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CHAPTER
4
MEC 451
Thermodynamics
Lecture Notes:
MOHD HAFIZ MOHD NOH
HAZRAN HUSAIN & MOHD SUHAIRIL
Faculty of Mechanical Engineering
Universiti Teknologi MARA, 40450
Shah Alam, Selangor
Second Law of
Thermodynamics
For students EM 220 and EM 221 only
Faculty of Mechanical Engineering, UiTM
Introduction
 A process must satisfy the first law in order to occur.
 Satisfying the first law alone does not ensure that the process will take
place.
 Second law is useful:
 provide means for predicting the direction of processes,
 establishing conditions for equilibrium,
 determining the best theoretical performance of cycles, engines
and other devices.
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A cup of hot coffee does
not get hotter in a cooler
room.
Transferring
heat to a paddle
wheel will not
cause it to
rotate.
Transferring heat to a wire
will not generate electricity.
These processes cannot occur
even though they are not in
violation of the first law.
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Second Law of Thermodynamics
Kelvin-Planck statement
 No heat engine can have a
thermal
efficiency
100
percent.
 As for a power plant to
operate, the working fluid
must exchange heat with the
environment as well as the
furnace.
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Heat Engines
 Work can easily be converted to other forms of
energy, but?
 Heat engine differ considerably from one
another, but all can be characterized :
o they receive heat from a high-temperature
source
o they convert part of this heat to work
o they reject the remaining waste heat to a lowtemperature sink atmosphere
o they operate on a cycle
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The work-producing
device that best fit
into the definition of a
heat engine is the
steam power plant,
which is an external
combustion engine.
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Thermal Efficiency
 Represent the magnitude of the energy wasted in order
to complete the cycle.
 A measure of the performance that is called the
thermal efficiency.
 Can be expressed in terms of the desired output and
the required input
Desired Result
 th 
Required Input
 For a heat engine the desired result is the net work
done and the input is the heat supplied to make
the cycle operate.
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The thermal efficiency is always less than 1 or less than
100 percent.
 th 
Wnet , out
Qin
where
Wnet , out  Wout  Win
Qin  Qnet
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 Applying the first law to the cyclic heat engine
Qnet , in  Wnet , out  U
Wnet , out  Qnet , in
Wnet , out  Qin  Qout
 The cycle thermal efficiency may be written as
 th 
Wnet , out
Qin
Qin  Qout

Qin
Qout
 1
Qin
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 A thermodynamic temperature scale related to the heat
transfers between a reversible device and the high and lowtemperature reservoirs by
QL
T
 L
QH TH
 The heat engine that operates on the reversible Carnot
cycle is called the Carnot Heat Engine in which its
efficiency is
 th , rev
TL
 1
TH
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Heat Pumps and Refrigerators
 A device that transfers heat from a low
temperature medium to a high temperature one is
the heat pump.
 Refrigerator operates exactly like heat pump
except that the desired output is the amount of
heat removed out of the system
 The index of performance of a heat pumps or
refrigerators are expressed in terms of the
coefficient of performance.
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MEC 451 – THERMODYNAMICS
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COPHP
QH
QH


Wnet , in QH  QL
QL
COPR 
Wnet , in
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Carnot Cycle
Process
Description
1-2
Reversible isothermal heat addition at
high temperature
2-3
Reversible adiabatic expansion from high
temperature to low temperature
3-4
Reversible isothermal heat rejection at
low temperature
4-1
Reversible adiabatic compression from low
temperature to high temperature
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Execution of Carnot cycle in a piston cylinder device
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 The thermal efficiencies of actual and reversible heat
engines operating between the same temperature limits
compare as follows
 The coefficients of performance of actual and reversible
refrigerators operating between the same temperature limits
compare as follows
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Example 4.1
A steam power plant
produces 50 MW of net
work while burning fuel
to produce 150 MW of
heat energy at the high
temperature. Determine
the
cycle
thermal
efficiency and the heat
rejected by the cycle to
the surroundings.
Solution:
 th 
Wnet , out
QH
50 MW

 0.333 or 33.3%
150 MW
Wnet , out  QH  QL
QL  QH  Wnet , out
 150 MW  50 MW
 100 MW
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MEC 451 – THERMODYNAMICS
Faculty of Mechanical Engineering, UiTM
Example 4.2
A Carnot heat engine receives 500 kJ of heat per cycle from a hightemperature heat reservoir at 652ºC and rejects heat to a lowtemperature heat reservoir at 30ºC. Determine :
(a) The thermal efficiency of this Carnot engine
(b) The amount of heat rejected to the low-temperature heat
reservoir
Solution:
TH = 652oC
QH
WOUT
HE
QL
TL = 30oC
TL
TH
(30  273) K
 1
(652  273) K
 0.672 or 67.2%
 th , rev  1 
QL TL

QH TH
(30  273) K
 0.328
(652  273) K
QL  500 kJ (0.328)

 164 kJ
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MEC 451 – THERMODYNAMICS
Faculty of Mechanical Engineering, UiTM
Example 4.3
An inventor claims to have developed a refrigerator that maintains
the refrigerated space at 2ºC while operating in a room where the
temperature is 25ºC and has a COP of 13.5. Is there any truth to his
claim?
Solution:
TH = 25oC
COPR 
QH
Win
R
QL
QL
TL

QH  QL TH  TL
(2  273) K
(25  2) K
 1196
.

TL = 2oC
- this claim is also false!
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Faculty of Mechanical Engineering, UiTM
Supplementary Problem 4.1
1. A 600 MW steam power plant, which is cooled by a river, has a thermal
efficiency of 40 percent. Determine the rate of heat transfer to the river
water. Will the actual heat transfer rate be higher or lower than this
value? Why?
[900 MW]
2. A steam power plant receives heat from a furnace at a rate of 280
GJ/h. Heat losses to the surrounding air from the steam as it passes
through the pipes and other components are estimated to be about 8
GJ/h. If the waste heat is transferred to the cooling water at a rate of
145 GJ/h, determine (a) net power output and (b) the thermal
efficiency of this power plant.
[ 35.3 MW, 45.4% ]
3. An air conditioner removes heat steadily from a house at a rate of 750
kJ/min while drawing electric power at a rate of 6 kW. Determine (a)
the COP of this air conditioner and (b) the rate of heat transfer to the
outside air.
[ 2.08, 1110 kJ/min ]
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4. Determine the COP of a heat pump that supplies energy to a house at
a rate of 8000 kJ/h for each kW of electric power it draws. Also,
determine the rate of energy absorption from the outdoor air.
[ 2.22, 4400 kJ/h ]
5. An inventor claims to have developed a heat engine that receives 700
kJ of heat from a source at 500 K and produces 300 kJ of net work
while rejecting the waste heat to a sink at 290 K. Is this reasonable
claim?
6. An air-conditioning system operating on the reversed Carnot cycle is
required to transfer heat from a house at a rate of 750 kJ/min to
maintain its temperature at 24oC. If the outdoor air temperature is
35oC, determine the power required to operate this air-conditioning
system.
[ 0.463 kW ]
7. A heat pump is used to heat a house and maintain it at 24oC. On a
winter day when the outdoor air temperature is -5oC, the house is
estimated to lose heat at a rate of 80,000 kJ/h. Determine the
minimum power required to operate this heat pump.
[ 2.18 kW ]
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Entropy
 The 2nd law states that process occur in a certain
direction, not in any direction.
 It often leads to the definition of a new property called
entropy, which is a quantitative measure of disorder
for a system.
 Entropy can also be explained as a measure of the
unavailability of heat to perform work in a cycle.
 This relates to the 2nd law since the 2nd law predicts
that not all heat provided to a cycle can be
transformed into an equal amount of work, some heat
rejection must take place.
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Entropy Change
 The entropy change during a reversible process is defined
as
 For a reversible, adiabatic process
dS  0
S2  S1
 The reversible, adiabatic process is called an isentropic
process.
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MEC 451 – THERMODYNAMICS
Faculty of Mechanical Engineering, UiTM
Entropy Change and Isentropic Processes
The entropy-change and isentropic relations for a process
can be summarized as follows:
i. Pure substances:
Any process: Δs = s2 – s1 (kJ/kgK)
Isentropic process: s2 = s1
ii. Incompressible substances (liquids and solids):
Any process: s2 – s1 = cav T2/T1 (kJ/kg
Isentropic process: T2 = T1
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iii. Ideal gases:
a) constant specific heats (approximate treatment):
for all process
T2
v2
s2  s1  Cv , av ln  R ln
T1
v1
T2
P2
s2  s1  C p , av ln  R ln
T1
P1
for isentropic process
 P2 
 v1 
 
 
 P1  s  const .  v2 
k
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MEC 451 – THERMODYNAMICS
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Example 4.5
Steam at 1 MPa, 600oC, expands in a turbine to 0.01 MPa. If the
process is isentropic, find the final temperature, the final enthalpy of
the steam, and the turbine work.
Solution:
mass balance : m1  m2  m
energy balance
Ein  Eout
m1h1  m2 h2  Wout
Wout  m  h1  h2 
State1
sup erheated
P1  1 MPa 
kJ
 h1  3698.6 kg
o
T1  600 C 
s1  8.0311 kgkJ. K
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 Since that the process is
isentropic, s2=s1
 Work of turbine
Wout  h1  h2
State 2
 3698.6  2545.6
P2  0.01 MPa 
 sat.mixture

s2  8.0311 kgkJ. K 
 x2  0.984
 1153 kJ
kg
h2  191.8  0.984  2392.1
 2545.6 kJ
kg
T2  Tsat @ P2  45.81o C
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Isentropic Efficiency for Turbine
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Isentropic Efficiency for Compressor
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Example 4.6
Steam at 1 MPa, 600°C,
expands in a turbine to 0.01
MPa. The isentropic work
of the turbine is 1152.2
kJ/kg. If the isentropic
efficiency of the turbine is
90 percent, calculate the
actual work.
Find the
actual
turbine
exit
temperature or quality of
the steam.
Solution:
 Theoretically:
isen ,T
wa h1  h2 a


ws h1  h2 s
wa  isen ,T  ws
 0.9 1153
 1037.7 kJ
kg
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State1
kJ
P1  1 MPa  h1  3698.6 kg

T1  600o C  s1  8.0311 kgkJ. K
State 2 s
sat.mixture
P2  0.01 MPa 

x2 s  0.984
kJ 
s2 s  s1  8.0311 kg . K 
 h  2545.6 kJ
2s
kg
 Obtain h2a from Wa
wa  h1  h2 a
h2 a  h1  wa
 2660.9 kJ
kg
State 2a
P2  0.01 MPa 
 sup erheated

o
h2 a  2660.9 kJ
T

86.85
C
kg 
 2a
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Faculty of Mechanical Engineering, UiTM
Example 4.7
Air enters a compressor
and
is
compressed
adiabatically from 0.1 MPa,
27°C, to a final state of 0.5
MPa. Find the work done
on the air for a compressor
isentropic efficiency of 80
percent.
Solution:
 From energy balance
Wc , s  m  h2 s  h1 
Wc , s 
Wc , s
m
 h2 s  h1
 For isentropic process of IGL
 T2 s

 T1
  P2 
 
  P1 
k 1
k
 0.5 
T2 s   27  273 

0.1


 475.4 K
0.4/1.4
 Then
Wc , s  1.005  475.4  300 
 176 kJ
kg
Wc ,a 
Wc , s
isen ,c
 CP T2 s  T1 
 220 kJ
kg
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MEC 451 – THERMODYNAMICS
Faculty of Mechanical Engineering, UiTM
Supplementary Problems 4.2
1.
The radiator of a steam heating system has a volume of 20 L and is
filled with the superheated water vapor at 200 kPa and 150oC. At
this moment both inlet and exit valves to the radiator are closed.
After a while the temperature of the steam drops to 40oC as a result
of heat transfer to the room air. Determine the entropy change of
the steam during this process.
[ -0.132 kJ/.K ]
2.
A heavily insulated piston-cylinder device contains 0.05 m3 of
steam at 300 kPa and 150oC. Steam is now compressed in a
reversible manner to a pressure of 1 MPa. Determine the work
done on the steam during this process.
[ 16 kJ ]
3.
A piston –cylinder device contains 1.2 kg of nitrogen gas at 120 kPa
and 27oC. The gas is now compressed slowly in a polytropic process
during which PV1.3=constant. The process ends when the volume is
reduced by one-half. Determine the entropy change of nitrogen
during this process.
[ -0.0617 kJ/kg.K ]
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4.
Steam enters an adiabatic turbine at 8 MPa and 500oC with a
mass flow rate of 3 kg/s and leaves at 30 kPa. The isentropic
efficiency of the turbine is 0.90. Neglecting the kinetic energy of
the steam, determine (a) the temperature at the turbine exit and
(b) the power output of the turbine.
[ 69.09oC,3054 kW ]
5.
Refrigerant-R134a enters an adiabatic compressor as saturated
vapor at 120 kPa at a rate of 0.3 m3/min and exits at 1 MPa
pressure. If the isentropic efficiency of the compressor is 80
percent, determine (a) the temperature of the refrigerant at the
exit of the compressor and (b) the power input, in kW. Also, show
the process on a T-s diagram with respect to the saturation lines.
[ 58.9oC,1.70 kW ]
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