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Entropy Rate Balance for Closed Systems
Example: An inventor claims that the device shown
generates electricity at a rate of 100 kJ/s while receiving a
heat transfer of energy at a rate of 250 kJ/s at a temperature
of 500 K, receiving a second heat transfer at a rate of 350
kJ/s at 700 K, and discharging energy by heat transfer at a
rate of 500 kJ/s at a temperature of 1000 K. Each heat
transfer is positive in the direction of the accompanying
arrow. For operation at steady state, evaluate this claim.
Q1  250 kJ/s
T1 = 500 K
Q 2  350 kJ/s
+
–
T2 = 700 K
T3 = 1000 K
Q 3  500 kJ/s
Entropy Rate Balance for Closed Systems
► Applying an energy rate balance dE 0
 0  Q1  Q 2  Q 3  W e
at steady state
dt
Solving We  250 kJ/s  350 kJ/s  500 kJ/s  100 kJ/s
The claim is in accord with the first law of thermodynamics.
► Applying an entropy rate balance dS 0
Q1 Q 2 Q 3
0


 
dt
T1 T2 T3
at steady state
Solving
 250 kJ/s 350 kJ/s 500 kJ/s 



700 K
1000 K 
 500 K
kJ/s
kJ/s
  0.5  0.5  0.5
 0.5
K
K
  
Since σ∙ is negative, the claim is not in accord with the
second law of thermodynamics and is therefore false.
Entropy Rate Balance for Control Volumes
► Like mass and energy, entropy can be transferred into or
out of a control volume by streams of matter.
► Since this is the principal difference between the closed
system and control volume entropy rate balances, the
control volume form can be obtained by modifying the
closed system form to account for such entropy transfer.
The result is
(Eq. 6.34)
 i si and m e se account, respectively, for rates of entropy
where m
transfer accompanying mass flow at inlets i and exits e.
Entropy Rate Balance for Control Volumes
► For control volumes at steady state, Eq. 6.34 reduces to
give
(Eq. 6.36)
► For a one-inlet, one-exit control volume at steady state,
Eq. 6.36 reduces to give
(Eq. 6.37)
where 1 and 2 denote the inlet and exit, respectively, and
is the common mass flow rate at these locations.
m
Entropy Rate Balance for Control Volumes
Example: Water vapor enters a valve at 0.7
bar, 280oC and exits at 0.35 bar. (a) If the
water vapor undergoes a throttling process,
determine the rate of entropy production
within the valve, in kJ/K per kg of water
p1 = 0.7 bar
vapor flowing. (b) What is the source of
T1 = 280oC
entropy production in this case?
(a) For a throttling process, there is no significant
heat transfer. Thus, Eq. 6.37 reduces to
0
Q j
 s1  s2    cv
0
 m s1  s2    cv → 0  m
Tj

j
p2 = 0.35 bar
h2 = h1
Entropy Rate Balance
for Control Volumes
Solving
 cv
m
 s 2  s1
p1 = 0.7 bar
T1 = 280oC
p2 = 0.35 bar
h2 = h1
From Table A-4, h1 = 3035.0 kJ/kg, s1 = 8.3162 kJ/kg∙K.
For a throttling process, h2 = h1 (Eq. 4.22). Interpolating
in Table A-4 at 0.35 bar and h2 = 3035.0 kJ/kg,
s2 = 8.6295 kJ/kg∙K.
 cv
 (8.6295 – 8.3162) kJ/kg∙K = 0.3133 kJ/kg∙K
Finally
m
(b) Selecting from the list of irreversibilities provided in
Sec. 5.3.1, the source of the entropy production here is
the unrestrained expansion to a lower pressure undergone
by the water vapor.
Entropy Rate Balance for Control Volumes
Comment: The value of the entropy production for a single
component such as the throttling valve considered here often
does not have much significance by itself. The significance of
the entropy production of any component is normally
determined through comparison with the entropy production
values of other components combined with that component to
form an integrated system. Reducing irreversibilities of
components with the highest entropy production rates may
lead to improved thermodynamic performance of the
integrated system.
p = 0.7 bar
p = 0.35 bar
Calculating Entropy Change
►The property data provided in Tables A-2
through A-18, similar compilations for other
substances, and numerous important relations
among such properties are established using the
TdS equations. When expressed on a unit mass
basis, these equations are
(Eq. 6.10a)
(Eq. 6.10b)
(Eq. 6.10a)
(Eq. 6.10b)
(Eq. 6.10a)
(Eq. 6.10b)
(Eq. 6.23)
Calculating Entropy Change
► As an application, consider a
change in phase from saturated
liquid to saturated vapor at
constant pressure.
► Since pressure is constant, Eq.
6.10b reduces to give
dh
ds 
T
► Then, because temperature is also constant during the
phase change
(Eq. 6.12)
This relationship is applied in property tables for
tabulating (sg – sf) from known values of (hg – hf).
Calculating Entropy Change
►For example, consider water vapor at 100oC
(373.15 K). From Table A-2, (hg – hf) = 2257.1 kJ/kg.
Thus
(sg – sf) = (2257.1 kJ/kg)/373.15 K = 6.049 kJ/kg∙K
which agrees with the value from Table A-2, as
expected.
►Next, the TdS equations are applied to two
additional cases: substances modeled as
incompressible and gases modeled as ideal
gases.
Calculating Entropy Change of an
Incompressible Substance
► The incompressible substance model assumes the specific
volume is constant and specific internal energy depends
solely on temperature: u = u(T). Thus, du = c(T)dT, where
c denotes specific heat.
► With these relations, Eq. 6.10a reduces to give
► On integration, the change in specific entropy is
► When the specific heat is constant
(Eq. 6.13)
Calculating Entropy Change of an Ideal Gas
► The ideal gas model assumes pressure, specific volume
and temperature are related by pv = RT. Also, specific
internal energy and specific enthalpy each depend solely
on temperature: u = u(T), h = h(T), giving du = cvdT and
dh = cpdT, respectively.
► Using these relations and integrating, the TdS equations
give, respectively
(Eq. 6.17)
(Eq. 6.18)
Calculating Entropy Change of an Ideal Gas
► Since these particular equations give entropy change on a
unit of mass basis, the constant R is determined from
R  R / M.
► Since cv and cp are functions of temperature for ideal gases,
such functional relations are required to perform the
integration of the first term on the right of Eqs. 6.17 and 6.18.
► For several gases modeled as ideal gases, including air,
CO2, CO, O2, N2, and water vapor, the evaluation of
entropy change can be reduced to a convenient tabular
approach using the variable so defined by
(Eq. 6.19)
where T ' is an arbitrary reference temperature.
Calculating Entropy Change of an Ideal Gas
► Using so, the integral term of Eq. 6.18 can be expressed as
► Accordingly, Eq. 6.18 becomes
(Eq. 6.20a)
or on a per mole basis as
(Eq. 6.20b)
► For air, Tables A-22 and A-22E provide so in units of
kJ/kg∙K and Btu/lb∙oR, respectively. For the other gases
mentioned, Tables A-23 and A-23E provide s o in units of
kJ/kmol∙K and Btu/lbmol∙oR, respectively.
Calculating Entropy Change of an Ideal Gas
Example: Determine the change in specific entropy, in
kJ/kg∙K, of air as an ideal gas undergoing a process from
T1 = 300 K, p1 = 1 bar to T2 = 1420 K, p2 = 5 bar.
► From Table A-22, we get so1 = 1.70203 and so2 = 3.37901,
each in kJ/kg∙K. Substituting into Eq. 6.20a
s2  s1  (3.37901  1.70203 )
kJ
kJ
 8.314  kJ   5 bar 



ln

1
.
215



kg  K  28 .97  kg  K   1 bar 
kg  K
Ideal Gas Properties of Air
T(K), h and u(kJ/kg), so (kJ/kg∙K)
when s = 0
Table A-22
T
250
260
270
280
285
290
295
300
305
310
h
250.05
260.09
270.11
280.13
285.14
290.16
295.17
300.19
305.22
310.24
u
178.28
185.45
192.60
199.75
203.33
206.91
210.49
214.07
217.67
221.25
so
1.51917
1.55848
1.59634
1.63279
1.65055
1.66802
1.68515
1.70203
1.71865
1.73498
pr
0.7329
0.8405
0.9590
1.0889
1.1584
1.2311
1.3068
1.3860
1.4686
1.5546
vr
979.
887.8
808.0
738.0
706.1
676.1
647.9
621.2
596.0
572.3
T
1400
1420
1440
1460
1480
1500
1520
1540
1560
1580
h
1515.42
1539.44
1563.51
1587.63
1611.79
1635.97
1660.23
1684.51
1708.82
1733.17
when s = 0
u
1113.52
1131.77
1150.13
1168.49
1186.95
1205.41
1223.87
1242.43
1260.99
1279.65
so
3.36200
3.37901
3.39586
3.41247
3.42892
3.44516
3.46120
3.47712
3.49276
3.50829
pr
450.5
478.0
506.9
537.1
568.8
601.9
636.5
672.8
710.5
750.0
vr
8.919
8.526
8.153
7.801
7.468
7.152
6.854
6.569
6.301
6.046
Calculating Entropy Change of an Ideal Gas
►Tables A-22 and A-22E provide
additional data for air modeled
as an ideal gas. These values,
denoted by pr and vr, refer only
to two states having the same
specific entropy. This case has
important applications, and is
shown in the figure.
Calculating Entropy Change of an Ideal Gas
► When s2 = s1, the following equation relates T1, T2, p1,
and p2
p2 pr (T2 )

(s1 = s2, air only) (Eq. 6.41)
p1
pr (T1 )
where pr(T ) is read from Table A-22 or A-22E, as
appropriate.
Ideal Gas Properties of Air
T(K), h and u(kJ/kg), so (kJ/kg∙K)
when s = 0
Table A-22
T
250
260
270
280
285
290
295
300
305
310
h
250.05
260.09
270.11
280.13
285.14
290.16
295.17
300.19
305.22
310.24
u
178.28
185.45
192.60
199.75
203.33
206.91
210.49
214.07
217.67
221.25
so
1.51917
1.55848
1.59634
1.63279
1.65055
1.66802
1.68515
1.70203
1.71865
1.73498
pr
0.7329
0.8405
0.9590
1.0889
1.1584
1.2311
1.3068
1.3860
1.4686
1.5546
vr
979.
887.8
808.0
738.0
706.1
676.1
647.9
621.2
596.0
572.3
T
1400
1420
1440
1460
1480
1500
1520
1540
1560
1580
h
1515.42
1539.44
1563.51
1587.63
1611.79
1635.97
1660.23
1684.51
1708.82
1733.17
when s = 0
u
1113.52
1131.77
1150.13
1168.49
1186.95
1205.41
1223.87
1242.43
1260.99
1279.65
so
3.36200
3.37901
3.39586
3.41247
3.42892
3.44516
3.46120
3.47712
3.49276
3.50829
pr
450.5
478.0
506.9
537.1
568.8
601.9
636.5
672.8
710.5
750.0
vr
8.919
8.526
8.153
7.801
7.468
7.152
6.854
6.569
6.301
6.046
Calculating Entropy Change of an Ideal Gas
► When s2 = s1, the following equation relates T1, T2, v1,
and v2
v 2 v r (T2 )

(s1 = s2, air only) (Eq. 6.42)
v1 v r (T1 )
where vr(T ) is read from Table A-22 or A-22E, as
appropriate.
Ideal Gas Properties of Air
T(K), h and u(kJ/kg), so (kJ/kg∙K)
when s = 0
Table A-22
T
250
260
270
280
285
290
295
300
305
310
h
250.05
260.09
270.11
280.13
285.14
290.16
295.17
300.19
305.22
310.24
u
178.28
185.45
192.60
199.75
203.33
206.91
210.49
214.07
217.67
221.25
so
1.51917
1.55848
1.59634
1.63279
1.65055
1.66802
1.68515
1.70203
1.71865
1.73498
pr
0.7329
0.8405
0.9590
1.0889
1.1584
1.2311
1.3068
1.3860
1.4686
1.5546
vr
979.
887.8
808.0
738.0
706.1
676.1
647.9
621.2
596.0
572.3
T
1400
1420
1440
1460
1480
1500
1520
1540
1560
1580
h
1515.42
1539.44
1563.51
1587.63
1611.79
1635.97
1660.23
1684.51
1708.82
1733.17
when s = 0
u
1113.52
1131.77
1150.13
1168.49
1186.95
1205.41
1223.87
1242.43
1260.99
1279.65
so
3.36200
3.37901
3.39586
3.41247
3.42892
3.44516
3.46120
3.47712
3.49276
3.50829
pr
450.5
478.0
506.9
537.1
568.8
601.9
636.5
672.8
710.5
750.0
vr
8.919
8.526
8.153
7.801
7.468
7.152
6.854
6.569
6.301
6.046
Entropy Change of an Ideal Gas
Assuming Constant Specific Heats
► When the specific heats cv and cp are assumed constant,
Eqs. 6.17 and 6.18 reduce, respectively, to
(Eq. 6.17)
(Eq. 6.18)
(Eq. 6.21)
(Eq. 6.22)
► These expressions have many
applications. In particular, they can be
applied to develop relations among T,
p, and v at two states having the same
specific entropy as shown in the figure.
Entropy Change of an Ideal Gas
Assuming Constant Specific Heats
► Since s2 = s1, Eqs. 6.21 and
6.22 become
► With the ideal gas relations
where k is the specific ratio
► These equations
can be solved,
respectively, to give
► Eliminating the
temperature ratio gives
(Eq. 6.43)
(Eq. 6.44)
(Eq. 6.45)
Calculating Entropy Change of an Ideal Gas
Example: Air undergoes a process from T1 = 620 K, p1 = 12 bar
to a final state where s2 = s1, p2 = 1.4 bar. Employing the ideal
gas model, determine the final temperature T2, in K. Solve using
(a) pr data from Table A-22 and (b) a constant specific heat ratio
k evaluated at 620 K from Table A-20: k = 1.374. Comment.
(a) With Eq. 6.41 and pr(T1) = 18.36 from Table A-22
 p2 
 1.4 bar 


pr T2   pr T1 
 18 .36
  2.142

 12 bar 
 p1 
Interpolating in Table A-22, T2 = 339.7 K
Ideal Gas Properties of Air
T(K), h and u(kJ/kg), so (kJ/kg∙K)
when s = 0
Table A-22
T
315
320
325
330
340
350
h
315.27
320.29
325.31
330.34
340.42
350.49
u
224.85
228.42
232.02
235.61
242.82
250.02
s
1.75106
1.76690
1.78249
1.79783
1.82790
1.85708
o
pr
1.6442
1.7375
1.8345
1.9352
2.149
2.379
vr
549.8
528.6
508.4
489.4
454.1
422.2
T
600
610
620
630
640
650
h
607.02
617.53
628.07
638.63
649.22
659.84
when s = 0
u
434.78
442.42
450.09
457.78
465.50
473.25
s
2.40902
2.42644
2.44356
2.46048
2.47716
2.49364
o
pr
16.28
17.30
18.36
19.84
20.64
21.86
vr
105.8
101.2
96.92
92.84
88.99
85.34
Calculating Entropy Change of an Ideal Gas
(b) With Eq. 6.43

T2  T1 

p2 

p1 
k 1 / k
 1.4 bar 
 620 K

 12 bar 
0.374/ 1.374
T2 = 345.5 K
Comment: The approach of (a) accounts for
variation of specific heat with temperature but
the approach of (b) does not. With a k value
more representative of the temperature interval,
the value obtained in (b) using Eq. 6.43 would be
in better agreement with that obtained in (a) with
Eq. 6.41.
Isentropic Turbine Efficiency
► For a turbine, the energy rate
balance reduces to
1
2
2


(V

V
)
1
2


0  Qcv  Wcv  m (h1  h2 ) 
 g ( z1  z2 )
2


► If the change in kinetic energy of flowing matter is negligible,
½(V12 – V22) drops out.
► If the change in potential energy of flowing matter is
negligible, g(z1 – z2) drops out.
► If the heat transfer with surroundings is negligible, Q cv drops
out.
W cv
 h1  h2
m
where
the left side is work developed per unit of mass flowing.
2
Isentropic Turbine Efficiency
►For a turbine, the entropy rate
balance reduces to
1
► If the heat transfer with surroundings is negligible, Q j
drops out.
 cv
m
 s2  s1  0
2
×

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