Thermodynamics I
Spring 1432/1433H (2011/2012H)
Saturday, Wednesday 8:00am - 10:00am &
Monday 8:00am - 9:00am
MEP 261 Class ZA
Dr. Walid A. Aissa
Associate Professor, Mech. Engg. Dept.
Faculty of Engineering at Rabigh, KAU, KSA
Chapter #1
September XX, 2011
Announcements:
Dr. Walid’s e-mail and Office Hours
[email protected]
Office hours for Thermo 01 will be every
Sunday and Tuesday from 9:00 – 12:00 am
in Dr. Walid’s office (Room 5-213).
Text book:
Thermodynamics An Engineering Approach
Yunus A. Cengel & Michael A. Boles
7th Edition, McGraw-Hill Companies,
ISBN-978-0-07-352932-5, 2008
Chapter 1
INTRODUCTION AND BASIC
CONCEPTS
Objectives of CH1: To
• Identify unique vocabulary associated with
thermodynamics.
• Review metric SI units.
• Explain basic concepts of thermodynamics
(system, state, state postulate, equilibrium,
process, cycle).
• Review concepts of temperature, temperature
scales, pressure, and absolute and gage
pressure.
Engineering sciences
1) Thermodynamics
…….
Heat & mass transfer
Fluid Mechanics
Chapter 1
INTRODUCTION AND BASIC CONCEPTS
1–1 ■ THERMODYNAMICS AND ENERGY
The name thermodynamics stems from the Greek
words therme (heat) and dynamis (power),
One of the most fundamental laws is the
conservation of energy principle. It simply states
that during an interaction, energy cannot be
created or destroyed.
Fig. 1–1 ■ Energy cannot be created or
destroyed
Change in energy content of a body (or any other
System) is equal to difference between energy input
and energy output, energy balance is expressed as
Ein - Eout = ∆E.
First law of thermodynamics is simply expression of
conservation of energy principle, and it asserts that
energy is a thermodynamic property.
Second law of thermodynamics asserts that
energy has quality as well as quantity, and actual
processes occur in the direction of decreasing quality
of energy. For example, a cup of hot coffee left on a
table eventually cools, but a cup of cool coffee in the
same room never gets hot by itself .
1–2 ■ Units
Primary (Fundamental) (Main) (base) quantities: SI Units
S
Quantity
Designation
Unit
Symbol
1
mass
m
kilogram
kg
2
length
L
meter
m
3
time
t
second
s
4
temperature
T
Kelvin
K
Secondary (Derived) quantities : SI Units
s
Quantity
Designation& Equation
Unit
1
velocity
V = L/t
m/s
2
Acceleration
a = L/t2
m/s2
3
Volume
V = L3
m3
* g (Gravitational acceleration) = 9.807 m/s2 ≈ 9.81 m/s2
SI Unit Prefixes
Factor Prefix Symbol Factor Prefix
Symbol
1012
tera
T
10-1
deci
d
109
giga
G
10-2
centi
C
106
mega
M
10-3
milli
m
103
kilo
k
10-6
micro
µ
102
hecto
h
10-9
nano
n
101
deka
da
10-12
pico
p
Secondary quantities : SI Units: (Continued)
Force
s
Quantity
4 Force
Designation& Equation
SI Symbol
F=ma
Kg m/s2
1N = Kg m/s2 , Newton 1 kgf = 9.807 N
Weight is gravitational force applied to
a body, its magnitude is determined
from Newton’s second law
W= mg
(N)
(1-2)
Mass of a body; m remains the same
regardless of its location in universe.
However, Its weight; W , changes with
change in gravitational acceleration, g.
For a body on sea level ( g = 9.807
m/s2 ) having a mass; m of 1 kg, Its
weight; W = 1 kg× 9.807 m/s2 =9.807
kg m/s2 =9.807 N=1 kgf.
Work, which is a form of energy, can
simply be defined as force times
distance. Therefore, it has the unit
“Newton-meter (N · m),” which is called
a joule (J).
1J= 1
(1-3)
A more common unit for energy in SI is
the kilojoule (1 kJ = 103 J).
Another energy unit is calorie Cal).
1 Cal = 4.1868 J
Dimensional Homogeneity
In engineering, all equations
must be dimensionally homogeneous.
EXAMPLE 1–1 Spotting Errors from
Unit Inconsistencies.
While solving a problem, a person
ended up with the following equation
at some stage:
E = 25 kJ + 7 kJ/kg
Where, E is the total energy and has
the unit of kilojoules. Determine how
to correct the error and discuss what
may have caused it.
Solution: The two terms on RHS do
not have the same units. Therefore,
they cannot be added to obtain the
total energy.
Multiplying ast term by mass will
eliminate the kilograms in the
denominator. The whole Eq. will
become dimensionally homogeneous;
i.e., every term in the Eq. will have the
same unit.
Obviously this error was caused by
forgetting to multiply the last term
by mass at an earlier stage.
EXAMPLE 1–3 What is the weight
of a mass of 454 grams in N.
Solution
W = m g = (454/1000) kg ∗ 9.807 m/s2
= 4.45 N.
1–3 ■ SYSTEMS AND CONTROL
VOLUMES
A system is defined as a
quantity of matter (or a
region in space) chosen
for study.
The mass (or region)outside the system
is called the surroundings.
The real (or imaginary) surface that
separates the system from its
surroundings is called the boundary.
System
Closed system
(Control mass)
Open system
(Control
volume)
Closed system:
Energy, in the form of
heat or work, can cross
the boundary; and the
volume of a closed system
does not have to be fixed.
If, as a special case, even energy is not
allowed to cross the boundary, that system
is called an isolated system.
A closed system with a moving boundary.
* A large number of engineering problems
involve mass flow in and out of a system and,
therefore, are modeled as control volumes
(CV).
(Open systems)
* A water heater, car radiator, turbine, and
compressor all involve mass flow and should
be analyzed as control volumes (open
systems) instead of as control masses
(closed systems).
* CV can also involve heat and work
interactions (just as a closed system), in
addition to mass interaction
An open
system (CV)
with one inlet
and one exit.
CV with fixed
and moving
boundaries.
Boundaries of CV (called control surfaces)
may be fixed, moving, real, and imaginary.
CV with real
and
imaginary
boundaries.
Boundaries of CV (called control surfaces)
may be fixed, moving, real, and imaginary.
In general, proper choice of the system may
greatly simplify the analysis.
1–4 ■ PROPERTIES OF A SYSTEM
Any C/C of a system is called a property
Intensive
properties {those
independent on
the size or extent
of the system}:
such as pressure
p, temperature T,
density; ρ)
Extensive
properties {those
dependent on the
size or extent of
the system}: such
as mass; m and
volume; V)
1–5 ■ DENSITY, SPECIFIC VOLUME,
SPECIFIC GRAVITY & SPECIFIC WEIGHT
Density; ρ is mass of a unit volume of
a substance
3)
(1-4)
(Kg/m
ρ = m /V
Specific volume; v is volume per unit
mass (reciprocal of density)
v = V /m
(m3/kg)
(1-5)
Specific gravity; SG (relative density)
defined as the ratio of the density of a
substance to the density of water at
4°C, ( ρH2O = 1000 kg/m3)
(1-6)
)at 4°C
Specific weight; γ is weight of a unit
volume of a substance
γ = W /V= mg/V= ρg (N/m3) (1-#1)
EXAMPLE 1–2 Obtaining Formulas
from Unit Considerations.
A tank is filled with oil whose density is
ρ = 850 kg/m3. If the volume of the
tank is V = 2 m3, determine the amount
of mass m in the tank..
Solution:
m=ρV
= (850 kg/m3)∗ (2 m3)
= 1700 kg
1–7 ■ PROCESSES AND CYCLES
Any change that a system undergoes from
one equilibrium state to another is called a
process.
A process between states 1 and 2 and the
process path.
The series of states through which a
system passes during a process is called
the path of the process.
To describe a process completely, one
should
(1) specify the initial and final states of the
process, as well as
(2) the path it follows, and
(3) the interactions with the surroundings.
When a process
proceeds in such a
manner that the
system remains
infinitesimally close
to an equilibrium
state at all times, it
is called a quasistatic, or quasiQuasi-equilibrium and
equilibrium,
non quasi-equilibrium
process.
compression processes
Process diagrams plotted by employing
thermodynamic properties as coordinates r
very useful in visualizing the processes.
Some common properties used as
coordinates r temperature; T, pressure ;p,
and volume; V (or specific volume; v).
p-V diagram of a compression process of a
gas is illustrated.
The P-V
diagram of a
compression
process.
The prefix iso- is often used to designate
a process for which a particular property
remains constant.
An isothermal process, for example, is
a process during which the temperature; T
remains constant.
An isobaric process is a process
during which the pressure; p remains
constant.
An isochoric (or isometric) process is a
process during which the specific
volume; v remains constant.
A system is said to have undergone a
cycle if it returns to its initial state
at the end of the process. i.e.,
for a cycle initial & final states r identical.
&
3
cycle
Final
state
The Steady-Flow Process
The term steady implies no change with
time. The opposite of steady is unsteady,
or transient.
A large number of engineering devices
operate for long periods of time under the
same conditions, and they r classified as
steady-flow devices.
steady-flow (Continued) That is, fluid
properties can change from point to point
within CV, but at any fixed point they
remain the same
during the entire
Process.
#The term uniform, however, implies no
change with location over a specified region.
1–8 ■ TEMPERATURE AND ZEROTH
LAW OF THERMODYNAMICS
When a body is brought into contact with
another body that is at a different
temperature, heat is transferred from the
body at higher temperature to the one at
lower temperature until both bodies attain the
same temperature.
At that point, the heat transfer stops, and the
two bodies r said to have reached thermal
equilibrium. The equality of temperature is
the only requirement for thermal equilibrium.
The zeroth law of thermodynamics
states that if two bodies are in thermal
equilibrium with a third body, they are also
in thermal equilibrium with each other.
# By replacing the third body with a
thermometer, the zeroth law can be
restated as:
two bodies r in thermal equilibrium if
both have the same temperature
reading even if they r not in contact.
Temperature Scales
a) absolute thermodynamic scale: K (SI units)
b) common scale: °C (SI units)
T (K ) = T (°C ) + 273.15
#When we r dealing with temperature
differences ∆T, the temperature interval is
the same.
∆T (K) = ∆T (°C).
# Some thermodynamic relations involve the
temperature T (such as a = bT),and often
the question arises of whether it is in K or °C.
# If the relation involves temperature
differences (such as a = b∆T), it makes no
difference and either (K or °C) can be used.
EXAMPLE 1–4 Expressing Temperature
Rise in Different Units
During a heating process, the temperature of
a system rises by 10°C. Express
this rise in temperature in K.
Solution:
∆T (K) = ∆T (°C) = 10
EXAMPLE 1–4(B) Expressing Temperature
value in Different Units
During a heating process, the temperature of a
system is 10°C. Express the temperature in K,.
Solution:
T (K) = T (°C) +273.15 =10 + 273.15=283.15
1-9 ■ Pressure
Pressure is defined as a normal
force exerted by a fluid per unit
area. We speak of pressure only
when we deal with a gas or a liquid.,
The counterpart of pressure in solids
is normal stress. Since pressure is
defined as force per unit area, it has
the unit of newtons per square meter
(N/m2), which is called a pascal (Pa).
Multiples of Pa, r:
kilopascal (1 kPa = 103 Pa) and
megapascal (1 Mpa = 106 Pa)
r commonly used.
Three other pressure units commonly
used in practice, r :
- bar (bar), - standard atmosphere (atm),
& kilogram-force per square centimeter
(kgf/cm2).
1 bar = 105 Pa = 0.1 MPa = 100 kPa
1 atm =101,325 Pa =101.325 kPa =1.01325 bars
1 kgf /cm2 = 9.807 N/cm2 = 9.807 ×
×104 N/m2
= 9.807 ×
×104 Pa = 0.9807 bar = (0.9807/
1.01325 ) atm = 0.9679 atm
EXAMPLE 1–#1
Calculate p)fat man & p)thin man
If m)fat man =136 kg & m)thin man=68 kg, A)feet= 0.0326 m2
Solution:
68 kg
136 kg
p)thin man =
20.46 kPa
p)fat man = 40.93 kPa
# W)fat man = m g = 136 kg ∗ 9.81
m/s2 = 1334.2 N.
p)fat man = W)fat man / A = 1334.2 N /
3 Pa= 40.93 kPa.
0.0326 m2 = 40.93
# W)thin man = m g = 68 kg ∗ 9.81 m/s2 =
667.1 N.
p)thin man = W)thin man / A = 667.1 N /
3 Pa= 20.46
0.0326 m2 = 20.46
kPa.
Absolute, gage, and vacuum pressures
Pgage
Movable datum
(zero gage)
Pvac
Pabs Patm
Fixed datum
(zero absolute)
Pabs
Pabs = Patm + Pgage
= Patm - Pvac
1 atm = 101,325 Pa = 101.325 kPa =1.01325 bars
=14.7 psi.
EXAMPLE 1–5 Absolute Pressure of a
Vacuum Chamber
A vacuum gage connected to a chamber
reads 5.8 psi at a location where the
atmospheric pressure is 14.5 psi.
Determine the absolute pressure in the
Chamber in psi, atm and bar.
Solution:
Pabs = Patm - Pvac = 14.5 psi - 5.8 psi = 8.7 psi
= (8.7/14.7) atm = 0.5918 atm =
8.7*(1.01325 / 14.7) = 0.6 bar
Variation of Pressure with Depth
Pressure in a fluid at
rest does not change in
the horizontal direction.
This is not the case in
the vertical direction in
a gravity field.
Pressure in a fluid increases with depth
because more fluid rests on deeper layers,
and the effect of this “extra weight” on a
deeper layer is balanced by an increase in
pressure.
To obtain a relation
for the variation of
pressure with depth,
consider a rectangular
fluid element of height
∆z, length ∆x, and unit
depth (into the page)
in equilibrium, as shown
in Figure:
p1
p2
F1
F2
Free-body diagram of a
rectangular
fluid element in equilibrium
1
∆z
F1
∆x
W
Assuming the density
of the fluid ρ to be
constant, Hence
F2
W = m g = [ρ (∆z × ∆x × 1)] g = ρ g ∆x ∆z
F2= p2 (∆x × 1)= p2 *∆x
F1 = p1 (∆x × 1)= p1 *∆x
From Newton’s 2nd law (force balance) in
vertical z-direction gives
+ve
Hence, F2 – (F1 + W)= 0
p2 *∆x – ( p1 *∆x + ρ g ∆x ∆z )= 0
(1-17)
Dividing both sides of Eq. by ∆x, we get
p2 – ( p1 + ρ g ∆z )= 0 Hence,
∆ p = p2 – p1 = ρ g ∆ z = γ ∆ z
(1-18)
From Eq. (1-18), we can conclude:
1) Pressure in a fluid increases linearly with
depth.
2) For a given fluid, the vertical distance ∆z is
sometimes used as a measure of pressure,
and it is called the pressure head.
3) For small to moderate distances, the
variation of pressure with height is negligible
for gases because of their low density.
4) The pressure in a tank containing a gas
e.g., can be considered to be uniform since
the weight of the gas is too small to make a
significant difference. Also, the pressure in a
room filled with air can be assumed to be
constant.
patm
h
patm
γh
p
p= patm + γh, Hence, pgage =p- patm = γh
(1-19)
p+dp
dz
dp= -ρ g dz or
p
dp/dz= - ρ g (1-20)
The -ve sign is due to our taking the +ve z
direction to be upward so that dp is -ve
when dz is +ve (since pressure decreases
in an upward direction).
When the variation of density
with elevation is known the
z
pressure difference between
points 1 and 2 can be
determined by integration to be.
z2
From Eq. (1-20) dp/dz= - ρ g
i.e.
z1
Pascal’s law:
Lifting of a large weight by
a small force by the
application of Pascal’s
law.
Hence,
Hence,
But,
Hence,
1–10 ■ THE MANOMETER
Owing to Pascal’ law
patm
p1=p2=patm+ρgh
p2)gage=p2-patm=ρgh
A
Fluid of density; ρ
The basic manometer
h
Fluid of
density; ρ
W = mg =(ρV) g =(ρAh) g
Hence;;
p2)gage= W /A= ρAh) g /A= ρgh
EXAMPLE 1–6 Measuring Pressure with a
Manometer
A manometer is used to measure the
pressure in a tank. The fluid used has a
specific gravity of 0.85, and the manometer
column height is 55 cm. If the local
atmospheric pressure is 96 kPa, determine
the absolute pressure within the tank
Solution:
p=p2=p1=patm+ρgh
ρ =SG* ρ water=0.85*1000 kg/m3
2
1
ρ =850 kg/m3
Schematic for Example 1–6.
Hence;
P=96 kPa+{[850*9.81*(55/100)]/1000} kPa
P=96 kPa+4.586 kPa = 100.586 kPa
In stacked-up fluid layers
×1
×2
3
p1=patm+ρ1gh1
×
p2=p1+ρ2gh2 =(patm+ρ1gh1)+ ρ2gh2=
patm+ρ1gh1 +ρ2gh2
p3=p2+ρ3gh3=(patm+ρ1gh1)+ρ2gh2 )+ρ3gh3
= patm+ρ1gh1 +ρ2gh2 +ρ3gh3
Differential manometer
Used to measure
pressure differential
pA=pB
P1+ρ1g(h+a) =p2+ρ1ga +ρ2gh
Hence,
manometer
p1- p2 =(ρ1ga +ρ2gh)- ρ1g(h+a) reading
i.e., ∆p=p1- p2 =ρ2gh- ρ1gh=(ρ2-ρ1) gh
Multifluid Manometer
EXAMPLE 1–7 Measuring pressure with
a Multifluid Manometer
The water in a tank is pressurized by
air, and the pressure is measured by
a multifluid manometer. The tank is
located on a mountain at an altitude
of 1400 m where the atmospheric
pressure is 85.6 kPa. Determine the
air pressure in the tank if h1 = 0.1 m,
h2 =0.2 m, and h3 = 0.35 m.
Take the densities of water, oil, and
mercury to be 1000 kg/m3,850 kg/m3,
and 13,600 kg/m3, respectively.
×6
5
4
3
1
Solution:
p3=p1=p2+ρ mercury gh3=patm+ρ mercury gh3 (1-#2)
p3=p4+ρ oil gh2 Hence,
p4=p3-ρ oil gh2 (1-#3)
Combining Eqs. (1-#2) & (1-#3) to get
p4=[patm+ρ mercury gh3]-ρ oil gh2 (1-#4)
p5=p4 =[patm+ρ mercury gh3]-ρ oil gh2 (1-#5)
p6=p5 =[patm+ρ mercury gh3]-ρ oil gh2 (1-#6)
But, p6=pair +ρ water gh1 (1-#7)
Hence, pair =p6 - ρ water gh1 (1-#8)
Combining Eqs. (1-#6) & (1-#8) to get
pair ={[patm+ρ mercury gh3]-ρ oil gh2}-ρwater gh1
Hence,
pair ={[85.6 kPa + (13,600×9.81×
0.35/1000)]-(850× 9.81 × 0.2) /1000}[(1000× 9.81 × 0.1)/1000]
pair =85.6 kPa + 46.696 kPa -1.668 kPa 0.981 kPa = 129.65 kPa ≈ 130 kPa
Other Pressure Measurement Devices
# Bourdon tube to measure gage pressure
# Pressure transducers r used to measure
gage, absolute, and differential pressures
- Strain-gage pressure transducer.
- Piezoelectric transducers, also
called solid-state pressure transducers.
1–11 ■ THE BAROMETER AND
ATMOSPHERIC PRESSURE
patm=pD=pB=0
+ρ Mercury gh
W/A=mg/A=
(ρMercuryV) g/A =
(ρMercuryAh) g/A=
ρ Mercury gh
i.e.,
×D
Patm=ρ Mercury gh
EXAMPLE 1–8 Measuring Atmospheric
Pressure with a Barometer
Determine the atmospheric pressure at a location
where the barometric reading is 740 mm Hg and
the gravitational acceleration is g = 9.81 m/s2.
Assume the temperature of mercury to be 10°C,
at which its density is 13,570 kg/m3..
Solution:
Patm=[(13,570× 9.81×(740mm/1000)]/1000
=98.51 kPa
EXAMPLE 1–9 Effect of Piston Weight on
Pressure in a Cylinder
+ve
The piston of a vertical piston–
cylinder device containing a gas
has a mass of 60 kg and a
cross-sectional area of 0.04 m2.
The local atmospheric pressure
is 0.97 bar, and the gravitational
acceleration is 9.81 m/s2.
(a) Determine the pressure
inside the cylinder.
(b) If some heat is transferred to
the gas and its volume is
doubled, do you expect the
pressure
inside the cylinder to change?
∑Fz = 0
Solution:
+ve
(a) ∑Fz = 0
pA- patmA –W = 0
Hence,
pA= patmA +W i.e., p = patm +(W /A) (1-#7)
patm = 0.97 bar
From Slide 69:
1 atm = 101,325 Pa = 101.325 kPa =1.01325 bars
1 bar = (101.325/ 1.01325) kPa = 100 kPa
patm = 0.97 bar= (0.97 * 100) kPa= 97 kPa
W = m g = (60 × 9.81)/1000 kN= 0.5886 KN
From Eq. (1-#7),
p = patm+(W /A)= 97 kPa+(0.5886 KN/0.04 m2)
i.e., p = 97 kPa+14.715 kPa = 111.715 kPa=
(111.715 /100)=1.11 bar
(b) If the piston moved , then the piston
balance will be the same. Hence, p is the
same.
Homework
Group 1: Thermodynamics : 1–3C.
Group 2: Mass, Force, and Units : 1–5C, 1–7
,1–8, 1–9, 1–12.
Group 3: Systems, Properties, State, and
Processes: 1–15C , 1–16C, 1–18C, 1–20C,
1–23C, 1–24C.
Group 4: Temperature
1–26C , 1–27C, 1–29, 1–31.
Group 5: Pressure, Manometer, and Barometer
1–34C , 1–39C, 1–40, 1–42, 1–43, 1–44, 1–45,
1–48, 1–49, 1–51, 1–52, 1–53, 1–55 , 1–57, 1–
59, 1–61, 1–62, 1–66, 1–67, 1–73, 1–75.
Group 6: Review Problems
1–85 , 1–86, 1–106, 1–107, 1–110.