
Tr →1
which due to limTr →1 (8Tr ) = 2(4 − α2 ) simplifies to:
∆v hrVdW = −2α 34TTrr − 94 = 3α .
16 Or, we can write the enthalpy expression on a temperature explicit
form:
p
p
(14.23)
lim ∆v hrVdW = 3α = 3 4 − 4Tr = 3 1 − Tr .
793 / 1776
14 Simple . . .
We can ignore all higher order terms of the kind αβ, α2 and β2 :
3(α+β)
6
0 = α + 12 β − (−β − 21 α) + 1 − 2+α−β
4−2α+2β−αβ
6
1
≃ 1 + 1 − 2+α−β
2+β−α
= 4 − (β − α)2 + 2 + (β − α) − 6
(14.20)
≃ β − α.
8
The conclusion is that β ≈ α, which shows that the vapour and
liquid densities are symmetrical about the point xc = yc = 1.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
02 September 2013
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Upper temperature limit (2)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
14 Substituting x = 1 + α and y = 1 − β where limTr →1 (β) = α into
Eq. 14.16 yields
3Tr
3Tr
9
lim ∆v hrVdW = −(α + β) (2−α)(2+β)
− 94 = −2α 4−α
2 − 4 ,
15
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Upper temperature limit (4)
(14.22)
11 Here we should note that both TrVdW and prVdW are simple functions
of α2 .
12 In other words, the phase envelope prsat (ρr ) calculated in Figure 14.6 is parabolic with the critical point as its origin.
§ 115
§ 114
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 116
14 Simple . . .
Use the information available to calculate limTr →1 (pr · ∆v v ).
Afterwards, calculate the derivative in Eq. 14.1 at the
beginning of this chapter. Compare the result with
Eqs. 14.2 and 14.4.
Tr →1
Tr →1
13
4
If we combine this with the Clausius–Clapeyron vapour pressure
Eq. 14.2–14.4 we get the slope limTr ≪1 b = −∆v hr = − 27
8 .
.
2
At low temperatures the liquid molar density approaches a constant
value of limTr ≪1 (x ) = 3 while the corresponding molar vapour density y
tends to zero.
News Clausius–Clapeyron Waals min A
[a] The critical volume of a fluid can be estimated using straight-line extrapolation of
the sub-critical vapour and liquid densities, while the critical pressure and temperature
can be measured directly.
8Tr
3
x −1
1
β
3
In this state there is no fundamental difference between a vapour
and a liquid.
4
This idea is supported by Figure 14.6 on page 782, which shows
that the density of the two equilibrium phases is symmetrical about xc =
yc = 1 close to the critical point.
[a] Molar entropy displays a similar symmetry; see Figure 18.9 on page 975.
lim (prVdW ) =
≈
Near the critical point, the reduced densities can be expressed as
5
x = 1 + α and y = 1 − β, where α and β are small positive numbers. If
we substitute these values into the equilibrium Eq. 14.15 we get
3(α+β)
(1+α)(2+β)
6
= 0.
ln (1−β)(2−α) + 1 − 2+α−β
(2−α)(2+β)
9
This is an important result, which has also been confirmed experimentally[a][a]. Confer also the previous Figure 14.4 on page 767 and
the next Figure 14.6 (cf. Program 1.11 in Appendix 31).
Tr →1
1
3−α
As the temperature increases, the liquid phase gradually becomes
less significant, until it ceases to exist entirely at the critical point where
limTr →1 (x ) = limTr →1 (y ) = 1.
14 Simple . . .
2(1+α) 2(1−α)
2−α − 2+α
−
Upper temperature limit
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
(1+α)2 −(1−α)2
1
β
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
10 By substituting into the temperature Eq. 14.14 we can obtain the
limit value
lim (8TrVdW ) = 6
pr
β
∆µ=0
Tore Haug-Warberg (Chem Eng, NTNU)
1
02 September 2013
Upper temperature limit (3)
Tr →1
Tr ≪1
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Figure 14.5: Enthalpy of vaporization measured at the normal boiling point of
a few selected substances compared with ∆v hrVdW = 27
8 = 3.375.
Tb
[K]
lıq
− vr =
Combined further with Eq. 14.1:
dln p VdW
c
= − pRT
lim d(1/Trr )
· Tprr ·
c vc
3
02 September 2013
14 Simple . . .
Combined with the Van der Waals Eq. 14.6 and the temperature
limit 14.18 we get:
Tr ≪1
Termodynamic methods TKP4175
§ 114 Clapeyron’s equation I (2)
2
In the same limit, the volume of the vapour phase will completely
dominate the vaporization volume:
Use the information available to calculate limTr ≪1 (pr · ∆v v ).
Afterwards, calculate the derivative in Eq. 14.1 at the
beginning of this chapter. Compare the result with
Eqs. 14.2 and 14.4.
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
17 The well-known correlation due to Watson[a] states that
∆v h Watson = a (1 − Tr )b where b ≈ 0.38 for ordinary nonpolar organic
compounds, goes a long way towards confirming this result.
[a] K. M. Watson. Ind. Eng. Chem., 35:398, 1943.
The enthalpy of vaporization has a similar representation.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
§ 115 Clapeyron’s equation II
News Clausius–Clapeyron Waals min A
14 Simple . . .
1
Close to the critical point, the reduced densities of the two phases
are x = 1 + α and y = 1 − β:
4
In particular you should note the expression to the far right which
tells us that the vaporization volume is proportional to ∆v hr /pr .
2
From here, the derivation closely follows that given in the main text
above:
vap
ˆ vr
lim (∆v vrVdW ) =
Tr →1
3
lıq
− vr =
1
1−β
−
1
1−α
≈
2α
(1−α)2
=
2α
pr
=
2
3
·
∆v hr
pr
See Eqs. 14.22 and 14.23 for more details.
The central issue is that the right-hand side of Eq. 14.1 approaches
7
a constant value even though the volume of vaporization and the enthalpy of vaporization are both tending towards zero:
dln p VdW
2
3
8
c
v hr
lim d(1/Trr )
= − pRT
· Tprr · ∆
∆v vr ≈ − 3 · Tr · 2 = −4Tr = −(4 − α ) .
c vc
5
The denominator cancels thereby with the multiplication of pr
lim (pr · ∆v vrVdW ) =
Tr →1
2∆v hr
3
,
6
which is the direct cause for that the vapour pressure curve in Figure 14.1 looks almost straight.
Tr →1
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Slope of vapour pressure
News Clausius–Clapeyron Waals min A
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
3
14 Simple . . .
1
In other words, the predicted slope of the vapour pressure curve
varies in the range −4 < −b < − 27
8 .
pr =
ˆ p /pc
News Clausius–Clapeyron Waals min A
The Van der Waals equation therefore gives us a qualitatively ac3
curate representation of the vapour pressure curve, although from experience we know that the model is insufficiently accurate for quantitative purposes.
2
This is what the theory predicts but at the small scale of Figure 14.1
the variation in the slope is hardly noticeable for the experimental data
points either, and the graph appears to be a straight line.
2
1
0
As seen in Figure 14.6 the one-phase region is not adequately
represented either, but once again it is qualitatively accurate .
0
0.5
1
1.5
2
2.5
ρr =
ˆ vc /v
4
Expanded footnote [a]:
[a] W. Duschek, R. Kleinrahm, and W. Wagner. J. Chem. Thermodynamics, 22:841–
864, 1990.
∆µ=0
8
If we combine this with the Clausius–Clapeyron vapour pressure
Eq. 14.2–14.4 we get the slope limTr →1 b = −4Tr2 ≥ −4.
Figure 14.6: The saturation pressure of CO2 along with selected isotherms
in the one-phase region (280 K, 313 K and 330 K respectively). Note the symmetry of the saturation pressure prsat around the critical density ρr = 1.
9
From Section 14.6 we know that limTr ≪1 b = −∆v hr = −27/8. The
difference is surprisingly small.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
Near-critical density
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
1
As such, only small changes are needed to improve the predictive
accuracy.
2
In 1949, Redlich and Kwong[a] introduced the first improvement by
making the a term temperature-dependent.
3
There are now around a hundred closely related models, with the
Redlich–Kwong–Soave[a] (SRK) and Peng–Robinson[a] (PR) models
being two of the best-known ones.
4
They are all cubic functions of the volume (and thereby of density),
and their qualitative properties are the same.
5
This means that the Van der Waals equation of state still remains
an important conceptual model — 140 years after it was formulated the
first time.
803 / 1776
[a] Otto Redlich and J. N. S. Kwong. Chem. Rev. (Washington, D. C.), 44:233–244,
1949.
[a] Giorgio Soave. Fluid Phase Equilib., pages 1197–1203, 1972.
02 September 2013
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Tr →1
r
This makes the logarithm of the saturation volume of the liquid
phase
p
lıq
lim (ln(vr )VdW ) ≃ −2 1 − Tr ,
2
Tr →1
while the Rackett equation, which forms the basis for several semilıq
empirical correlations, states that ln(vr )Rackett = ln(zc ) · (1 − Tr )2/7
where ln(zc ) ≈ −1.2 for nonpolar substances.
Expanded footnote [a]:
[a] Harold G. Rackett. J. Chem. Eng. Data., 15(4):514–518, 1970.
3
The parameters in the two equations are different, but qualitatively
there is still a real correspondence between theory and practice[a].
4
Confer also Table 14.7 to get an overview of the main results obtained in this section.
[a] The Van der Waals equation indicates that the function should take the form a (1 −
Tr )b and also gives us good initial values that can be adjusted to reflect experimental
data (both the proportionality constant ln(zc ) and the exponent 2/7 in the Rackett
equation are empirical).
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
02 September 2013
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
2nd order phase transition (2)
Two phases, Tr < 1 :
804 / 1776
ρvap
r
Tr
pr
3−α
β
−
−
vrlıq
1
3−α
vrvap
1
β
pr · ∆v vr
−
∆v hr
∂ ln pr
∂(1/Tr )
−
−
Tore Haug-Warberg (Chem Eng, NTNU)
Approximation
Definition
3−α
9
α
(14.27)
Approximation
2nd order phase transition
02 September 2013
809 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
1
It has been theoretically proven — and experimentally confirmed —
that the energy surface close to the critical point of a vapour–liquid
form, which is parametrised by what we call
equilibrium is of a universal
an order parameter.
2
The order parameter reflects the characteristic behaviour of the
fluid in phase transitions.
14 Simple . . .
1
In Theme 49 on page 328 we showed that the Van der Waals equa- 3 By definition it is zero at the critical point and in the surrounding
tion can be written in its reduced form without the use of any arbitrary one-phase region.
model parameters.
4
In two-phase regions it gradually increases with the distance from
the critical point.
2
3
This produces two non-linear equations that contain the model pa- 6 There is absolutely no reason for it to do so, since there is no
rameters a and b .
thermodynamic distinction between a vapour phase and a liquid phase.
They are both fluids and nothing more.
8
For vapour–liquid equilibria, it is natural to use ρr − 1 as an order
5
This results in a useful form of the equation of state, which we parameter[a], provided that it produces symmetrical phase transitions,
have used in this chapter to analyse the fluid’s behaviour in a number which it indeed does.
of interesting limit cases.
9
In fact, that order parameter is precisely what explains the rectilin6
Unfortunately, Figure 14.6 shows that the theory is by no means ear density graph mentioned in conjunction with Figure 14.4.807
universally applicable.
/ 1776
7
In this section we will look in greater detail at what happens close to
the critical point, and quantify to what extent the Van der Waals equation
[a] The theory applies to all phase equilibria that have a single (scalar) order parameter.
fails to predict experimental results.
For example, ferromagnetic materials display relatively analogous behaviour, which
806 / 1776 is described by the order parameter |Mr − 1|, where M is the magnetisation of the
substance.
1
The closer we measure towards the critical point, the more anomalous the p (T , V ) behaviour of the fluid.
2
Let us assume we can do measurements along the fluid’s critical
isochore moving towards the critical point.
3
In the direction of that point, but still outside the two-phase region,
it is possible to observe a gradual phase transition.
1−α
tion.
α
−
−
1 − α2
6
In the literature for this field, which incidentally covers a large and
exciting range of research , the critical anomalies of the fluid are expressed using a clever choice of 4 asymptotically divergent correlations:
1
1−α
1+α
−
2α
)
exp (−3
α
27 ·
1
α
3 1+ 3
1
4
1
1−α
· exp (α3)
3α
α
27
8 1− 3
−
− 27
8
−
Termodynamic methods TKP4175
(4 − α2 )
5
The name λ transition comes from the variation in heat capacity at
constant volume as a function of the distance to the critical point: When
Tr − 1 is used as the measure of distance from the critical point, the
shape of the graph is reminiscent of the Greek letter lambda.
Expanded footnote [a]:
[a] Kenneth G. Wilson received the Nobel prize in 1982 for his contribution to the
development of renormalisation groups applied to, amongst other things, critical phase
transitions.
1+α
γ
1
1.237(2)
1.3960(9)
1.3960(9)
0.10(0)
1.19(0)
δ
3
4.789(2)
4.783(3)
4.783(3)
4.35(0)
Termodynamic methods TKP4175
(14.24)
∝ (Tr − 1) ,
(14.25)
|pr − 1| ∝ |ρr − 1| +δ ,
(14.26)
∂pr Tr
Critical, Tr = 1 :
805 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
0.35(5)
815 / 1776
-γ
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
β
0.5
0.326(5)
0.3689(3)
0.3689(3)
02 September 2013
∝ (Tr − 1)-α ,
One phase, ρr = 1 :
−(4 − α2 )
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
α
0
0.110(1)
−.1336(2)
−.1336(2)
cv
∂ρr One phase, ρr = 1 :
3α
02 September 2013
This behaviour is called a λ transition or a 2nd order phase transi-
4
02 September 2013
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Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
ENGLISH text is missing.Isochoric heat capacity
(α)
Looking back
1
1
From our knowledge of thermodynamic theory, and particularly
from chapters 4 and 8 on Legendre transformations and on the thermodynamics of systems of constant composition, we know the general relationships (∂A /∂T )V ,n = −S and (∂S /∂T )V ,n =
ˆ CV /T .
ENGLISH text is missing.Universal scaling
ENGLISH text is missing.Correlation length
2
These equations can also be expressed in reduced coordinates as
In the critical state, there is a balance between the thermal energy
1
Our next observation is loosely related to statistical thermodynam- 1
(∂ar /∂Tr )vr = −s /R and (∂s /∂Tr )vr =
ˆ cv /Tr .
ics and to the knowledge that some models are easier to deal with than and the potential (inter-molecular) energy of the fluid.
3
Eliminating s gives us cv = −RTr (∂2 ar /∂Tr ∂Tr )vr . In other words,
others.
2
This results in the energy surface stretching out and losing its curthe heat capacity of the fluid gives us the second derivative of Helmholtz
2
The simplest special case is a gas in which the molecules are con- vature. Both the first and second order contributions vanishes thereby.
energy.
stantly moving around without colliding or interacting with each other in 3 Consequently, it is impossible for there to be a classic static equi4
For a Van der Waals fluid, we can obtain the Helmholtz energy by
any way. That is what we mean by an ideal gas.
librium, as there appears to be at other pressures and temperatures,
ENGLISH text is missing.Critical point measurecombining Eq. 14.8 with equations 14.6 and 14.7.
3
We will assume that the correlation length of the gas is zero, even judging by observations.
ments
5
Double differentiation of ar with respect to Tr gives us
at the risk of claiming to know something that we haven’t verified.
4
The compressibility diverges (tends to infinity) and the fluid appears
2 ◦
1
Table 14.8 gives a summary of what we have learned so far about
ENGLISH text is missing.Critical exponents
∂ µ cvVdW = −RTr ∂Tr ∂Tr r + T1r = cp◦ − R = cv◦ (T ) .
4
The second special case is a system in which the molecules (or to be constantly moving, without ever quite coming to rest.
phase transitions and how they relate to the Van der Waals equation.
1
In the next four sections we will derive the classical values of the
atoms) are arranged in a repeating lattice with long-range order. That 5 One practical consequence of these conditions is that the force of
exponents α, γ, δ and β, based on the properties of the Van der Waals
2
As can clearly be seen from the table, the equation of state does
6
Note that the temperature derivative of µ◦ gives us cp◦ rather than ENGLISH text is missing.Bulk modulus (γ)
is what we mean by a crystalline structure.
gravity becomes relatively more important, leading to relatively large
fluid.
not accurately describe what happens in real life, but we knew that
cv◦ .
1
The Van der Waals equation expressed in reduced Tr , vr coordi5
In one sense, the correlation length of the crystal is infinite. We do variations in density that can distort experimental results.
already.
2
However, the results that we will obtain are not only true for the
7
This is because the standard state is defined at constant p◦ , which nates was discussed in Theme 49. Here we need the same equation, ENGLISH text is missing.Critical isotherm (δ)
not say that the oscillations of the particles are strongly correlated but 6 Measurements have therefore been taken under zero-gravity conVan der Waals equation, they are also valid for all explicit equations of
3
What is surprising is that the experimental values do not diverge;
implies that v◦ (T , p◦ ) is a variable with respect to temperature.
but this time as a function of Tr and ρr :
1
We will start out from the Van der Waals equation expressed as a
that the position of each single particle is known once the unit cell of ditions in order to validate current theories relating to the field.
state, regardless of whether or not they are cubic.
instead they suggest that physical quantities exhibit certain universal
8Tr ρr
r
function of Tr and ρr , just as we did for the bulk modulus.
the crystal is known.
8
This topic is discussed more thoroughly in the context of Eq. 7.9 on
− v32 =
− 3ρ2r
prVdW = 3v8rT−1
ˆ 3−ρ
behaviour during phase transitions.
7
All observations seem to indicate that theory and practice agree,
r
3
This is because we are only interested in the exponents in the
r
ENGLISH text is missing.Nature is non-analytical
page 279.
2
Along the critical isotherm, the equation of state can be factored as
above correlations. The proportionality factors are of no significance in
4
Here we should note that all the exponents α, γ, δ and β vary 1 What we can say for sure is that there is no equation with an ana- 6 The third special case only arises in conjunction with phase transi- but that the Van der Waals equation (like other equations of state with
2
Let us differentiate the pressure with respect to density, and rearfollows:
9
We can therefore conclude that αVdW = 0 in Eq. 14.24.
analytical solutions) produces inaccurate estimates of the critical expothis respect.
surprisingly little, both in response to changes in the chemistry of the lytical solution that accurately describes critical phenomena in nature. tions.
range the resulting equation in such a way as to isolate Tr − 1 in one of
8Tr ρr
nents.
lim (prVdW − 1) = 3−ρ
− 3ρ2r − 1
system and to the physical interpretation of the order parameter.
4
It turns out that the exponents are the same, whether you start out 10 This assumes that the heat capacity in the standard state behaves the numerators:
r
2
In other words, the critical exponents are neither integers (z ∈ Z) 7 Here the correlation length varies over several orders of magnitude,
Tr →1
ρr
from the Van der Waals equation or from a more advanced (modern) “normally” close to the critical temperature of the fluid, but this is very
∂pr VdW
Tr −1
3
1
5
Intuitively you would expect there to be a big difference between nor simple fractions (q ∈ Q). That is what they would need to be for which means that this phenomenon is much more significant than the 8 However, the principle that the critical phenomena display univer(ρ3r − 3ρ2r + 3ρr − 1)
= 3−ρ
lim
= 24 (3−ρ
= 6(Tr − 1)
2 + (3−ρ )2 − 4
r
r)
r
likely to be the case, given that cv◦ refers to an ideal gas state.
sal scaling behaviour holds true, and there is in fact a general rule that
ρr →1 ∂ρr Tr
equation of state.
vapour–liquid phase transitions and ferromagnetism, but it turns out there to be any hope of finding an equation of state with an analytical contributions from the intra-molecular states of the particles.
| {z }
3
(ρr − 1)3
= 3−ρ
ENGLISH text is missing.Binodal curve (β)
r
→0
that the critical exponents of these phenomena are roughly the same. solution capable of describing the critical phase transition behaviour.
8
At the critical point, the correlation length increases to a macro- applies to critical phase transitions in all substances, as stated in equa5
The reason is that a Taylor series expansion of the critical proper- 11 After all, the critical state is a result of the non-ideal nature of the
1
Based on our discussion of phase equilibrium problems in Sections 14.24–14.27. Moreover, we also know that
ties gives the same classic exponents independent of the exact form of fluid, and the standard state doesn’t tell us anything about non-ideal 3 The exponent of the expression on the right side of the equation is 3 That gives us δVdW = 3 in Eq. 14.26. In other words, the critical
6
We are clearly standing at the threshold of a new understanding of 3 This is because a series expansion of such an equation would scopic level and to thereby interfere with the wavelength of visible light.
tion 14.5, and with particular reference to Eq. 14.22, it follows that:
conditions.
α = 2−β(δ + 1)
the equation of state.
1 which means that γVdW = 1 (not −1) in Eq. 14.25.
isotherm describes a third-degree curve with its origin at the critical
the world.
always produce exponents that are either rational numbers or integers. 9 This phenomenon is called critical opalescence.
2 (ρr − 1) = ±(1 − Tr )0.5 .
γ=
β(δ − 1)
12 Although c ◦ is a function of temperature, it does never display any
6
This classification is also known as the Landau[a] theory.
v
4
The expression above is written for the bulk modulus while point.
7
However, the theory that underpins it is unfortunately beyond our 4 Nature is not like that, and we must therefore give up on trying 10 It induces a characteristic light scattering that results in an array of
-α
810 / 1776 temperature divergence of the type (Tr − 1) .
Eq. 14.25 is valid for the isothermal compressibility. These are mu- 4 This is why the Van der Waals equation and other equations of the 2 The vapour pressure curve describes a parabola with the order pa- reach, and we will therefore have to limit ourselves to a verbal discus- to find an all-encompassing universal equation to describe the laws of beatifully tinted colours when white light hits a sample that is in a near- The exponents summarised in Table 14.8 are judiciously adjusted (using the least square principle) to fulfill these relations.
13 c ◦ can therefore be considered a constant in this context.
rameter ρr − 1, which means that the exponent βVdW = 0.5 in Eq. 14.27. sion of the topic.
tually resiprocal quantities which explains the sign shift of the exponent. same familiy are sometimes called “cubic equations of state”.
nature.
critical state.
v
819 / 1776
811 / 1776
813 / 1776
814 / 1776
812 / 1776
816 / 1776
817 / 1776
818 / 1776
14 Simple . . .
[a] Lev Davidovich Landau, 1908–1968. Russian theoretical physicist.
Saturation state:
dp dT ∆µ=0
=
s vap −s lıq
v vap −v lıq
=
ˆ
∆v s
∆v v
∂ ln p p →0 ∂(1/T ) ∆µ=0
=
∆v h
T ∆v v
= − ∆Rv h
2
Clausius–Clapeyron: lim
3
Rectilinear density: ρr + ρr ≈ 1
R V vap
Maxwell’s area rule: V lıq (p − psat )(dV )T ,N = 0
4
5
lıq
vap
Van der Waals equation of state:
p VdW
pc
=
ˆ prVdW =
µVdW
=
ˆ µVdW
= µ◦r − Tr ln pr◦ + Tr ln
r
RTc
Termodynamic methods TKP4175
ENGLISH text is missing.Order parameter
ENGLISH text is missing.Near-critical states
The underlying theory is based on the critical properties of the fluid
under the conditions (∂p /∂v )T = 0 and (∂2 p /∂v ∂v )T = 0, calculated at 5 For reasons of symmetry, we wouldn’t expect the order parameter
T = Tc and v = vc , and of course also at p = pc .
to differ between vapour–liquid and liquid–vapour transitions.
4
Conventionally, the equations are solved with respect to a and b 7
as functions of Tc and pc .
1+α
[†] J. M. H. Levelt Sengers and J. V. Sengers. Phys. Rev. A, 12(6):2622–2627, 1975.
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
1−β
Figure 14.8: Critical exponents measured experimentally for selected compounds compared with the classical values calculated from the Van der Waals
equation of state.
System
Van der Waals
Ising 3D [†]
Heisenberg 3D [†]
Ni, Fe, Gd2 BrC, Gd2 IC,
Tl2 Mn2 O7 , . . . [†]
3
He, 4 He, Xe, CO2 ,
H2 O, O2 [†]
802 / 1776
1+α
9
8
α
9
Tr → 1
02 September 2013
)
· exp (−3
α
News Clausius–Clapeyron Waals min A
14 Simple . . .
|ρr − 1| ∝ (1 − Tr )+β .
lıq
ρr
Definition
Tr ≪ 1
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
Figure 14.7: Asymptotic solutions of the vapour–liquid phase equilibrium.
Property
Tore Haug-Warberg (Chem Eng, NTNU)
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
[a] Ding-Yu Peng and Donald B. Robinson. Ind. Eng. Chem. Fundam., 15(1):59–64,
1976.
1
The volumes of the saturated phases can be expressed as x =
ˆ
(vr )-1 = 1 ± α where α is a small number, and from Eq. 14.21 we can
then derive the asymptotic temperature function:
p
1
1
lim (vrVdW ) = 1±α
≃ 1 ∓ 2 1 − Tr .
= 1± √4−4
T
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Clausius–Clapeyron Waals min A
14 Simple . . .
ENGLISH text is missing.Cubic equations of state
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
8Tr
3vr −1
−
3
vr2
8Tr 3vr −1
Termodynamic methods TKP4175
+
Tr
3vr −1
−
9
4vr
02 September 2013
820 / 1776
News Clausius–Clapeyron Waals min A
Phase boundary Tr ≪ 1 Tr → 1 Criticality Olds
Looking back (2)
6
New concepts
Critical point behaviour:
cv
∂ρr One phase, ρr = 1 :
∝ (Tr − 1)-α
∂pr Tr
∝ (Tr − 1)-γ
Part 15
|pr − 1| ∝ |ρr − 1| +δ
Critical, Tr = 1 :
Two phases, Tr < 1 :
Multicomponent phase equilibrium
|ρr − 1| ∝ (1 − Tr )+β
Classical values for the exponents: α = 0, γ = 1, δ = 3 and β = 0.5
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Minimum Gibbs energy I
821 / 1776
1
At thermodynamic equilibrium the system has reached a state of
minimum energy.
2
This far-reaching postulate is of tremendous importance for our
understanding of matter and energy, but it remains to discuss which of
the energy functions that are being minimised.
3
In Chapter 16 on page 851 it is for example proved that Gibbs
energy of a closed system decreases (due to chemical reactions) till
the minimum value minn G (n)T ,p is reached.
This is generally true when the temperature and pressure are constant.
4
5
If the volume is kept constant and the pressure is varied the
Helmholtz energy will be minimised.
6
Chapter 14 on page 741 illustrates this for the case of a simple
vapour–liquid equilibrium.
7
The other energy functions U , H , etc. are minimised at constant
values of their respective canonical variables.
824 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
In general we shall investigate a closed system consisting of π ∈ {α,
β, . . . , ψ, ω} phases and a set of i ∈ {1, 2, . . . , n} components which are
common to all the phases .
1
Expanded footnote [a]:
[a] Phase α is usually the low temperature phase and β, . . . , ψ, ω are phases that become stable at successively increasing temperatures.
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
ENGLISH text is missing.Minimum energy principle
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
14 Simple . . .
One phase, ρr = 1 :
7
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Minimum Gibbs energy I (2)
822 / 1776
15 Multicomponent . . .
Pω
4
From the component balances it follows that π=α Niπ = Ni and
P
π
even ω
π=α dNi = 0 because the total moles Ni are constant in the
system as a whole.
Expanded footnote [a]:
[a] Reacting systems with (maybe) disjoint sets of the components in each of the
phases is taken up in Chapter Phase Reactions.
15 Multicomponent . . .
1
Thermodynamic phases
2
Minimum (Gibbs) energy
3
Fugacity versus activity coefficient
4
Direct substitution
5
Newton–Raphson iteration
6
K values
7
Fugacity coefficients
8
Activity coefficients
9
Order of convergence
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
Minimum Gibbs energy II
823 / 1776
Convergence orde
15 Multicomponent . . .
Elimination of dNiβ makes (dG )T ,p ,n =
1
Convergence orde
Pn
α
i =1 (µi
− µβi ) dNiα = 0.
2
In the vicinity of an equilibrium point all the differentials dNiα are
freely varying quantities which means the equilibrium relationship is
equivalent to:
Expanded footnote [a]:
2
When combined, this information is sufficient for deriving an invertable set of equations for calculating the equilibrium state.
5
For the special case of given temperature and pressure
(dG )T ,p ,nα +nβ +··· =
ω P
n
P
π=α i =1
µπi dN πi = 0 ,
7
6
To ensure that the problem size does not escalate unreasonably
the main concern of this chapter is a two-phase system where α and
β may include vapour–liquid, liquid–liquid, solid–liquid and solid–solid
equilibrium.
(dG )T ,p =
We note that dN πi are not independent variables in the differential
of G , but that they are interconnected by the n component balances of
the system.
3
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 116
§ 115
§ 117
825 / 1776
Convergence orde
15 Multicomponent . . .
Show that the necessary criteria for multiphase
equilibrium at given total entropy S , total volume V and
total composition Ni for each component i , are T α =
T β = · · · = T ω , p α = p β = · · · = p ω and µα = µβ = · · · =
µω . It is assumed that all the components i ∈ [1, n] are
present in all the phases α, β, . . . , ω. Hint: Start from
n
P
i =1
µαi dNiα +
n
P
i =1
µiβ dNiβ = 0 ,
µαn
dNiα + dNiβ = 0 .
Tore Haug-Warberg (Chem Eng, NTNU)
ENGLISH text is missing.
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 116
§ 115
(2)
§ 117
826 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
15 Multicomponent . . .
Termodynamic methods TKP4175
T
s =
ˆ ( S
§ 116 Minimum internal energy
α
S
β
nTi =
ˆ ( N αi N i
β
··· S
ω
π=α
Convergence orde
15 Multicomponent . . .
),
),
· · · Nω
i
3
Note that U has no absolute minimum.
4
Due to its extensive properties the energy will diverge to ±∞ when
the system size goes towards ∞.
π=α
ω
P
∀i ∈ [1, n] ,
π=α
ω
P
π=α
828 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
i =1
π=α
The variations in S π , V π and Niπ must therefore be constrained
such that the total entropy, volume and mole numbers (of each component) are conserved:
ω
P
dS π = 0 ,
5
define the extensive co-ordinates of the phases.
ev = V ,
02 September 2013
827 / 1776
In this (stationary) state the differential of U must be zero for all
1
feasible variations in S π , V π and Niπ :
n
ω
ω P
P
P
dU π =
µπi dNiπ = 0 .
T π dS π − p π dV π +
(dU )S ,V ,n =
vT =
ˆ ( Vα Vβ · · · Vω ) ,
es = S ,
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
where e =
ˆ ( 1 1 · · · 1 ) has as many elements
as there are phases in the system, and where the
other vectors
s,v,ni
Tore Haug-Warberg (Chem Eng, NTNU)
=
(15.1)
β
µn
2
The internal energy of the system is minimised in the equilibrium
state.
Ueq = min U (s, v, n1 , . . . , nn ) ,
eni = Ni ,
β
µα1 = µ1
...
The thermodynamic equilibrium criterion is then simplified to:
is a necessary condition for thermodynamic equilibrium.
Tore Haug-Warberg (Chem Eng, NTNU)
[a] At equilibrium dN αi are fluctuations beyond our control. These fluctuations are quite
inevitable from the laws of quantum mechanics and will be observed in every physically
acceptable system. Thermodynamic equilibria are in other words dynamic, not static.
829 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
dV π = 0 ,
dNiπ = 0 ,
Termodynamic methods TKP4175
∀i ∈ [1, n] .
02 September 2013
830 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 116 Minimum internal energy (2)
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Activity–activity
Substituted into the differential of U :
ψ h
i
n
P
P
(dU )S ,V ,n =
(T π − T ω ) dS π − (p π − p ω ) dV π + (µπi − µiω ) dNiπ = 0 .
7
9
8
In the neighbourhood of an equilibrium point the quantities
ψ
dS α , . . . , dS ψ and dV α , . . . , dV ψ and dNα1 , . . . , dNn are truly independent variables.
If (dU )S ,V ,n = 0 then it must be true that:
β
ψ
Termodynamic methods TKP4175
02 September 2013
Fugacity–Fugacity
831 / 1776
15 Multicomponent . . .
+ RT ln
ENGLISH text is missing.
µi◦ (T , p◦ ) + RT ln
ϕαi xiα p p◦
ϕi xi p p◦
= µ◦i (T , p◦ ) + RT ln
ENGLISH text is missing.
(15.6)
β β
ϕi xi p p◦
Tore Haug-Warberg (Chem Eng, NTNU)
(15.5)
02 September 2013
Fugacity–Activity
833 / 1776
15 Multicomponent . . .
3
In practise this means that the fugacity coefficient ϕi can be replaced by the activity coefficient γi defined by:
15 Multicomponent . . .
ENGLISH text is missing.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
ϕi (T ,p ,n)
ϕ⋆
(T ,p )
i
Ideal gas
1
µi (T , p , n) =
β
ϕαi = ϕi ,
µ◦i (T , p◦ )
|
+ RT ln
{z
Reference potential
ϕi xi + RT ln
}
ϕ⋆
i
ENGLISH text is missing.
(15.9)
=
ˆ µ⋆
i (T , p ) + RT ln [γi (T , p , n)xi ] .
ENGLISH text is missing.
(7.12)
ENGLISH text is missing.
hi◦ (T )
− hi◦ (T◦ )
=
RT
cp◦,i (τ) dτ ,
(7.14)
RT
cp◦,i (τ) dln τ .
(7.15)
T◦
si◦ (T , p◦ ) − si◦ (T◦ , p◦ ) =
,
Convergence orde
µ◦i (T , p◦ ) = ∆f hi◦ (T◦ ) + ( hi◦ (T ) − hi◦ (T◦ ) ) − Tsi◦ (T , p◦ ) ,
ENGLISH text is missing.
ϕ⋆
p
i
p◦
834 / 1776
15 Multicomponent . . .
(15.8)
.
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
1
If the chemical potentials of the pure components are the same in
both phases, at the temperature and pressure of the system, it can be
concluded that the phases are of the same nature.
ENGLISH text is missing.
⇒
Termodynamic methods TKP4175
Convergence orde
6.3 Blandet bruk av fugasitets- og aktivitetsmodell for
damp og væske ved forholdsvis lave trykk som vist i
figur 15.3.
6.4 To ulike aktivitetsmodeller for to ulike faste faser
eventuelt fast fase og væske som vist i figur 15.4.
β
= Ni ,
γi (T , p , n) =
ˆ
β β
β
Activity–activity (2)
2
If this is the case it is no longer necessary to integrate all the way
from the ideal gas in Eq. 13.12 but rather refer to the properties of pure
component -i - as a reference.
(15.7)
.
ϕiα xiα = ϕi xi .
xiα = 1 ∧ xi = 1
β
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
ENGLISH text is missing.
µi =
+
β
Ni
6.1 Én felles fugasitetsmodell for damp og væske,
eventuelt for to fluider med et kritisk punkt for
avblanding, som vist i figur 15.1.
6.2 Én felles aktivitetsmodell for to faste faser,
eventuelt for to væskefaser, med et kritisk punkt for
avblanding som vist i figur 15.2.
(15.4)
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
µ◦i (T , p◦ )
β
β
µnα = µ n = · · · = µ n = µ ω
n .
1
15 Multicomponent . . .
Niα
(15.3)
µα1 = µ 1 = · · · = µ 1 = µ ω
1 ,
..
.
Tore Haug-Warberg (Chem Eng, NTNU)
nn ) dµ = 0 .
(15.2)
pα = p β = · · · = p ψ = p ω ,
ψ
s dT − v dp + ( n1 n2 · · ·
By and large, all we can do is to write algorithms that treat certain
subclasses of the problems we encounter. Vapour–liquid equilibria at
low to medium high pressure— maybe up to a few hundred bar— is one
such subclass. In the next two sections we will look at two competing
methods, called direct substitution and Newton–Raphson iteration, to
solve this kind of equilibria. It should be stressed that the methods do
work for other types of phase equilibria as well, but the vapour–liquid
equilibria are very common in the chemical process industry and it does
not hurt to have these in mind.
832 / 1776
ENGLISH text is missing.
Tα = Tβ = · · · = Tψ = Tω ,
β
The Gibbs–Duhem’s equation
6
µiα (T , p , N1α , N2α , . . . , Nnα ) = µi (T , p , N1 , N2 , . . . , Nn ) ,
i =1
π=α
2
5
Systems with non-negative degrees of fredom are in principle solvable, but it is nevertheless not so easy to find a good solution algorithm
in each case.
ENGLISH text is missing.
1
6
From these n + 2 balance equations we can eliminate dS ω , dV ω
and dNiω for all the components i ∈ [1, n].
1
The equations above show that temperature, pressure and chemical potentials are uniform in all phases at thermodynamic equilibrium.
This does not imply that we can specify these variables directly.
3
The number of independent intensive variables in a system is
therefore F = dim(n) + 2 − dim(e), or in other words F = N + 2 − P ,
also known as the Gibbs phase rule for un-reacting systems.
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
ENGLISH text is missing.Gibbs’ phase rule
removes one degree of freedom per phase.
4
The practical consequence of this law is that it we are not allowed
to calculate the equilibrium of a system with F < 0 degrees of freedom.
Doing so will lead to an over-specified system which eventually will lead
to a singular set of equations to be solved.
ENGLISH text is missing.
T◦
(15.10)
4
because the pure component reference ϕi⋆ is the same in both
phases.
RT ln γi (T , p , n) =
ˆ RT ln ϕi − RT ln ϕi⋆
= RT ln γi (T , p⋆ , n) +
Rp ¯
vi (T , π, n) − vi⋆ (T , π) dπ .
p⋆
(15.11)
ENGLISH text is missing.
T ≪ 100 K
RT | ln γi | ≪ 1000 J mol-1
p ≫ 10 MPa
ENGLISH text is missing.
γi (T , p , n) ≈ γi (T , n) ,
ENGLISH text is missing. ENGLISH text is missing.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Condensed phase
1
835 / 1776
Convergence orde
15 Multicomponent . . .
µi⋆ (T , p ) = µi⋆ (T , p⋆ ) +
Rp
p⋆
vi⋆ (T , π) dπ .
(15.12)
ENGLISH text is missing.
= µ◦i (T , p◦ ) +
V (Rp⋆ ) ∞
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Condensed phase (2)
836 / 1776
ϕi⋆ p⋆
p◦
RT
ν
−
µi (T , p , n) =
µ⋆
i ( T , p⋆ )
15 Multicomponent . . .
+ RT ln [γi (T , p⋆ , n)xi ] +
Rp
p⋆
Tore Haug-Warberg (Chem Eng, NTNU)
(15.14)
ENGLISH text is missing.
dν − RT ln
p◦ V (p⋆ )
NRT
β,sat
µi
(T , pisat ) = µα,sat
(T , pisat ) ,
i
µ◦i (T , p◦ ) + RT ln
ϕisat pisat p◦
= µ⋆
i ( T , p⋆ ) +
µi (T , p , n) = µi◦ (T , p◦ )
(15.13)
02 September 2013
838 / 1776
pisat
R
p⋆
vi⋆ (T , π) dπ .
ENGLISH text is missing.
.
pisat ϕsat
i
p◦
+
pR⋆
pisat
vi⋆ (T , π) dπ
+ RT ln [γi (T , p⋆ , n)xi ] +
Termodynamic methods TKP4175
Convergence orde
ENGLISH text is missing.
+ RT ln
Tore Haug-Warberg (Chem Eng, NTNU)
837 / 1776
15 Multicomponent . . .
,
∂π ∂Ni T ,ν,Nj ,i
02 September 2013
Saturation state
1
v̄i (T , π, n) dπ .
Termodynamic methods TKP4175
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
ENGLISH text is missing.
ENGLISH text is missing.
µi⋆ (T , p⋆ ) = µi◦ (T , p◦ ) + RT ln
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
839 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Rp
p⋆
Termodynamic methods TKP4175
v̄i (T , π, n) dπ .
02 September 2013
840 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Saturation state (2)
15 Multicomponent . . .
ENGLISH text is missing.
sat
ϕi yi p ≈ ϕsat
i pi γi (T , p⋆ , n)xi exp
ENGLISH text is missing.
yi = 1 ∧ xi = 1
vi (p −pisat )
RT
(15.15)
=
i
ENGLISH text is missing.
ϕi p = ϕisat pisat .
⇒
T
ENGLISH text is missing.
,
15 Multicomponent . . .
∆tr µ⋆
(T )
i
T
ENGLISH text is missing.
−
(Ti⋆ )
∆tr µ⋆
i
Ti⋆
=
| {z }
Phase transition point (2)
Convergence orde
15 Multicomponent . . .
ENGLISH text is missing.
∆tr µi⋆
∆tr µi⋆ T
T T ⋆
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
ENGLISH text is missing.
d
i
Phase transition point
1




p
 1 R ⋆

ϕi yi p ≈ ϕisat pisat γi (T , p⋆ , n)xi exp  RT
vi (T , π) dπ .
 p sat

ENGLISH text is missing.
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
p
= ∆tr hi⋆ d
∆tr hi⋆ T
T T ⋆
i
∆tr hi⋆ (T )
T
−
−
RT
Ti⋆
1
T
⋆
⋆
∆tr µ⋆
i (T ) = ∆tr hi (Ti )+
RT
Ti⋆
∆tr cp⋆,i (τ) dτ− TT⋆ ·∆tr hi⋆ (Ti⋆ )− T
i
RT
∆tr cp⋆,i (τ)
τ
Ti⋆
dτ
ENGLISH text is missing.
d∆tr hi⋆
τ
∆tr hi⋆ (Ti⋆ )
Ti⋆
−
∆tr µi⋆ (T ) = ∆tr hi⋆ (Ti⋆ ) 1 −
RT
Ti⋆
∆tr cp⋆,i
τ
T
Ti⋆
+
RT
Ti⋆
∆tr cp⋆,i (τ) 1 −
T
τ
dτ .
(15.16)
dτ
0
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
K -value
841 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Vapour–liquid equilibria are frequently calculated by an iterative
method which is due to Rachford and Rice .
1
Expanded footnote [a]:
[a] H. H. Rachford and J. D. Rice. Trans. Am. Inst. Min., Metall. Pet. Eng., 195:327–
328, 1952.
K -value (2)
∀i ∈ [1, n] ,
xiβ
Ni
N α +Ki (N −N α )
=
zi
z α + Ki z β
Tore Haug-Warberg (Chem Eng, NTNU)
(15.18)
.
(15.17)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
xiα =
This is also known as the K -value method because the equilibrium
relations are solved as a set of K -value problems on the form
xiα
842 / 1776
6
Elimination of N β produces xiα N α + xiβ (N − N α ) = Ni and a subsequent substitution of xiβ from Eq. 15.17 yields:
2
xiβ =
ˆ Ki xiα ,
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Termodynamic methods TKP4175
Direct substitution
843 / 1776
Convergence orde
15 Multicomponent . . .
The corresponding expression for xiβ is
1
xiβ =
Ki Ni
N α +Ki (N −N α )
=
K i zi
z α + Ki z β
(15.19)
,
ˆ N α /(N α +N β ) and z β =
ˆ N β /(N α +N β ) are the phase fractions
where z α =
and zi =
ˆ Ni /N is the total (feed) mole fraction of component i .
2
Note that the right hand sides of Eqs. 15.18 and 15.19 contain a
single unknown variable z α , or equivalently z β = 1 − z α .
where
and
are the mole fractions of component i in α (liquid) and
β (vapour) respectively.
3
We shall later learn that there is a one-to-one relationship between
Ki and the chemical potentials of component i in the two phases.
The total component balance is N α + N β = N and the individual
5
component balances can be written xiα N α + xiβ N β = Ni .
4
This tells us that the K -value method can be applied even when
α and β are two arbitrary phases, but prior to the discussion of exotic
phase equilibria we shall derive a calculation scheme (algorithm) for
Eq. 15.17.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Direct substitution (2)
844 / 1776
15 Multicomponent . . .
P
3
Definition-wise in xi = 1 in both phases and it is possible to solve
for z α from Eq. 15.18 or 15.19, but practical experience proves that
the algorithm is more stable when the symmetric condition f (z α ) =
Pn
α
β
i =1 (xi − xi ) = 0 is being used.
n
P
i =1
(1−Ki )zi
z α + Ki z β
fi =
ˆ
=
1−Ki
z α + Ki z β
n
P
i =1
fi zi = 0 ,
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
4
Summing up xiα and xiβ from Eqs. 15.18 and 15.19 yields a closed
expression for f (z α )
f (z α ) =
Tore Haug-Warberg (Chem Eng, NTNU)
K -value algorithm
See Function 2.6 in Appendix 31.
2
Note that the function interface is more elaborate than has been
asked for in the text.
The syntax has notably been extended to TpKvalue(n, k, u, U, v , V)
where U and V are data structures transmitting model specific parameters used by u and v .
3
4
This makes it possible to iterate on non-ideal equilibrium states by
supplying two fugacity (activity) coefficient models u and v .
§ 117
5
The update of Ki assumes for each iteration that ln kk +1 = ln k +
1
Computerise the function [xα , xβ , z α ] = TpKvalue(n, k) using u(xα , U) − v (xβ , V). Constant Ki values will be assumed if the functions
Eq. 15.20 as your reference algorithm.
u and v are expelled from the argument list.
848 / 1776
849 / 1776
1
A fixed derivative sign means that there is at most one solution in
the domain z α ∈ [0, 1].
2
Note that ∂f /∂z α < 0 in Eq. 15.20 except for the degenerated case
Ki = 1 ∀i ∈ [1, n] then ∂f /∂z α = 0.
First, Eq. 15.20 is solved with respect to z α at given or estimated
Ki (inner loop).
5
,
which is easily solved with respect to z α using a Newton–Raphson iteration:
P
-1
n
-1
z α,k +1 = z α,k − ∂∂zfα f = z α,k +
fi2 zi f .
(15.20)
i =1
Tore Haug-Warberg (Chem Eng, NTNU)
3
Similar to the Newton–Rapshon method, which is derived in Section 15.5, the number of floating point operations in the update of x α is
proportional to n, i.e. the computation time increases linearly with the
number of components in the system.
4
These are the strengths of the K -value method, but it has also an
inherit weakness in that non-ideal equilibria must be solved by nested
iteration.
The phase compositions are then updated by the Eqs. 15.18
and 15.19 before new Ki -values can be calculated (outer loop).
6
7
The double iteration procedure is repeated until the phase compositions xiα and xiβ have converged.
8
The Newton–Raphson method in Section 15.5 is without this flaw
because ∆µi (equivalent to Ki ) is updated in every iteration avoiding the
outer loop[a].
[a] Also applies to the K -value problem if disregarding the monotonic form of f (z α ).
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
§ 117 Rachford–Rice
1
845 / 1776
Linearization of chemical potential
846 / 1776
Convergence orde
15 Multicomponent . . .
1
Contrary to the K -value method of Rachford–Rice, the Newton –
Raphson s method fulfills the component balance in every iteration.
Expanded footnote [a]:
[a] Sir Isaac Newton, 1642–1727 by the Julian calendar. English physicist and mathematician.
Expanded footnote [a]:
[a] Joseph Raphson, ca. 1648–1715. English mathematician.
3
2
This requires a feasible starting point where nα + nβ = n is fulfilled,
and which of course is as close to the solution point as possible.
The Newton–Raphson iteration of Eq. 15.1 is then,
µiα +
n
P
j =1
∂µαi ∂Njα
∆Njα = µiβ +
T ,p ,Nl ,j
n
P
j =1
−∆Njβ = ∆Njα ,
∂µiβ ∂Njβ
∆Njβ ,
T ,p ,Nl ,j
or on matrix form:
µα + Gα ∆nα = µβ + Gβ ∆nβ ,
−∆nβ = ∆nα .
The recursion formula makes a new z α,k +1 available, which on
back-substitution into Eqs. 15.18 and 15.19 produces new values of
xiα and xiβ .
5
6
These updates (hopefully) improve the Ki -values, and subsequent
iterates in Eq. 15.20 converge finally to the equilibrium state.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
847 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
850 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
851 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
Convergence orde
ENGLISH text is missing.Ideal mixture Hessian
1
The calculation of H may be approximated by the Hessian of an
ideal mixture.
Inversion
15 Multicomponent . . .
The matrix G =
ˆ {(∂2 G /∂Ni ∂Nj )T ,p } is the Hessian of Gibbs energy
k
+1
and ∆n =
ˆ n
− nk is the composition difference between two (subsequent) iterations k + 1 and k .
1
Expanded footnote [a]:
[a] Ludwig Otto Hesse, 1811–1874. German mathematician.
The Newton–Raphson equations can be combined into
2
Inversion (2)
15 Multicomponent . . .
5
If the relations are violated it is mandatory to shorten the step size
according to
∆nα = −τ H-1 ∆µ ,
(15.21)
6
where τ ∈ h0, 1] is calculated such that all the updated mole numbers are positive.
7
In this way we can ensure that the component balance is fulfilled in
every iteration.
8
The Newton–Raphson method converges to a state where ∆µ = 0,
but note that the phase models have not been taken into account yet.
9
It must therefore be anticipated that ∆µ is calculated from an equation of state, or an activity coefficient model, that describes the system
with sufficient accuracy.
It must be realised, however, that the algorithm has to be guided.
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Ki versus µi
See Function 2.7 in Appendix 31.
2
Note that the function interface is more elaborate than has been
asked for in the text.
3
The syntax has notably been extended to TpNewton(n, k, u, U, v , V)
where U and V are data structures transmitting model specific parame- ENGLISH text is missing.Computation time
ters used by u and v .
1
It should be mentioned that the number of floating point operations
4
This makes it possible to iterate on non-ideal equilibrium states by in the update is proportional to n.
supplying two fugacity (activity) coefficient models u and v .
2
Hence, the computation time increases linearly with the number of
5
The update of ∆µ assumes that for each iteration ∆µ = ∆µıd + components in the mixture.
r,p
ıd
r,p
∆µ where ∆µ = ln k + ln x − ln x and ∆µ = u(x , U) − v (x , V). 3 However, if we had not spent time on the Sherman–Morrison for1
Computerise the function [xα , xβ , z α ] = TpNewton(n, k). Use 6 Constant ∆µr,p = 0 will be assumed if the functions u and v are mula, but rather used numerical inversion of H in each iteration, the
Eq. 15.24 as your reference algorithm.
expelled from the argument list.
computation time would be proportional to n3 .
859 / 1776
861 / 1776
860 / 1776
§ 119
α
β
α
β
The Rachford–Rice iteration spends time on calculating K -values
while the Newton–Raphson method deals directly with chemical potentials.
1
However, because the two methods aim at solving the same equilibrium problem there must be a rational connection between them.
β
µi = µiα
µi◦ (T , p◦ ) + RT ln
Tore Haug-Warberg (Chem Eng, NTNU)
ϕβi xiβ p p◦
⇓
= µi◦ (T , p◦ ) + RT ln
Termodynamic methods TKP4175
(15.1)
ϕiα xiα p ,
p◦
§ 120
§ 119
§ 121
NRT ∂(Ni /N )
Ni
∂Nj
N
δ
ij
Ni
−
1
N
1
The matrix (nα )-D + (nβ )-D has diagonal elements (Niα )-1 + (Niβ )-1 =
Ni /(Niα Niβ ) and the factor (N α )-1 + (N β )-1 can be rewritten to (N α +
N β )/(N α N β ).
k ,j
,
2
15 Multicomponent . . .
This makes the following factorisation of Hıd possible:
diag α
β
Ni
Nα Nβ
Hıd = RT NNα+NNβ
− eeT ,
Niα Niβ (N α +N β )
N α +N β
RT
Hıd =
1
zα zβ
D-1 =
ˆ diag
(D-1 − eeT ) ,
zi ,
xαxβ
i
(15.22)
i
H-1 with an analytical formula as shown below.
8
4
From the definition H =
ˆ Gα + Gβ it follows:
h(n
(
ee i
α -D
T
) + (nβ )-D − N α )-1 + (N β )-1
.
Hıd = RT
854 / 1776
Eq. 15.22 indicates that H is a rank one update of D.
5
The inverse of H can then be calculated from the Sherman–MorriD
-1
son formula -1 − eeT = D + (1 − eT De)-1 DeeT D, or, equivalently, if
§ 118 Sherman–Morrison
§ 118
we define d =
ˆ De:
Straightforward substitution.
D
H -1
1
The Sherman–Morrison formula is widely applicable and not at all 1
ıd
RT
(15.23)
= z α z β + (1 − eT d)-1 ddT .
N α +N β
limited to thermodynamic problems. Verify the formula by proving that 2 The general formula for a rank one update of the matrix is: (A −
H-1 H = I. Attempt to find a more general formula valid for (A − uvT )-1 . uvT )-1 = A-1 + A-1 u(1 − vT A-1 u)-1 vT A-1 .
855 / 1776
857 / 1776
856 / 1776
[a] Leopold Kronecker, 1823–1891. German mathematician.
Note that z α , z β , d, D and ∆µ/RT are dimensionless variables.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Ki versus µi (2)
853 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
where the left side stands for µβi and the right side stands for µiα . From
Eq. 15.17 there is the alternativ formulation of the same equilibrium
criterion:
β
xi = Ki xiα .
(15.17)
A thermodynamic equilibrium constant depends on temperature, and
nothing but temperature, whereas Ki is in fact a non-linear function in
temperature, pressure and the compositions of both phases (and depends therefore on two thermodynamic states simultaneously). If not,
the two alternative formulations would not be equivalent.
4
But it should be stressed that in the context of being an equilibrium
constant the K -value is quite misleading.
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Termodynamic methods TKP4175
Fugacity–fugacity
858 / 1776
Convergence orde
15 Multicomponent . . .
1
On the basis of the discussion above, the quantity ∆µi /RT which
appears in Eq. 15.24 can be defined as
µα −µβ
K eos x α ∆µi
(15.25)
ˆ iRT i = ln i x β i ,
RT =
i
where Kieos is defined as the ratio between the fugacity coefficients in
the phases α and β:
Kieos (T , p , nα , nβ ) =
ϕiα (T ,p ,nα )
ϕβi (T ,p ,nβ )
(15.26)
.
2
It is used an excessively obvious notation in this case to emphasize
that Ki is not a thermodynamic equilibrium constant.
(15.7)
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
=
= RT
[a] This is just another example on the special properties of extensive functions.
2
3
The purpose of this section is to show that (and how) Eq. 15.1 on
page 794 can be used to derive some useful relations between Ki and
∆µi .
ENGLISH text is missing.Sherman–Morrison inverse
T ,p ,Nk ,j
T ,p ,Nk ,j
e = ( 1 1 . . . 1 )T ,
Expanded footnote [a]:
Convergence orde
15 Multicomponent . . .
§ 119 Newton–Raphson
1
852 / 1776
From the definition of an ideal mixture we can write,
µ
∂µiıd Gıd
= ∂∂Nj i⋆ + RT ln NNi
ij = ∂Nj
α
α
β
6
Here, n-D is an (inverted) diagonal matrix having 1/Ni along the where z and z are phase fractions, xi and
main diagonal.
3
As usual xiβ are component mole fractions, and zi is the feed mole
7
The simple structure of this matrix makes it possible to calculate fraction of component i .
4
We may therefore scale the system such that N α + N β = 1 without
affecting the mole fractions in the phases α and β during the iteration.
In particular the mole number updates must be checked for
= nα,k + ∆nα > 0 and nβ,k +1 = nβ,k − ∆nα > 0.
nα,k +1
Tore Haug-Warberg (Chem Eng, NTNU)
4
where the Kronecker[a] delta δij = 1 if i = j and δij = 0 else.
3
This makes the iteration sequence independent of the system size
for a given total composition z.
or even better: ∆nα = −H-1 ∆µ where H =
ˆ Gα + Gβ and ∆µ =
ˆ µα − µβ .
3
3
In summary it means that ideal mixing is assumed in the calculation
of H while ∆µ is calculated rigorously.
5
On matrix form this formula is written — see also Theme 12 on
page 86 in Chapter 3 on page 57:
n
or,
Gıd = RT -D − N -1 eeT ,
1
From Eqs. 15.21 and 15.23 it is possible to express the simplified
two-phase Newton–Raphson iteration as:
∆µ
∆nα
= −z α z β D + 1−d1 T e ddT RT .
(15.24)
N α +N β
2
(Gα + Gβ )∆nα = −(µα − µβ ) ,
4
Update formula
2
The convergence properties will deteriorate close to the solution
compared to the rigorous implementation of H = Gα + Gβ , but the simplicity of the model makes it interesting in its own right.
862 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Termodynamic methods TKP4175
§ 120 Natural gas (T , p )
863 / 1776
Convergence orde
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
The plot is given a high colour density where the liquid and vapour
phases coexist in approximately equal amounts, and a light colour near
the single-phase region (vapour or liquid).
3
4
The dense ridge running along the left edge of the diagram combined with the shallow basin to the right is typical for methane-rich natural gases.
5
The bubble point is dominated by the large methane content while
the dew point is determined by the trace components comprising the
“heavy tail” of the gas.
6
The Newton–Raphson method does an excellent job in this
case[a], but note that the dew-point calculations are sensitive to the
start estimate.
7
Increasing the liquid composition of the medium components by
less than one tenth of a mole fraction causes the rightmost part of the
phase boundary to be left out.
8
The K -value method has even larger problems in this part of the
diagram, and needs some type of guidance which is not implemented
here.
[a] Far better than anticipated. In fact, it does a better job than for a rigorous implementation of the Hessian. The reason for this somewhat strange behaviour is that the
eigenvalues of the simplified Hessian are strictly positive while the eigenvalues of the
rigorous Hessian change sign wherever thermodynamic instability occurs, see Chapter 21. This is known to cause numerical problems in Newton–Raphson methods.
Pressure [psia]
[a] Mario H. Gonzalez and Anthony L. Lee. J. Chem. Eng. Data, 13(2):172–176, April
1968.
The calculations have been using TpNewton as shown in Program 1.22 in Appendix 31.
Find experimental data for a typical hydrocarbon vapour–
liquid system and see how close the RK equation
of state matches the measurements. Calculate the
phase diagram using TpKvalue or TpNewton.
700
Expanded footnote [a]:
2
Convergence orde
c.p.
800
The phase diagram of a typical, albeit synthetic, natural gas is
illustrated in Figure 15.1 on the next page.
864 / 1776
900
15 Multicomponent . . .
1
02 September 2013
600
500
400
N = 2913
300
200
100
0
−200
−160
−120
−80
Temperature [F]
Figure 15.1: Phase diagram of nitrogen–methane–ethane–propane–n-butane put together from a total of 6093 triangular patches using a recursive
divide-and-conquer algorithm (each triangle is split into four smaller ones until all the vertices are in the same phase domain).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
865 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
866 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
867 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Activity–activity
2
One of these simplifications is known as the Lewis mixing rule.
§ 121 Lewis[1] mixing rule
1
This is always true for an ideal gas mixture and it is valid to a
high degree of approximation for real gases at low pressures or high
temperatures. The approximation also holds at high pressures if V ex ≈
0 in the entire pressure and composition region.
4
V ex = 0 means that V =
Pn
i =1 vi (T , p )Ni .
2
The fugacity coefficient is defined by the Eqs. 13.12 and 13.13, see
page 718. Substituted for the Lewis rule we get
Rp § 121
RT ln ϕiLewıs = vi (T , π) − RT
π dπ = f (T , p ) ,
5
In the case of ϕi (T , p , x) ≈ ϕ⋆
(T , p ) then Ki will be constant in 1
0
Show that the Lewis mixing rule is exact if Leduc??–Amagat[a]s
i
Eq. 15.26 (at a given temperature and pressure).
law V ex = 0 is valid over the entire composition and pressure [0, p ] which shows that ϕi depends on temperature and pressure only because the partial molar volume is independent of composition.
6
The K -value method is then superior to the Newton–Raphson iter- range.
870 / 1776
869 / 1776
ation because it solves the entire equilibrium problem in R and not in
Rn .
868 / 1776
[a] Émile Hilaire Amagat, 1841–1915. French chemist.
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
ENGLISH text is missing.Some physical chemistry
1
In the ages before the modern computer era it was mandatory to
make physical simplifications before any numerical calculation was attempted.
3
Under certain circumstances the fugacity coefficient ϕi will be almost independent of the mixture composition.
[a] Gilbert Newton Lewis, 1875–1946. American chemist.
§ 122
1
If the chemical potentials of the pure components are the same in
both phases, at the temperature and pressure of the system, it can be
concluded that the phases are of the same nature, but we cannot use
the terms vapour and liquid in a strict manner.
§ 121
§ 123
Convergence orde
15 Multicomponent . . .
[b] It is intuitive that the density of the vapour phase has to be lower than the density
of the liquid phase, but this is wrong. At high pressures the system H2 –He will in fact
invert because the vapour phase is dominated by He which has a higher molecular
mass than H2 .
ϕiβ
≡
β,⋆
ϕαi
·
ϕα,⋆
i
ϕi
ϕiβ
·
ϕα,⋆
i
ϕβ,⋆
i
=
ˆ
γαi
1
·
1
γiβ
·
ϕα,⋆
i
ϕiβ,⋆
=
γiα
γiβ
·
ϕα,⋆
i
ϕiβ,⋆
15 Multicomponent . . .
.
Expanded footnote [b]:
[b] Ternary systems. In Liquid–Liquid Equilibrium Data Collection, volume V, part 2,.
DECHEMA, Frankfurt/Main, 1980. p. 115.
is
3
The diagram was successfully calculated using TpNewton as Program 1.23 in Appendix 31 shows.
Find experimental data for a ternary liquid–liquid
system with known NRTL, Van Laar or Margules model
parameters. Select a system which has a fairly large
solubility region and a critical end-point inside the ternary
region. Use TpNewton to calculate the phase diagram
(TpKvalue will experience convergence problems).
6
In practise this means that the fugacity coefficient ϕi can be replaced by the activity coefficient γi from Eq. 15.8:
ϕαi
§ 122 Cycloalkanes–methanol (x1 , x2 )
Convergence orde
1
The phase diagram of cyclohexane–cyclopentane–methanol
illustrated in Figure 15.2 on the following page.
2
In Eq. 15.26 it is common that α is the liquid phase and β the vapour
phase, but this is just one out of many possibilities.
3
We could alternatively assume that α and β are two near-critical
fluids being something in between a vapour and a liquid, or two welldefined liquid phases.
4
There are also examples on supercritical gas–gas equilibria in the
literature.
5
In a thermodynamic sense there is no fundamental difference between a vapour phase and a liquid phase[b], but it is still possible to decide whether the phases are of the same type by looking at the pure
component states in the system.
Kieos =
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
2
It is a classic liquid–liquid diagram — quite symmetric and with a
critical end-point inside the triangle.
Still, the K -value method experiences serious problems close to
6
the critical point.
4
In this case the measurements were used as start values and the
calculations converged without difficulty even in the vicinity of the critical
point of the mixture.
5
The agreement between the measured and the calculated values
is otherwise extremely good.
7
Provided the same activity model is used to describe both phases
the Ki -values are
ϕiα,⋆ = ϕβ,⋆
i
⇒
Kieos → Kiex =
ˆ
γiα
γiβ
,
8
because the pure component reference ϕ⋆
is the same in both
i
phases.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
of c
ycl
ope
nta
ne
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
871 / 1776
Termodynamic methods TKP4175
Fugacity–activity
c.p.
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
872 / 1776
Convergence orde
15 Multicomponent . . .
rac
tion
1
For mixtures of polar substances it may be appropriate to employ
an equation of state for the vapour phase and an activity model for the
condensed phase.
The starting point is this time Eq. 15.15 on page 807 on page807
3
which gives the approximate formula:
sat
v (T )(p −pisat )
,
(15.15)
ϕi yi p ≈ ϕisat pisat γi (T , p⋆ , n)xi exp i
RT
Mo
le f
0
Tore Haug-Warberg (Chem Eng, NTNU)
2
The phase diagram will in general not be “closed” at the critical
point but the hybrid models have greater flexibility and are favourable in
the modelling of complex liquids at low pressures and crystal phases in
general.
0.2
0.4
0.6
0.8
1
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Poynting factor
873 / 1776
Convergence orde
15 Multicomponent . . .
1
∆µi /RT from the Newton–Raphson algorithm in Eq. 15.24 may
now be written on the same form as in Eq. 15.25 provided Ki is calculated as
v sat (T )·(p −pisat (T ))
ϕsat (T ) pisat (T ) γi (T ,p⋆ ,nα )
,
(15.27)
Kivle ≈ i
exp i
RT
ϕ (T ,p ,nβ ) p
i
⋆
The exponential term is known as the Poynting factor of Ki .
Expanded footnote [b]:
[b] John Henry Poynting, 1852–1914. English physicist.
2
This factor is often neglected at low pressures, but only if Ki 1.
3
If this is not true the Poynting-factor may dominate Ki down to 5–
10 bar.
4
Confer Section 2.2 for a more detailed discussion of this topic.
Mole fraction of methanol
Figure 15.2: Phase diagram of cyclohexane–cyclopentane–methanol at
298.15 K and 760 mmHg. The iterations are started at interpolated points
in the diagram (ensure safe convergence).
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 123
§ 122
§ 124
874 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Termodynamic methods TKP4175
§ 123 Hexane–toluene (T , x )
875 / 1776
Convergence orde
Expanded footnote [b]:
[b] Aliphatic hydrocarbons: C4 –C6 . In Vapor–Liquid Equilibrium Data Collection, volume
I, part 6a,. DECHEMA, Frankfurt/Main, 1980. p. 593.
2
The calculations have been using TpNewton as shown in Program 1.24 in Appendix 31.
Find experimental data for a close to ideal binary vapour–
liquid system with known NRTL, Wilson, Van Laar
or Margules model parameters. Use TpKvalue or
TpNewton to calculate the phase diagram. Comment
on the practical applicability of Raoult’s law versus
the accuracy of pure component data.
Correcting for non-ideality makes some improvement (compare σ1
and σ2 in the left subfigure), but the largest error is actually hidden in
the vapour pressure of toluene.
4
Termodynamic methods TKP4175
02 September 2013
02 September 2013
Antoine vapour pressure
120
876 / 1776
Convergence orde
Corrected vapour pressure
120
ss1=0.356
ss2=0.157
ss3=0.265
ss4=0.0953
100
100
80
80
3
The system is almost ideal, but there are nevertheless some notable deviations between the experimental and the calculated values.
By increasing the vapour pressure 2% the agreement is improved
even further (compare σ3 and σ4 in the right subfigure), while Raoult’s
law still shows a significant deviation from the measurements.
60
0
0.5
1
Mole fraction of hexane
60
0
0.5
1
Mole fraction of hexane
5
6
This shows that Raoult’s law is of limited value even for nearly ideal
systems.
7
In fact, the pure component parameters are maybe more important
for the mixture properties than the phase model itself.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
15 Multicomponent . . .
The phase diagram of hexane–toluene is illustrated in Figure 15.3
on the following page.
1
Tore Haug-Warberg (Chem Eng, NTNU)
Temperature [◦ C]
Tore Haug-Warberg (Chem Eng, NTNU)
877 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
878 / 1776
Figure 15.3: Phase diagram of hexane–toluene at 760 mmHg. Calculated
curves show Raoult’s law (stipled) and Van Laar activity (solid). The right
subfigure shows the effect of increasing the vapour pressure of toluene by
2%. The sum-of-squares are denoted σ1 , σ2 , σ3 and σ4 .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
879 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Raoult[2]s law
Convergence orde
15 Multicomponent . . .
1
For an approximately pure liquid component i it is reasonable to
set ϕi ≃ ϕsat
, γi ≃ 1 and p − pisat ≃ 0.
i
Expanded footnote :
[a] François-Marie Raoult, 1830–1901. French chemist.
2
The mixture is by definition ideal and Ki appears to be a simple
function of temperature and pressure:
KiRaoult = lim Kivle =
xi →1
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Henry[3]s law
15 Multicomponent . . .
1
In the same mixture it may also be appropriate to use a hypothetical vapour pressure Hij for the component i infinitly diluted in the solvent j , the value of which is chosen such that the phase equilibrium is
conventionally reproduced by:
Expanded footnote :
[b] William Henry, 1775–1836. English chemist.
Henry
Ki
pisat (T )
p
= lim Kivle =
xi →0
xj →1
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
Hij (T ,p )
p
Acitivity–activity
15 Multicomponent . . .
1
When α and β are condensed phases it is often appropriate to
neglect the influence of pressure on the system properties.
3
In general, pressure explicit equations of state are not conceivable
for these systems and activity models are often used for both phases.
2
This is quite typical for metallurgical melts and refractory systems
where the process conditions are close to atmospheric.
Eq. 15.14 yields for the phase equilibrium between β and α:
4
β β
β,⋆
µi (T ) + RT ln (γi xi ) ≈ µiα,⋆ (T ) + RT ln (γiα xiα ) .
(15.28)
ENGLISH text is missing.
β,⋆
∆tr µi⋆ (T ) =
ˆ µiα,⋆ (T , p⋆ ) − µi (T , p⋆ ) ≈ ∆tr hi⋆ (Ti⋆ ) 1 −
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Acitivity–activity (2)
880 / 1776
Convergence orde
15 Multicomponent . . .
5
∆µi /RT required by the Newton–Raphson algorithm in Eq. 15.24
can finally be written as in Eq. 15.25 provided Ki is calculated as:
γα (T ,nα )
∆tr hi⋆ (Ti⋆ ) 1
1
.
(15.30)
Kisle ≈ γiβ (T ,nβ ) exp
R
T − T⋆
i
i
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 124
§ 123
Convergence orde
§ 125
881 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
. (15.29)
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
Termodynamic methods TKP4175
T
Ti⋆
§ 124 Gold–copper (T , x )
882 / 1776
Convergence orde
15 Multicomponent . . .
1
Figure 15.4 on page 841 illustrates the high-temperature portion
of the gold–copper phase diagram, see Program 1.25 in Appendix 31.
Expanded footnote [d]:
[d] H. Okamoto, D. J. Chakrabarti,
Diagrams, 8(5):454–473, 1987.
D. E. Laughlin, and T. B. Massalski. Bull. Alloy Phase
The agreement between the measured and calculated values is
generally good even though it is not possible to calculate the entire
diagram — neither with TpNewton nor with TpKvalue.
2
Find experimental data for a binary alloy (or a binary salt)
with full solubility in the solid phase. Choose a system
with fitted Margules or Redlich–Kister model parameters.
Calculate the phase diagram using TpNewton or TpKvalue.
How important is the temperature dependency of ∆tr si⋆
in this context?
3
The problem is quite persistent, and although the starting values
are chosen judiciously there is a substantial region around the congruence point where things go wrong.
8
The Hessian matrix for the assumed ideal mixture will then overestimate the curvature of the energy surface and therefore yield a conservative step size in TpNewton (and a safe, albeit very slow, convergence).
4
The flaw is due to the (large) negative deviations from ideality which
is observed in the two phases.
5
This does in turn cause an oscillatory iteration sequence.
6
Quite surprisingly, maybe, because the phase calculations in Figure 15.2 on page 828 converges nicely over the entire composition
range.
7
However, the latter system benefits from having a positive deviation
from ideality.
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
§ 124 Gold–copper (T , x ) (2)
883 / 1776
Convergence orde
Tore Haug-Warberg (Chem Eng, NTNU)
15 Multicomponent . . .
9
With negative deviations from ideality the curvature of the enrgy
surface is under-estimated and the step size may eventually grow so
large that the iterations start oscillating around the solution point .
Expanded footnote [d]:
[d] An exactly calculated Hessian would stabilise the Newton–Raphson method.
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Temperature [◦ C]
Tore Haug-Warberg (Chem Eng, NTNU)
884 / 1776
Convergence orde
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
885 / 1776
Convergence orde
1100
First order convergence
1050
1
The K -value method is typically of first order, but close to the critical
point the convergence factor h → 1 which means xk +1 ≃ xk .
1
While azeotropes are common for vapour–liquid phase diagrams,
the analogous phenomena of congruent melting is quite rare among the
alloys— at least when disconnected from the formation of stoichiometric
or the homogenous phase splitting for one of the
solid compounds,
phases[d].
15 Multicomponent . . .
[d] Even though the equilibrium conditions 15.26, 15.27 and 15.30 look quite differently.
2
A sequence of iterative calculations has (convergence) order m and
(convergence) factor h if the sequence approximates limk →∞ log kxk +1 −
x∞ k = log(h ) + m log kxk − x∞ k.
1000
3
Thus, convergence is severely hampered and it may be difficult to
decide whether the sequence converges or not.
950
900
ENGLISH text is missing.Phase diagram morphology
2
It has nevertheless been chosen to present an exotic system here
to stress the fact that phase diagrams are classified according to their
ENGLISH text is missing.Consistency
morphology, and that the calculation method is quite independent of the
1
Concerning the accuracy of the calculations it is not easy to argue
phase models.
that ∆tr si⋆ varies over the temperature domain.
3
The thermodynamics of phase equilibria does not make any differ2
The simplification which is part of the pure component reference
ence between vapour, liquid and solid phases in this regard[d], but in
is dominated by the activity model, and the calculated phase diagram
practise the number of empirical model parameters is often higher in
looks very promising.
the description of crystalline phases as compared to the same descrip3
But, a single phase diagram is not sufficient to claim any kind of tion of vapour and liquid phases of simple organic cmpounds.
thermodynamic consistency. To do this we have to validate all types of
4
In the gold–copper case there are e.g. six binary parameters
calorimetric quantities, vapour pressures, cryoscopic measurements,
needed for the solid phases, while only three is needed for the liquid.
etc.
889 / 1776
4
A systematic study at this level would indeed reveal the noted inconsistency, as well as several minor ones.
◦
888 / 1776 [d] The gold–copper system exhibits solid phase separation at T < 400 C.
0
0.2
0.4
0.6
0.8
Mole fraction of copper
4
To broaden the applicability of first order methods it has been suggested to accelerate the iteration step along the strongest eigendirection of the Jacobi matrix. This is known as the Dominant Eigenvalue
Method, see Appendix 28.
1
Expanded footnote [d]:
[d] Carl Gustav Jacob Jacobi, 1804–1851. German mathematician.
Figure 15.4: Phase diagram of gold–copper showing a negative deviation
from ideality. Both TpNewton and TpKvalue has convergence problems close
to the azeotrope.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
886 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
887 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
890 / 1776
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
0
Second order convergence
h1 = 0.675
each 4th iteration
hn = 0.965
each 24th iteration
k +1
log10 ∆µ
−1
−2
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
15 Multicomponent . . .
The Newton–Raphson method is of second order — details concerning the convergence factor is discussed in Appendix 27.
1
−3
2
This means that the number of significant digits will double in each
iteration provided that iteration k is sufficiently close to the solution
point.
−4
−5
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
Second order convergence (2)
Convergence orde
15 Multicomponent . . .
6
The iteration sequence approaches a line with slope 1, which is
typical for first order methods, and for the near-critical point the convergence factor also approaches 1, in the sense that hn = 0.965 is pretty
close to unity in this context.
4
But, the simplifications in Section 15.5 makes m = 1 unless for
ideal mixtures where the Hessian is exact.
3
The local convergence properties are irreproachable, but the
method needs quite good start estimates.
−6
−6
−4
k −2
log10 ∆µ
0
Figure 15.5: Convergence properties of the quasi-Newton–Raphson method
when using an ideal Hessian in the calculation of the bottom (h1 ) and top (hn )
tie-lines in Figure 15.2. The latter is close to the critical point. Solid lines
illustrate 1st order convergence. Note that h → 1 for the near-critical point.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Looking back
1
891 / 1776
The general phase equilibrium criterion:
Tα = Tβ = · · · = Tψ = Tω ,
β
ψ
β
ψ
p = p = ··· = p = p ,
2
=
β
µn
ψ
µn
= ··· =
The Gibbs–Duhem equation on vector form:
s dT − v d p + ( n 1 n 2 · · ·
3
=
µω
n
K -value: xiβ =
ˆ Ki xiα ,
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Looking back (4)
14
15
894 / 1776
Convergence orde
02 September 2013
Looking back (2)
∆µi
RT
5
Fugacity coefficient: Kieos =
6
Activity coefficient: Kieos =
7
Pure component reference: ϕα,⋆
= ϕβ,⋆
i
i
8
Poynting factor: Kivle =
=
ˆ
RT
= ln
Kieos xiα
xiβ
Chemical potential:
ϕαi
ϕαi
ϕiβ
=
ˆ
pisat γi
ϕsat
i
ϕi p
12
10
Henry’s law: Ki
11
Phase transition enthalpy: Kisle =
xi →1
= lim Kivle =
xi →0
γiα
visat (p −pisat )
RT
13
γiβ
exp
Convergence: limk →∞
log kxk +1
−
x∞ k
= log(h ) +
m log kxk
−
n
P
i =1
fi zi = 0 ,
= z α,k +
i =1
fi2 zi
-1
f
−∆nβ = ∆nα
Termodynamic methods TKP4175
∆tr hi⋆ (Ti⋆ )
R
1
Ti
−
1
T
02 September 2013
895 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News The elements Reactions min G
Calculation Le Chateliér Olds
Part 16
Chemical equilibrium
02 September 2013
896 / 1776
Calculation Le Chateliér Olds
New concepts
x∞ k
P
n
µα + Gα ∆nα = µβ + Gβ ∆nβ ,
Hij
p
γαi
∂f -1
f
∂z α
=
Two-phase equilibrium solved by Newton–Raphson iteration:
pisat
p
15 Multicomponent . . .
Assume ideal mixture when calculating G:
∆µ
∆nα
= −z α z β D + 1−d1 T e ddT RT
N α +N β
n
P
(1−Ki )zi
z α + Ki z β
i =1
1−Ki
,
z α + Ki z β
z α,k +1 = z α,k −
γiβ
Convergence orde
Two-phase equilibrium solved by direct substitution:
f (z α ) =
Kieos = Kiex =
893 / 1776
15 Multicomponent . . .
fi =
ˆ
⇒
02 September 2013
Looking back (3)
γiβ ϕiβ,⋆
exp
News The elements Reactions min G
Termodynamic methods TKP4175
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
Convergence orde
γαi ϕiα,⋆
Raoult’s law: KiRaoult = lim Kivle =
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
ϕiβ
9
Henry
892 / 1776
15 Multicomponent . . .
µαi −µβi
4
nn ) dµ = 0
∀i ∈ [1, n]
Termodynamic methods TKP4175
News Activity versus fugacity Reference states Direct substitution Newton–Raphson Ki vs µi
ω
µα1 = µ 1 = · · · = µ 1 = µ ω
1 ,
..
.
µαn
Tore Haug-Warberg (Chem Eng, NTNU)
Convergence orde
15 Multicomponent . . .
α
5
Figure 15.5 illustrates what influence the simplification has on the
calculation of the bottom and top tie-lines of Figure 15.2.
16 Chemical . . .
1
The periodic table
2
Stable elements
3
Chemical reactions
4
Reaction equilibria
5
Minimum Gibbs energy
6
Le Chateliér’s principle
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
897 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
898 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News The elements Reactions min G
News The elements Reactions min G
Calculation Le Chateliér Olds
News The elements Reactions min G
Calculation Le Chateliér Olds
Calculation Le Chateliér Olds
ENGLISH text is missing.Chemistry
1
Chemical equilibrium theory introduces the concept of chemical
bonds into thermodynamics, but without considering the physical forces
that hold the compounds together. Neither is the geometry of the chemical bond an issue in this context.
2
Instead, it takes as its starting point a thermodynamic interpretation
of the nature of chemical compounds, and looks at what net reactions
are possible between them.
The elements
ENGLISH text is missing.The recipients
1
Although it is an environmental sink in the same way as large bodies of water and bedrock, the atmosphere occupies a unique position.
3
Practical experiments have thereby shown that all chemical compounds are built up of one or more unique elements (atoms)[a]each
with their own characteristic physical properties.
4
These elements are the unchangeable components of life on Earth,
which manifest themselves through millions of different chemical compounds and an equally large number of chemical reactions.
5
They are what we call conserved quantities in accordance with the
principle of conservation of mass[a].
6
The exceptions to the conservation principle are the nuclear reactions that take place in nuclear power stations, in the Earth’s crust
(granite rocks) and in the core of the sun.
16 Chemical . . .
7
These are governed by Einstein’s equation e = mc 2 , which states
that ultimately mass and energy are equivalent concepts.
8
In this exceptional case, new elements can arise to replace old
ones that disappear.
9
Approximately 112 elements have so far been identified, 80 of
which are considered stable.
10 The stable elements are all of the ones from hydrogen (1) to
lead (82), with the exception of technetium (43) and promethium (61).
11 If we include all of the isotopes[a] of the various elements, the
number of stable nuclei (species of nucleus) rises to 256.
12 This nuclear palette provides the basis for all life on Earth, and for
all of the minerals, natural products and chemicals that we know from
everyday life.
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[a] From the Greek átomos, which means indivisible.
[a] Demonstrated by Antoine-Laurent de Lavoisier in 1789. This paved the way for
stoichiometry, which in turn led to the final rupture between chemistry and alchemy.
[a] Two isotopes of the same element always have the same number of electrons and
protons in their nuclei, but different numbers of neutrons.
1
The big span in physical properties is clearly illustrated in the table
below. It shows the melting points of the stable elements rounded to
the nearest degree Celsius:
2
The physical properties of the elements vary greatly — from being
gases at room temperature (air) and cryogenic liquids down to 0 K (helium) to being solids that are virtually impossible to melt (carbon).
He
H2
Ne
O2
F2
N2
Ar
Kr
Xe
Cl2
nil
−259
−248
−223
−220
−210
−189
−157
−112
−101
Hg
Br2
Cs
Ga
Rb
P
K
Na
I2
S
−39
−7
29
30
40
44
63
98
114
115
In
Li
Se
Sn
Tl
Cd
Pb
Zn
Te
Sb
157
181
221
232
304
321
328
420
450
631
Tore Haug-Warberg (Chem Eng, NTNU)
Mg
Al
Ba
Sr
Ce
As
Eu
Yb
Ca
La
650
660
729
769
798
817
822
824
839
920
Pr
Ge
Ag
Nd
Au
Sm
Cu
Mn
Be
Gd
931
938
961
1016
1065
1072
1085
1246
1287
1312
Tb
Dy
Si
Ni
Ho
Co
Er
Y
Fe
Sc
Termodynamic methods TKP4175
News The elements Reactions min G
1357
1407
1410
1453
1470
1495
1522
1526
1535
1539
Tm
Pd
Ti
Lu
Pt
Zr
Cr
V
Rh
Hf
1545
1552
1660
1663
1772
1852
1857
1902
1966
2227
Ru
B
Ir
Nb
Mo
Ta
Os
Re
W
C
02 September 2013
2250
2300
2443
2468
2617
2996
3027
3180
3422
3675
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16 Chemical . . .
4
Here we will use the definition µi =
ˆ (∂G /∂Ni )T ,p ,Nj,i , and show
how it leads us to a standard formulation of chemical equilibria in ideal
gases.
2
Put simply, you can say that the change in the chemical potential of
a component (mixed with several other components in the same phase)
indicates the work required to transfer the chemical compound from one
thermodynamic state to another.
3
The change of state in question can be described in several different ways, and there is a significant amount of freedom in the choice of
suitable system variables, cf. the alternative definitions of chemical potential
µi =
ˆ ∂∂NU
=
ˆ ∂∂NA
=
ˆ ∂∂NH
=
ˆ ∂∂NG
i
T ,V ,Nj ,i
i
S ,p ,Nj ,i
i
formed under anaerobic conditions in ??
16 Chemical . . .
J. Chem., 23(4):725–735, 1970.
4
2
Based on the evidence of countless experiments, it has been found
that in many compounds (albeit not all), elements combine in simple
ratios to form stable structures, such as CH4 , H2 O, etc.
3
This observation is incredibly important, as it gives us an accurate
mathematical basis for predicting what will happen in a given situation.
Chemical equilibrium
CH4 (gas) + 2 O2 (gas) ⇔ CO2 (gas) + 2 H2 O(gas)
(16.1)
reliably predicts what will happen.
It is so accurate that in-situ measurements of the flame geometry,
gas velocity and methane-to-air ratio are superfluous .
5
Expanded footnote [a]:
[a] Except in relation to small, unburned particles and pollutants, which form in varying
amounts depending on the physical conditions in the flame.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
3
2
Amongst other things, we will see that there is a graphic link between the concept of equilibrium and minimum Gibbs energy, before
leaving the graphic analysis behind in Chapter 17 in favour of an analysis involving multivariate functions.
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a µ A + b µ B = c µC + d µ D .
Tore Haug-Warberg (Chem Eng, NTNU)
§ 125
The differential equation gives us a necessary (but insufficient) condition for Gibbs energy having a possible minimum value.
2
4
Using the components involved in the reaction Eq. 16.2, the system’s total Gibbs energy can be expressed as:
3
The word possible complicates matters in the general case, but for
ideal gases both expressions produce the same answer[a].
[a] This is related to the fact that the Gibbs energy of an ideal gas mixture is a convex
function of the number of moles at a given T and p .
G = U − TS + pV =
ˆ
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News The elements Reactions min G
02 September 2013
911 / 1776
We have a closed system with four mole fraction variables and one
reaction equation. Let us use NB , NC and ND to describe the overall
composition of the system before the reaction takes place.
1
2
This leaves NA as a free variable, which tells us something about
the extent of reaction of the system. We can express the minimisation
problem using the following differential
0 = (dG )T ,p =
and similarly for the constraints:
Tore Haug-Warberg (Chem Eng, NTNU)
D
P
A
µi dNi ,
(16.5)
µi Ni .
Termodynamic methods TKP4175
News The elements Reactions min G
16 Chemical . . .
A
§ 124
02 September 2013
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Calculation Le Chateliér Olds
§ 126
16 Chemical . . .
(16.4)
5
The challenge is to determine Geq = min(G )T ,p from the Eqs. 16.2
and 16.4, either by minimising G , or indirectly by finding a state at which
(dG )T ,p = 0. Here we will choose the second approach.
Calculation Le Chateliér Olds
§ 125 Reaction equilibrium
D
P
Termodynamic methods TKP4175
(16.3)
Eq. 16.3 is an example of a chemical equilibrium
equation. Derive this equation by applying the principle
of minimum Gibbs energy to Eq. 16.4 at a given T and
p , and using the reaction in 16.2 as a constraint. Do
the phases of the components affect your calculation
in any way?
T ,p ,Nj ,i
Imagine, for example, a chemical equilibrium in a closed reactor
ˆ (∂A /∂Ni )T ,V ,Nj,i would be a
with a constant volume. In that case µi =
more natural choice.
(16.2)
4
For a closed system at a given temperature and pressure, the equilibrium distribution of A , B , C and D will be such that
1
Eq. 16.3 agrees with the assumption that the point of minimum
Gibbs energy corresponds to the equilibrium state, i.e. Geq = min(G )T ,p
or (dG )T ,p = 0.
6
◦ and N ◦ indicate that the amounts 6
8
The fixed mole numbers NC◦ , NH
This makes the analysis of the system more challenging than the
O
of carbon, hydrogen and oxygen in the system remain constant — they results from Chapter 8, which deals with closed systems with a constant
are what we call reaction invariants.
composition.
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serve as an example.
from Chapter 4.
5
This is the right choice for our purposes here, but there are alternative formulations derived from the Legendre transformations in Chapter 4.
7
Here we can immediately conclude that NCH4 +NCO2 = NC◦ , 4NCH4 + 5 On the other hand, the system gains an internal degree of freedom,
◦ and 2N
◦
2NH2 O = NH
O2 + 2NCO2 + NH2 O = NO based on the principle of because the associated extent of reaction both can and will vary with
conservation of mass.
temperature and pressure.
News The elements Reactions min G
16 Chemical . . .
16 Chemical . . .
2
The compounds in the mixture (the phase) are often referred to as
components.
4
We say that the reaction stoichiometry of the mixture is a function 2 The system thus loses an external degree of freedom for each
of the observed component stoichiometries.
reaction that reaches equilibrium.
aA + bB ⇔ cC + dD
Calculation Le Chateliér Olds
Minimum Gibbs energy
1
When two or more compounds are combined in a homogeneous
mixture (as in Eq. 16.1) they constitute a thermodynamic phase.
3
For chemical reactions in a closed system, the composition of the ENGLISH text is missing.Degrees of freedom
mixture is determined by the stoichiometry of the individual compo- 1 Meanwhile, it is also the case that the reaction stoichiometry of the
nents.
system is determined by several atomic properties beyond our control.
Let the reaction equation
6
However, it is by no means true that chemical conversions always
follow the same pattern and always yield the same products.
7
In practice there may be competing reactions and alternative reaction pathways that affect the outcome, but for equilibrium reactions we
can predict what products will be formed and how far the reaction will
proceed.
8
We can do that by performing a thermodynamic analysis of the final
state — the mechanism that brings about that state is irrelevant to the
analysis.
02 September 2013
ENGLISH text is missing.Thermodynamic phases
5
If we know the stoichiometry of the components we can predict the 3 This means we cannot alter all of the composition variables indepossible changes in the composition of the mixture without needing to pendently.
perform any further experiments.
4
One consequence of this is that the Gibbs’ phase rule must be
6
This is a very powerful principle, which is illustrated by the combus- changed from F = C + 2 − P to F = C + 2 − P − R , where R is the
tion reaction shown in Eq. 16.1.
number of chemical equilibrium reactions in the system.
1
The postulate that the equilibrium state is equivalent to a state of
minimum energy will be a key concept when we now look at chemical
reaction equilibria.
For example, when methane burns in air, the equation
News The elements Reactions min G
1
The symbol µi refers to the chemical potential of component i , a
name that gives us a clear hint that it is a quantity related to work.
S ,V ,Nj ,i
discovered (1869).
1
The mathematical description of chemical reactions is closely
linked to the concept of stoichiometry and to the related changes in
the system’s composition. Stoichiometry is the study of the composition of chemical compounds.
Calculation Le Chateliér Olds
Chemical potential
i
Stoichiometry
2
The total mass of the atmosphere constitutes no more than
1 kg cm-2 , which is approximately equivalent to a 10 m layer of condensed material, but its composition is all the more interesting, as in addition to small amounts of water, methane, ammonia and carbon diox1
In nature, the elements are mainly found in various chemical comide, it contains of a number of elements in their free states: 78% nipounds, with the most obvious examples being water, dissolved salts in
trogen, 20.95% oxygen, 0.9% argon and trace amounts of other noble
the oceans and minerals in the Earth’s crust.
gases.
2
In some places there are native deposits of metals such as copper,
3
The noble gases are constantly being produced by radioactive prosilver and gold, and elsewhere you can find a few elements like sulphur,
cesses in the Earth’s crust and leak slowly into the atmosphere.
graphite and diamond in their free states, but these are the exceptions
4
Helium has a comparatively low mass and will eventually escape ENGLISH text is missing.Chemical reactions
that prove the rule.
into outer space, while the other gases take part in the big geological 1 In addition to the substances we are familiar with from nature, hu3
The discovery of the elements and their classification in the pericycles on Earth.
mans have managed to produce a number of new compounds in laboodic table [a] therefore played a key role in the emergence of modern
5
However, there is no equivalent inorganic explanation for why nitro- ratory conditions.
chemistry.
gen and oxygen dominate the atmosphere to the extent that they do.
2
The process that converts chemical compounds into other (new) ENGLISH text is missing.Extent of reaction
ENGLISH text is missing.Heterogenous reactions
4
However, although the classification system itself may be a rela1
Chemical reactions allow us to synthesise new compounds and to 1 A chemical reaction always involves at least two compounds (reaccompounds is called a chemical reaction.
tively recent innovation, a surprising number of methods for producing 6 The thermodynamic conditions on Earth are such that these elements should only be found as nitrates dissolved in water and as iron 3 This transformation involves breaking the existing bonds between harness the bond energy of the reactants as thermal energy (through tant and product) in their gas, liquid or solid phases.
chemical reactions have existed for thousands of years.
combustion) or as electrical energy (in a battery).
oxide in various ores.
the atoms in the original compounds and forming new bonds under
2
If there is only one macroscopic phase in the system, the reaction
5
The depth of knowledge in classical antiquity is amongst other
2
There are countless variations on the theme, but all chemical reac- is referred to as being homogeneous.
different conditions.
things evidenced by the fact that all of the known metals of that time — 7 Actually, the relative abundance of both of them in their free states
tions can nevertheless be defined in terms of two basic properties: The
4
The compounds that are transformed are called reactants, while
silver Á, copper Ã, mercury Â, iron Ä, tin Å and lead Æ — with the excep- in the atmosphere is entirely due to biological activity on Earth[a].
3
A heterogeneous reaction, on the other hand, takes place across
extent and rate of reaction.
tion of gold À, were mined from ore and chemically processed rather 8 The atmosphere is in other words dominated by the waste products the ones that are formed are called products.
the phase boundaries within the system, but the system as a whole
3
The extent tells us how far the reaction has proceeded, while the must be closed, or in some other way be consistent with the principles
than excavated from native deposits.
from living organisms.
5
The number of atoms of each element is the same before and afrate tells us how quickly it is progressing.
of mass and energy conservation.
6
Our modern description of the associated metallurgical processes 9 Life involves many other elements, such as hydrogen, carbon, iron, ter the chemical reaction, but the properties of the products may never4
The above description sets out the fundamental principles that un- 4 That is essential if we wish to formulate the problem in thermois so complex and extensive that you have to wonder whether trial and sulphur and phosphorous, but only nitrogen, oxygen (and under certain theless be completely different from those of the reactants.
error were the only tools they had at their disposal.
conditions sulphur) are produced in their free states in large quantities. 6 For example, if solid carbon reacts with chlorine gas, it produces derpin chemical reactions, and although the theoretical study of rates dynamic terms, and that is precisely our goal: To establish a physicoof reaction is perhaps unpredictable, all reactions will eventually reach mathematical description of chemical reaction equilibria based on the
902 / 1776
903 / 1776 carbon tetrachloride[a], a colourless liquid at ordinary conditions.
a point at which the rate of reaction is zero.
laws of conservation of mass and energy, as well as on the additional
904 / 1776
5
Once equilibrium is reached, the extent of reaction has attained a assumption that equilibrium will be reached when the system is at the
[a] The Russian chemist Dmitri Ivanovich Mendeleev (1834–1907) is credited with the [a] Nitrogen is formed by denitrifying bacteria under anaerobic conditions in the soil:
thermodynamic limit value, which we will learn to calculate.
lowest possible energy state.
discovery of the periodic table. He was’nt the only person working on the problem, but 2 NO3− + C6 H12 O6 = N2 +3 O2 , whereas oxygen is a waste product of photosynthesis
905 / 1776
906 / 1776
he was the first person able to predict the properties of elements that had not yet been in green plants: CO2 + H2 O = CH2 O + O2 . Similarly, sulphur and copper can also be [a] J. D. Blackwood and B. D. Cullis. Kinetics of the direct chlorination of carbon. Aust.
ENGLISH text is missing.Chemical compounds
02 September 2013
912 / 1776
16 Chemical . . .
Since dNA is a free variable, and a is an arbitrary coefficient, the
8
above equation is equivalent to
3
We can also say that aNB − bNA , aNC + cNA and aND + dNA are
reaction invariants, i.e. they are conserved during the chemical reaction.
4
Eq. 16.2 thus results in a principle of conservation quite similar to
the underlying principle for Eq. 16.1.
5
This indicates that mass balance invariants and chemical reactions
are equivalent concepts, and that it is sufficient to know either the component stoichiometry or the reaction stoichiometry of the system.
6
The truth isn’t quite that simple, and Chapter 17 gives a more detailed analysis of this relationship using methods taken from linear algebra.
7
For the moment it will suffice to say that combining the above constraints and 16.5 gives us
b
a
µB dNA −
c
a
µC dNA −
= (c µC + d µD − a µA − b µB ) dNa A .
d
a
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Calculation Le Chateliér Olds
§ 125 Reaction equilibrium (2)
0 = µA dNA +
Tore Haug-Warberg (Chem Eng, NTNU)
µD dNA
(16.6)
c µC + d µD = a µ A + b µB ,
(16.7)
cf. Eq. 16.3.
We have derived this equilibrium condition solely using general
thermodynamic relationships, and no assumptions have been made
about behaviour specific to a particular state.
9
02 September 2013
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Calculation Le Chateliér Olds
Ideal gas
16 Chemical . . .
1
This is based on the chemical potential of a component of an ideal
ıg
gas mixture as expressed by the formula µi = µi◦ + RT ln(Ni p /Np◦ )
where µ◦i is the standard chemical potential of the component in its
pure form at temperature T and standard pressure p◦ .
2
In order to calculate the equilibrium state we must find the point
at which the chemical potentials balance in accordance with the stoichiometry of the system (cf. Eq. 16.7), taking into consideration any external constraints such as temperature, pressure, volume, energy, etc.
3
In general this requires iteration involving N + 2 state variables, i.e.
T , p and n or T , V and n.
4
It is only possible to solve the equilibrium analytically in a few
cases: For example if there is one reaction in an ideal gas mixture at a
given T and p .
5
But even in a simple case like that, you need to have a complete
understanding of the standard chemical potential µ◦i .
In practice, µ◦i is purely a function of temperature because the current convention is that p◦ is equal to 1 bar (1 atm prior to 1982).
6
10 This relationship therefore applies to all homogeneous and heterogeneous reactions in the gas, liquid and solid phases.
dNB = + ba dNA ,
dNC = − ac dNA ,
Tore Haug-Warberg (Chem Eng, NTNU)
dND = − da dNA .
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News The elements Reactions min G
News The elements Reactions min G
Calculation Le Chateliér Olds
Standard state
16 Chemical . . .
§ 126
For most practical purposes µ◦i is determined by consulting a table or using an explicit function of temperature. A common calculation
◦
◦
◦
◦
◦
Rmethod is µi (T , p◦ ) = hi (T ) − Tsi (T , p◦ ) where hi (T ) = ∆f hi (T◦ ) +
cp◦,i dT .
§ 125
§ 127
News The elements Reactions min G
Calculation Le Chateliér Olds
16 Chemical . . .
ENGLISH text is missing.Numerical solution
1
If you support Eq. 16.3 with an explanation of chemical potential the
result is a simple, but nonetheless general, description of the criterion
for chemical equilibrium.
2
That leaves us with the question of how to solve this equilibrium
problem in practice.
3
Naturally there are several answers to such an open question, and
we must therefore be willing to make use of various numerical techniques.
4
In this chapter it is natural to solve polynomial expressions,
whereas in Chapter 17 on simultaneous reaction equilibria it will be
more appropriate to use matrix-based methods.
918 / 1776
For reactions in ideal gas mixtures, the equilibrium 16.3
can be expressed in the format Q (y) · f (p ) = K (T ).
Here we refer to Q as the reaction quotient and K as
the equilibrium constant of the reaction. The mole
fraction vector y includes all the components in the
systema — including the ones not participating in the
reaction. It is a function y(ξ) of the extent of reaction
ξ. Determine the functionals Q , f and K .
The formation enthalpy ∆f hi◦ plays a particularly important role in
this calculation.
4
Under the IUPAC convention, hi = 0 for the chemical elements in
their stable configurations at T◦ = 298.15 K and p◦ = 1 bar.
3
It is a measure of the energy level of component i relative to that of
the elements that make up the compound.
5
The difference between the formation enthalpy of the products and
reactants determines how far a chemical reaction can proceed before
reaching equilibrium.
6
This contrasts with the criterion for phase equilibrium discussed in
Chapter 14, which makes no reference to the enthalpy level.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News The elements Reactions min G
02 September 2013
Tore Haug-Warberg (Chem Eng, NTNU)
16 Chemical . . .
Finally we simplify and take the exponential of both sides:
◦ +d µ◦ −a µ◦ −b µ◦
yCc yDd
−(c µC
p c +d −a −b
D
A
B)
· p◦
= exp
RT
yAa yBb
|{z} | {z }
−∆rx G◦ (T )
=
ˆ exp
RT
Q (y)
f (p )
=
ˆ K (T )
Extent of reaction
Termodynamic methods TKP4175
News The elements Reactions min G
02 September 2013
(16.10)
921 / 1776
16 Chemical . . .
Substitute a = b = c = d = 1 into Eq. 16.10 together with the
chemical symbols from the above question, i.e. yA = NCO2 /N , yB =
NH2 /N etc:
!
NH O 1
2
N
!1
!1
NCO
NH
2
2
N
N
CO
N
1
·
p 1+1−1−1
p◦
Note that the total number of moles N of all of the components
remains unchanged by the reaction, since c + d − a − b = 0. This
implies that f (p ) = 1.0, which in turn means that the equilibrium is the
same at all pressures.
3
The next step is to substitute the mole balances from above into
the equation.
◦ = 1 mol and N ◦
The initial conditions are NCO
H
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
2O
4
For permanent gases at constant pressure and for condensed
phases in general, pressure has little impact on the result and so we
can use the approximation Qeq ≈ Kc (T ), which states that the reaction
quotient Q must balance the constant Kc at the equilibrium point.
5
Note, however, that Kc has dimensional units and should not be
thought of as a true thermodynamic equilibrium constant!
922 / 1776
3
If we look slightly more closely at the fraction Q (y(ξ)) we can see
that the reaction in fact must go to the right if Q < Kc and to the left if
Q > Kc .
This result is known as the Guldberg –Waage law of mass action.
Tore Haug-Warberg (Chem Eng, NTNU)
Expanded footnote [a]:
Expanded footnote [a]:
[a] Cato Maximilian Guldberg, 1836–1902. Norwegian chemist and mathematician.
[a] Peter Waage, 1833–1900. Norwegian chemist.
Termodynamic methods TKP4175
02 September 2013
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16 Chemical . . .
Let ξ ∈ [0, 1] be the dimensionless extent of reaction of the system.
◦ − ξ mol, N
For an arbitrary value of ξ we have NCO = NCO
H2 O =
◦
−
ξ
mol,
=
ξ
mol
and
=
ξ
mol,
which
gives rise to the
NH
N
N
H
CO
2
2
2O
equation
6
(1−ξ)(3−ξ)
ξ2
= 0.696 .
7
The physical solution of the second degree equation gives us ξ =
0.798, or
NCO2 = 0.798 mol ,
N H2 = 0.798 mol ,
N CO = 0.202 mol ,
NH2 O = 2.202 mol .
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§ 127
§ 126
§ 128
02 September 2013
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Calculation Le Chateliér Olds
16 Chemical . . .
ENGLISH text is missing.Polynomial solution
1
In order to solve Eq. 16.10 as a mathematical problem we must
know the exact stoichiometry of the system, as expressed by the coefficients a , b , c and d . We must also know the values for the equilibrium
constant K (T ), the pressure p and last but not least: The initial composition expressed by the numbers of moles NA◦ , NB◦ , NC◦ and ND◦ .
2
Next we must determine the roots of a polynomial of the order
n = max(a · b , c · d ).
3
That is because Q (yi ) is a rational function of the mole fractions
yA , yB , yC and yD , expressed such that when the function is rewritten to
a polynomial form n becomes the highest exponent for any yi that takes
part in the fraction.
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T [K]
ln(K )
T [K]
ln(K )
298.15
400
600
−11.55
−7.348
−3.345
1000
1800
3400
−0.363
1.336
2.073
Let A = CO2 , B = H2 , C = CO and D = H2 O in
Eq. 16.2. Calculate the reaction equilibrium at 1000 K.
Assume that the molar ratio of unreacted H2 O to CO
is 3 : 1. Equilibrium data are given in the table on the
right. What happens in this case if you change the
pressure, or temperature?
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
925 / 1776
Calculation Le Chateliér Olds
Changes in pressure
16 Chemical . . .
1
A change in the total pressure does not affect the water-gas equilibrium, since the reaction has the same number of moles of compounds
on each side of the reaction arrow, but this result only applies to ideal
gases.
2
The key factor in a real system is whether the products occupy a
volume that is bigger than, equal to or smaller than the volume of the
reactants.
3
Depending on the change in volume, any increase in pressure will
result in the reaction going to the left, remaining unchanged or going to
the right respectively.
4
This is a result of Le Chateliér’s principle, which is discussed in
Section 16.6.
= 3 mol.
02 September 2013
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Calculation Le Chateliér Olds
§ 127 Shift reaction (2)
5
Tore Haug-Warberg (Chem Eng, NTNU)
News The elements Reactions min G
As the reaction only has one degree of freedom, it must either go
to the right or left to reach equilibrium.
3−ξ
( 1−ξ
4 )( 4 )
=
( 4ξ )( 4ξ )
= e -0.363 = 0.696 .
2
4
More often than not, it will require numerical iteration.
3
A closely related equation describes equilibria in other ideal systems such as dilute solid in liquid solutions, but in that case the contribution of pressure is negligible.
News The elements Reactions min G
1
N
919 / 1776
16 Chemical . . .
1
Eq. 16.10 is the classic equilibrium equation for a single reaction
in an ideal gas mixture. The problem is thereby reduced to calculating
y(ξ) such that the given equation is fulfilled.
2
We start by substituting the values for an ideal gas into Eq. 16.7:
h
h
i
i
pN ◦ + RT ln pN D 0 = c µ◦C + RT ln NpC◦ + d µD
Np◦
i
i
h
h
pN pN (16.8)
− a µ◦A + RT ln NpA◦ − b µ◦B + RT ln NpB◦ .
Calculation Le Chateliér Olds
ENGLISH text is missing.Reaction quotient
16 Chemical . . .
3
Then we multiply out the brackets and substitute mole fractions into
the equation:
py py − c µ◦C + d µ◦D − a µ◦A − b µ◦B = cRT ln p◦C + dRT ln p◦D
pyA pyB − aRT ln p◦ − bRT ln p◦ .
(16.9)
2
Calculation Le Chateliér Olds
§ 127 Shift reaction
02 September 2013
2
1
Eq. 16.10 also explains what happens if we choose to start an
experiment with a composition where Q , Kc .
4
Tore Haug-Warberg (Chem Eng, NTNU)
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Calculation Le Chateliér Olds
§ 126 Reactions in ideal gas mixtures (2)
4
917 / 1776
§ 126 Reactions in ideal gas mixtures
1
We will now show that the equilibrium equation can be separated
into three constituent parts that are functions of composition, pressure
and temperature.
1
2
Calculation Le Chateliér Olds
Tore Haug-Warberg (Chem Eng, NTNU)
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927 / 1776
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928 / 1776
News The elements Reactions min G
Calculation Le Chateliér Olds
Changes in temperature
16 Chemical . . .
§ 128
2
ln(K )
Gibbs–Helmholtz’ equation 4.46 on page 161, applied to ln K =
−∆rx G◦ /(RT ), gives us the slope (∂ ln K /∂(1/T )) = −∆rx H◦ /R .
−4
−8
−10
1
4
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02 September 2013
929 / 1776
1000
T [K]
Tore Haug-Warberg (Chem Eng, NTNU)
For the water gas shift reaction in an ideal gas mixture, equilibrium
will assert itself independently of pressure. This is due to the fact that
the total number of moles of components is preserved during the reaction.
1
Bearing in mind that the Gibbs free energy should reach a minimum, we are able to write
2
G = C1
+ ξ(µ◦CO2 + RT ln ξ) + ξ(µ◦H2 + RT ln ξ)
02 September 2013
5
3
where C1 (p , N ) = {N }RT ln(p /(p◦ {N })) = −4RT ln 4 is a constant
that does not affect the result.
4
Admittedly the total number of moles (and the equilibrium pressure)
does affect the vertical position of the graph, but shifting the graph vertically will not alter where the minimum value of G (ξ) is on the ξ axis.
930 / 1776
16 Chemical . . .
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02 September 2013
933 / 1776
932 / 1776
Calculation Le Chateliér Olds
25
G = C1 + µ◦CO + 3µ◦H2 O + ξ(µ◦CO2 + µ◦H2 − µ◦CO − µ◦H2 O )
|
{z
}
|
{z
}
−∆rx G◦ =RT ln K
h
i
+ RT ln ξξ ξξ (1 − ξ)1−ξ (3 − ξ)3−ξ ,
minvalue
20
15
10
which, if we introduce the new constant C2 (T ) can be written
h
i
G = C2 − 0.363RT ξ + RT ln ξξ ξξ (1 − ξ)1−ξ (3 − ξ)3−ξ .
5
This means that the position of the minimum value of G (ξ) will
move with temperature. G (ξ) is illustrated in Figure 16.2 using Program 31:1.26.
0
0.5
xi
1
6
Here C2 results in a further vertical shift of G (ξ), while RT ln K =
−0.363RT is incorporated in combination with ξ, altering the slope of
the graph.
Tore Haug-Warberg (Chem Eng, NTNU)
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Calculation Le Chateliér Olds
02 September 2013
30
By simplyfying the function further we obtain
constant
Termodynamic methods TKP4175
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8
We notice that the function has a minimum at ξ = 0.798, which is
in agreement with Theme 127.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
Calculation Le Chateliér Olds
§ 128 Minimum Gibbs energy I (2)
7
+ (1 − ξ)(µ◦CO + RT ln(1 − ξ)) + (3 − ξ)(µ◦H2 O + RT ln(3 − ξ)) ,
3
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16 Chemical . . .
2
Figure 16.1: van’t Hoff’s equation (∂ ln K /∂(1/T )) = −∆rx H◦ /R ; also see the
derivation of Gibbs–Helmholtz’ equation in Theme 28, applied to the system
CO2 –H2 –CO–H2 O at 1000 K. The heat of reaction is a function of temperature, and the slope of the graph therefore varies over the domain of definition.
Calculation Le Chateliér Olds
§ 128 Minimum Gibbs energy I
16 Chemical . . .
2
This means that it is possible to change the size of the system at
constant T and p , i.e. by multiplying all of the numbers of moles by the
same factor, without disturbing the equilibrium.
3
What happens if a single component is added is slightly more complicated.
4
One specific example is discussed in Section 16.6 in conjunction
with the discussion of Le Chateliér’s principle.
931 / 1776
Sketch the graph of G (NCO , NH2 O , NCO2 , NH2 ) as a
function of ξ ∈ [0, 1] for Ni = Ni (ξ), where ξ = 0
represents the unreacted system and ξ = 1 is where
the limiting reactant has been used up. Show that the
minimum value of the function agrees with the result
from Theme 127.
−6
3
In other words, you would expect ln K as a function of 1/T to produce a virtually linear graph, provided that ∆rx H◦ does not change sign
within the domain of definition.
[a] Jacobus Henricus van’t Hoff, 1852–1911. Dutch chemist.
§ 129
1
Any change in the overall composition will also potentially affect the
equilibrium, but only if the reaction quotient Q is changed.
−∆rx H◦ /R
−2
2
5
This relationship is known as the van’t Hoff[a] equation — originally
expressed as (∂ ln K /∂T ) = ∆rx H◦ /(RT 2 ).
§ 127
Calculation Le Chateliér Olds
ENGLISH text is missing.Changes in composition
0
A change in temperature causes the equilibrium constant K (T ) to
change, and hence the equilibrium composition will also vary.
1
Figure 16.1 illustrates this for the water gas shift reaction. The
graph is clearly curved, but not so curved that there is any doubt about
the relationship between changes in ln K and changes in the reciprocal
temperature 1/T .
News The elements Reactions min G
Calculation Le Chateliér Olds
kjoule
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02 September 2013
934 / 1776
Figure 16.2: Gibbs energy for the system CO2 –H2 –CO–H2 O as a function of
the extent of reaction ξ. The short branch of the graph shows the Taylor series
to the second order. Note that the function’s minimum value agrees with the
solution of the classical equilibrium equation in Theme 127.
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Calculation Le Chateliér Olds
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935 / 1776
Calculation Le Chateliér Olds
ENGLISH text is missing.Mathematical analysis
1
In the above equation, Gibbs–Helmholtz’ equation 4.46 on
page 161 has been applied to the first of the derivatives on the right
side, and (∂ ln N /∂ ln Nj ) = (Nj /N ) (∂N /∂Nj ) = yj has been applied to
the other derivative.
2
In general terms, the changes to the composition of the system can
expressed as
ˆ Nj-1 dNj = Nj-1 (dNj◦ + νj dξ) ,
dln Nj =
but since the system is closed dNj◦ = 0 ∀j .
3
This means that dln Nj = νj Nj-1 dξ, from which it follows in the
normal way that the differential of Γ is:
P
µ dΓ = νi d RTi
i
=
P
i
νi
hi◦
R
d
1
T
+
P
i
νi dln p +
P
i
νi
νi
Ni
dξ −
P
i
νi
P
j
yj
νj
Nj
dξ ,
but for the sake of the subsequent explanation we will rewrite the expression in vectorial[a] form
√y)2
◦
] dξ ,
(16.11)
dΓ = ∆rxRH d T1 + (P νi ) dln p + [qT q − (qT
i
where qi =
ˆ νi /√Ni og yi =
ˆ Ni /N .
4
The first and second terms in this equation have clear physical
interpretations, whereas the meaning of the third term is less clear.
5
But if we apply the chain rule to Γ we get the result
P P ∂µi /RT P ∂Γ PP
d Nj =
νi ∂Nj
νj =
νi
dξ =
∂Nj
dΓ
dξ T ,p
j
T ,p
j
=
T ,p
i
d2 G /RT dξdξ T ,p
=
ˆ νT Gν ,
j
i
∂2 G /RT ∂Ni ∂Nj T ,p
νj
and by inserting the values for an ideal gas we find that νT Gν = qT q −
√y)2
(qT
where G is the matrix of second order derivatives of the Gibbs
free energy with respect to the mole numbers.
ENGLISH text is missing.Adding mass
Le Chateliér principle
6
It is also known as the Hessian of G .
7
In other words, the third term in the Eq. 16.11 corresponds to the
1
The following example reveals part of the problem: Assume that
second order derivative of G /RT with respect to the extent of reaction
the equilibrium reaction AB2 ⇔ A + 2B is described by Q =
ˆ yA yB2 /yAB2
ξ.
and f (p ) = 1, where Qeq = Kc cf. Eq. 16.10.
8
Figure 16.2 demonstrates that this is in fact true in practice.
2
The definition of the reaction quotient Q shows us that Q < Kc ENGLISH text is missing.Closed system changes
moves the reaction to the right, whereas Q > Kc moves the reaction to 1 In other words, adding a component to the mixture is not equivalent 9 Here the truncated graph shows a second order Taylor series[a] for
to imposing an external change on the system.
G (ξ) around the minimum, i.e. G = Geq + (νT Geq ν)(ξ − ξeq )2 + O(ξ3 ).
the left.
ENGLISH text is missing.Disturbances
1
By taking as our starting point the reaction aA + bB ⇔ cC + dD
in Sections 16.4 and 16.5 we have seen that the equilibrium principle
min G (n(ξ))T ,p results in the condition a µA + b µB = c µC + d µD .
2
The assumption is that ξ [mol] refers to the extent of reaction of the
system — selected such that any increase in ξ moves the reaction from
the left to the right. The composition of the reacting system can then at
all times be expressed as
It now remains to be seen what happens in a closed system.
10 Around the equilibrium point, the graph clearly bends upwards, and
Our task is to give the principle a precise mathematical definition, we can actually apply this result to all reactions in ideal gas mixtures
over the whole domain of definition.
4
Based on Le Chaterlier’s principle we might draw the overly hasty so that we can remove all doubt as to what conditions apply.
T √y|
≤ ||q|| ·
conclusion that the added component A must always be consumed, as 4 We will take as our starting point the dimensionless Gibbs energy 11 The Cauchy[a]–Schwarz[a] inequality states that |q
√y|
≤ ||q||. From this it
||√y||, but as ||√y|| = (Pi |yi |)0.5 = 1, then |qT
P
this would counteract the addition of the component.
Γ=
ˆ (RT )-1 ∆rx G =
ˆ (RT )-1 νi µi ,
follows that
i
5
The requirement would then be that (∂Q /∂NA ) > 0 for all values of
√y)2
where νi represents the stoichiometric coefficients a , b , etc.
qT q − (qT
≥ qT q − qT q = 0 ,
Kc , but this is not true.
3
If we add e.g. component A to the system the equilibrium will be 2
disturbed, but in which direction will the reaction then move?
3
16 Chemical . . .
5
At equilibrium Γeq = 0 and for changes in the state of equilibrium which proves that νT Gν ≥ 0 across the entire composition range.
6
In fact, if we differentiate the reaction quotient we get
dΓeq = 0.
NA = NA◦ − a ξ , NC = NC◦ + c ξ ,
N2
NA N 2
§ 129 Minimum Gibbs energy II
∂Q 12 For random changes in the state of equilibrium dΓeq = 0, and if we
= N -2 NABB − 2N -3 NAB B
∂ NA
2
2
1
The necessary condition for a minimum value of G is that the
NB = NB◦ − b ξ , ND = ND◦ + d ξ .
6
We will first derive the differential dΓ using appropriate coordinates substitute the above result into Eq. 16.11 it is easy to calculate the shift
= NA-1 (1 − 2yA )Q ,
gradient ∂G /∂ξ = 0.
and then study the effect of changing the temperature, pressure and in the equilibrium composition as a result of changes in 1/T and ln p :
3
By varying ξ it is possible to find a minimum value of G (n(ξ)) which
∆ H
which shows that the change in Q is governed by the stoichiometry of composition with one reservation: In this problem it is practical to use
dξ 2
Alternatively we can choose to divide by RT in order to obtain a
= −(νT Gν)-1 rxR ◦ ,
satisfies the equilibrium condition, cf. Figure 16.2.
d(1/T ) eq
the system (the derivative changes sign at yA = 0.5) and not by the d(1/T ) and dln p as free variables rather than dT and dp .
dimensionless expression. This does not alter the necessary condition
dξ T
-1 P
νi .
4
But what happens to the equilibrium if the external conditions ENGLISH text is missing.Le Chateliers principle
equilibrium condition.
dln p eq = −(ν Gν)
for a minimum, as the temperature of the system remains constant.
7
Our point of departure is the chemical potential of an ideal gas,
i
change, i.e. if the temperature, pressure or compositions of NA◦ , NB◦ , 1 For example, any rapid increase in pressure will cause the reacENGLISH text is missing.Graphical solution
7
Consequently, Le Chateliér’s principle does not apply if you add which is the same as in Theme 126 on page 862.
3
If we subsequently differentiate with respect to ξ, we get
NC◦ and ND◦ are altered?
tion, which has now been driven away from equilibrium, to move in a di1
Figure 16.2 illustrates the equilibrium principle in an easy-to-grasp
ıg
943 / 1776
more mass of a component to an equilibrium system.
1−ξ
3−ξ
8
In a dimensionless form it can be written (RT )-1 µi = (RT )-1 µi◦ +
0 = − 0.363 + 2 ln ξ + 2ξ
way, but it is only possible to illustrate systems with one degree of freeξ − ln(1 − ξ) − 1−ξ − ln(3 − ξ) − 3−ξ
5
The famous Le Chateliér’s principle[a][a] states that: if you impose rection that results in the system taking up less space, i.e. the pressure
8
The direction of the reaction is in this case determined by 1−2yA ,eq , ln(Ni p /Np◦ ).
ξ2
dom like this.
an external change on a system at equilibrium, then the equilibrium gradually decreases as the equilibrium is reasserted.
= − 0.363 + ln (1−ξ)(3−ξ)
.
P νi
P P νj
P
2
as shown in the table below:
9
The total differential is:
[a] The summation
i νi Ni −
i νi
j yj N can be expressed as
i (νi /√Ni ) −
shifts to reduce or eliminate the effect of the imposed external change. 2 This simple principle is incredibly elegant, but it can easily be mis2
Here we should remember that our system has a total of four chemP 2j
P
√
√
It is now easy to rearrange the expression to show that it is identical to
ıg
P ∂ ln N ˆ νi /√Ni .
µ ∂µi◦ /RT i (νi /√Ni )√Ni /N ·Pj (νj /pNj )pNj /N or
i qi −(Pi qi yi )·(Pj qj yj ) where qi =
1
939 / 1776 interpreted.
(∂∂NQA )
Case yA ,eq yB ,eq yAB2 ,eq Kc
Direction
√y)2
ical composition variables, which means that the graph is an example § 129
dln Nj
d RTi = ∂(1/
the equilibrium equation solved in Theme 127. Clearly, the tangent of
T ) d T + dln p + dln Ni −
∂ ln Nj
In vector form the expression reads qT q − (qT
.
j
of a one-dimensional manifold in four-dimensional space.
i
2/3
1/6
1/6
1/9 −1/18
→
1
Find an analytical expression for min G (n(ξ))T ,p , based on the the graph has a zero slope in the minimum point. Besides, the graph
3
History also tells us that Le Chateliér presented his ideas through
[a] Simplifies to a constant and a second order term because the first order term is
P
h◦
ii
1/2
1/4
1/4
1/8
0
−
= Ri d T1 + dln p + dln Ni − yj dln Nj .
zero, in accordance with the equilibrium criterion dG / dξ = 0.
a number of works in 1884, 1888, 1908 and 1933, which also suggests
3
In other words, describing chemical equilibria geometrically is com- same function as the one used in Theme 128. Show that the result itself is convex in the same point. These are the necessary and suffij
1/3
1/3
1/3
1/9 +1/9
←
iii
[a] Henri Louis Le Chatelier, 1850–1936. French chemist.
agrees with Theme 127.
that the principle is not quite perfect after all.
cient conditions of a function minimum.
plicated.
[a] Augustin Louis Cauchy, 1789–1857. French matematician.
940 / 1776
938 / 1776 [a] Aschehougs konversasjonsleksikon. Aschehoug & co., 5th edition, 1974. In Norwe937 / 1776
941 / 1776
936 / 1776
942 / 1776
These equations conclude Le Chateliér’s principle for a chemically
equilibrated, closed system.
1
2
A summary of the principle, as it applies to ideal gas mixtures, is
set out below:
Criterion
∆rx H◦ > 0
∆rx H◦ < 0
P
ν >0
Pi i
i νi < 0
Tore Haug-Warberg (Chem Eng, NTNU)
Description
endothermic
exothermic
expansion
contraction
Looking back
16 Chemical . . .
Looking back (2)
16 Chemical . . .
[a] Hermann Amandus Schwarz, 1843–1921. German mathematician.
gian.
dT
+
+
0
0
dp
0
0
+
+
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Direction
→
←
←
→
02 September 2013
1
256 non-radioactive nuclei (80 different elements altogether)
2
Standard states: He(gas), O2 (gas), C(sol), Br2 (lıq), etc.
3
Chemical reactions: νA A + νB B + · · · = 0
4
5
8
Le Chateliér’s principle:
dξ d(1/T ) eq
dξ dln p eq
Ideal gas: Q (y) · f (p ) = K (T )
n
P
νi µi = 0
Reaction equilibrium:
= −(νT Gν)-1
= −(νT Gν)
∆rx H◦
R
-1 P
i
νi
i =1
944 / 1776
∂ ln K ∂(1/T )
= − ∆rxRH◦
6
van’t Hoffs slope:
7
Minimum Gibbs energy: Geq = min G (n(ξ)))T ,p
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945 / 1776
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946 / 1776
News Stoichiometry min G
Lagrange Null space Combustion Flame temperature
News Stoichiometry min G
Olds
Lagrange Null space Combustion Flame temperature
New concepts
Part 17
Simultaneous reactions
News Stoichiometry min G
Olds
17 Simultaneous . . .
1
Atom balances
2
Formula matrix
3
Null space
4
Independent reactions
5
Lagrange multipliers
6
Extent of reaction
7
Newton–Raphson iteration
8
Adiabatic (flame) temperature
Lagrange Null space Combustion Flame temperature
Linear algebra
Olds
17 Simultaneous . . .
1
From Chapter 16 we know that one single reaction equation can be
solved using a polynomial function of the mole fractions, but in order to
solve several simultaneous reactions there is a need for more abstract
methods.
Below we shall investigate the combustion of methane (CH4 ) in air
4
(0.78N2 , 0.21O2 , 0.01Ar) which gives the reaction products CO, CO2 ,
H2 O, H2 , OH, H and NO.
2
It is indeed quite possible to write down explicit equations for all of
the concurrent reactions using the same format as explained in Chapter 16, but the challenge is to select the correct (physical) root of each
polynomial (there is one for each independent reaction).
3
Linear algebra turns out to be a good option in this respect, as it
provides us with an opportunity to structure the equations in a good
way, and at the same time is capable of solving the general equilibrium
problem without having to make arbitrary choices between several possible solutions.
see also Part-Contents
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Linear algebra (2)
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Olds
17 Simultaneous . . .
§ 130
§ 129
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§ 131
948 / 1776
17 Simultaneous . . .
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 130 Atom balances
949 / 1776
Olds
17 Simultaneous . . .
1
At stationary conditions there are exactly the same number of
moles of C, H, N, O and Ar leaving the burner as there are entering
the burner.
CO2 , H2 O, CO, etc.
6
It is of course possible to choose a more traditional approach involving stoichiometric coefficients and chemical symbols, but it is not
clear that doing so would make it any easier for us.
7
If the matrix notation is unfamiliar, it is advisable to use both approaches side-by-side until you are sufficiently experienced with it.
N2 , O2 , Ar
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Olds
5
Altogether there are 11 compounds involved, but matrix notation
simplifies the mathematical representation considerably.
CH4
Tore Haug-Warberg (Chem Eng, NTNU)
Write down the atom balances for the burner in Figure 17.1
on the preceding page. Assume stationary conditions.
Is there any connection between the atom balances
and the mass balance of the system?
2
The atom balance for the apparatus is hence
~
bın = ~
bout ,
3
where ~
b is the molar flow rate of all of the various atoms involved
in the problem:
Expanded footnote [a]:
[a] Hint: Assume that the symbols C and H are assigned the correct atomic masses in
the chemical formula CH4 . By inserting ∗ and + in the appropriate order you will see
that C + H ∗ 4 is an algebraic expression for the molecular mass of methane.
~
~C
b=( N
~O N
~ Ar )T .
~H N
~N N
N
Figure 17.1: Sketch of an adiabatic gas burner
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§ 130 Atom balances (2)
950 / 1776
News Stoichiometry min G
Olds
17 Simultaneous . . .
4
Making use of the system’s atom matrix (atoms × components )
(also called the stoichiometric matrix)
CH4 N2 O2 Ar CO CO2 H2 O H2 OH H NO






A = 




1
4
0
0
0
0
0
2
0
0
0
0
0
2
0
0
0
0
0
1
1
0
0
1
0
1
0
0
2
0
0
2
0
1
0
0
2
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
1
1
0











N
(17.1)
O
Ar
02 September 2013
Lagrange Null space Combustion Flame temperature
951 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Olds
17 Simultaneous . . .
§ 131
§ 130
§ 132
02 September 2013
Lagrange Null space Combustion Flame temperature
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Olds
17 Simultaneous . . .
ENGLISH text is missing.Atom balance
1
An atom balance represents a general principle that can be applied to all chemical reaction problems with the exception of nuclear
processes in atomic reactors (fission) and inside the sun (fusion).
2
There must the atom balance principle give way to the more general principle of conservation of the nucleon number in the reactor.
3
At even higher energy densities we have to reformulate this principle too, until we finally end up with Einstein’s equation e = mc 2 .
4
At that level, the mass and energy balances do not apply separately
anymore.
955 / 1776
The atom balance is thus
6
(17.3)
A~
nın = A~
nout .
7
8
The relationship with the mass balance is simple.
Let mw be the molar mass of the atoms.
~
bout
H
Termodynamic methods TKP4175
§ 130 Atom balances (3)
9
C
and the molar flow rate of all the components in the mixture
~ CO N
~ CO N
~H N
~H N
~N N
~O N
~ Ar N
~H O N
~ OH N
~ NO T , (17.2)
~ CH N
~
n= N
4
2
2
2
2
2
5
Tore Haug-Warberg (Chem Eng, NTNU)
~ = mT ~
The mass flow rate is then M
w b, which when substituted into
~
~
~
= bın gives Mın = Mout .
10 From this we can conclude that the mass balance is taken care of
by the principle of atom conservation.
11 This is a straightforward consequence of the mass balance being
a scalar (inner) product over the (vectorial) atom balance.
Set out the component balance for the burner in
Figure 17.1 on page 882. Assume stationary conditions.
Is there any link between the system’s component
and mass balances?
12 Hence the principle does not work the other way around (the atom
balance is not given by the mass balance).
it is possible to express the molar flow rate of the atoms as ~
b=
ˆ A~
n.
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Lagrange Null space Combustion Flame temperature
Mass balance principle
News Stoichiometry min G
Olds
17 Simultaneous . . .
From Theme 130 we know that any arbitrary change in the number
of moles of the components must fulfil the conditions
1
~
bout − ~
bın = A(~
nout − ~
nın ) =
ˆ A∆~
n = 0.
(17.4)
The atom matrix in Eq. 17.1 has 5 rows and 11 columns.
2
Lagrange Null space Combustion Flame temperature
Mass balance principle (2)
News Stoichiometry min G
Olds
17 Simultaneous . . .
5
Finally we solve the set of equations with respect to ∆~
n1 =
A ∆~
n2 , which gives us the solution vector:
−A-1
1 2
A
−A-1
1 2 ∆~
n2 =
ˆ Nξ̊ .
(17.6)
∆~
n=
I
N = {νij } is the system’s stoichiometric matrix, where νij is the stoi7
chiometric coefficient of component i in reaction j .
Lagrange Null space Combustion Flame temperature
§ 131 Component balances
Olds
17 Simultaneous . . .
1
If we combine equations 17.5 and 17.6, it is clear that AN = 0. We
can say that N lies in the null space of A.
3
The reverse is only true to a limited extent: Given a matrix N we
can always calculate a matrix A′ with a rank equal to the number of
reaction invariants in the system.
2
This is an important result, which amongst other things means that
the mass and atom balances are part of the component balance.
n1 × dim ~
n1 whereas
6
Note that the first I matrix has dimension dim ~
the second I matrix has dimension dim ~
n2 × dim ~
n2 in the equations
above.
3
This means that Eq. 17.4 has 5 specified values and 11 unknowns,
giving 11 − 5 = 6 degrees of freedom.
Expanded footnote [a]:
[a] The partitioning of A = ( A1 A2 ) and ∆~
n = ( ∆~
nT1 ∆~
nT2 )T )T requires A1 to
n1 must also be equal to the number
be quadratic, and the number of elements in ∆~
of elements in ~
b. Moreover, A1 must be invertible, but we don’t have space to go
into the details of that here. Suffice to say that it will always be possible to perform
the necessary operations, albeit through a more laborious process involving partial
pivoting of the coefficient matrix.
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Lagrange Null space Combustion Flame temperature
§ 131 Component balances (2)
957 / 1776
17 Simultaneous . . .
6
The atom matrix is surjective, whereas the stoichiometric matrix is
only injective:
sur
A⇄N
inj
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Termodynamic methods TKP4175
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Lagrange Null space Combustion Flame temperature
§ 132 Reaction equations
5
For example, this is true of the equilibrium 2 NO2 ⇔ N2 O4 . Here
the ratio between nitrogen and oxygen is 1:2 on both sides of the reaction arrow, which means that (dim A = 2) > (rank A = rank A′ = 1).
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
7
If we substitute in the figures from Theme 130, the final result we
get is (check that AN = 0):


0 
 0 −4 −4 −2 −2


 0
0
0
0
0 −4 



 −4 −6 −2 −5 −1 −4 


 0
0
0
0
0
0 


 −8
4
4
2
2
0 



1

∆~
n = 8  8
(17.7)
0
0
0
0
0  ξ̊ .


 0
8
0
0
0
0 



0
8
0
0
0 
 0


 0

0
0
8
0
0


 0
0
0
0
8
0 



0
0
0
0
0
8
Tore Haug-Warberg (Chem Eng, NTNU)
8
The elements of vector ξ̊ =
ˆ ∆~
n2 are the rates of reaction of the
independent reactions.
960 / 1776
17 Simultaneous . . .
In this case there are 11 − 5 = 6 independent reactions. More
generally, this number is given by dim(n) − rang(A).
1
2
A set of independent reaction equations can be inferred directly
from matrix N in Eq. 17.7:
Termodynamic methods TKP4175
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Lagrange Null space Combustion Flame temperature
§ 131 Component balances (3)
958 / 1776
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News Stoichiometry min G
Olds
17 Simultaneous . . .
9
Note that the system’s rates of reaction are wholly determined by
the components included in the ∆~
n2 vector, which in this case means
CO2 , H2 O, H2 , OH, H and NO.
10 It is possible to choose a different basis by changing the order in
the n vector, but the columns in A must be in the same order as the
rows in n.
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Lagrange Null space Combustion Flame temperature
§ 132 Reaction equations (2)
961 / 1776
§ 132
§ 131
§ 133
1
3
17 Simultaneous . . .
3
This is just one of the many options, and there are countless combinations of these reaction equations — we say that they constitute a
basis of the vector space.
4 O2 + 8 CO = 8 CO2
Olds
Isn’t it odd, then, that the fourth row of N is 0?
4
What is the physical explanation for this? Hint: Which compound
is fourth in the atomic number vector, and how reactive is it?
.
962 / 1776
Find a (mathematical) set of independent reaction
equations for the system in Theme 131. What are
your options?
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Lagrange Null space Combustion Flame temperature
Minimum Gibbs energy
963 / 1776
Olds
17 Simultaneous . . .
1
Finding the energy minima of systems is essential to all thermodynamic equilibrium calculations.
2
This also applies to chemical reactions in systems where the temperature has been measured, or estimated by some other means, and
where the pressure is known.
3
For a closed system at constant T and p , the minimum energy
state is:
4 CH4 + 2 O2 = 4 CO +8 H2
Geq = min G (T , p , n) ,
2 CH4 + 5 O2 = 2 CO +8 OH
n
2 CH4 + 1 O2 = 2 CO +8 H
A∆n = 0
4 N2 + 4 O2 = 8 NO
Termodynamic methods TKP4175
959 / 1776
17 Simultaneous . . .
As previously mentioned, A1 must be invertible.
2
This means that the rows and columns in the matrix must be linearly independent.
5
02 September 2013
Lagrange Null space Combustion Flame temperature
ENGLISH text is missing.Inert compounds
News Stoichiometry min G
Olds
4 CH4 + 6 O2 = 4 CO +8 H2 O
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
8
The factor 1/8 has been used in order to give us integers in the
matrix.
News Stoichiometry min G
Olds
4
But A′ will not necessarily be of the same shape as the atom matrix, as for the latter dim A > rank A if some of the atoms are always
present in specific ratios.
Inert compound
We can therefore solve the set of equations with respect to 5 molar
flow rates, while the remaining 6 must be considered independent variables. First we partition the atom matrix and flow rate vector, and then
:
we left-multiply by A-1
1
~ ~ ∆n1
∆n 1
( A1 A2 )
∼ ( I A-1
= 0.
(17.5)
A2 )
1
~
∆n2
∆~
n2
4
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
965 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
⇔
∆n = Nξ .
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Lagrange Null space Combustion Flame temperature
Minimum Gibbs energy (2)
News Stoichiometry min G
Olds
17 Simultaneous . . .
§ 133
§ 132
Lagrange Null space Combustion Flame temperature
§ 134
News Stoichiometry min G
Olds
17 Simultaneous . . .
4
The only difference between these two formulations is in the mass
balance term, but the difference is big enough to mean that different
solution algorithms are required.
Lagrange Null space Combustion Flame temperature
About Lagrange’s method
Olds
17 Simultaneous . . .
1
In brief, the method involves finding the stationary points of the
Lagrange function
2
The Lagrange[a] method is a topic that belongs under multivariate
analysis, and so we will limit ourselves to a quick glance at the underlying principles.
[a] Joseph Louis Lagrange, alias Giuseppe Luigi Lagrangia, 1736–1813. French mathematician.
5
There was a breakthrough in this field in the 1960s, as a direct
result of the US space programme.
6
The historical development of the algorithms is discussed in subchapters 17.4 and 17.5, but before we embark on that discussion we
must first derive the equilibrium conditions in a form that is conducive to
solving the equations numerically.
L (λ, n) =
ˆ G (n) − λT A∆n ,
AT λ
Minimize G (n) in order to show that µ =
and
NT µ = 0 are two equivalent criteria for multicomponent
chemical equilibria based on Lagrange multipliers and
the reaction stoichiometry of the system respectively.
where λ is a vector of the Lagrange multipliers.
3
Based on classical optimisation theory it can be demonstrated that
all Gibbs energy minima must satisfy the following conditions:
∂L
∂n
∂L
∂λ
=
∂G
∂n
=
∂G
∂λ
−
−
∂(λT A∆n)
∂n
= 0,
∂(λT A∆n)
∂λ
= 0,
=
where ∂/∂n
=
( ∂/∂N1 ∂/∂N2 . . . ∂/∂Nn )T and 0
( 0 0 · · · 0 )T . Our next task is to rewrite the partial derivatives of L in a form that is relevant to thermodynamics.
Tore Haug-Warberg (Chem Eng, NTNU)
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Lagrange Null space Combustion Flame temperature
§ 133 Lagrange multipliers
967 / 1776
News Stoichiometry min G
Olds
17 Simultaneous . . .
We immediately recognise ∂G /∂n as being the chemical potential
vector µ, and if we give it a little bit of thought it is clear that ∂G /∂λ = 0,
since G (T , p , n) is not dependent on λ.
1
Now what we need to do is to differentiate λT A∆n with respect to
λ and n.
2
3
Let us start by writing the inner product in summation notation
PP
Aij λi ∆Nj
f = λT An =
i
j
before differentiating.
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Lagrange Null space Combustion Flame temperature
§ 133 Extent of reaction
ENGLISH text is missing.Multipliers
1
In thermodynamics, Lagrange’s method of undetermined multipliers is not as obscure as you might think, and in the example above λ
can be seen as a vector of the chemical potentials associated with the
atoms that comprise matrix A; see Theme 136 on page 909 for a comprehensive discussion of this.
973 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
970 / 1776
Alternatively, the criterion for equilibrium can be derived by looking
at the independent extents of reaction as free variables.
For a system with n compounds and m atom balances, there will
be r = n − m independent reactions.
2
3
For each independent reaction there is precisely one extent of reaction, as shown in Theme 131 on page 887.
968 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
§ 133 Extent of reaction (2)
Termodynamic methods TKP4175
News Stoichiometry min G
17 Simultaneous . . .
Lagrange Null space Combustion Flame temperature
Tore Haug-Warberg (Chem Eng, NTNU)
Olds
4
Finally the results can be brought together in the following matrix–
vector expressions



 


 ∂f   P
 ∂f   P A1j ∆Nj 
 ∂N1   Ai 1 λi 
 ∂λ1   j





  P
  P
i
 ∂f  
 ∂f  

A2j ∆Nj 


  Ai 2 λi 
 








∂f
∂f
 = AT λ , ∂λ
 = A∆
=  ∂N2  =  i .
=  ∂λ2  =  j
∂n


..
..  
 ..  

.



 .   P .
 .   P .






 
 
 ∂f   Ain λi 
 ∂f   Amj ∆Nj 




i
j
∂Nn
∂λm
News Stoichiometry min G
1
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 133 Lagrange multipliers (2)
Olds
17 Simultaneous . . .
Termodynamic methods TKP4175
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Olds
17 Simultaneous . . .
The conditions for a stationary point can be summarised as follows:
5
∂L
∂n
∂L
∂λ
=0
=0
⇒
⇒
AT λ = µ ,
(17.8)
A∆n = 0 ,
(17.9)
with the two equations representing the criterion for equilibrium and the
mixture’s atom balance respectively.
6
Together, they define the conditions for reaction equilibrium in a
closed system at a given T and p .
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
4
The extents of reaction are linearly independent, and the minimum
Gibbs energy can therefore be calculated by finding the stationary point
of the implicit function G (n(ξ)). The chain rule for partial differentiation
gives us


 P ∂G ∂Ni 


 i ∂Ni ∂ξ1 
 P

∂G ∂Ni 



 i ∂Ni ∂ξ2 
∂G 
= 
 = 0 .
∂ξ T ,p


.


..


 P ∂G ∂Ni 


i ∂Ni ∂ξr
Lagrange Null space Combustion Flame temperature
§ 133 Lagrange multipliers (3)
Olds
17 Simultaneous . . .
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Lagrange Null space Combustion Flame temperature
§ 133 Extent of reaction (3)
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Olds
17 Simultaneous . . .
This gives us the following criterion for equilibrium:
6
NT µ = 0 .
(17.10)
7
It is worth noting that the solution obtained using Lagrange multipliers 17.8 also satisfies the condition for equilibrium 17.10.
Since N is designed to ensure that AN = 0, this will be true for all
9
possible (equilibrium) values of λ.
8
We can show this by combining the two equations, which gives us
NT AT λ = 0.
Next we introduce the stoichiometric matrix NT =
ˆ {∂Ni /∂ξj } and the
chemical potential vector µ =
ˆ {(∂G /∂Ni )T ,p }.
5
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Lagrange Null space Combustion Flame temperature
Newton–Raphson–Lagrange iteration
News Stoichiometry min G
Olds
17 Simultaneous . . .
In order to obtain a numerical solution to the equilibrium Eq. 17.8,
µ and λ must be iterated simultaneously subject to the constraints given
by the atom balance 17.9.
1
4
2
There are a number of algorithms for solving this type of problem,
but it is natural to use the Newton–Raphson method of iteration if the
mixture has a moderate number of chemical components.
3
The weakness of the Newton–Raphson method is that it requires
you to invert a matrix with n2 elements, but for systems with 10–30
compounds this isn’t a problem in practice.
Linearisation of Eq. 17.8 gives us:
T
T
A λk + A (λk +1 − λk ) = µk + G(nk +1 − nk ) .
Lagrange Null space Combustion Flame temperature
Newton–Raphson–Lagrange iteration (2)
News Stoichiometry min G
Olds
17 Simultaneous . . .
10 Combining the linear version of the condition for equilibrium with
Eq. 17.9, which is naturally linear, gives us the following iterative model
for the equilibrium problem:

 


 Nk Gk AT   1 [nk +1 − nk ]   −1 [µ − AT λk ] 
 RT
 =  RT k
  Nk
 .
(17.11)

 


−λ ]  
A
0   −1 [λ
0
7
The symmetry of G results in n(n − 1)/2 Maxwell relations, but in
this case they are of no practical importance.
8
However, the fact that Hessian matrices are symmetrical is important in relation to convergence analysis and optimisation theory.
9
The discussion of the Newton–Raphson method of iteration in Appendix 27 is an example of this.
RT
k +1
The expression has been written in dimensionless form, in order
to emphasise the fact that the iteration sequence is independent of the
size of the system .
6
The matrix, and other (symmetrical) matrices containing second
derivatives of scalar functions of state, are called Hessian matrices.
In order to start the iteration we need an initial value n◦ > 0 consistent with the system’s atom balance. There are two ways to approach
the matter:
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Termodynamic methods TKP4175
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977 / 1776
Newton–Raphson–Lagrange iteration (3)
Olds
17 Simultaneous . . .
components of which there are zero moles are given an arbitrary small
value ǫ and the inaccuracy in the atom balance Ank − b is moved to the
right-hand side of the set of equations in the hope that the discrepancy
will become smaller in the next iteration.
use of linear programming to calculate a feasible n◦ > 0 that is
consistent with the atom balance.
1
2
k
The matrix G =
ˆ (∂µ/∂n)T ,p is often referred to as the Jacobian of
µ, but since µ = (∂G /∂n)T ,p we can alternatively express it as G =
(∂2 G /∂n∂n)T ,p .
5
Lagrange Null space Combustion Flame temperature
vmode (bug): hmode expected
15 The first option is perhaps more intuitive, but the discrepancy in the
atom balance can in some cases lead to the iterations diverging.
16 The second option requires greater mathematical expertise, but
has the benefit of having better convergence properties.
17 Incidentally, Appendix 29 gives a brief introduciton to linear programming.
11
Expanded footnote [a]:
[a] It is easy to see that Nk-1 [nk +1 − nk ] is a dimensionless measure of the change
in the composition of the system, and that this vector actually expresses the change
in the mole fractions in the mixture if the number of moles remains constant during
the reaction: Nk +1 = Nk . For the matrix Nk (RT )-1 Gk the situation is more complex,
and here it is important to understand how the Gibbs energy G (n) relates to Euler’s
homoegeneous function theorem. From Chapter 5 we know that G is a homogeneous
function of degree one, whereas µ is a function of degree zero. By differentiating
yet again we can obtain G, which is a homogeneous function of degree minus one.
Multiplying by Nk thus gives us a matrix that is independent of the size of the system.
The effect of this scaling on the thermodynamic properties in the set of equations is
invaluable to our basic understanding of thermodynamic equilibrium calculations.
12
13 In practice this will be the composition of the reactor feed stream,
or the mixture of reactants if the reaction is taking place in a closed
space.
14 As such, the initial estimate is not a big problem, but in certain
contexts we only know the system’s atom balance, or the number of
moles of some of the components may be zero.
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
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Lagrange Null space Combustion Flame temperature
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Olds
02 September 2013
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Olds
ENGLISH text is missing.RAND (ideal gas)
1
Working out what to do next with Eq. 17.12 requires a certain
amount of cunning and a bit of imagination.
2
What is obvious is that we cannot perform a normal Gaussian elimination on the coefficient matrix, as the upper left-hand block yk-D − eeT
is singular. This can easily be shown by multiplying by ykT .
3
However, it is still possible to multiply the first row of the set of
§ 134 RAND (method)
equations by ykD , and then substract the result from the last row.
1
The equilibrium is not affected by the state space model used, but
4
This results in the modified equation
if some of the mole fractions become very small (which they typically 
 


1
yD
AT   N1k nk +1  
µ⋆ + ln yk ] 
−ykD [ RT
do in combustion equilibria), the inverse matrix coefficient can become  I − yk eT
k
 = 

 
 ,
 

  −1

1
numerically unstable.
λk +1   Ayk + AykD [ RT
µ⋆ + ln yk ] 
Ayk eT −AykD AT   RT
2
This is because the diagonal elements in Gk are inversely propor- from where the set of equations can be split into two sub-problems. The
tional to yi when yi → 0.
first row of the above equation gives us the expression
3
We are therefore going to rearrange the set of equations in order
to make it more stable.
1
Nk
nk +1 = yk
1
Nk
1
1
eT nk +1 + ykD [ RT
AT λk +1 − ( RT
µ⋆ + ln yk )] . (17.13)
T
§ 136
§ 135
ˆ Nk +1
4
When an ideal gas is introduced into Eq. 17.11 the chemical po- 5 Our job is then to determine the inner product e nk +1 =
independently of nk +1 .
tential vector and Hessian matrix can be written
6
On the face of it this is impossible, but it turns out that multiplying
⋆
1
1
RT µk = RT µ + ln yk ,
the first row of the equation by eT allows us to split it, because the
Nk
-D
T
RT Gk = yk − ee ,
equation can now be rewritten as:



 
where yk =
ˆ Nk-1 nk and µ⋆ =
ˆ µ◦ + RT ln(p /p◦ ).
1
 0
µ⋆ + ln yk ] 
ykT AT   N1k eT nk +1  
ykT [ RT

 
 .
 = 


 
5
The reference potential µ⋆ is in this case independent of the itera Ay AyD AT   1 λ
1
µ⋆ + ln yk ] 
Ayk + AykD [ RT
k
RT k +1
k
tion sequence, as the equilibrium requires T and p to be constant.
(17.14)
6
Also note that y-D yk ≡ e and that eT yk =
ˆ 1, which means that
7
The elements in the vector on the right-hand side can alternatively
Gk yk = 0.
ıg
ıg
be written as gk /RT and bk + AykD µk /RT . The result will be the same.
7
This relationship is a consequence of the homogeneity of G , cf.
8
A trivial change of rows and columns moves the zero element from
Theme 34 on page 188, and is true of all Gibbs energy functions, althe top left down to the bottom right corner, which ensures that Gausthough it is easiest to verify for an ideal gas.
sian elimination is stable for the coefficient matrix without it first need8
In the above example, the homogeneity means that Gk [yk +1 −yk ] = ing to be pivoted.
Gk yk +1 .
9
The presence of small mole fractions does not result in any major
9
This enables us to solve the equations for the new state nk +1 with- problems during inversion, at least if a row transformation is performed
out having to use the update nk +1 − nk .
on A to give [I B], with the minor components being placed in matrix B.
§ 137
10 Furthermore, the update λk +1 − λk can be replaced by the solution 10 The equilibrium problem itself has incidentally been split into two
λk +1 .
sub-problems.
17 Simultaneous . . .
§ 135 RAND (implementation)
1
See Function 2.2 in Appendix 31. This is a one-to-one correspondence between this function and Eq. 17.14.
2
To enable us to handle components of which there are zero moles,
11 This is made possible by subtracting AT λk from both sides of the 11 First we need to solve Eq. 17.14 with respect to the Lagrange mulwe give them an arbitrary, small value before starting to iterate. Meanequation, and is a general property of all minimisation problems with tipliers λk +1 and the size of the system eT nk +1 .
while, the inaccuracy in the mass balance is moved to the right-hand
linear constraints.
12 Then these results are substituted into Eq. 17.13, which gives the
side of the set of equations.
12 All of this means that the equation can be rewritten as:
new composition vector nk +1 .


 

3
This simple strategy works well in many cases, but not for stoichio y-D − eeT AT   1 nk +1   −1 µ − ln yk 
13 What is interesting is that this calculation always converges[a] prometric mixtures in which the equilibrium is far to the right or left.
  Nk
 .
 =  RT ⋆
 k
(17.12)
  −1

 

§ 134
vided that we ensure that none of the numbers of moles in Eq. 17.13
λk +1  
Ayk
A
0   RT
4
In those cases the precision of the computer may affect the out1
Eq. 17.11 represents the classical non-stoichiometric formulation
are negative.
come of the calculation.
of an equilibrium problem in a homogeneous phase, i.e. in a gas, liquid Unlike in the case of Eq. 17.11, the solution to Eq. 17.12 obtained by 14 In practice we can do that by specifying that τ ≤ 1, and hence that
§ 135
5
Problems associated with limitations in computer precision can
or solid solution. Rewrite the equation in such a way that it explicitly iteration is only valid for an ideal gas.
the mole number vector nk + τ(nk +1 − nk ) > 0.
Computerise the function [n] = MinGibbs(n◦ , T , p , µ◦ , A) using largely be eliminated by the transformation A = ( I B ), with the miapplies to an ideal gas. Use the structure of the model to simplify the 13 In spite of its limited validity, this equation is a good starting point
982 / 1776 1
for learning more about the homogeneity of the Gibbs energy function.
Eq. 17.14 as your starting point.
nor components being placed in matrix B.
set of equations as much as possible.
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[a] Minimising the Gibbs energy of an ideal gas is an example of a convex problem.
§ 136 Chemical potentials of atoms
17 Simultaneous . . .
From the equation µ = AT λ it follows that µ is located in the column
space of AT .
1
3
In order to avoid using random basic variables, it is also possible to
use the least square solution
-1
λ = AAT Aµ .
2
This is an overspecified problem because dim(µ) > dim(λ), but
since the equilibrium state does in fact exist, any invertible sub-system
can be used to calculate λ from µ.
A dimensional analysis of µ = AT λ in Eq. 17.8 reveals
that λ can be interpreted as a chemical potential vector
for the atoms in the system. Assume that µ is known
from a previous equilibrium calculation. How can you
then determine λ, and what does this mean in practice?
Expanded footnote [a]:
[a] Or, in a more general context, the system’s independent extensive variables.
News Stoichiometry min G
Termodynamic methods TKP4175
02 September 2013
Lagrange Null space Combustion Flame temperature
Newton–Raphson null space iteration
17 Simultaneous . . .
6
The equilibrium potential is by definition π = BT λ and the minor
mole fractions become
πi −π◦ p
yi = p◦ exp RT i ,
assuming that the mixture is an ideal gas. It is now easy to decide
whether any of the new compounds should be included in the minimisation.
4
After minimising the Gibbs energy you can check whether any new
(and more important) minor components have been formed that were
not originally expected.
Let the stoichiometric matrix for the new compounds be B.
5
Tore Haug-Warberg (Chem Eng, NTNU)
§ 136 Chemical potentials of atoms (2)
985 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
17 Simultaneous . . .
Any numerical solution to the equilibrium Eq. 17.10 on page 905
must be found in the null space of the stoichiometric matrix A. Linearising the equilibrium equation gives us
1
Termodynamic methods TKP4175
02 September 2013
Lagrange Null space Combustion Flame temperature
Newton–Raphson null space iteration (2)
7
2
The Newton–Raphson–Lagrange method used in Eq. 17.11 also
has this property, but only as a result of a loose assumption that theoretically does not always hold.
3
This time we are going to solve NT µ = 0 in such a way that the
mass balance is correct after each iteration.
5
The composition is updated by nk +1 = nk + τN ∆ξ, where τ ≤ 1 is
used to ensure that nk +1 > 0.
6
Alternatively we can write the equation in dimensionless form, to
emphasise the fact that the iteration sequence is independent of the
size of the system.
986 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Stoichiometry min G
Olds
17 Simultaneous . . .
Meanwhile, the solution ∆ξ is substituted into the update of nk +1 :
-1
T Nk
1
1
NT RT
µk .
(17.15)
Nk nk +1 = yk − N N RT Gk N
Combustion
02 September 2013
Lagrange Null space Combustion Flame temperature
2
This factor must be weighed up against the fact that Eq. 17.11 has
one matrix product and one inverse, whereas Eq. 17.15 has four matrix
products and one inverse.
3
The stoichiometric formulation therefore involves 2–3 times more
operations than the non-stoichiometric formulation if A and N have
roughly the same rank.
4
However, the iteration sequence for Eq. 17.15 is the same as
for 17.11, provided that the latter starts with the constraints provided
by the atom balance being met.
5
This may appear to be a surprising result, but the two algorithms
§ 137 Null space (implementation)
1
Olds
17 Simultaneous . . .
ENGLISH text is missing.Null space vs Lagrange
1
In terms of calculation efficiency, the non-stoichiometric formulation
generally gives you fewer equations to solve if n >> rang A, whereas the
stoichiometric formulation is more efficient if n is a big number and the
number of reaction equations is strongly limited by kinetic conditions,
i.e. when rang N << n.
987 / 1776
See Function 2.3 in Appendix 31.
2
There is a one-to-one correspondence between this function and
Eq. 17.15, but for the sake of efficiency the basis for N, i.e. [I B]T , is
only updated if there are significant changes in the composition.
3
During each iteration the mole fractions are sorted by increasing
value, and are compared with the previous sorted vector.
4
If one or more elements in the new vector differ from those in the
previous vector by a factor of at least 106 , the new list of components is
used to calculate a new basis for N.
5
Components of which there are zero moles are handled with the
aid of linear programming, but note that the initialisation requires more
code than the equilibrium algorithm itself.
have the same starting point and use the same linearisation.
§ 137
6
It is often the case that trivial matters such as initialisation are more
6
There is therefore no reason why their iteration sequences should 1
Computerise the function [n] = MinGibbsNull(n◦ , T , p , µ◦ , A) from vulnerable to numerical problems and require more time and effort than
be different.
Eq. 17.15.
the main calculation.
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1
Below we will assume that combustion takes place at a temperature of 1500 K. Once again we have two options: Either we can use a
set standard temperature of 298.15 K
2
For high-temperature combustion it is realistic to assume ideal gas
behaviour, such that µi = µi◦ + RT ln(pyi /p◦ ).
3
We will use the JANAF tables[a] to determine the standard potential µi◦ (1500 K) for all of the compounds mentioned in Theme 130 on
page 883.
[a] JANAF thermochemical
Suppl., 14(1), 1985.
tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,
8
Eq. 17.15 shows the classic iteration sequence for the stoichiometric formulation of an equilibrium problem, but unlike in the case of the
equivalent non-stoichiometric formulation in Eq. 17.11, assuming ideal
gas properties will not significantly simplify the result.
9
In practice Eq. 17.15 will run into problems if any of the mole fractions are very small.
10 The reason for this is the same as for the RAND algorithm in
Theme 134 on page 909.
11 In order to preempt potential problems, it is helpful to perform a
column transformation on N, to give us [I B]T , where the minor components are placed in I.
NT µk + NT Gk (nk +1 − nk ) = 0 .
i)
Here nk +1 − nk has to occupy the null space of A, which means that
nk +1 − nk = N ∆ξ.
4
1500
R K
µ◦i (1500 K) = ∆f hi◦ (298.15 K) +
cp◦,i
298.15 K
|
{z
hi◦ (1500 K)

1500
R K

− 1500si◦ (298.15 K) +
|
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
}
hi◦ (1500 K)−hi◦ (298.15 K)
|
This leads us to the equation
-1
∆ξ = − NT Gk N NT µk .
( T ) dT
{z
298.15 K
{z
si◦ (1500 K)
Termodynamic methods TKP4175
cp◦,i (T )
T
}


dT  ,
(17.16)
}
02 September 2013
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News Stoichiometry min G
Lagrange Null space Combustion Flame temperature
Combustion (2)
News Stoichiometry min G
Olds
17 Simultaneous . . .
ii)
=
∆f µ◦i (1500 K) .
(17.17)
CH4
N2
O2
Ar
CO
CO2
H2 O
H2
OH
H
NO
4
The JANAF tables contain the data that is summarised in Table 17.2 on the following page.
5
The left-hand column is equivalent to option ii) and the right-hand
column to option i).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Stoichiometry min G
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 138 Reference state
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∆f hi◦ (T◦ )
74918
0
0
0
−243740
−396288
−164376
0
16163
136522
71425
−74873
0
0
0
−110527
−393522
−241826
0
38987
217999
90291
News Stoichiometry min G
Option ii) uses a changing reference state in which the stable state
of each element at temperature T and 1 bar has zero enthalpy, whereas
option i) uses a fixed reference temperature of 298.15 K.
1
∆f µ◦i (T )
Tore Haug-Warberg (Chem Eng, NTNU)
Olds
17 Simultaneous . . .
RT
T◦
cp◦,i
dT
78153
38405
40599
24982
38850
61705
48151
36290
36839
24982
38729
si◦ (T )
279.763
241.880
258.068
188.427
248.426
292.199
250.620
178.846
232.602
148.299
262.703
Termodynamic methods TKP4175
7
§ 138
−416364.5
−324415.0
−346503.0
−257658.5
−444316.0
−770115.5
−569605.0
−231979.0
−273077.0
20532.5
−265034.5
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Lagrange Null space Combustion Flame temperature
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Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
17 Simultaneous . . .
§ 139
8
Since NT µ = 0 at equilibrium, and as we know that N is located in
the null space of A, we can conclude that
NT µ⋆ = NT µ + NT AT ∆λ = 0 .
9
In other words, a change in the zero point energy of the atoms (or
elements) does not affect the system’s criterion for equilibrium.
[a] Germain Henri Hess, 1802–1850. Russian physician and chemist.
News Stoichiometry min G
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17 Simultaneous . . .
Termodynamic methods TKP4175
0
The results for selected compounds are shown on the left-hand
side of Figure 17.3 on the following page.
4
Note the transition from reducing to oxidising conditions at the stoichiometric point.
The similarity with an acid–base titration curve is striking.
Lagrange Null space Combustion Flame temperature
§ 140
10
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
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Olds
17 Simultaneous . . .
Termodynamic methods TKP4175
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02 September 2013
Lagrange Null space Combustion Flame temperature
Stationary–adiabatic
§ 140 Le Chateliér[1]’s principle
1
The pressure doesn’t affect the NO concentration much if there is
excess air in the system.
2
In that case the main reaction is N2 + O2 = 2 NO, which maintains
the number of moles of the reactants.
3
If there is a shortage of air, the dominant reaction is N2 +2 H2 O =
2 NO +2 H2 , and the quantity of NO declines with increasing pressure,
§ 140
1
Repeat the calculation that you did in Theme 139, but this time as shown on the right-hand side of Figure 17.3 on the previous page.
1003 / 1776
use p = 20 bar. Compare the NO concentration with your previous
calculation and comment on the results.
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Olds
17 Simultaneous . . .
[a] Henri Louis Le Chatelier, 1850–1936. French chemist.
1
An energy balance for stationary, adiabatic conditions shows that
the energy entering the system per time unit, which is a combination of
matter and pV work, must also leave the burner at the same rate:
−5
−6
~ out ,
~ ın = H
H
~ out .
~ ın = M
M
10
−10
0.9
1
1.1
Air–methane stoichiometry
6
That is not really surprising, when you think about it, as the
methane gas is being titrated with oxygen — in a solution of nitrogen.
998 / 1776
NO (1 and 20 bar)
10
10
02 September 2013
Lagrange Null space Combustion Flame temperature
CO, O2 , H, OH
We will use data from Table 17.2 on page 916.
3
5
10
Mole fraction
2
Tore Haug-Warberg (Chem Eng, NTNU)
Olds
The Gibbs energy has been minimised using the Function 2.2; see
Program 1.27 in Appendix 31 for details.
1
02 September 2013
(17.18)
Lagrange Null space Combustion Flame temperature
§ 139 Combustion
§ 138
Termodynamic methods TKP4175
Assume that the pressure in the burner is 1 bar and that
the methane–air ratio varies ±0.1 about the stoichiometric
mixture ratio. Calculate the composition of the smoke
as a function of the methane–air ratio, and represent
the result graphically in a suitable diagram. Explain
what happens to the concentrations of the various
components as you approach the stoichiometric point.
(17.19)
4
In other words, the standard state affects the values of ∆f µ◦i ,
whereas the values of ∆rx Gr◦ for the equilibrium reactions remain unchanged.
5
This is related to Hess[a]’s law, which states that the total heat of
formation of a system is equal to the sum of the heats of formation of
all of the intermediate reactions.
Termodynamic methods TKP4175
17 Simultaneous . . .
The two µ◦i functions 17.16 and 17.17 that are shown
in Table 17.2 are very different, but they nevertheless
result in the same equilibrium distribution. Explain
why this is. Use practical calculations to show that
the equilibrium distribution is indeed the same.
6
The most direct way to interpret this is to add to the enthalpy of all
of the compounds through a virtual change ∆λ in the zero point energy
of the individual atoms:
Tore Haug-Warberg (Chem Eng, NTNU)
§ 139
Olds
Then let us substitute µ⋆ into the equilibrium Eq. 17.10.
3
µ⋆ = µ + AT ∆λ .
§ 137
Lagrange Null space Combustion Flame temperature
µi◦ (T ) =
hi◦ (T ) − Tsi◦ (T )
§ 138 Reference state (2)
2
The difference between the two options lies in the nature of the
formation reactions.
However, the energies of the chemical elements involved in each of
the formation reactions cancel each other out when the standard Gibbs
energy is calculated for the equilibrium reactions, and therefore have no
impact on the outcome of the calculation.
News Stoichiometry min G
Olds
Figure 17.2: Thermodynamic data from JANAF at T = 1500 K.
or we can use a “variable” reference temperature T :
µi◦ (1500 K)
Lagrange Null space Combustion Flame temperature
10
−8
0.9
1
1.1
Air–methane stoichiometry
~ is the enthalpy rate, i.e. the amount of enthalpy carried by the feed
H
and product streams per time unit.
5
The only thing that we do know is that the enthalpy rates into and
out of the burner are the same, but we know nothing about the absolute
quantity of energy (enthalpy) that flows with the mass over the same
period of time.
~ being [J s-1 ], we cannot use the term power
2
In spite of the unit of H
to describe it.
3
Power is energy per time unit supplied in the shape of heat or work,
~
in other words energy transfers that do not involve mass, whereas H
involves the physical movement of mass.
4
As it happens, the zero point energy is unknown, and hence the
~ is also unknown.
value of H
Figure 17.3: Concentrations of CO, O2 , H, OH (left sub-figure) and NO (right
sub-figure) as functions of the excess combustion air. The NO content has
been calculated at both 1 and 20 bar to show the pressure dependence of the
equilibrium.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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News Stoichiometry min G
§ 141
§ 140
Lagrange Null space Combustion Flame temperature
§ 142
News Stoichiometry min G
Olds
17 Simultaneous . . .
Lagrange Null space Combustion Flame temperature
§ 141 Adiabatic temperature I
News Stoichiometry min G
Olds
17 Simultaneous . . .
By selecting the time frame in such a way that the mass entering
the burner exactly matches the mass of the components on the lefthand side of the total reaction equation
1
2
Enthalpy and mass are conserved quantities, but how can we combine the two balance equations?
3
A molar analysis won’t work because the number of moles of components leaving the burner are greater than the number entering it.
4
One approach is to use specific values, but that is not really based
on chemistry.
5
The solution is to take the middle road.
a CH4 +b O2 +c N2 + · · · ⇒ d CO2 +e CO +f H2 O + · · ·
Write the energy balance for an adiabatic burner in
such a way that the flame temperature is an explicit
variable. Assume stationary conditions.
Nın,i h̄i (Tın ) =
CH4
NO
P
Nout,i h̄i (Tout ) ,
§ 141 Adiabatic temperature I (2)
Olds
17 Simultaneous . . .
6
Note that for an ideal gas, the partial molar enthalpy is independent
of the composition:
ıg
h̄i (Tout ) = ∆f hi◦ (298.15 K) +
TRout
cp◦,i (T ) dT .
(17.21)
298.15 K
the energy balance can be expressed on an ordinary molar basis as
Ar
P
Lagrange Null space Combustion Flame temperature
(17.20)
CH4
where Ni is the number of moles of each compound in accordance with
the total reaction.
7
The combustion is adiabatic, and the outlet temperature Tout is by
definition equal to the adiabatic flame temperature.
8
This quantity can thus be calculated from the energy balance, provided that the inlet temperature and feed and product streams are
known.
9
If the product stream is unknown, the energy balance must be
solved along with the equilibrium problem in an adiabatic equilibrium
calculation.
10 We will come back to that later, but for the moment it will suffice
to say that equations 17.20 and 17.21 together represent one equation
with one unknown, assuming that Tın and all Nın,i and Nout,i are known
quantities.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News Stoichiometry min G
§ 142
§ 141
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 143
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News Stoichiometry min G
Olds
17 Simultaneous . . .
Termodynamic methods TKP4175
News Stoichiometry min G
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 142 Adiabatic temperature II
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
17 Simultaneous . . .
02 September 2013
Lagrange Null space Combustion Flame temperature
Simplified energy balance (2)
1007 / 1776
Olds
17 Simultaneous . . .
3
Meanwhile, let us simplify the composition of the air to 20% O2 and
80% N2 , in order to make the stoichiometry as simple as possible:
CH4 +2 O2 +8 N2 = CO2 +2 H2 O +8 N2
|
{z
} |
{z
}
reactants
CO2 , H2 O, N2
products
2
The adiabatic combustion temperature Tout is indpendent of the
flow rate and we can, for instance, use one mole of methane as our
basis, i.e. Nın,CH4 = 1 mol.
1008 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Stoichiometry min G
Olds
17 Simultaneous . . .
Termodynamic methods TKP4175
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 142 Adiabatic temperature II (2)
1009 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
17 Simultaneous . . .
Since the heat capacity in this case is a monotonic function of tem3
perature, the upper temperature limit can be estimated by using cbp◦,i ≈
cp◦,i (298.15 K), giving an estimate of 2677 K.
2
In order to find a lower and upper limit for the flame temperature,
we need an upper and lower limit respectively for the individual cbp◦,i in
the product stream.
When substituted into the temperature equations, the following numerical results are obtained:
Tout
−393522 − 2 · 241826 + 74873
< 298.15 −
= 2677 ,
[K]
37.129 + 2 · 33.590 + 8 · 29.124
Tout
−393522 − 2 · 241826 + 74873
> 298.15 −
= 2021 .
[K]
61.802 + 2 · 54.723 + 8 · 36.801
◦
◦
◦
◦
,
− ∆f hCH
+ 2∆f hH
− 2∆f hO
∆rx H◦ (Tın ) = ∆f hCO
2
2O
4
2
|
{z
}
6
calculated at Tın
c◦ = cb◦
+ 2cbp◦,H O + 8cbp◦,N .
C
P
p ,CO2
2
2
|
{z
}
calculated at Tın <T <Tout
1011 / 1776
Termodynamic methods TKP4175
News Stoichiometry min G
Olds
02 September 2013
Lagrange Null space Combustion Flame temperature
§ 142 Adiabatic temperature II (3)
11
7
This simplified procedure enables us to estimate a conservative
temperature range, but no more than that.
8
An adiabatic temperature calculation is and will always be a nonlinear problem, and using the mean heat capacity doesn’t make the
problem any simplier.
9
It is possible to narrow the temperature range by making successive improvements to the mean heat capacity, but in practice it
is easier to calculate the outlet enthalpy per 100th of a degree Celsius and compare the result with the reaction enthalpy ∆rx H◦ (Tın ) =
−393522 − 2 · 241826 + 74873 = −802301 J.
10 When the sum of the integral of the heat capacity and the reaction
enthalpy is zero, we have reached our goal.
1010 / 1776
Olds
17 Simultaneous . . .
We can use the JANAF tables to produce the following table:
T [K]
2000
4
The lower temperature limit follows from the conservative assumption that cbp◦,i ≈ cp◦,i (2700 K) > cp◦,i (2677 K) > cp◦,i (298.15 K), which gives
an estimate of 2021 K.
,
02 September 2013
Termodynamic methods TKP4175
ˆ (Tout − Tın )-1 T cp◦,i dT is the mean heat capacity for the
where cbp◦,i =
ın
temperature range T ∈ [Tın , Tout ].
P
Termodynamic methods TKP4175
1006 / 1776
R Tout
where the following quantities have been used:
Tore Haug-Warberg (Chem Eng, NTNU)
Lagrange Null space Combustion Flame temperature
Simplified energy balance
CH4 , N2 , O2
Introducing the mean heat capacity of the reaction products enables us to solve the energy balance explicitly with respect to the outlet
temperature
∆rx H◦ (Tın )
c◦
C
02 September 2013
Given the abovementioned assumptions, the energy balance for
~ ın = H
~ out , and from equations 17.20
the burner can be written as H
and 17.21 on page 926 we get the expression
i
h
P
P
Nın,i ∆f hi◦ (Tın ) =
Nout,i ∆f hi◦ (Tın ) + cbp◦,i (Tout − Tın ) ,
1
Tout = Tın −
Termodynamic methods TKP4175
1
In Themes 139 and 140, the reaction temperature has
been set to 1500 K, but by solving the mass and energy
balances simultaneously, the actual temperature can
be calculated. First we will simply estimate an upper
and lower bound for the (adiabatic) flame temperature.
Assume a stoichiometric mixture of methane and air,
and set the inlet temperature to 25 ◦ C. Simplify the
composition of the air to 80% N2 and 20% O2 , and
assume complete combustion to CO2 and H2 O. Hint:
The heat capacity of the gas rises monotonically with
the temperature.
Tore Haug-Warberg (Chem Eng, NTNU)
Tore Haug-Warberg (Chem Eng, NTNU)
RT
298.15
CP ,prod dτ
T [K]
RT
298.15
CP ,prod dτ
686115
2400
868381
2100
731226
2500
914501
2200
776756
2600
960817
2300
822454
2700
999304
5
These calculations require the enthalpy values shown in Table 17.2
and heat capacities taken from JANAF[b].
[b] JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data,
Suppl., 14(1), 1985.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1012 / 1776
12 Linear interpolation of
2256 K.
13 Note that this temperature is safely within the estimate range Tout ∈
[2021, 2677] K.
14 For the sake of comparison, the actual flame temperature[b] has
been measured to be approximately 2150 K.
[b] Lange’s Handbook of Chemistry. McGraw-Hill, 5th edition, 1961. p. 825.
Tore Haug-Warberg (Chem Eng, NTNU)
RT
298.15
CP ,prod dτ = 802301 J gives us Tout =
Termodynamic methods TKP4175
02 September 2013
1013 / 1776
Lagrange Null space Combustion Flame temperature
Flame temperature
News Stoichiometry min G
Olds
17 Simultaneous . . .
In order to calculate the exact flame temperature, we need to simultaneously solve the energy balance 17.20 and equilibrium criteria 17.8–
17.9, and if necessary equilibrium Eq. 17.10.
In the example above, the difference between the rigorous solution
(2246 K) and the interpolated value at constant product composition
(2256 K) is insignificant, and as such it is a poor example, as the dissociation equilibria are negligible at temperatures below approximately
2500 K.
6
News Stoichiometry min G
Termodynamic methods TKP4175
02 September 2013
Lagrange Null space Combustion Flame temperature
Looking back (2)
11
2
−8
1000
2000
2500
Olds
Nk
RT
-1
Gk N NT
1
RT
T
-1
17 Simultaneous . . .
T
10 If it is the former, then all of the numbers of moles will be constant,
[b] A pragmatic solution to the problem is clearly to calculate the enthalpy for a number
but if it is the latter, the composition may vary as a result of chemical of temperatures (e.g. for every one hundred degrees) until the outlet enthalpy exceeds
reactions in the gas (the number of moles of each atom, however, will the inlet enthalpy, and then repeat the process with increasingly small temperature
increments.
remain constant).
[b] In the range 3000–6000 K there is a transition from low to high enthalpy, and the
11 The effective heat capacity of the system is therefore affected by
internal degrees of freedom of the system result in the characteristic hump shape of
the reaction as well as by CP .
(dH / dT )p ,b as shown in 17.4. The proposed Newton–Raphson algorithm requires
1015 / 1776 (d2 H / dT 2 )p,b to have a constant sign in the range where we are iterating. This condition ceases to be met at approximately 4000 K, resulting in divergence.
2
2
Atom (mass) balance: A∆n = 0
3
Null space solution: AN = 0
Independent reactions: AT N has full rank
1.5
4
0.5
1000
5000
9000
T k [K]
T [K]
Figure 17.4: The left sub-figure shows the iteration sequence of an adiabatic calculation where an upper limit of 400 K has been applied to the step
length. The enthalpy values along the ordinate axis represent the combustion
of 10 mol methane in air. The right sub-figure shows (dH / dT )p ,b along the
upper curve and CP =
ˆ (∂H /∂T )p ,n along the lower curve (normalized values).
Tore Haug-Warberg (Chem Eng, NTNU)
T
Atom (or formula) matrix: A = [ atoms x compounds ]
Extent of reaction: ∆n = Nξ
5
1500
(dH )p ,b
1
1
Termodynamic methods TKP4175
02 September 2013
1016 / 1776
6
Reaction equilibrium: min G (n)T ,p ⇔ min G (n(ξ))T ,p
7
Linearization: µk +1 ≈ µk + Gk (nk +1 − nk )
An=b
ξ
Lagrange functional: L(n, λ) =
ˆ G (n) − λT (An − b)

 N

 1

T
k

 1 µ 
Gk A   Nk ∆n 
N–R iteration:  RT
 = −  RT k 
 


0
A
0
λk +1
8
9
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
Termodynamic methods TKP4175
02 September 2013
1018 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
New concepts
µk
Adiabatic (flame) temperature: h (T , p , n)out = h (T , p , n)ın
12
Looking back
4
This is because in a closed system (dn)b = N dξ, as shown in
Eq. 17.7. Substituting this into the above expression gives us the relation
ENGLISH text is missing.Iteration sequence
N
-1
1
The graph on the left-hand side of Figure 17.4 on the previous page
(dξ) = T GN NT s̄ dT ,
p ,b
shows the iteration sequence for an adiabatic temperature calculation
which tells us how the extent of reaction changes with temperature, in which Gibbs energy is minimised at each step.
assuming complete chemical equilibrium.
2
However, it is not entirely typical in that there is an upper limit on
5
In order to calculate the change in enthalpy we need the differential the step length of 400 K, in order to ensure a suitable distribution of the
points[b].
dH = CP dT + ∂∂Hp
dp + h̄T dn ,
T ,n
3
In reality the calculation converges much quicker, because the
and provided that (dn)b = N dξ we can then say that (dH )p ,b = CP dT + heat capacity of the combustion gas is virtually constant in the releh̄T N dξ or
vant range.
= CP + h̄ N(N GN) N s̄ .
4
According to the graph on the right-hand side, this is true up to
dT
temperatures of 2–3000 K, at which point the dissociation equilibria be6
Linearisation gives us the following Newton–Raphson iteration for
come important[b] and the reaction starts to make the dominant contrithe flame temperature:
bution to (dH / dT )p ,b .
H k − Hın
T k +1 = T k − h
.
5
Note that constant heat capacities have been used in these calcuk
CP + h̄T N(NT GN)-1 NT s̄i
lations, and that the non-linearity is solely due to the varying composition of the combustion gas.
ˆ CP , but
7
You may have noticed that (dH / dT )p ,b , (∂H /∂T )p ,n =
6
The reaction equilibria are, incidentally, the same as in Theme 139,
what is the physical explanation for this difference?
and once again the calculations have been performed using Pro8
The answer lies in the internal degrees of freedom of the system.
gram 31:1.27.
9
It therefore makes a big difference whether it is n or b that is con1017 / 1776
stant.
2.5
1
Olds
2
Any change in the equilibrium state must satisfy NT dµ = 0, and if
we substitute in the total differential of the chemical potential we get
h−¯
NT dµ = NT s dT + v̄ dp + G dni = 0 ,
where G =
ˆ {∂2 G /∂Ni ∂Nj }.
3
The equation is generally valid, both for open and closed systems,
but in the latter the component balance puts significant constraints on
the behaviour of the system.
17 Simultaneous . . .
Reaction manifold: NT µ = 0
N–R iteration: N1k ∆n = −N NT
10
1014 / 1776
3
−4
−6
3
This means that T and n must be updated simultaneously, or that
T must be calculated in an external loop based on the properties of a
system that is always at chemical equilibrium (convergeng in an internal
loop).
4
The first of these methods is preferable for simulations where robustness and flexibility are very important, but there are pedagogical
advantages to the second method: It highlights the difference between
the heat capacity CP at constant composition and the change in enthalpy (dH / dT )p ,b along the equilibrium manifold (see below).
5
In certain circumstances there can be a significant difference between their values, and in such cases the temperature calculation will
give highly divergent results whether or not the product composition is
assumed to be constant.
Tore Haug-Warberg (Chem Eng, NTNU)
5−10
4
Lagrange Null space Combustion Flame temperature
§ 142 Adiabatic temperature III
1
An iterative calculation of the flame temperature can be derived
from equilibrium Eq. 17.10, in other words NT µ = 0.
3
−2
H k [MJ]
It should be remembered that heat capacities are temperature dependent, and that the dissociation equilibria become increasingly important at higher temperatures.
News Stoichiometry min G
Olds
Heat capacity
0
Hın
1
2
Lagrange Null space Combustion Flame temperature
Iteration sequence
(dH / dT )p ,b
News Stoichiometry min G
Part 18
Heat engines
18 Heat . . .
1
Thermodynamic cycles
2
Work and heat diagrams
3
The basic 6: Otto, Brayton, Carnot, Stirling, Ericsson & Rankine
4
Thermodynamic efficiency
5
Thermal power plants
6
The classic steam engine
7
Modern steam turbines
see also Part-Contents
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1019 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
Heat power
ENGLISH text is missing.Energy consumption
dustrial growth must sooner or later come to a halt, but the technologi- Earth, this would mean 10 TW of demand, which is in turn equivalent to 4 The first step of this tried and tested technology is combustion in
cal insight that we have developed since the emergence of modern sci- 4.5?? times the total amount of hydroelectricity generated in the world, a boiler or burner, while the second stage involves transforming the
ence in the 17th century will always inform our understanding of the lim- or 3.3 times the energy dissipated by tides, or 1.1 times the energy released heat in a heat engine (a mechanical device that converts heat
its on human activity and development.
produced by photosynthesis on Earth.
into work).
1022 / 1776
1023 / 1776
1024 / 1776
02 September 2013
1020 / 1776
The discovery of the “mechanical equivalent of heat”, and hence
of the principle that work could be converted into heat, and vice versa,
was an important prerequisite for the industrial revolution of the 19th
century.
1
Heat power (2)
18 Heat . . .
15 The six fundamental cycles have been given names that link them
to more or less famous historical figures:
8
5
while the 2nd law tells us that the entropy must also be conserved
in a reversible change of state.
H
H
[a] Since (dU )n = T dS − p d V , then p dV = T dS . The projected area of the
cycle is thus the same in T , s and p , v coordinates, provided that the physical units
form part of a coherent system of units, as is the case in Figure 18.2 on page 948 and
Figure 18.3 on page 950.
Put another way, δQ rev = T dS and δW rev = p dV .
7
In a closed system, the entropy is a function of only two state variables i.e. s (T , p ), s (T , v ) or s (p , v ).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1025 / 1776
1021 / 1776
11 We are not going to say much about the practical implications of
these coordinate systems but in principle the four branches of the cycle
can be separated either in time or in space.
18 Heat . . .
External combustion:
Internal combustion:
Limiting case:
Vapour–liquid:
10 An ideal heat engine is defined by the fact that it has a rectangular
cycle in one of these systems of coordinates.
Under isentropic conditions ds = 0, which can be used to calculate
the temperature functions T (s , v ) and T (s , p ), and the pressure functions p (T , s ) and p (v , s ).
6
Expanded footnote [a]:
02 September 2013
Heat power (3)
9
Together, there are four primary variables s , v , T and p giving rise
to 4 · 3/2 = 6 different coordinate systems, as shown in Appendix 25.
12 Which means the cycle will show either a time-dependent state describing a closed loop over the period τ or it will show four time independent states existing simultaneously at different locations in the machinery.
2
This discovery also gave a big impetus to the subject of thermodynamics, and particularly to our understanding of thermodynamic cycles.
3
Here we will look at a small selection of closed cycles, where an
ideal gas is used as the working medium. The 1st and 2nd laws of
thermodynamics are important underlying principles, as is the concept
of reversibility.
Termodynamic methods TKP4175
News Thermodynamic cycles Power plants The steam engine Olds
Based on Eq. 3.8 in Chapter 3:3.5, the 1st law can be written
H
H
(dU )n = (δQ − δW ) = 0 ,
4
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
ENGLISH text is missing.Fossil fuels
1
Our standard of living is a direct function of our energy consump- 1 Since the farming and slaving societies of the past, we have retion.
alised that we can (more than) replace raw muscle power by converting
2
In practice, consumption of 100 W per capita is the lower limit for chemical and thermal energy into mechanical energy.
human survival. This is equivalent to the energy we require in the form 2 Ultimately, chemically bound energy is most important to us, and
above all we exploit it through the combustion of fossil fuels, i.e. oil, coal
of food.
1
Our modern industrial society requires huge amounts of mechani- 3 Consumption of 104 W per capita is in many ways an upper limit on and gas.
cal and electrical energy.
what is possible, albeit unsustainable at a global level. This is equivalent 3 The development of an alternative technology based on directly
converting chemical energy in fuel cells is still at the experimental stage,
2
Ready access to raw materials and cheap energy has fuelled eco- to Euro-American lifestyles today.
nomic growth and ever-improving standards of living.
4
Idealistically, if we want a sustainable society, 1000 W per capita and in practice the transformation of the chemical energy in the fuel still
3
Given the limits on available natural resources, it is clear that in- would be a reasonable compromise. Assuming 10 billion people on involves two steps.
ENGLISH text is missing.Energy
Termodynamic methods TKP4175
Expanded footnote :
Expanded footnote :
Expanded footnote :
Expanded footnote :
Expanded footnote :
Expanded footnote :
[a] Robert Stirling, 1790–1878. Scottish reverend.
[b] John Ericsson, 1803–1889. Swedish engineer.
[c] Nikolaus August Otto, 1832–1891. German engineer.
[d] George Brayton, 1830–1892. American engineer.
[e] Nicolas Leonard Sadi Carnot, 1796–1832. French engineer.
[f] William John Macquorn Rankine, 1820–1872. Scottish physicist.
Stirling[1] (T , V ) and Ericsson[2] (T , p )
Otto[3] (S , V ) and Brayton[4] (S , p )
Carnot [5] (T , S )
Rankine[6] (p , V )
13 The two alternatives lead to a dynamic or a stationary process description, respectively, but such details will not be of prime concern for
our understanding of the matter.
14 It is clearly possible to design countless alternative, non-trivial cycles, but that lies outside the scope of this chapter.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1026 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1027 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
News Thermodynamic cycles Power plants The steam engine Olds
Efficiency
18 Heat . . .
The thermodynamic efficiency of these cycles is defined as being
η=
ˆ W /Qh .
1
4
2
For a real, physical machine, W and Qh are measured under controlled conditions in a laboratory.
3
By looking at an ideal cycle, we avoid having to take measurements, but it is important to realise that the additional definitions set
out below only apply to reversible changes of state and frictionless machines:
H
W=
ˆ p dV ,
(18.1)
R
R
H
ˆ T d S + T dS = T dS .
Qh − Qc =
(18.2)
dS >0
dS <0
Note that Qc is positive if heat is removed from the cycle.
Maximum efficiency
1
Qh .
News Thermodynamic cycles Power plants The steam engine Olds
ENGLISH text is missing.Inexact differentials
1
Also note that T (dS )n and p (dV )n in isolation are inexact differentials.
2
This is in spite of the fact that both S and V are state functions, so
dS and dV can actually be written as exact (total) differentials.
3
18 Heat . . .
The explanation for this lies in the system’s degrees of freedom.
4
At constant composition, the cycles have two degrees of freedom — let us call them x and y — which means that each of the integrands T (x , y ) and p (x , y ) has one degree of freedom that is not being used if only the differential change in (dS )n or (dV )n is known.
5
Generally we can say that there is no state function f (x , y ) with a
total differential of df = g (x , y ) dx or df = h (x , y ) dy .
6
Since the system has two degrees of freedom, we must instead
expect the total differential to be of the form df = g (x , y ) dx + h (x , y ) dy .
7
It is, in other words, impossible to express all possible variations in
f (x , y ) as functions of only dx or dy .
8
Thus, the closed path integrals in equations 18.1 and 18.2 do not
represent state functions, and their values are therefore non-zero.
9
Conversely, e.g. T (dS )V ,n =
ˆ CV and p (dV )S ,n are exact differentials, because the degrees of freedom are locked in such a way that the
changes take place in the T , S and p , V planes respectively.
T and p are themselves
state functions, which means that
H
H
T (dS )V ,n = 0 and p (dV )S ,n = 0, as each of the closed path integrals can be simplified to two line integrals, one in each direction.
1029 / 1776
10
3
From the above definitions we can see that Qh > Qc > 0, and that
η → 1 if Qh → ∞ and Qc → 0.
5
What happens then is graphically illustrated in Figure 18.1.
7
This cycle is independent of the working medium, since its design
means that the heat integrals are only dependent on the temperatures
on the cold
R and hot sides (the two isentropes do not contribute to the
integral T dS ).
8
Provided that the temperatures do not alter the composition, the
Carnot cycle will achieve a given efficiency regardless of the medium
used.
9
No other cycles have this property, but then again the Carnot cycle
cannot be achieved in practice, since isothermal heat transfer requires
the system to have heat exchange properties that are physically impossible (infinite heat exchange surface or infinite heat conductivity).
1028 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
Thermodynamic cycles
18 Heat . . .
4
A frictionless drive or a resistance-free source of voltage are two
examples of this. Exploiting the potential energy of a lossless hydropower reservoir is an example with a thermodynamic dimension.
5
The hydrostatic pressure in the system can be used to drive an
ideal turbine generator that converts p , v work into electrical energy.
6
This is a flow process, which has little in common with closed cycles, but we can still imagine using hydrostatic pressure to drive a hypothetical cyclic process with a rectangular path in p , v coordinates.
7
The efficiency of this cycle is η = W /Wh = (Wh − Wl )/Wh = (ph −
pl )/ph , assuming that
R
ˆ
Wh =
p dV = ph (vh − vl ) and
dV >0
R
dV <0
Termodynamic methods TKP4175
02 September 2013
1030 / 1776
p dV = pl (vh − vl ) .
1032 / 1776
To perform a mathematical analysis of a thermodynamic cycle, we
first project the cycle onto T , s and p , V diagrams.
1
4
Interestingly, the slopes of the graphs T (s ) and p (v ), as seen in
the T , s and p , v diagrams, always satisfy the criteria
∂T ∂T =
ˆ cTv
>
=
ˆ cTp > 0 ,
(18.3)
∂s v
∂s p
−cp
∂p
∂p
−1
=
ˆ cv v β <
=
ˆ vβ < 0
(18.4)
∂v s
∂v T
Thermodynamic cycles (2)
18 Heat . . .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1031 / 1776
§ 143
§ 142
§ 144
18 Heat . . .
8
This broad conclusion is related to the fact that cv > 0, cp > 0 and
β > 0 are three fundamental properties of any stable thermodynamic
system.
9
Equations 18.3–18.4 will significantly assist our discussion of the
heat and work diagrams.
6
In other words, an isochor is always steeper than an isobar, and an
isentrope is always steeper than an isotherm in the two diagrams.
5
Figure 18.1: Optimal design of a closed cycle in T , s coordinates. The heat
integrals satisfy the relation Ai +1 > Ai . Combined with the physical constraints,
this implies that Qh,i +1 > Qh,i and Qc,i +1 < Qc,i . Consequently, ηi +1 > ηi where
η∞ =
ˆ ηCarnot .
News Thermodynamic cycles Power plants The steam engine Olds
2
This projection, or perhaps it is better to call it a mapping, transforms the cycle’s original system of coordinates to curvilinear[g] coordinate systems, one of which describes the heat leaving and entering
the system, and the other of which describes the work delivered and received.
3
In each of these coordinate systems, the cycle can be represented
by a closed graph in which the temperature and pressure vary in a manner that is characteristic of the cycle and working medium.
[g] The curvilinear coordinates are derived from the original (orthogonal) coordinates
using a (locally) invertible transformation (one–to–one mapping) at each point along
the graph. This means that each point in the orthogonal coordinate system can be
transformed to just one point in the curvilinear coordinate system, and vice versa. As
the word curvilinear implies, the original coordinate axes are curved by the mapping
process.
S2
S1
News Thermodynamic cycles Power plants The steam engine Olds
ENGLISH text is missing.Mechanical energy
1
To explain the Carnot cycle we assume the existence of a thermal
reservoir.
2
For reasons of symmetry, it should also be possible to define a
mechanical reservoir, but does one actually exist?
3
The answer is yes, if we assume the existence of an external
source of mechanical work that is not affected by its interaction with
the system.
ˆ
Wl =
Tc
This limit is of no practical significance, but if the machine is of finite size (meaning the entropy undergoes a finite change during the cycle), and there is also a maximum and minimum temperature for heat
exchange with the surroundings, the question becomes more interesting.
Consequently the optimal design of the cycle has a rectangular
path when represented on a T , s diagram, with η = (Th − Tc )/Th ; this is
commonly known as the Carnot cycle.
02 September 2013
A1
A0
4
6
Termodynamic methods TKP4175
A∞
A2
2
Naturally this value, and particularly its maximum, is of great practical and theoretical interest.
This goes against the normal sign convention, but it has the practical advantage that all heat flows are considered positive, whether they
are entering or leaving the cycle.
5
Tore Haug-Warberg (Chem Eng, NTNU)
Th
Thermodynamic efficiency is defined by: η =
ˆ W /Qh = (Qh − Qc )/
Draw realistic T , s and p , v diagrams for the ideal
(frictionless) cycles postulated by Otto, Brayton, Carnot,
Stirling, Ericsson and Rankine. Let the working medium
be one mole of an ideal, diatomic gas. Two of the
cycles are incompatible (in practice) with the working
fluid being an ideal gas. Which two?
regardless of the working medium.
7
Here it does not make any difference whether the medium is an
ideal gas or some other fluid (or even a solid).
Tore Haug-Warberg (Chem Eng, NTNU)
3
0
170
8
6
4
2
170
7
6
5
4
3
2
170
180
180
180
190
190
190
p [bar]
6
20
2.2
1.8
Brayton
W = 22.9
30
40
1
10
20
40
4
1
30
40
50
1
v [dm3 /mol]
5
10
50100
18 Heat . . .
Expanded footnote [g]:
[g] The change in volume is, incidentally, a direct measure of the size of the machine,
whereas the change in entropy is an indirect measure of the machine’s heat exchange
area.
2
The six cycles have been drawn in Figures 18.2 and 18.3 using the
Program 1.30 in Appendix 31.
Any attempt to specify both v and s makes these diagrams overspecified from a thermodynamic point of view.
500
These diagrams also show that the Carnot and Rankine cycles are
incompatible with using an ideal gas as the working medium.
4
Figure 18.2: T , s diagrams (left) and p , v diagrams (right) for one mole of
ideal gas (γ = 1.40). The cycles move through their respective states in the
clockwise direction. The characteristics of the cycle are: Compression of cold
gas (green), external heating (red), expansion of hot gas (yellow) and external
cooling (blue). The displaced volume of gas, and its entropy change, is the
same in all four cases except for the p , v diagram for the Carnot cycle, which
for practical reasons is drawn on a logarithmic scale.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
1036 / 1776
5
The integration ranges are 3.5 decades of pressure and 1 decade
of temperature respectively, so it is difficult to avoid considerable friction
or heat losses.
6
It is therefore worth knowing that, with a simple modification, the
Stirling and Ericsson cycles can achieve the same efficiency as the
Carnot cycle (see Theme 144), while the Rankine cycle is equivalent
to the classical steam engine (see Subchapter 18.4).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Termodynamic methods TKP4175
02 September 2013
1035 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
3
W = 93.9
0.4
0.04
Carnot
1034 / 1776
Note that the changes in volume and entropy are the same in all
of the diagrams, with the exception of the p , v diagram for the Carnot
cycle, and the T , s diagram for the Rankine cycle.
50
1.4
200
200
3
0
10
200
§ 143 Otto . . . Rankine
W = 73.7
6
02 September 2013
News Thermodynamic cycles Power plants The steam engine Olds
Otto
9
p [bar]
9
Otto
Qh = 169
Qc = 95.6
Brayton
Qh = 135
Qc = 112
Carnot
Qh = 174
Qc = 80.1
s [J/K mol]
p [bar]
T [K/100]T [K/100]T [K/100]
News Thermodynamic cycles Power plants The steam engine Olds
12
Termodynamic methods TKP4175
1037 / 1776
7
6
5
4
3
2
170
7
6
Stirling
Qh = 144
Qc = 104
Ericsson
Qh = 131
180
190
3
2
170
13
9
5
1
150
Qc = 113
Rankine
Qh = 293
Qc = 257
s [J/K mol]
180
170
190
190
210
4
3
2
1
20
2.2
Ericsson
30
40
50
W = 18.2
1.8
1.4
1
10
200
Stirling
W = 39.8
5
0
10
200
5
4
p [bar]
1033 / 1776
p [bar]
02 September 2013
p [bar]
Termodynamic methods TKP4175
T [K/100]T [K/100]T [K/100]
Tore Haug-Warberg (Chem Eng, NTNU)
20
2.2
Rankine
30
40
50
W = 36.2
1.8
1.4
1
10
v [dm3 /mol]
20
30
40
50
Figure 18.3: T , s diagrams (left) and p , v diagrams (right) for one mole of
ideal gas (γ = 1.40). The cycles move through their respective states in the
clockwise direction. The characteristics of the cycle are: Compression of cold
gas (green), external heating (red), expansion of hot gas (yellow) and external
cooling (blue). The displaced volume of gas, and its entropy change, is the
same in all four cases except for the T , s diagram for the Rankine cycle, which
for practical reasons is drawn on a different scale.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1038 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
§ 144
§ 143
News Thermodynamic cycles Power plants The steam engine Olds
§ 145
Th
18 Heat . . .
d)
Tc
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
Qc
1039 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
ıg
=
T (dS )T = T
5
∂T V
dV =
18 Heat . . .
=
dV .
7
The heat integral for the hot side can now be calculated as follows:
R
ıg
ˆ
Qh =
T dS ıg
=
dS >0
T
Rh ıg
CV dT
Tc
+
V
R2
V1
NRTh
V
cıg (T − T ) + NRT ln
dV =
ˆ C
c
h
h
V
Finally we can calculate the integral for the cold side:
R
ıg
Qc =
(−T ) dS ıg
ˆ
v2
02 September 2013
1040 / 1776
T
Rh
V2 V1
,
6
ıg
ıg
For your own benefit you should check that W ıg = Qh − Qc .
ıg
CV dT +
V
R2
V1
NRTc
V
18 Heat . . .
cıg (T − T ) + NRT ln
dV =
ˆ C
c
c
h
V
V2 V1
.
=
and with the same sign in both
and
ıg
Qc .
If this heat is temporarily stored in a recuperator, the consumption
ıg
of Qh is reduced by an equivalent amount, and assuming perfect heat
recovery the efficiency rises to
V2 8
ıg
ηStirling →
NR (Th −Tc ) ln V
1
V NRTh ln V2
1
=
Th −Tc
Th
V
R2
V1
NRTh
V
dV +
V
R1
V2
NRTc
V
dV = NR (Th − Tc ) ln
V2 V1
,
but in order to integrate T dS we need to know the differential of S as a
function of the free variables T and V :
C
∂p ˆ TV dT + ∂T V dV .
dS = ∂∂ST V dT + ∂∂VS T dV =
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1041 / 1776
= ηCarnot
Heat recovery
18 Heat . . .
1
An equivalent argument also applies to the Ericsson cycle, where
ıg
cıg (T − T ) + NRT ln p2 and Q ıg = C
cıg (T − T ) + NRT ln p2 .
Qh = C
c
c
c
h
h
h
c
p1
p1
P
P
2
Here Qh contradicts the simple definition contained in Eq. 18.2 on
page 942, and a new graphical representation (that only applies in the
event of perfect heat recovery) follows from Figure 18.4d .
cıg (T − T ) appears in the same context
Also note that the term C
c
h
V
ıg
Qh
The work integral is consequently defined by
H
W ıg =
ˆ p ıg dV
News Thermodynamic cycles Power plants The steam engine Olds
§ 144 Stirling & Ericsson (3)
Tc
NRT
V
Termodynamic methods TKP4175
2
The projections onto the T , S and p , V planes are shown in Figures 18.4b and 18.4c,d .
News Thermodynamic cycles Power plants The steam engine Olds
Along the various isopleths, it is also true that
ıg
CV dT ,
∂p ıg
W
v1
Figure 18.4: Figure a illustrates the Stirling cycle in the natural variables T
and v . Figures b, c and d show the projected work and heat diagrams where
W = Qh − Qc . The cycles move Hthrough their respective states in the clockwise
direction (which means W = p dv > 0). The heat integral is divided into
four sections as shown in Figure c, the coloured parts of which cancel for an
ideal gas with perfect heat recovery (because the heat capacity is then only
a temperature function). The net (external) heat requirement is therefore as
shown in Figure d.
02 September 2013
§ 144 Stirling & Ericsson (2)
ıg
T (dS )V
3
b)
s
18 Heat . . .
1
The Stirling cycle describes a rectangle in the T , V plane, as
shown in Figure 18.4a .
v2
p
c)
Tc
News Thermodynamic cycles Power plants The steam engine Olds
4
v1
s
Qh
Show that the Stirling and Ericsson cycles can achieve
the same efficiency as the Carnot cycle if the working
medium is an ideal gas, and assuming perfect heat
recovery in the non-isothermal part of the process.
§ 144 Stirling & Ericsson
a)
Tc
Th
News Thermodynamic cycles Power plants The steam engine Olds
Th
3
cıg (T − T ).
Here the heat that needs to be recovered is C
c
h
P
4
We can draw the conclusion that the Stirling and Ericsson cycles
reach the same efficiency as the Carnot cycle, provided that there is
perfect heat recovery in the heating and cooling steps, and that the
working medium is an ideal gas:
ıg
ıg
ηStirling and ηErıcsson → ηCarnot (Th ) .
(18.5)
(18.6)
5
The Carnot cycle has the same theoretical efficiency whatever the
working medium, but unlike the Stirling and Ericsson cycles, it would not
be possible to implement it using an ideal gas as the working medium.
dS <0
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1042 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
§ 145
§ 144
§ 146
Termodynamic methods TKP4175
02 September 2013
1043 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
§ 145 Otto & Brayton
Find an expression for the thermodynamic efficiency of
the Otto cycle (an idealised petrol engine) as a function
of the temperature Th at the end of the compression
stroke, i.e. immediately before ignition. Set out an
equivalent expression for the Brayton cycle (a gas
turbine). In both cases assume that the working medium
is an ideal gas. Compare your results with the Carnot
cycle.
18 Heat . . .
§ 145 Otto & Brayton (2)
3
=
T
Ra
Th
=
ıg
CV dT +
ıg
CV (Ta
T
Rc
Te
ıg
ıg
ˆ
Qh =
R
− Th ) −
Th
ıg
ˆ
ηOtto =
− Tc ) .
T
Ra
ıg
ıg
CV dT = CV (Ta − Th ) ,
which gives us the efficiency
CV dT
ıg
CV (Te
T dS ıg =
dS >0
W ıg
ıg
Qh
=1−
Te −Tc
Ta −Th
,
where Ta , Th , Te and Tc are the respective temperatures after combustion, compression, expansion and cooling. From this we can conclude
ıg
that Ta = Te Th /Tc .
4
The Otto cycle uses isentropic compression, where sc = sh , and
isentropic expansion, where sa = se , and as the volume ratio is the
same for these subprocesses, we get the relation
Th ıg
V 1−γ
Ta ıg
,
= V21
Tc S ,n = Te
in line with Theme 61 on page 382.
The final expression for the efficiency is thus
ıg
02 September 2013
18 Heat . . .
The associated heat integral is
ηOtto = 1 −
Termodynamic methods TKP4175
1044 / 1776
2
The ideal Otto cycle describes a rectangle in the S , V plane, but all
we need to know is the projection of this rectangle in the T , S plane.
5
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
News Thermodynamic cycles Power plants The steam engine Olds
This is because, assuming constant heat capacity, the work integral
can be solved as follows:
H
H
W ıg = p ıg dV = T dS ıg
1
Termodynamic methods TKP4175
1045 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1046 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Te −Tc
Th Te
Tc −Th
=
Th −Tc
Th
=1−
Termodynamic methods TKP4175
V2 1−γ
V1
.
02 September 2013
1047 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
§ 145 Otto & Brayton (3)
18 Heat . . .
6
The efficiency is the same for a hypothetical Carnot cycle in which
heat is added at (an assumed) constant temperature Th immediately
after the completion of the compression stroke.
8
7
An equivalent conclusion can be drawn for the Brayton cycle, by
replacing cv with cp in our calculation:
ıg
p (γ−1)/γ
.
ηBrayton = 1 − TTh Te −e Tc = ThT−hTc = 1 − p21
Tc −Th
News Thermodynamic cycles Power plants The steam engine Olds
News Thermodynamic cycles Power plants The steam engine Olds
Figure 18.5: The thermodynamic efficiency of the six ideal (rectangular) cy∆c x
. ∆h x and
cles described in the main text can be expressed as η = 1 − ∆
hx
∆c x denote the change in the state variable x over the hot and the cold sides
of the cycle. The definition of x for each of the cycles is shown in the table below.
Figure 18.6: Thermodynamic work W = ∆h x − ∆c x for the six ideal (rectangular) cycles described in the main text. ∆h x and ∆c x denote the change in
the state variable x over the hot and the cold sides of the cycle. The definition
of x for each of the cases is shown in the table below.
x
We can draw the conclusion that
ıg
ıg
ηOtto = ηBrayton = ηCarnot (Th )
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
(18.7)
02 September 2013
1048 / 1776
Stirling
Ericsson
(∆u)v − T (∆s )T
(∆h )p − T (∆s )T
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
Thermal power plants
18 Heat . . .
3
The adiabatic (isentropic) changes of state do not contribute to the
integral of T dS .
8
This approach gives us general expressions for the work performed 4 The Rankine cycle, on the other hand, can be seen as a purely
mechanical device that merely exploits pressure–volume forces without
6
This approach gives us general expressions for the efficiency of by the cycles; see Table 18.6.
the cycles, without taking into consideration the working medium itself; 9 The table also states values for the Carnot and Rankine cycles, them necessarily being linked to a thermal change of state.
5
Examples ??
which come directly from the T , s and p , v diagrams respectively.
see Table 18.5.
1050 / 1776
1052 / 1776
1049 / 1776
simple difference in the function values of the corners of the cycle.
Otto
(∆u)v
Carnot
Brayton
(∆h )p
Rankine
Termodynamic methods TKP4175
x
§ 146
§ 145
If no electricity is being generated, and assuming adiabatic conditions, the temperature of the flue gases will be Ta , but when the plant is
doing work, the temperature falls.
3
Consequently, the efficiency is lower than the Carnot efficiency implied by the adiabatic flame temperature Ta .
(∆a )v
Otto
(∆u)v
Carnot
T (∆s )T
T
Ericsson
(∆g )p
Brayton
(∆h )p
Rankine
p (∆v )p
(∆u)v − (∆h )p
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1054 / 1776
§ 147
18 Heat . . .
18 Heat . . .
6
The efficiency of a single-engine gas-fired power plant with a cooling water temperature of 300 K is shown in Figure 18.7 on page 969, cf.
Program 1.28 in Appendix 31.
=
Th −(Th −Tc )
Th2
·
Ta −Th
Ta −Tc
−
Th −Tc
Th
Tc
Th
8
We can simplify the equation to
slightly differently:
·
1
Ta −Tc
02 September 2013
= 0.
√
√
Ta − Tc
√
Ta
·
=
√
√
T − T
√ a √ c
Ta + Tc
=
ˆ
2
where Ta is the adiabatic flame temperature.
5
The overall efficiency of the power plant is therefore lower than the
maximum theoretical Carnot efficiency:
4
Note that Qtot ≥ Qh and that the difference between the two heat
flows can be attributed to the emission of hot flue gases from the plant.
1055 / 1776
√
Tc , and
Topt −Tc
Topt +Tc
Tore Haug-Warberg (Chem Eng, NTNU)
·
=
Tore Haug-Warberg (Chem Eng, NTNU)
W
Qtot
=
Th −Tc
Th
·
Ta −Th
Ta −Tc
=
ˆ ηCarnot · ηthermal .
Termodynamic methods TKP4175
02 September 2013
1056 / 1776
Curzon–Ahlborn optimum
18 Heat . . .
1
According to the definition in the above equation, ηopt gives the
overall efficiency of the plant.
2
How does this compare with the Carnot efficiency of the plant?
3
The effective Carnot efficiency of the plant is given by (Th − Tc )/Th .
4
If we substitute in the optimisation condition from Eq. 18.8, this can
√
be expressed as ηCA =
ˆ 1 − Tc /Ta , following Curzon and Ahlborn .
(Ta − Th ) = Th − Tc , or put
[g] Adrian Bejan. Advanced Engineering Thermodynamics. Wiley, 1988.
p
Ta Tc .
The optimal efficiency can also be expressed as a function of Topt =
Ta Tc from Eq. 18.8. At the limit case when ∆T =
ˆ Topt − Tc is small,
we get
(18.8)
7
√
9
If we substitute this into the expression for the total efficiency ηtot ,
we get
√
Ta Tc −Tc
√
Ta Tc
W
∆rx H
Expanded footnote [g]:
ˆ (Th )opt =
Topt =
ηopt =
Qh ,
3
The total heat content of the flue gases is Qtot = ∆rx H = CP (Ta −
Tc ) where CP is the heat capacity of the mixture.
18 Heat . . .
10 Let us multiply the numerator and the denominator by
then factorise the right-hand side:
ηopt =
Th −Tc
Th
News Thermodynamic cycles Power plants The steam engine Olds
§ 146 Single stage power plant (3)
√
√
√
Ta ( Ta − Tc )
√
√
√
√
( Ta − Tc )( Ta + Tc )
18 Heat . . .
Qh = CP (Ta − Th ) ,
News Thermodynamic cycles Power plants The steam engine Olds
Still assuming constant heat capacity, we can obtain an approxi7
mation for (Th )opt , and this closely matches the optimal operating temperature of the boiler:
∂ηtot
∂Th
Termodynamic methods TKP4175
1053 / 1776
The work performed by the Carnot engine can be expressed as
W=
A thermal power plant produces flue gases of temperature
Ta and has one Carnot engine available. Heat is added
to the engine at the boiler temperature Th and is removed
at the cooling water temperature Tc . Calculate the
overall efficiency of the power plant as a function of
Th . Determine the optimal boiler temperature. The
flue gases have constant heat capacity.
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
§ 146 Single stage power plant
1
News Thermodynamic cycles Power plants The steam engine Olds
§ 146 Single stage power plant (2)
Termodynamic methods TKP4175
News Thermodynamic cycles Power plants The steam engine Olds
ηtot =
ˆ
Tore Haug-Warberg (Chem Eng, NTNU)
x
Stirling
1
At a thermal power plant, a proportion of the energy in the fuel will
be lost with the hot flue gases produced by the plant.
2
x
x
News Thermodynamic cycles Power plants The steam engine Olds
ENGLISH text is missing.Corner co-ordinates II
1
In addition to knowing the efficiency of each of the cycles, it is also
interesting to know how much work they can perform.
H
2
For the s , v and s , p cycles, W = T dS is equal to the changes
(∆u)v and (∆h )p on the hot and cold sides, cf. the integration of Eq. 5.
ENGLISH text is missing.Corner coordinates
1
The two preceeding sections show that equations 18.1 and 18.2 4 For the T , v and T , p cycles the calculation is more complicated, ENGLISH text is missing.Thermomechanical encan be integrated analytically if the working medium is an ideal gas.
but the problem can be solved by integration by parts.
gines
R
R
H
H
2
However, it is likely that the approach we have used becomes un- 5 From T ds = Ts − s dT we can say that T ds = − s dT . 1 We can draw the conclusion that the four cycles — Otto, Brayton,
necessarily complex if the ideal gas is replaced with a real fluid.
Note that the Ts term is a state function that makes a zero contribution Stirling and Ericsson— are abstract representations of thermomechanical engines that convert changes in internal energy, enthalpy, Helmholtz
3
If we think about it, all of the cycles discussed in this chapter are to the closed path integral. H
H
energy and Gibbs energy into mechanical work.
rectangular in their respective characteristic coordinate systems.
6
Similarly, we know that p dv = − v dp from the work diagram,
2
The Carnot and Rankine cycles are not related to any of the canon4
It should therefore be possible to express the efficiency of each but that is of no importance here.
ical thermodynamic potentials; rather they can be considered purely
cycle as a function of its corner coordinates alone.
7
All we now need are the partial differentials
thermal or mechanical engines.
5
All we need are analytical expressions for the partial differentials
(dA )V ,n = (dG )p ,n = −S dT ,
3
Here it should be added that the Carnot cycle is independent of
H
(dU )V ,n = (dH )p ,n = T dS ,
so that we can integrate them to obtain (−S ) dT expressed as a simple the working medium, and that it could alternatively be conceived as a
H
thermoelectric or thermomagnetic engine.
which can then be integrated in such a way that T dS is given as a difference in the function values of the corners of the cycle.
x
√
Ta − Ta Tc
Ta −Tc
Termodynamic methods TKP4175
5
then
At the limit case when x =
ˆ (Ta − Tc )/Ta approaches a small value,
lim ηopt =
Ta →Tc
6
···
√1−
1− √1−x
1+
x
=
x
4−x
Here we have used the expansion
≃
√1
1
4
· ηCarnot (Ta ) .
3 3
+ x = 1 + 21 x − 81 x 2 + 48
x −
lim ηopt =
Topt →Tc
.
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1058 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
1
2
·
Topt −Tc
Topt
=
1
2
· ηCarnot (Topt ) .
Termodynamic methods TKP4175
02 September 2013
(18.9)
1059 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
0.6
ηopt =
News Thermodynamic cycles Power plants The steam engine Olds
§ 147
Topt −Tc
Topt +Tc
§ 146
§ 148
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
ENGLISH text is missing.Exhaust heat
1
This means that a thermal power plant that extracts its heat from
flue gases, or more generally from a hot process stream, will have an
efficiency that is lower than if it had extracted the heat from a heat
reservoir of the same temperature as the flue gas.
2
According to the above equations, the efficiency will be 25% or 50%
of the equivalent Carnot efficiency, depending on whether you base
your calculation on the flame temperature or the optimal boiler temperature.
3
We should hasten to add that these statements are only valid for
low process temperatures.
4
At high temperatures, the differences between the various efficiencies decreases, until finally they disappear when T → ∞.
5
This, however, is a purely mathematical limit, which is of no practical relevance.
1060 / 1776
2
As previously mentioned, the Carnot engines are connected in series on the hot side, and have a common cold reservoir.
η
3
1.4 1
2
Show that if the power plant in Theme 146 has n > 2
engines available, then optimal operation will occur
p
when Ti = Ti +1 Ti −1 , where i − 1, i and i + 1 represent
three engines connected in series on the hot side. What
will the maximum available work be when n → ∞?
6
20
400
800 1200 1600 2000 2400
T [K]
3
The total amount of work that they can perform is the sum of their
individual contributions
Wn =
Termodynamic methods TKP4175
02 September 2013
1061 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
4
Tc is the temperature of the common reservoir on the cold side of
the engine.
5
The optimal operating conditions are given by
1 ∂Wn
CP ∂Ti
6
=
Ti −(Ti −Tc )
Ti2
(Ti −1 − Ti ) −
Ti −Tc
Ti
+
Ti +1 −Tc
Ti +1
Termodynamic methods TKP4175
W∞ = −CP
T
Rc
Ta
gral.
Tore Haug-Warberg (Chem Eng, NTNU)
18 Heat . . .
T −Tc
T
dT = −CP
T
Rc
dT + Tc CP
Ta
T
Rc
Ta
dT
T
= CP (Ta − Tc ) − Tc CP ln
Ti +1 Tc (Ti −1 − Ti ) − Ti Ti +1 (Ti − Tc ) + Ti2 (Ti +1 − Tc ) = 0 .
8
1062 / 1776
9
The fraction (Ti − Tc )/Ti then takes the variable value (T − Tc )/T ,
while the factor Ti −1 − Ti can be seen as a differential temperature
change − dT :
= 0.
= ∆H − T c ∆ S .
This can be expressed as the tridiagonal system of equations Ti =
p
Ti −1 Ti +1 ; show that this has the general solution Tin+1 = Tci Tan+1−i .
CP (Ti −1 − Ti ) +
1064 / 1776
(18.11)
0
300
10 Constant heat capacity has been assumed in the above derivation,
but in Chapter 19 we showed that W∞ = ∆H − Tc ∆S (called the exergy
of the flue gases) is also valid as a general result, based on a thermodynamic analysis of flow processes.
5
0
9
James Watt
c)
40
80
120
160
5
0
9
Ideal steam turbine
e)
40
80
120
160
5
0
40
80
120
160
02 September 2013
1065 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
ENGLISH text is missing.Horror vacui
1063 / 1776
400
500
600
Vapour–liquid cycles
[g] As so often in history, the results came first, and the theory was developed subsequently.
[g] Aristotle, 384–322 f.Kr. Greek philosopher.
[g] Using ∆S = CP ln(Ta /Tc ) = CP ln(1 + x ) and ∆H = CP (Ta − Tc ) = Tc CP x , and
3
5
1
ln(1 + x ) = 2+x x + 13 2+x x + 15 2+x x + · · · .
[g] Heron fra Alexandria, ca. 10–70. Greek engineer and mathematician.
[g] Evangelista Torricelli, 1608–1647. Italian physicist and mathematician.
2
substituting in the series
Figure 18.8: Thermodynamic efficiencies of Newcomen (blue) and Watt
(green) steam engines, and of a steam turbine with superheated steam (red).
The dashed curve shows the corresponding efficiency of the Carnot cycle,
where the boiler and the condenser (40 ◦ C) act as thermal reservoirs. The
measurements
years 1740–1960.
Tore
Haug-Warberg (Chemare
Eng, from
NTNU) theTermodynamic
methods TKP4175
02 September 2013
1072 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
ENGLISH text is missing.Savery, Newcomen &
Watt
1
The first true steam engine was not developed until the 18th century, first as a low-pressure steam engine, which means that atmospheric pressure — rather than the steam pressure — performs the mechanical work.
1
Savery had understood the need to separate the boiler from the
pump, but The Miner’s Friend was still dreadfully inefficient.
2
Papin proposed (1706) that the steam should be separated from
2
Heron’s æolipile, on the other hand, had been a high-pressure enthe water with a hollow metal piston.
gine, and high pressure had later been used in another context to produce effective guns, but in the early 17th century people started to un3
This was a significant improvement, and also meant that the pump
derstand that vacuums were also associated with great forces, and that
would only use positive pressure, but once again he was out of luck.
ENGLISH text is missing.Heron’s æolipile
these forces could be harnessed to perform mechanical work.
4
That was because the previous year, Newcomen[g] had reached
1
The trusty old steam engine has long-since become passé for pracan agreement with Savery, and had patented the first working atmotical applications, but it retains its exalted position, high above all of the 3 In order to properly appreciate this rebellious thought, we must
refer back to Aristotle[g] and his thesis that nature harbours a “fear of
spheric steam engine based on a combination of Papin’s original steam
other machines, in the engineering hall of fame.
the empty space” (lat. horror vacui).
engine, and Savery’s boiler.
2
Not only did it pave the way for Western industrialisation, but it also
ENGLISH text is missing.Dennis Papin
5
Subsequent improvements included directly injecting water to coninspired the emergence and development of thermodynamics[g] in the 4 He considered the vacuum (or aether) to be a fifth, divine element
alongside earth, air, fire and water.
1
All that remained was a small practical difficulty — how could you dense the steam, although the temperature in the cylinder still varied.
early 19th century.
create the negative pressure?
6
It was only once Watt[g] patented the condenser in 1769 that the
3
However, from a thermodynamic point of view, the development of 5 Aristotle’s works disappeared in the West, and were only rediscovered in the 12th century, but his ideas were a potential threat to Catholic
2
A solution to this intricate problem was first demonstrated by Pa- steam engine achieved its final design, with the boiler, engine and conthe steam engine was something of a coincidence.
ENGLISH text is missing.Otto von Guericke
ENGLISH text is missing.Limit case
ENGLISH text is missing.High pressure engines
doctrine.
pin[g] in 1690.
denser becoming three separate units.
1
In 1654, Guericke[g] demonstrated his “Magdeburg hemispheres”,
1
The efficiency of the plant can be expressed as η∞ =
ˆ W∞ /∆H = 4 The story starts with Heron[g]’s æolipile, and ends with the closely
1
However, Watt stuck to the principle of using atmospheric pressure,
1 − Tc · ∆S /∆H , where ∆H is the heat needed to raise the temperature related reaction turbine 2,000 years later, but there is no direct link 6 It was the Church’s great philosopher and theologian Thomas thereby showing that it was possible to use the atmosphere to control 3 By alternately boiling and condensing water (steam) in a metal 7 Watt worked systematically to further improve the engine, and in
Aquinas who reconciled Christian doctrine with the Aristotelian world- forces that were far greater than human muscle power.
cylinder tightly sealed with a piston, he created a periodic negative 1782 he introduced partial (isentropic) expansion of the steam in the without realising that this left huge room for improvement.
from Tc to Ta .
between these two machines.
view.
gauge pressure that was able to raise a mechanical load using a spool, final part of the piston stroke.
2
The problem was that an improvement of this kind required a high
2
This demonstration came at a time when there was a great deal of
2
If we assume constant heat capacity, η∞ can be expanded[g] in x 5 Heron’s invention was far ahead of his time, and remained a mere
but the idea did not catch on in the contemporary scientific community.
working pressure, which in turn would require multiple expansion stages
7
This reinforced Aristotle’s position in the West, and the dogma of interest in scientific innovation in Europe.
where x =
ˆ (Ta − Tc )/Tc :
8
This doubled the theoretical efficiency in the relevant temperature
curiosity.
and a far stronger construction of the boiler.
4
Savery[g], on the other hand, was granted a patent in 1698 for his range 373–383 K, as shown in Figure 18.8.
the horror vacui remained unchallenged right until 1643, when Torriln(1+x )
x
x 3
2
x 5
2
3
Meanwhile, the mining industry in England needed greater pumpη∞ = 1 − x
= 2+
6
There are many possible explanations for this, but the most imporx − 3x 2+x − 5x 2+x − · · ·
3
The development of multiple expansion engines therefore took a
celli[g] created the first barometer, demonstrating that the hydrostatic
steam-driven water pump designed to “raise water by fire”.
ing capacity.
9
The coloured spots reflect technological progress[g][g] in this area.
tant one is probably that in Antiquity people kept slaves, and didn’t need
At the limit case where x is small, we get
pressure of the atmosphere could balance the weight of a column of
long time, culminating around the same time that Parsons[g] launched
5
This was a pistonless device, in which the steam came into direct
1071 / 1776
machines to perform mechanical work.
4
The mine water was causing major technical and financial prob−Tc
1
the first reaction turbine in 1884.
mercury.
lim η∞ = 2+x x = TTaa+
contact with the water that it was lifting, which was first raised 4–5 meTc ≃ 2 · ηCarnot (Ta ) .
Ta →Tc
lems, and in view of the scientific progress that had recently been made,
7
It is unclear whether the æolipile had any practical application other
8
Only then did people develop a scientific understanding of how
4
Shortly afterwards (1887), de Laval[g] created the first impulse turtres using negative pressure, and then pumped 10–12 metres using
it was natural that an attempt was made to remove the water using neg3
At low process temperatures the efficiency is twice that of a single than as a “miracle of nature” in one of the temples in Alexandria.
ancient inventions such as the water pump and the siphon worked.
bine, and in 1890 he equipped it with a supersonic nozzle.
positive pressure.
[g] Thomas Newcomen, ca. 1664–1729. English blacksmith.
ative gauge pressure, either directly or in combination with some device
1067 / 1776
Carnot engine, cf. the result on page 968.
1068 / 1776
1073 / 1776
1070 / 1776
that could be used to pump the water in the traditional way.
[g] James Watt, 1736–1819. Scottish engineer.
1066 / 1776
1069 / 1776
[g] Otto von Guericke, alias Gericke, 1602–1686. German engineer and poltician.
[g] Denis Papin, 1647–1714. French physician.
[g] R. L. Hills. Power from Steam, pages 59,103–112,131. Cambridge Univ. Press,
1989.
[g] Charles Algernon Parsons, 1754–1931. Irish engineer.
[g] Thomas Savery, ca. 1650–1715. English inventor.
[g] Adrian Bejan. Advanced Engineering Thermodynamics. Wiley, 1988.
18 Heat . . .
[g] Carl Gustaf Patrik de Laval, 1845–1913. Swedish engineer.
James Watt
d)
For a long time the steam engine remained the gold standard for
railways and mechanical industry, while the steam turbine was universally adopted in ships and for electricity generation.
Ideal steam turbine
f)
2
This rapid transition was due to the fact that steam turbines are
smaller, have a higher operating temperature and can make effective
use of the isentropic expansion of superheated steam.
26.02
10
1
0.5
0.01
0.018
26.02
10
0.5
v [dm3 /mol]
s [J/K mol]
Figure 18.9: T , s diagrams (left) and p , v diagrams (right) for a single mole of
water and steam. The cycles that describe the Newcomen and Watt engines
make use of saturated steam whereas the turbine works with superheated
steam. The volumes of water and steam at 1 bar and 40 ◦ C are marked on the
abscissas in the p , V diagrams. The enlarged portion of the diagrams focus
on the details on the liquid side. Otherwise the notation is the same as in
Figures 18.2 and 18.3.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1074 / 1776
3
02 September 2013
Vapour–liquid cycles (2)
18 Heat . . .
10
500
7
Termodynamic methods TKP4175
0.5
500
7
3
Thomas Newcomen
b)
0.01
0.018
p [bar]
3
p [bar]
7
500
p [bar]
T [K/100] T [K/100] T [K/100]
Thomas Newcomen
a)
Termodynamic methods TKP4175
Boiler temperature [K]
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
9
(18.10)
0.2
Ta Tc
When n → ∞, the sum of Eq. 18.10 can be expressed as an inte02 September 2013
CP (Ti − Ti +1 ) + · · ·
0.4
[g] Refers to the structure when written in matrix form.
Termodynamic methods TKP4175
Ti +1 −Tc
Ti +1
0.6
Expanded footnote [g]:
Tore Haug-Warberg (Chem Eng, NTNU)
CP (T1 − T2 )
News Thermodynamic cycles Power plants The steam engine Olds
§ 147 Exergy (3)
Multiplying by Ti2 Ti +1 gives us
7
02 September 2013
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
Ti −Tc
Ti
T2 −Tc
T2
if Ti −1 is the inlet temperature for engine i and Ti is the outlet temperature of the same engine (which in turn is equal to the inlet temperature
for the next engine).
News Thermodynamic cycles Power plants The steam engine Olds
§ 147 Exergy (2)
CP (Ta − T1 ) +
+ ··· +
Figure 18.7: Overall efficiency of a single-engine natural gas-fired power
plant (with temperature-dependent ideal gas heat capacity) with a variable
air : methane ratio. 1.0 is the stoichiometric ratio.
Tore Haug-Warberg (Chem Eng, NTNU)
T1 −Tc
T1
η
0.2
0
18 Heat . . .
1
Let us assume that the heat capacity of the flue gases is constant
(in practice calculated as the effective average value).
ηCarnot
0.4
0
§ 147 Exergy
0.01
0.018
26.02
This latter quality is particularly important and, as shown in Figure 18.8, it raises the efficiency significantly above that of a steam engine using saturated steam.
3
5
The theoretical difference between a basic steam engine — with
and without steam expansion — and a one-stage steam tubine can be
seen from the T , s - and p , s diagrams in Figure 18.9.
6
In the literature the general term Rankine cycle is used regardless
of the engine, but note that only the Newcomen cycle has a rectangular
path in p , V coordinates.
7
The Watt cycle is more like the Carnot cycle, while the steam turbine cycle (as shown here) is really a Brayton cycle, with a rectangular
path in p , S coordinates.
4
In fact, using superheated steam results in an engine that is theoretically perfect; see Theme 150 on page 979.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1075 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1076 / 1776
News Thermodynamic cycles Power plants The steam engine Olds
State diagrams
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
ENGLISH text is missing.Practical calculations
1
All of the diagrams (including Figure 18.8) have been drawn using
the Program 1.31 in Appendix 31.
2
This is based on rigorous calculations involving an advanced equation of state[g] with 232 parameters.
3
One practical disadvantage of this equation of state is that it has an
unknown number of volumes that satisfy the expression f (T , p , v ) = 0
at a given pressure and temperature.
4
By comparison, the Van der Waals equation has either one or three
real solutions.
5
You therefore need good initial values for the calculations — particularly along the phase equilibrium curve and close to the critical point.
6
In this particular case there are correlations between the saturation
properties of water and steam[g] that get us out of our predicament, but
in general the problem of obtaining adequate initial values is difficult to
circumvent.
1077 / 1776
[g] A. Saul and W. Wagner. J. Phys. Chem. Ref. Data, 18(4):1537–1564, 1989.
[g] A. Saul and W. Wagner. J. Phys. Chem. Ref. Data, 16(4):893–901, 1987.
§ 150
§ 149
§ 151
News Thermodynamic cycles Power plants The steam engine Olds
18 Heat . . .
§ 149 Watt’s engine
§ 148 Newcomen’s engine
1
The heat added is qh = (∆u)v + (∆h )p ; also see Figure 18.9a and
Table 18.5 on page 961.
1
The heat requirement is the
R same as in Theme 148, but the cycle
performs the expansion work p (dv )s in addition to the compression
work (∆p )v (∆v )p .
2
Under normal operating conditions, the isochoric heat is much less
2
The expansion takes place when the saturation curve for the steam
than the isobaric heat (of vaporization), and qh ≈ ∆h is therefore a good
and dv are dominated by the change in p .
approximation of the total heat requirement.
3
This means that p (dv )s ≈ p ∂∂pv dp ≈ v ıg dp .
b
T
1
Derive a simplified expression for the efficiency of Newcomen’s at- 3 The work performed is w = (∆p )v (∆v )p ; see Figure 18.9 , but
4
The Clausius–Clapeyron relation in Chapter 14 gives
mospheric steam engine. The cylinder is completely filled with satu- since the pressure in the condenser is much lower than atmospheric
R us dp =
p (dv )s ≈
∆h /(RT ∆v ) dT ≈ ∆h /(Tv vap ) dT , which means that
and the steam is virtually an ideal gas, we can say that w ≈ § 149
rated steam, and both evaporation and (residual) condensation occur pressure,
sat vap
1
Repeat Exercise 148, but this time for Watt’s steam engine, includ- −∆h dln T = ∆h ln(Th /Tc ) and ηWatt ≈ ηNewcomen + ln(Th /Tc ).
at constant pressure (and temperature), while heating and cooling oc- ph vh ≈ RTh .
ing partial (isentropic) expansion of the saturated steam.
cur at constant volume.
4
This means that the efficiency is ηNewcomen ≈ RTh /∆h .
5
This trebles the efficiency of the engine at Tc ≈ 313 K.
1079 / 1776
1081 / 1776
1082 / 1776
1080 / 1776
§ 148
Firstly, it is important to note that the saturation states are in different locations in the diagrams — the steam engines convert liquid to
saturated steam at a high temperature, whereas the steam turbine converts saturated steam to liquid at a low temperature.
1
§ 150 The ideal steam turbine
Figure 18.9e and Table 18.5 imply that w = (∆h h )p − (∆c h )p .
1
2
At first glance the difference between the p , V diagrams 18.9b,d,f
appears to be relatively insignificant, and we will therefore use the rest
of this chapter to discuss some important details.
2
All of the heat is removed at constant (low) temperature i.e.
(∆c h )p = Tc (∆c s )p , but from Figure 18.9e it follows that ∆c s = ∆h s , so
the work performed can therefore be expressed as w = ∆h − Tc ∆s , in
line with Eq. 18.11 (the ∆ changes refer to the limits of the cycle).
Show that the maximum work that can be performed by
an ideal steam turbine is equal to the exergy content
w = ∆h − Tc ∆s of the superheated steam.
In the case of the two steam engines, both heating and cooling take
place within the two-phase region, whereas the steam turbine exploits
isentropic compression and expansion in the one-phase region.
3
18 Heat . . .
4
Consequently, the steam turbine can make use of some of the
energy contained in the fuel at a higher temperature than the boiler
temperature.
6
The efficiency of the turbine therefore exceeds the Carnot efficiency of the system based on the boiler temperature.
5
This can be seen clearly from the T , s diagram in Figure 18.9e .
7
The steam turbine is really a Brayton cycle with alternate isobaric
and isentropic changes of state, but as can be seen from Figures 18.9e
and 18.9f , the isentropic compression of the water produces so little
change in volume as to make it, to all intents and purposes, isochoric.
8
The steam engine, meanwhile, is limited by the Carnot efficiency,
but there is still a big difference between Newcomen’s and Watt’s designs.
9
Figure 18.8 shows that Watt’s steam engine is almost as good as a
theoretical Carnot engine, which is pretty good considering that it was
developed half a century before Carnot set out his ideas and a whole
century before the emergence of modern thermodynamics.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1078 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News Thermodynamic cycles Power plants The steam engine Olds
1
2
3
4
5
6
7
18 Heat . . .
H
H
ıg
NR (Th −Tc ) ln V
1
V NRTh ln V2
1
Tore Haug-Warberg (Chem Eng, NTNU)
=
Th −Tc
Th
Termodynamic methods TKP4175
1
Reversible heat
2
Clausius’s inequality
3
Entropy production
4
Lost work
√
√
T − T
√ a √ c
Ta + Tc
W
∆rx H
Th −Tc
Th
·
Ta −Th
Ta −Tc
Optimal design: ηopt =
Entropy production and available work
11
Infinite number of stages: W∞ = −CP
T
Rc
Ta
T −Tc
T
dT = ∆ H − T c ∆ S
= ηCarnot
see also Part-Contents
02 September 2013
1085 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1086 / 1776
Chemical energy
7
Exergy
Termodynamic methods TKP4175
02 September 2013
1087 / 1776
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
19 Entropy . . .
In a fuel cell, the ionic transport through the electrolyte is coupled
to the chemical reactions on the surface of a membrane and to the
transfer of an equivalent number of electrons in an external circuit .
4
Available energy
Tore Haug-Warberg (Chem Eng, NTNU)
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
Chemical energy can be converted directly into work at the molecular level, as a coupled electron–ion transport process in a fuel cell, or
indirectly by first transforming it to thermal energy in a thermal power
station.
Gouy–Stodola’s theorem
Part 19
=
ˆ ηCarnot ·
Topt −Tc
Topt +Tc
10
=
ˆ
=
1
6
1084 / 1776
ıg
9
Single stage power plant: ηtot =
ˆ
ηthermal
19 Entropy . . .
5
02 September 2013
Otto (and Brayton) after compression stroke:
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
New concepts
Termodynamic methods TKP4175
18 Heat . . .
ıg
Carnot efficiency: ηCarnot = (Th − Tc )/Th
ηStirling →
Tore Haug-Warberg (Chem Eng, NTNU)
ηOtto = ηBrayton = ηCarnot (Th )
Thermodynamic efficiency: η =
ˆ W /Qh = (Qh − Qc )/Qh
Stirling (and Ericsson) with recuperator:
V2 1083 / 1776
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
Looking back (2)
8
(dU )n = (δQ − δW ) = 0
H
Work = heat: p dV = T dS
R
Heating: Qh =
ˆ dS >0 T k dS k
R
ˆ dS <0 T k dS k
Cooling: Qc =
Closed cycle:
02 September 2013
News Thermodynamic cycles Power plants The steam engine Olds
Looking back
H
Termodynamic methods TKP4175
Chemical energy (2)
19 Entropy . . .
10 We will build on a number of concepts that we have previously studied: Mass and energy balances, state functions, conditions for equilibrium, reference states and Carnot efficiency.
11 In addition, we will introduce some new concepts: Entropy production and energy availability.
2
Fuel cell technology is undoubtedly the most elegant option, but
it has to be acknowledged that power stations are much more widely
used.
3
The two methods of conversion are not only different at a practical
level — the very concepts behind them are also different.
6
In a thermal power station, the chemical energy is first transformed
into thermal energy, which is then converted into mechanical energy
using suitable machinery.
5
It should be pointed out that by itself our analysis only tells us what
processes are theoretically possible — there is no explanation of the
nature of the process
13
12 The reward for our efforts will be a more general energy equation that allows high-level analysis of chemical and mechanical energy
transformation.
Expanded footnote [a]:
14 or, if you prefer, its physical design.
[a] In general we cannot say how long the process takes, whether it requires a large
amount of space, or if there are any other physical limitations. However, we can ensure 15 The difference between analysis and design is not self-evident, and
that the 2nd law of thermodynamics is fulfilled, and that the mass and energy balance one of the subsidiary aims of this chapter is precisely to clarify the disequations can be solved.
tinction.
In principle, this involves a lossless conversion process.
9
This chapter is about thermal energy in an abstract sense — the
exact details of the processes involved are beyond its scope.
7
The first stage of this transformation is in theory lossless, whereas
the second stage is subject to the 2nd law of thermodynamics, which
limits the potential efficiency of the process.
8
The fact that combustion reactions are irreversible means that
losses are inevitable during the conversion process.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1088 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1089 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1090 / 1776
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
The Clausius inequality
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
19 Entropy . . .
The Clausius inequality applies to changes in thermal energy in a
closed
control volume, and put simply it states that the maximum value
R
of T -1 δQ occurs in a reversible process.
1
H
Expanded footnote [a]:
[a] Rudolf Julius Emanuel Clausius, 1822–1888. German physicist.
ırr
2
This is an independent thermodynamic hypothesis, which doesn’t
affect our understanding of the thermodynamic state itself.
3
What we are interested in here is the interaction between the system and its surroundings.
2
rev
4
The Clausius inequality (2)
1
Figure 19.1: Two alternative paths illustrating the Clausius inequality.
This can be expressed in terms of equations as
H
H
T -1 δQ
T -1 δQrev
=
ˆ
H
≤0 ,
dS = 0 .
T -1 δQ =
R2
19 Entropy . . .
T -1 δQrev +
R1
2
1
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
1
2
9
Up to this point everything seems normal but we have to be extra
careful if we are ever planning to change the direction of integration in
the inequality above.
10 The change of sign on both sides of the inequality sign gives
−
R1
19 Entropy . . .
1
Based on the integral inequality we can, for differential changes in
entropy, conclude that T dS ≥ δQ .
T -1 δQırr ≤ 0 .
8
The integral of the reversible heat transfer (divided by the temperature) defines entropy as a thermodynamic state variable, and we can
state that
R2
R1
R1
T -1 δQırr ≤ − T -1 δQrev = T -1 δQrev = S1 − S2 =
ˆ ∆2,1 S .
2
Entropy is time’s arrow
T -1 δQırr ≥ ∆1,2 S ,
2
R2
but we cannot on Hthis basis conclude that 1 T -1 δQırr ≥ ∆1,2 S . This
would then imply T -1 δQ ≥ 0 which is contradictory to the Clausius
inequality.
11 The reason of the contradiction stems from the fact that the irreversible integral has only a single direction of integration which is physically meaningful, see footnote on the previous page.
12 To avoid confusion it is therefore better to write
R
T -1 δQırr ≤ ∆S
and at the same time make clear that the inequality is valid for two processes being physically meaningful, and sharing the same end states,
while having entirely different process routes.
Note that 1/T plays the role of an integrating factor for δQrev in the
last expression.
5
3
For an isolated system, or a closed system that only experiences
adiabatic changes of state, there is the special condition δQ = 0 which
leads to
2
Agreeably, it is tacitly assumed that T > 0 but this does hardly
cause any frustration in the current context.
(T dS )isolated ≥ 0 .
4
Since T > 0 and dt > 0, then it is also true that (dS / dt )isolated ≥ 0.
The entropy of an isolated system therefore increases monotonically with time.
5
6
This has given rise to the claim that entropy is time’s arrow [a],
and to speculation about the link between time and irreversible physical events[a].
[a] Arthur Stanley Eddington. The Nature of the Physical World. MacMillan, 1928.
[a] What if time is inversely proportional to the frequency of physical events and the “Big
Bang” is a singularity in the grand scheme of things? If so, creation took place infinitely
long ago. Alternatively, the universe may have a finite age, but in that case what was
its state prior to creation? One critical discussion of the 2nd law of thermodynamics
and how it relates to our understanding of time is presented by Jos Uffink. Stud. Hist.
Phil. Mod. Phys., 32(3):305–394, 2001.
The thermodynamic cycle in Figure 19.1 has one reversible and
one irreversible path, which leads to the inequality :
7
6
A more thorough explanation of this concept is given on page 109
under the discussion of Pfaff?? differentials.
Expanded footnote [a]:
R2
P
ˆ limn→∞ ni Ti-1 ∆Qi+ , where ∆Qi+ = Q~i+ dti+ .
[a] From calculus we have 1 T -1 δQrev =
For a process to be reversible, the integral must have the same absolute value, but
the opposite sign, when the process is reversed to establish the initial state 1 from the
final state 2. In other words, for each heat term ∆Qi+ = Q~i+ dti+ in the forward process, there is an equivalent and opposite heat term ∆Qi− = Q~i− dti− in the reverse
process, and hence all temperatures Ti assuming i ∈ [1, n ] stay the same in the
two processes. The net effect of the reversible heat added, and subtracted, from
P
P
+
the system is in Ti-1 ∆Qi+ + ni Ti-1 ∆Qi− , and since ∆Q − =
R 2−∆Q this can
R 1 be written
Pn -1
+
-1
-1
-1
i (Ti − Ti )∆Qi = 0. On integral form we can say that 1 T δQrev + 2 T δQrev =
0. However, the return to the initial state along the irreversible route does not have
this property. For an irreversible process, it is generally imposssible to force it in a
backward direction. Here we must assume that the physical events occur in reverse
order and have the opposite signs for the reversed process. Which means time itself is being reversed. This is not amenable to real
R 2 for our hypothetiR 1 physics but
cally reversed process we may consider to write 2 T -1 δQırr = 1 T -1 δQırr , because
∆Qi− = −Q~i+ (− dti+ ) = ∆Qi+ when time is reversed. The point being that the integral
along the irreversible route has a natural direction of integration and that the rule of
sign reversal, valid for ordinary integrals, does not apply to this process.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
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The Gouy–Stodola theorem
02 September 2013
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19 Entropy . . .
4
A loss must relate to something that is transient, and in thermodynamics we speak about lost work.
2
Everyone has a subjective notion of what loss involves[a], but what
should really be considered a loss in thermodynamic terms?
3
Whatever it is, it isn’t the loss of energy, mass, impulse or charge,
as they are all subject to fundamental conservation laws.
[a] E.g. financial loss, memory loss, loss of youth etc.
The Gouy–Stodola theorem (2)
19 Entropy . . .
8
The Gouy–Stodola theorem can be expressed as an equation:
W̊lost = T◦ S̊ırr . However, it is not yet clear how the entropy production term can (or should) be interpreted.
10 There is every reason to fear that a derived quantity such as entropy production will only make things worse, but in some ways it is easier to conceive that entropy can be produced in a physical process than
it is to ascribe any deep philosphical meaning to the entropy as an abstract subject.
[a] We are not forgetting about the Boltzmann equation S = k ln Ω. There is a finite
relationship between entropy on the one hand, and the number of microstates of the
system on the other, but this is deeply rooted in statistical–mechanical theory, and it
isn’t designed to shed light on the concept of entropy at a macroscopic level.
This refers to a reduction in the system’s ability to perform mechanical work.
5
6
According to the Gouy –Stodola theorem , W̊lost is proportional
to the irreversible entropy production S̊ırr in the control volume.
Expanded footnote [a]:
Expanded footnote [a]:
Expanded footnote [a]:
[a] Louis Georges Gouy, 1854–1926. French physicist.
[a] Aurel Boleslav Stodola, 1859–1942. Slovenian physicist and engineer.
[a] G. Gouy. Journal de Physique, 8:501–518, 1889.
[a] A. Stodola. Z. VDI, 32(38):1086–1091, 1898.
7
The factor of proportionality is the temperature T◦ of the surroundings, or more specifically: The temperature of the thermal reservoir that
exchanges heat with the system.
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02 September 2013
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Towards equilibrium
19 Entropy . . .
The blocks have uniform temperatures, and individually they produce no entropy.
~
Q
The thermal resistance is to blame!
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
1095 / 1776
Towards equilibrium (2)
19 Entropy . . .
3
It drains valuable thermal energy from the system without producing any work, and without the thermal energy being used to produce
anything other than less valuable heat.
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dS
dt
=
dSh
dt
+
d Sc
dt
§ 151
§ 150
§ 152
19 Entropy . . .
ENGLISH text is missing.Simplest case
1
In order to illustrate the Gouy–Stodola theorem we will study a simple control volume with two thermodynamic states, and we will limit ourselves to verifying that the entropy production in this system is actually
a measure of the lost work.
2
At the same time we will demonstrate that the concept of entropy
production is underpinned by an intricate model that soon has us solving complex partial differential equations.
1096 / 1776
Let us consider an isolated system consisting of two
solid blocks of temperatures Th and Tc . Each block
is a diabatic heat conductor, with a constant heat
capacity and volume. Heat is transferred from the hot
block to the cold block in the presence of a thermal
resistance with negligible (thermal) capacity. At t = 0
the temperatures of the two blocks are Th (0) and Tc (0)
respectively. Your task is to find an expression for ddSt and
to explain how it relates to the lost work of the system.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
Towards equilibrium (3)
1097 / 1776
19 Entropy . . .
6
The energy balances for the whole system and for each of the
individual blocks can be stated as follows:
System :
dUtot
dt
= ddUth + ddUt c = 0 ,
dUh
dt
Hot :
dUc
dt
Cold :
dS
dt
CV
T
For each block it is true that (dU )V ,n = T dS , where (dS )V ,n =
ˆ
dT , which means that (dU )V ,n = CV dT assuming the volume of the
block does not change with temperature (it said to be constant in this
case).
Termodynamic methods TKP4175
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~,
= −Q
~.
= Q
When they are substituted into the entropy equation we get
.
5
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
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4
As the thermal resistance has negligible (thermal) capacity, we can
leave it out of the balance equations, and hence the change in entropy
of the system can be expressed as follows:
Tc
Figure 19.2: Production of entropy caused by heat conduction in an isolated
control volume.
2
02 September 2013
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1
Th
Termodynamic methods TKP4175
Termodynamic methods TKP4175
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9
Entropy is inherently a difficult concept, and facile claims that entropy is a measure of disorder or chaos, or that entropy is an indicator
of time, do not help us to understand it any better[a], and if anything
they further confuse the matter.
Expanded footnote [a]:
Tore Haug-Warberg (Chem Eng, NTNU)
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In all real processes there are irreversible losses.
1
Termodynamic methods TKP4175
02 September 2013
1099 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
=
~
Q
Tc
−
~
Q
Th
.
Termodynamic methods TKP4175
02 September 2013
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Differential equation
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19 Entropy . . .
Differential equation (2)
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19 Entropy . . .
~ to the temperature gradient
2
What remains to be done is to relate Q
between the two blocks.
2
Integration between the two states Th (0) and Th gives us the intermediate result
h −b
= −at .
ln aTaT
h (0)−b
Let us first determine Tc by substituting the differentials dUh = Ch
dTh and dUc = Cc dTc into the energy balance of the isolated system.
That yields Ch dTh + Cc dTc = 0 which can be integrated to give Tc =
Tc (0) + Ch Cc-1 (Th (0) − Th ).
3
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Termodynamic methods TKP4175
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4
Now what remains to be done is to insert realistic values for a
physical system into the final equation from the previous section, and
it probably makes most sense to use values that tell us about typical
system behaviour, rather than getting hung up in complexities relating
to geometry and mass distribution.
1
We will therefore choose the symmetric state description Cc =
Ch = 2κτ, where τ is the system’s time constant.
3
2
Our motto is maximum insight with the minimum of effort.
That gives us the manageable temperature equation Th =
+ Th (0)) + 21 (Tc (0) − Th (0)) exp( −τt ) which can be further simT=
ˆ 21 (Th (0) + Tc (0)) and ∆T =
ˆ 12 (Th (0) − Tc (0)).
plified by introducing b
4
1
2 (Tc (0)
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
19 Entropy . . .
Finally we will show that the control volume fulfils the Gouy–
Stodolas theorem W̊lost = T◦ S̊ırr = T◦ ddSt .
5
1102 / 1776
Let us first replace the conductor with a perfect insulator.
The final result is
Cc Tc (0)+Ch Th (0)
Cc +Ch
19 Entropy . . .
Tore Haug-Warberg (Chem Eng, NTNU)
[a] One little extra exerciseRfor anyone who is interested: S is a state function and the
∞
increase in entropy ∆S = 0 ( ddSt ) dt from the initial state to the final state can also be
b
b
expressed as ∆S = Ch ln( b T ) + Cc ln( b T ) = 2κτ ln(1 − (∆T /b
T )2 )-1 . Verify that
T +∆T
T −∆T
the two alternative expressions for ∆S are identical.
~ = κ(Th − Tc ) = 2κ∆T
6
If we combine these with the heat flux Q
exp( −τt ) and substitute them into the entropy equation dS / dt = Q /Tc −
Q /Th , we get
#
"
1
−t
1
dS
.
−
=
2κ∆
T
exp(
)
−t
−t
dt
τ
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
§ 151 Production of entropy (3)
Expanded footnote [a]:
Tc = b
T − ∆T exp( −τt ) .
b
T −∆T exp( τ )
Termodynamic methods TKP4175
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19 Entropy . . .
dS
dt
dS
dt
4κ
=
b
T
∆T exp( −τt )
!2
−1
∈ h0, ∞i .
8
It is important to note that 0 < ∆T ≤ b
T and 0 < exp( −τt ) ≤ 1.
Applied to the above equation this demonstrates that ddSt ∈ h0, ∞i.
9
Remember that the control volume is entirely isolated from its surroundings.
10 Hence the change in the system’s entropy must be attributable to
the irreversible production of entropy in the control volume S̊ırr = ddSt .
b
T +∆T exp( τ )
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Lost work (2)
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19 Entropy . . .
Lost work (3)
~ c = 0, the
~h + Q
8
If the two engines are synchronised such that Q
system as a whole will behave in the same way as the isolated control
volume , except that it will produce the following amount of net work
19 Entropy . . .
Th
[a] But unlike in the case of the isolated system, in this case the surroundings will be
affected.
~ h > 0 flows past a Carnot engine.
On its way, the heat Q
5
~ h.
~ h = (1 − T◦ T -1 )Q
For Th > T◦ the power of this engine will be W
h
per unit time:
~c =
~h−W
~ Carnot = W
W
~
~ = T◦ TQc − TQh
= T◦
dS
dt
Th −T◦
Th
~h −
Q
Tc −T◦
Tc
~h
Q
~c
Q
η=
.
~ c < 0 flowing between the reservoir and the block is in
7
The heat Q
~ c.
~ c = (1 − T◦ T -1 )Q
this case W
c
1107 / 1776
~h
W
Th −Tc
Th
~◦
Q
A similar argument can be made with respect to the cold block.
02 September 2013
exp −κ(Ch-1 + Cc-1 )t .
Expanded footnote [a]:
4
Termodynamic methods TKP4175
Cc (Th (0)−Tc (0))
Cc +Ch
7
The final expression for
is obtained by finding a common denominator for the fractions and then dividing both the numerator and
t
denominator by ∆T 2 exp( −2
τ ):
Th = b
T + ∆T exp( −τt ) ,
Next we will allow heat transfer from the hot block to a thermal
reservoir of temperature Tc < T◦ < Th .
Tore Haug-Warberg (Chem Eng, NTNU)
+
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Bearing these two definitions in mind, we can state that
3
6
02 September 2013
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1
2
Termodynamic methods TKP4175
§ 151 Production of entropy (2)
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Lost work
[a] In spite of the fact that in general it is a stated aim of this book to avoid shortcuts of
the kind “from which it trivially follows that. . . ”
From symmetry reasons we expect a similar result for Tc but the exact
formula is not so important for the current discussion.
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19 Entropy . . .
3
If we write the above equation in exponential form, we get an explicit expression for Th = Th (t ), and a bit of routine manipulation is all
we need to determine a and b , but as the intermediate steps are not of
any great value we will leave them out[a].
Th =
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
§ 151 Production of entropy
19 Entropy . . .
1
This is a separable differential equation that can be expressed in
the form dTh /(aTh − b ) = − dt .
1
The thermal resistance is directly proportional to the temperature
difference, which means that the heat flux fulfils the empirical equation
~ = κ(Th − Tc ).
Q
~ =
4
Substituting the expression for Tc into the heat flux equation Q
~ into the energy balance of the hot
κ(Th − Tc ), and then substituting Q
block, produces an equation that describes the change in the state of
Th :
h
ii
h
κ(Th −Tc )
Ch
dTh
h
= − Cκh (1 + C
dt = −
Ch
Cc )Th − Tc (0) + Cc Th (0) .
Time–temperature curve
T◦
Figure 19.3: Extraction of valuable heat in a Carnot engine.
Tore Haug-Warberg (Chem Eng, NTNU)
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Lost work (4)
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19 Entropy . . .
9
In other words, T◦ ddSt represents the mechanical work that can be
done using two external Carnot engines and one thermal reservoir.
10 If this work is not performed, it will be permanently lost as a result
of the irreversible heat transfer within the control volume.
Process analysis
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19 Entropy . . .
To analyse the available energy in a dynamic system, or the exergy
of a flowing fluid, we need to look at the energy, entropy and mass
balances together.
1
Process analysis (2)
19 Entropy . . .
8
The pressure in the surrounding atmosphere has been given the
symbol p◦ .
Initially we will therefore limit ourselves to looking at a single chem4
ical component.
2
At first sight this may appear to be a case of mixing apples and
oranges, but in fact it is possible to combine the equations if they are
first rewritten in a common form (in this case energy per unit of time).
3
This section deals with a rather esoteric topic, so it makes sense
to take on the challenges that it poses gradually, even if that makes the
book a bit longer.
5
Moreover, the potential and kinetic energy contributions can be
ignored, which means that the energy balance can be written ddUt =
~ −W
~ out + Q
~.
~ ın − U
U
6
The work performed can be split into pressure–volume work and
~ s + pV
~ =W
~ + p◦ dV .
mechanical work, i.e. W
dt
7
The final term allows for the fact that the size (volume) of the control
volume can change.
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Termodynamic methods TKP4175
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1110 / 1776
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Energy balance
02 September 2013
1111 / 1776
Introducing the enthalpy function H = U + pV makes it possible to
express the energy balance in the form
1
~ −W
~ out + Q
~ ın − H
~s.
=H
Entropy balance
19 Entropy . . .
In the case of reversible processes, entropy will be a conserved
quantity, but in the general case there is no such conservation law.
1
In order to concoct an entropy balance, we therefore need to rely
on the source term S̊ırr , which quantifies the entropy production of the
irreversible subprocesses in the control volume.
For high efficiency, S̊ırr must be as low as possible (optimal design),
4
while all potential processes that result in S̊ırr < 0 can immediately be
rejected (based on the 2nd law).
3
Here the difference between design and analysis becomes impor-
tant.
It is therefore also valid for multicomponent systems.
In addition to entropy production, we need to take into account any
~ out ) and any change in entropy
~ ın and S
contribution from convection (S
due to heat transfer.
5
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Termodynamic methods TKP4175
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Mass balance
If the inflows and outflows consist of one, and only one, chemical
component, the mass balance can alternatively be expressed by the
~ out .
~ ın − N
mole balance dN = N
1
dt
As we did with the entropy equation, we will now scale this equation, but this time by the factor µ◦ , 0:
2
dN
dt
~ out .
~ ın − µ◦ N
= µ◦ N
Termodynamic methods TKP4175
Available entropy
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Entropy balance (2)
19 Entropy . . .
This gives us the following balance equation:
-1 nT ~
q dΩ + S̊ırr .
T
Ω
R6
dS
dt
~ ın − S
~ out +
= S
8
In this case n ∈ R3 is the unit normal vector to the surface element
dΩ [m-2 ], and ~
q [W/m2 ] is the associated heat flux.
7
The integral in this equation sums the contributions from the
~ = nT ~
changes in heat δQ
q dΩ, weighted by T -1 , over the surface area
of heat transfer Ω.
9
Note that the entropy equation can be multiplied by the ambient
temperature T◦ , 0 without it changing the nature of the equation in
any way:
R
~ + T◦ S̊ırr .
~ ın − T◦ S
~ out + T◦ T -1 δQ
T◦ ddSt = T◦ S
Ω
Like the energy balance, the entropy balance is also valid for multicomponent systems.
12
10 What this operation means (and the motivation behind performing
it) will become clearer in due course.
11 Here let it suffice to say that the equation has not fundamentally
changed as a result of being scaled up.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1115 / 1776
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19 Entropy . . .
For the moment we will stick to our goal of combining the energy,
entropy and mass balances.
1
2
Most of the groundwork has already been done , but it will help
~ =
into two parts using the identity Q
R the heat flux integral
R
Rus to split
~ + T◦ T -1 δQ
~.
~ ≡ (1 − T◦ T -1 ) δQ
δQ
Expanded footnote [a]:
[a] Also check that the three equations have the same physical units. The real reason
why we multiplied the entropy and mass balances by T◦ and µ◦ respectively was to
achieve this!
3
3
In due course it will become clear that µ◦ can be viewed as a reference for chemical potential, but right now that is of secondary importance.
Tore Haug-Warberg (Chem Eng, NTNU)
02 September 2013
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19 Entropy . . .
µ◦
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2
2
Note that the energy balance does not require information about
the chemical components involved in the flows of material.
3
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19 Entropy . . .
d(U +p◦ V )
dt
Termodynamic methods TKP4175
Available entropy (2)
19 Entropy . . .
5
We will also choose to subtract the mass balance, in spite of the
fact that none of the terms cancel, as it gives us a more balanced result
from a thermodynamic point of view:
d(U −T◦ S +p◦ V −µ◦ N )
dt
~ − µ◦ N
~ − µ◦ N
~ − T◦ S
~ )ın − (H
~ − T◦ S
~ )out
= (H
R
-1
~ −W
~ s − T◦ S̊ırr .
+ (1 − T◦ T ) δQ
Ω
If we substitute this into the entropy equation we get
R
~ − (1 − T◦ T -1 ) δQ
~ + T◦ S̊ırr .
~ ın − T◦ S
~ out + Q
T◦ ddSt = T◦ S
Ω
By subtracting this result from the energy balance, we can elimi~.
nate the heat flux Q
4
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Termodynamic methods TKP4175
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Euler homogeneity
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Euler homogeneity (2)
19 Entropy . . .
In Chapter 5, specifically on page 176, we used homogeneous
functions to which Euler’s theorem applies to derive the integral expressions U = TS − pV + µN and H = TS + µN .
1
2
Applied to the above equation, this allows us to define a new state
ˆ F (x) − gT◦ x = (g − g◦ )T x.
function F̃ (x) =
3
This assumes that x is the homogeneous set of variables of the
original function F , and that the gradient vector g◦ defines the tangent
plane to F at the reference point x◦ .
4
This shows us that F̃ has the same properties as an ordinary thermodynamic energy function, except that the function’s gradient has
changed from g to g − g◦ .
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Termodynamic methods TKP4175
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19 Entropy . . .
§ 152
With this in mind, we can choose to express the combined energy,
7
entropy and mass balance as
R
-1
dŨ
~
~
(19.1)
dt = H̃ın − H̃out + (1 − T◦ T ) δQ − Ws − T◦ S̊ırr .
Stationary and adiabatic imply that dŨ / dt = 0 and ~
q = 0 respectively.
R
~
2
When the heat flux is zero, the heat flux integral (1 − T◦ T -1 ) δQ
8
Here Ũ is the available energy of the system, and H̃ is the exergy
~ , but
of the process flows (admittedly we have lost the rate symbol of H
even LATEX has a limited selection of function attributes).
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
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Tore Haug-Warberg (Chem Eng, NTNU)
Since T◦ > 0 and S̊ırr ≥ 0, the maximum power can be given as
~ s ) = H̃ın − H̃out .
max(W
3
Let us next assume tha that the outlet temperature Tout is equal to
the temperature T◦ of the process’s surroundings. If we also select
µ◦ = µout , then H̃out = 0.
4
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News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
19 Entropy . . .
§ 153
§ 152
§ 154
19 Entropy . . .
ENGLISH text is missing.A little bit of philosophy
1
First we assume that the outlet temperature is the same as the
ambient temperature and then we conclude that the result is generally
valid if the outflow has no exergy.
2
But what does this really mean? Are we really entirely free to define
the zero point of an energy exchange process as whatever we want?
3
Yes, we can indeed, provided that we stick to independent, possible
states. We, and not the principles of thermodynamics, decide where
zero lies.
4
This is also at the heart of the difference between analysis and
design.
5
It is only possible to perform a thermodynamic analysis on the process once all of the premises for the design are in place.
6
This is particularly true of practical issues such as the choice of
temperature and pressure levels, the exchange of heat and work in the
process, and the characterisation of the product and waste streams.
1124 / 1776
5
It follows that the exergy H̃ın is a direct measure of the maximum
achievable power:
~ s ) = H̃ın .
max(W
Ω
~ s = H̃ın − H̃out − T◦ S̊ırr .
will also be zero, which in turn means that W
19 Entropy . . .
Find the maximum achievable work per unit time
(power) that can be extracted from a stationary control
volume under adiabatic conditions.
Ω
§ 152 Process work (2)
1
§ 153
6
When performing a thermodynamic analysis of an open control volume in contact with a thermal, mechanical and chemical reservoir, this
property is important, because it defines the absolute amount of available energy in the system.
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
19 Entropy . . .
§ 151
5
Whereas the ordinary energy functions U , H , G , A etc. have an arbitrary zero point, in the case of Ũ , H̃ , G̃ , Ã etc. it is defined exactly.
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
§ 152 Process work
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
6
This result is generally valid if the outflow can be viewed as a waste
stream with no exergy.
Determine the entropy production of a stationary,
adiabatic control volume that does not perform any
mechanical work.
8
The only thing we need to do is to (re)define T◦ and µ◦ in such a
way that their values reflect the outflow state.
7
Logically, the inflow would then have positive exergy, and it is impossible to do better than to use 100% of it.
Expanded footnote [a]:
[a] Yes, we can indeed choose our own zero point. For each independent variable we
are free to select one zero point for either the enthalpy, internal energy, Gibbs energy
or any other energy function. Hence, in the case of a one-component system we can
always say that H̃out = 0. However, it is only if Tout = T◦ that we also get the simple
result µout = µ◦ .
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1122 / 1776
Tore Haug-Warberg (Chem Eng, NTNU)
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
§ 153 Isenthalpic flow)
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News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
19 Entropy . . .
§ 154
§ 153
§ 155
2
But the process parameters also mean that the control volume is
~ ın = H
~ out .
isenthalpic, which means that H
~ ın = N
~ out .
Moreover, we also know that N
We can therefore conclude that the entropy production T◦ S̊ırr =
~ ın − S
~ out ) is equal to the difference between the
(H̃ın − H̃out )H~ ,N~ = T◦ (S
exergy flows into and out of the control volume.
4
19 Entropy . . .
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§ 154 Carnot work
19 Entropy . . .
For a closed process we know that H̃ın = H̃out = 0.
Rτ
Since the process is cyclical, it is also true that 0 (dŨ / dt ) dt = 0,
where τ is the period of the cycle.
R
~ − T◦ S̊ırr .
~ s = (1 − T◦ T -1 ) δQ
3
That leaves us with the expression W
Ω
1
2
Determine the maximum possible work (per cycle)
for a closed process if heat is added at a constant
temperature Th > T◦ and removed at the ambient
temperature Tc = T◦ .
4
The amount of work performed per cycle can be determined by
~ s over the period τ.
integrating W
5
We have been told that all of the heat is added to the process at a
temperature of Th > T◦ and is removed at a temperature of Tc = T◦ .
5
It is important to note that this difference is independent of the
~ via H
~ ).
reference state of H (also note how H̃ is transformed into S
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
~ s ) as set out in
1
One formal answer is that T◦ S̊ırr = max(W
Theme 152.
3
Tore Haug-Warberg (Chem Eng, NTNU)
6
Let us therefore divide the cycle into one period of time t ∈ [0, τ1 ]
during which heat is added to the process, and another period of time
t ∈ hτ1 , τ] during which heat is removed.
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02 September 2013
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News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
News The Clausius inequality The Gouy–Stodola theorem Available energy Olds
§ 155 Boiler
1
If all of the outlet parameters are adhered to, the maximum power
~ s ) = − dŨ .
is max(W
dt
2
Integrated over the whole duration of the process, this gives us the
following simple result for available work[a]:
max(Ws ) =
R∞
0
~ s dt = −
W
R0
dŨ = Ũ
Ũ
=
ˆ (T − T◦ )S − (p − p◦ )V + (µ − µ◦ )N .
3
We are now in a position to determine how much work the boiler
can perform.
§ 154 Carnot work (2)
19 Entropy . . .
8
Moving heat from within the system to the surroundings does nothing, since there is no temperature difference. Hence we can concentrate on what happens during the addition of heat.
7
If substituted into the heat flux integral, the latter will not contribute
any useful work.
9
We know that T = Th > T◦ for this part of the cycle, and since
T◦ S̊ırr ≥ 0 the maximum work that can be performed by the cycle is:
τR1 R
τR1
~ dt = (1 − T◦ T -1 )Q
~ dt = Th −T◦ Qh .
max(Ws ) =
(1 − T◦ Th-1 ) δQ
h
Th
0 Ω
0
It shouldn’t surprise anyone that this result is identical to the work
done by the Carnot cycle, at least if you give it a little bit of thought, but it
is worth noting that we have found a formula that covers both stationary
flow processes (previous section) and cyclical closed processes (this
section).
10
Figure 19.4: Excerpts from the steam tables of water. The reference state of
zero enthalpy and zero entropy is liquid water at its triple point.
4
For waste streams that go straight to landfill, or that are released
into the air or water, it is natural to choose T◦ ≈ 280 K and p◦ ≈ 1 atm.
5
The waste stream from our boiler fits into that category, and the
optimal solution will be to depressurise and cool the water (reversibly)
until it finally can be released into the recipient at T◦ and p◦ .
6
Note that we do not have any degrees of freedom with respect to
the choice of µ◦ .
7
The chemical potential is uniquely determined if both the temperature and pressure are known.
Looking back
8
Now what remains to be done is to determine the values of p , S , µ
and ρ at 225 ◦ C and µ◦ at 280 K and 1 atm.
9
The steam table 19.4 on the previous page contains some information that will help us to do that.
10 The values for enthalpy and entropy in the table relate to liquid
water at the triple point.
11 In the above equation, on the other hand, the entropy has 0 K as
its reference, which means that we need to rewrite the expression.
12 To do so, we will apply the fundamental relation µ = h − Ts , in this
instance using specific values.
ENGLISH text is missing.Efficiency
19 Entropy . . .
The thermodynamic efficiency of the boiler can be defined as the
13 We will also take care to state the final expression in terms of vol- 1
ENGLISH text is missing.Energy minimum
work performed (delivered) divided by the energy supplied (electricity),
~ − µ◦ N
~) =
~ − T◦ S
~ we can, by differ- ume, since we have been asked to find the available energy in a cubic
1
From the definition H̃ (~
S , p, N
ˆ H
but this immediately raises the question: Should the energy supplied be
metre of water:
S )p ,~N = (T − T◦ ) and (∂H̃ /∂~
N )p ,~S = (µ − µ◦ ).
entiating, derive (∂H̃ /∂~
calculated assuming stationary or dynamic conditions?
Ũ = (T − T◦ )S − (p − p◦ )V + (µ − µ◦ )M
2
In other words, at fixed ambient pressure p◦ , H̃ has a stationary
2
In this case we will assume the former, since boilers normally op= TsM − T◦ sM − (p − p◦ )V + (h − Ts )M − (h◦ − T◦ s◦ )M ,
point at T = T◦ and µ = µ◦ , with the associated function H̃ = 0.
erate under stationary conditions, but it should be emphasised that this
max(Ws )
= ρ(h − h◦ ) − ρT◦ (s − s◦ ) − (p − p◦ ) .
V
is not by definition true.
3
It can be shown that this stationary point is a local minimum, see
~ >0
Chapter 21 on page 1312, and since (∂H̃ /∂p )S~ ,~N = (∂~
H /∂p )S~ ,~N = V
For an electric boiler, the energy supplied is equal to the electric
14 Based on the data from Table 19.4, the maximum available energy 3
it follows that H̃ = 0 is actually a global minimum for the depressuring is thus V -1 max(W ) = 833.90 · [966890 − 29410 − 280.15 · (2564.1 − power supplied, and from the energy balance we can deduce that Wel ≈
s
ENGLISH text is missing.Process dynamics
process[a].
V ρ(h − h◦ ).
106.3)] − (25.501 − 1.01325) · 105 = 205.13 MJ m-3 .
1
If no mass is added, andR the process takes place under adiabatic
~ = 0.
4
Our best bet is to ensure that the outlet conditions remain at all 15 This can be converted to 57 kW h m-3 , which is equivalent to the 4 The approximately equal to symbol has been used to emphasise
conditions, then H̃ın = 0 and Ω (1 − T◦ T -1 ) δQ
times T = T◦ , p = p◦ and µ = µ◦ (T◦ , p◦ ).
dŨ
daily electricity consumption of a normal Norwegian detached house. the fact that the equation is holds if: i) there is no heat loss during
~
1
Steam is used for heating and as an energy buffer in many chem- 2 It follows that Ws = − dt − H̃out − T◦ S̊ırr .
5
This is equivalent to requiring reversible mass and energy ex- 16 Also note that a (microscopic) error has been introduced into these heating and compression, ii) conditions are stationary; and iii) that H̃ın =
ical process plants. But what is the energy density of a boiler, and what 3 As in the previous two sections T◦ S̊ırr ≥ 0, and ideally the process
change with the surroundings[a]. By way of contrast, imagine that T , calculations by using the saturation properties of water at 0.010012 bar, 0.
is its thermodynamic efficiency if it is heated with electricity? Start with should be reversible.
T◦ . The temperature difference could in that case be used by an exter- rather than the actual properties at 1 atm.
5
Based on these assumptions, the efficiency is η = max(Ws )/Wel =
the maximum amount of work that can be performed under the adia~ s = − dŨ − H̃out ?
4
But what next? How can we maximise W
nal heat engine, which would mean that our process was not optimal.
dt
1134 / 1776 max(Ws )/V ρ(h − h◦ ) = 205.13/781.76 = 0.262.
batic depressuring and emptying of a pressurised system. Let the vol1133 / 1776
6
Given this poor theoretical efficiency, and bearing in mind the low
ume of the boiler be V = 1 m3 , and assume that initially it is filled with 5 This is an open question, which can be solved in a variety of ways,
water at saturation point at a temperature of T = 225 ◦ C. Use data from one of which is to find the lower limit value for H̃out .
energy density, there is no reason to use the boiler as anything other
[a] Note that Ũ (t → ∞) = 0 in the final state. This is a direct result of the self-imposed
~ s ) can be calculated by integration. [a] Depressuring to atmospheric pressure requires the condition p > p◦ to be met.
Table 19.4.
6
If there is such a limit, max(W
limits T = T◦ , p = p◦ and µ = µ◦ at t → ∞. Hence it is not necessary to empty the than a thermal energy store.
1135 / 1776
1130 / 1776
1132 / 1776 [a] It is the outlet from the energy transformation process that needs to meet this tank completely — it is sufficient for there to be no available energy in the contents of
§ 155
requirement — not the outlet from the boiler.
the tank.
H
T
[◦ C]
p
[bar]
0.01
0.00611
0.0048
999.80
9.1575
0
2501.6
0
2
7
0.01001
0.0078
999.90
8.9762
0.1063
2514.4
29.41
3
25
0.03166
0.0230
997.11
8.5592
0.3670
2547.3
104.77
4
100
1.01325
0.5977
958.13
7.3554
1.3069
2676.0
419.06
Gouy–Stodola’s theorem: W̊lost = T◦ S̊ırr
5
225
25.501
12.763
833.89
6.2461
2.5641
2801.2
966.89
Available energy: Ũ =
ˆ U − T ◦ S + p◦ V − µ ◦ N
6
285
69.186
741.45
36.062
5.8220
3.1146
2774.5
1263.2
325
120.56
654.07
70.472
5.4969
3.5008
2688.0
1194.0
374.15
221.20
ρ
[kg/m3 ]
315.46
s
[kJ/K kg]
4.4429
h
[kJ/kg]
1
2107.4
7
Clausius’s inequality: T -1 δQ ≤ 0
H
H
ˆ dS
Reversible heat: T -1 δQrev =
Entropy production: (T dS )isolated ≥ 0
~ − µ◦ N
~ − T◦ S
~
Exergy: H̃ =
ˆ H
Available energy balance equation:
R
-1
dŨ
~
~
dt = H̃ın − H̃out + (1 − T◦ T ) δQ − Ws − T◦ S̊ırr
Ω
11 These two extremes are relatively simple to describe, but it is also
possible to analyse dynamic processes, as we will do below.
Tore Haug-Warberg (Chem Eng, NTNU)
Termodynamic methods TKP4175
02 September 2013
1129 / 1776
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Termodynamic methods TKP4175
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