1
Thermodynamics
Lecture 01:
Introduction to Thermodynamics
Thermodynamics
Definition
-- Thermodynamics is the science of energy and entropy
-- Thermodynamics: the branch of physics dealing with
the relationships between heat, work, and forms of
energy.
From the Greek:
- θερμη, therme, meaning “heat“ → Energy
- δυναμις, dynamis, meaning “power“→ Transport
To the Engineer…Thermodynamics means:
- Understanding the 4 Laws of Thermodynamics
- Learn to work in with four different temperature
scales.
- Learn to balance energy, heat, and work with
respect to open and closed systems.
- Learn about common thermodynamcis devices
and applications and how the principles can be
used to predict system performance and
efficiencies.
Thermo….a Quick Survey
-- Properties of systems:
Temperature, Pressure, Specific Volume,
Phase, Quality, Density, Enthalpy, Entropy
-- Processes and Cycles:
State to state transitions.
Carnot cycle, Rankine cycle, Otto cycle,
Diesel cycle, Brayton cycle, Refrigeration cycle
-- Work-Energy-Heat Balance:
Applying the 1st Law of thermodynamics
Thermo…A Quick Survey…continued
-- Entropy Production and Accounting:
Applying the 2nd Law of thermodynamics
-- Thermodynamics devices:
Turbines, Heat Exchangers, Condensers, Engines
Pumps, Cooling Towers, Compressors, Diffusers
-- Psychrometrics:
HVAC, heating, ventilation, air conditioning,
and humidity
-- Combustion and Power Production:
Chemical energy production and balances
A basic concept: System
• A System is whatever it is that we want to study.
• In thermodynamics, the first step in defining any
problem is to define exactly what is to be
monitored, examined, measured, etc.
Environment
Interactions
System
Boundary
System Definition
Radiation
Radiation
System Definition
Radiation
Radiation
System Definition
Gas
Q
System Definition
Gas
Gas
Q
W
Types of Systems
• Isolated Systems – matter and energy may not
cross the boundary.
• Adiabatic Systems – heat may not cross the
boundary.
• Diathermic Systems - heat may cross boundary.
• Closed Systems – matter may not cross the
boundary.
• Open Systems – heat, work, and matter may cross
the boundary
(more often called a Control Volume (CV)).
Types of Systems
• Isolated Systems – matter and energy may not
cross the boundary.
• Adiabatic Systems – heat must not cross the
boundary.
• Diathermic Systems - heat may cross boundary.
• Closed Systems – matter may not cross the
boundary.
• Open Systems – heat, work, and matter may cross
the boundary
(more often called a Control Volume (CV)).
Closed System
• Matter (m) may not cross the boundary.
• Heat (Q) and Work (W) may cross the boundary
which change the energy (ΔE) of the system.
Po, Vo, To
Combustion
CxHy+A O2  B CO2 + C H2O
Pf, Vf, Tf
Q
∆V →W
Control Volume / Open System
• Heat (Q), work (W), and matter (m) may cross
the boundary
Ga
s
Q
System Properties
• The State of a system is defined by its properties (T, V,
P, E, ρ, v, u, h, s)
• Extensive properties – Depends on mass
• Intensive properties – Does not depend on mass.
•
Give some examples of
Extensive Properties
Intensive properties
Process vs. Cycle.
• A Process is when properties of the system
undergo a change. A process moves from one
“state” to another “state”.
• If State i = State f then system is said to be
Steady State.
• If a Process undergo changes that eventually
bring it back to the original state, these
transitions together are called a Cycle.
Process:
Cycle:
P
P
S1
S2
S2
S1
S4
S3
v
v
Unit Review: Weight and Mass
Weight
Mass
U.S. Customary
pound , lbf
pound of mass, lbm
or
slug
Metric
Newton, N
kilogram, kg
W =m g
where
g = 9.81 m/s2
g = 32.2 ft/s2
1 N = 0.2248 lbf
1 lbf = 4.4482 N
1 kg = 2.205 lbm
1 lbm = 0.4536 kg
1 slug = 32.2 lbm = 71.0 kg
Relationships:
Conversions:
Unit review: Density and Specific Volume
Density
ρ
mass basis
U.S.
Customary
[lbm / ft3 ]
molar basis
mass basis
Specific Volume
v or v
[ ft3 /lbm]
[ ft3 /mol]
[kg/m3]
[ m3/kg]
Metric
molar basis
[m3/mol]
Relationships
m = mass
V = Volume

m
V
M = molecular weight
n = number of moles
n
m
M

1

 M
Unit review: Pressure
Pressure =
U.S. Customary
Presure head
[psi ] or [lbf /in2 ]
[psf] or [lbf /ft2 ]
[in of Hg ]
[ in of H20 ]
[ft of H20]
Metric
[N/m2]
[ Pa]
[bar]
[mm of Hg ]
[ mm of H20 ]
[cm of H20]
Other
atm
Relationships:
1 atm = 14.69 psi = 1.01325 bar
= 100 324 Pa = 760 mm of Hg
= 29.92 in of Hg = 33.96 ft of H2O
Gage vs. Absolute
pressure
pabs = patm + pgage
Vacuum vs. absolute
pressure
pabs = patm - pvac
Unit review: Temperature
Metric
U.S.
Relative
Temperature
Celcius
[ oC ]
Fahrenheit
[ o F]
Absolute
Temperature
Kelvin
[K]
Rankine
[ oR ]
Relationships
K = oC + 273.15
oR = oF + 459.67
oC =( oF - 32)/1.8
oF= 1.8 oC + 32
oR = 1.8 K
Example 1: An object whose mass is 35 lb is subjected to an
applied upward force of 15 lbf. The only other force acting
on the object is the force of gravity. Determine the net
acceleration of the object in ft/s2 assuming the acceleration
of gravity is constant (g = 32.2 ft/s2). Is the net acceleration
up or down?
FAF= 15 lbf
mass= 35 lbm
Fg= 15 lbf
Example 2: A system consists of N2 in a piston-cylinder
assembly, initially at P1 = 20 psi, and occupying a volume of
2.5 ft3. The N2 is compressed to P2 = 100 psi and a final
volume of 1.5 ft3. During the process, the relationship
between P and V is linear. Determine the P in psi at an
intermediate state where the volume is 2.1 ft3 and sketch
the process on a graph of P vs V.
P1= 20 psi
V1 =2.5 ft3
P2= 100 psi
V2 =1.5 ft3
Example 3: A monometer is attached to a tank of gas in which the pressure is
104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3.
If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate
(a) The difference in mercury levels in the manometer, in cm.
(b) The gage pressure of the gas in kPa and bar
(c) The absolute pressure of the gas kPa, atm, and psi
Chapter 2
Lecture 02:
Work and Energy
Sec 2.1: Reviewing Mechanical Concepts of Energy
Work, Heat, and Energy
Energy is conserved, but can be converted to different types
Ways to Transfer Energy Into or Out of A System
Work – transfers by applying a force and causing a
displacement of the point of application of the force.
Mechanical Waves – allow a disturbance to propagate
through a medium.
Heat – is driven by a temperature difference between two
regions in space.
Matter Transfer – matter physically crosses the boundary
of the system, carrying energy with it.
Electrical Transmission – transfer is by electric current.
Electromagnetic Radiation – energy is transferred by
electromagnetic waves
W = PE = KE = U
26
Sec 2.2: Broadening Our Understanding of Work
Thermodynamics “Work”
Physics definition of work is W = F s
But, in thermodynamics often we are working with
fluids (non-solids), so we need a broader definition.
“Work is done by a system on
its surroundings if the sole
effect on everything external
to the system could have
been done by raising (or
dropping) a weight.”
27
Sec 2.2: Broadening Our Understanding of Work
Joule’s Experiment (1845-Salford, England)
Joule dropped a known mass and measured the change in
temperature of the water.
The experiment was conducted in the basement of his family’s
brewery, where there was a constant ambient temperature.
The friction of the water molecules rubbing together caused the
temperature to increase.
Joule’s Equipment - Manchester
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Sec 2.2.1: Sign Convention
29
Work Sign Convention
W > 0 : Work done BY the system
W < 0 : Work done ON the system
Sign is not inherently important, but this is the convention.
Vf
W 
Vi
W < 0 : Work done ON the system
(System is compressed)
PdV
W > 0 : Work done BY the system
(System expands)
Sec 2.2 .2: Power
Power = Rate of Energy Transfer
Book’s convention: Dot above symbol represents the rate.
The rate of work can be expressed as
30
Sec 2.2 : Work
31
Work Properties
Work is NOT a property of a system like V, T, or E.
Work occurs when the system undergoes a process.
A differential of a property is exact.
V  
Vf
Vi
dV  V f  Vi
The differential of work depends upon the path.
W 
Vf
V
Vi
pdV
Sec 2.2 : Work
32
But, work depends on the process.
For “Bobby” work
depends on the path,
since friction is a nonconservative force.
Sec 2.2 : Work
So, we need to have a PV relationship for
the process.
The process of changing volume is NOT
necessarily in equilibrium.
- He balloon popping, gas does not
instantly mix with air
- Gas cylinder rupture, pressure inside is
higher then outside for some time, t
For this class, we will used an idealized process, that are
completely reversible. We call this type of process
- quasi-equilbirium
- quasi-static
33
Sec 2.2 : Work
34
V
V
V
Non-quasi-static Process
Consider a box initially divided in half.
- Initially, one is filled with gas, the other a vacuum.
- The divider is then removed.
- The gas takes some time to fill the new volume.
During that time, there are different local values for P
in the volume. There is also likely some heat
generated, as the process is irreversible.
Thermodynamics ≠ Kinetics
Sec 2.2 : Work
35
V
V
V
V
V
V
V
V
Quasi-static
Now we move the wall slowly, such that the gas is able to
adjust instantly. This is a reversible quasi-static process.
36
V
Example (2.34): Air contained within a piston-cylinder assembly
undergoes three processes in series. Evaluate W.
Process 1-2: Compression at constant
pressure from p1=10 psi, V1=4.0 ft3 to state 2
Process 2-3: Constant volume heating to state 3, where p3=50 psi
Process 3-1: Expansion to the initial state, during which the p-V
relationship is pV = constant.
50
P psi 
10
 
V ft
3
4
37
Sec 2.3: Broadening Our Understanding of Energy
38
Energy
Physics energy types
Kinetic Energy: Energy of objects in motion
Potential Energy: Energy of objects in a field (g,E,B)
Internal Energy
Spring
Chemical
Pressure
Pressure can be a form of energy if P> Patm
Thus, the general energy
equation is
E  PE  KE  U
P
Patm
39
Example Problem (2.37) A 10 V battery supplies a constant current of
0.5 amp to a resistance for 30 min.
a) Determine the resistance, in ohms.
b) For the battery, determine the amount
of energy transfer to work, in kJ.
Solution:
40
Example Problem (2.31)
Air contained within a piston-cylinder assembly is slowly heated. As
shown in Fig P2.31, during the process the pressure first varies linearly
with volume and then remains constant. Determine the total work in
kJ.
P (kPa)
Solution:
2
150
100
3
1
50
0.030 0.045
0.070
V (m3)
Chapter 2:
Lecture 03:
1st Law of Thermodynamics
and Introduction to Heat Transfer
Sec 2.4: Energy Transfer by Heat
Heat, Q: An interaction which causes a change in energy
due to differences in Temperature.
Heat Flow Rate, Q: the rate at which heat flows into or
out of a system, dQ/dt.
Heat flux, q: the heat flow rate per unit surface area.
Q   Q dt
42
Sec 2.4.2: Heat Transfer Modes
43
3 Types of Heat Transfer
Conduction
Radiation
Convection
Conduction:
Heat transfer through a stationary media due to collision
of atomic particles passing momentum from molecule to
molecule.
Fourier’s Law
dT
Q   A
dx
where κ is the thermal conductivity of the material.
Sec 2.4.2: Heat Transfer Modes
44
Types of Heat Transfer
Convection:
Heat transfer due to movement of
matter (fluids). Molecules carry
away kinetic energy with them as
a fluid mixes.
Newton’s Law of Cooling:
Q  hc A Tb  T f

hc = coefficient of convection
(An empirical value, that depends on the material, the velocity, etc.)
Sec 2.4.2: Heat Transfer Modes
45
Types of Heat Transfer
Radiation :
Heat transfer which occurs as
matter exchanges
Electromagnetic radiation with
other matter.
Stefan-Boltzmann Law:
Q   ATb4
Tb = absolute surface temperature
ε = emissivity of the surface
σ = Stefan-Boltzmann constant
Sec 2.4.2: Heat Transfer Modes
46
Summary of Heat Transfer Methods
Conduction:
where A is area
κ is thermal conductivity
dT/dx is temperature gradient
dT
Q   A
dx
Convection:
Q  hc A Tb  T f
Radiation:
Q   AT
4
b

where A is area
hc is the convection coefficient
Tb -Tf is the difference between the
body and the fluid temp.
where Tb is absolute surface temperature
ε is emissivity of the surface
σ is Stefan-Boltzmann constant
A is surface area
47
Example (2.45): An oven wall consists of a 0.25” layer of steel
(S=8.7 BTU/(hftoR) )and a layer of brick (B=0.42 BTU/(hftoR) ). At
steady state, a temperature decrease of 1.2oF occurs through the steel
layer. Inside the oven, the surface temperature of the steel is 540oF. If the
temperature of the outer wall of the brick must not exceed 105 oF,
determine the minimum thickness of brick needed.
48
49
Example Problem (2.50)
At steady state, a spherical interplanetary electronics laden probe having a
diameter of 0.5 m transfers energy by radiation from its outer surface at a rate
of 150 W. If the probe does not receive radiation from the sun or deep space,
what is the surface temperature in K? Let ε=0.8.
Sec 2.4: Energy Transfer by Heat
50
Recall from yesterday: by convention
Work, W:
W > 0 : Work done BY the system
W < 0 : Work done ON the system
(This is reversed from the sign convention for work often used in
Physics. It is an artifact from engine calculations.)
Heat, Q:
Q > 0 : Heat transferred TO the system
Q < 0 : Heat transferred FROM the system
system
+Q
+W
Sec 2.5: Energy Balance for Closed Systems
51
First Law of Thermodynamics
Energy is conserved
“The change in the internal energy of
a closed system is equal to the sum
of the amount of heat energy
supplied to the system and the
work done on the system”
[
E within
the system
][ ][ ]
=
net Q
input
+ net W
output
Esystem  PE  KE  U  Qin  Wout
dE  Q  W
where  denotes path dependent derivatives
Sec 2.5: Energy Balance for Closed Systems
52
The First Law of Thermodynamics
The 1st Law of Thermodynamics is an expanded form of the
Law of Conservation of Energy, also known by other name an
Energy Balance.
Esystem  Qin  Wout
ΔE
Qin
system
Wout
53
Example (2.55): A mass of 10 kg undergoes a process during with there is
heat transfer from the mass at a rate of 5 kJ per kg, an elevation decrease
of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific
internal energy decreases by 5 kJ/kg and the acceleration of gravity is
constant at 9.7 m/s2. Determine the work for the process, in kJ.
54
Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.
55
56
Example (2.70): A gas is contained in a vertical piston-cylinder assembly
by a piston weighing 1000 lbf and having a face area of 12 in2. The
atmosphere exerts a pressure of 14.7 psi on the top of the piston. An
electrical resistor transfers energy to the
Patm=14.7 psi
gas in the amount of 5 BTU as the elevation
of the piston increases by 2 ft. The piston and
h = 2 ft
cylinder are poor thermal conductors and
Apiston = 12 in2
friction can be neglected. Determine the
Wpiston = 1000 lbf
change in internal energy of the gas, in BTU,
assuming it is the only significant internal
energy change of any component present.
Welec= - 5 BTU
Chapter 2: Sections 6 and 7
Lecture 04:
Energy Analysis of Cycles
Quiz Today?
Sec 2.4: Energy Transfer by Heat
59
Recall past concepts:
Work, W:
W > 0 : Work done BY the system
W < 0 : Work done ON the system
Heat, Q:
Q > 0 : Heat transferred TO the system
Q < 0 : Heat transferred FROM the system
1st Law of Thermodynamics:
Esystem  PE  KE  U  Qin  Wout
system
Qin
ΔE
Wout
Sec 2.2.1: Sign Convention
60
Work done by Gas System
W > 0 : Expansion of Gas
W < 0 : Compression of Gas
Vf
W 
Vi
PdV
Run
Animation
Sec 2.6: Energy Analysis of Cycles
Energy Analysis of Cycles
A cycle is any process that returns to its original state.
Ecycle= Qcycle - Wcycle
therefore: Ecycle= 0
and
Qcycle = Wcycle
61
Sec 2.6: Energy Analysis of Cycles
62
Schematic models of Cycles:
Power cycle
Refrigeration
(Heat pump) Cycle
Sec 2.6.2: Power Cycles
63
Power Cycle
Wcycle = Qcycle = Qin - Qout
Qin : Heat in from hot body
(chemical reaction,
geothermal)
Qout : Heat out to cold body
(surroundings)
Qin > Qout 
Thermal efficiency

Wcycle
Qin
W>0
Qin  Qout
Qout

 1Qin
Qin
when η = 1, efficiency = 100% so want to minimize Qout
Sec 2.6.3: Refrigeration and Heat Pump Cycles
64
Refrigeration Cycle
Wcycle = Qcycle = Qout - Qin
Qout : Heat out to warm body
Qin : Heat in from cold body
W>0
Refrigeration Coefficient
of Performance

Qout > Qin
Qin
Qin


Wcycle Qout  Qin
Sec 2.6.3: Refrigeration and Heat Pump Cycles
65
Heat Pump Cycle
Wcycle = Qcycle = Qout - Qin
Qout : Heat out to warm body
Qin : Heat in from cold body
W>0
Heat Pump Coefficient
of Performance

> Qout > Qin
Qout
Qout


Wcycle Qout  Qin
66
A quick comparison
Refrigeration Cycle
Objective is to do Work to
Remove heat from system.
Qremoved

Wdone by
objective
COP 
effort
Heat Pump Cycle
Objective is to do Work
to Add heat to a system.
Qout

Win
67
Example 1 (Problem 2.76): A gas within a piston-cylinder assembly undergoes
a thermodynamic cycle consisting of three processes:
Process 1-2 : Constant volume, V = 0.028 m3, U2-U1=26.4 kJ
Process 2-3: Expansion with PV = constant, U3 = U2
Process 3-1: Constant pressure, P = 1.4 bar, W31 = -10.5 kJ
There are no significant changes in kinetic or potential energy.
(a) Sketch to cycle on a PV diagram
(b) Calculate the net work for the cycle, in kJ
(c) Calculate the heat transfer for process 2-3, in kJ
(d) Calculate the heat transfer for process 3-1, in kJ
Is this a power cycle or a refrigeration cycle?
68
69
Example 2 (Problem 2.83):
A power cycle has a thermal efficiency of 40% and generates electricity at a
rate of 100 MW. The electricity is valued at $0.08 per kWh. Based on the cost
of fuel, the cost to supply Qin is $4.50 per GJ. For 8000 hours of operation
annually, determine, in $
(a) the value of electricity generated per year and
(b) the annual fuel cost
(c) Is operation profitable?
70
71
Example 3 (Problem 2.91):
A heat pump maintains a dwelling at 68oF. When operating steadily, the
power input to the heat pump is 5 hp, and the heat pump receives energy
by heat transfer from 55oF well water at a rate of 500 BTU/min.
a) Determine the COP.
b) Evaluating electricity at $0.10 per kW-hr, determine the cost of
electricity in a month when the heat pump operate for 300 hr.
72
Chapter 2: Review
Lecture 05:
Chapter 2 Review
Quiz Today?
Main Concepts of Chap 2.
• Work done by compressed gas system:
• Heat Transfer
▫ Conduction
▫ Convection
▫ Radiation
W   pdV
• 1st law of Thermodynamics: E  Qin  Wout
• Power cycle, Refrigeration cycle, Heat Pump cycle
• Thermal Efficiency and Coefficient of Performance.
Reading Assignment:
• Read Chap 3: Sections 1-5
Homework Assignment:
From Chap 2: 46, 59, 82,97
75
Recall Chapter 2 concepts:
1st Law of Thermodynamics:
Esystem  PE  KE  U  Qin  Wout
system
Qin
ΔE
for most Thermodynamics applications:
U  Qin  Wout
Wout
ΔKE = ΔPE = 0 then
76
Work done by Gas System
W > 0 : Expansion of Gas
W < 0 : Compression of Gas
W 
Vf

Vi
PdV
= Area under the p-V graph
over process
Run
Animation
77
Heat Transfer
Conduction:
where A is area
κ is thermal conductivity
dT/dx is temperature gradient
dT
Q   A
dx
Convection:
Q  hc A Tb  T f
Radiation:
Q   AT
4
b

where A is area
hc is the convection coefficient
Tb -Tf is the difference between the
body and free steam fluid temp.
where Tb is absolute surface temperature
ε is emissivity of the surface
σ is Stefan-Boltzmann constant
A is surface area
Thermodynamic Cycle:
E  Qcycle  Wcycle = 0
P
S1
Clockwise around the cycle:
Work is done by the system.
S2
Power Cycle:
Wcycle > 0
Counter clockwise around the cycle:
Work is done on the system.
S4
S3
Refrigeration Cycle:
v
Wcycle < 0
Sec 2.6: Energy Analysis of Cycles
Cycle Models:
Power cycle:
79
Effectiveness 
Result
Effort
Refrigeration cycle:
Qin  Wcycle  Qout
Wcycle  Qin  Qout
Wcycle
Qin

Wcycle
Qin

Qout  Qin


Qin
Qin  Qout
Qin
Heat Pump Cycle
Wcycle  Qin  Qout
Qout

Wcycle
Qout

Qout  Qin
80
Concept Questions:
True or False:
a) In principle, expansion work can be evaluated using ∫pdV for both
actual and quasi-equilibrium expansion processes. False
b) The change in the internal energy of a system between two states is
the change in the total energy of the system between the two states
less the change of the system’s kinetic and gravitation potential
energies between these states.
True
c) The change in gravitational potential energy of a 2 lb mass whose
elevation decreases by 40 ft where g = 32.2 ft/s2 is -2576 ft-lbf . True
d) The rate of heat transfer from a hot baked potato to the ambient
air is greater with forced convection than natural convection. True
81
Example 3 (Problem 2.91):
A heat pump maintains a dwelling at 68oF. When operating steadily, the
power input to the heat pump is 5 hp, and the heat pump receives energy
by heat transfer from 55oF well water at a rate of 500 BTU/min.
a) Determine the COP.
b) Evaluating electricity at $0.10 per kW-hr, determine the cost of
electricity in a month when the heat pump operates for 300 hr.
82
83
Example (2.70): A gas is contained in a vertical piston-cylinder assembly by a
piston weighing 1000 lbf and having a face area of 12 in2. The atmosphere exerts a
pressure of 14.7 psi on the top of the piston. An electrical resistor transfers energy
to the gas in the amount of 5 BTU as the elevation
Patm=14.7 psi
of the piston increases by 2 ft. The piston and
cylinder are poor thermal conductors and friction
h = 2 ft
can be neglected. Determine the change in internal
Apiston = 12 in2
energy of the gas, in BTU, assuming it is the only
Wpiston = 1000 lbf
significant internal energy change of any component
present.
Welec= - 5 BTU
84
85
Example (2.83): A power cycle has a thermal efficiency of 40% and
generates electricity at a rate of 100 MW. The electricity is valued at $0.08
per kWh. Based on the cost of fuel, the cost to supply Qin is $4.50 per GJ.
For 8000 hours of operation annually, determine, in $
(a) the value of electricity generated per year
(b) the waste heat returned to the environment
(c) the annual fuel cost
(d) Is operation profitable?
Qin
Fuel
Wout =100 MW
Air
Qout
86
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Example Problem (2.63)
A gas is compressed in a piston cylinder assembly form p1 = 2 bar to p2 = 8 bar,
V2 = 0.02 m3 in a process during which the relation between pressure and
volume is pV1.3 = constant. The mass of the gas is 0.2 kg . If the specific internal
energy of the gas increase by 50 kJ/kg during the process, deter the heat
transfer in kJ. KE and PE changes are negligible.
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