The document discusses various concepts and calculations related to the first law of thermodynamics, focusing on processes involving heat transfer, work, and internal energy changes in thermodynamic systems. It includes specific computations for adiabatic processes, superheated steam, and interpolations using thermodynamic tables. The calculations illustrate principles such as heat loss, energy conservation, and temperature changes in rigid containers under different conditions.

1. FOUR
The First Law of Thermodynamics
4.1FE
B
4.2FE
A
4.3FE
C
Because the container is insulated we assume the heat transfer is zero. For a rigid container:
Q  electric  v 
W
mc T
    
( 400 0 60) 10 10.085(T2 
20). T2  
29 C
Table B.2 was used to find cv .
4.4FE
A
For a rigid container the boundary work is zero:
q w   2  1 2961 
u u u
2553.6 
407 kJ/kg
where we used v2  1 and interpolated for
v
u2 in Table C.3 as follows:
v2 v1 
0.4625 
2961 kJ/kg
 u2 
T2 400
C 
4.5FE
C
For this adiabatic process we use
(k  / k
1)
 
p
 = m(u2  
W
 
u 
and T2  1  2 
T
293  0.2857 
6
488.9 K
p1 

  v (T2  1 )  
W mc
T
2 0.717 
(489 
293)  kJ
281
4.6FE
C
This is an adiabatic process so that


Then
 = m(u2 u   v (T2  1 ) 
W
  W mc


T
         .  
( 10 100 10 50) (5 60) 2 717 T
T 313.8 
C

V2

1 0.287  ln 0.5 
373
74.2 kJ .
V1
The minus sign indicates that heat is rejected.
Q 
W mRT ln
4.7FE
D
4.8FE
D
 mcv   
W
T 5 0.717  
100 358.5 kJ
4.9FE
A
p2 = p1 = 400 kPa. Use Table C.3 for this superheated steam:
V1
1 0.7726 
0.7726 m 3 .
V
2
V 1  
2 0.05 
0.6726 m 3 and v2 
0.6726 m 3 / kg.
Interpolate at 400 kPa and 0.6726 m3/kg:
0.7726 
0.6726
T2 400 
(400 
300)  
315 C
0.7726 
0.6548
4.10FE D
Use Table C.3 at 315 and interpolate to find h2:
C
Q  
m h 1 (3098 
3273)  kJ
175
31

2. 4.11FE A
Use Tables C.3 and C.2 with W = 0:
Q  (u3  2 )  
m
u
1 (1245 
2829) = 1584 kJ
We interpolated at 315 in Table C.3 to find u2 = 2829 kJ/kg and used v2 = v3 =
C
0.672 m3/kg which is in the quality region. Then using Table C.2,
x3 = 0.3963 and u3 = 1245 kJ/kg.
4.12FE D
If T = const,
Q 
W mRT ln
4.13FE C
p1
100
 
10 0.287  
333 ln

1987 kJ
p2
800
For p = const,
q h2  1 
h 3273 
604.7 
2668 kJ/kg
4.14FE C
For p = const,
Q  (h2  1 ).
m
h
170   h2 
1 (
3108).
h2 
3278 and T2  
402 C
We interpolated at 320 in Table C.3 to find h1 and then to find T2.
C
4.15FE D
For p = const,
W  ( v2  1 )   
mp
v
1 400 (0.7749 
0.6784) 
38.6 kJ
We interpolated in Table C.3 to find v1 and
v2 and the two temperatures.
4.16FE D
In Table C.3 we observe that u =2949 kJ/kg is between p = 1.6 MPa and 1.8 MPa but
closer to 1.6 MPa so the closest answer is 1.7 MPa.
4.17FE C
Interpolate in Table C.3 and find h = 3459 kJ/kg.
4.18FE C
Use a central difference for accuracy:
 3213.6 
h
2960.7
cp 


2.53 kJ/kg 
蚓

T
400 
300
4.19FE C
We use q because the mass is not known:
q  p (T2  1 ) 
c
T
2.254  
300 676.2 kJ/kg
4.20FE B
The copper gains heat and the water losses an equal amount of heat:
mc c p ,c (T2   w c p , w (30  2 ).
0) m
T
4.21FE B
20 
0.38  2  
T 10 4.18   2 ). T2 
(30 T
25.4
C
First the ice melts and then the ice water heats up:
mi ( i  p , w  i )  w c p , w  w .
h c
T
m
T
10 
[333 
4.18(T2   
0)] 60 4.18   2 ) .
(20 T
4.22FE C
For process 1:
Q = W =100 = a.
For process 2:
b = 40 + 60 = 100.
W or 100      .
100 40 100 60 c
Q 
32
c 
80
T2 = 5.85蚓

3. 4.23FE B
For the T = const process:
v
0.8
w1-2 RT1 ln 2 
0.287 
278.7 
ln

166.4 kJ/kg
v1
0.1
We used pv = RT to find T1 = (800)(0.1)/0.287 = 278.7
C.
4.24FE C
For the adiabatic process:
 3-1  v (T1  3 ) 
w
c
T
0.717(278.7 
121.3)  kJ/kg
113
k
1
0.4
 
v
0.1
where T3  1  1  
T
278.7   
  121.3 K .
v
0.8
 
3 
4.25FE D
The net heat transfer is equal to the net work. The work for the 2-3 process is zero.
Therefore, using the results from the above two problems, the net q is
qnet w1-2 w2-3  3-1 = 166.4 
w
113 = 53 kJ/kg
4.26FE A
k k
1/
4.27FE D
 
p
T2  1  2 
T
p
1 
4.28FE B
For the Q = 0 process:
293  0.2857 
8
530.7 K or 258 .
C
W  mcv (T2  1 )  

T
2 0.717 
(622 
373)  358 kJ

k k
1/
 
p
where T2  1  2 
T
p
1 
4.29FE A
  0.2857 
373 6
622.3 K .
100 400
 

Q  p  or 
mc T
  15        T
0.3  60 (10 20 3 1.2) 1.00 
 3600

  
T 14.3 C
We assumed a p = const process since there are openings where air is allowed to escape
or enter.
4.30FE D
4.31FE D
This term includes the flow work rate term due to the pressure force moving due to the
velocity.
4.32FE B
4.33FE C
4.34FE D
V 2  12
V
202  2
200
0  p  2
c T
.
1.00 
T
.
   .
T 19.8 C
2
2
1000
The factor 1000 is needed to convert J to kJ since cp includes the kJ unit.
h2 h1 
3500.9 kJ/kg.
T2 = 508
C
p2  MPa. Interpolate in Table C.3 at 0.8 MPa and find
0.8
33

4. k
1/1
4.35FE D
 
p
T2  1  2 
T
p
1 
4.36FE B
p
 1
m A1V1  1 A1V1 .
RT1
4.37FE D
The heat that leaves the steam enters the cooling water:
  0.2857 
293 8
530.7 K or 258 .
C
100
2
  2 1
 0.1 V
0.287 
293
V1 
53.5 m/s
 h mw
ms  s c p  w
T
Use hfg from Table C.2:
10 
2373  
400 4.18  T .

4.38FE C
   .
T 14.2 C
Assume p = const in the tube:
  T 100 1.00 (25 20) 500 kJ/min
Q mc p      
4.39FE C
For liquid:
p
100
 m  4 6000  
WP    
23.6 kW

1000
4.40FE B
The enthalpy of the fluid when it enters the tank is assumed equal to the enthalpy
of the fluid in the pipe on the upstream side of a valve connecting the tank and pipe.
4.41FE C
Q1  c1 A1 1 1
h
T t
and Q2  c 2 A2  2 2 . But, Q2  1 , h2  h1 ,  2 1 , and A2 A1.
h
T t
Q
2
T
T
So,
t2 Q2 h1 A1 1 1
T
t

 . t2  1  s
20
t1 Q1h2 A2  2 2
T
2
4.42FE C
 q

q  B or k A
A
TA2  A1
T
T 
T
 B B 2 B1
k
LA
LB
or
If LB LA / 2 and TB1  TA1 and TB 2  TA2
2
2
TA2  A1 2(TA2  A1 )
T
T

LA
2 LA
and 1 = 1 so C is the answer.
34
TA2  A1 TB 2  B1
T
T

LA
4LB
then

5. 4.1
Qnet  net
W
100 m 
9.81  m 
3.
3.398 kg
4.2
1
1
Qnet  net . Qnet  mV 2  
W
1500  2 
30
675000 J
2
2
4.3
 0.
F
pA  patm A 
mg
0
p  2  
 0.05 10 9.81  000   2.  p  490 Pa
100
 0.05
112
Q   .   
W
U
U 300 (112 490   2 )  
 0.05
0.2 123.3 J
4.4
        24)]  ft-lbf or 
U Q W
2 778 [ 600 (2 1/
381
0.49 Btu
4.5
a)    20   kJ E2  E2  kJ
E Q W
5 15
7.
22
b) Q   6   kJ. E2   E2  kJ
E W
3 3
8 6.
14
c) W   E 40   25 kJ.     kJ
Q 
(30 15)
E 30 15 15
d) W   E    kJ. 20   1 . E1  kJ
Q 
10 20
30
10 E
10
e) Q       kJ.     kJ
E W
8 6 10
4
E
8 6
14
4.6
a  1-2  1-2  U    kJ
W
Q

200 0
200
( )cycle 0. 0   
U
c 400 1200  c 
0.
800 kJ
b  2-3  2-3  U    kJ
W
Q
 800 800 0
d  3-4   3-4 400  
Q
U W
600 1000 kJ
e  4-1  4-1  U  
W
Q

0 ( 1200) 
1200 kJ or we could use Qnet  net
W
4.7
a  1-2   1-2   200 kJ
Q
E W
100 100
( )cycle  100  
E
0.
c 200  c  kJ
0.
100
b  2-3   2-3    kJ
Q
E W
100 50 50
d  3-1  3-1  E  
W
Q
 100 ( 200) 
300 kJ or we could use Qnet  net
W
35

6. 4.8
First, find the initial pressure in the cylinder:
W
60 
9.81
p  patm 
 000  000 Pa
100
119
A
 2
0.1
Now, apply the 1st law adding the work of the pressure force and the work to
compress the spring:
1
Q   .
W
U
Q  pA   Kx 2 ) 
(
h
U
2
1
 200 
U
(119 000   2 
 0.1 0.05   000  2 )  J
50
0.05
49
2
4.9
W1-2   6     000 J.
VI t
5 (20 60) 36
Q1-2  1-2  2-1 W2-1   )cycle 0. Q2-1  1-2  kJ
W
Q
( U
W
36
4.10
Q    000       600 J or 377.6 kJ
U W
400
12 3 (6 60 60) 377
4.11
Q  U   300 000      84 000 J or  kJ
 W 
12 10 (30 60) 
84
4.12
Q   
U W 8000      /1055 
110 15 (2 60 60)
3260 Btu
w e t f t 15 cne s tBus
hr h a o 05 ovr Jo t .
e e cr
t
’
4.13
The only transfer of energy across the boundary of the system is via the
electrical wires leading to the refrigerator. The 1st law is
Q   .    ) 2 
W
U
U
W
( VI t
0.746   2686 kJ
(30 60)
V
0.3
(u2  1 ) 
u
(3655.3 
2621.9) 
1505 kJ
v
0.206
4.14
Q W   
m u
4.15
a) Q W   
m u
V
(u2  1 )
u
v
0.2
1000 
{u2 
[669.9 
0.8(2567.4 
669.9)]}
0.0011 
0.8(.3157 
.0011)
u2 
3452 kJ/kg and v2 
0.2528 m 3 /kg . We must find where in Table C.3
this state exists. After checking the table we interpolate for the following:
p2  MPa, v2 
1.5
0.2528, u2 
3216 : T2   
556 C
686
C
 T2 
p2  MPa, v2 
2.0
0.2528, u2 
3706 : T2   
826 C
b) TK solution:
Rule Sheet
;This is a closed, constant-volume system so no work is done and the first law is
Q = m * (u2 -u1)
; First law
V1 = m * v1
; Definition of specific volume
v2 = v1 ; For a constant-volume system
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George S.
Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
36

7. Variable Sheet
Input
Name
Output
Unit
`
600
0.8
0.253
1000
0.2
4.16
T1
p1
h1
s1
v1
x1
phase1
u1
159
T2
p2
h2
s2
v2
x2
phase2
u2
688
1750
3890
7.99
Q
m
V1
2340
5.79
0.253
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
'SAT
2190
kJ/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
'mngless
'SH
3450
0.791
kJ/kg
kJ
kg
m^3
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-15.tkw Problem 4.15
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Internal energy
Temperature (starting guess needed)
Pressure
Enthalpy
Entropy
Specific Volume Transfer value to input)
Quality
Phase
Internal energy
*THUNITS.TKW Units for thermo
Heat transfer
Mass of steam
Volume
a L t fsf dh m s We s T ble C.1E:
) e siti t as
’ r n e
. ue a
v1 
0.016 
0.5(203 
0.016) 
101.5 ft 3 /lbm. m 
V
2


0.0197 lbm
v 101.5
The 1st law is Q W m(u2  1 ) or 8 
0.0197(u2  1 ) :
u
u
u1 
87.99 
0.5(961.9) 
568.9 Btu/lbm . u2 
975 Btu/lbm
This is less than u g so that state 2 is in the wet region with v2 
101.5 ft 3 /lbm.
This requires trial-and error:
At T2  F :
140
101.5 0.016 x2 (122.9 
0.016). x2 
0.826
975  
108 948.2 x2 .
37
x2 
0.914

8. At T2  F :
150
vg 
96.99.
slightly superheat
975  
118 941.3 x2 .
x2 
0.912
State 2 lies between 140 F and 150 F . Since the quality is insensitive to the
internal energy, we find T2 such that v2 
101.5 ft 3 /lbm:
101.5 
96.99
T2  
150
  F
10 148
122.88 
96.99
A temperature slightly less than this provides the answer: T2  F .
147
b) TK solution:
Rule Sheet
;This is a closed, constant-volume system so no work is done and the first law is
Q = m * (u2 -u1)
; First law
V1 = m * v1
; Definition of specific volume
v2 = v1
; For a constant-volume system
; 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
Comment
`
Thermal Sciences, Potter & Scott
*STEAM8E.TKW Steam, 1-8 States, English units
P4-16.tkw
Problem 4.16
120
T1
F
Temperature
p1
1.69
psi
Pressure
h1
601
B/lbm
Enthalpy
s1
1.05
B/(lbm*R) Entropy
v1
101
ft^3/lbm
Specific Volume
0.5
x1
Quality
phase1 'SAT
Phase
101
8
2
T2
p2
h2
s2
v2
x2
phase2
Q
m
u2
u1
V1
144
3.21
1030
1.73
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
0.913
'SAT
0.0197
975
569
B
lbm
B/lbm
B/lbm
ft^3
Temperature (starting guess needed)
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*thunits.tkw Units for thermodynamics
Heat transfer
Mass of steam
Internal energy
Internal energy
Initial volume
38

9. V 100 / 1728


0.01633 lbm
v
3.544
120 
144
100
Q m(u2  1 )  
u
W 0.01633{1371.5 
[312.3 
0.95(796)]} 
(7.208 
)
778
1728

6.28 Btu
4.17 v1 
0.0179 
0.95(3.73 
0.0179) 
3.544. m 
4.18
First, find the mass: m 
V 1 0.004


0.0302 kg . The 1st law is Q  m(u2  1 )
W
u
v1
0.1325
which takes the form
40 
1500 
0.03019( v2 
0.1325) 
0.0302(u2 
2598) or 124.4  v2 
45.3
0.0302u2
The above equation has two variables but the steam tables represent the 2nd equation.
'
This requires trial-and-error (let u2 be the u2 from the above equation):
'
At T2 
600, p2  : v2 
1.5
.2668, u2 
3294. u2 
3720 

'
At T2 
780, p2  : v2 
1.5
.3230, u2 
3622. u2 
3636  T2  
785 C
'
At T2 
800, p2  : v2 
1.5
.3292, u2 
3627. u2 
3720 

4.19
100 /1728
a) Q m(h2  1 ) 
h
{1531.5 
[312.7 
0.95(878.5)]} 
6.27 Btu
3.544
b) TK solution:
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressure
becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or
Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;and George
S. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8E.TKW
Variable Sheet
Input Name
`
120
0.95
1000
120
T1
p1
h1
s1
v1
x1
phase1
T2
p2
Output
341
1150
1.53
3.54
'SAT
Unit
Comment
Thermal Sciences, Potter & Scott
*STEAM8E.TKW Steam, 1-8 States, English units
P4-19.tkw Problem 4.19
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm
Specific Volume
Quality
Phase
F
Temperature
psi
Pressure
39

10. h2
s2
v2
x2
phase2
100
4.20
1530
B/lbm
1.9
B/(lbm*R)
7.2
ft^3/lbm
'mngless
'SH
Q
V1
m
6.28
B
in^3
lbm
0.0163
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
Initial volume
Mass of steam
0.004
40 
(h2 
2797).
h2 4120 kJ/kg .
0.1325
At p2  MPa we interpolate T2 787
1.5
C.
b) TK solution:
a) Q m .
h
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
; 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
`
200
1500
4
1500
40
T1
p1
h1
s1
v1
x1
phase1
V1
m
T2
p2
h2
s2
v2
x2
phase2
Q
2800
6.45
0.132
'mngless
'SH
0.0302
785
4120
8.28
0.324
'mngless
'SH
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
L
kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-20.tkw Problem 4.20
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Initial volume
Mass of steam
Temperature (starting guess is needed)
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
40

11. 4.21
From Table D.1 we find h1  
62 0.8(190.3) 214.2 kJ/kg. Then
q h2  1 .
h
80 h2 
214.2.
h2 294.2 kJ/kg
We interpolate at p2 0.4 MPa in Table D.3:
294.2 
291.8
T2 
(10)  
50 52.5
C
301.5 
291.8
4.22
a) Q m(h2  1 ) 2(3278 
h
209) 
6138 kJ
b) Q m(h2  1 ) 2(4044 
h
3278) 
1531 kJ
c) TK solution:
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressure becomes
Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) for each process or
Q12 = m * (h2 - h1)
; for the process 1-2
Q23 = m * (h3 - h2)
; for the process 2-3
; 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
`
50
100
400
100
750
100
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
T3
p3
h3
s3
v3
x3
phase3
209
0.704
0.00101
'mngless
'CL
3280
8.54
3.1
'mngless
'SH
4040
9.45
4.71
'mngless
'SH
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-22.tkw Problem 4.22
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
41

12. 2
4.23
m
Q12
Q23
6140
1530
*THUNITS.TKW Units for thermo
Mass of steam
Heat transfer, process 1-2
Heat transfer, process 2-3
kg
kJ
kJ
h1 
1008 
0.8(1796) 
2445. v1 
0.0012 
0.8(0.06668 
0.0012) 
0.5359
V1
1.2
T
m


22.39 kg
v1
0.05359
a)
Q m(h2  1 )
h
(b)
3000 22.39(h2 
2445). h2 2579
h2 2579 

 T2 234 C
p2  MPa 
3
1 (a)
v
b) Q m(h2  1 ) . 30 000 22.39(h2 
h
2445). h2 
3785
h2 
3785 

 T2 645 C
p2  MPa 
3
4.24
v1 
0.0012 
0.9(0.1274 
0.0012) 
0.1148 m 3 /kg and
v2  
3 0.1148 
0.3444 m 3 /kg
u1 
850.6 
0.9(2595.3 
850.6) 2421 kJ/kg
Interpolate in Table C.3:
v2 
0.3444 
0.3444 
0.2608
0.8 
(0.2) 
0.617 MPa
  p2 
T2 
200 C 
0.3520 
0.2608
T 200 
C
0.617 
0.6
Also, 2
(2638.9 
2630.6) 2638 kJ/kg
 u2 2638.9 
0.8 
0.6
p2 
0.617 
To find the heat transfer we must know the work. It is found by using graph paper
and plotting p vs. v and graphically integrating. The work is twice the area since
m = 2 kg. Doing this, we find
W 2  456 kJ.
228
4.25
Q m (u2  1 )  2(2638 
u
W
2421) 
456 
890 kJ
 3489 
h
3076.5
a) c p 

2.06 kJ/kg 
K

T
200
 3488.1 
h
3074.3
b) c p 

2.07 kJ/kg 
K

T
200
 2821.4 
h
2151
c) c p 


13.4 kJ/kg 
K

T
50
42

13. d) TK solution:
Rule Sheet
; cp = the slope of a constant pressure line on an hT diagram or approximately.
;cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 398C and T1 = 402C
; at 10 kPa. Likewise, we define states 3 and 4 at these temperatures on a 100-kpa line and
;states 5 and 6 at these temperatures ona 30,000-kPa line.. Then
cp1 = (h2 - h1) / (T2-T1) ; at 10 kPa, 400 kPa
cp2 = (h4 - h3) / (T4 - T3) ; at 100 kPa, 400 kPa
cp3 = (h6 - h5 ) / (T6 - T5) ; at 30000 kPa, 400 kPa
; 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
`
398
10
402
10
398
100
402
100
398
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
T3
p3
h3
s3
v3
x3
phase3
T4
p4
h4
s4
v4
x4
phase4
T5
3280
9.6
31
'mless
'SH
3280
9.61
31.2
'mless
'SH
3270
8.54
3.09
'mnless
'SH
3280
8.55
3.11
'mless
'SH
Unit
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-25.tkw Problem 4.25
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
43

14. 30000 p5
h5
s5
kPa
2120 kJ/kg
4.43 kJ/(kg*K)
0.0026
v5
m^3/kg
9
x5
'mless
phase5 'SH
402 T6
C
30000 p6
kPa
h6
2180 kJ/kg
s6
4.51 kJ/(kg*K)
v6
0.0029 m^3/kg
x6
'mless
phase6 'SH
cp1
cp2
cp3
4.26
2.07
2.07
13.9
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
kJ/(kg*K) Approx cp at 10 kPa, 400 C
kJ/(kg*K) Approx cp at 100 kPa, 400 C
kJ/(kg*K) Approx cp at 30000 kPa, 400 C
 1373.9 
u
1219.3
a) cv 

0.386 Btu/lbm-
R

T
400
 1373.7 
u
1218.6
b) cv 

0.388 Btu/lbm-
R

T
400
 1156.5 
u
960.7
c) cv 


1.96 Btu/lbm-
R

T
100
4.27
 117.6 
h
126.4
a) c p 


0.22 Btu/lbm-
R

T
120 
80
 115.5 
u
107.6
cv 


0.20 Btu/lbm-
R

T
120 
80
b) TK solution:
Rule Sheet
; cp = the slope of a constant-pressure line on an hT diagram or approximately:
; cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 98 F and T2 = 102 F
; at 30 psia. Then cp = (h2 - h1) / (T2-T1) ; at 30 psia, 100 F
; cv = the slope of a constant-volume line on a uT diagram or approximately
;cv = (u4 - u3) / (T4 - T3) where states 3 and 4 are close together, say T3 = 98 F and T4 = 102 F at
the specific volume of the state of 30 psia, 100 F: 1.89 ft^3/lbm. Then
cv = (u4 - u3) / (T4 - T3)
;*R1348e.tkw, R134a, 1-8 States, English units
; R134a tables based on 'Thermodynamic Properties of HFC-134a' DuPont Technical Information,
which is based upon the Modified Benedict-Webb-Rubin equation of state.
R1348E.tkw
44

15. Variable Sheet
Input
Name
Output
Unit
`
98
30
102
30
98
1.89
102
1.89
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
T3
p3
h3
s3
v3
x3
phase3
u3
T4
p4
h4
s4
v4
x4
phase4
u4
cp
cv
4.28
122
0.257
1.88
'mngless
'SH
123
0.259
1.9
'mngless
'SH
29.9
122
0.257
'mngless
'SH
111.98
30.1
123
0.258
'mngless
'SH
112.74
0.213
0.189
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
B/lbm
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
Comment
Thermal Sciences, Potter & Scott
*R1348e.tkw, R134a, 1-8 States, English
P4-27.tkw Problem 4.27
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
B/lbm
B/(lbm*R)
B/(lbm*R)
*THUNITS.TKW Units for thermo
Constant-pressure specific heat
Constant-volume specific heat
a)   p (T2  1 ) 
h c
T
1.006(700 
300) 402.4 kJ/kg
700
b)   (0.946 
h
0.213   T 
10 3
0.031   T 2 )dT 417.7 kJ/kg
10 6
300
c) Using the gas table F.1 we find  h2  1 713.3 
h
h
300.2 413.1 kJ/kg
45

16. d) The error in (a) is 
2.6% and that in (b) is 1.1% assuming that of (c) is the
best answer. All three methods are acceptable for the temperature range of
this problem.
4.29
a)  mc p (T2  1 ) 2 
H
T
.006 
(600 
400) 402 kJ


.213  
10 3
.031  
10 6
b)  2   
H
.946 200
(6002  2 ) 
400
(6003  3 )  418 kJ
400 
2
3


c)  m(h2  1 ) 2(607 
H
h
401) 412 kJ
4.30
 water mc p  2 
H
T
4.18(60  418 kJ
10)
 ice mc p  2 
H
T
1.86[   kJ
10 ( 60)] 186
where we averaged the (c p )ice between   and   in Table B.4.
10 C
60 C
4.31
Assume that all of the ice melts. The ice warms up to 0 , melts at 0 , then
C
C
warms up to the final temperature T2 . Fr, t f dh m so t ice:
it e si t as fh
s l’ n e
e

6
V 16  
8 10
mi  

0.1174 kg
v
0.00109
where we found v in Table C.5. Energy is conserved so that the heat gained by the
ice equals the heat lost by the water:
mi  p )i ( )i  if  c p ) w (  )iw  mw (c p ) w (  ) w
(c
T
h (
T 

T

0.1174    
2.1 10 330 4.18(T2   (1000   ) 
0) 
10 3 4.18(20  2 )
T
T2 
9.07
C
4.32
a) Assume the ice melts and then the water is heated:
Q m(c p )i ( )i  if  (c p ) w (  ) w
T
mh m
T
2000 2.3    T2  T2  
1.89 60 330 4.18 .
102 C
b) Assume the ice melts and then the water is heated:
Q mhif  (c p ) w ( ) w
m
T
2000 2.3  T2  T2 
330 4.18 .
129.1
C
This is in the superheat region so the above temperature is too high. The water
must first vaporize at T2 
as e n m a r
120.2 ,o ht t f at pr ue
C s t ’ h i le e t .
46

17. 4.33
First, the temperature of the ice is raised, then the ice is melted, then the
temperature is raised to the boiling point, then the water is vaporized, then it is
superheated. This requires the following:
T
Q mc p ( )i    p ( )w  fg  (h2  g )
T
m h mc T
mh
m
h
 
10 0.486   
32 143 1.0(250   
32) 945 (1334 
1164)
2
=14,900 Btu
where for ice c p 
0.47 
0.001T and we used an average
1
temperature of 16  .
C
4.34
v
The heat gained by the ice cubes is lost by the cola. The mass of the ice is
mc 4       0.0734 kg
2 2 5 10 6 917
a) Assume all of the ice melts:
mi c p ( )i  i   i c p (  ) w mc (c p )c (  )c
T
m h m
T
T
0.0734(2.0    T2 ) 2   
20 330 4.18
10 3 4.18 
1000(20  2 ). T2  
T
16.1 C
b) Assume that all of the ice does not melt:
mi c p ( )i x mc (c p )c ( )c
T
h
T
0.0734   x  
2.0 20
330 0.25   
10 3 1000 
4.18 
20. x 
0.0561 kg
0.0561
 
100 76.4%
0.0734
4.35
The heat lost by the copper is gained by the water:
mc (c p )c ( )c mw (c p ) w ( ) w
T
T
Use an average value of c p for copper from Table B.4:
5
0.39 
(300  2 ) 20   
T
10 3 1000 
4.18 T2 
(
0). T2 
6.84
C
4.36
The heat lost by the copper is gained by the water:
mc (c p )c (  )c mw (c p ) w (  ) w
T
T
  v (T2  1 ) 4 
u c
T
0.717(400 
100) 
860 kJ
4.37
The heat lost by the iron is gained by the copper:
mc (c p )c ( )c mi (c p )i ( )i
T
T
50  T2  
0.39
100 0.45(200  2 ). T2 
T
139.5
C
4.38
  p (T2  1 ) 4 
h c
T
1.00(400 
100) 
1200 kJ
  v (T2  1 ) 4 
u c
T
0.717(400 
100) 
860 kJ
47

18. 4.39
a)   0, W   kJ,
U
H
Q 60
60 0.4 
0.287 
373ln
T1  
100 C,
0.4 
0.287 
373

0.856 m3 ,
50
V2
0.856
p1 p2

50
203 kPa
V1
0.211
V
0.856
. V 1 
0.211 m3 ,
V1
2
b) If V const, W    
0.
H 0.4 1.00(300  1 ) 
T
60. T1  
150 C
 0.4 
U
0.717(300 
150) 43 kJ, Q  43 kJ,
U
T
423
0.4 
0.287 
573
p1 p2 1 200

147.6 kPa, V 2 V 1 

0.329 m 3
T2
573
200
c) Q mc p  .   kJ  
T
H 100
0.4 1.00(T2 
200). T2  
450 C,
p2 
500 kPa,
T
0.4 
0.287 
723
473


0.166 m3 , V 1 V 2 1 
0.166

0.109 m3
500
T2
723
W p ( V 2 V 1 ) 
500(0.166 
0.109) 28.5 kJ
Also,  0.4 
U
0.717(450 
200) 
71.7 kJ
V
2
k
1
0.4
V
0.48
 1
 
d) T2  1      
T
323
605 K or 332
C,
V2
0.1 


 0.4 
U
0.717(332  
50) 80.9 kJ, W  U 

80.9 kJ,
0.4 
0.287 
323
 0.4 
H
1.00(332   kJ, p1 
50) 113

77.3 kPa,
0.48
0.4 
0.287 
605
p2 

695 kPa
0.1
3.42
4.40
a) Q  (    ). 60 
m u m pdv
0.4(2512  1   ) . Integrate graphically selecting
u
pdv
v1
various initial states. We find W 49.4 kJ,  
U 10.2 kJ,  
H 11.8 kJ,
T1  
100 C,
p1  kPa,
100
V
2

1.37 m3 ,
V 1 0.662 m3
b) To locate state 1 using the t mt l ivr d f u . ol’asm t th
s a a e s e ii l S , t s eh t
e
bs
y fc t
es u
a e
steam acts almost as an ideal gas and use the constants from Table B.2:
If V const, W    
0.
H 0.4 1.872(300  1 ) 
T
60. T1 220 .
C
 0.4 
U
1.411(300 
220) 45 kJ, Q  45 kJ,
U
T
493
0.4 
0.462 
573
p1 p2 1 200
 kPa, V 2 V 1 
172
0.529 m3
T2
573
200
c) Q  m(h2  1 ).   kJ  h2 
H
h
H 100
0.4(
2855). h2 
3105, p2  MPa.
0.5
Interpolate and find T2  
320 C. Then V 2  
0.4 0.542 
0.217 m3 and
V 1 0.4 
0.4249 
0.170 m3 and W 
500(0.217 
0.170) 23.5 kJ
Also,  0.4(2835 
U
2643) 
76.8 kJ
48

19. 4.41
Q   mRT ln
U W
4.42
m
p1
20
2 
(53.5 / 778)  ln
560

176.7 Btu
p2
200
p
pV
200 
2
800

0.596 kg. T2  1 2 
T
323

1292 K
RT
2.077 
323
p1
200
Q W mcv  0.596 
T
3.116(1292 
323) 
1800 kJ
4.43
Q m(h2  1 )
h
h2 
1832

1551F
 T2 
p2 60 psia 
h2 
1939 
b) 1000 2.0(h2 
1439). h2 
1939 kJ/kg.
1741
F
 T2 
p2  psia 
60
a) 1000 2.0(h2 
1332). h2 
1832 kJ/kg.
c) TK solution:
Rule Sheet
;*P4-43.tkw Problem 4.43
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or, for part (a),
Q = m * (h2 - h1)
;Similarly, for part (b):
Q = m * (h4 - h3)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;and
George S. Kell, Hemisphere Publishing Corp., 1984. ; STM8E.tkw
Input Name
`
600 T1
60 p1
h1
s1
v1
u1
x1
phase1
T2
60 p2
h2
s2
v2
u2
Output
1330
1.82
10.4
1220
'mngless
'SH
1830
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English
P4-43.tkw Problem 4.43
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
F
Final temperature, part (a)
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
49

20. x2
phase2
815 T3
60 p3
h3
s3
v3
u3
x3
phase3
T4
60 p4
h4
s4
v4
u4
x4
phase4
1440
1.91
12.6
1300
'mngless
'SH
1740
1940
2.2
21.7
1700
'mngless
'SH
1000 Q
2
m
4.44
Q mc p (T2  1 ) ,
T
Quality
Phase
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
F
Final temperature, part (b)
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
*THUNITS.TKW Units for thermo
B
Heat transfer
lbm
Mass of steam
m
p1 V 1
400 
0.2


1.021 kg
RT1
0.287 
273
a) 50 
1.021  T2 
1.0(
0). T2 49
C
b) 50 
1.021  T2 
1.0(
200). T2 249
C
4.45
4.46
p1 V 1 800 
8000  
10 6
Q W mcv (T2  1 ). m 
T

0.05978 kg
RT1
0.287 
373
p
200
a) Q 0.05978 
0.717(93.2 
373) 
11.99 kJ where T2  1 2 
T
373

93.2 K
p1
800
p
3000
b) Q 
T
373

1399 K
0.05978 
0.717(1399 
373) 43.98 kJ where T2  1 2 
p1
800
We assume a quasiequilibrium process with q = 0 so that
k k
1/
 
p
10
 000 
T2  1  2   
T
373

1390 K

p
 100 
1 
q   . w v (T2  1 ) 
w
u
c
T
0.717(1390 
373)  kJ/kg
729
0.4 / 1.4
50

21. n n
1/
4.47
 
p
100
 
For the polytropic process T2  1  2    
T
373
276.7 K
p
600
 
1 
p v  1v1 R(T2  1 ) 0.287(276.7 
p
T
373)
w 2 2



138.2 kJ/kg
1
n
1
n
1
1.2
0.2 /1.2
q  v  
c T w 0.745(276.7 
373) 
138.2 
66.5 kJ/kg
4.48
100   
(3 5 2.4)
Q mc p (T2  1 ) 
T

1.00(27  
10) 753 kJ
0.287 
283
4.49
Q  m .
W
u
pA  paddle mcv (T2  1 )
h W
T
The pressure is found from a force balance on the piston:
W
175
p  patm 

14.7 
18.18 psia
A
 2
4
The mass is found from the ideal-gas law:
p1 V 1 (18.18 
144)   2  /1728
 4 10
m


0.0255 lbm
RT1
53.3 
560
The temperature at state 2 is
p V 2 (18.18 
144)   2  /1728
 4 15
T2  2

 R
840
mR
0.0255 
53.5
Finally,
Wpaddle 
18.18   2  /12 
 4 5
0.0255 
0.171  
778 (840 
560) 
1331 ft-lbf
4.50
a) v1 
.0011 
.5(.4625 
.0011) 
0.2318 m 3 /kg
u1 
604.3 
.5(2553.6 
604.3) 
1579 kJ/kg
0.15
Q W m(u2  1 ). 800 
u
(u2 
1579). u2 2815 kJ/kg
0.2318
Now, search Table C.3 for the location of state 2. It takes a couple interpolations:
u2 
2815 
314 and p2 
C
1.14 MPa
 T2 
v2 
0.2318 
We used:
p2 
1.0

248 and u2 
C
2707 kJ/kg
 T2 
v2 
0.2318 
p2 
1.5

489 and u2 
C
3102 kJ/kg
 T2 
v2 
0.2318 
p
(a)
(b)
1
v
51

22. b) Use some values from part (a):
0.15
Q W m(u2  1 ). 200 
u
(u2 
1579). u2 
1888 kJ/kg
0.2318
Try several guesses for T2 . We tried T2   and T2   . Finally
150 C
160 C
u2 
1888 
154 and p2 
C
0.533 MPa
T 
v2 
0.2318  2
4.51
60  
144 3
a) m 
0.9925 lbm. Q W m(u2  1 ) 
u
0.9925(u2 
83.5)
53.3 
490
u2 285 Btu/lbm and T2 
1595R or 1135F
600  
144 3
b) m 

3.80 lbm. Q W m(u2  1 ) 
u
3.80(u2 
244)
53.3 
1280
u2 297 Btu/lbm and T2 
1655R or 1195F
4.52
a) Q mc p   
T 5 1.00(313  
20) 1465 kJ
V2
293  
2 586 K or 313  .
C
V1
b) Q W mcv   
T 5 0.717(313  
20) 1050 kJ
where we used T2  1
T
p2
293  
2 586 K or 313  .
C
p1
p
1
mRT ln 1  
5 0.287 
293ln  kJ
291
p2
2
where we used T2  1
T
c) Q  mRT ln
W
V2
V1
d) Q mc p   
T 5 1.00(586 
293) 
1465 kJ
4.53
0.5
m

1.35 kg, h1 
604.7  
.8 2133.8 2312 kJ/kg
.00108 
.8(0.4625 
.00108)
a) Q m(h2  1 ) 
h
1.35(3485 
2312) 
1584 kJ
b) Q m(h2  1 ) 
h
1.35(3870 
2312) 2104 kJ where h2 is found by interpolation.
c) TK solution:
Rule Sheet
This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) at constant pressure becomes
Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or
Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
W = m * p1* (v2 - v1) ; from w = INT pdv for a constant pressure process
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and George S.
Kell, Hemisphere Publishing Corp., 1984.
; STM8SI.tkw
52

23. Variable Sheet
Input
Name
Output
Unit
`
0.8
0.5
675
400
4.54
144
Q
m
W
400
T1
p1
h1
s1
v1
x1
phase1
V1
T2
p2
h2
s2
v2
x2
phase2
2110
1.35
390
2310
5.87
0.37
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
'SAT
3870
8.64
1.09
'mngless
'SH
m^3
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
kg
kJ
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-53.tkw
Problem 4.53(b)
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Initial volume
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
Mass of steam
Work
0.005
1.495
1.688
mf 
4.713 kg, mg 

1.688 kg. x1 

0.2637
0.001061
0.8857
1.688 
4.713
v1  2 
v 0.001061 
0.2637(0.8857 
0.001961) 
0.234 m 3 / kg
u1 
504.5 
0.2637(2025) 
1038 kJ/kg
a) Find u2 in Table C.2: Q m(u2  1 ) 
u
6.401(2578 
1038) 
9850 kJ
b)
T2   
400 C
2953 and Q  (u2  1 ) 
m
u
6.401(2953 
1038)  260 kJ
12
 u2 
v2 
0.234 
c)
p2  MPa 
0.8
2530 and Q  (u2  1 ) 
m
u
6.401(2530 
1038) 
9550 kJ
 u2 
v2 
0.234 
4.55
Q mc p  . 10 4 
T
0.171  T .  

T 14.62F and  4 
H
0.24 
14.62 
14.04 Btu
4.56
Q  mRT ln
W
p1
p
200
p1 V 1 ln 1 200 
(8000   ) ln
10 6

2.22 kJ
p2
p2
800
53

24. k k
1/
4.57
 
p
a) T2  1  2 
T
p
1 
0.4 /1.4
15000 

473 

 400 

1332 K or 1059
C
W  v   
mc T
2 0.717(1059 
200) 
1230 kJ
k k
1/
 
p
15000 

b) T2  1  2  623 
T

1755 K or 1482
C

p
 400 
1 
W  v   
mc T
2 0.717(1482 
200) 
1620 kJ
0.4 /1.4
k k
1/
4.58
 
p V 1 100   2 
p
 0.3 0.8
T1  1


394 K and T2  1  2 
T
mR
0.2 
0.287
p
1 
  0.2857 
394 50
1205 K
W  v (T2  1 )  
mc
T
0.2 0.717(1205 
394)  kJ
116
4.59
15 100   
10 75 150
a) Q mc p  . 400 
T
1000  

0.24  T .  49.4F

T
60
53.3 
530
15 100   
10 75 150
b) Q mcv  . 400 
T
1000  

0.171  T .  

T 69.4
F
60
53.3 
530
c) Probably the constant pressure assumption since there are air passages under the doors
and through the air vents so the volume is not constant. The passages allow the pressure
to be essentially constant.
4.60
200 
2
Q  mcv  . 200 
W
T

0.717(T2 
100). T2 
174.7 
C
0.287 
373
mRT 3.737 
0.287 
447.7
p2 

240 kPa where m = 3.737 kg.
V2
2
4.61
Maximum increase would be for an insulated container so that Q = 0:
Q  mc p  .   
W
T
( 2 10 9.81)    
10 10 3 1000  T .  4.69

T
C
4.62
Q  paddle ) mc p 
( W
T
400 
0.1
a) Q 

1.00(1092 
273)    /1000 
10 100 45
373.1 kJ
0.287 
273
V2
where we used T2  1
T
273  
4 1092 K . The factor of 1000 in the above
V1
equation changed J to kJ.
54

25. 400 
0.1
b) Q 

1.00(2292 
573)    /1000 
10 100 45
373.1 kJ
0.287 
573
V2
where we used T2  1
T
  2292 K
573 4
V1
4.63
0.8 
53.3 
1260
V1

6.218 ft 3
60 
144
V1
V3
6.218
 
60 144(10 
6.218)  
0.8 53.3 
1260 ln

7150 kJ
10
7150
 net 
W

9.19 Btu
778
a) Wnet  1-2 W2-3  3-1 p1 ( V 2 V 1 ) 
W
W
mRT ln
Qnet
b) Wnet  1-2 W2-3  3-1 p1 ( V 2 V 1 )  v (T1  3 )
W
W
mc
T
 
60 144(10 
6.218)  
0.8 0.171(1042 
1260) 
9480 kJ
9480
Qnet  net 
W

12.2 Btu
778
k
1
0.4
V
6.218 
 1

where we used T3  1   
T
1260 
1042 K.
 
V
 3
 10 
4.64
The temperatures and V 3 are
100 
0.08
800 
0.08
T1 
278.7 K. T2  3 
T
2230 K
0.1 
0.287
0.1 
0.287
p
800
V 3 V 2 2 0.8

0.64 m3
p3
100
p
Wnet W1-2  2-3  3-1 mRT ln 2 p1 ( V 1 V 3 )
W
W
p3
 
0.1 0.287 
2230 ln
For a cycle Qnet  net 
W
77.1 kJ
4.65
800

100(0.08 
0.64) 
77.1 kJ
100
a) Wnet  1-2 W2-3  3-4 W4-1  1 ( v2  1 )  3 ( v4  3 )
W
W
mp
v
mp
v

10.04 
4000(0.04978 
0.00125) 
47.39(0.00125 
0.04978) 1930 kJ

V2
0.5
where we used m 


10.04 kg.
v2
0.04978
Qnet  net 
W
1930 kJ
55

26. b) TK solution
Rule Sheet
;For this cycle of a closed system, assume that all processes are frictionless. Then
Qnet = Wnet ; = W12 + W23 + W34 + W 41 =m * (INT pdv + 0 + INT pdv + 0) and
Wnet = m * (p1 * (v2 - v1) + p3 * (v4 - v3)) ; for the two constant pressure processes
m = V2 / v2
v3 = v2
v4 = v1
; 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
`
4000
0
4000
1
80
0.0498
80
0.00125
0.5
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
T3
p3
v3
phase3
T4
p4
v4
phase4
V2
m
Qnet
Wnet
Variable Sheet
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw, Steam, 1-8 States, SI Units
P4-65.tkw Problem 4.65
250
C
Temperature
kPa
Pressure
0.00125 m^3/kg Specific Volume
Quality
'SAT
Phase
250
C
Temperature (starting guess needed)
kPa
Pressure
0.0498
m^3/kg Specific Volume (transfer value to input)
Quality
'SAT
Phase
*THUNITS.tkw
Units for thermo
C
47.4
kPa
m^3/kg Specific volume (transfer value to input)
'SAT
C
47.4
kPa
m^3/kg Specific volume (transfer value to input)
'SAT
m^3
Volume at state 2
10
kg
Mass of steam
1930
kJ
Net heat transfer for the cycle
1930
kJ
Net work for the cycle
k
1
4.66
100 
0.02
V
 1
0.02 
T3 
232.3 K. T1  3   232.3 
T
583.5 K
 
0.03 
0.287
V
0.002 
 3

0.4
V2
 v (T1  3 )
mc
T
V3
0.02



0.03  0.287 
583.5ln

0.717(583.5 
232.3)  4.014 kJ


0.002


Wnet  1-2 W2-3  3-1 mRT ln
W
W
56

27. 4.67
4.68
A1
 2
5
 
100
25 m/s
A2
 2
10
V2  1
V
0.02
 1 1
m AV1  A2V2 . V2 

0.0025 m/s
2
1000 
(800 
0.01 
0.001)
L
0.60
t  
240 s or 4 min
V 0.0025
A1
 2
1
 
150

37.5 fps
A2
2 
4 0.5
4.69
V2  1
V
4.70
4000
 1 1
m AV1 
  ) 
(10 10 4 150 3.65 kg/s
0.287 
573

m
3.65
V2 

 m/s
195
400
 A2

4
2
  )
(50 10
0.287 
373
4.71
a) The continuity equation with one inlet is
dm
d
AV1  c.v. V
1 1
dt
dt
where V is the volume of the tank. Then, using   v1 ,
1/
1
d  A1V1  2 
0.05 20



0.0409 kg/m 3 
s
dt
V v1
10 
0.3843
b) TK solution:
Rule Sheet
; State 1: State of entering steam in the pipe; State 2: State of steam in the tank. A mass
;balance on the tank is: Rate of mass flow in = rate of mass increase in the tank. This is
mdot1 = dm2%dt ; where dm2%dt = dm2/dt
mdot1 = A1* V1/ v1
A1 = pi() * d1^2/4
drho%dt = dm2%dt/Volume2
; 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
Engineering Thermo, Jones & Dugan
*Stm8si.tkw Steam, 1-8 States, SI Units
P4-71.tkw Problem 4.71
`
57

28. 400
800
20
10
10
4.72
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
mdot1
A1
V1
d1
Volume2
dm2%dt
drho%dt
  m3
m1 m2 .
C
kPa
m^3/kg
C
kPa
m^3/kg
0.384
'mngless
'SH
0.409
0.00785
Temperature in inlet pipe
Pressure in inlet pipe
Specific Volume in inlet pipe
Quality
Phase
Temperature
Pressure
Specific Volume
Quality
Phase
Mass flow rate in
Cross-sectinal area of pipe
Velocity in pipe
Diamter of pipe
Volume of tank
Rate of iuncreas of mass in tank
Rate of increase of density in tank
kg/s
m^2
m/s
cm
m^3
kg/s
kg/(m^3*s)
0.409
0.0409
2000
150
  2 
 .04 125
  2  
 .05 40 m2
.287 
623
.287 
423
 6.64 kg/s, V2 
m2 
6.64
450
  2
 0.05
0.287 
473
4.73

255 m/s

m
15
Vavg 


1.442 fps
 62.4   24
A
4 1/
 
m  V dA
 y2
15  max  2
V
1

h
 h
h

4
62.4  Vmax
4
 dy 

h
 y3 
4 1/
  48 
y
62.4  Vmax 
4
 2  
 3 

 3h h

Vmax 2.16 fps
4.74
1
1
4.95

a) m  A1V1 
  2 
 0.25 30 4.95 kg/s, V2 

109 m/s
v1
0.5951
2 
 0.1252 / 1.08
b) TK solution:
Rule Sheet
;For steady flow, mdot1 = 2 * mdot2, with state 1 in large pipe, state 2 in each smaller one.
A1 * V1 / v1 = 2 * A2 * V2 / v2
A1 = pi() * d1^2/4
A2 = pi() * d2^2 /4
mdot2 = A2 * V2 / v2
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
58

29. Variable Sheet
Input
Name
Output
Unit
`
250
400
30
50
200
200
25
T1
p1
h1
s1
v1
x1
phase1
V1
d1
T2
p2
h2
s2
v2
x2
phase2
V2
d2
A1
A2
mdot2
2960
7.38
0.595
'mngless
'SH
2870
7.51
1.08
'mngless
'SH
109
0.196
0.0491
4.95
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
m/s
cm
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
m/s
cm
m^2
m^2
kg/s
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Velocity in large pipe
Diameter of large pipe
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Velocity in each small pipe
Diameter of each small pipe
*THUNITS.TKW Units for thermo
Area of large pipe
Area of each small pipe
Flow rate in each small pipe
4.75
1
1
10
2.18

m  A1V1 
  
800 2.18 lbm/sec, V1 
 fps
182
v1
25.43 144
(2 / 144) / 1.1583
4.76
a) V
d
d
1
1
AV1  2 A2V2 . 10


  2 
 0.04 20
  2
 0.06 10
1 1
dt
dt
0.1996
0.3066
d


0.01348 kg/m3 
s
dt
b) TK solution:
Rule Sheet
; State 1 is defined as the state of steam in the entrance.pipe. State 2 is defined as the
; state of steam in the exit pipe. m is the mass of steam in the tank at a specified instant.
; A mass balance on the tank is therefore dm/dt = mdot1 - mdot2 or
dm%dt = A1* V1 /v1 - A2* V2 / v2
A1 = pi() * d1^2/4
A2 = pi() * d2^2/4
drho%dt = dm%dt/Volumet
59

30. ; 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
`
600
2000
400
1000
20
8
10
10
12
4.77
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
A1
V1
d1
Volumet
A2
V2
d2
dm%dt
drho%dt
0.2
'mngless
'SH
0.307
'mngless
'SH
0.00503
0.0113
0.135
0.0135
C
kPa
m^3/kg
C
kPa
m^3/kg
m^2
m/s
cm
m^3
m^2
m/s
cm
kg/s
kg/(m^3*s)
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P4-76.tkw
Problem 4.76
Temperature in inlet pipe
Pressure in inlet pipe
Specific Volume in inlet pipe
Quality
Phase
Temperature in exit pipe
Pressure in exit pipe
Specific Volume
Quality
Phase
Cross-sectinal area of pipe
Velocity in pipe
Diamter of pipe
Volume of tank
Rate of iuncreas of mass in tank
Rate of increase of density in tank
Use Eq. 4.58 with   :
1
2
0.006

r2
AV1 A2V2 . 
0.0062   Vmax 
0.8
1
1
2
0
 0.006

2 r
1.6
 dr. Vmax  m/s

 1 1
m AV1 
1000  
 0.0062  
0.8 0.0905 kg/s
4.78
1
1
 1 4
 
m  2(1  2 )2  r dr 
r
10 4
1000      10  
4 

0.314 kg/s
0
2
 4
0.314 
1000  
 0.00252  2 . V2  m/s
V
16
4.79
60 
144  2
2
 1 1
m AV1 

 2.526 lbm/sec
100
53.3 
560 144
70 
144  2
2
 A2V2 

V2 . V2 
116.3 fps.
2
53.3 
760 144
60

31. If we neglect  , Q p (T2  1 ) 
T
2.526 
0.24(300 
100) 
121.2 Btu/sec
KE  mc
V 2  12
V
116.32  2
100
Note:  m 2
KE 
2.526 
4453 ft-lbf/sec or 5.72 Btu/sec
2
2
4.80
a) Across a valve   . From Table D.2 at 0.8 MPa we find h1 
h 0
93.42 :
h2 h1 
93.42 
3.46 x2 (221.3).  x2 
0.4065
u2  f x2 (u g  f ) 
u
u
3.41 
0.4065(206.1 
3.41) 
85.8 kJ/kg
b) TK solution:
Rule Sheet
;This is a throttling process. w = 0. Assume q = 0 and change in ke is negligible. Therefore,
h2 = h1 ; from the first law.
; R134a tables based on 'Thermodynamic Properties of HFC-134a'
; DuPont Technical Information, which is based upon the Modified
; Benedict-Webb-Rubin equation of state.
R1348si.tkw
h2 = u2 + p2 * v2
Variable Sheet
Input
Name
Output
Unit
`
30
800
T1
p1
h1
s1
v1
x1
phase1
T2
60
p2
93.602 h2
s2
v2
x2
phase2
u2
4.81
93.602
0.34761
0.00084403
'mngless
'CLQ
-36.91
0.39624
0.12501
0.40047
'SAT
86.101
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ/kg
Comment
Thermal Sciences, Potter & Scott
*R1348si.tkw, R134a, 1-8 States, SI units
P4-80.tkw
Problem 4.80
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy (transfer value to input)
Entropy
Specific Volume
Quality
Phase
Internal energy
*THUNITS.TKW Units for thermo
Neglecting kinetic energy changes and using h1 h f 
1344 kJ/kg (from Table C.1),
p2 0.6 MPa. h2 h1 
1344 
670.6 x2 (2086.3).  x2 
0.3228
u2  f x2 (u g  f ) 
u
u
669.9 
0.3228(2567.4 
669.9) 
1282 kJ/kg
61

32. 4.82
Neglecting kinetic energy changes and using h1 
3634 kJ/kg (from Table C.3),
a)
p2 0.6 MPa 
569
C
 T2 
h2 
3634 kJ/kg 
 
b) With V2  1 and m2 m1 we have
V
A
1
1
A2 V2 
A1 V1 .  2 22.3
0.9695
0.04341
A1
c) For air   T2  1  
h 0.
T 600 C
4.83
4.84
 W
Using the energy equation in the form of Eq. 4.65 we have, with Q    and
0
S
V2  1 and z2 z1 ,
V
p
p
(450  
30) 144
h2 h1 or u2  2  1  1 . u2 
u
38.09 

39.34 Btu/lbm


62.4 
778
2
1
The factor 778 converts ft-lbf to Btu.
The energy equation in the form of Eq. 4.68 is used. First, the velocities are

m
200
200
V1 

6.366 m/s. V2 

17.68 m/s
2
 1 1000  
A
 0.1
1000   2
 0.06
 2  12 

V V
p

WP  2
m
 g  
z

 2

2
2

4 000 000 
17.68 
6.366
 
200


 827 000 W or 1110 hp
2
1000 

4.85
 W

 0.021 kg/s
Q   mc p (T2  1 ).   m 
T
( 5)  1.00(530.7 
293). m 
S
k k
1/
 
p
where we used T2  1  2 
T
p
1 
4.86
0.4 /1.4
400
 
293  
50 

530.7 K .
Fr, t f dh m s f x
it e si t asl :
s l’ n e
u
500
 2
m  A2V2 
  2  2.72 kg/s
 0.05 100
0.287 
503
0.4 /1.4
500
 
where T2 293  

503 K .
80 
The energy equation gives (the kinetic energy change is negligible)
 ( W


Q ) mc p (T2  1 ). WC 2.72 
T
1.00(503 
293) 
558 kW
C
4.87
( W


a) Q ) m(h2  1 ). Q 0.01 
h
(2839 
2592)  
6
3.53 kJ/s
C
The negative sign indicates a heat loss.
62

33. b) TK solution:
Rule Sheet
Qdot = mdot * (h2 - h1) + Wdot ; First law for steady flow through compressor with negligible
; changes in kinetic and potential energy.
; 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
`
50
1
200
800
0.01
-6
4.88
T1
p1
h1
x1
phase1
T2
p2
h2
x2
phase2
Qdot
mdot
Wdot
12.3
2590
C
kPa
kJ/kg
'SAT
2840
'mngless
'SH
-3.52
C
kPa
kJ/kg
kJ/s
kg/s
kW
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
Temperature
Pressure
Enthalpy
Quality
Phase
Temperature
Pressure
Enthalpy
Quality
Phase
*THUNITS.TKW Units for thermo
Rate of heat addition, i.e., heat is removed
Mass flow rate
Power output
We use the energy equation in the form of Eq. 4.68:
p p
(2000   
2) 144
  1  2000 
)   2
( WP m 




 2594 ft-lbf/sec or 4.72 hp
 1/ v  3600  1/ 0.01623 
4.89
Use Eq. 4.58 with   :
2
1
A2V2 AV1 0.1 m3 /s
1
0.1
0.1
V1 

50.93 m/s and V2 

12.73 m/s
2

0.025
 2
0.05
The volume flow rate is the product AV . Assuming a 100% efficient pump for
minimum power,
 2  12 p2 p1 
V V

) m  2
( WP


 
 2

12.732 
50.932 4000  
200
 
0.1 1000 


 258.4 kW or 346 hp
1000
1000 
 2
T e os n “00 i t dnm nt o t k ec nrye cne s t k
h cnt t10”nh eo i o fh i t ee t m ovr W o W.
a
e
ar e ni
g r
t
63

34. 4.90
Assume a 100% efficient turbine:
 2  12
V V
p p1 
0
300

WT  AV1  2
 1
 2

20

6000 kW
 1000  
 2
 
1000


We neglected the kinetic energy change because information was not given and
because it is most often negligible. Since the pressure is in kPa the work rate is in kW.
4.91
Assume a 100% efficient turbine with the pressure of zero gage on the surface of the
backwater and at the turbine outlet:
 2  12

V V
p p1

WT  AV1  2
 1
 2
g ( z2  1 ) 
z
 2




    240 W or 39.24 kW
100 9.81 ( 40) 39
4.92
Assume a 100% efficient turbine with the pressure of zero gage on the surface of the
backwater and at the turbine outlet:
 2  12

V V
p p1

WT  AV1  2
 1
 2
g ( z2  1 ) 
z
 2





1000    9.81  21200 W or 21.2 kW
(0.6 1.2) 1.5 
( 2) 
4.93
3414
W
 10

a) Q   m(h2  1 ). Q  000 
h

30(1131 
1512).
T
3600
 1954 Btu/sec
Q 
b) TK solution:
Rule Sheet
Wdot = mdot * (h1 - h2) + Qdot ; First law for steady flow with no changes in potential or
; kinetic energy
;*Stm8e.tkw Steam, 1-8 States, English units
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,
; and Kell, Hemisphere Publishing Corp., 1984. See Comment sheet.
Input
Name
`
1000
800
5
T1
p1
h1
v1
x1
phase1
T2
p2
h2
Variable Sheet
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English units
F
Temperature
psi
Pressure
1510
B/lbm
Enthalpy
1.05
ft^3/lbm Specific Volume
'mngless
Quality
'SH
Phase
162
F
Temperature
psi
Pressure
1130
B/lbm
Enthalpy
64

35. 1
10
30
4.94
4.95
v2
x2
phase2
Wdot
mdot
Qdot
73.5
ft^3/lbm
'SAT
MW
lbm/s
B/s
-1950
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Power output
Mass flow rate
Heat is removed at a rate of 1950 B/s
1000
 m(
WT h2  1 ) 
h
(1116.1 
1623.8) 
8462 Btu/sec or 11,970 hp
60

m
1000 / 60
V

230 fps
A  2 /173.75
2
where we used  v.
1/
h1 
3445.2 kJ/kg, h2 h f x2 h fg 251.4 
0.9(2358.3) 2374 kJ/kg
 m(
WT h2  1 ) 
h
6(2374 
3445) 
6430 kW

m
6
V2 


82.2 m/s
2
A 
0.4 /[0.001 
0.9(7.649 
0.001)]
4.96
4.97
600
  2
 0.05 100
0.287 
373
 A2V2 AV1 . V2 

18.17 m/s
2
1 1
140
2
 
 0.2
0.287 
253
600
 1 1
m AV1 
  2  4.402 kg/s
 0.05 100
0.287 
373

V 2  12 
V
18.17 2  2 
100
 m
WT  c p (T2  1 )  2
T

4.402   
1.00( 20 100) 



 539 kW
2 
2
1000 


T eat “00 it dnm nt o t k ec nrye cne s t k
h f o 10”n h eo i o fh i t ee t m ovr W o W.
cr
e
ar e ni
g r
t
800
 1 1
m AV1 
  2 
 0.05 20 0.9257 kg/s
0.287 
473
 
Q mc (T  )   0.9257 
T W
1.0(200   
20) ( 400)  kJ/s
233
p
2
1
S
The negative sign indicates a heat loss, as expected.
4.98
450
 
Q mc p (T2  1 )   
T WS
  2   
 .06 150 1.0 (20 100)  
500
70.5 kJ/s
.287 
373
65

36. 4.99
Fr, t cl leh vl ie:
it e s a u tt e ci
s l’ c a e o ts

m
30
30
V1 


5.509 fps, V2 

137.7 fps
2
A1 62.4   /144
 2
62.4   2 /144
 0.4
1
2 V2
   V  1 p2 p1 
Q WS m 2

 
 2

137.7 2 
5.5092 (14.7  1 )  
P 144
0 
30

.
142
  p1  psia
32.2
62.4
 2

The factor 32.2 above converts slugs to lbm.
4.100 The energy equation: V22  12  c p (T1  2 ) 202  
V
2
T
2 1.0(293  2 )
T
p1
AV1  A2V2
1
2
RT1
80
2
4
V2 

 
20 5.384 kg/m 2 
s
2
0.287 
293  2
3
Assume an adiabatic expansion using  v :
1/
Continuity equation: AV1  A2V2 .
1 1
2
k-1
T2  2 

T2
293

298.9
  or 0.4 
T1  1 


[80 / 0.287  0.4
293]
2
The above three equations include three unknowns , V2 , and T2 . They are solved
2
by trial-and-error to give

3.47 kg/m3 , V2 
1.55 m/s and T2 219
C
2
k /k
1
 
p
4.101 a) Assume an adiabatic process: T2  1  2 
T
p
1 
0.4 /1.4
585
 
  
468
85 

812 K or 539
C
 2  12
 2  2

V V
V 100
b) 0 2
 p (T2  1 )  2
c
T 

1.0(812 
468)  V2 
.
835 m/s
1000
 2
 2 

p
p
585  2  
0.1 100 812
c) 2 ( d 22 / 4)V2  1 12  1 . d22 

r V
. d2 
0.239 m
RT2
RT1
185   / 4
468 835
p
80

4.102 a) m  AV 
 2  
0.05 200 1.672 kg/s
RT
0.297 
253
V22  12
V
152  2
200
b) 0 
 p (T2  1 ). 0 
c
T

1.042(T2 
20). T2  
0.91 C
2
2
1000
V 2  12
V
502  2
600
4.103 0  2
 2  1. 0 
h h
2 
h 1154.
2
2
1000
P2 20 psia 

 T2 238 F
h2 
1161 
66
h2 
1161 Btu/lbm

37. 4.104
The heat transfer rate for the steam, assuming no pressure drop through the condenser, is
 
Q m (h  ) 600 
h
(94.02 
1116.1)  200 Btu/min
613,
s
s
2s
1s
This equals the energy gined by the water:
 
 40,900 lbm/min
Qw mw c p (T2 w  1w ). 613, 200    
T
mw 1.00 (15). mw 
4.105
a) The heat lost by the air is gained by the water:



ma c pa (T2 a  1a ) mw c pw (T2 w  1w ). 5   mw 
T
T
1.0 200  4.18 
10. mw 23.9 kg/s
 
b) Q m c  23.9 
T
4.18  
200 1000 kJ/s
w
4.106
w pw
With Q 0 and mi 0 there results u f h1 since m f m1 . Across a valve the enthalpy
is constant so that h1 hline 
3674.4 kJ/kg . The final pressure is p f  MPa and
4
uf 
3674.4 kJ/kg. The temperature is interpolated from Table C.3 to be
3674.4 
3650.1
Tf 
   
50 800 813 C
3650.1 
3555.5
The final specific volume is
812.8 
800
vf 
(0.1229 
0.1169) 
0.1229 
0.1244 m 3 /kg
50
V f
4
The mass of steam in the tank is then m f 


32.15 kg
vf
0.1244
4.107
a)
pf 
800 psia 
1587
F
 T f 
uf  1 
h 1620 
b) Interpolate and find v f 
1.512 m 3 /kg. m f 
V f
50


33.1 lbm
vf
1.512
pVi
250 
3
80 
3
4.108 mi  i


8.769 kg. m f 

2.806 kg
RTi
0.287 
298
0.287 
298

Q m u  u  h
m
m
f
f
i i
2 2
2.806 
0.717  
298 8.769 
0.717  
298 (8.769 
2.806)   
1.0 298 503 kJ
13937
90 
5
800 
5
4.109 mi 

5.351 kg. m f 

0.287 
293
0.287  f
T
Tf
0  f m f  i mi  1 h1  f m f 
u
u
m
u
0.717  
293 5.351  m f 
(
5.351)  
1.0 353
Then 0 
0.717  f
T
13937

1124  m f 
(
5.351) 
353. m f 
30.48 kg
Tf
13937
Finally, T f 
457 or 184 m1 
C.
30.48 
5.351 25.1 kg
30.48
67

38. 4.110 Since Q  and mi  , u f h1 
0
0
0.24  
530 0.171T f . T f   or 284F
744 R
p f V f 12  
144 10
mf 


4.36 lbm
RT f
53.3 
744
k k
1/
0.4 /1.4
p f Tf
p
f 
 95 
4.111 a)

. Tf  i    
T
303
127
146
  K or  C
V i pi RTi
piTi
2000 

pi 
T
303
b) For V 
const p2 p1  2   
95
227 kPa
T1
127
V f p f RT f
800 
4
4.112 mi  f 0.02   kg. mi 
m
300 6

36.8 kg. m f 
30.8 kg
0.287 
303
k
1
0.4
m
 f 
30.8
 
a) T f  i      282 or 9
T
303
C
m
36.8
 
 i 
m RT
30.8 
0.287 
282
b) p f  f f 

624 kPa
V f
4
1/ k 
1
T
f 
c) m f mi  
T
i 
2.5
253
 

36.8   23.44 kg
303
 
 
m 13.36 0.02  t 60.  

t 11.13 min
1.8
0.2
4.113 mi 
158.9
C


1726 kg . a) T 
0.001043 1.694
1
1
b) m f 


911.4 kg.  
m 814.5 kg
0.001101 0.3157
0.5
1.5
 d
c) m f 

458.9 kg. Q  (um)  2 m2  f m2  2 m2
h  u  h 
0.001101 0.3157
dt
10000
10 000 (u f  2 )m2 . m2 
h  

1.878 kg/min
2567 
2757
1.878 
t 1725.9 
458.9.  
t 675 min
4.114 a) Assume steady state with the heat sink area equal to the transistor area:
 Q  Q
Q  Q2 
1
3
Tj  
T
4
4

  j  )   j  sur )
h A (T T  A (T T
1
0.5 
h
A
Tj 
325
4
1
   Tj        Tj  4 )
15 10 4 (
325) 0.5 5.67 10 8 10 4 (
300
1
0.5 
15  
10 4
4
2.835   Tj 
10 12
0.003Tj  
2 0
68

39. By trial and error,
Tj = 568 K (295 >> 125
C)
C
The transistor exceeds its maximum allowable temperature of 125
C
b)
  A (T 4 T 4
Qrad     j  surr )        4  4 ) 
0.5 5.67 10 8 10 4 (568 300
0.27 W
0.27
fraction 

0.27
1
4.115 Assume the temperature to be uniform at any instant, negligible radiation, and
constant properties.
Air
T = 20
C
h = 8 W/m2·K
3 cm
7.5 cm
h
L
Bi  c
k
a)
where Lc is the effective length and is defined as Volume/Surfce area. For the
pancake,


(0.075)2 
(0.003)
4
LC 

0.0015 m

2
2 
(0.075)  
 0.0075 
0.003
4
Thus,
h  8
L
0.0015
Bi  c 

0.048  0.1

k
0.25
The lumped capacitance assumption is valid.
b) Energy balance:
 out = Estored
E
 s   )  V
h A (T T
c
h
A
dT
dT
or

 s dt
dt
T 
T
V
c
Let  T  . Then,
=
T
h
A
h
A
d
T 
T

 s dt. ln

 s 
t

V
c
Ti  
T
V
c
V T 
c
T
500 
3000 
0.0015 25 
20
t

ln


ln

647 s
h s Ti  
A
T
8
80 
20
69

40. 4.115 Assume steady-state, one-dimensional conduction, with negligible contact
resistance and constant properties.
1
1
1
1
Roconv 


0.07 
C/W, Riconv 


0.21 
C/W
ho w 20.8 
A
0.6
hi w 8 
A
0.6
t panel
0.005
R panel 


0.07 
C/W
k panel  panel 0.12 
A
0.6
t plaster
0.015
R plaster 


0.015 
C/W
k plaster  plaster 0.17 
A
0.6
t
t
0.2
0.2
Rstud  stud


312.5 
C/W, Rins  ins 

11.91 
C/W
k stud stud 0.16 
A
0.06
kins ins 0.03 
A
0.56
tsiding
0.025
Rsidings 


0.443 
C/W
ksiding siding 0.094 
A
0.6
R  ins
R
Rtotal Roconv  panel  plaster  stud
R
R
 sidings  iconv
R
R
Rstud ins
R
T
20 
5
15
  
Q 
 
16.67 W
312.5 
11.91
Rtotal 0.07 
0.9
0.07 
0.015 

0.443 
0.21
312.5 
11.91
4.117 Neglect radiation.
Tair, h
D2, D1
lpipe
Rpipe
Rair
Tinside
Tair

T
 T
a) Q  inside air 
R pipe  air
R
Tinside  air
T
200 
10


9.98 kW
ln r2 / r1
1
ln 4 / 3.75
1


 16.5 2 .04  

7 30
2 l k 2 r2  2 7 

 l h
Tinside  air
T
200 
10


536 W
ln 4 / 3.75
ln 9 / 4
1
R pipe  insulation  air
R
R


2 7 
 16.5 2 7 
 0.055 2 0.09  

7 30

b) Q 

 Q
% reduction  1  insulation


Q


 536 


94%
 100  1 

 100 
 9980 

70

41. 4.118 Assume the system to be at steady state.
r2 = 3.5 cm
r1 = 2.5 cm
Q total = 500W
1 cm
Ti = 400oC
Rcond
Tamb
Rconv
T
T
T 
T

Qtot  i amb . Rcond  i amb  conv
R

Rcond  conv
R
Qtot
T T
1  1   i  amb
1
1 

     2 
4 1 r2   Qtot
k r
4 h 
r
1 1
1
1


r1 r2
0.025 0.035
k


8.23 W/m 
K
20
1
 i  amb


T T
1  4 400  
4


   2 
4
0.0352  
100
 500
4 h 
r
 Qtot
71