This document contains 20 multiple choice questions regarding concepts of thermodynamics and the second law of thermodynamics. It includes calculations of efficiency, COP, temperatures, enthalpies, and entropy involving heat engines and refrigerators. Key concepts covered are the Carnot cycle, Carnot efficiency, ideal gas processes, heat transfer, work, and isentropic processes.

1. FIVE
The Second Law of Thermodynamics
5.1FE
D
5.2FE
D
5.3FE
A
T
283
  L 
1
1

0.05667.
max
TH
300
5.4FE
C

T
293
W
10
  L  
1
1

0.3805 

.
 W
 10
TH
473
QL   QL 
5.5FE
B
5.7FE
A
 16.3 kW
QL 
C
5.6FE
T
293
COP  H 

7.325.
TH  L 293 
T
253
Q
20
COP  L  
2.
W 10
T
293
From Eq. 5.11, COP  L


2.
TH  L TH 
T
293
5.8FE
C
T
293
Use Eq. 5.10: COP  H 

7.325.
TH  L 293 
T
253
Q
2000
COP 
7.325  H 
.
W
W
5.9FE
A
TH 
439.5 K
W 
273 kJ/hr
T
293
  L  
1
1

0.214
TH
373
5.10FE C
T2
573

1 0.717 
ln

0.481 kJ/K
T1
293
5.11FE C
  v ln
S mc
5.12FE D
  v (T2  1 )  
W mc
T
10 0.717 
(356.6 
773) 
2986 kJ
k k
1/
 
p
where T2  1  2 
T
p
1 
5.13FE D
0.2857
400 
 
773

6000 


356.6 K
The heat gained by the ice is lost by the water:  melt  i  w  p , w  w . (We
h
m m c
T
assume not all the ice melts.) The entropy change is:
T
Q
5.02 
333
273
 i  S w   w c p , w ln 2 
S 
m
 
20 4.18 
ln

0.213 kJ/K
Ti
293
273
293
72

2. 5.14FE A
The final temperature is 0 The heat lost by the iron is:
C.
Q  Fe c p (T1  2 )  
m
T
10 0.448  
300 1344 kJ
  Fe c p ln
S m
T2 Q
273 1344
  
10 0.448 
ln


0.881 kJ/K
T1 Tice
673 273
5.15FE C
k k
1/
5.16FE D
5.17FE D
 
p
2000 

First, find T2: T2  1  2 
T
 
673

1066 K .

p
400 
1 
With Q = 0:   v (T2  1 )  
W mc
T
2 0.717 
(1066 
673)  kJ
563
0.2857
State 1 is at the inlet and state 2 at the outlet. Use enthalpies found in Tables C.3 and C.2:
wT  1  2 
h h 3658 
2637 
1021 kJ/kg
5.18FE C
State 2 at the same entropy as state 1:
is
s2 s1 
7.1677 
1.0259  2' 
x 6.6441.
h2' 
317.6 
0.9244 
2319 
2462 kJ/kg.
x2' 
0.9244
wT ,max 
3658 
2462 
1197 kJ/kg
5.19FE B
Ish t pr ue f aoi t n on i T b C2
t t e e t o vpr ao fud n al .
’ e m ar
zi
e .
5.20FE: A
wT ,actual
1020



0.85
wT ,isentropic 1200
5.1
30 000 / 3600
output

COP 
. 4
. W 2.083 kW or 2.79 hp

input
W
5.2
output 75000 / 3600
COP 


5.21
input
4
5.3

T
500
2545
 W 10 
  H  
1
1

0.256. QH  
 400 Btu/hr
99,
TL
672

0.256
5.4
output
500


0.3
input 100 
6000 / 3600
5.5
V  
100 1000 / 3600 27.78 m/s
1
1
 
1.23 27.782  
2 0.28 
27.78
V
V
Drag power 2  2 ACD 
2



0.519

Inputpower
W
1 100 
1000  
10 3

 
740 9000
13
3600
73

3. 5.6
A Kelvin-Planck violation is sketched as the first block in the figure below. The
dashed box encloses a refrigerator which is a violation of the Clausius statement:
QH1
QH 1  . QH 2   L  H 1
W
W Q Q
Let QH 2  H 1  H
Q
Q
Then QL  QH with no work required.
This is a violation of the Clausius
statement of the 2nd law.
QH2
W
QL
5.7
No. It is not operating on a cycle.
5.8
QH1
QH2
Impossible.
Assume WR  C . Then (QL ) net  (QH ) net .
W
WC
WR
QL1
5.9
Assume WR  C . Then (QL ) net  (QH ) net .
W
QL2
Impossible. It violates the Clausius statement.
The maximum temperature drop for the seawater is 17
  T
Q mc  20 
4.18  
17 1421 kW
H
p
The efficiency of the proposed engine is

W 100
 

0.0704 or 7.04%

Q 1421
The efficiency of the Carnot engine would be
T
283
  L  
1
1

0.0567 or 5.67%
TH
300
T enet ’c i ii ps b s c iecesh C roe iciency.
h i n r lm sm os l i etxed t a tf
v os a
ie n
e n f
5.10
The maximum possible efficiency is the Carnot efficiency:
T
293
  L  
1
1

0.2038
TH
368
assuming the water is rejected at atmospheric temperature. Then
  T 0.2 4.18   
Q mc   
(95 20) 62.7 kW
H
Then
p
  0.2038 
Wmax  QH 
62.7 
12.8 kW
74

4. 5.11
T
473
a)   L  
1
1

0.679
TH
1473
QL
T
473
b) COP 
 L


0.473
QH  L TH  L 1000
Q
T
QH
T
1473
c) COP 
 H 

1.473
QH  L TH  L 1000
Q
T
5.12
First, we find the power produced:
80 
1000
 Q
W    40 
QL

17.78 kW
H
3600

W 17.78
283
 
0.4444   . TH 
1
509.4 or 236.4
C

40
TH
QH
5.13
Let T be the unknown intermediate temperature. Then
T
560
 
1
and   
1
1
2
1060
T
It is given that   2   . Substitute and find
 0.2 2
1
T
 560 
1
   T 2  T 
1.2 1
.
212
712,300  T  R or 285F
0.
745
1060
T 

5.14
q
T
20
  L   0.6.  L 0.4
1
1
qH
50
TH
For the adiabatic process (see Fig. 5.8)
k
1
TL  2 
v
v
  .  2  1/ 0.4 
0.4
0.1012

TH  3 
v
v3
Then v2 
0.1012 v3 
0.1012  
10 1.012 m 3 /kg. The high temperature is then
pv
200 
1.012
TH  2 2 

705 K or 432.2
C
R
0.287
The low temperature is
TL 0.4TH 0.4 
705.2 282.1 K or 9.1
C
5.15
5.16
The maximum possible efficiency is
T
293
output 43 
0.746
  L  
1
1

0.75.  


0.77. Impossible
actual
TH
1173
input
2500 / 60
T
510
a)   L  
1
1

0.0642 or 6.42%
TH
545
QH
T
545
b) COP 
 H 

15.57
QH  L TH  L 545 
Q
T
510
75

5. 5.17
T
T
T
   1 ,    2 ,    1    1 )(1  2 )  1  2  1
1
1
1
1 (1 
   2
1
2
3
T2
T3
T3
As an example let T1  C, T2 200 and T3   Then
100
C,
300 C.
373
473
373
 
1

0.2114,   
1

0.1745,   
1

0.349
1
2
3
473
573
573

0.2114 
0.1745 
0.2114 
0.1745 
0.349
3
or
5.18
T
273

1 L 
0.75. TH 

1092 K
engine  
TH
1
0.75
T
273
COP  L


0.333
TH  L 1092 
T
273
5.19
T
313
  2 . 1  2   . T22 773 

1
313. T2 492 K or 219
C
1
773
T2
5.20
T
293
T
COP  L .

. T 2 
138600 or T 
372 K =99
C
TH  L T 
T
293 473 
T
5.21
pv
200 
0.03
20.9
TL  4 4 

20.9 K.  
1

0.956 or 95.6%
R
0.287
473
w  qH 0.956  28.7 kJ/kg

30
5.22
p3 V3 15   /1728
144 250


586.3R
mR
0.01 
53.3
400   /1728
144 25
TH 

1173R
0.01 
53.3
586
 
1
0.500 or 50%
1173

p 
p
W  H  L mR H ln 1  L ln 3 
Q
Q
T
T
p2
p4 

Refer to Fig. 5.7: TL 
300
15 

0.01 
53.3 
1173ln
 ln
586
 ft-lbf
177
170.2
26.44 


where
k / k
1
 
T
p2 p3  2 
T
3 
3.5
3.5
1173 
586 


 
15
170.2 psia, p4  
300
 
 26.44 psia
586 
1173 


76

6. 0.4 /1.4
5.23
1
 
TL 600  
15
 
273.4
273.4 K.  
1

0.544
600
3.5
0.287 
273.4
273.4 
v3 

0.7847 m 3 /kg. p4 
4698 
300 kPa

 
100
600 
3.5
600 
v
p2  
100 
1566 kPa. p1 p2 2 
1566  
3 4698 kPa
 
273.4 
v1

p
p
4698
100 

wnet RTH ln 1  L ln 3 
RT
0.287  ln
600

273.4 ln
 kJ/kg
103
p2
p4
1566
300 


5.24
T
530
COPpump  H 
26.5
TH  L 530 
T
510

QH
75, 000
 72,170 Btu/hr


. QL 
  75, 000 
Q
Q
Q
H
L
L
 

QL mwater c p  . 72,170   
T
mwater 4.18 
12. mwater 
6014 lbm/hr

Q
75, 000

COP  H . 26.5 
. W 2830 Btu/hr or 1.11 hp


W
W
5.25
5.26
 1800 / 60
T
Q
293
 4.61 kW or 6.18 hp
COP  H 

6.511  H 
. W 


TH  L 293 
T
248
W
W

T
2
 1800
QL 
 (296  L )
T
60 20 
( 25) 3
2
(296  L )
T

TL
QL
TL
COPac 
 . 
3
. TL2  TL  620 
599
87
0

TH  L W
T
296  L
T
4.61

W 4.61 kW.
TL 
345 K or 72
C
5.27
5.28

Q
500 / 60
a) COPcool  L 
2.36

550
W 5  / 778
 500 / 60   / 778
Q
5 550
b) COPheat  H 

3.36

5  / 778
550
W
5
Q
T
10
2 
10 6
COPref  L  L .

. W 
1465 kJ
W TH  L
T
W
293   
2 10 6
77

7. 5.29
5.30
T
255
COPmax  L 

6.71
TH  L 293 
T
255
 3000 / 60
Q
COPactual  L 

3.73.
Yes, it's possible

0.746
W 10 

(Q )
278
COP1 

13.9  L 1

298 
278
W1

(Q )
253
COP2 

6.325  L 2 .

298 
253
W
 W
But W2  
1
2


(QL ) 2 (QL )1
12

 0.455   000 

. (QL )2 
0.455(QL )1 
5
1.055  kJ/s
8
6.325 13.9
360
5.31
 100
T
373
 W
  L  
1
1

0.707. QH  

141.4 kW.
TH
1273
 0.707
 Q W 
QL     141.4  41.4 kW
100
H
 t
Q 
141.4  
20 60
 H  H 
S

133.3 kJ/K
TH
1273
 t
Q  41.4  
20 60
L  L 
S

133.2 kJ/K
TL
373
 net 
S
133.2 
133.3  kJ/K which is zero except for round-off error.
0.1
5.32
5.33
5.34
5.35
Q
Q

100
100
a)  R  R 
S

0.0932 kJ/K. b)  S  S 
S

0.341 kJ/K
TR 1073
TS
293
Q
200
COP  L . 10 
. W 20 kJ. QH 220 kJ
W
W

200
220
a)  L 
S

0.76 kJ/K and b)  H 
S
0.76 kJ/K
263
289.3
263
where COP  
10
and TH 289.3 K.
TH 
263
 50, 000 / 3600
Q
530
COP  H 
4.91 
. TL 422.1
R
 4  / 778
550
530  L
T
W
50, 000  6
1/

39,820  6
1/
S 
S

15.72 Btu/  and  L 
R
S

15.72 Btu/ 
R
530
422.1
 Q W 
where QL     50, 000  
4 (550 / 778) 
3600 
39,820 Btu/hr.
H
From Problem 5.28 W 
1465 
0.00001 
1465 kJ.
1465 kJ. QH 
1465
 
10 5
 H 
S
 kJ/K and  S 
5
S
 kJ/K
5
293
2 
10 6
78

8. 5.36
T
v
0  v ln 2  ln 2
c
R
T1
v1
 / cv
R
 
T
v
or ln 2   2 
ln
T1
v
1 
T
R v
or ln 2 
 ln 2
T1
c v v1
k 
( 1)
R
T  
v
But
  (see Eq. 4.32).  2  2 
k 1

cv
T1  1 
v
To show Eq. 5.29, use Eqs. 5.28:
(k k
1)/
k
1
1  2 
p
v

   
v
p
2   1 
(k
1)
p  
v
 2  1

p1  2 
v
.
k
1
 
v
 1 .

v
2 
k
k
1
k
 
v
 1

v
2 
p1 V 1 200 
0.8

278.7 K
mR
2
0.287
 T
p 
773
 m p ln 2  ln 2  2  
S
c
R
 1.0 ln
2.04 kJ/K
p1 
278.7
 T1


5.37
T1 
5.38
mRT1
V1
p1
1/ k
and V
2
 
p
V 1  1 
p
2 
0.2 
0.287 
313
a) V 1 
=0.1198 m3 ,
150
1/1.4
V
0.2 
0.1889 
313
b) V 1 
=0.0788 m3 ,
150
k k
1/
5.39
5.40
 
p
T2  1  2 
T
p
1 
0.4 /1.4
2000 

 
520

14.7 


0.0445 m3
1/1.289
V
0.2 
0.2968 
313
c) V 1 
=0.1239 m 3 ,
150
0.2 
4.124 
313
d) V 1 
=1.769 m3 ,
150
2
150
 

0.1198  
600
 
2
150
 
0.0788  
600
 
2
150
 

0.1239  
600
 

0.0269 m3
1/1.4
V

0.0460 m3
1/1.409
V
2
150
 

1.769  
600
 

0.246 m3
2117R or 1657 F

p
T
v 
  v ln 2  ln 2 
S m c
R
,
T2  1 2 since V 
T
const

T1
v1 
p1


1500
600
a)  2 
S
0.717 ln
2.31 kJ/K since T2  
300

1500 K
300
120
1500
600
b)  2 
S
0.653ln
2.10 kJ/K since T2  
300

1500 K
300
120
1500
600
c)  2 
S
0.745ln
2.40 kJ/K since T2  
300

1500 K
300
120
1500
600
d)  2 
S
10.08ln

32.4 kJ/K since T2  
300

1500 K
300
120
79

9. 5.41
Q  mRT ln
W
 mc p ln
S
5.42
p1 p1 V 1
p
6000

RT1 ln 1 
6000    ln
500 10 6

10.2 kJ
p2
p2
200
RT 1
T2
p V 1 p2
6000   
500 10 6
200
1
ln

ln

9.51 kJ/K
T1
T1
p1
1073
6000
 T
p V1
v
m 1
,   v ln 2  ln 2
S m c
R

RT1
T1
v1

27.2  
144 10
a) m 

1.592 lbm,
53.3 
470
27.2  
144 10
b) m 
2.42 lbm,
35.1 
470
27.2  
144 10
c) m 

1.54 lbm,
55.15 
470
27.2  
144 10
d) m 

0.111 lbm,
766.4 
470
5.43
5.44
a)
b)
c)
d)
5.45

T

 mcv ln 2

T1

100

0.349 Btu/ 
R
27.7
100
 2.42 
S
0.156 ln

0.485 Btu/ 
R
27.7
100
 
S 1.54 
0.178ln

0.352 Btu/ 
R
27.7
100
 
S 0.111 
2.40 ln

0.342 Btu/ 
R
27.7
 
S 1.592 
0.171ln
400 
0.2
 mcv  .  
W
T
( 200) 

0.717(T2 
313). T2 
626.2 K
0.287 
313

T
v 
T
400 
.2
626.2
  v ln 2  ln 2  mcv ln 2 
S m c
R


.717 ln

0.443 kJ/K

T1
v1 
T1 .287 
313
313


 T
p1 V 1
V2
,  m v ln 2 R ln
S
c

RT1
V1
 T1
80 
4
400
V2

3
m

3.72 kg, 5.2 
3.72  ln
.717

.287 ln
.
  V 2 254 m
.287 
300
300
4 

80 
4
400
V2

m

5.64 kg, 5.2 
5.64 
.653ln

.189 ln
.
195 3
 V 2  m
.189 
300
300
4 

80 
4
400
V2

3
m

3.59 kg, 5.2 
3.59 
.745ln

.297 ln
.
  V 2 255 m
.297 
300
300
4 

80 
4
400
V2

3
m

.259 kg, 5.2 
.259 
10.08ln

4.12 ln
.
  V 2 259 m
4.12 
300
300
4 

m
p1 V 1
, V 1  r2h   2  

 0.1 0.2 0.00628 m3 , Q mc p 
T
RT1
T
T
V 1 2 , W m p ( V 2 V 1 ),  mc p ln 2
S
T1
T1
m
V
2
80

10. 200 
0.00628
a) m 

0.0137 kg, 9 
0.0137  T2 
1.0(
320). T2 
978 K
0.287 
320
978
V2
.00628

.0192 m3 , W 
.0137 
200(.0192 
.00628) 
0.0354 kJ
320
978
 0.0137  ln
S
1.0

0.154 kJ/K
320
200 
0.00628
b) m 

0.0208 kg, 9 0.0208 
0.824(T2 
320). T2 
843 K
0.189 
320
834
V2
.00628

.0164 m3 , W 
.0208 
200(.0164 
.00628) 
0.0420 kJ
320
834
 0.0208 
S
0.842 ln

0.168 kJ/K
320
200 
0.00628
c) m 

0.0132 kg, 9 0.0132 
1.04(T2 
320). T2 
976 K
0.297 
320
976
V2
.00628

.0191 m 3 , W 
.0132 
200(.0191 
.00628) 
0.0340 kJ
320
976
 0.0132  ln
S
1.04

0.153 kJ/K
320
200 
0.00628
d) m 
0.000953 kg, 9 0.000953 
14.21(T2 
320). T2 
984 K
4.124 
320
984
V2
.00628

.0193 m3 , W 
.000953 
200(.0193 
.00628) 
0.00258 kJ
320
984
 0.000953 
S
14.21ln

0.0152 kJ/K
320
5.46
q w  ln
RT
p2
,
p1
  ln
s
R
a) q w 
0.287  ln
300
p2
p1
4000
 kJ/kg
377
50
4000

1.26 kJ/kg 
K
50
4000
b) q w 
0.189  ln
300
 kJ/kg
248
50
4000
 
s
0.189 ln

0.828 kJ/kg 
K
50
 
s
0.287 ln
c) q w 
0.297  ln
300
 
s
0.297 ln
4000
 kJ/kg
390
50
4000

1.30 kJ/kg 
K
50
81

11. d) q w 
4.124  ln
300
 
s
4.124 ln
5.47
q  p ,
c T
4000

5420 kJ/kg
50
4000

18.1 kJ/kg 
K
50
  p ln
s c
T2
T1
1360

0.231 Btu/lbm-
R
520
1360
b) q 
0.202 
(900   Btu/lbm,  
60) 170
s 0.202 
ln

0.194 Btu/lbm-
R
520
1360
c) q 
0.248 
(900  208 Btu/lbm,  
60)
s 0.248 
ln

0.238 Btu/lbm-
R
520
1360
d) q 
3.42 
(900  2870 Btu/lbm,  
60)
s 3.42 
ln

3.29 Btu/lbm-
R
520
a) q 
0.24 
(900  202 Btu/lbm,  
60)
s 0.24 
ln
5.48
5.49
p1 V 1
100 
2
50

2.38 kg,   
S
2.38 0.287 ln

0.473 kJ/K
RT1
0.287 
293
100
V1
2
where p2 p1
   kPa .
100
50
V2
4
m
a) m 
p1 V 1
100 
2

2.11 kg,
RT1
0.287 
330
Q  mcv 
W
T

( 400) 2.11 
.717(T2 
330). T2 
594 K.  2.11 
S
.717 ln
594

0.889 kJ/K
330
100 
2
b) m 
 kg, Q  mcv   
3.2
W
T 3.2 0.653(T2 
300). T2  K
511
0.189 
330
511
  
S 3.2 .653ln

0.914 kJ/K
330
100 
2
c) m 
2.04 kg, Q  mcv  2.04 
W
T
.745(T2 
300). T2 
593 K
.297 
330
593
 2.04 
S
.745ln
0.891 kJ/K
330
100 
2
d) m 

.147 kg, Q  mcv  
W
T .147 
10.1(T2 
300). T2 
600 K
4.12 
330
600
 
S 0.147 
10.1ln

0.886 kJ/K
330
82

12. 5.50
p2
2000

1.2147 

165. Interpolate from Table F.1E:
p1
14.7
Compare with the constant specific heat prediction:
pr 2 pr1
k k
1/
 
p
T2  1  2 
T
p
1 
5.51
0.2857
2000 

 
520

14.7 

T2 
1974R.
2117R, an error of 7.24%.
p1 V 1
200 
2

4.36 kg. Q  m .
W
u
RT1
0.287 
320
 
40 40
a) 500 
  4.36 
10 60
0.717(T2 
320). T2 
787 K
1000
T
787
 mcv ln 2 4.36 
S
0.717 
ln
2.81 kJ/K
T1
320
b) From Table F.1 u1 228.4 kJ/kg and  
1.767 kJ/kg 
K.
1
 
40 40
500 
  4.36(u2 
10 60
228.4). u2 
563 kJ/kg .
1000
T
764
T2 
764 K and  2.667. Then p2 p1 2 200 
478 kPa
2
T1
320
m
478 

 4.36 
S
2.667 
1.767 
0.287 ln
2.83 kJ/K
200 


5.52
a) Q W mcv  .
T
k / k
1
200, 000
2 
0.171(T2 
520). T2 
1272R or 812
F
778
 
T
1272 

p2 p1  2   
16
366 psia
 
T
520 

1 
200, 000
b)
2 u2 
(
88.62). u2 217.2 Btu/lbm. T2 
1245R or 785
F
778
27.5
pr1 
1.215, p2 27.5.  p2  
16

362 psia
1.215
5.53
3.5
600   
200 10 6
V2
1000
m

0.001427 kg . T2  1
T
293 

1465 K
0.287 
293
V1
200
a) Q mc p  
T 0.001427 
1.0(1465 
293) 
1.67 kJ
b) Find the enthalpies in Table F.1:
Q m(h2  1 ) 0.001427 
h
(1593.7 
293.2) 
1.86 kJ
83

13. k k
1/
5.54
 
p
2000 

a) T2  1  2   
T
300
670.2 K
 
p
120 
1 
W  
0.2 0.717(670.2 
300)  kJ
53.1
2000
b) ( 
1.702) 
0.287 ln
.  2.51 kJ/kg  u2 481 kJ/kg
K.
2
2
120
W  
0.2 (481 
214) 
53.4 kJ
k k
1/
5.55
 
p
a) T2  1  2 
T
p
1 
0.2857
0.2857
500
 
  
500
20 
 K. W  
199
4 .717(199 
500) 
863 kJ
20
.  
1.296. T2 200 K
2
500
W  
4 (143 
359) 
864 kJ
b)  
2.220 
0.287 ln
2
5.56
a) s1 2.047 
0.85(4.617) 
5.971, h1  
721 0.85(2048) 2462
p2  
800
8.877
 s2 
h2 4462 
 
s 8.877 
5.971 2.91 kJ/kg  and T2  
K
829 C
b) TK solution:
q h2  1 . 2000 h2 
h
2462. h2 4462.
Rule Sheet
; This is a closed system, so the first law is q = u2 - u1 + p2*v2 - p1 * v1 or
q = h2 - h1
dels = s2 - s1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar,
; John S. Gallagher, and George S. Kell, Hemisphere Publishing Corp., 1984.
; STEAM STEAM8SI.TKW
Variable Sheet
Status Input
Name
Output
Unit
`
800
0.85
800
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
170
2460
5.97
0.204
'SAT
927
4460
8.87
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
84
Comment
Thermal Sciences, Potter & Scott
P5-56.tkw Problem 5-56
*STEAM8SI.TKW Steam, 1-8 States, SI Units
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature (Enter a guess value.)
Pressure
Enthalpy
Entropy

14. 2000
5.57
v2
x2
phase2
q
dels
0.691
'mngless
'SH
2.9
m^3/kg
kJ/kg
kJ/(kg*K)
Specific Volume
Quality
Phase
Heat transfer
Entropy change
p1 600 psia, T1  2 486F, s1 
T
0.672 
0.4(0.774) 
0.982
u1 
506.6 
0.4(608.4) 
750 kJ/kg, v1 
0.0201 
0.4(0.7501) 
0.320 m 3 / kg
p2 
300 psia 
1.56, u2 
1153, v2 
1.73
 s2 
T2   
486 F
   
S m s 2 (1.56 
0.982) 
1.158 Btu/lbm-
R
Q 
m u W
600 
144(0.77 
0.32)  
450 144(1.733 
0.77)

2 (1153 
750)  
2

1066 Btu
778
Note: The work was estimated using graphical integration (a straight line was assumed
between the saturated vapor point and state 2).
5.58
a) v1  f at 20 w p( v2  1 ) 
v
C.
v
400(0.7726 
0.001) 
309 kJ/kg
q  2  1  2964.4 
u u w
83.9  
309 3189 kJ/kg
or
q h2  1 
h 3273  
84 3189 kJ/kg
 s2  1 
s
s 7.899 
0.2965 7.604 kJ/kg 
K
b) TK solution:
Rule Sheet
;Assume that the water is in a closed system. The first law is q = u2 - u1 + p2*v2 - p1 * v1 or
q = h2 - h1
dels = s2 - s1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George
S. Kell, Hemisphere Publishing Corp., 1984.
;STEAM STEAM8SI.TKW
Variable Sheet
Input
Name
Output
Unit
`
20
400
T1
p1
h1
s1
v1
x1
phase1
84.2
0.293
0.001
'mngless
'CL
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P5-58.tkw Problem 5-58
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
85

15. 400
400
3270
7.9
0.773
'mngless
'SH
q
dels
5.59
T2
p2
h2
s2
v2
x2
phase2
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
3190
7.6
kJ/kg
kJ/(kg*K)
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer per unit mass
Entropy change
V 6 
10 3


0.001017  1 (7.671 
x
0.001). x1 
0.0002585
m
2
u1 251.1 
0.0002585(2456.6 
251.1) 251.7 kJ/kg
Q
1000
Q m(u2  1 ) W . u2  1  251.7 
u
u

751.7 kJ/kg
m
2
v2  1 
v 0.003 m 3 / kg and u2 
752 kJ/kg. Locate state 2 by trial-and-error:
v1 
Guess T2   : 0.003 
170 C
0.0011 x2 (0.2428 
0.001). x2 
0.00786
751.7 718.3 x2 (2576.5 
718.3). x2 
0.0178
Guess T2 
177 C: 0.003 
0.0011 x2 (0.2087 
0.0011). x2 0.00915

751.7 750.0 x2 (2581.5 
750.0).  x2 
0.00093

A temperature of 176 C is chosen. We interpolate to find
0.003 0.0011 x2 (0.2136 
0.0011).  x2 
0.00894.
S 2 m( s f x2 s fg ) 2 
(2.101 
0.00894 
4.518) 4.28 kJ/K
5.60
a) Q  m . m  m . 0.1(3674 
W
u
h W
u
1087)  
W 0.1(3279 
1082)
W  kJ and  
39
S 0.1(7.370 
2.797) 
0.457 kJ/K
b) TK solution:
Rule Sheet
;This is a closed system, so the work of a frictionless process is W = INT pdV, and for a
; constant pressure process this becomes
W = m* p1 * (v2 - v1)
delS = m* (s2 - s1)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George
S. Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
Variable Sheet
Input
Name
Output
Unit
`
T1
250
C
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P5-60.tkw Problem 5-60
Temperature
86

16. 4000
0
600
4000
0.1
5.61
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
1090
2.8
0.00125
'SAT
3670
7.37
0.0988
'mngless
'SH
W
m
delS
39
0.457
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
kg
kJ/K
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Work
Mass of steam
Entropy change of system
u1 2507 kJ/kg, s1 s2 
7.356 
0.832 x2 (7.077). x2 
0.922.
u2 251 
0.922(2205) 2284 kJ/kg
W m(u1  2 ) 2(2507 
u
2284) 447 kJ
5.62
2
a) v1   
0.4 0.001 
3.992 x1 . x1 
0.1. Then s1 
1.69 and u1 
533
5
p2  MPa 
5
570 kJ/kg. W m(u1  2 ) 
u
5(570 
533)  kJ
185
 u2 
s2 s1 
1.69 
b) TK solution:
Rule Sheet
;This is a closed system, so for an adiabatic process, W = U1 - U2 or
W = m * (u1 - u2)
; First law
v1 = V1/m
; Definition of specific volume
s2 = s1
; for an isentropic process
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George
S. Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
Variable Sheet
Status Input
Name
Output
Unit
`
40
T1
p1
h1
s1
75.9
550
1.69
C
kPa
kJ/kg
kJ/(kg*K)
87
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P5-62.tkw Problem 5-62
Temperature
Pressure
Enthalpy
Entropy

17. 0.4
5000
5
2
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
W
m
u1
u2
V1
m^3/kg
0.1
'SAT
136
577
1.69
0.00107
'mngless
'CL
-190
534
572
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
kg
kJ/kg
kJ/kg
m^3
Specific Volume (transfer value to input)
Quality
Phase
Temperature (starting guess needed)
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Work
Mass
Internal energy, state 1
Internal Energy, state 2
Specific volume, state 1
5.63
Q m 
h 10(1150 
8.02)  420 Btu.  
11,
S 10(1.757 
0.016) 
17.4 Btu/ 
R
5.64
The heat that enters the ice leaves the water:
a) Assume that all the ice does not melt:
20      
1.9 5 330 m 10 4.18(20). m 
1.96 kg and T2  
0C
T
Q
T
 mi (c p )i ln 2i  i  w (c p )w ln 2 w
S
m
T1i Ti
T1w
273 1.96
273

  
330 10 4.18 
ln

0.064 kJ/K
253 273
293
b) Assume that all the ice melts:
 
5 1.9 ln
20      
1.9 5 330 5 5 4.18(T2  40 
0)
4.18(20  2 ). T2  
T
8.0 C
T
Q
T
T
 mi (c p )i ln 2i  i  iw (c p )w ln 2iw  w (c p )w ln 2 w
S
m
m
T1i Ti
T1iw
T1w
  ln
5 1.9
5.65
273 5 
330
281
281


5 4.18ln
 
40 4.18ln

0.378 kJ/K
253
273
273
293
5  /1728
1.2
mi 
0.199 lbm,
mw  lbm
1
0.01745
The heat that enters the ice leaves the water. Assume that not all of the ice melts:
0.199 
0.49    melt     
(32 0) m
143 1 1.0 (60 32). mmelt 
0.178 lbm
492 0.178 
143
492
 0.199 
S
0.49 
ln

 
1 1.0 ln

0.166 Btu/lbm
460
460
520
88

18. 5.66
T
293
  L  
1
1

0.489 or 48.9%. w qH  L  H   L  
q T s T s 300 kJ/kg
TH
573
300
 
s

1.071 
5.705  1 . s1 4.634 kJ/kg 
s
K
573 
293
s1 4.634 
3.254 x1 (2.451). x1 
0.563
5.67
a) For the cycle, the work output equals the heat input:
W  . 500 (275.6 
T s
45.8)( s2 
3.027). s2 
5.203 kJ/kg 
K
At 6 MPa 5.203 
3.027 x2 (2.863). x2 
0.760
b) TK solution:
Rule Sheet
;The net work of a Carnot cycle is the enclosed area on a Ts diagram. The adiabatic ompression
(process 1-2) is entirely within the saturated vapor region, so the specified pressures determine the
upper and lower temperatures.
w = (T2 - T1)*(s3 - s2)
; work = Ts diagram area
s1 = s2
T3 = T2
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp.,
1984. Stm8si.tkw
Input
Name
Output
`
10
3.027
6000
0
275.6
5.203
500
T1
p1
s1
v1
x1
phase1
T2
p2
s2
v2
x2
phase2
T3
p3
s3
v3
x3
phase3
w
45.83
4.637
0.317
'SAT
275.6
3.027
0.001319
'SAT
6000
0.02498
0.7603
'SAT
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P5-67.tkw
Problem 5.67
C
Temperature
kPa
Pressure
kJ/(kg*K)
Entropy (transfer value to input)
m^3/kg
Specific Volume
Quality
Phase
C
Temperature
kPa
Pressure
kJ/(kg*K)
Entropy
m^3/kg
Specific Volume
Quality
Phase
C
Temperature (transfer value to input)
kPa
Pressure
kJ/(kg*K)
Entropy (transfer value to input)
m^3/kg
Specific Volume
Quality
Phase
*THUNITS.tkw Units for thermo
kJ/kg
89

19. 5.68
s1 2.046 
0.15(7.077) 2.738 kJ/kg   
K. s 6.664 
2.738 
3.93 kJ/kg 
K
wnet qnet  
s T 3.93(170.4 
60.1) 433 kJ/kg
60.1 
273
 
1

0.249 or 24.9%
170.4 
273
5.69
a) s4 
0.704 
0.2(7.372) 2.178 kJ/kg  s2 
K.
5.704 kJ/kg 
K
qH  H   
T s 573 (5.704 
2.178) 2020 kJ/kg
b) TK solution:
Rule Sheet
s3 = s2
s1 = s4
q12 = T1 * (s2 - s1)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Variable Sheet
Input Name
Output
Unit
Comment
`
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P5-69.tkw Problem 5.69
300
T1
C
Temperature
p1
kPa
Pressure
s1
2.18
kJ/(kg*K)
Entropy
x1
Quality
phase1
Phase
300
T2
C
Temperature
p2
8580
kPa
Pressure
s2
5.7
kJ/(kg*K)
Entropy
1
x2
Quality
phase2 'SAT
Phase
50
T3
C
Temperature
p3
kPa
Pressure
s3
5.7
kJ/(kg*K)
Entropy
x3
Quality
phase3
Phase
50
T4
C
Temperature
p4
12.3
kPa
Pressure
s4
2.18
kJ/(kg*K)
Entropy
0.2
x4
Quality
phase4 'SAT
Phase
*THUNITS.tkw Units for thermo
q12
2020
kJ/kg
90

20. 5.70
QL
T
q
253
COP 
 L


3.614  L where w 
s T
QH  L TH  L 323 
Q
T
253
w
s1 s4 0.704 
0.15(7.372) 
1.81, s3 s2 
5.704 kJ/kg 
K
 Q
 s T 0.02 
a) Wout   m 
(5.704 
1.81)(300  
50) 19.5 kW
net
b) 5.704 
0.704 x3 (7.372).
5.71
x3 
0.678
a) A refrigeration cycle is a reversed power cycle. Heat is added to the R134a from
4 to 1 and rejected from 2 to 3:
QL
T
q
253
COP 
 L


3.614  L where wnet 
s T
QH  L TH  L 323 
Q
T
253
wnet
wnet   
[50 ( 20)] [0.901 
0.434] 
32.7 kJ/kg
qL COP  net 
w
3.614 
32.7  kJ/kg
118
s4 s3 0.434 
0.0996 x4 (0.9332 
0.0996). x4 
0.401
c) TK solution:
Rule Sheet
q23 = h3 - h2
; for the constant pressure process
q41/ q23 = T4/T3 ; for a Carnot refrigerator
s4 = s3
; R134a tables based on 'Thermodynamic Properties of HFC-134a'
; DuPont Technical Information, which is based upon the Modified
; Benedict-Webb-Rubin equation of state.
R134a8SI.tkw
Variable Sheet
Input
Name
Output
Unit
`
-20
50
1
50
T1
p1
s1
x1
phase1
T2
p2
h2
s2
x2
phase2
T3
p3
h3
s3
C
kPa
kJ/(kg*K)
1320
276
0.913
C
kPa
kJ/kg
kJ/(kg*K)
'SAT
1320
124
0.442
C
kPa
kJ/kg
kJ/(kg*K)
Comment
Thermal Sciences, Potter & Scott
*R1348si.tkw, R134a, 1-8 States, SI units
P5-71.tkw Problem 5.71
Temperature
Pressure
Entropy
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
91

21. 0
x3
phase3
-20
T4
p4
0.442 s4
x4
phase4
q23
q41
'SAT
133
C
kPa
kJ/(kg*K)
0.402
'SAT
-152
-119
kJ/kg
kJ/kg
Quality
Phase
Temperature
Pressure
Entropy (transfer value to input)
Quality at beginning of heat addition process
Phase
*THUNITS.tkw
Units for thermo
Amount of heat rejected
Amount of heat gained from refrigerated space
5.72
Refer to the numbers in Problem 5.12:
QH QL 40  
40 60 80 000
 

 which verifies the inequality of Clasius.
0
TH TL
509.4
283
5.73
wnet  . 350  (250.4 
s T
s
75.9).  2.006 kJ/kg 
s
K
qH  H  
T s (250.4 
273)(2.006) 
1050 kJ/kg
qL  L  
T s (75.9 
273)(2.006) 
700 kJ/kg
qH qL 1050
700
 

0.0002 which is essentially zero.
TH TL 523.4 348.9
qH qL
30 1.326
 

 .
0
TH TL 473 20.9
5.74
Using values from Problem 5.21:
5.75
a) Using values from Problem 5.66, we have
qH qL 613.5 313.5
300
 

 using qH 
0,

613.5 kJ/kg.
TH TL
573
293
0.489
b) TK solution:
Rule Sheet
;Problem 5.66
eta = (T1 - T4) /T1
q12 = wnet/ eta
q12 = T1 * (s2 - s1) ; the sought value of s1 is shown on the Variable Sheet
Problem 5.75: The cyclic integral of deltaq/T = q12/T1 + 0 + q34/T3 + 0 , but q12/T1 = - q34T3;
therefore , the cyclic integral of delta q/T = 0, as called for by the Clausius inequality for a
reversible cycle.
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Input
Name
Output
`
300
T1
C
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P5-75.tkw Problem 5.75
Temperature
92

22. p1
s1
x1
T2
p2
s2
x2
T3
p3
s3
x3
T4
p4
s4
300
1
20
20
614
300
5.76
8580
eta
q12
wnet
4.63
kPa
kJ/(kg*K)
0.489
0.563
8580
5.7
C
kPa
kJ/(kg*K)
C
kPa
kJ/(kg*K)
C
kPa
kJ/(kg*K)
kJ/kg
kJ/kg
Pressure
Entropy (transfer value to input)
Quality
Temperature
Pressure
Entropy
Quality
Temperature
Pressure
Entropy
Quality
Temperature
Pressure
Entropy
*THUNITS.tkw
Units for thermo
Thermal efficiency
Heat input to cycle
Net work of cycle
a) mc (c p )c  c mw (c p ) w  w . 5 
T
T
0.093(200  2 )   T2 
T
10 1.0(
50). T2 
56.7 
F
T2
516.7

5 0.093ln

0.1138 Btu/ 
R
(T1 )c
660
T
516.7
 w mw (c p ) w ln 2   ln
S
10 1.0

0.1305 Btu/ 
R
(T1 ) w
510
 c mc (c p )c ln
S
 universe  c  S w 
S
S 
0.1138 
0.1305 
0.0167 Btu/ 
R
b) TK solution:
Rule Sheet
;mc * cc * (Tc - T2) = mw * cw * (T2 - T1) ; This is the first law applied to system of copper and
;
water, under the assumption that there is no heat transfer.
delSuniv = delSc + delSw
delSc = mc* cc * ln(T2/Tc)
; from delSc = INT (delQ/T) = INT(m * cc*dT/T)
delSw = mw * (s2 -s1)
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,
; and Kell, Hemisphere Publishing Corp., 1984.
Stm8e.tkw
Variable Sheet
Input
Name
`
50
0
T1
h1
s1
x1
phase1
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English units
P5-76.tkw
Problem 5.76
F
Initial temperature of water
18.1
B/lbm
Enthalpy
0.0361 B/(lbm*R) Entropy
Quality
'SAT
Phase
93

23. T2
h2
s2
x2
phase2
56.6
F
Final temp of water and copper (starting guess needed)
24.7
B/lbm
Enthalpy
0.0489 B/(lbm*R) Entropy
0
Quality
'SAT
Phase
*THUNITS.tkw Units for thermo
5
mc
lbm
Mass of copper
0.092 cc
B/(lbm*R) Specific heat of copper
200 Tc
F
Initial temperature of copper
10
mw
lbm
Mass of water
delSuniv 0.0159 B/R
Entropy change of the universe
delSc
-0.113 B/R
Entropy change of the copper
delSw
0.129 B/R
Entropy change of the water
1
cw
B/(lbm*R) Specific heat of water
5.77
a)  universe 0.264 
S
0.156 0.104 Btu/ 
R
0.2
v1   m 3 /kg and from Tables C.1 and C.2 we observe that p2  MPa
0.1
2
2
and T1 212.4 s1 
C,
6.342 kJ/kg  and u1 2600 kJ/kg . Since for a rigid volume
K,
trial-and-error provides
v2  1
v
At p2  MPa: v2 
0.4
0.0011 
0.2(0.4625 
0.0011) 
0.0934
At p2  MPa: v2 
0.3
0.0011 
0.2(0.6058) 
0.122
Obviously, at v2  state 2 lies between 0.3 and 0.4 MPa. Interpolate:
0.1
0.1 
 0.122 
p2 

0.3 0.377 MPa

 0.1  
0.122 
0.0934 

Interpolate to find s2 and u2 :
s2 
1.753  
0.2 5.166 2.786 kJ/kg  u2   
K,
594 0.2 (2551 
594) 
986 kJ/kg
Then
Q W  (u2  1 ) 2 
m
u
(986 
2600) 
3230 kJ (heat flows to surr.)
3230
 universe msystem  S surr 2 
S
s

(2.786 
6.342) 

3.55 kJ/K
30 
273
b) TK solution:
Rule Sheet
v1 = V1 / m
; Definition of specific volume
v2 = v1
; Volume of steam is constant
Q12 = m * (u2 - u1) ;heat added to steam
Q12 = - Qsurr
delS = m* (s2 - s1)
; Entropy change of steam
delSsurr = Qsurr/Tsurr ; entropy change of surroundings
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
94

24. Input
Name
`
0.1
1
0.1
0.2
0.2
2
30
5.78
5.79
T1
p1
s1
v1
x1
phase1
T2
p2
s2
v2
x2
phase2
u1
u2
V1
m
Q12
delS
delSsurr
Qsurr
Tsurr
delSuniv
Variable Sheet
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P5-77.tkw Problem 5.77
212
C
Initial temperature of steam
1990
kPa
Pressure
6.34
kJ/(kg*K)
Entropy
m^3/kg
Specific Volume (transfer value to input)
Quality
'SAT
Phase
141
C
Temperature
371
kPa
Pressure
2.78
kJ/(kg*K)
Entropy
m^3/kg
Specific Volume (transfer value to input)
Quality
'SAT
Phase
2600
kJ/kg
Initial internal energy
985
kJ/kg
Final internal energy
*THUNITS.tkw Units for thermo
m^3
Volume of steam
kg
Mass of steam
-3230
kJ
Heat added to steam
-7.11
kJ/K
Entropy change of steam
10.7
kJ/K
Entropy change of the surroundings
3230
kJ
Heat added to surroundings
C
Temperature of the surroundings
3.54
kJ/K
Entropy change of the universe
30  
144 6
T1 
486.3R, T2 
1459R , Q  
1 0.24(1459 
486.3) 233 Btu
1
53.3
1459
a)  air  
S
1 0.24 ln

0.264 Btu/ 
R
486.3

233
b)  surr 
S

0.156 Btu/ 
R
1460
c)  universe 0.264 
S
0.156 0.104 Btu/ 
R
2
0.287 
573
120
p1 

164.5 kPa, T2  
573
418 K
2.0
164.5
418
a)  air 2 
S
0.717 ln

0.452 kJ/K
573
222
b) Q 2 
0.717(573 
418) 222 kJ.  universe 
S
0.452 

0.289 kJ/K
300
95

25. 5.80
a) Q m 
h 3(2793 
852) 
5821 kJ.  steam 
S
3(2.331 
6.433) 
12.31 kJ/K
5821
 universe 
S
12.31 

7.56 kJ/K
293
b) TK solution:
Rule Sheet
Q12 = m * (h2 - h1) ; from the first law, Q12 = U2 - U1 + W = m* (u2 - u1 + p1 * (v2 - v1)
; for a quasiequilibrium constant pressure process of a closed system
delSuniv = delSsys+ delSsurr
Q12 = -Q12surr
delSsurr = Q12surr/Tsurr
delSsys = m * (s2 - s1)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Input
Name
Output
`
200
1
200
0
3
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
m
delSuniv
delSsys
1554
2793
6.431
0.1273
'SAT
1554
852.4
2.331
0.001156
'SAT
7.553
-12.3
delSsurr
20
19.85
Q12
Q12surr
Tsurr
-5820
5820
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
P5-80.tkw
Problem 5.80
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K)
Entropy
m^3/kg
Specific Volume
Quality
Phase
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K)
Entropy
m^3/kg
Specific Volume
Quality
Phase
kg
Mass of steam in system
kJ/K
Entropy change of the universe
kJ/K
Entropy change of the system
Entropy change of the
kJ/K
surroundings
kJ
Heat added to the system
kJ
Heat added to the surroundings
C
Temperature of the surroundings
96

26. 5.81
s1 
1.53 
0.8(5.598) 
6.008. u1 
504.5 
0.8(2025) 2124 kJ/kg
p2  MPa
0.8

650
C, u 2 
3389, s2 
8.391
 T2 
v2  1 
v .0011 
.8(.8846) 
.709 
400  
10 6
Q
(3389 
2124) 
0.714 kJ
0.709
400  
10 6

0.714
 universe 
S
(8.391 
6.008) 

6.11   kJ/K
10 4
0.709
973
5.82
The enthalpy leaving the heater equals the enthalpy entering the heater:
 1283  
 1.678 lbm/sec
262.2(8 ) ms 
ms
8 48.1. ms 
  S
S S 
out
in
(8 
1.678) 
0.4273  
8 0.0933 
1.678 
1.768 
0.432 Btu/sec-
R
5.83
 
a) Qt m(h2  1 )   2 
h WT
(2609.7 
3658.4) 
2000 
97.4 kJ/s

Qsurr  
QT 97.4 kJ/s

Q
97.4

 m(
S prod Sc.v. s2  1 )  surr  
s
0 2(7.909 
7.168) 

1.80 kW/K
Tsurr
303
b) TK solution:
Rule Sheet
Wdot = mdot * (h1 - h2) +Qdot ; first law for steady flow turbine with zero change in ke and pe
Qdot = - Qdotsurr
delSuniv = delSsys + delSsurr = 0 + delSsurr
delSsurr = mdot* (s2 - s1) + Qdotsurr/Tsurr
Input
Name
Output
`
600
6000
20
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
3660
7.17
0.0652
'mngless
'SH
60.1
2610
7.91
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P5-83.tkw Problem 5.83
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K) Entropy
m^3/kg
Specific Volume
Quality
Phase
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K) Entropy
97

27. v2
x2
phase2
1
2000
2
0
7.65
m^3/kg
'SAT
Wdot
mdot
Qdot
delSuniv
delSsys
-98.3
0.001
kW
kg/s
kJ/s
kJ/(K*s)
kJ/(K*s)
delSsurr
30
1.8
kJ/(K*s)
Qdotsurr
Tsurr
98.3
kJ/s
C
Specific Volume
Quality
Phase
*THUNITS.tkw Units for thermo
Poer output
Mass rate of flow
Rate of heat transfer to the system
Rate of entropy increase of the universe
Rate of entropy increase of the system
Rate of entropy increase of the
surroundings
Rate of heat transfer to the surroundings
Temperature of the surroundings
0.4 /1.4
5.84
T
323
100 
p2 p1 2  
100

118.3 kPa. T2  
323

T1
273
118.3 


308 K
1000 
(323 
308) 
173.8 m/s
V3  2c p (T2  3 )  2 
T
Note: The factor of 1000 converts kJ to J, so the units will work out.
0.4 /1.4
5.85
V22
100
 
T2   
300
272.5 K.

1.0(300 
272.5). V2 
234.5 m/s
140
2
1000
 
100

m

( 0.01252 ) 
234.5 
0.147 kg/s
0.287 
272.5
Note: We assumed p2  kPa since the exiting pressure was not given.
100
0.4 / 1.4
5.86
85
 
T2 423  
130
 
5.87
90 
a) T2 
1173  
800
 

374.6 K.
V22  2
40

1.0(423 
374.6). V2 
309 m/s
2
1000
0.4 /1.4
628.4 K. wT 
1.0(628.4 
1173) 
545 kJ/kg
90
.  2.525 kJ/kg  h2 
K.
682 kJ/kg
2
800
wT h2  1 ) 
(
h
(682 
1246) 
564 kJ/kg
b)  
3.152 
0.287 ln
2
5.88
a) s1 s2 
1.636, h1 
1180,
p2 
800 psia 
1449 kJ/kg
 h2 
s2 
1.636 
 
WC m(h2  1 ) 
h
6(1449 
1180)  / 550 2280 hp
778
Note: The factor 778 converts Btu to ft-lbf, and 550 converts ft-lbf/sec to hp.
98

28. b) TK solution:
Rule Sheet
Wdotin = mdot * (h2 - h1) ; First law, assuming negligible change in ke and pe and no heat
transfer
s2 = s1 ; for a reversible adiabatic process
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher, ; and Kell, Hemisphere
Publishing Corp., 198
Variable Sheet
Input
Name
Output
Unit
`
300
1
800
1.64
6
5.89
T1
p1
h1
s1
x1
phase1
T2
p2
h2
s2
x2
phase2
Wdotin
mdot
67
1180
1.64
'SAT
888
1450
F
psi
B/lbm
B/(lbm*R)
F
psi
B/lbm
B/(lbm*R)
'mngless
'SH
2280
hp
lbm/s
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English units
P5-88.tkw
Problem 5.88
Temperature
Pressure
Enthalpy
Entropy
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy (transfer value to input)
Quality
Phase
*THUNITS.tkw Units for thermo
Power input to compressor
Mass flow rate
s2 s1 
7.168 
0.649 x2 (7.502). x2 
0.869
 h
h2  
192 0.869(2393) 2271 kJ/kg. WT m 2(3658 
2271) 
2774 kW
5.90
a) s2 s1 
1.681 
0.175 x2 (1.745). x2 
0.863
h2  
94 0.863(1022) 
976 Btu/lbm. wT  
h 1512  
976 536 Btu/lbm
 3000  / 778
W
550
wactual  T 

382 Btu/lbm
 20, 000 / 3600
m
w
382
  actual 

0.713 or 71.3%
T
ws
536
99

29. b) TK solution:
Rule Sheet
;State 1 = turbine throttle state. State 2 = Ideal (isentropic turbine exhaust state. State 3 = actual
turbine exhaust state
wi = h1 - h2
; first law for ideal turbine with negligible changes in ke and pe
s2 = s1
; for isentropic turbine. Note that the rule below for a state identified by p2
and ;s2 in model Stm8e.tkw is modified so that it is unnecessary to transfer value of s2 from
output to input manually..
if and (given('p2),known('s2)) then call pands(p2,s2;T2,h2,v2,x2,phase2)
wa = Wdot/ mdot ; actual turbine work
etat = wa / wi ; definition of turbine efficiency
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,
; and Kell, Hemisphere Publishing Corp., 1984. Stm8e.tkw
Input
Name
Output
`
1000
800
2
2
T1
p1
h1
s1
x1
phase1
T2
p2
h2
s2
x2
phase2
T3
p3
h3
s3
x3
phase3
wi
wa
3000 Wdot
20000 mdot
etat
1510
1.68
'mngless
'SH
126
976
1.68
0.863
'SAT
536
382
0.712
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English
P5-90.tkw Problem 5.90
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
Quality
Phase
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
Quality
Phase
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
Quality
Phase
*THUNITS.tkw Units for thermo
B/lbm
Work of ideal turbine
B/lbm
Work of actual turbine
hp
Power output of actual turbine
lbm/h
Mass rate of flow
Turbine efficiency
100

30. 5.91
s2 s1 
7.839 
0.832 x2 (7.077). x2 
0.990. h1 
3694 kJ/kg
 3.5(3694 
h2 251 
0.990(2358) 2585 kJ/kg. WT 
2585) 
3880 kW
5.92
T1 600
C

3677,
 h1 
s2 s1 7.435 
 

 0.198 kg/s
WT m(h1  2 ). 200 m(3678 
h
2666). m 
b) TK solution:
a) h2 2666, x2 
1.0,
Rule Sheet
Wdot = mdot * (h1 - h2) ;
First law, assuming negligible change in ke and pe
s1 = s2
; for isentropic turbine. Note that the rule below for a state identified by T1 and s1 in
;model Stm8e.tkw is altered to make it unnecessary to transfer value of s1 from output to input.
if and (given('T1),known('s1)) then call Tands(T1,s1;p1,h1,v1,x1,phase1)
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,
; and Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Input
Name
Output
`
600
T1
p1
h1
s1
x1
phase1
T2
p2
h2
s2
x2
phase2
80
1
200
5.93
Wdot
mdot
3530
3680
7.43
'mngless
'SH
93.5
2670
7.43
'SAT
0.197
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI uniits
P5-92.tkw Problem 5.92
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K)
Entropy
Quality
Phase
C
Temperature
kPa
Pressure
kJ/kg
Enthalpy
kJ/(kg*K)
Entropy
Quality
Phase
*thunits.tkw
Units for thermo
kW
Power output
kg/s
Mass rate of flow
h1 
1984 kJ/kg, s2 s1 
6.710 
0.261 x2 (8.464). x2 
0.762. h2 a 
2534
2923-2534
Then h2 s 73.5 
.762(2460) 
1948 kJ/kg. 

.399 or 39.9%
2923 
1948
5.94
a) h1 
1474, s1 s2 
1.759 
0.2198 x2 (1.6426), x2 
0.937
h2 s 
120.9 
0.937(1006) 
1064. wT 
0.85(1474 
1064) 
348 Btu/lbm
 
 348. m 
 6.09 lbm/sec
b) W mw . 3000  / 778 m 
550
T
T
101

31. 5.95
The efficiency is the actual kinetic energy increase divided by the maximum possible
increase:
 a (V22  12 )a
KE
V

 2
 s (V2  12 ) s
KE
V
0.2857
T2 s
 
p
1 2 
T
p1 
0.2857
100
 
293 
 
115
 
281.5 K
V22  12  p (T2  1 )  2  
V
c
T
10 2 1000 
(281.5 
293).
1502  2
10
 2
0.97
152  2
10
5.96
or
V2  m/s
152
97%
The 1st law:
V22  12 2(h1  2 ) 2 
V
h
(2874  2 ) 
h
1000
.
To find h2 we use s2 = s1 = 7.759 kJ/kg K and p2 = 100 kPa. Therefore h2 = 2841 kJ/kg.
(V 2  2 )
V
V22  2
20
 22 12 a .
0.85 
.
V2 
238 m/s
2
(2874 
2841) 
1000
(V2  1 ) s
V
102