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CHAPTER
4MEC 451Thermodynamics
Second Law of Thermodynamics
Lecture Notes:MOHD HAFIZ MOHD NOH HAZRAN HUSAIN & MOHD SUHAIRIL Faculty of Mechanical EngineeringUniversiti Teknologi MARA, 40450 Shah Alam, Selangor
For students EM 220 and EM 221 only
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Faculty of Mechanical Engineering, UiTM
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MEC 451 – THERMODYNAMICS
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.
MEC 451 – THERMODYNAMICS
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A cup of hot coffee does not get hotter in a cooler room.
Transferring heat to a wire will not generate electricity.
Transferring heat to a paddle wheel will not cause it to rotate.
These processes cannot occur even though they are not in violation of the first law.
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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 low-temperature sink atmosphere
o they operate on a cycle
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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
th Desired Result
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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MEC 451 – THERMODYNAMICS
The thermal efficiency is always less than 1 or less than 100 percent.
thnet out
in
W
Q ,
W W W
Q Q
net out out in
in net
,
where
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MEC 451 – THERMODYNAMICS
Applying the first law to the cyclic heat engine
Q W U
W Q
W Q Q
net in net out
net out net in
net out in out
, ,
, ,
,
The cycle thermal efficiency may be written as
thnet out
in
in out
in
out
in
W
Q
Q Q
Q
Q
Q
,
1
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MEC 451 – THERMODYNAMICS
A thermodynamic temperature scale related to the heat transfers between a reversible device and the high and low-temperature reservoirs by
Q
Q
T
TL
H
L
H
The heat engine that operates on the reversible Carnot cycle is called the Carnot Heat Engine in which its efficiency is
th revL
H
T
T, 1
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MEC 451 – THERMODYNAMICS
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
Faculty of Mechanical Engineering, UiTM
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MEC 451 – THERMODYNAMICS
COPQ
W
Q
Q QHPH
net in
H
H L
,
COPQ
WRL
net in
,
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
Execution of Carnot cycle in a piston cylinder device
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MEC 451 – THERMODYNAMICS
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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:
thnet out
H
W
Q
MW
MW
,
.50
1500 333 or 33.3%
W Q Q
Q Q W
MW MW
MW
net out H L
L H net out
,
,
150 50
100
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MEC 451 – THERMODYNAMICS
A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at 652ºC and rejects heat to a low-temperature 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
Example 4.2
QL
WOUT
QH
TH = 652oC
TL = 30oC
HE
th revL
H
T
T
K
K
or
,
( )
( )
. .
1
130 273
652 273
0 672 67 2%
Q
Q
T
T
K
K
Q kJ
kJ
L
H
L
H
L
( )
( ).
( . )
30 273
652 2730 328
500 0 328
164
Solution:
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MEC 451 – THERMODYNAMICS
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?
Example 4.3
Solution:
QL
Win
QH
TH = 25oC
TL = 2oC
R
COPQ
Q Q
T
T T
K
K
RL
H L
L
H L
( )
( )
.
2 273
25 2
1196
- this claim is also false!
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
Entropy Change
The entropy change during a reversible process is defined as
For a reversible, adiabatic process
dS
S S
0
2 1
The reversible, adiabatic process is called an isentropic process.
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MEC 451 – THERMODYNAMICS
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/kgK)
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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MEC 451 – THERMODYNAMICS
iii. Ideal gases:
a) constant specific heats (approximate treatment):
s s CT
TR
v
vv av2 12
1
2
1
, ln ln
2 22 1 ,
1 1
ln lnp av
T Ps s C R
T P
for isentropic process
2 1
1 2.
k
s const
P v
P v
for all process
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MEC 451 – THERMODYNAMICS
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:
1 2
1 1 2 2
1 2
:
in out
out
out
mass balance m m m
energy balance
E E
m h m h W
W m h h
11
1
1 .
1
sup1
3698.6600
8.0311
kJkgo
kJkg K
State
erheatedP MPa
hT C
s
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MEC 451 – THERMODYNAMICS
2
2
2 . 2
2
2 @
2
0.01 .
8.0311 0.984
191.8 0.984 2392.1
2545.6
45.81
kJkg K
kJkg
osat P
State
P MPa sat mixture
s x
h
T T C
Since that the process is isentropic, s2=s1
Work of turbine
1 2
3698.6 2545.6
1153
out
kJkg
W h h
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Isentropic Efficiency for Turbine
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Isentropic Efficiency for Compressor
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MEC 451 – THERMODYNAMICS
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:
1 2,
1 2
,
0.9 1153
1037.7
a aisen T
s s
a isen T s
kJkg
w h h
w h h
w w
Theoretically:
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MEC 451 – THERMODYNAMICS
11
1 1 .
2
22 1 .
2
1
3698.61
600 8.0311
2
.0.01
0.9848.0311
2545.6
kJkg
o kJkg K
skJs kg K kJ
s kg
State
hP MPa
T C s
State s
sat mixtureP MPa
xs s
h
Obtain h2a from Wa
1 2
2 1
2660.9
a a
a a
kJkg
w h h
h h w
2
2 2
2
0.01 sup
2660.9 86.85okJa akg
State a
P MPa erheated
h T C
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MEC 451 – THERMODYNAMICS
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
, 2 1
,, 2 1
2 1
c s s
c sc s s
P s
W m h h
WW h h
mC T T
For isentropic process of IGL
1
2 2
1 1
0.4/1.4
2
0.527 273
0.1
475.4
k
ks
s
T P
T P
T
K
Then
,
,,
,
1.005 475.4 300
176
220
c s
kJkg
c s kJc a kg
isen c
W
WW
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MEC 451 – THERMODYNAMICS
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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MEC 451 – THERMODYNAMICS
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 ]