Second Law of Thermodynamics(YAC Ch.5)
•Identifies the direction of a process. (e.g.: Heat can only
spontaneouslytransfer from a hot object to a cold object, not vice
versa)
•Used to determine the “Quality” of energy. (e.g.: A high-temperature
energy source has a higher quality since it is easier to extract energy
from it to deliver useable work.)
•Used to exclude the possibility of constructing 100% efficient heat
engine and perpetual-motion machines. (violates the Kevin-Planckand
the Clausius statements of the second law)
•Used to introduce concepts of reversible processesand
irreversibilities.
•Determines the theoretical performance limitsof engineering systems.
(e.g.: A Carnot engine is theoretically the most efficient heat engine; its
performance can be used as a standard for other practical engines)
Second-law.ppt
Modified 10/9/02
Second Law(cont)
•A process can not happen unless it satisfies both the first and
second laws of thermodynamics. The first law characterizes the
balance of energy which defines the “quantity” of energy. The
second law defines the direction which the process can take place
and its “quality”.
•Define a “Heat Engine”: A device that converts heat into work
while operating in a cycle.
Heat engine
QH
QL
TH
TL
Wnet
DQ-Wnet=DU (since DU=0 for a cycle)
net=QH-QL
Thermal efficiency, hth is defined as
hth=Wnet/QH=(QH-QL)/QH
=1-(QL/QH)
Question: Can we produce an 100% heat engine, i.e. a heat engine where
QL=0?
Steam Power Plant
•A steam power plantis a good example of a heat engine where
the working fluid, water, undergoes a thermodynamic cycle
Wnet= Wout-Win= Qin-Qout
Qinis the heat transferred from the high temp. reservoir, and is generally referred to as
QH
Qoutis the heat transferred to the low temp. reservoir, and is generally referred to as QL
Thermal efficiency
hth= Wnet/QH= (QH-QL)/QH =1-(QL/QH)
Thermal Reservoir
Ahypothetical body with a very large thermal capacity (relative to the system beig
examined) to/from which heat can be transferred without changing its temperature.
E.g. the ocean, atmosphere, large lakes.
Back
Steam Power Plant
•A steam power plantis a good example of a heat engine where
the working fluid, water, undergoes a thermodynamic cycle
Wnet= Wout-Win= Qin-Qout
Qinis the heat transferred from the high temp. reservoir, and is generally referred to as
QH
Qoutis the heat transferred to the low temp. reservoir, and is generally referred to as QL
Thermal efficiency
hth= Wnet/QH= (QH-QL)/QH =1-(QL/QH)
Thermal Reservoir
Ahypothetical body with a very large thermal capacity (relative to the system beig
examined) to/from which heat can be transferred without changing its temperature.
E.g. the ocean, atmosphere, large lakes.
Back
Kevin-Planck Statement
•The Kelvin-Planck Statement is another expression of the second law of
thermodynamics. It states that:
It is impossiblefor any device that operates on a cycleto receive heat from
a single reservoirand produce net work.
•This statement is without proof, however it has not been violated yet.
•Consequently, it is impossible to built a heat engine that is 100%.
Heat engine
QH
TH
Wnet
•A heat engine has toreject some energy into a lower temperature sink in order
to complete the cycle.
•TH>TL in order to operate the engine. Therefore, the higher the temperature,
TH, the higher the qualityof the energy source and more work is produced.
Impossible because it violates the Kelvin-Planck Statement/Second
Law Clausius Statement
•The Clausius Statement is another expression of the second law of
thermodynamics. It states that:
It is impossibleto construct a device that operates in a cycle and produces no
effect other than the transfer of heat from a lower-temperature body to a
higher-temperature body.
•Similar to the K-P Statement, it is a negative statement and has no proof, it is
based on experimental observations and has yet to be violated.
•Heat can not be transferred from low temperature to higher temperature
unless external work is supplied.
Heat pump
QH
QL
TH
TL
Therefore, it is impossible to build a heat pump or a refrigerator without
external work input.
Equivalence of the Two Statements
It can be shown that the violation of one statement leads to a violation
of the other statement, i.e. they are equivalent.
A 100% efficient heat engine; violates K-PStatement
Heat pump
QL
QL
TH
TL
Heat transfer from low-temp body to high-temp body without work; A
violation of the Clausiusstatement
Heat pump
QH+QL
QL
TH
TL
Wnet
=QH
Heat engine
QH
Perpetual-Motion Machines(YAC: 5-5)
Imagine that we can extract energy from unlimited low-temperature energy
sources such as the ocean or the atmosphere (both can be thought of as thermal
reservoirs).
Heat
engine
Heat
pump
QL
QH
QH
Win= QH-QL
Wnet=QL
TH
Ocean TL
It is against the Kevin-Planck statement: it is impossible
to build an 100% heat engine.
Perpetual Motion Machines, PMM, are classified into
two types:
PMM1-Perpetual Motion Machines of the First
Kind: They violate the First Law of Thermodynamics
PMM2 -Perpetual Motion Machines of the Second
Kind: Violate the Second Law of Thermodynamics
Reversible Processes and
Irreversibilities(YAC: 5-6)
•A reversible process is one that can be executed
in the reverse direction with no net change in the
system or the surroundings.
•At the end of a forwards and backwards reversible
process, both system and the surroundings are
returned to their initial states.
•No real processes are reversible.
•However, reversible processes aretheoretically
the most efficientprocesses.
•All real processes are irreversible due to
irreversibilities. Hence, real processes are less
efficient than reversible processes.
Common Sources of Irreversibility:
•Friction
•Sudden Expansion and compression
•Heat Transfer between bodies with a finite
temperature difference.
•A quasi-equilibrium process, e.g. very slow,
frictionless expansion or compression is a
reversible process.
Reversible Processes and Irreversibilities
(cont’d)
•A work-producingdevice which employs
quasi-equlibrium or reversible processes
produces the maximum amount of
worktheoretically possible.
•A work-consumingdevice which employs
quasi-equilibrium or reversible processes
requires the minimumamount of
worktheoretically possible.
•One of the most common idealized cycles
that employs all reversible processes is called
the Carnot Cycleproposed in 1824 by Sadi
Carnot.
Chapter 5: The Second Law of Thermodynamics
In this chapter we consider a more abstract approach to heat engine, refrigerator
and heat pump cycles, in an attempt to determine if they are feasible, and to obtain
the limiting maximum performance available for these cycles. The concept of
mechanical and thermal reversibility is central to the analysis, leading to the ideal
Carnot cycles. (Refer to Wikipedia: Sadi Carnot a French physicist,
mathematician and engineer who gave the first successful account of heat engines,
the Carnot cycle, and laid the foundations of the second law of
thermodynamics). For more information on this subject, refer to a paper: A
Meeting between Robert Stirling and Sadi Carnot in 1824 presented at the 2014
ISEC.
We represent a heat engine and a heat pump cycle in a minimalist abstract format
as in the following diagrams. In both cases there are two temperature reservoirs
TH and TL, with TH > TL.
In the case of a heat engine heat QH is extracted from the high temperature source
TH, part of that heat is converted to work W done on the surroundings, and the rest
is rejected to the low temperature sink TL. The opposite occurs for a heat pump, in
which work W is done on the system in order to extract heat QL from the low
temperature source TL and "pump" it to the high temperature sink TH. Notice that
the thickness of the line represents the amount of heat or work energy transferred.
We now present two statements of the Second Law of Thermodynamics, the first
regarding a heat engine, and the second regarding a heat pump. Neither of these
statements can be proved, however have never been observed to be violated.
The Kelvin-Planck Statement: It is impossible to construct a device which
operates on a cycle and produces no other effect than the transfer of heat from a
single body in order to produce work.
We prefer a less formal description of this statement in terms of a boat extracting
heat from the ocean in order to produce its required propulsion work:
The Clausius Statement: It is impossible to construct a device which operates on
a cycle and produces no other effect than the transfer of heat from a cooler body to
a hotter body.
Equivalence of the Clausius and Kelvin-Planck Statements
It is remarkable that the two above statements of the Second Law are in fact
equivalent. In order to demonstrate their equivalence consider the following
diagram. On the left we see a heat pump which violates the Clausius statement by
pumping heat QL from the low temperature reservoir to the high temperature
reservoir without any work input. On the right we see a heat engine rejecting heat
QL to the low temperature reservoir.
If we now connect the two devices as shown below such that the heat rejected by
the heat engine QL is simply pumped back to the high temperature reservoir then
there will be no need for a low temperature reservoir, resulting in a heat engine
which violates the Kelvin-Planck statement by extracting heat from a single heat
source and converting it directly into work.
Mechanical and Thermal Reversibility
Notice that the statements on the Second Law are negative statements in that they
only describe what is impossible to achieve. In order to determine the maximum
performance available from a heat engine or a heat pump we need to introduce the
concept of Reversibilty, including both mechanical and thermal reversibility. We
will attempt to clarify these concepts in terms of the following example of a
reversible piston cylinder device in thermal equilibrium with the surroundings at
temperature T0, and undergoing a cyclic compression/expansion process.
For mechanical reversibility we assume that the process is frictionless, however we
also require that the process is a quasi-equilibrium one. In the diagram we notice
that during compression the gas particles closest to the piston will be at a higher
pressure than those farther away, thus the piston will be doing more compression
work than it would do if we had waited for equilibrium conditions to occur after
each incremental step. Similarly, thermal reversibility requires that all heat transfer
is isothermal. Thus if there is an incremental rise in temperature due to
compression then we need to wait until thermal equilibrium is established. During
expansion the incremental fall in temperature will result in heat being
transferred from the surroundings to the system until equilibrium is established.
In summary, there are three conditions required for reversible operation:



All mechanical processes are frictionless.
At each incremental step in the process thermal and pressure equilibrium
conditions are established.
All heat transfer processes are isothermal.
Carnot's Theorem
Carnot's theorem, also known as Carnot's rule, or the Carnot principle, can be
stated as follows:
No heat engine operating between two heat reservoirs can be more efficient
than a reversible heat engine operating between the same two reservoirs.
The simplest way to prove this theorem is to consider the scenario shown below, in
which we have an irreversible engine as well as a reversible engine operating
between the reservoirs TH and TL, however the irreversible heat engine has a
higher efficiency than the reversible one. They both draw the same amount of heat
QH from the high temperature reservoir, however the irreversible engine produces
more work WI than that of the reversible engine WR.
Note that the reversible engine by its nature can operate in reverse, ie if we use
some of the work output (WR) from the irreversible engine in order to drive the
reversible engine then it will operate as a heat pump, transferring heat Q H to the
high temperature reservoir, as shown in the following diagram:
Notice that the high temperature reservoir becomes redundent, and we end up
drawing a net amount of heat (QLR - QLI) from the low temperature reservoir in
order to produce a net amount of work (W I - WR) - a Kelvin-Planck violator - thus
proving Carnot's Theorem.
Corollary 1 of Carnot's Theorem:
The first Corollary of Carnot's theorem can be stated as follows:
All reversible heat engines operating between the same two heat reservoirs
must have the same efficiency.
Thus regardless of the type of heat engine, the working fluid, or any other factor if
the heat engine is reversible, then it must have the same maximum efficiency. If
this is not the case then we can drive the reversible engine with the lower
efficiency as a heat pump and produce a Kelvin-Planck violater as above.
Corollary 2 of Carnot's Theorem:
The second Corollary of Carnot's theorem can be stated as follows:
The efficiency of a reversible heat engine is a function only of the respective
temperatures of the hot and cold reservoirs. It can be evaluated by replacing
the ratio of heat transfers QL and QH by the ratio of temperatures TL and
TH of the respective heat reservoirs.
Thus using this corollary we can evaluate the thermal efficiency of a reversible
heat engine as follows:
Notice that we always go into "meditation mode" before replacing the ratio of
heats with the ratio of absolute temperatures, which is only valid for reversible
machines. The simplest conceptual example of a reversible heat engine is the
Carnot cycle engine, as described in the following diagram:
Obviously a totally impractical engine which cannot be realized in practice, since
for each of the four processes in the cycle the surrounding environment needs to be
changed from isothermal to adiabatic. A more practical example is the ideal
Stirling cycle engine as described in the following diagram:
This engine has a piston for compression and expansion work as well as a displacer
in order to shuttle the working gas between the hot and cold spaces, and was
described previously in Chapter 3b. Note that under the same conditions of
temperatures and compression ratio the ideal Carnot engine has the same efficiency
however a significantly lower net work output per cycle than the Ideal Stirling
cycle engine, as can be easily seen in the following diagram:
When the reversible engine is operated in reverse it becomes a heat pump or a
refrigerator. The coefficient of Performance of these machines is developed as
follows:
__________________________________________________________________
______
Solved Problem 5.1 - Reversible Home Air Conditioner and Hot Water
Heater
__________________________________________________________________
______
Problem 5.2 - A heat pump is used to meet the heating requirements of a
house and maintain it at 20°C. On a day when the outdoor air temperature drops to
-10°C it is estimated that the house looses heat at the rate of 10 kW. Under these
conditions the actual Coefficient of Performance (COPHP) of the heat pump is 2.5.





a) Draw a diagram representing the heat pump system showing the flow of
energy and the temperatures, and determine:
b) the actual power consumed by the heat pump [4 kW]
c) the power that would be consumed by a reversible heat pump under these
conditions [1.02 kW]
d) the power that would be consumed by an electric resistance heater under
these conditions [10 kW]
e) Comparing the actual heat pump to the reversible heat pump determine if
the performance of the actual heat pump is feasible,
Derive all equations used starting from the basic definition of COPHP.
__________________________________________________________________
______
Problem 5.3 - During an experiment conducted in senior lab at 25°C, a
student measured that a Stirling cycle refrigerator that draws 250W of power has
removed 1000kJ of heat from the refrigerated space maintained at -30°C. The
running time of the refrigerator during the experiment was 20min. Draw a diagram
representing the refrigerator system showing the flow of energy and the
temperatures, and determine if these measurements are reasonable [COPR = 3.33,
COPR,rev = 4.42, ratio COPR/COPR,rev = 75% > 60% - not feasible]. State the
reasons for your conclusions. Derive all equations used starting from the basic
definition of the Coefficient of Performance of a refrigerator (COPR).
__________________________________________________________________
______
On to Chapter 6: Entropy - A New Property
___________________________________________________________________________
___________
Engineering Thermodynamics by Israel Urieli is licensed under a Creative
Commons Attribution-Noncommercial-Share Alike 3.0 United States License
Module 4. Second law of thermodynamics
Lesson 8
ENTROPY AND AVAILABILITY
8.1 Entropy Change for an Irreversible Process in a Closed System
First of all, it may be made clear here that, entropy is a property and any change in it between
two states is always constant irrespective of the type of process between these states.
Let us consider a reversible cycle 1A2B1 formed by two reversible processes A and B
between state 1 & 2 and an irreversible cycle 1A2C1 formed by one reversible process A and
other irreversible process C between states 1 & 2 as shown in Fig 8.1.
Fig. 8.1 Irreversible cycle
Applying Classius equality & inequality to both cycle (1A2B1)rev & (1A2C1)irr
� �����������������������������������������
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��������������������������������� ���������
�������������������������������������� ����
����������� ��� (i)
�����������
0� ����������������������������������������
������������������������������������������
������������� �����������������������������
���������������������������������� ��������
����������������������� ��������������� ���
��������� (ii)
From (i) and (ii)
It is due to �ve sign otherwise the numerical value of
is high.
Now
Thus for the irreversible process C, change in entropy
So even if Q =0 for irreversible process, the entropy will always increase. It can be taken
mathematically as
Where, δI is the rise in entropy due to irreversibility factor
By integrating the total change in entropy during an irreversible process is
������������������������������������������
������������������������������������������
����������� �������������������������������
�������������������������������� ����������
������������������������������������� ���
(Eq. 8.1)
The effect of this irreversible term �I� is always to increase some entropy whether the heat
is added or rejected or not exchanged.
Conclusion
For an Irreversible Process
1. If heat is added, entropy increases due to heat addition and irreversibility.
2. If there is no heat transfer, the entropy still increases due to irreversibility.
3. If heat is rejected, entropy decreases due to heat rejection and increases due to
irreversibly. Net effect may be +ve or �ve.
General mathematical statement for change in entropy between two states is
�����������
��������������������������������
���������������������������������
������������� �������������������
������������������ ��������� �����
���� ������������������ ����������
���� (Eq. 8.2)
Where
�=� sign is for Reversible process between the given states.
�>� Sign is for Irreversible Process between the given states.
It can be understood more clearly with the help of fig. 8.2 as given below:
Fig. 8.2 Relationship of change in entropy with heat exchange during reversible and
irreversible processes
8.2 Carnot Cycle on Temperature- Entropy Chart
As the entropy is a property, it can be tabulated for a pure substance as a function of two
other independent properties.
Also, as it is a property it can be plotted on a chart with another property. Most common
charts that are used in thermodynamics are T-s charts. Area under the curve on T-s chart
gives directly the heat added or rejected for a reversible process.
Thus Carnot cycle can also be represented more conveniently on a T-s chart because it is
comprised of two isothermal (constant temperature) processes and two reversible adiabatic
(constant entropy) processes. It is shown in Fig. 8.3 �as 1-2-3-4. Four processes of Carnot
cycle are as follows:
Process 1-2: Isothermal heat absorption. Heat absorption is given by area under line 1-2.
Process 2-3: Reversible adiabatic expansion Process, δQ = 0, so dS = 0
Process 3-4: Isothermal Heat Rejection. Heat rejection is given by area under line 3-4.
Process 4-1: Reversible adiabatic compression, δQ =0, so dS = 0
Heat Absorption, Qh = Th∆S
(Area under line 1-2)
Heat Rejection, Qc = Tc∆S
(Area under line 3-4)
So, Net work of cycle, W = Qh � Qc = (Th-Tc) ∆S (Area covered by cycle 1-2-3-4 on T-S
chart)
Thermal efficiency of Carnot cycle is given as
Thus it is easier to draw, analyze the Carnot cycle and find its thermal efficiency on a T-S
chart.
8.3 Availability
It is the work potential of any source of energy at some specified state. It is also referred to
as exergy. It is represented as available energy, which is the maximum possible work that can
be extracted from a heat source at temperature, Th by introducing a reversible heat engine
taking heat from it and rejecting remaining heat to surroundings at temperature, Tc. (Fig. 8.4)
If Q is the amount of heat available with heat reservoir at temp Th, then
Available Energy,
A.E = (
.
And Unavailable Energy, U.E = Q - A.E.
From fig 8.3
Area 1- 2- S1- S2, = Qh = Heat absorbed
Area 1-2-3-4 = W = Available Energy (A.E.)
Area----3- 4- S1- S2 = QC = Unavailable Energy (U.E.)
Thus,
�����������
������������������������������������������
������������������������������������������
�������� ���������������������������������
�������������� ����������������������� ����
������������������������������� ����������
� ����������������������� ��� (Eq. 8.3)
���������������
� �����������������������������������������
������������������������������������������
��������������������������������� ���������
������������������������������������������
���������������������������� ��������������
����������������� ��� (Eq. 8.4)
If Th increases A.E increases & if Th decreases, A.E decreases. (Fig.8.5)
8.3.1 Real conditions affecting the availability
Condition-1: If the temperature of heat source does not remain same while supplying a given
amount of heat, Qh to the system, then the available energy reduces as shown on T-S diagram
in figure 8.4
Fig. 8.6 Reduction in A.E. due to decrease in temperature of heat source
If, we apply a number of reversible engines, each taking heat, δQ from the source at constant
temperature T but which is different for each engine and vary between Ta to Tb for different
engines. Then the A.E can be determined by integrating δQ � TC.dS between the initial and
final temperatures of heat sources.
Where, m & C are the mass and specific heat of matter working as heat source.
It will be clear by solving the numerical problem given below.
Total energy supplied by heat source, Q is shown by area �a-b-e-f� on T-S chart shown in
fig 8.6.
Available Energy, A.E. is shown by area a-b-c-d on T-S chart shown in fig 8.6
Unavailable Energy, U.E. is shown by area c-d-e-f on T-S chart shown in fig 8.6
Example: Estimate the A.E contained in 1 Kg of water at 1000C and 1 atm pressure with
respect to surroundings at 300C.
Solution: Let the water be cooled from 1000C (373K) to 300C (303K) while supplying heat
in a number of steps to reversible heat engines, operating in series and rejecting heat to
surroundings at temperature 30�C. So temperature, T of heat source (water) is varying. And
temperature, TC of surroundings (heat sink) is 30�C.
Then available energy of a reversible heat engine absorbing heat δQ from heat source at
temperature, T and rejecting at Tc is given by
Therefore total maximum work, which can be available from water cooled from 1000C to
300C will be
�
������
�����������
����������
Thus in this case available energy is only about 10% of total energy.
Condition-2
In real conditions, heat is transferred from heat source to working substance only when there
is some finite temperature difference between both. So, there is loss of available energy due
to heat transfer through a finite temperature difference.
If there is no temperature difference between heat reservoir and working substance of heat
engine and heat given by reservoir at temperature Th is Q when the surrounding is at
temperature,� Tc then,
But, if there is finite temperature difference, such that working substance is at
temperature
lower than Th of heat source then for the same heat Q given to heat engine,
the entropy change of working substance in heat engine will be more as shown in Fig. 8.7.
As Q = Th ∆S = Th′ ∆S′
So if Th′ < Th then ∆S′ > ∆S
And the available energy, A.E., in case the heat engine absorbs heat at temperature
will be
Loss in A.E
�
For the same heat absorbed,
�����������
��������� (Shown by
shaded region in Fig. 8.7)
Example: In a boiler for generating steam, water is evaporated at 200oC while the
combustion gases are cooled from 1350oC to 350oC. The specific heat of gases at constant
pressure is 1.1 kJ/ kg K. The surroundings are at 30oC. Determine the loss in A.E.
Solution: Had the heat transfer from hot gases to water been reversible and water/steam
would have been heated from �b� to �a� and expanded from �a� to �3� as shown in
fig 8.6 then the resulting U.E would have been as shown by area:
���
Fig. 8.8 Loss in A. E. and Increase in U. E.
But in actual the water is being heated at 2000C from 1 to 2.
So, Heat Rejected by hot gases (Area a-b-c-d) = Heat absorbed by water (Area 1-2-e-c)
Due to decrease in temperature of hot gases, there is direct loss in A.E shown by shaded area
a-0-b.
Due to the finite temperature difference between the gases and water, there will be loss in
A.E or gain in U.E shown by shaded area
To calculate this, we require finding the difference between entropy change of gases and
water.
Entropy increase for 1 kg of water
Where, hfg = Latent heat of water at saturation temperature of 200o C
By doing energy balance in boiler, mass of hot gases used per kg of water can be calculated
as:
mg (1.1). (1350-350) = 1938.50
or mgas = 1.762 kg / kg of water
Entropy decrease of gases
�����������
�����������
Difference in entropy change of water and hot gases, ∆S = 4.1 � 1.8557 = 2.244 kJ/kg K
= 303x 2.244 = 679.93 kJ / kg of water
So, due to temperature difference, increase in Un-available Energy
���� = 303x 2.338
������ ��= 708.414
kJ/ kgK (Ans)
Module 4. Second law of thermodynamics
Different forms of energy and are convertible in to each other. During this conversion, law of
conservation of energy i.e. first law of thermodynamics is followed. But this law has
limitation in depicting the fraction of heat energy of a system or supplied to system which can
be converted to work. Also it does not specify the conditions under which conversion of heat
in to work is possible. Second law of thermodynamics removes this limitation and tells under
what conditions, in what direction of heat flow and how much of it can maximum be
converted in to work.
7.2 Statements of Second Law of Thermodynamics
7.2.1 Kelvin-Planck statement
�It is impossible to construct a thermodynamic system or device which operates in a cycle
and produce no effect other than the production of work by exchange of heat with a single
reservoir�. Or in simple terms it states that all the heat from a single heat reservoir cannot
be converted to work.
In detail, the meaning of statement is that there is no such device possible which can
continuously take heat from heat reservoir on one side and convert all of it into work on the
other side. But only a part of heat energy while flowing from high temperature reservoir to
low temperature reservoir can be converted to work and the remaining part must be rejected
to low temperature reservoir i.e. atmosphere. Therefore only a part of heat energy while in
transition from high temperature to low temperature is possible to be converted in to work.
7.2.2 Classius statement
This statement is regarding the conversion of work in to heat and it states that (�it is
impossible to construct a thermodynamic system or device which, while operating in cycle
(i.e. working continuously), transfers heat from low temperature reservoir to high
temperature reservoir without taking help or absorbing work from some external agency.)
In detail, the meaning of statement is that heat can be made to flow from low temperature to
high temperature only by applying external work.
7.3 Heat Engine
It is a thermodynamic system or device which can continuously convert heat energy into
work energy or we can say thermal energy in to mechanical energy. We know that to work
continuously, anything has to operate in a cycle. Therefore heat engine is also a
thermodynamic device operating in a cycle. The performance of a heat engine is measured in
terms of its thermal efficiency which is the ratio of work output to heat absorbed by engine,
i.e.
Where W = Rate of mechanical work done by engine
Q = Heat absorbed by engine or rate of heat supplied to engine.
7.3.1 Reversible heat engine
A heat engine which operates through a reversible cycle is called Reversible Heat Engine. As
per second law of thermodynamics, heat engine absorbs heat Qh from a high temperature
source, converts a part of it into Mechanical work �W� and rejects the remaining part of
heat �Qc� to a low temperature heat sink as shown in Fig. 7.1. If the heat engine is
reversible i.e. all its processes are reversible, then it can be operated on the reverse cycle in
reverse direction with the same performance. Then it will start taking heat �Qc� back from
low temperature heat source by absorbing same amount of work �W� from some external
agency and reject the sum of heat absorbed and work absorbed in the form of heat �Qh� to
high temperature heat source. This reversed heat engine will be called heat pump as shown
in Fig. 7.1. If this heat pump is used for the purpose of extracting heat from a low
temperature body, it is called a refrigerator.
7.3.2 Corollary of 2nd law of thermodynamics
No heat engine operating between two heat reservoirs, always operating at constant
temperature, can be more efficient than a reversible heat engine operating between the same
temperature limits. Also all types of reversible heat engines operating between same
temperature limits will have the same efficiency. It can also be proved with a simple logic.
Let us say, there is an irreversible engine having more efficiency than that of a reversible
engine operating between same temperature limits. Let irreversible engine produces work
�W�irr and reversible engine produces work Wrev (such that Wirr>Wrev) by absorbing heat
Qh from heat source at temperature Th and rejecting heat Qc to heat reservoir at temperature
Tc. Now if we operate reversible engine in reverse direction like a heat pump taking work
�Wrev� from the work produced by irreversible engine and absorbing heat Qc back from
heat reservoir at temperature Tc and rejecting back heat Qh to heat reservoir at temperature
Th as shown in fig 7.2. We will find that a net positive work, W = Wirr - Wrev should be
produced continuously without any effect or any net heat exchanged with reservoirs which is
completely opposite to the law of conservation of energy i.e. energy cannot be produced
without its expenditure.
Fig. 7.2 Reversible and irreversible heat engines
In this way, our assumption that an irreversible engine is more efficient than a reversible
engine is totally wrong. Hence
Wrev > Wirrev
One more fact about a reversible heat engine is that it does not exist in reality. But the idea of
reversible heat engine is completely hypothetical in which the heat exchange process is
thought as reversible without any change in temperature. Otherwise heat exchanged in a
medium is irreversible and taken as Q = m.C.ΔT,
m = mass
C = Specific heat
ΔT = temperature change.
When some heat flows from a high temperature body to low temperature body, change in
their temperatures occurs i.e. hot body becomes somewhat cool and cool body becomes
somewhat hot, but now this heat cannot come back from cold body to hot body i.e. it is an
irreversible process.
So, with ΔT = 0 i.e. heat exchanged without change in temperature can only be visualized in
a way that the heat reservoirs and working medium in the reversible heat engine, which is
exchanging heat with heat reservoirs, both are of infinite heat capacity and there is no change
in their temperature and the amount of heat exchanged depends on the absolute temperature
of reservoirs at which heat is being exchanged i.e. Q ∝ T. So in a reversible heat exchange
process happening at constant temperature, heat exchanged is proportional to absolute
temperature. This is also named as CLASSIUS statement.
7.4 Carnot Cycle/Carnot Engine
Carnot suggested a reversible cycle comprising of two reversible isothermal heat exchange
processes and two reversible adiabatic expansion/compression processes as shown on P-V
and T-S charts in Fig 7.3 (a)
Fig. 7.3 (b) Carnot cycle
Carnot Engine is the reversible heat engine working on Carnot cycle 1-2-3-4 as explained
below:
Process 1-2: Reversible isothermal heat addition: Heat, Qh is transferred to the working
substance from the high temperature reservoir at temperature Th= T1 =T2. The heat transfer
is reversible and isothermal. Expansion of gas takes places i.e. heat energy is converted to
work but the internal energy of system remains constant.
Process 2-3: Reversible adiabatic expansion: During the expansion process, the system is
thermally insulated so that Q = 0. The temperature of the working substance decreases from
high temperature, Th to low temperature, Tc = T3 = T4. Expansion of gas takes place at the
expense of its own internal energy.
Process 3-4: Reversible isothermal heat rejection: Heat Qc is transferred from the working
substance (gas) to low temperature heat reservoir (sink) at constant temperature �Tc�.
Heat transfer is isothermal & reversible. Gas is compressed by spending of external work
and equivalent heat to this work is rejected to heat sink. Internal energy remains constant.
Process 4-1: Reversible adiabatic compression: During the compression process, the
system is thermally insulated, so Q =0. Temperature of working substance increases from
Tc to Th. So internal energy of the system increases by an equal amount to the compression
work done on the system.
7.4.1 Carnot efficiency
An engine operating on the Carnot cycle has maximum efficiency.
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���������������� ����������������������
������������ ��� (Eq. 7.1)
Now,����������� Qh ∝Th, Absolute temperature of hot reservoir
Qc ∝ Tc, Absolute temperature of cold reservoir
Thus,
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������������������������� ����������� �
��������������������������������� ��� (Eq.
7.2)
Thus, the Carnot efficiency does not depend on the type of working substance but only on
the absolute temperature of hot & cold reservoirs.
7.5 Classius Equality & Inequality
As per corollary of 2nd law of thermodynamics all the reversible engines operating between
two heat reservoirs at the same temperature, have the same efficiency, irrespective of the
working substance used. That means between the temperature th & tc the value of Qh & Qc are
always same for any reversible engine.
The function �f� proposed by Kelvin and subsequently accepted is given by the simple
relation.
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���������������������������������
���������������������������������
���������� ��� (i)
Where �C� is a constant and Th & Tc are the absolute or thermodynamic temperatures on
thermodynamic temperature scale.
Or
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or
It is called Classius equality for a reversible cycle.
In case of an irreversible cycle for a heat engine,
Wirrev < Wrev
Or������� Qh - Qirr < Qh � Qrev
(for the same amount of heat absorbed, Qh in both cases)
Or����� Qc irr > Qc rev
Or ������
,
which is called Classius inequality.
Thus Classius Equality and In- equality statements can be combined in mathematical terms
as:
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��������������������������������������
���������������������� (Eq. 7.3)
7.6 Concept of Entropy
As the first law of thermodynamics introduces a property named as internal energy. Second
law of thermodynamics, when applied to a process, introduces the property named as entropy,
which is of extensive type.
The physical significance of entropy is somewhat difficult to imagine, but if we start from the
very basic, it will definitely give an idea of entropy and also its importance in thermodynamic
calculations.
Now consider a reversible cycle �1A2B1� having two reversible processes A & B between
the states 1 & 2 and another reversible cycle �1A2C1� having two reversible processes A
and C between the same states 1 & 2, as shown in fig 7.4.
Fig. 7.4 Reversible cycles
Applying Classius equality to reversible cycles 1A2B1 and 1A2C1
From the above two equations:
Thus, we see that
for any reversible process between state 1 & 2 is same i.e. independent
of path followed B or C or any other path (process) and depends only on states 1 & 2. Thus it
is a point function and hence is a property called entropy (S), such that a change in this
property,
� Hence change in entropy during a process 1-2 is given as:
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������������������������� �������������
��������������������������������������
�������� ����������������������� ��� (Eq. 7.4)
Thus, entropy of a system may be defined as a property such that change in it from one state
to other is always equal to integral of heat exchanged divided by absolute temperature, at
which heat is exchanged, during any reversible process between the states.
7.6.1 Specific entropy
Entropy per unit mass is called specific entropy. It is denoted by small letter �s�.
Conclusion: For a reversible process in a closed system
a) Entropy increases if heat is added
b) Entropy decrease if heat is rejected
c) Entropy remains constant if there is no heat transfer.