Thermodynamic Chapter 4 Second Law Of Thermodynamics

CHAPTER
4
MEC 451
Thermodynamics
Second Law of
Thermodynamics
Lecture Notes:
MOHD HAFIZ MOHD NOH
HAZRAN HUSAIN & MOHD
SUHAIRIL
Faculty of Mechanical Engineering
Universiti Teknologi MARA, 40450
Shah Alam, Selangor
For students EM 220 and EM 221 only
1

Faculty of Mechanical Engineering, UiTM
2
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.

3
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.

4
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.

5
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

6
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.

7
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
ηth =
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.

8
The thermal efficiency is always less than 1 or less than
100 percent.
ηth
net out
in
W
Q
=
,
W W W
Q Q
net out out in
in net
, = −
≠
where

9
 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
ηth
net out
in
in out
in
out
in
W
Q
Q Q
Q
Q
Q
=
=
−
= −
,
1

10
 A thermodynamic temperature scale related to the heat
transfers between a reversible device and the high and low-
temperature reservoirs by
Q
Q
T
T
L
H
L
H
=
 The heat engine that operates on the reversible Carnot
cycle is called the Carnot Heat Engine in which its
efficiency is
ηth rev
L
H
T
T
, = −1

11
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.

12

13
COP
Q
W
Q
Q Q
HP
H
net in
H
H L
= =
−,
COP
Q
W
R
L
net in
=
,

14
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

15
Execution of Carnot cycle in a piston cylinder device

16

17
 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

18
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:
ηth
net out
H
W
Q
MW
MW
=
= =
,
.
50
150
0 333 or 33.3%
W Q Q
Q Q W
MW MW
MW
net out H L
L H net out
,
,
= −
= −
= −
=
150 50
100

19
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 = 652o
C
TL = 30o
C
HE
ηth rev
L
H
T
T
K
K
or
,
( )
( )
. .
= −
= −
+
+
=
1
1
30 273
652 273
0 672 67 2%
Q
Q
T
T
K
K
Q kJ
kJ
L
H
L
H
L
=
=
+
+
=
=
=
( )
( )
.
( . )
30 273
652 273
0 328
500 0 328
164
Solution:

20
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 = 25o
C
TL = 2o
C
R
COP
Q
Q Q
T
T T
K
K
R
L
H L
L
H L
=
−
=
−
=
+
−
=
( )
( )
.
2 273
25 2
1196
- this claim is also false!

21
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
]

22
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 24o
C. If the outdoor air temperature is
35o
C, 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 24o
C. On a
winter day when the outdoor air temperature is -5o
C, 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 ]

23
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.

24
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.

25
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

26
iii. Ideal gases:
a) constant specific heats (approximate treatment):
s s C
T
T
R
v
v
v av2 1
2
1
2
1
− = +, ln ln
2 2
2 1 ,
1 1
ln lnp av
T P
s s C R
T P
− = −
for isentropic process
2 1
1 2.
k
s const
P v
P v=
   
=   
   
for all process

27
Example 4.5
Steam at 1 MPa, 600o
C, 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
massbalance m m m
energybalance
E E
m h m h W
W m h h
= =
=
= +
= −
& & &
& &
&& &
& &
1
1
1
1 .
1
sup
1
3698.6
600
8.0311
kJ
kgo
kJ
kg K
State
erheated
P MPa
h
T C
s
= 
=
= 
=

28
( )
2
2
2 . 2
2
2 @
2
0.01 .
8.0311 0.984
191.8 0.984 2392.1
2545.6
45.81
kJ
kg K
kJ
kg
o
sat 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
kJ
kg
W h h= −
= −
=

29
Isentropic Efficiency for Turbine

30
Isentropic Efficiency for Compressor

31
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 a
isen T
s s
a isen T s
kJ
kg
w h h
w h h
w w
η
η
−
= =
−
= ×
=
=
 Theoretically:

32
11
1 1 .
2
2
2 1 .
2
1
3698.61
600 8.0311
2
.
0.01
0.984
8.0311
2545.6
kJ
kg
o kJ
kg K
skJ
s kg K kJ
s kg
State
hP MPa
T C s
State s
sat mixture
P MPa
x
s s
h
== 

= =
= 
=
= =  =
 Obtain h2a from Wa
1 2
2 1
2660.9
a a
a a
kJ
kg
w h h
h h w
= −
= −
=
2
2 2
2
0.01 sup
2660.9 86.85okJ
a akg
State a
P MPa erheated
h T C
= 

= =

33
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 s
c s s
P s
W m h h
W
W h h
m
C T T
= −
= = −
= −
& &
&
&
 For isentropic process of IGL
( )
1
2 2
1 1
0.4/1.4
2
0.5
27 273
0.1
475.4
k
k
s
s
T P
T P
T
K
−
   
= ÷  ÷
   
 
= +  ÷
 
=
 Then
( ),
,
,
,
1.005 475.4 300
176
220
c s
kJ
kg
c s kJ
c a kg
isen c
W
W
W
η
= −
=
= =

34
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 150o
C. At
this moment both inlet and exit valves to the radiator are closed.
After a while the temperature of the steam drops to 40o
C 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 150o
C. 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 27o
C. 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 ]

35
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.
[ 69.09o
C,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.
[ 58.9o
C,1.70 kW ]

Thermodynamic Chapter 4 Second Law Of Thermodynamics

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