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PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-1 
 
Solutions Manual 
for 
Heat and Mass Transfer: Fundamentals & Applications 
5th Edition 
 Yunus A. Cengel & Afshin J. Ghajar 
McGraw-Hill, 2015 
 
 
 
 
 
Chapter 10 
BOILING AND CONDENSATION 
 
 
 
 
 
 
 
PROPRIETARY AND CONFIDENTIAL 
 
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protected by copyright and other state and federal laws. By opening and using this Manual the user 
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should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to 
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and may not be distributed to or used by any student or other third party. No part of this Manual 
may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, 
without the prior written permission of McGraw-Hill. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-2 
 
Boiling Heat Transfer 
 
10-1C Boiling is the liquid-to-vapor phase change process that occurs at a solid-liquid interface when the surface is heated 
above the saturation temperature of the liquid. The formation and rise of the bubbles and the liquid entrainment coupled with 
the large amount of heat absorbed during liquid-vapor phase change at essentially constant temperature are responsible for the 
very high heat transfer coefficients associated with nucleate boiling. 
 
 
10-2C The different boiling regimes that occur in a vertical tube during flow boiling are forced convection of liquid, bubbly 
flow, slug flow, annular flow, transition flow, mist flow, and forced convection of vapor. 
 
 
10-3C Both boiling and evaporation are liquid-to-vapor phase change processes, but evaporation occurs at the liquid-vapor 
interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature, and it involves no 
bubble formation or bubble motion. Boiling, on the other hand, occurs at the solid-liquid interface when a liquid is brought 
into contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid. 
 
 
10-4C Boiling is called pool boiling in the absence of bulk fluid flow, and flow boiling (or forced convection boiling) in the 
presence of it. In pool boiling, the fluid is stationary, and any motion of the fluid is due to natural convection currents and the 
motion of the bubbles due to the influence of buoyancy. 
 
 
10-5C The boiling curve is given in Figure 10-6 in the text. In the natural convection boiling regime, the fluid motion is 
governed by natural convection currents, and heat transfer from the heating surface to the fluid is by natural convection. In the 
nucleate boiling regime, bubbles form at various preferential sites on the heating surface, and rise to the top. In the transition 
boiling regime, part of the surface is covered by a vapor film. In the film boiling regime, the heater surface is completely 
covered by a continuous stable vapor film, and heat transfer is by combined convection and radiation. 
 
 
10-6C In the film boiling regime, the heater surface is completely covered by a continuous stable vapor film, and heat transfer 
is by combined convection and radiation. In the nucleate boiling regime, the heater surface is covered by the liquid. The 
boiling heat flux in the stable film boiling regime can be higher or lower than that in the nucleate boiling regime, as can be 
seen from the boiling curve. 
 
 
10-7C The boiling curve is given in Figure 10-6 in the text. The burnout point in the curve is point C. The burnout during 
boiling is caused by the heater surface being blanketed by a continuous layer of vapor film at increased heat fluxes, and the 
resulting rise in heater surface temperature in order to maintain the same heat transfer rate across a low-conducting vapor film. 
Any attempt to increase the heat flux beyond maxq will cause the operation point on the boiling curve to jump suddenly from 
point C to point E. However, the surface temperature that corresponds to point E is beyond the melting point of most heater 
materials, and burnout occurs. The burnout point is avoided in the design of boilers in order to avoid the disastrous explosions 
of the boilers. 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
10-3 
 
10-8C Pool boiling heat transfer can be increased permanently by increasing the number of nucleation sites on the heater 
surface by coating the surface with a thin layer (much less than 1 mm) of very porous material, or by forming cavities on the 
surface mechanically to facilitate continuous vapor formation. Such surfaces are reported to enhance heat transfer in the 
nucleate boiling regime by a factor of up to 10, and the critical heat flux by a factor of 3. The use of finned surfaces is also 
known to enhance nucleate boiling heat transfer and the critical heat flux. 
 
 
10-9C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance. 
 
 
 
 
10-10 Water is boiled at Tsat = 120C in a mechanically polished stainless steel pressure cooker whose inner surface 
temperature is maintained at Ts = 130C. The heat flux on the surface is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 120C are (Tables 10-1 and A-9) 
44.1Pr
CJ/kg 4244cN/m 0550.0
skg/m 10232.0kg/m 121.1
J/kg 102203kg/m 4.943
33
33





l
pl
lv
fgl h



 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a 
mechanically polished stainless steel surface (Table 10-3). Note that 
we expressed the properties in units specified under Eq. 10-2 in 
connection with their definitions in order to avoid unit manipulations. 
Analysis The excess temperature in this case is C10120130sat  TTT s which is relatively low (less than 30C). 
Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be 
 
2kW/m 228.4
























 





 


2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m400,228
44.1)102203(0130.0
)120130(4244
0550.0
1.121)-9.81(943.4
)10)(220310232.0(
Pr
)()(
n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
 
 
 
Heating 
 
130C 
120C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-4 
 
10-11 The nucleate pool boiling heat transfer rate per unit length and the rate of evaporation per unit length of water being 
boiled by a rod that is maintained at 10°C above the saturation temperature are to be determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100°C are σ = 0.0589 N/m (Table10-1) and, from Table 
A-9, 
 ρl = 957.9 kg/m
3
 hfg = 2257 × 10
3
 J/kg 
 ρv = 0.5978 kg/m
3
 μl = 0.282 × 10
−3
 kg/m·s 
 Prl = 1.75 cpl = 4217 J/kg·K 
Also, Csf = 0.0130 and n = 1.0 for the boiling of water on platinum 
surface (Table 10-3). 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 10°C, 
which is less than 30°C for water from Fig. 10-6. Therefore, nucleate 
boiling will occur. The heat flux in this case can be determined from the 
Rohsenow relation to be 
 
25
3
3
2/1
33
3
sat
2/1
nucleate
 W/m10408.1
)75.1)(102257)(013.0(
)10(4217
0589.0
)5978.09.957(81.9
)102257)(10282.0(
Pr
)()(















 








 





 


n
lfgsf
splvl
fgl
hC
TTcg
hq



 
Finally, the nucleate pool boiling heat transfer rate per unit length is 
 W/m4420 ) W/m10408.1)(m 010.0(/ 25nucleateboiling  qDLQ 
 
The rate of evaporation per unit length is 
 mkg/s 101.96 3 


 
J/kg 102257
mJ/s4420)/(
3
boilingnevaporatio 
h
LQ
L
m
fg

 
Discussion The value for the rate of evaporation per unit length indicates that 1 m of the platinum-plated rod would boil water 
at a rate of about 2 grams per second. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-5 
 
10-12 Water is boiled at a saturation (or boiling) temperature of Tsat = 120C by a brass heating element whose temperature is 
not to exceed Ts = 125C. The highest rate of steam production is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling since C5120125sat  TTT s which is in the nucleate boiling range of 5 to 30C for water. 
Properties The properties of water at the saturation temperature of 120C are (Tables 10-1 and A-9) 
44.1Pr
N/m 0550.0
kg/m 12.1
kg/m 4.943
3
3




l
v
l



 
CJ/kg 4244
m/skg 10232.0
J/kg 102203
3
3




pl
l
fg
c
h
 
Also, sfC 0.0060 and n = 1.0 for the boiling of water on a brass surface (Table 10-3). Note that we expressed the 
properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit manipulations. 
Analysis Assuming nucleate boiling, the heat flux in this case can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m300,290
44.1)102203(0060.0
)120125(4244
0550.0
)12.19.81(943.4
)10)(220310232.0(
Pr
)()(
















 








 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The surface area of the heater is 
 2m 04084.0m) m)(0.65 02.0(  DLAs 
Then the rate of heat transfer during nucleate boiling becomes 
 W856,11) W/m300,290)(m 04084.0( 22nucleateboiling  qAQ s 
 
(b) The rate of evaporation of water is determined from 
 kg/h 19.4







h 1
s 3600
J/kg 102203
J/s 856,11
3
boiling
nevaporatio
fgh
Q
m

 
Therefore, steam can be produced at a rate of about 20 kg/h by this heater. 
 
Ts=125C Water 
120C 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-6 
 
10-13 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100C in a mechanically 
polished stainless steel pan whose inner surface temperature is maintained at Ts = 110C. The rate of heat transfer to the water 
and the rate of evaporation of water are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note 
that we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit 
manipulations. 
Analysis The excess temperature in this case is C10100110sat  TTT s which is relatively low (less than 30C). 
Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m700,140
75.1)102257(0130.0
)100110(4217
0589.0
0.60)-9.8(957.9
)10)(225710282.0(
Pr
)()(

























 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The surface area of the bottom of the pan is 
 222 m 07069.04/m) 30.0(4/  DAs 
Then the rate of heat transfer during nucleate boiling becomes 
 W9945 ) W/m700,140)(m 07069.0( 22nucleateboiling qAQ s 
 
(b) The rate of evaporation of water is determined from 
 kg/s 0.00441


J/kg 102257
J/s 9945
3
boiling
nevaporatio
fgh
Q
m

 
That is, water in the pan will boil at a rate of 4.4 grams per second. 
 
Heating 
P = 1 atm 
110C 
100C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-7 
 
10-14 The nucleate pool boiling heat transfer coefficient of water being boiled by a horizontal platinum-plated rod is to be 
determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100°C are σ = 0.0589 N/m (Table 10-1) and, from Table 
A-9, 
 ρl = 957.9 kg/m
3
 hfg = 2257 × 10
3
 J/kg 
 ρv = 0.5978 kg/m
3
 μl = 0.282 × 10
−3
 kg/m·s 
 Prl = 1.75 cpl = 4217 J/kg·K 
Also, Csf = 0.0130 and n = 1.0 for the boiling of water on platinum 
surface (Table 10-3). 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 10°C, 
which is less than 30°C for water from Fig. 10-6. Therefore, nucleate 
boiling will occur. The heat flux in this case can be determined from the 
Rohsenow relation to be 
 
25
3
3
2/1
33
3
sat
2/1
nucleate
 W/m10408.1
)75.1)(102257)(013.0(
)10(4217
0589.0
)5978.09.957(81.9
)102257)(10282.0(
Pr
)()(















 








 





 


n
lfgsf
splvl
fgl
hC
TTcg
hq



 
Using the Newton’s law of cooling, the boiling heat transfer coefficient is 
 )( satnucleate TThq s  → 
sat
nucleate
TT
q
h
s 


 
 K W/m14,100 2 



K )100110(
 W/m10408.1 25
h 
Discussion Heat transfer coefficient on the order of 10
4
 W/m
2
·K can be obtained in nucleate boiling with a temperature 
difference of just 10°C. 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
10-8 
 
10-15 The nucleate boiling heat transfer coefficient and the value of Csf for water being boiled by a long electrical wire are to 
be determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation 
temperature of 100°C are σ = 0.0589 N/m (Table 10-1) 
and, from Table A-9, 
 ρl = 957.9 kg/m
3
 hfg = 2257 × 10
3
 J/kg 
 ρv = 0.5978 kg/m
3
 μl = 0.282 × 10
−3kg/m·s 
 Prl = 1.75 cpl = 4217 J/kg·K 
Also, n = 1.0 is given. 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 28°C, which is less than 30°C for water from Fig. 10-6. 
Therefore, nucleate boiling will occur. The nucleate boiling heat transfer coefficient can be determined using 
 )( satboiling TThq s  → 
sat
boiling
TT
q
h
s 


 
Also, we know 
 W/m4100/ boilingboiling  qDLQ 
  
 
26
boiling W/m10305.1
)m 001.0(
 W/m4100 W/m4100

D
q 
Hence, the nucleate boiling heat transfer coefficient is 
 K W/m46,600 2 



K )100128(
 W/m10305.1 6
h 
The value of the experimental constant Csf can be determined from the Rohsenow relation to be 
 
3
sat
2/1
boiling
Pr
)()(







 





 

n
lfgsf
splvl
fgl
hC
TTcg
hq


 
or 
 
3/1
3
sat
2/1
boiling Pr
)()(

















 





 

n
lfg
splvlfgl
sf
h
TTcg
q
h
C



 
 
0.0173























 




3/1
3
3
2/1
6
33
)75.1)(102257(
)100128(4217
0589.0
)5978.09.957(81.9
10305.1
)102257)(10282.0(
sfC 
Discussion The boiling heat transfer coefficient of 46,600 W/m
2
·K is within the range suggested by Table 1-5 for boiling and 
condensation (2500 to 100,000 W/m
2
·K). 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-9 
 
10-16 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C by a 
stainless steel heating element. The surface temperature of the heating element and its power rating are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the coffee maker are negligible. 3 The boiling regime is 
nucleate boiling (this assumption will be checked later). 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a stainless steel surface (Table 10-3 ). Note that we expressed the 
properties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit manipulations. 
Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18C is nearly 1 
kg. The rate of energy transfer needed to evaporate half of this water in 25 min and the heat flux are 
 
22
2
2
 W/m299,360=kW/m 299.36=
m 0.002513
kW 7523.0
m 002513.0m) m)(0.20 004.0(
kW 7523.0
s) 60(25
kJ/kg) kg)(2257 5.0(
 







s
s
fg
fg
A
Q
q
DLA
t
mh
QmhtQQ



 
The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. 
Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0130.0
)100(4217
0589.0
0.60)9.81(957.9
)10)(225710282.0(360,299















 
  s
T
 
It gives 
 C112.9sT 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
 The specific heat of water at the average temperature of (14+100)/2 = 57C is cp = 4.184 kJ/kgC. Then the time it 
takes for the entire water to be heated from 14C to 100C is determined to be 
 min 7.97=s 478
kJ/s 0.7523
C14)C)(100kJ/kg kg)(4.184 1(
 




Q
Tmc
tTmctQQ
p
p 
 
P = 1 atm 
1 L 
Water, 100C 
Coffee 
maker 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-10 
 
10-17 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C by a 
copper heating element. The surface temperature of the heating element and its power rating are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the coffee maker are negligible. 3 The boiling regime is 
nucleate boiling (this assumption will be checked later). 
Properties The properties of water at the saturation temperature of 100C are 
(Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a copper surface (Table 10-3 ). Note that we expressed the 
properties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit manipulations. 
Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18C is nearly 1 
kg. The rate of energy transfer needed to evaporate half of this water in 25 min and the heat flux are 
 
22
2
2
 W/m299,360=kW/m 299.36=
m 0.002513
kW 7523.0
m 002513.0m) m)(0.20 004.0(
kW 7523.0
s) 60(25
kJ/kg) kg)(2257 5.0(
 







s
s
fg
fg
A
Q
q
DLA
t
mh
QmhtQQ



 
The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. 
Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0130.0
)100(4217
0589.0
0.60)9.81(957.9
)10)(225710282.0(360,299















 
  s
T
 
It gives 
 C112.9sT 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
 The specific heat of water at the average temperature of (14+100)/2 = 57C is cp = 4.184 kJ/kgC. Then the time it 
takes for the entire water to be heated from 14C to 100C is determined to be 
min 7.97=s 478
kJ/s 0.7523
C14)C)(100kJ/kg kg)(4.184 1(
 




Q
Tmc
tTmctQQ
p
p 
 
P = 1 atm 
1 L 
Water, 100C 
Coffee 
maker 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-11 
 
10-18 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C in a 
teflon-pitted stainless steel pan placed on an electric burner. The water level drops by 10 cm in 30 min during boiling. The 
inner surface temperature of the pan is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the pan are negligible. 3 The boiling regime is nucleate 
boiling (this assumption will be checked later). 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0058 and n = 1.0 for the boiling of water on a teflon-pitted stainless steel surface (Table 10-3). Note that we 
expressed theproperties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit 
manipulations. 
Analysis The rate of heat transfer to the water and the heat flux are 
 
22
222
evap
23
evap
evap
 W/m240,200=)m 42 W)/(0.0317547(/
m 03142.04/m) 20.0(4/
kW 547.7kJ/kg) kg/s)(2257 03344.0(
kg/s 003344.0
s 6015
m) 0.10 /4m) 0.2()(kg/m 9.957(












s
s
fg
AQq
DA
hmQ
t
V
t
m
m





 
 The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be 
used to determine the surface temperature when the heat flux is given. Assuming nucleate boiling, the temperature of the inner 
surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0058.0
)100(4217
0589.0
0.60)9.8(957.9
)10)(225710282.0(200,240















 
  s
T
 
It gives 
 Ts = 105.3C 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
Heating 
P = 1 atm 
Ts 
100C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-12 
 
10-19 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C in a 
polished copper pan placed on an electric burner. The water level drops by 10 cm in 30 min during boiling. The inner surface 
temperature of the pan is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the pan are negligible. 3 The boiling regime is nucleate 
boiling (this assumption will be checked later). 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a copper surface (Table 10-3). Note that we expressed the 
properties in units specified under Eq. 10-2 connection with their definitions in order to avoid unit manipulations. 
 Analysis The rate of heat transfer to the water and the heat flux are 
 
22
222
evap
23
evap
evap
 W/m240,200=)m 42 W)/(0.0317547(/
m 03142.04/m) 20.0(4/
kW 547.7kJ/kg) kg/s)(2257 03344.0(
kg/s 003344.0
s 6015
m) 0.10 /4m) 0.2()(kg/m 9.957(












s
s
fg
AQq
DA
hmQ
tt
m
m




 V
 
 The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be 
used to determine the surface temperature when the heat flux is given. Assuming nucleate boiling, the temperature of the inner 
surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0130.0
)100(4217
0589.0
0.60)9.8(957.9
)10)(225710282.0(200,240















 
  s
T
 
It gives 
 Ts = 111.9C 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
Heating 
P = 1 atm 
Ts 
100C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-13 
 
10-20 Water is boiled at Tsat = 120C in a mechanically polished stainless steel pressure cooker whose inner surface 
temperature is maintained at Ts = 132C. The boiling heat transfer coefficient is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 120C are (Tables 10-1 and A-9) 
44.1Pr
CJ/kg 4244N/m 0550.0
skg/m 10232.0kg/m 121.1
J/kg 102203kg/m 4.943
33
33





l
pl
lv
fgl
c
h



 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically 
polished stainless steel surface (Table 10-3). Note that we expressed the 
properties in units specified under Eq. 10-2 in connection with their 
definitions in order to avoid unit manipulations. 
Analysis The excess temperature in this case is C12120132sat  TTT s which is relatively low (less than 30C). 
Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m600,394
44.1)102203(0130.0
)120132(4244
0550.0
1.121)-9.81(943.4
)10)(220310232.0(
Pr
)()(

























 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The boiling heat transfer coefficient is 
CkW/m 32.9 2 



 C W/m880,32
C)120132(
 W/m600,394
)( 2
2
sat
nucleate
satnucleate
TT
q
hTThq
s
s

 
 
Heating 
 
132C 
120C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-14 
 
10-21E Water is boiled at a temperature of Tsat = 250F by a nickel-plated heating element whose surface temperature is 
maintained at Ts = 280F. The boiling heat transfer coefficient, the electric power consumed, and the rate of evaporation of 
water are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling since T T Ts     sat F280 250 30 which is in the nucleate boiling range of 9 to 55F for water. 
Properties The properties of water at the saturation temperature of 250F are (Tables 10-1 and A-9E) 
43.1Pr
lbm/s 1208.0lbf/ft 003755.0
lbm/ft 0723.0
lbm/ft 82.58
2
3
3




l
v
l



 
FBtu/lbm 015.1
hlbm/ft 0.556slbm/ft 10544.1
Btu/lbm 946
4




pl
l
fg
c
h
 
Also, g = 32.2 ft/s
2
 and sfC 0.0060 and n = 1.0 for the boiling of water on a nickel plated surface (Table 10-3). Note that 
we expressed the properties in units that will cancel each other in boiling heat transfer relations. 
Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be 
 
2
31/2
3
sat,
2/1
nucleate
ftBtu/h 221,475,3
43.1)946(0060.0
)250280(015.1
1208.0
)0723.032.2(58.82
))(946556.0(
Pr
)()(






 





 








 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
Then the convection heat transfer coefficient becomes 
FftBtu/h 115,840
2 





F)250280(
ftBtu/h 3,475,221
 )(
2
sat
sat
TT
q
hTThq
s
s

 
(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from 
 
Btu/h) 3412 =kW 1 (since =
Btu/h 811,909)ftBtu/h 221ft)(3,475, 2ft 12/5.0()( 2
 kW 266.7
  qDLAqQW se 

 
(c) Finally, the rate of evaporation of water is determined from 
 lbm/h 961.7
Btu/lbm 946
Btu/h 811,909boiling
nevaporatio
fgh
Q
m

 
Ts=280F 
Water 
250F 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-15 
 
10-22E Water is boiled ata temperature of Tsat = 250F by a platinum-plated heating element whose surface temperature is 
maintained at Ts = 280F. The boiling heat transfer coefficient, the electric power consumed, and the rate of evaporation of 
water are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling since F30250280sat  TTT s which is in the nucleate boiling range of 9 to 55F for water. 
Properties The properties of water at the saturation temperature of 250F are (Tables 10-1 and A-9E) 
43.1Pr
lbm/s 1208.0lbf/ft 003755.0
lbm/ft 0723.0
lbm/ft 82.58
2
3
3




l
v
l



 
FBtu/lbm 015.1
hlbm/ft 0.556slbm/ft 10544.1
Btu/lbm 946
4




pl
l
fg
c
h
 
Also, g = 32.2 ft/s
2
 and sfC 0.0130 and n = 1.0 for the boiling of water on a platinum plated surface (Table 10-3). Note 
that we expressed the properties in units that will cancel each other in boiling heat transfer relations. 
Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be 
 
2
3
3
1/2
3
sat,
2/1
nucleate
ftBtu/h 670,341
43.1)101208.0(0130.0
)250280(015.1
1208.0
)0723.032.2(58.82
))(946556.0(
Pr
)()(
















 








 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
Then the convection heat transfer coefficient becomes 
FftBtu/h 11,390
2 





F)250280(
ftBtu/h 341,670
 )(
2
sat
sat
TT
q
hTThq
s
s

 
(b) The electric power consumed is equal to the rate of heat transfer to the water, and is determined from 
 
Btu/h) 3412 =kW 1 (since =
Btu/h 449,89)ftBtu/h 0ft)(341,67 2ft 12/5.0()( 2
 kW 26.2
  qDLAqQW se 

 
(c) Finally, the rate of evaporation of water is determined from 
 lbm/h 94.6
Btu/lbm 946
Btu/h 449,89boiling
nevaporatio
fgh
Q
m

 
Ts=280F 
Water 
250F 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-16 
 
10-23E Prob. 10-22E is reconsidered. The effect of surface temperature of the heating element on the boiling heat 
transfer coefficient, the electric power, and the rate of evaporation of water is to be investigated. 
 Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
T_sat=250 [F] 
L=2 [ft] 
D=0.5/12 [ft] 
T_s=280 [F] 
 
"PROPERTIES" 
Fluid$='steam_IAPWS' 
P_sat=pressure(Fluid$, T=T_sat, x=1) 
rho_l=density(Fluid$, T=T_sat, x=0) 
rho_v=density(Fluid$, T=T_sat, x=1) 
sigma=SurfaceTension(Fluid$, T=T_sat)*Convert(lbf/ft, lbm/s^2) 
mu_l=Viscosity(Fluid$,T=T_sat, x=0) 
Pr_l=Prandtl(Fluid$, T=T_sat, P=P_sat+1) "P=P_sat+1 is used to get liquid state" 
C_l=CP(Fluid$, T=T_sat, x=0) 
h_f=enthalpy(Fluid$, T=T_sat, x=0) 
h_g=enthalpy(Fluid$, T=T_sat, x=1) 
h_fg=h_g-h_f 
C_sf=0.0060 "from Table 10-3 of the text" 
n=1 "from Table 10-3 of the text" 
g=32.2 [ft/s^2] 
 
"ANALYSIS" 
"(a)" 
q_dot_nucleate=mu_l*h_fg*(((g*(rho_l-rho_v))/sigma)^0.5)*((C_l*(T_s-T_sat))/(C_sf*h_fg*Pr_l^n))^3 
q_dot_nucleate=h*(T_s-T_sat) 
"(b)" 
W_dot_e=q_dot_nucleate*A*Convert(Btu/h, kW) 
A=pi*D*L 
"(c)" 
m_dot_evap=Q_dot_boiling/h_fg 
Q_dot_boiling=W_dot_e*Convert(kW, Btu/h) 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-17 
 
 
Ts 
[F] 
h 
[Btu/h.ft
2
.F] 
eW
 
[kW] 
evapm 
[lbm/h] 
260 
262 
264 
266 
268 
270 
272 
274 
276 
278 
280 
282 
284 
286 
288 
290 
292 
294 
296 
298 
300 
12919 
18603 
25321 
33073 
41857 
51676 
62528 
74413 
87332 
101285 
116271 
132290 
149343 
167430 
186550 
206704 
227891 
250111 
273366 
297653 
322974 
9.912 
17.13 
27.2 
40.6 
57.81 
79.3 
105.5 
137 
174.2 
217.6 
267.6 
324.8 
389.6 
462.5 
543.9 
634.4 
734.4 
844.4 
964.8 
1096 
1239 
35.77 
61.82 
98.17 
146.5 
208.6 
286.2 
380.9 
494.5 
628.8 
785.3 
965.9 
1172 
1406 
1669 
1963 
2290 
2650 
3047 
3482 
3956 
4472 
 
 
260 265 270 275 280 285 290 295 300
0
50000
100000
150000
200000
250000
300000
350000
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
Ts [F]
h
 
[B
tu
/h
-f
t2
-F
]
W
e
 
[k
W
]
h
We
 
260 265 270 275 280 285 290 295 300
0
500
1000
1500
2000
2500
3000
3500
4000
4500
Ts [F]
m
e
v
a
p
 
[l
b
m
/h
]
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-18 
 
10-24 Hot mechanically polished stainless steel ball bearings are cooled by submerging them in water at 1 atm. The rate of 
heat removed from a ball bearing at the instant it is submerged in the water is to be determined. 
Assumptions 1 Steady operating conditions exist at the instant of submersion. 2 Surface temperature is uniform. 3 The boiling 
regime is nucleate boiling since ΔTexcess = Ts − Tsat = 25°C, which is in the nucleate boiling range of 5 to 30°C for water. 
Properties At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are σ = 
0.0589 N/m (Tables 10-1) and, from Table A-9, 
 
75.1Pr
kg/m 5978.0
kg/m 9.957
3
3



l
v
l


 
KJ/kg 4217
skg/m 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, Csf = 0.0130 and n = 1 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). 
Analysis The instant a ball bearing is submerged in the water, with ΔTexcess = 25°C, nucleate boiling would occur. The heat 
flux can be determined from Rohsenow relation to be 
 
26
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m101998.2
75.1)102257(0130.0
)100125(4217
05890
)597809957(819
)10)(225710282.0(
Pr
)()(
















 








 





 


.
...
hC
TTcg
hq
n
lfgsf
slpvl
fgl



 
The heat transfer surface area is 
 222 m 007854.0)m 05.0(  DAs 
The rate of heat removed from a ball bearing at the instant it is submerged in the water is 
 kW 17.3 ) W/m101998.2)(m 007854.0(
262
nucleateboiling qAQ s 
 
Discussion Note that a 5-cm-diameter stainless steel ball can release 17.3 kW of heat at the instant it is submerged in water. 
This high value of heat rate removal, even with a temperature difference of only 25°C, is a result from nucleate boiling. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-19 
 
10-25 A polished copper tube is used to generate 1.5 kg/s of steam at 270 kPa. The surface temperature of the tube, with the 
interest to minimize the excess temperature, is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The boiling regime is nucleate boiling since ΔTexcess is to be minimized 
(this assumption will be verified). 
Properties At 270 kPa, the saturation temperature of water is Tsat = 130°C. The properties of water at Tsat = 130°C are σ = 
0.05295 N/m (Tables 10-1) and, from Table A-9, 
 
33.1Pr
kg/m 496.1
kg/m 6.934
3
3



l
v
l


 
KJ/kg 4263
skg/m 10213.0
J/kg 102174
3
3




pl
l
fg
c
h
 
Also, Csf = 0.013 and n = 1 for the boiling of water on a mechanically polished copper surface (Table 10-3). 
Analysis The heat flux can be determined from the rateof vaporization to be 
 
26
3
vaporvapor
nucleate W/m10384.1
m) m)(15 05.0(
)J/kg 102174)(kg/s 5.1(



DL
hm
A
hm
q
fg
s
fg

 
In the interest of minimizing the excess temperature, the boiling regime would be nucleate boiling. The heat flux can be 
expressed using the Rohsenow relation, 
 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
 
3
3
1/2
3326
33.1)102174(0130.0
)130(4263
052950
)496.16.934(819
)10)(217410213.0( W/m10384.1















 
  s
T
.
.
 
Solving for Ts yields 
C147sT 
Discussion With Ts = 147°C, the excess temperature would be since ΔTexcess = Ts − Tsat = 17°C, which is in the nucleate 
boiling range of 5 to 30C for water. Thus the assumption of nucleate boiling regime is verified. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-20 
 
10-26 A long hot mechanically polished stainless steel sheet is being cooled in a water bath. The temperature of the 
stainless steel sheet leaving the water bath is to be determined whether or not it has the risk of thermal burn hazard. 
Assumptions 1 Steady operating conditions exist. 2 Surface temperature is uniform. 3 The boiling regime is nucleate boiling 
since ΔTexcess = Ts − Tsat = 25°C, which is in the nucleate boiling range of 5 to 30°C for water. 
Properties The specific heat and the density of stainless steel are given as cp,ss = 450 J/kg∙K and ρss = 7900 kg/m
3
, 
respectively. 
At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are σ = 0.0589 N/m 
(Tables 10-1) and, from Table A-9, 
 
75.1Pr
kg/m 5978.0
kg/m 9.957
3
3



l
v
l


 
KJ/kg 4217
skg/m 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, Csf = 0.013 and n = 1 for the boiling of water on a 
mechanically polished stainless steel surface (Table 10-3). 
Analysis The mass of the stainless steel sheet being conveyed 
enters and exits the water bath at a rate of 
 Vwtm ss 
The rate of heat that needs to be removed from the sheet so that it leaves the water bath below 45°C is 
 )( outin,removed TTcmQ ssp  
 
Then, 
 
 W10422.1
K )45125(K)J/kg 450)(m 005.0)(m 5.0)(m/s 2)(kg/m 7900(
)(
6
3
outin,removed


 TTVwtcQ sspss

 
With ΔTexcess = 25°C, nucleate boiling would occur in the water bath. The heat flux can be determined from Rohsenow 
relation to be 
 
26
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m101998.2
75.1)102257(0130.0
)100125(4217
05890
)597809957(819
)10)(225710282.0(
Pr
)()(
















 








 





 


.
...
hC
TTcg
hq
n
lfgsf
slpvl
fgl



 
The heat transfer surface area of the sheet submerged in the water bath is 
 2m 01.1)m 005.0)(m 1(2)m 5.0)(m 1(2 sA 
The rate of heat that could be removed from the sheet in the water bath is 
 W10422.1) W/m101998.2)(m 01.1(
6
removed
262
nucleateboiling  QqAQ s
 W102.222 6 
Discussion The rate of heat that could be removed from the stainless steel sheet in the water bath via nucleate boiling is 
greater than the heat that needs to be removed from the sheet so that it leaves the water bath below 45°C. Thus, there is no 
risk of thermal burn on the stainless steel sheet as it leaves the water bath. 
Note that this analysis is simplified to steady state conditions, but the actual cooling process is transient. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-21 
 
10-27 Water is boiled at the saturation (or boiling) temperature of Tsat = 90C by a horizontal brass heating element. The 
maximum heat flux in the nucleate boiling regime is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 90C 
are (Tables 10-1 and A-9) 
 
96.1Pr
CJ/kg 4206cN/m 0608.0
skg/m 10315.0kg/m 4235.0
J/kg 102283kg/m 3.965
33
33





l
pl
lv
fgl h



 
Also, sfC 0.0060 and n = 1.0 for the boiling of water on a brass heating (Table 10-3). Note that we expressed the 
properties in units specified under Eqs. 10-2 and 10-3 in connection with their definitions in order to avoid unit 
manipulations. For a large horizontal heating element, Ccr = 0.12 (Table 10-4). (It can be shown that L* = 1.58 > 1.2 and thus 
the restriction in Table 10-4 is satisfied). 
Analysis The maximum or critical heat flux is determined from 
2kW/m 873.2


2
4/123
4/12
max
 W/m873,200
)]4235.03.965()4235.0(81.90608.0)[102283(12.0
)]([ vlvfgcr ghCq 
 
qmax 
Water, 90C 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-22 
 
10-28 The electrical current at which a nickel wire would be in danger of burnout in nucleate boiling is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100°C are σ = 0.0589 N/m (Table 10-1) and, from Table 
A-9, 
 ρl = 957.9 kg/m
3
 
hfg = 2257 × 10
3
 J/kg 
 ρv = 0.5978 kg/m
3
 
Analysis The danger of burnout occurs when the heat 
flux is at maximum in nucleate boiling, which can be 
determined using 
 4/12max )]([ vlvfgcr ghCq   
Using Table 10-4, the parameter L* and the constant Ccr are determined to be 
 1997.0
0589.0
)5978.09.957(81.9
)0005.0(
)(
*
2/12/1





 





 


 vlgLL 
which correspond to 
 1795.0)1997.0(12.0*12.0 25.025.0  LCcr 
Hence, the maximum heat flux is 
 
26
4/123
max
 W/m10519.1
)]5978.09.957()5978.0)(81.9)(0589.0)[(102257)(1795.0(

q
 
We know that 
 
DL
RI
A
RI
q
s 
22
 
Thus, the electrical current at which the wire would be in danger of burnout is 
 A192

















2/1
262/1
max
/m 129.0
)m 001.0() W/m10519.1(
)/(

LR
Dq
I

 
Discussion The electrical current at which burnout could occur will decrease if the resistance of the wire increases. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-23 
 
10-29 Water is boiled at a temperature of Tsat = 100°C by hot gases flowing through a tube submerged in water. The 
maximum rate of vaporization is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at Tsat = 100°C are σ = 0.0589 N/m (Tables 10-1) and, from Table A-9, ρl = 957.9 kg/m
3
, 
ρv = 0.5978 kg/m
3
, hfg = 2257 × 10
3
 J/kg. 
Analysis The maximum rate of vaporization occurs at the maximum heat flux. 
For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
cylinder) large thusand 1.2 > * (since 12.0
1.2 >98.9
0589.0
)5978.09.957(81.9
)2/050.0(
)(
2
*
2/12/1
LC
gD
L
cr
vl






 





 



 
Then the maximum heat flux is determined from 
 
264/123
4/12
max
 W/m100155.1)]5978.09.957()5978.0(81.90589.0)[102257(12.0
)]([

 vlvfgcr ghCq 
 
Theheat transfer surface area is 
 2m 5708.1m) m)(10 05.0(  DLAs 
Then, the rate of heat transfer during nucleate boiling becomes 
 W105952.1) W/m100155.1)(m 5708.1(
6262
maxboiling  qAQ s 
 
The maximum rate of vaporization of water is determined from 
 kg/s 0.707



J/kg 102257
J/s 105952.1
3
6
boiling
vapor
fgh
Q
m

 
Discussion The rate of vaporization can be increased by increasing the tube diameter, thereby increasing the heat transfer 
surface area. 
 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-24 
 
10-30 Water is boiled at a temperature of Tsat = 160°C by a 3 m × 3 m horizontal flat heater heated by hot gases flowing 
through an array of tubes embedded in it. The maximum rate of vaporization is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 
Properties The properties of water at Tsat = 160°C are σ = 0.0466 N/m (Tables 10-1) and, from Table A-9, ρl = 907.4 kg/m
3
, 
ρv = 3.256 kg/m
3
, hfg = 2083 × 10
3
 J/kg. 
Analysis The maximum rate of vaporization occurs at the maximum heat flux. 
For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
heater)flat large thusand 27 > * (since 149.0
27 >1309
0466.0
)256.34.907(81.9
)3(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
Then the maximum heat flux is determined from 
 26
4/123
4/12
max
 W/m105252.2
)]256.34.907()256.3(81.90466.0)[102083(149.0
)]([


 vlvfgcr ghCq 
 
The heat transfer surface area is 
 2m 9m 3 m 3  LLAs 
Then, the rate of heat transfer during nucleate boiling becomes 
 W102727.2) W/m105252.2)(m 9(
7262
maxboiling  qAQ s 
 
The maximum rate of vaporization of water is determined from 
 kg/s 10.9



J/kg 102083
J/s 102727.2
3
7
boiling
vapor
fgh
Q
m

 
Discussion The rate of vaporization can be increased by increasing the surface area of the plate. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-25 
 
10-31 Water is to be boiled at 1 atm by a spherical heater and a square horizontal flat heater of the same surface area, and 
which heater geometry would produce higher maximum rate of vaporization is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heaters are negligible. 
Properties At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are σ = 
0.0589 N/m (Tables 10-1) and, from Table A-9, ρl = 957.9 kg/m
3
, ρv = 0.5978 kg/m
3
, hfg = 2257 × 10
3
 J/kg. 
Analysis The maximum rate of vaporization occurs at the maximum heat flux. 
For a spherical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
sphere) large thusand 4.26 > * (since 11.0
4.26 >7.199
0589.0
)5978.09.957(81.9
)2/1(
)(
2
*
2/12/1
LC
gD
L
cr
vl






 





 



 
Then the maximum heat flux is determined from 
 
254/123
4/12
max
 W/m103092.9)]5978.09.957()5978.0(81.90589.0)[102257(11.0
)]([

 vlvfgcr ghCq 
 
The heat transfer surface area for a sphere is 
 222 m 142.3)m 1(  DAs 
The maximum rate of vaporization of water is determined from 
 kg/s 1.296



J/kg 102257
) W/m103092.9)(m 142.3(
3
252
maxboiling
vapor
fg
s
fg h
qA
h
Q
m

 (spherical heater) 
The width of a square with the same surface area as the spherical heater is 
 22 m 142.3 LAs  m 7726.1L 
For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
heater)flat large thusand 27 > * (since 149.0
27 >8.707
0589.0
)5978.09.957(81.9
)7726.1(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
Then the maximum heat flux is determined from 
 
264/123
4/12
max
 W/m10261.1)]5978.09.957()5978.0(81.90589.0)[102257(149.0
)]([

 vlvfgcr ghCq 
 
The maximum rate of vaporization of water is determined from 
 kg/s 1.755



J/kg 102257
) W/m10261.1)(m 142.3(
3
262
maxboiling
vapor
fg
s
fg h
qA
h
Q
m

 (square heater) 
Discussion For the same surface area, the square heater would produce about 35% higher in the maximum rate of 
vaporization than the spherical heater. This is because, for the same surface area, the square heater has a Ccr coefficient that is 
about 35% higher than that of the spherical heater. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-26 
 
10-32 Water is boiled at a temperature of Tsat = 180°C by a 3 m × 3 m nickel plated flat heater that is heated by hot gases 
flowing through an array of tubes embedded in it. The surface temperature that produced the maximum rate of steam 
generation is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 3 The boiling regime is 
nucleate boiling. 
Properties The properties of water at Tsat = 180°C are σ = 0.0422 N/m (Tables 10-1) and, from Table A-9, 
 
983.0Pr
kg/m 153.5
kg/m 3.887
3
3



l
v
l


 
KJ/kg 4410
skg/m 10150.0
J/kg 102015
3
3




pl
l
fg
c
h
 
Also, Csf = 0.0060 and n = 1 for the boiling of water on a nickel surface (Table 10-3). 
Analysis The maximum rate of steam generation occurs at the maximum heat flux. 
For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
heater)flat large thusand 27 > * (since 149.0
27 >1359
0422.0
)153.53.887(81.9
)3(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
Then the maximum heat flux is determined from 
 
264/123
4/12
max
 W/m109794.2)]153.53.887()153.5(81.90422.0)[102015(149.0
)]([

 vlvfgcr ghCq 
 
The heat flux for nucleate boiling can be expressed using the Rohsenow relation to be 
 
3
sat,
2/1
nucleatemax
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hqq


 
 
3
3
1/2
3326
983.0)102015(0060.0
)180(4410
04220
)153.53.887(819
)10)(201510150.0( W/m109794.2















 
  s
T
.
.
 
Solving for Ts yields 
C187.5sT 
The convection heat transfer coefficient can be determined as 
 )( satTThq s   K W/m103.97
25 



K )1805.187(
 W/m109794.2 26
h 
Discussion Note that a heat transfer coefficient of about 400 kW/m
2
∙K can be achieved in nucleate boiling with a temperature 
difference of less than 10°C. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-27 
 
10-33 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100C by a mechanically 
polished stainless steel heating element. The maximum heat flux in the nucleate boiling regime and the surface temperature of 
the heater for that case are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from 
the boiler are negligible. 
Properties The properties of water at the saturation temperature of 
100C are (Tables 10-1 and A-9) 
 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l


 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note 
that we expressed the properties in units specified under Eqs. 10-2 and 10-3 in connection with their definitions in order to 
avoid unit manipulations. For a large horizontal heating element, Ccr = 0.12 (Table 10-4). (It can be shown that L* = 3.99 > 
1.2 and thus the restriction in Table 10-4 is satisfied). 
Analysis The maximum or critical heat flux is determined from 
 
2 W/m1,017,000


4/123
4/12
max
)]60.09.957()6.0(8.90589.0)[102257(12.0
)]([ vlvfgcr ghCq 
 
The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into the Rohsenow relation 
together with other properties gives 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0130.0
)100(4217
0589.0
0.60)-9.8(957.9
)10)(225710282.0(000,017,1
















  s
T
 
It gives 
 C119.3sT 
Therefore, the temperature of the heater surface will be only 19.3C above the boiling temperature of water when burnout 
occurs. 
P = 1 atm 
qmax 
Ts = ? Water, 100C 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-28 
 
10-34 Prob. 10-33 is reconsidered. The effect of local atmospheric pressure on the maximum heat flux and the 
temperature difference Ts –Tsat is to be investigated. 
 Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
D=0.03 [m] 
P_sat=101.3 [kPa] 
 
"PROPERTIES" 
Fluid$='steam_IAPWS' 
T_sat=temperature(Fluid$, P=P_sat, x=1) 
rho_l=density(Fluid$, T=T_sat, x=0) 
rho_v=density(Fluid$, T=T_sat, x=1) 
sigma=SurfaceTension(Fluid$, T=T_sat) 
mu_l=Viscosity(Fluid$,T=T_sat, x=0) 
Pr_l=Prandtl(Fluid$, T=T_sat, P=P_sat+1[kPa]) 
c_l=CP(Fluid$, T=T_sat, x=0) 
h_f=enthalpy(Fluid$, T=T_sat, x=0) 
h_g=enthalpy(Fluid$, T=T_sat, x=1) 
h_fg=h_g-h_f 
C_sf=0.0130 "from Table 10-3 of the text" 
n=1 "from Table 10-3 of the text" 
C_cr=0.12 "from Table 10-4 of the text" 
g=9.8 [m/s^2] “gravitational acceleraton" 
"ANALYSIS" 
q_dot_max=C_cr*h_fg*(sigma*g*rho_v^2*(rho_l-rho_v))^0.25 
q_dot_nucleate=q_dot_max 
q_dot_nucleate=mu_l*h_fg*(((g*(rho_l-rho_v))/sigma)^0.5)*((c_l*(T_s-T_sat))/(C_sf*h_fg*Pr_l^n))^3 
DELTAT=T_s-T_sat 
 
 
Psat 
[kPa] 
maxq 
[kW/m
2
] 
T 
[C] 
70 871.9 20.12 
71.65 880.3 20.07 
73.29 888.6 20.02 
74.94 896.8 19.97 
76.59 904.9 19.92 
78.24 912.8 19.88 
79.88 920.7 19.83 
81.53 928.4 19.79 
83.18 936.1 19.74 
84.83 943.6 19.7 
86.47 951.1 19.66 
88.12 958.5 19.62 
89.77 965.8 19.58 
91.42 973 19.54 
93.06 980.1 19.5 
94.71 987.2 19.47 
96.36 994.1 19.43 
98.01 1001 19.4 
99.65 1008 19.36 
101.3 1015 19.33 
 
 
70 75 80 85 90 95 100 105
850
885
920
955
990
1025
19.3
19.4
19.5
19.6
19.7
19.8
19.9
20
20.1
20.2
Psat [kPa]
q
m
a
x
 
[k
W
/m
2
]

T
 
[C
]
Heat
Temp. Dif.
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-29 
 
10-35 A 10 cm × 10 cm flat heater is used for vaporizing refrigerant-134a at 350 kPa. The surface temperature of the heater 
is given as 25°C and the heater is subjected to a heat flux of 0.35 MW/m
2
. The coefficient Csf is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 3 The boiling regime is 
nucleate boiling since ΔTexcess = Ts − Tsat = 20°C. 
Properties At 350 kPa, the saturation temperature of R134a is 5°C (Table A-10). The properties of R134a at Tsat = 5°C are 
from Table A-10, 
 
802.3Pr
kg/m 12.17
kg/m 1278
3
3



l
v
l


 
N/m 0.01084
KJ/kg 1358
skg/m 10589.2
J/kg 108.194
4
3







pl
l
fg
c
h
 
Also, n = 1.7 for the boiling of R134a is given. 
Analysis The heat flux for nucleate boiling can be expressed using the Rohsenow relation to be 
 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
Using the Rohsenow relation to solve for Csf yields 
 
0.00772















 
















 


3/1
6
346/1
7.13
3/1
nucleate
6/1
sat,
1035.0
)108.194)(10589.2(
01084.0
)12.171278(81.9
)802.3)(108.194(
)525(1358
)(
Pr
)(
q
hg
h
TTc
C
fglvl
n
lfg
slp
sf 



 
For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
heater)flat large thusand 27 > * (since 149.0
27 >8.106
01084.0
)12.171278(81.9
)1.0(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
The maximum heat flux in the nucleate boiling regime can be determined from 
 
225
4/123
4/12
max
MW/m 35.0 W/m10087.4
)]12.171278()12.17(81.901084.0)[108.194(149.0
)]([


 vlvfgcr ghCq 
 
Discussion Since q̇max > 0.35 MW/m
2
, the Rohsenow relation for nucleate boiling is appropriate for this analysis. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-30 
 
10-36 Water is boiled at a temperature of Tsat = 150C by hot gases flowing through a mechanically polished stainless steel 
pipe submerged in water whose outer surface temperature is maintained at Ts = 165C. The rate of heat transfer to the water, 
the rate of evaporation, the ratio of critical heat flux to current heat flux, and the pipe surface temperature at critical heat flux 
conditions are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling since C15150165sat  TTT s which is in the nucleate boiling range of 5 to 30C for water. 
Properties The properties of water at the saturation temperature of 150C are (Tables 10-1 and A-9) 
16.1Pr
N/m 0488.0
kg/m 55.2
kg/m 6.916
3
3




l
v
l



 
CJ/kg 4311
m/skg 10183.0
J/kg 102114
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). Note 
that we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit 
manipulations. 
Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m060,384,1
16.1)102114(0130.0
)150165(4311
0488.0
)55.29.81(916.6
)10)(211410183.0(
Pr
)()(
















 








 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The heat transfer surface area is 
 2m 854.7m) m)(50 05.0(  DLAs 
Then the rate of heat transfer during nucleate boiling becomes 
 kW 10,870 W400,870,10) W/m060,384,1)(m 854.7( 22nucleateboiling qAQ s 
 
(b) The rate of evaporation of water is determined from 
 kg/s 5.142
kJ/kg 2114
kJ/s 870,10boiling
nevaporatio
fgh
Q
m

 
 (c) For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
cylinder) large thusand 1.2 > * (since 12.0
1.2 >7.100488.0
)55.26.916(8.9
)025.0(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
Then the maximum or critical heat flux is determined from 
24/123
4/12
max
 W/m1,852,000)]55.26.916()55.2(8.90488.0)[102114(12.0
)]([

 vlvfgcr ghCq 
 
Therefore, 
1.338
060,384,1
000,852,1
current
max
q
q


 
(d) The surface temperature of the pipe at the burnout point is determined from Rohsenow relation at the critical heat flux 
value to be 
 
C166.5















 








 





 


crs
crs
n
lfgsf
crslpvl
fgl
T
T
hC
TTcg
hq
,
3
3
,
1/2
33
3
sat,,
2/1
crnucleate,
16.1)102114(0130.0
)150(4311
0488.0
)55.29.8(916.6
)10)(211410183.0(000,852,1
Pr
)()(



 
Water, 150C 
Boiler 
Hot 
gases 
Vent 
Ts,pipe = 165C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-31 
 
10-37 Water is boiled at a temperature of Tsat = 160C by hot gases flowing through a mechanically polished stainless steel 
pipe submerged in water whose outer surface temperature is maintained at Ts = 165C. The rate of heat transfer to the water, 
the rate of evaporation, the ratio of critical heat flux to current heat flux, and the pipe surface temperature at critical heat flux 
conditions are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling since C5160165sat  TTT s which is in the nucleate boiling range of 5 to 30C for water. 
Properties The properties of water at the saturation temperature of 160C are (Tables 10-1 and A-9) 
09.1Pr
N/m 0466.0
kg/m 256.3
kg/m 4.907
3
3




l
v
l



 
CJ/kg 4340
m/skg 10170.0
J/kg 102083
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3 ). Note 
that we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit 
manipulations. 
Analysis (a) Assuming nucleate boiling, the heat flux can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m390,61
09.1)102083(0130.0
)160165(4340
0466.0
)256.39.81(907.4
)10)(208310170.0(
Pr
)()(
















 








 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The heat transfer surface area is 
 2m 854.7m) m)(50 05.0(  DLAs 
Then the rate of heat transfer during nucleate boiling becomes 
 W482,200 ) W/m390,61)(m 854.7( 22nucleateboiling qAQ s 
 
(b) The rate of evaporation of water is determined from 
 kg/s 0.2315
kJ/kg 2083
kJ/s 2.482boiling
nevaporatio
fgh
Q
m

 
(c) For a horizontal cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
cylinder) large thusand 1.2 > * (since 12.0
0.12 >9.10
0466.0
)256.34.907(81.9
)025.0(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
Then the maximum or critical heat flux is determined from 
264/123
4/12
max
 W/m10.0342)]256.34.907()256.3(81.90466.0)[102083(12.0
)]([

 vlvfgcr ghCq 
 
Therefore, 
33.13


2
26
current
max
 W/m390,61
 W/m10.0342
q
q


 
(d) The surface temperature of the pipe at the burnout point is determined from Rohsenow relation at the critical heat flux 
value to be 
 
C176.1















 








 





 


crs
crs
n
lfgsf
crslpvl
fgl
T
T
hC
TTcg
h
,
3
3
,
1/2
33
3
sat,,
2/1
6
09.1)102083(0130.0
)160(4340
0466.0
)256.39.81(907.4
)10)(208310170.0(
Pr
)()(
10034.2



 
Water, 160C 
Boiler 
Hot 
gases 
Vent 
Ts,pipe = 165C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-32 
 
10-38 Steam is generated by a 1 m × 1 m flat heater boiling water at 1 atm with an excess temperature above 300°C. The 
range of the steam generation rate that can be achieved by the heater is to be determined. 
Assumptions 1 Steady operating conditions exists. 2 Heat losses from the heater are negligible. 3 The boiling regime is film 
boiling since ΔTexcess > 300°C, which is much larger than 30°C. 
Properties At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are σ = 
0.0589 N/m (Tables 10-1) and, from Table A-9, ρl = 957.9 kg/m
3
, ρv = 0.5978 kg/m
3
, hfg = 2257 × 10
3
 J/kg. 
Analysis For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
heater)flat large thusand 27 > * (since 149.0
27 >4.399
0589.0
)5978.09.957(81.9
)1(
)(
*
2/12/1
LC
g
LL
cr
vl






 





 



 
The range of steam generation rate not exceeding the burnout point can be determined from the minimum and maximum 
boiling heat fluxes. 
The minimum rate of vaporization occurs at the minimum heat flux, which can be determined from 
 
2
4/1
2
3
4/1
2min
 W/m19021
)5978.09.957(
)5978.09.957)(81.9)(0589.0(
)102257)(5978.0(09.0
)(
)(
09.0























vl
vl
fgv
g
hq



 
The maximum rate of vaporization occurs at the maximum heat flux, which can be determined from 
 
264/123
4/12
max
 W/m10261.1)]5978.09.957()5978.0(81.90589.0)[102257(149.0
)]([

 vlvfgcr ghCq 
 
The heat transfer surface area is 
 2m 1m 1 m 1  LLAs 
Then, the rate of heat transfer during boiling is 
 qAQ s 
 boiling 
Thus, the range of the steam generation rate in the film boiling regime is 
 
fgfg h
Q
m
h
Q max boiling,
vapor
min boiling,



 
 
fg
s
fg
s
h
qA
m
h
qA max
vapor
min



 
 
J/kg 102257
) W/m10261.1)(m 1(
J/kg 102257
) W/m19021)(m 1(
3
262
vapor3
22




m 
or kg/s 0.559mkg/s 0.00843 vapor   
Discussion The maximum rate of steam generation is more than 66 times larger than the minimum rate of steam generation. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-33 
 
10-39 Water is boiled at Tsat = 90C in a brass heating element. The surface temperature of the heater is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 90C 
are (Tables 10-1 and A-9) 
 
96.1Pr
CJ/kg 4206cN/m 0608.0
skg/m 10315.0kg/m 4235.0
J/kg 102283kg/m 3.965
33
33





l
pl
lv
fgl h



 
Also, sfC 0.0060 and n = 1.0 for the boiling of water on a brass heating (Table 10-3). 
Analysis The minimum heat flux is determined from 
2
4/1
2
3
4/1
2min
 W/m715,13
)4235.03.965(
)4235.03.965)(81.9)(0608.0(
)102283)(4235.0(09.0
)(
)(
09.0























vl
vl
fgv
g
hq



 
The surface temperature can be determined from Rohsenow equation to be 
C92.3
























 





 


s
s
n
lfgsf
slpvl
fgl
T
T
hC
TTcg
hq
3
3
1/2
332
3
sat,
2/1
nucleate
96.1)102283(0060.0
)90(4206
0608.0
0.4235)-9.81(965.3
)10)(228310315.0( W/m715,13
Pr
)()(



 
qminWater, 90C 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-34 
 
10-40 The power dissipation per unit length of a metal rod submerged horizontally in water, when electric current is passed 
through it, is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). 
The properties of vapor at the film temperature of 
Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16, 
 ρv = 0.3831 kg/m
3
 cpv = 1997 J/kg·K 
 μv = 2.045 × 10
−5
 kg/m·s kv = 0.04345 W/m·K 
Analysis The excess temperature in this case is ΔT = Ts 
− Tsat = 400°C, which is much larger than 30°C for 
water from Fig. 10-6. Therefore, film boiling will occur. 
The film boiling heat flux in this case can be determined 
from 
25
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10152.1
)400(
)400)(002.0)(10045.2(
)]400)(1997(4.0102257)[3831.09.957)(3831.0()04345.0(81.9
62.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
The radiation heat flux is determined from 
 24444284sat
4
rad W/m9573K )373773)(K W/m1067.5)(5.0()( 
TTq s 
Then the total heat flux becomes 
 
25225
radfilmtotal W/m10224.1) W/m9573(
4
3
 W/m10152.1
4
3
 qqq  
Finally, the power dissipation per unit length of the metal rod is 
 W/m769 ) W/m10224.1)(m 002.0(/ 25totaltotal  qDLQ 
 
Discussion The contribution of radiation to the total heat flux is about 8%. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-35 
 
10-41E Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 212F by a horizontal 
polished copper heating element whose surface temperature is maintained at Ts = 788F. The rate of heat transfer to the water 
per unit length of the heater is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 212F are 3lbm/ft 82.59l and Btu/lbm 970fgh 
(Table A-9E). The properties of the vapor at the film temperature of F5002/)788212(2/)( sat  sf TTT are (Table 
A-16E) 
FftBtu/h 02267.0
FBtu/lbm 4707.0
hlbm/ft 0.04561slbm/ft 10267.1
lbm/ft 02571.0
5
3





v
pv
v
v
k
c


 
Also, g = 32.2 ft/s
2
 = 32.2(3600)
2
 ft/h
2
. Note that we expressed the properties in units that will cancel each other in boiling 
heat transfer relations. Also note that we used vapor properties at 1 atm pressure from Table A-16E instead of the properties 
of saturated vapor from Table A-9E since the latter are at the saturation pressure of 680 psia (46 atm). 
Analysis The excess temperature in this case is F576212788sat  TTT s , which is much larger than 30C or 54F. 
Therefore, film boiling will occur. The film boiling heat flux in this case can be determined to be 
 
2
4/1
32
sat
4/1
sat
3
film
ftBtu/h 600,18
)212788(
)212788)(12/5.0)(04561.0(
)]212788(4707.04.0970)[02571.082.59)(02571.0()02267.0()3600(2.32
62.0
)(
)(
)](4.0)[(
62.0
























 TT
TTD
TTchgk
q s
sv
satspvfgvlvv



 
 The radiation heat flux is determined from 
 
2
44428
4
sat
4
rad
ftBtu/h 4.190
R) 460212(R) 460788()RftBtu/h 101714.0)(05.0(
)(




TTq s
 
Note that heat transfer by radiation is very small in this case because of the low emissivity of the surface and the relatively 
low surface temperature of the heating element. Then the total heat flux becomes 
 
2
radfilmtotal ftBtu/h 743,184.190
4
3
600,18
4
3
 qqq  
Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat 
transfer surface area, 
 
Btu/h 2453


)ftBtu/h ft)(18,743 1ft 12/5.0(
)(
2
totaltotaltotal

 qDLqAQ s 

 
P = 1 atm 
Water, 212F 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-36 
 
10-42E Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 212F by a horizontal 
polished copper heating element whose surface temperature is maintained at Ts = 988F. The rate of heat transfer to the water 
per unit length of the heater is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 212F are 3lbm/ft 82.59l and Btu/lbm 970fgh 
(Table A-9E). The properties of the vapor at the film temperature of F6002/)988212(2/)( sat  sf TTT are, by 
interpolation, (Table A-16E) 
FftBtu/h 02640.0
FBtu/lbm 4799.0
hlbm/ft 0.05099slbm/ft 10416.1
lbm/ft 02395.0
5
3





v
pv
v
v
k
c


 
Also, g = 32.2 ft/s
2
 = 32.2(3600)
2
 ft/h
2
. Note that we expressed the properties in units that will cancel each other in boiling 
heat transfer relations. Also note that we used vapor properties at 1 atm pressure from Table A-16E instead of the properties 
of saturated vapor from Table A-9E since the latter are at the saturation pressure of 1541 psia (105 atm). 
Analysis The excess temperature in this case is F776212988sat  TTT s , which is much larger than 30C or 54F. 
Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 
 
2
4/1
32
sat
4/1
sat
3
film
ftBtu/h 147,25
)212988(
)212988)(12/5.0)(05099.0(
)]212988(4799.04.0970)[02395.082.59)(02395.0()0264.0()3600(2.32
62.0
)(
)(
)](4.0)[(
62.0
























 TT
TTD
TTChgk
q s
sv
satspvfgvlvv



 
 The radiation heat flux is determined from 
  
2
44428
4
sat
4
rad
ftBtu/h 3.359
R) 460212(R) 460988()RftBtu/h 101714.0)(05.0(
)(




TTq s
 
Note that heat transfer by radiation is very small in this case because of the low emissivity of the surface and the relatively 
low surface temperature of the heating element. Then the total heat flux becomes 
 
2
radfilmtotal ftBtu/h 416,253.359
4
3
147,25
4
3
 qqq  
Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat 
transfer surface area, 
 
Btu/h 3327


)ftBtu/h ft)(25,416 1ft 12/5.0(
)(
2
totaltotaltotal

 qDLqAQ s 

 
P = 1 atm 
Water, 212F 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-37 
 
10-43 The initial heat transfer rate from a hot metal sphere that is suddenly submerged in a water bath is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The metal sphere has uniform initial surface temperature. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = 400°Care, from Table A-16, 
 ρv = 0.3262 kg/m
3
 cpv = 2066 J/kg·K 
 μv = 2.446 × 10
−5
 kg/m·s kv = 0.05467 W/m·K 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 600°C, which is much larger than 30°C for water from Fig. 
10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 
25
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10052.1
)600(
)600)(02.0)(10446.2(
)]600)(2066(4.0102257)[3262.09.957)(3262.0()05467.0(81.9
67.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
The radiation heat flux is determined from 
 
24
444428
4
sat
4
rad
 W/m10729.3
K )373973)(K W/m1067.5)(75.0(
)(




TTq s
 
Then the total heat flux becomes 
 
252425
radfilmtotal W/m10332.1) W/m10729.3(
4
3
 W/m10052.1
4
3
 qqq  
Finally, the initial heat transfer rate from the submerged metal sphere is 
 W167 2252totaltotal )m 02.0() W/m10332.1( DqQ 
 
Discussion The contribution of radiation to the total heat flux is about 21%, which is significant and cannot be neglected. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-38 
 
10-44 The initial heat transfer rate from a hot steel rod that is suddenly submerged in a water bath is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The steel rod has uniform initial surface temperature. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16, 
 ρv = 0.3831 kg/m
3
 cpv = 1997 J/kg·K 
 μv = 2.045 × 10
−5
 kg/m·s kv = 0.04345 W/m·K 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 400°C, which is much larger than 30°C for water from Fig. 
10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 
24
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10476.6
)400(
)400)(02.0)(10045.2(
)]400)(1997(4.0102257)[3831.09.957)(3831.0()04345.0(81.9
62.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
The radiation heat flux is determined from 
 
24
444428
4
sat
4
rad
 W/m10723.1
K )373773)(K W/m1067.5)(9.0(
)(




TTq s
 
Then the total heat flux becomes 
 
242424
radfilmtotal W/m10768.7) W/m10723.1(
4
3
 W/m10476.6
4
3
 qqq  
Finally, the initial heat transfer rate from the submerged steel rod is 
 W976 )m 2.0)(m 02.0() W/m10768.7( 24totaltotal DLqQ 
 
Discussion The contribution of radiation to the total heat flux is about 17%, which is significant and cannot be neglected. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-39 
 
10-45 Water is boiled at Tsat = 100C by a spherical platinum heating element immersed in water. The surface temperature is 
Ts = 350C. The rate of heat transfer is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100C are (Table A-9) 
3
3
kg/m 9.957
J/kg 102257


l
fgh

 
The properties of water vapor at (350+100)/2 = 225C are (Table A-16) 
C W/m03581.0
CJ/kg 1951
skg/m 10749.1
kg/m 444.0
5
3





v
pv
v
v
k
c


 
Analysis The film boiling occurs since the temperature difference between the surface and the fluid. The heat flux in this case 
can be determined from 
 
 
 
2
4/1
5
33
sat
4/1
sat
sat
3
film
 W/m207,25
)100350(
)100350)(15.0)(10749.1(
)100350)(1951(4.0102257)444.09.957)(444.0()03581.0)(81.9(
67.0
)(
)(
)(4.0)(
67.0


























TT
TTD
TTchgk
q s
sv
spvfgvlvv



 
The radiation heat transfer is 
   24484sat4rad W/m745)273100()273350()1067.5)(10.0()(  TTq s 
The total heat flux is 
 
2
radfilmtotal W/m766,25)745(
4
3
207,25
4
3
 qqq  
Then the total rate of heat transfer becomes 
 W1821 ) W/m766,25()15.0( 22totaltotal qAQ 
 
350C 
100C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-40 
 
10-46 Cylindrical stainless steel rods are heated to 700°C and then suddenly quenched in water at 1 atm. The convection heat 
transfer coefficient and the total rate of heat removed from a rod at the instant it is submerged in the water are to be 
determined. 
Assumptions 1 Steady operating conditions exist at the instant of submersion. 2 Surface temperature is uniform. 3 The boiling 
regime is film boiling since ΔTexcess = Ts − Tsat = 600°C, which is much larger than 30°C. 
Properties At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are hfg = 2257 
× 10
3
 J/kg and ρl = 957.9 kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = (100 + 
700)/2 = 400°C are, from Table A-16, 
 
skg/m 10446.2
kg/m 3262.0
5
3



v
v


 
K W/m05467.0
KJ/kg 2066


v
pv
k
c
 
Analysis The film boiling heat flux can be determined from 
 
2
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m92097
)100700(
)100700)(025.0)(10446.2(
)]100700)(2066(4.0102257)[3262.09.957)(3262.0()05467.0(81.9
62.0
)(62.0
)(
)(
)](4.0)[(



























cylindershorizontalCwhere
TT
TTD
TTchgk
Cq
film
s
sv
spvfgvlvv



 
Thus, the convection heat transfer coefficient is 
 )( satfilm TThq s   K W/m153.5
2 


K )100700(
 W/m92097 2
h 
The radiation heat flux is determined from 
 
2
444428
4
sat
4
rad
 W/m14917
K )373973)(K W/m1067.5)(3.0(
)(




TTq s
 
Then the total heat flux becomes 
 2522radfilmtotal W/m100328.1) W/m14917(
4
3
 W/m92097
4
3
 qqq  
The total rate of heat removed from a rod at the instant it is submerged in the water is 
 W2028 ) W/m100328.1)(m 25.0)(m 025.0()( 25totaltotal  qDLQ 
 
Discussion Convection heat transfer coefficient in film boiling is generally lower than that of nucleate boiling, because the 
excess temperature of film boiling is much larger than that of nucleate boiling. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-41 
 
10-47 A long cylindrical stainless steel rod with mechanically polished surface is being quenched in a water bath. The 
temperature of the rod leaving the water bath is to be determined whether or not it has the risk of thermal burn hazard. 
Assumptions 1 Steady operating conditions exist. 2 Surface temperature is uniform. 3 The boiling regime is film boiling since 
ΔTexcess = Ts − Tsat = 700°C − 100°C = 600°C, which is much larger than 30°C. 
Properties The specific heat and the density of stainless steel are given as cp,ss = 450 J/kg∙K and ρss = 7900 kg/m
3
, 
respectively. 
At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are hfg = 2257 × 10
3
 J/kgand ρl = 957.9 kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = (100 + 700) = 400°C 
are, from Table A-16, 
 
skg/m 10446.2
kg/m 3262.0
5
3



v
v


 
K W/m05467.0
KJ/kg 2066


v
pv
k
c
 
Analysis With ΔTexcess = 600°C, film boiling would occur in the water bath. The heat flux can be determined from 
 
2
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m92097
)100700(
)100700)(025.0)(10446.2(
)]100700)(2066(4.0102257)[3262.09.957)(3262.0()05467.0(81.9
62.0
)(62.0
)(
)(
)](4.0)[(



























cylindershorizontalCwhere
TT
TTD
TTchgk
Cq
film
s
sv
spvfgvlvv



 
The radiation heat flux is determined from 
 
24444284
sat
4
rad W/m14917K )373973)(K W/m1067.5)(3.0()( 
TTq s 
Then the total heat flux becomes 
 2522radfilmtotal W/m100328.1) W/m14917(
4
3
 W/m92097
4
3
 qqq  
The rate of heat that could be removed from the rod in the water bath is 
 W24335) W/m100328.1)(m 3)(m 025.0()( 25totaltotaltotal   qLDqAQ s 
 
The mass of the stainless steel rod being conveyed enters and exits the water bath at a rate of 
 )4/( 2DVm ss  
The rate of heat that needs to be removed from the rod so that it leaves the water bath below 45°C can be determined using 
 )()4/()( outin,
2
outin,total TTcDVTTcmQ sspssssp  
 
Thus, the speed of the rod conveying through the water bath is 
 
m/hr 76.7




m/s 0213.0
K)45700(K)J/kg 450](4/)m 025.0()[kg/m 7900(
 W24335
)()4/(
23
outin,
2
total

 TTcD
Q
V
sspss

 
Discussion To ensure that the stainless steel rod leaves the water bath below 45°C, in order to prevent thermal burn hazard, 
the speed of the rod conveying through the water bath should be about 77 m/hr or slower. 
Note that this analysis is simplified to steady state conditions, but the actual quenching process is transient. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-42 
 
10-48 A cylindrical heater is used for boiling water at 1 atm. The film boiling convection heat transfer coefficient at the 
burnout point is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 3 The boiling regime is film 
boiling. 
Properties At 1 atm, the saturation temperature of water is Tsat = 100°C. The properties of water at Tsat = 100°C are σ = 
0.0589 N/m (Tables 10-1) and, from Table A-9, ρl = 957.9 kg/m
3
, ρv,sat = 0.5978 kg/m
3
, hfg = 2257 × 10
3
 J/kg. 
The properties of vapor at the film temperature of Tf = 1150°C are, from Table A-16, 
 
skg/m 10283.5
kg/m 1543.0
5
3



v
v


 
K W/m1588.0
KJ/kg 2571


v
pv
k
c
 
Analysis For a cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
cylinder) large thusand 1.2 > * (since 12.0
1.2 >997.1
0589.0
)5978.09.957(81.9
)2/01.0(
)(
2
*
2/12/1
sat,
LC
gD
L
cr
vl






 







 



 
The burnout point occurs at the maximum heat flux, which is 
 
264/123
4/1
sat,
2
sat,max
 W/m100155.1)]5978.09.957()5978.0(81.90589.0)[102257(12.0
)]([

 vlvfgcr ghCq 
 
To determine the film boiling convection heat transfer coefficient, the knowledge of Ts is needed, which can be determined 
from the heat transfer in the film boiling region: 
 
)(62.0
)(
4
3
)(
)(
)](4.0)[(
4
3 4
sat
4
sat
4/1
sat
3
filmradfilmtotal
cylindershorizontalCwhere
TTTT
TTD
TTchgk
Cqqq
film
ss
sv
satspvfgvlvv












 



 
Substituting the values, 
 
])373()273T)[(1067.5)(3.0(
4
3
)100T(
)100T)(01.0)(10283.5(
)]100T)(2571(4.0102257)[1543.09.957)(1543.0()1588.0(81.9
62.0100155.1
44
s
8
s
4/1
s
5
s
33
6












 
Solving for the surface temperature yield Ts = 2231°C 
The film boiling heat flux is 
 
25
4/1
5
33
sat
4/1
sat
3
filmfilm
 W/m101419.5
)1002231(
)1002231)(01.0)(10283.5(
)]1002231)(2571(4.0102257)[1543.09.957)(1543.0()1588.0(81.9
62.0
)(
)(
)](4.0)[(


























TT
TTD
TTchgk
Cq s
sv
satspvfgvlvv



 
Thus, the film boiling convection heat transfer coefficient is 
 K W/m241.3 2 





K )1002231(
 W/m101419.5 25
sat
film
TT
q
h
s

 
Discussion Note that the film temperature Tf = (2231 + 100)/2 = 1166C, is close to the assumed value of 1150C for the 
evaluation of vapor properties. Therefore, 1150°C is a reasonable film temperature for the vapor properties. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-43 
 
10-49 A cylindrical heater is used for boiling water at 100°C. The boiling convection heat transfer coefficients at the 
maximum heat flux for nucleate boiling and film boiling are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 
Properties The properties of water at Tsat = 100°C are σ = 0.0589 N/m (Tables 10-1) and, from Table A-9, 
 
75.1Pr
kg/m 5978.0
kg/m 9.957
3
sat,
3



l
v
l


 
KJ/kg 4217
skg/m 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, Csf = 0.0130 and n = 1 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3). 
The properties of vapor at the film temperature of Tf = 1150°C are, from Table A-16, 
 
skg/m 10283.5
kg/m 1543.0
5
3



v
v


 
K W/m1588.0
KJ/kg 2571


v
pv
k
c
 
Analysis For a cylindrical heating element, the coefficient Ccr is determined from Table 10-4 to be 
 
cylinder) small thusand 1.2 * (since 1364.0*12.0
1.2599.0
0589.0
)5978.09.957(81.9
)2/003.0(
)(
2
*
25.0
2/12/1
sat,






 







 

 LLC
gD
L
cr
vl


 
The maximum heat flux can be determined as 
 
264/123
4/1
sat,
2
sat,max
 W/m101543.1)]5978.09.957()5978.0(81.90589.0)[102257(1364.0
)]([

 vlvfgcr ghCq 
 
(a) The surface temperature Ts for nucleate boiling at q̇max can be solved as 
 
3
sat,
2/1
sat,
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
Substituting the values, 
 
3
3
1/2
336
75.1)102257(0130.0
)100(4217
05890
)597809957(819
)10)(225710282.0(101543.1















 
  s
T
.
...
 
  Ts = 120.2°C 
Thus, the nucleate boiling convection heat transfer coefficient is 
 K W/m57,144 2 





K )1002.120(
 W/m101543.1 26
sat
nucleate
nucleate
TT
q
h
s

 
 
(b) The surface temperature Ts for film boiling at q̇max can be solved as 
 )(
4
3
)(
)(
)](4.0)[(
4
3 4
sat
4
sat
4/1
sat
3
filmradfilmtotal TTTT
TTD
TTchgk
Cqqq ss
sv
satspvfgvlvv











 


 
Substituting the values, 
 
])373()273)[(1067.5)(3.0(
4
3
)100(
)100)(003.0)(10283.5(
)]100)(2571(4.0102257)[1543.09.957)(1543.0()1588.0(81.9
62.0101543.1
448
4/1
5
33
6














ss
s
s
TT
T
T
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-44 
 
  Ts = 2192°C 
The film boiling heat flux is 
 
25
4/1
5
33
film
sats
4/1
satsv
satspvfgvlv
3
v
filmfilm
W/m 108366.6)1002192(
)1002192)(003.0)(10283.5(
)]1002192)(2571(4.0102257)[1543.09.957)(1543.0()1588.0(81.9
62.0
)cylindershorizontal(62.0Cwhere
)TT(
)TT(D
)]TT(c4.0h)[(gk
Cq

























 
Thus, the film boiling convection heat transfer coefficient is 
 K W/m327 2 





K )1002192(
 W/m108366.6 25
sat
film
TT
q
h
s

 
Discussion The nucleate boiling convection heat transfer coefficient is about 175 times higher than that of film boiling. This 
is because the vapor film surrounding the heater surface during film boiling impedes convection heat transfer. 
Note that the film temperature Tf = (2192 + 100)/2 = 1146C, is close to the assumed value of 1150C used in film boiling for 
the evaluation of vapor properties. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-45 
 
10-50 Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 100C by a horizontal nickel 
plated copper heating element. The maximum (critical) heat flux and the temperature jump of the wire when the operating 
point jumps from nucleate boiling to film boiling regime are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 5978.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0060 and n = 1.0 for the boiling of water on a nickel plated surface (Table 10-3 ). Note that we expressed the 
properties in units specified under Eqs. 10-2 and 10-3 in connection with their definitions in order to avoid unit 
manipulations. The vapor properties at the anticipated film temperature of Tf = (Ts+Tsat )/2 of 1000C (will be checked) 
(Table A-16) 
skg/m 10762.4
CJ/kg 2471
C W/m1362.0
kg/m 1725.0
5
3





v
pv
v
v
c
k 

 
Analysis (a) For a horizontal heating element, the coefficient Ccr is 
determined from Table 10-4 to be 
 
1364.0)5990.0(12.0*12.0
1.2< 5990.0
0589.0
)5978.09.957(81.9
)0015.0(
)(
*
25.025.0
2/12/1






 





 

LC
g
LL
cr
vl


 
Then the maximum or critical heat flux is determined from 
2 W/m1,154,000


4/123
4/12
max
)]5978.09.957()5978.0(81.90589.0)[102257(1364.0
)]([ vlvfgcr ghCq 
 
The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into the Rohsenow relation 
together with other properties gives 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0060.0
)100(4217
0589.0
0.5978)-9.81(957.9
)10)(225710282.0(000,154,1
















  s
T
 
It gives C109.3sT 
(b) Heat transfer in the film boiling region can be expressed as 
)(
4
3
)(
)(
)](4.0)[(
62.0
4
3 4
sat
4
sat
4/1
sat
3
radfilmtotal TTTT
TTD
TTchgk
qqq ss
sv
satspvfgvlvv











 


 
Substituting, 
 44428
4/1
5
33
)273100()273()K W/m1067.5)(5.0(
4
3
)100(
)100)(003.0)(10762.4(
)]100(24714.0102257)[1725.09.957)(1725.0()1362.0(81.9
62.0000,154,1














ss
s
s
TT
T
T
 
Solving for the surface temperature gives Ts = 1999C. Therefore, the temperature jump of the wire when the operating point 
jumps from nucleate boiling to film boiling is 
Temperature jump: C1890 1091999crit,film s, sTTT 
Note that the film temperature is (1999+100)/2=1050C, which is close enough to the assumed value of 1000C for the 
evaluation of vapor paroperties. 
P = 1 atm 
qmax 
Ts Water, 100C 
Heating element 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-46 
 
10-51 Prob. 10-50 is reconsidered. The effects of the local atmospheric pressure and the emissivity of the wire on the 
critical heat flux and the temperature rise of wire are to be investigated. 
 Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=0.3 [m] 
D=0.003 [m] 
epsilon=0.5 
P=101.3 [kPa] 
 
"PROPERTIES" 
Fluid$='steam_IAPWS' 
T_sat=temperature(Fluid$, P=P, x=1) 
rho_l=density(Fluid$, T=T_sat, x=0) 
rho_v=density(Fluid$, T=T_sat, x=1) 
sigma=SurfaceTension(Fluid$, T=T_sat) 
mu_l=Viscosity(Fluid$,T=T_sat, x=0) 
Pr_l=Prandtl(Fluid$, T=T_sat, P=P+1) 
c_l=CP(Fluid$, T=T_sat, x=0)*Convert(kJ/kg-C, J/kg-C) 
h_f=enthalpy(Fluid$, T=T_sat, x=0) 
h_g=enthalpy(Fluid$, T=T_sat, x=1) 
h_fg=(h_g-h_f)*Convert(kJ/kg, J/kg) 
C_sf=0.0060 "from Table 10-3 of the text" 
n=1 "from Table 10-3 of the text" 
 
T_vapor=1000-273 "[C], assumed vapor temperature in the film boiling region" 
rho_v_f=density(Fluid$, T=T_vapor, P=P) "f stands for film" 
c_v_f=CP(Fluid$, T=T_vapor, P=P)*Convert(kJ/kg-C, J/kg-C) 
k_v_f=Conductivity(Fluid$, T=T_vapor, P=P) 
mu_v_f=Viscosity(Fluid$,T=T_vapor, P=P) 
 
g=9.81 [m/s^2] “gravitational acceleraton" 
sigma_rad=5.67E-8 [W/m^2-K^4] “Stefan-Boltzmann constant" 
 
"ANALYSIS" 
"(a)" 
"C_cr is to be determined from Table 10-4 of the text" 
C_cr=0.12*L_star^(-0.25) 
L_star=D/2*((g*(rho_l-rho_v))/sigma)^0.5 
q_dot_max=C_cr*h_fg*(sigma*g*rho_v^2*(rho_l-rho_v))^0.25 
q_dot_nucleate=q_dot_max 
q_dot_nucleate=mu_l*h_fg*(((g*(rho_l-rho_v))/sigma)^0.5)*((c_l*(T_s_crit-T_sat))/(C_sf*h_fg*Pr_l^n))^3 
"(b)" 
q_dot_total=q_dot_film+3/4*q_dot_rad "Heat transfer in the film boiling region" 
q_dot_total=q_dot_nucleate 
q_dot_film=0.62*((g*k_v_f^3*rho_v_f*(rho_l-rho_v_f)*(h_fg+0.4*c_v_f*(T_s_film-
T_sat)))/(mu_v_f*D*(T_s_film-T_sat)))^0.25*(T_s_film-T_sat) 
q_dot_rad=epsilon*sigma_rad*((T_s_film+273)^4-(T_sat+273)^4) 
DELTAT=T_s_film-T_s_crit 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-47 
 
 
P 
[kPa] 
maxq 
[kW/m
2
] 
T 
[C] 
70 
71.65 
73.29 
74.94 
76.59 
78.24 
79.88 
81.53 
83.18 
84.83 
86.47 
88.12 
89.77 
91.42 
93.06 
94.71 
96.36 
98.01 
99.65 
101.3 
994682 
1004096 
1013373 
1022516 
1031531 
1040422 
1049194 
1057848 
1066391 
1074825 
1083153 
1091379 
1099505 
1107535 
1115471 
1123316 
1131072 
1138742 
1146328 
1153832 
1865 
1871 
1876 
1881 
1886 
1891 
1896 
1901 
1905 
1910 
1914 
1919 
1923 
1927 
1931 
1935 
1939 
1943 
1947 
1951 
 
 
 maxq 
[kW/m
2
] 
T 
[C] 
0.1 
0.15 
0.2 
0.25 
0.3 
0.35 
0.4 
0.45 
0.5 
0.55 
0.6 
0.65 
0.7 
0.75 
0.8 
0.85 
0.9 
0.95 
1 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
1153832 
2800 
2574 
2418 
2300 
2205 
2127 
2060 
2002 
1951 
1905 
1864 
1827 
1793 
1762 
1733 
1706 
1681 
1657 
1635 
 
 
70 75 80 85 90 95 100 105
980000
1000000
1.020x106
1.040x106
1.060x106
1.080x106
1.100x106
1.120x106
1.140x106
1.160x106
1860
1880
1900
1920
1940
1960
P [kPa]
q
m
a
x
 
[W
/m
2
]
D
T
 
[C
]
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1600
1800
2000
2200
2400
2600
2800
5.0x105
7.5x105
1.0x106
1.3x106
1.5x106
e 
D
T
 
[C
]q
m
a
x
 
[W
/m
2
]
 
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preparation. If you are a student using this Manual, you are using it without permission. 
10-48 
 
Condensation Heat Transfer 
 
10-52C Condensation is a vapor-to-liquid phase change process. It occurs when the temperature of a vapor is reduced below 
its saturation temperature Tsat. This is usually done by bringing the vapor into contact with a solid surface whose temperature 
Ts is below the saturation temperature Tsat of the vapor. 
 
 
10-53C In film condensation, the condensate wets the surface and forms a liquid film on the surface which slides down 
under the influence of gravity. The thickness of the liquid film increases in the flow direction as more vapor condenses on the 
film. This is how condensation normally occurs in practice. In dropwise condensation, the condensed vapor forms droplets 
on the surface instead of a continuous film, and the surface is covered by countless droplets of varying diameters. Dropwise 
condensation is a much more effective mechanism of heat transfer. 
 
 
10-54C The presence of noncondensable gases in the vapor has a detrimental effect on condensation heat transfer. Even small 
amounts of a noncondensable gas in the vapor cause significant drops in heat transfer coefficient during condensation. 
 
 
10-55C The modified latent heat of vaporization *fgh is the amount of heat released as a unit mass of vapor condenses at a 
specified temperature, plus the amount of heat released as the condensate is cooled further to some average temperature 
between Tsat and sT . It is defined as )(68.0 sat
*
splfgfg TTchh  where cpl is the specific heat of the liquid at the average 
film temperature. 
 
 
10-56C During film condensation on a vertical plate, heat flux at the top will be higher since the thickness of the film at the 
top, and thus its thermal resistance, is lower. 
 
 
10-57C The condensation heat transfer coefficient for the tubes will be the highest for the case of horizontal side by side 
(case b) since (1) for long tubes, the horizontal position gives the highest heat transfer coefficients, and (2) for tubes in a 
vertical tier, the average thickness of the liquid film at the lower tubes is much larger as a result of condensate falling on top 
of them from the tubes directly above, and thus the average heat transfer coefficient at the lower tubes in such arrangements is 
smaller. 
 
 
10-58C In condensate flow, the wetted perimeter is defined as the length of the surface-condensate interface at a cross-
section of condensate flow. It differs from the ordinary perimeter in that the latter refers to the entire circumference of the 
condensate at some cross-section. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-49 
 
10-59 The hydraulic diameter Dh for all 4 cases are expressed in terms of the boundary layer thickness  as follows: 
(a) Vertical plate: 

4
44

w
w
p
A
D ch 
(b) Tilted plate: 

4
44

w
w
p
A
D ch 
(c)Vertical cylinder: 


4
44

D
D
p
A
D ch 
(d) Horizontal cylinder: 

4
2
)2(44

L
L
p
A
D ch 
(e) Sphere: 


4
44

D
D
p
A
D ch 
Therefore, the Reynolds number for all 5 cases can be expressed as 
 
l
ll
l
llh
l
llc
l
VVD
p
VA
p
m







444
Re 

 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-50 
 
10-60 The local heat transfer coefficients at the middle and at the bottom of a vertical plate undergoing film condensation are 
to be determined. 
Assumptions 1 Steady operating condition exists. 2 The plate surface has uniform temperature. 3 The flow is laminar. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρv = 0.5978 
kg/m
3
 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 90°C are, from Table A-9, 
 ρl = 965.3 kg/m
3
 cpl = 4206 J/kg·K 
 μl = 0.315 × 10
−3
 kg/m·s kl = 0.675 W/m·K 
 νl = μl / ρl = 0.326 × 10
−6
 m
2
/s 
Analysis The modified latent heat of vaporization is 
 
J/kg 102314
)80100)(4206(68.0102257
)(68.0
3
3
sat


 splfgfg TTchh
 
The local heat transfer coefficient can be calculated using 
 
KW/m
1
4008
)80100)(10315.0(4
)675.0)(102314)(5978.03.965)(3.965)(81.9(
)(4
)(
2
4/1
4/1
3
33
4/1
sat
3































 
x
x
xTT
khg
h
sl
lfgvll
x


 
The local heat transfer coefficient at the middle of the plate (x = 0.1 m) is 
 K W/m7130 2 











 KW/m
1.0
1
4008KW/m
1
4008 2
4/1
2
4/1
x
hx 
The local heat transfer coefficient at the bottom of the plate (x = 0.2 m) is 
 K W/m5990 2 











 KW/m
2.0
1
4008KW/m
1
4008 2
4/1
2
4/1
x
hx 
Discussion The assumption that the flow is laminar is verified to be appropriate: 
 1800176
5990
675.0
)10315.0(3
)3.965)(81.9(4
3
4
Re
3
23
23
2
2

















Lx
l
l
l
h
kg


 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-51 
 
10-61 The necessary surface temperature of the plate used to condensate saturated water vapor at a desired condensation rate 
is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The plate surface has uniform temperature. 3 The film temperature is 
90°C. 
Properties Based on the problem statement, we take film temperature to be Tf = (Tsat + Ts)/2 = 90°C and the surface 
temperature to be Ts = 80°C. The properties of liquid water at the film temperature of Tf = 90°C are, from Table A-9, 
 ρl = 965.3 kg/m
3
 cpl = 4206 J/kg·K 
 μl = 0.315 × 10
−3
 kg/m·s kl = 0.675 W/m·K 
 νl = μl / ρl = 0.326 × 10
−6
 m
2
/s 
The properties of water at the saturation temperature of Tsat = 100°C are hfg = 2257 kJ/kg (Table A-2) and ρv = 0.5978 kg/m
3
 
(Table A-9). 
Analysis The calculation of the modified latent heat of vaporization requires the knowledge of the Ts. Hence, we assume Ts = 
80°C, and iterate the solution, if necessary, until good agreement with the calculated value of Ts is achieved: 
 
J/kg 102314
)80100)(4206(68.0102257
)(68.0
3
3
sat


 splfgfg TTchh
 
The Reynolds number is 
 3.406
)skg/m 10315.0)(m 5.0(
)kg/s 016.0(44
Re
3




lp
m


 
which is between 30 and 1800, and thus the flow is wavy-laminar. The heat transfer coefficient is 
 
K W/m7558
)/sm 10326.0(
m/s 81.9
2.5)3.406(08.1
)K W/m675.0)(3.406(
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1 wavyvert,























l
l gkhh

 
Hence, the surface temperature can be calculated using 
 
 fgss hmTThA )( sat → 
s
fg
s
hA
hm
TT



sat 
 C80.4



22
3
)m 5.0)(K W/m7558(
)J/kg 102314)(kg/s 016.0(
C100sT 
Discussion The assumed Ts = 80°C and Tf = 90°C are good, thus the solution does not require iteration. 
 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-52 
 
10-62 Saturated ammoniaat a saturation temperature of Tsat = 30C condenses on vertical plates which are maintained at 10
C. The average heat transfer coefficient and the rate of condensation of ammonia are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the 
entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of ammonia at the saturation temperature of 30C are hfg = 114410
3
 J/kg and v = 9.055 kg/m
3
. 
The properties of liquid ammonia at the film temperature of  2/)( sat sf TTT (30 + 10)/2 = 20C are (Table A-11), 
 
C W/m4927.0
CJ/kg 4745
/sm10489.2/
skg/m10519.1
kg/m 2.610
27
4
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 101209=
C10)C(30J/kg 47450.68+J/kg 101144
)(68.0
3
3
sat
*


 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
0.307
)s/m 10489.2(
m/s 8.9
)J/kg 101209)(skg/m 10519.1(
C)1030(C) W/m4927.0(m) 1.0(70.3
81.4
)(70.3
81.4ReRe
82.0
3/1
227
2
34
820.0
3/1
2*
sat
wavyvertical,






































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined to be 
 
C W/m7032 2 






















3/1
227
2
22.1
3/1
222.1wavyvertical,
)/sm 10489.2(
m/s 8.9
2.5)307(08.1
C) W/m4927.0(307
2.5Re08.1
Re
l
l gkhh

 
The total heat transfer surface area of the plates is 
 2m 75.0m) m)(0.10 25.0(30  LWAs 
Then the rate of heat transfer during this condensation process becomes 
 W480,105C)1030)(m 75.0)(C W/m7032()( 22sat  ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 0.0872


J/kg 101209
J/s 480,105
3*oncondensati
fgh
Q
m

 
 
m 
10 cm 
10C 
Ammonia 
30C 
 
25 cm 
 
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preparation. If you are a student using this Manual, you are using it without permission. 
10-53 
 
10-63 Saturated steam at atmospheric pressure thus at a saturation temperature of Tsat = 100C condenses on a vertical plate 
which is maintained at 90C by circulating cooling water through the other side. The rate of heat transfer to the plate and the 
rate of condensation of steam are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the 
entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.60 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (100 + 90)/2 = 95C are (Table A-9), 
 
C W/m677.0
CJ/kg 4212
/sm10309.0/
skg/m10297.0
kg/m 5.961
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 102,286=C90)C(100J/kg 42120.68+J/kg 102257
)(68.0
33
sat
*

 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
1113
)s/m 10309.0(
m/s 81.9
)J/kg 102286)(skg/m 10297.0(
C)90100(C) W/m677.0(m) 3(70.3
81.4
)(70.3
81.4ReRe
820.0
3/1
226
2
33
820.0
3/1
2*
sat
wavyvertical,






































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined to be 
 
C W/m6279
)/sm 10309.0(
m/s 81.9
2.5)1113(08.1
C) W/m677.0(1113
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1wavyvertical,























l
l gkhh

 
The heat transfer surface area of the plate is 
 2m 24m) m)(8 3(  LWAs 
Then the rate of heat transfer during this condensation process becomes 
 kW 1507 W1,506,960C)90100)(m 24)(C W/m6279()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 0.659


J/kg 102286
J/s 960,506,1
3*oncondensati
fgh
Q
m

 
m 
3 m 
90C 
1 atm 
Steam 
8 m 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-54 
 
10-64 Saturated steam at a saturation temperature of Tsat = 100C condenses on a plate which is tilted 60 from the vertical 
and maintained at 90C by circulating cooling water through the other side. The rate of heat transfer to the plate and the rate 
of condensation of the steam are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the 
entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.60 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (100 + 90)/2 = 95C are (Table A-9), 
 
C W/m677.0
CJ/kg 4212
/sm10309.0/
skg/m10297.0
kg/m 5.961
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 102,286=
C90)C(100J/kg 42120.68+J/kg 102257
)(68.0
3
3
sat
*


 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from the vertical plate 
relation by replacing g by g cos where  = 60 to be 
 
8.920
)s/m 10309.0(
60cos)m/s 81.9(
)J/kg 102286)(skg/m 10297.0(
C)90100(C) W/m677.0(m) 3(70.3
81.4
60cos)(70.3
81.4ReRe
82.0
3/1
226
2
33
820.0
3/1
2*
sat
wavytilted,






































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined from 
 
C W/m5198
)/sm 10309.0(
60cos)m/s 81.9(
2.5)8.920(08.1
C) W/m677.0(8.920
cos
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1wavytilted,























l
l gkhh


 
The heat transfer surface area of the plate is 
2m 24m) m)(8 3(  LWAs . 
Then the rate of heat transfer during this condensation process becomes 
 kW 1248 W520,247,1C)90100)(m 24)(C W/m5198()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 0.546


J/kg 102286
J/s 520,247,1
3*oncondensati
fgh
Q
m

 
Discussion Using the heat transfer coefficient determined in the previous problem for the vertical plate, we could also 
determine the heat transfer coefficient from 4/1vertinclined )(coshh  . It would give 5280 W/m
2
C, which is 1.6% different 
than the value determined above. 
m
 
3 m 
90C 
1 atm 
Steam 
8 m 
60 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-55 
 
10-65 Saturated steam at a saturation temperature of Tsat = 100C condenses on a plate which is tilted 40from the vertical 
and maintained at 80C by circulating cooling water through the other side. The rate of heat transfer to the plate and the rate 
of condensation of the steam are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the 
entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.60 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (100 + 80)/2 = 90C are (Table A-9), 
 
C W/m675.0
CJ/kg 4206
/sm10326.0/
skg/m10315.0
kg/m 3.965
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 102,314=C0)8C(100J/kg 42060.68+J/kg 102257
)(68.0
33
sat
*

 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from the vertical plate relation by replacing g by cosg 
where  = 40 to be 
 
1197
)s/m 10326.0(
40cos)m/s 81.9(
)J/kg 102314)(skg/m 10315.0(
C)80100(C) W/m675.0(m) 2(70.3
81.4
cos)(70.3
81.4ReRe
82.0
3/1
226
2
33
820.0
3/1
2*
sat
wavytilted,






































lfgl
sl g
h
TTLk



 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined from 
 
C W/m5440
2 






















3/1
226
2
22.1
3/1
222.1wavytilted,
)/sm 10326.0(
40cos)m/s 81.9(
2.5)1197(08.1
C) W/m675.0(1197
cos
2.5Re08.1
Re
l
l gkhh


 
The heat transfer surface area of the plate is: 
2m 4m) m)(2 2(  LwA . 
Then the rate of heat transfer during this condensation process becomes 
 W435,200C)80100)(m 4)(C W/m5440()( 22sat  sTThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 0.188


J/kg 102314
J/s 200,435
3*oncondensati
fgh
Q
m

 
Discussion We could also determine the heat transfer coefficient from 
4/1
vertinclined )(coshh  . 
2 m Inclined 
plate 
80C 
Condensate 
40 
Vapor 
100C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-56 
 
10-66 Prob. 10-65 is reconsidered. The effects of plate temperature and the angle of the plate from the vertical on the 
average heat transfer coefficient and the rate at which the condensate drips off are to be investigated. 
 Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
T_sat=100 [C] 
L=2 [m] 
theta=40 [degrees] 
T_s=80 [C] 
 
"PROPERTIES" 
Fluid$='steam_IAPWS' 
T_f=1/2*(T_sat+T_s) 
P_sat=pressure(Fluid$, T=T_sat, x=1) 
rho_l=density(Fluid$, T=T_f, x=0) 
mu_l=Viscosity(Fluid$,T=T_f, x=0) 
nu_l=mu_l/rho_l 
c_l=CP(Fluid$, T=T_f, x=0)*Convert(kJ/kg-C, J/kg-C) 
k_l=Conductivity(Fluid$, T=T_f, P=P_sat+1) 
h_f=enthalpy(Fluid$, T=T_sat, x=0) 
h_g=enthalpy(Fluid$, T=T_sat, x=1) 
h_fg=(h_g-h_f)*Convert(kJ/kg, J/kg) 
g=9.8 [m/s^2] 
 
"ANALYSIS" 
"(a)" 
h_fg_star=h_fg+0.68*c_l*(T_sat-T_s) 
Re=(4.81+(3.7*L*k_l*(T_sat-T_s))/(mu_l*h_fg_star)*((g*Cos(theta))/nu_l^2)^(1/3))^0.820 
h=(Re*k_l)/(1.08*Re^1.22-5.2)*((g*Cos(theta))/nu_l^2)^(1/3) 
Q_dot=h*A*(T_sat-T_s) 
A=L^2 
"(b)" 
m_dot_cond=Q_dot/h_fg_star 
 
 
Ts 
[C] 
h 
[W/m
2
.C] 
condm 
[kg/s] 
40 
42.5 
45 
47.5 
50 
52.5 
55 
57.5 
60 
62.5 
65 
67.5 
70 
72.5 
75 
77.5 
80 
82.5 
85 
87.5 
90 
4073 
4132 
4191 
4253 
4317 
4384 
4453 
4526 
4602 
4682 
4767 
4857 
4954 
5059 
5174 
5300 
5441 
5601 
5787 
6009 
6286 
0.4027 
0.3926 
0.3821 
0.3712 
0.3599 
0.3482 
0.3361 
0.3236 
0.3106 
0.2971 
0.2832 
0.2688 
0.2538 
0.2383 
0.2222 
0.2055 
0.1881 
0.17 
0.151 
0.1311 
0.11 
 
 
40 50 60 70 80 90
4000
4500
5000
5500
6000
6500
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Ts [C]
h
 [
W
/m
2
-C
]
m
c
o
n
d
 
[k
g
/s
]
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-57 
 
 
 
[degrees] 
h 
[W/m
2
.C] 
condm 
[kg/s] 
0 
3 
6 
9 
12 
15 
18 
21 
24 
27 
30 
33 
36 
39 
42 
45 
48 
51 
54 
57 
60 
5851 
5849 
5842 
5831 
5816 
5796 
5771 
5742 
5708 
5670 
5626 
5577 
5522 
5462 
5396 
5323 
5243 
5156 
5061 
4957 
4842 
0.2023 
0.2022 
0.202 
0.2016 
0.2011 
0.2004 
0.1996 
0.1986 
0.1974 
0.196 
0.1945 
0.1928 
0.1909 
0.1889 
0.1866 
0.1841 
0.1813 
0.1783 
0.175 
0.1714 
0.1674 
 
 
 
0 10 20 30 40 50 60
4800
5000
5200
5400
5600
5800
6000
0.165
0.17
0.175
0.18
0.185
0.19
0.195
0.2
0.205
 [degrees]
h
 [
W
/m
2
-C
]
m
c
o
n
d
 
[k
g
/s
]
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-58 
 
10-67 Saturated steam condenses outside of vertical tube. The rate of heat transfer to the coolant, the rate of condensation and 
the thickness of the condensate layer at the bottom are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 3 The tube can be treated as a vertical plate. 4 
The condensate flow is wavy-laminar over the entire tube (this assumption will be verified). 5 Nusselt’s analysis can be used 
to determine the thickness of the condensate film layer. 6 The density of vapor is much smaller than the density of liquid, 
lv   . 
Properties The properties of water at the saturation temperature of 30C are hfg = 243110
3
 J/kg and v = 0.03 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (30 + 20)/2 = 25C are (Table A-9), 
 
C W/m607.0
CJ/kg 4180
/sm10894.0/
skg/m10891.0
kg/m 0.997
26
3
3







l
pl
lll
l
l
k
c



 
Analysis (a)The modified latent heat of vaporization is 
 
J/kg 102459=C0)2C(30J/kg 41800.68+J/kg 102431
)(68.0
33
sat
*

 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
3.157
)s/m 10894.0(
m/s 8.9
)J/kg 102459)(skg/m 10891.0(
C)2030(C) W/m607.0(m) 2(70.3
81.4
)(70.3
81.4ReRe
82.0
3/1
226
2
33
820.0
3/1
2*
sat
wavyvertical,






































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined to be 
 
C W/m4302
)/sm 10894.0(
m/s 8.9
2.5)3.157(08.1
C) W/m607.0(3.157
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1wavyvertical,























l
l gkhh

 
The heat transfer surface area of the tube is 2m 2513.0m) m)(2 04.0(  DLAs . Then the rate of heat transfer during 
this condensation process becomes 
 W10,811 C)2030)(m 2513.0)(C W/m4302()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 104.40
3-


J/kg 102459
J/s 811,10
3*oncondensati
fgh
Q
m

 
(c) Combining equations llL hk / and Lhh )3/4( , the thickness of the liquid film at the bottom of the tube is 
determined to be 
 mm 0.2


 m 100.188=
 C) W/m4302(3
C) W/m607.0(4
3
4 3-
2h
k lL 
Discussion The assumption of wavy laminar flow is verified since Reynolds number is between 30 and 1800. The assumption 
that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube is verified since the 
thickness of the liquid film is 0.2 mm, which is much smaller than the diameter of the tube (4 cm). 
Steam 
30C 
Condensate L = 2 m 
D = 4 cm 
20C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-59 
 
10-68 The rate of condensation and the heat transfer rate for a vertical pipe, with specified surface temperature, are to be 
determined. 
Assumptions 1 Steady operating condition exists. 2 The surface has uniform temperature. 3 The pipe can be treated as a 
vertical plate. 4 The condensate flow is wavy-laminar over the entire tube (this assumption will be verified). 5 Nusselt’s 
analysis can be used to determine the thickness of the condensate film layer. 6 The density of vapor is much smaller than the 
density of liquid, ρv << ρl. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρv = 0.5978 
kg/m
3
 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 90°C are, from Table A-9, 
 ρl = 965.3 kg/m
3
 cpl = 4206 J/kg·K 
 μl = 0.315 × 10
−3
 kg/m·s kl = 0.675 W/m·K 
 νl = μl / ρl = 0.326 × 10
−6
 m
2
/s 
Analysis The modified latent heat of vaporization is 
 
J/kg 102314
)80100)(4206(68.0102257
)(68.0
3
3
sat


 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
7.729
)10326.0(
81.9
)102314)(10315.0(
)80100)(675.0)(1(70.3
81.4
)(70.3
81.4Re
820.0
3/1
2633
820.0
3/1
2
sat
 wavyvert,







































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined to be 
 
K W/m6633
)/sm 10326.0(
m/s 81.9
2.5)7.729(08.1
)K W/m675.0)(7.729(
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1 wavyvert,























l
l gkhh
 
Then the rate of heat transfer during this condensation process becomes 
 
 W104.168
4


K )80100)(K W/m6633)(m 1)(m 1.0(
)(
2
sat

 sTTDLhQ

 
The rate of condensation of steam is determined from 
 kg/s 0.018



 J/kg 102314
W101684
3
4
oncondensati
 .
h
Q
m
fg

 
Discussion Combining equations LlL hk / and Lhh )3/4( , the thickness of the liquid film at the bottom of the tube is 
determined to be 
 mm 100mm 136.0
)K W/m6633(3
)K W/m675.0(4
3
4
2




h
kl
L 
Since δL << D, the pipe can be treated as a vertical plate. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-60 
 
10-69 Saturated steam at a saturation temperature of Tsat = 55C condenses on the outer surface of a vertical tube which is 
maintained at 45C. The required tube length to condense steam at a rate of 10 kg/h is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 3 The vertical tube can be treated as a vertical 
plate. 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of water at the saturation temperature of 55C are hfg = 237110
3
 J/kg and v = 0.1045 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (55 + 45)/2 = 50C are (Table A-9), 
 
C W/m644.0
CJ/kg 4181
/sm10554.0/
skg/m10547.0
kg/m 1.988
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 102399=C)45C(55J/kg 41810.68+J/kg 102371
)(68.0
33
sat
*

 splfgfg TTchh
 
The Reynolds number is determined from its definition to be 
5.215
)skg/m 10m)(0.547 (0.03
kg/s) 3600/10(44
Re
3



lp
m
 
which is between 30 and 1800. Therefore the condensate flow is wavy laminar, and the condensation heat transfer coefficient 
is determined from 
 
C W/m5644
)/sm 10554.0(
m/s 8.9
2.5)5.215(08.1
C) W/m644.0(5.215
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1wavyvertical,























l
l gkhh

 
The rate of heat transfer during this condensation process is 
 W664,6)J/kg 10kg/s)(2399 3600/10( 3*  fghmQ 
 
Heat transfer can also be expressed as 
 ))(()( satsat sss TTDLhTThAQ  
 
Then the required length of the tube becomes 
 m 1.21




C)4555)(m 03.0()C W/m5844(
 W6664
))((
 
2
sat  sTTDh
Q
L

 
Discussion Combining equations llL hk / and Lhh )3/4( , the thickness of the liquid film at the bottom of the tube is 
determined to be 
 mm 0.15m 100.147=
 C) W/m5844(3
C) W/m644.0(4
3
4 3-
2




h
kl
L 
The assumption that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube is verified 
since the thickness of the liquid film is 0.15 mm, which is much smaller than the diameter of the tube (3 cm). Also, the 
assumption of wavy laminar flow is verified since Reynolds number is between 30 and 1800. 
Steam 
55C 
Condensate Ltube = ? 
D = 3 cm 
45C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-61 
 
10-70 Saturated steam at a saturation temperature of Tsat = 55C condenses on the outer surface of a horizontal tube which is 
maintained at 45C. The required tube length to condense steam at a rate of 10 kg/h is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 
Properties The properties of water at the saturation temperature of 55C are hfg = 237110
3
 J/kg and v = 0.1045 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (55 + 45)/2 = 50C are (Table A-9), 
 
C W/m644.0
CJ/kg 4181
/sm10554.0/
skg/m10547.0
kg/m 1.988
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 102399=C)45C(55J/kg 41810.68+J/kg 102371
)(68.0
33
sat
*

 splfgfg TTchh
 
Noting that the tube is horizontal, the condensation heat transfer coefficient is determined from 
 
C. W/m135,10
m) C(0.03)4555(s)kg/m 10547.0(
)C W/m644.0)(J/kg 102399)(kg/m 10.01.988)(kg/m 1.988)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
The rate of heat transfer during this condensation process is 
 W664,6)J/kg 10kg/s)(2399 3600/10( 3*  fghmQ 
 
Heat transfer can also be expressed as 
 ))(()( satsat sss TTDLhTThAQ  
 
Then the required length of the tube becomes 
 m 0.70




C)4555)(m 03.0()C W/m135,10(
 W6664
))((
 
2
sat  sTTDh
Q
L

 
Steam 
55C 
Condensate 
45C 
Cooling 
water 
Ltube = ? 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-62 
 
10-71 Saturated vapor condenses on the outer surface of a 1.5-m-long vertical tube at 60°C that is maintained with a surface 
temperatureof 40°C. The rate of heat transfer to the tube and the required tube diameter to condense 12 kg/h of steam are to 
be determined. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tube surface. 3 The vertical tube can be treated as a vertical 
plate (this assumption will be verified). 4 The condensate flow is wavy-laminar over the entire plate (this assumption will be 
verified). 5 The density of vapor is much smaller than the density of liquid, ρv << ρl. 
Properties The properties of water at the saturation temperature of 60°C are hfg = 2359  10
3
 J/kg and ρv = 0.1304 kg/m
3
 
(Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (60 + 40)/2 = 50°C are (Table A-9) 
 
K W/m644.0
KJ/kg 4181
/sm10554.0/
skg/m10547.0
kg/m 1.988
26
3
3







l
pl
lll
l
l
k
c



 0 
Analysis The modified latent heat of vaporization is 
 J/kg 104159.2K )4060)(KJ/kg 4181(68.0J/kg 102359)(68.0
63
sat
*  splfgfg TTchh 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
73.454
)/sm 10554.0(
m/s 81.9
)J/kg 104159.2)(skg/m 10547.0(
K )4060)(K W/m644.0)(m 5.1)(70.3(
81.4
)(70.3
81.4ReRe
820.0
3/1
226
2
63
820.0
3/1
2*
sat
wavyvertical,






































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined to be 
 
K W/m5.4937
)/sm 10554.0(
m/s 81.9
2.5)73.454(08.1
)K W/m644.0)(73.454(
2.5Re08.1
Re
2
3/1
226
2
22.1
3/1
222.1wavyvertical,























l
l gkhh

 
The rate of heat transfer to the tube during this condensation process is 
 W8053 )J/kg 104159.2)(kg/s 3600/12(
6*
fghmQ 
 
Heat transfer can also be expressed as 
 ))(()( satsat sss TTDLhTThAQ  
 
Then the required diameter of the tube becomes 
 m 0.0173




K )4060)(m 5.1()K W/m5.4937(
 W8053
)()(
 
2
sat  sTTLh
Q
D

 
To verify the assumption that vertical tube can be treated as a vertical plate (D >> ), calculate  from 
m 0.0173=D m 10× 1.79= K) W/m(4937.5 ) W/m·K)/(34(0.664=)/3h(4k = -42l  
Thus our assumption of D >>  is verified. 
Discussion With diameter known, the Reynolds number can also be verified to be wavy- laminar flow 
 448
)skg/m10547.0)(m 0173.0(
)kg/s 3600/12(44
Re
3



 lD
m
 
Note that the Re value obtained based on equation 10-27 will not exactly match the above calculated value of Re since 
equation 10-27 is based on experimental data and several approximations. However, the two values are very close (within 
2%). 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-63 
 
10-72 Repeat Prob.10-71 for a horizontal tube. Saturated vapor condenses on the outer surface of a 1.5-m-long horizontal 
tube at 60°C that is maintained with a surface temperature of 40°C. The rate of heat transfer to the tube and the required tube 
diameter to condense 12 kg/h of steam are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tube surface. 
Properties The properties of water at the saturation temperature of 60°C are hfg = 2359  10
3
 J/kg and ρv = 0.1304 kg/m
3
 
(Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (60 + 40)/2 = 50°C are (Table A-9) 
 
K W/m644.0
KJ/kg 4181
/sm10554.0/
skg/m10547.0
kg/m 1.988
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 104159.2K )4060)(KJ/kg 4181(68.0J/kg 102359
)(68.0
63
sat
*

 splfgfg TTchh
 
The rate of heat transfer to the tube during this condensation process is 
 W8053 )J/kg 104159.2)(kg/s 3600/12(
6*
fghmQ 
 
Heat transfer can also be expressed as 
 ))(()( satsat sss TTLDhTThAQ  
  
))(( sat sTTLD
Q
h




 
Noting that the tube is horizontal, the condensation heat transfer coefficient is determined from 
 
))(()(
)(
729.0
sat
4/1
sat
3*
horiz
ssl
lfgvll
TTLD
Q
DTT
khg
hh














 
 
Solving for the required tube diameter, 
 
m 0.00694

















 

























3/4
4/1
3
36332
3/4
4/1
sat
3*
sat
K)4060)(skg/m10547.0(
)K W/m644.0)(J/kg 104159.2(kg/m)1304.01.988)(kg/m 1.988)(m/s 81.9(
J/s 8053
K)4060)(m 5.1(
729.0
)(
)())((
729.0



sl
lfgvlls
TT
khg
Q
TTL
D

 
Discussion When placed vertically, the required tube diameter is about 2.5 times larger than that of a horizontal tube. Due to 
the higher heat transfer coefficient for a horizontal tube, in comparison with a vertical tube, the horizontal tube requires a 
smaller diameter for the same length to achieve the same rate of condensation. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-64 
 
10-73 Saturated ammonia vapor at a saturation temperature of Tsat = 10C condenses on the outer surface of a horizontal tube 
which is maintained at -10C. The rate of heat transfer from the ammonia and the rate of condensation of ammonia are to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 
Properties The properties of ammonia at the saturation temperature of 10C are hfg = 122610
3
 J/kg and v = 4.870 kg/m
3
 
(Table A-11). The properties of liquid ammonia at the film temperature of  2/)( sat sf TTT (10 + (-10))/2 = 0C are 
(Table A-11), 
 
C W/m5390.0
CJ/kg 4617
/sm102969.0
skg/m10896.1
kg/m 6.638
26
4
3







l
pl
l
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 101288=C)]10(C[10J/kg 46170.68+J/kg 101226
)(68.0
33
sat
*

 splfgfg TTchh
 
 Noting that the tube is horizontal, the condensation heat transfer coefficient is determined from 
 
C. W/m7390
m) C(0.02)]10(10[s)kg/m 10896.1(
)C W/m5390.0)(J/kg 101288)(kg/m 870.4)(638.6kg/m 6.638)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
4
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
 The heat transfer surface area of the tube is 
 2m 1.885= m) m)(15 04.0(  DLAs 
Then the rate of heat transfer during this condensation process becomes 
 W278,600 C)]10(10)[m 885.1)(C. W/m7390()( 22sat ss TThAQ
 
(b) The rate of condensation of ammonia is determined from 
 kg/s 0.216


J/kg 101288
J/s 600,278
3*oncondensati
fgh
Q
m

 
Ammonia 
10C 
Condensate 
-10C 
Dtube = 4 cm 
Ltube = 15 m 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-65 
 
10-74 A spherical tank containing cold fluid is causing condensation of moist air on the outer surface. The rate of 
moisture condensation is to be determined whether or not risk of electrical hazard exists. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tank surface. 3 Film condensation occurs on the tank surface. 
Properties The properties of water at the saturation temperature of 25°C are hfg = 2442  10
3
 J/kg and ρv = 0.0231 kg/m
3
 
(Table A-9). The properties of liquid water at the filmtemperature of Tf = (Tsat + Ts)/2 = (25 + 5)/2 = 15°C are (Table A-9) 
 
K W/m589.0
KJ/kg 4185
/sm10139.1/
skg/m10138.1
kg/m 1.999
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 104989.2K )525)(KJ/kg 4185(68.0J/kg 102442
)(68.0
63
sat
*

 splfgfg TTchh
 
The film condensation heat transfer coefficient for a sphere is determined from 
 
K W/m1.2384
)m 3(K)525)(skg/m10138.1(
)K W/m589.0)(J/kg 104989.2(kg/m)0231.01.999)(kg/m 1.999)(m/s 81.9(
815.0
)(
)(
815.0
2
4/1
3
36332
4/1
sat
3*
horiz
























DTT
khg
hh
sl
lfgvll


 
Thus, the rate of film condensation is 
 
kg/s 5.0
J/kg 104989.2
K )525()m 3()K W/m1.2384(
))((
6
22
*
sat
2
*






kg/s 0.5395


fg
s
fg h
TTDh
h
Q
m


 
Discussion The rate of condensation from the tank surface is greater than the capability of the system in removing the 
condensate. Thus, there is a risk of excess condensate coming in contact with the high voltage device and cause electrical 
hazard. To prevent electrical hazard, the preventive system should be capable of removing more than 0.54 kg/s of condensate. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-66 
 
10-75 There is film condensation on the outer surfaces of N horizontal tubes arranged in a vertical tier. The value of N for 
which the average heat transfer coefficient for the entire tier be equal to half of the value for a single horizontal tube is to be 
determined. 
Assumptions Steady operating conditions exist. 
Analysis The relation between the heat transfer coefficients for the two cases is given to be 
 
4/1
 tube1 ,horizontal
 tubesN ,horizontal
N
h
h  
Therefore, 
 16 N
Nh
h
 
1
2
1
4/1
 tube1 ,horizontal
 tubesN ,horizontal
 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-67 
 
10-76 Saturated steam at a saturation temperature of Tsat = 50C condenses on the outer surfaces of a tube bank with 33 tubes 
in each column maintained at 20C. The average heat transfer coefficient and the rate of condensation of steam on the tubes 
are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. 
Properties The properties of water at the saturation temperature of 50C are hfg = 238310
3
 J/kg and v = 0.0831 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (50 + 20)/2 = 35C are (Table A-9), 
 
C W/m623.0
CJ/kg 4178
/sm10724.0/
skg/m10720.0
kg/m 0.994
26
3
3







l
pl
lll
l
l
k
c



 
Analysis (a) The modified latent heat of vaporization is 
 
J/kg 102468=
C0)2C(50J/kg 41780.68+J/kg 102383
)(68.0
3
3
sat
*


 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
C W/m8425
m) C(0.015)2050(s)kg/m 10720.0(
)C W/m623.0)(J/kg 102468)(kg/m 08.0994)(kg/m 994)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 33-tube high vertical tier becomes 
 C W/m3515 2  C) W/m8425(
33
11 2
4/1 tube1 horiz,4/1 tubesN horiz,
h
N
h 
The surface area for all 33 tubes per unit length is 
 2total m 1.555= m) m)(1 015.0(33  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
 W000,164C)2050)(m 555.1)(C W/m3515()( 22sat  ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 0.0664


J/kg 102468
J/s 000,164
3*oncondensati
fgh
Q
m

 
Steam 
50C 
20C 
Condensate 
flow 
33 tubes in 
a column 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-68 
 
10-77 Saturated steam at a pressure of 4.25 kPa and thus at a saturation temperature of Tsat = 30C (Table A-9) condenses on 
the outer surfaces of 100 horizontal tubes arranged in a 1010 square array maintained at 20C by circulating cooling water. 
The rate of heat transfer to the cooling water and the rate of condensation of steam on the tubes are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are 
isothermal. 
Properties The properties of water at the saturation temperature 
of 30C are hfg = 243110
3
 J/kg and v = 0.03 kg/m
3
. The 
properties of liquid water at the film temperature of 
 2/)( sat sf TTT (30 + 20)/2 = 25C are (Table A-9), 
 
C W/m607.0
CJ/kg 4180
/sm10894.0/
skg/m10891.0
kg/m 0.997
26
3
3







l
pl
lll
l
l
k
c



 
Analysis (a) The modified latent heat of vaporization is 
 
J/kg 102,459=C0)2C(30J/kg 41800.68+J/kg 102431
)(68.0
33
sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
C. W/m8674
m) C(0.03)2030(s)kg/m 10891.0(
)C W/m607.0)(J/kg 102459)(kg/m 03.0997)(kg/m 997)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 10-pipe high vertical tier becomes 
 C W/m4878C) W/m8674(
10
11 22
4/1 tube1 horiz,4/1 tubesN horiz,
 h
N
h 
The surface area for all 100 tubes is 
 2total m 75.40= m) m)(8 03.0(100  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
kW 3678 W,678,0003C)2030)(m 40.75)(C. W/m4878()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 kg/s 1.496


J/kg 102459
J/s 000,678,3
3*oncondensati
fgh
Q
m

 
n = 100 tubes 20C 
L = 8 m 
P = 4.25 kPa 
Cooling 
water 
Saturated 
steam 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-69 
 
10-78 Prob. 10-77 is reconsidered. The effect of the condenser pressure on the rate of heat transfer and the rate of 
condensation of the steam is to be investigated. 
 Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
P_sat=4.25 [kPa] 
n_tube=100 
N=10 
L=8 [m] 
D=0.03 [m] 
T_s=20 [C] 
 
"PROPERTIES" 
Fluid$='steam_IAPWS' 
T_sat=temperature(Fluid$, P=P_sat, x=1) 
T_f=1/2*(T_sat+T_s) 
h_f=enthalpy(Fluid$, T=T_sat, x=0) 
h_g=enthalpy(Fluid$, T=T_sat, x=1) 
h_fg=(h_g-h_f)*Convert(kJ/kg, J/kg) 
rho_v=density(Fluid$, T=T_sat, x=1) 
rho_l=density(Fluid$, T=T_f, x=0) 
mu_l=Viscosity(Fluid$,T=T_f, x=0) 
nu_l=mu_l/rho_l 
c_l=CP(Fluid$, T=T_f, x=0)*Convert(kJ/kg-C, J/kg-C) 
k_l=Conductivity(Fluid$, T=T_f, P=P_sat+1) 
g=9.8 [m/s^2] 
 
"ANALYSIS" 
h_fg_star=h_fg+0.68*c_l*(T_sat-T_s) 
h_1tube=0.729*((g*rho_l*(rho_l-rho_v)*h_fg_star*k_l^3)/(mu_l*(T_sat-T_s)*D))^0.25 
h=1/N^0.25*h_1tube 
Q_dot=h*A*(T_sat-T_s) 
A=n_tube*pi*D*L 
m_dot_cond=Q_dot/h_fg_star 
 
 
Psat 
[kPa] 
Q 
[W] 
condm 
[kg/s] 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
1834387 
3374608 
4495906 
5397473 
6158471 
6820321 
7408006 
7937849 
8421171 
8866242 
9279144 
9664649 
10026498 
0.7471 
1.373 
1.828 
2.193 
2.5012.769 
3.006 
3.22 
3.415 
3.594 
3.76 
3.916 
4.061 
 
 
 
3 5 7 9 11 13 15
4.0x10
6
6.0x10
6
8.0x10
6
1.0x10
7
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Psat [kPa]
Q
 
[W
]
m
c
o
n
d
 
[k
g
/s
]
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-70 
 
10-79E Saturated steam at a saturation temperature of Tsat = 95F condenses on the outer surfaces of horizontal pipes which 
are maintained at 65F by circulating cooling water. The rate of heat transfer to the cooling water and the rate of condensation 
per unit length of a single horizontal pipe are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The pipe is isothermal. 3 There is no interference between the pipes (no 
drip of the condensate from one tube to another). 
Properties The properties of water at the saturation temperature of 95F are hfg = 1040 Btu/lbm and v = 0.0025 lbm/ft
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (95 + 65)/2 = 80F are (Table A-9E), 
FftBtu/h 352.0
FBtu/lbm 999.0
/hft 03335.0/
hlbm/ft 075.2slbm/ft 10764.5
lbm/ft 22.62
2
4
3






l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
Btu/lbm 1060=
F)65F)(95Btu/lbm 999.0(0.68+Btu/lbm 1040
)(68.0 sat
*

 splfgfg TTchh
 
Noting that we have condensation on a horizontal tube, the heat transfer coefficient is determined from 
 
FftBtu/h 1420
ft) F(1/12)6595)(hlbm/ft 075.2](s) 3600h/ 1[(
)FftBtu/h 352.0)(Btu/lbm 1060)(lbm/ft 0025.022.62)(lbm/ft 22.62)(ft/s 2.32(
729.0
)(
)(
729.0
2
4/1
2
3332
4/1
sat
3*
horiz























DTT
khg
hh
sl
lfgvll


 
The heat transfer surface area of the tube per unit length is 
 2ft 2618.0ft) ft)(1 12/1(  DLAs 
Then the rate of heat transfer during this condensation process becomes 
 Btu/h 11,150 F)6595)(ft 2618.0)(FftBtu/h 1420()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 lbm/h 10.5
Btu/lbm 1060
Btu/h 150,11
*oncondensati
fgh
Q
m

 
Steam 
95F 
................... 
65F 
Condensate 
flow 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-71 
 
10-80E Saturated steam at a saturation temperature of Tsat = 95F condenses on the outer surfaces of 20 horizontal pipes 
which are maintained at 65F by circulating cooling water and arranged in a rectangular array of 4 pipes high and 5 pipes 
wide. The rate of heat transfer to the cooling water and the rate of condensation per unit length of the pipes are to be 
determined. 
Assumptions 1 Steady operating conditions exist. 2 The pipes are isothermal. 
Properties The properties of water at the saturation temperature of 95F are hfg = 1040 Btu/lbm and v = 0.0025 lbm/ft
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (95 + 65)/2 = 80F are (Table A-9E), 
 
FftBtu/h 352.0
FBtu/lbm 999.0
/hft 03335.0/
hlbm/ft 075.2slbm/ft 10764.5
lbm/ft 22.62
2
4
3






l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
Btu/lbm 1060=
F)65F)(95Btu/lbm 999.0(0.68+Btu/lbm 1040
)(68.0 sat
*

 splfgfg TTchh
 
Noting that we have condensation on a horizontal tube, the heat transfer coefficient is determined from 
 
FftBtu/h 1420
ft) F(1/12)6595)(hlbm/ft 075.2](s) 3600h/ 1[(
)FftBtu/h 352.0)(Btu/lbm 1060)(lbm/ft 0025.022.62)(lbm/ft 22.62)(ft/s 2.32(
729.0
)(
)(
729.0
2
4/1
2
3332
4/1
sat
3*
horiz























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 4-pipe high vertical tier becomes 
 FftBtu/h 1004F)ftBtu/h 1420(
4
11 22
4/1 tube1 horiz,4/1 tubesN horiz,
 h
N
h 
The surface area for all 32 pipes per unit length of the pipes is 
 2total ft 8.378= ft) ft)(1 12/1(32  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
 Btu/h 252,300 F)6595)(ft 378.8)(FBtu/h.ft 1004()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 lbm/h 238
Btu/lbm 1060
Btu/h 300,252
*oncondensati
fgh
Q
m

 
 
Steam 
95F 
................... 
65F 
Condensate 
flow 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-72 
 
10-81 Saturated steam at a saturation temperature of Tsat = 100C condenses on the outer surfaces of a tube bank maintained 
at 80C. The rate of condensation of steam on the tubes are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.5978 kg/m
3
 
(Table A-9). The properties of liquid water at the film temperature of  2/)( sat sf TTT (100 + 80)/2 = 90C are (Table A-
9), 
 
C W/m675.0
CJ/kg 4206
skg/m10315.0
kg/m 3.965
3
3





l
pl
l
l
k
c


 
Analysis (a) The modified latent heat of vaporization is 
J/kg 102314=
C80)C(100J/kg 42060.68+J/kg 102357
)(68.0
3
3
sat
*


 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
C W/m8736
m) C(0.05)80100(s)kg/m 10315.0(
)C W/m675.0)(J/kg 102314)(kg/m 5978.03.965)(kg/m 3.965)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 4-pipe high vertical tier becomes 
 C W/m6177C) W/m8736(
4
11 22
4/1 tube1 horiz,4/1 tubesN horiz,
 h
N
h 
The surface area for all 16 tubes is 
 2total m 5.027= m) m)(2 05.0(16  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
 W000,621C)80100)(m 027.5)(C W/m6177()( 22sat  ss TThAQ
 
The rate of condensation of steam is determined from 
 kg/h 966

 kg/s 0.2684
J/kg 102314
J/s 000,621
3*oncondensati
fgh
Q
m

 
Steam 
100C 
80C 
Condensate 
flow 
4  4 array 
of tubes 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-73 
 
10-82 Saturated water vapor at a pressure of 12.4 kPa condenses on a rectangular array of 100 horizontal tubes at 30°C. The 
condensation rate per unit length of is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The tube surfaces are isothermal. 
Properties The properties of water at the saturation temperature of 50°C corresponding to 12.4 kPa are hfg = 2383 kJ/kg and 
ρv = 0.0831 kg/m
3
 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 40°C are, from 
Table A-9, 
 ρl = 992.1 kg/m
3
 cpl = 4179 J/kg·K 
 μl = 0.653 × 10
−3
 kg/m·s kl = 0.631 W/m·K 
Analysis The modified latent heat of vaporization is 
 
J/kg 102440
)3050)(4179(68.0102383
)(68.0
3
3
sat


 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
KW/m250,11
)008.0)(3050)(10653.0(
)631.0)(102440)(0831.01.992)(1.992)(81.9(
729.0
)(
)(
729.0
2
4/1
3
33
4/1sat
3
 tube1 horiz,

























 
DTT
khg
h
sl
lfgvll


 
Then the average heat transfer coefficient for a 5-tube high vertical tier becomes 
 KW/m7523)KW/m250,11(
5
11 22
4/1 tube1 horiz,4/1 tubes horiz,
 h
N
hh N 
The rate of heat transfer per unit length during this condensation process becomes 
 
W/m107813
K )3050)(K W/m7523)(m 008.0()100(
)(/
5
2
sattotal
 


.
TTDhNLQ s


 
The rate of condensation per unit length is 
 mkg/s 0.155 



 J/kg 102440
W/m10781.3/
3
5
oncondensati 
h
LQ
L
m
fg

 
Discussion Therefore, water vapor condenses at a rate of 155 g/s per meter length of the tubes. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-74 
 
10-83 Saturated water vapor at a pressure of 12.4 kPa condenses on an array of 100 horizontal tubes at 30°C. The 
condensation rates for (a) a rectangular array of 5 tubes high and 20 tubes wide and (b) a square array of 10 tubes high and 10 
tubes wide are to be determined. 
Assumptions 1 Steady operating condition exists. 2 The tube surfaces are isothermal. 
Properties The properties of water at the saturation temperature of 50°C corresponding to 12.4 kPa are hfg = 2383 kJ/kg and 
ρv = 0.0831 kg/m
3
 (Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = 40°C are, from 
Table A-9, 
 ρl = 992.1 kg/m
3
 cpl = 4179 J/kg·K 
 μl = 0.653 × 10
−3
 kg/m·s kl = 0.631 W/m·K 
Analysis The modified latent heat of vaporization is 
 
J/kg 102440
)3050)(4179(68.0102383
)(68.0
3
3
sat


 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
KW/m250,11
)008.0)(3050)(10653.0(
)631.0)(102440)(0831.01.992)(1.992)(81.9(
729.0
)(
)(
729.0
2
4/1
3
33
4/1
sat
3
 tube1 horiz,

























 
DTT
khg
h
sl
lfgvll


 
(a) For a rectangular array, the average heat transfer coefficient for a 5-tube high vertical tier becomes 
 KW/m7523)KW/m250,11(
5
11 22
4/1 tube1 horiz,4/1 tubes horiz,
 h
N
hh N 
The rate of heat transfer during this condensation process becomes 
 W107813K )3050)(K W/m7523)(m 1)(m 008.0()100()( 52sattotal  .TTDLhNQ s 
 
The rate of condensation is 
 kg/s 0.155



 J/kg 102440
W10781.3
3
5
oncondensati
 
h
Q
m
fg

 (rectangular array) 
(b) For a square array, the average heat transfer coefficient for a 10-tube high vertical tier becomes 
 KW/m6326)KW/m250,11(
10
11 22
4/1 tube1 horiz,4/1 tubes horiz,
 h
N
hh N 
The rate of heat transfer during this condensation process becomes 
 W101803K )3050)(K W/m6326)(m 1)(m 008.0()100()( 52sattotal  .TTDLhNQ s 
 
The rate of condensation is 
 kg/s 0.130



 J/kg 102440
W10180.3
3
5
oncondensati
 
h
Q
m
fg

 (square array) 
Discussion The condensation rate of the rectangular array tube bank is about 20% higher than that of the square array tube 
bank: 
 19.1
kg/s 130.0
kg/s 155.0
)square(
)rrectangula(
oncondensati
oncondensati 
m
m


 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-75 
 
10-84 Saturated refrigerant-134a vapor is condensed as it is flowing through a tube. With a given vapor flow rate at the 
entrance, the flow rate of the vapor at the exit is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The tube surfaces are isothermal. 
Properties The properties of refrigerant-134a at the saturation temperature of 35°C corresponding to 888 kPa are hfg = 168.2 
kJ/kg, ρv = 43.41 kg/m
3
, and μv = 1.327 × 10
−5
 kg/m·s (Table A-10). The properties of liquid refrigerant-134a at the film 
temperature of Tf = (Tsat + Ts)/2 = 25°C are, from Table A-10, 
 ρl = 1207 kg/m
3
 cpl = 1427 J/kg·K 
 μl = 2.012 × 10
−4
 kg/m·s kl = 0.0833 W/m·K 
Analysis The modified latent heat of vaporization for film condensation inside horizontal tube is 
 
J/kg 109.178
)1535)(1427(
8
3
102.168
)(
8
3
3
3
sat


 splfgfg TTchh
 
The Reynolds number associated with film condensation inside a horizontal tube is 
 000,35000,24
)skg/m 10327.1)(m 012.0(
)kg/s 003.0(44
Re
5
inlet ,
inlet
vapor 












v
v
v
vv
D
mDV 
 
Hence, the heat transfer coefficient for film condensation inside a horizontal tube can be determined using 
 
KW/m1293
)012.0)(1535)(10012.2(
)109.178()0833.0)(41.431207)(1207)(81.9(
555.0
)(
)(
555.0
2
4/1
4
33
4/1
sat
3
internal

























 
DTT
hkg
hh
sl
fglvll


 
The rate of heat transfer during this condensation process becomes 
 
W7.243
K )1535)(K W/m1293)(m 25.0)(m 012.0(
)(
2
sat
 



 sTTDLhQ

 
Then, the rate of condensation can be calculated as 
 kg/s 00136.0
J/kg 109.178
W7.243
3oncondensati




 
h
Q
m
fg

 
Applying the conservation of mass, the flow rate of vapor leaving the tube can be determined as 
 oncondensatioutlet ,inlet , mmm vv   → oncondensatiinlet ,outlet , mmm vv   
 kg/s 0.00164 kg/s 00136.0kg/s 003.0outlet ,vm 
Discussion About 45% of the refrigerant-134a vapor that entered the tube is condensed inside it. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-76 
 
10-85 Saturated ammonia vapor is condensed as it flows through a tube. With a given vapor flow rate at the exit, the flow rate 
of the vapor at the inlet is to be determined. 
Assumptions 1 Steady operating condition exists. 2 The tube surfaces are isothermal. 3 The Reynolds number of the vapor at 
the inlet is less than 35,000 (this assumption will be verified). 
Properties The properties of ammonia at the saturation temperature of 25°C corresponding to 1003 kPa are hfg = 1166 kJ/kg, 
ρv = 7.809 kg/m
3
, and μv = 1.037 × 10
−5
 kg/m·s (Table A-11). The properties of liquid ammonia at the film temperature of Tf = 
(Tsat + Ts)/2 = 15°C are, from Table A-11, 
 ρl = 617.5 kg/m
3
 
cpl = 4709 J/kg·K 
 μl = 1.606 × 10
−4
 kg/m·s 
kl = 0.5042 W/m·K 
Analysis The modified latent heat of vaporization for film condensation inside horizontal tube is 
 
J/kg 101201
)525)(4709(
8
3
101166
)(
8
3
3
3
sat


 splfgfg TTchh
 
Assuming Revapor < 35,000 and the heat transfer coefficient for film condensation inside a horizontal tube can be determined 
using 
 
KW/m5091
)025.0)(525)(10606.1(
)101201()5042.0)(809.75.617)(5.617)(81.9(
555.0
)(
)(
555.0
2
4/1
4
33
4/1
sat
3
internal

























 
DTT
hkg
hh
sl
fglvll


 
The rate of heat transfer during this condensation process becomes 
 
W3998
K )525)(K W/m5091)(m 5.0)(m 025.0(
)(
2
sat
 



 sTTDLhQ

 
Then, the rate of condensation can be calculated as 
 kg/s 00333.0
J/kg 101201
W3998
3oncondensati




 
h
Q
m
fg

 
Applying the conservation of mass, the flow rate of vapor leaving the tube can be determined as 
 oncondensatioutlet ,inlet , mmm vv   
 kg/s 0.00533 kg/s 00333.0kg/s 002.0inlet ,vm 
Discussion The Reynolds number associated with film condensation inside a horizontal tube is 
 000,35200,26
)skg/m 10037.1)(m 025.0(
)kg/s 00533.0(44
Re
5
inlet ,
inletvapor 












v
v
v
vv
D
mDV 
 
Thus, the Revapor < 35,000 assumption is appropriate for this problem. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-77 
 
10-86 A copper tube transporting cold coolant has a surface temperature of 5°C and condenses moist air at 25°C. The rate of 
condensation is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tube surface. 
Properties The required property of water at the saturation temperature Tsat = 25°C is hfg = 2442  10
3
 J/kg (Table A-9). The 
required property of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (25 + 5)/2 = 15°C is cpl = 4185 J/kg∙K (Table 
A-9). 
Analysis The modified latent heat of vaporization is 
 
J/kg 104989.2K )525)(KJ/kg 4185(68.0J/kg 102442
)(68.0
63
sat
*

 splfgfg TTchh
 
The dropwise condensation heat transfer coefficient, for 22°C < Tsat < 100°C, is determined to be 
 
K W/m204,102)C25(2044104,51
2044104,51
2
satdropwise

 Thh
 
The rate of heat transfer to the tube during this condensation process is 
 
 W106054.1
K )525)(m 10)(m 025.0()K W/m204,102(
))(()(
6
2
satsat




 sss TTDLhTThAQ

 
Thus, the rate of condensation is 
 kg/s 0.6424



J/kg 104989.2
J/s 106054.1
6
6
*
fgh
Q
m

 
Discussion The heat transfer rate during the dropwise condensation can be increased by increasing the surface area of the 
tube, and thereby increasing the rate of condensation as well. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-78 
 
10-87 A copper tube that is used for transporting cold fluid is causing condensation of moist air on its outer surface. The 
rate of moisture condensation is to be determined so that a system for removing the condensate can be sized to alleviate the 
risk of electrical hazard. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tube surface. 
Properties The required property of water at the saturation temperature Tsat = 27°C is hfg = 2438  10
3
 J/kg (Table A-9). The 
required property of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (27 + 3)/2 = 15°C is cpl = 4185 J/kg∙K (Table 
A-9). 
Analysis Since dropwise condensation can occur, and it will have higher rate of heat transfer and therefore higher 
condensation rate than film condensation. The system must be able to handle the dropwise condensation rate. 
The modified latent heat of vaporization is 
 
J/kg 105063.2K )327)(KJ/kg 4185(68.0J/kg 102438
)(68.0
63
sat
*

 splfgfg TTchh
 
The dropwise condensation heat transfer coefficient, for 22°C < Tsat < 100°C, is determined to be 
 
K W/m292,106)C27(2044104,51
2044104,51
2
satdropwise

 Thh
 
The rate of heat transfer during the condensation process is 
 
 W100036.2
K )327)(m 10)(m 025.0()K W/m292,106(
))((
)(
6
2
sat
sat





 s
ss
TTDLh
TThAQ
 
Thus, the rate of dropwise condensation is 
 kg/s 0.799



J/kg 105063.2
J/s 100036.2
6
6
*
fgh
Q
m

 
Discussion In order to alleviate the risk of electrical hazard, the system must be able to remove the condensate at a rate of 0.8 
kg/s. Note that dropwise condensation rate are higher than film condensation rate (by 10 times or more), thus a system 
capable of removing the condensate at a rate of 0.8 kg/s can also handle the condensate from film condensation. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-79 
 
10-88 Steam condenses at 60°C on a copper tube that is maintained with a surface temperature of 40°C. The condensation 
rates during film condensation and dropwise condensation are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Isothermal tube surface. 
Properties The properties of water at the saturation temperature of 60°C are hfg = 2359  10
3
 J/kg and ρv = 0.1304 kg/m
3
 
(Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (60 + 40)/2 = 50°C are (Table A-9) 
 
K W/m644.0
KJ/kg 4181
/sm10554.0/
skg/m10547.0
kg/m 1.988
26
3
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 104159.2K )4060)(KJ/kg 4181(68.0J/kg 102359
)(68.0
63
sat
*

 splfgfg TTchh
 
(a) The film condensation heat transfer coefficient is determined from 
 
K W/m7.8937
)m 025.0(K)4060)(skg/m10547.0(
)K W/m644.0)(J/kg 104159.2(kg/m)1304.01.988)(kg/m 1.988)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
3
36332
4/1
sat
3*
horizfilm
























DTT
khg
hh
sl
lfgvll


 
Thus, the rate of film condensation is 
 
kg/s 0.08717





J/kg 104159.2
K )4060)(m 15)(m 025.0()K W/m7.8937(
))((
6
2
*
satfilm
*film


fg
s
fg h
TTDLh
h
Q
m


 
(b) The dropwise condensation heat transfer coefficient, for 22°C < Tsat < 100°C, is determined from 
 
K W/m744,173)C60(2044104,51
2044104,51
2
satdropwise

 Th
 
Thus, the rate of dropwise condensation is 
 
kg/s 1.695





J/kg 104159.2
K )4060)(m 15)(m 025.0()K W/m744,173(
))((
6
2
*
satdropwise
*dropwise


fg
s
fg h
TTDLh
h
Q
m


 
Discussion The dropwise condensation rate is 19.4 times greater than that of film condensation. This is because the 
convection heat transfer coefficient for dropwise condensation is greater than that of film condensation at the same factor, 
hdropwise/hfilm = 19.4. 
With dropwise condensation, there is no liquid film to impede heat transfer. Therefore, a much higher heat transfer 
coefficient can be achieved in dropwise condensation than in film condensation. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-80 
 
Special Topic: Non-Boiling Two-Phase Flow Heat Transfer 
 
 
10-89 The flow quality of a non-boiling two-phase flow in a tube with 300/ gl mm  is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Analysis The flow quality is given as 
 
gl
g
mm
m
x



 
Hence, the equation can be rearranged as 
 
1/
1
/)(
/






glggl
gg
gl
g
mmmmm
mm
mm
m
x




 
Thus, the flow quality is 
 0.00332


1300
1
x 
Discussion The flow quality is a dimensionless parameter. 
 
 
 
 
10-90 The flow quality and the mass flow rates of the gas and the liquid for a non-boiling two-phase flow, where Vsl = 3Vsg, 
are to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The densities of the gas and liquid are given to be ρg = 8.5 kg/m
3
 and ρl = 855 kg/m
3
, respectively. 
Analysis The mass flow rate of gas can be calculated using 
 
kg/s 0.0556

4
m) 1020(
)m/s 8.0)(kg/m 5.8(
4
2
3
2
.
D
VAVm sggcsggg


 
Then, the mass flow rate of liquid is (with Vsl = 3Vsg) 
 
kg/s 16.8

4
m) 1020(
)m/s 8.0)(kg/m 855(3
4
3
2
3
2
.
D
VAVm sglcslll


 
Thus, the flow quality is 
 0.00330



0556.08.16
0556.0
gl
g
mm
m
x


 
Discussion The total mass flow rate of gas and liquid for this two-phase flow is simply 
 kg/s 86.16kg/s 0556.0kg/s 8.16tot  gl mmm  
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-81 
 
10-91 The mass flow rate of air and the superficial velocities of air and engine oil for a non-boiling two-phase flow in a tube 
are to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The densities of air and engine oil at the bulk mean temperature Tb = 140°C are ρg = 0.8542 kg/m
3
 (Table A-15) 
and ρl = 816.8 kg/m
3
 (Table A-13), respectively. 
Analysis The flow quality is given as 
 
1/
1
/)(
/






glggl
gg
gl
g
mmmmm
mm
mm
m
x




 
or 
 
xm
m
g
l 11


 → 
x
x
m
m
g
l 
1


 
With known liquid (engine oil) mass flow rate and flow quality, the gas (air) mass flow rate is determined using 
 kg/s0.00189







)kg/s 9.0(
101.21
101.2
1 3
3
lg m
x
x
m  
From the gas and liquid mass flow rates, the superficial gas and liquid velocities can be calculated: 
 m/s 4.51
232 )m 025.0()kg/m 8542.0(
kg/s) 00189.0(44
 D
m
A
m
V
g
g
g
g
sg

 
 m/s 2.25
232 )m 025.0()kg/m 8.816(
)kg/s 9.0(44
 D
m
A
m
V
l
l
l
l
sl

 
Discussion The superficial velocity of air is twice that of the engine oil. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-82 
 
10-92 Starting with the two-phase non-boiling heat transfer correlation, the expression that is appropriate for the case when 
only water is flowing in the tube is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Analysis The two-phase non-boiling heat transfer correlation is given as 
  































 








25.0*
25.025.04.01.0 1
1
55.01 I
Pr
Pr
F
F
x
x
Fhh
g
l
l
g
p
p
pltp


 
where 
 
2
2
1
)(
)(
tan
2
)1(






















 
gl
lgg
p
gD
VV
F



 and 
gl
g
mm
m
x



 
For the situation when the air flow is shut off and only water is flowing in the pipe, we have 0gm and 0 . Hence, we 
get 
 10)01( pF and 0x 
Thus, the two-phase non-boiling heat transfer correlation becomes 
     ll
g
l
l
g
lαtp hhI
Pr
Pr
hh 





























 







 01)1(
1
11
01
0
55.01)1(
25.0*
25.025.04.01.0
0,


 
The liquid phase heat transfer coefficient is calculated using: 
 
14.0
3/15/4 PrRe027.0 














s
ll
lll
D
k
h


 
where the in situ liquid Reynolds number is 
 
D
m
D
m
D
m
l
l
l
l
l
l
l

 4
01
4
1
4
Re 



 
Therefore, we have 
 
14.0
3/15/4
0, PrRe027.0 














s
ll
lllαtp
D
k
hh


 
Discussion When only water is flowing in the tube, the two-phase non-boiling heat transfer correlation is reduced to a familiar 
equation for internal forced convection. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-83 
 
10-93 Air-water slug flows through a 25.4-mm diameter horizontal tube in microgravity condition. Using the non-boiling two-
phase heat transfer correlation, the two-phase heat transfer coefficient (htp) is to be determined 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The properties of water (liquid) are given to be μl = 85.5×10
-5
 kg/m·s, μs = 73.9×10
-5
 kg/m·s, ρl = 997 kg/m
3
, kl = 
0.613 W/m·K, and Prl = 5.0. The properties of air (gas) are given to be μg = 18.5×10
-6
 kg/m·s, ρg = 1.16 kg/m
3
, and Prg = 
0.71. 
Analysis From the superficial gas and liquid velocities, and void fraction, the gas and liquid velocities can be calculated as 
 m/s 11.1
27.0
m/s 3.0


sg
g
V
V 
 m/s 745.0
27.01
m/s 544.0
1






sl
l
V
V 
The gas and liquid mass flow rates are calculated as 
 kg/s 1076.1
4
m) 02540(
)m/s 3.0)(kg/m 16.1(
4
4
2
3
2

.D
VAVm sggcsggg  
 kg/s 275.0
4
m) 02540(
)m/s 544.0)(kg/m 997(
4
2
3
2

.D
VAVm sllcslll  
Using the gas and liquid mass flow rates, the quality is determined to be 
 4
4
4
1040.6
1076.1275.0
1076.1 








gl
g
mm
m
x


 
The flow pattern factor (Fp) can be calculated using 
 
730.0
)kg/m 16.1kg/m 997()m 0254.0)(m/s 81.9(
)m/s 745.0m/s 11.1)(kg/m 16.1(
tan
2
)27.0()27.01(
)(
)(
tan
2
)1(
2
332
23
1
2
2
1


















































gl
lgg
p
gD
VV
F
 
The liquid phase heat transfer coefficient is calculated using: 
 
K W/m2995
skg/m109.73
skg/m105.85
m 0254.0
K W/m613.0
)0.5()18870(027.0
PrRe027.0
2
14.0
5
5
3/15/4
14.0
3/15/4
















 


















s
ll
lll
D
k
h


 
where the in situ liquid Reynolds number is 
 18870
)m 0254.0)(skg/m 105.85(27.01
)kg/s 275.0(4
1
4
Re
5





 D
m
l
l
l

 
The inclination factor (I
 *
) has a value of one for horizontal tube (θ = 0). Thus, using the general two-phase heat transfer 
correlation, the value for htp is estimated to be 
 
 
9369.0
skg/m105.18
skg/m105.85
0.5
71.0
730.0
730.01
1040.61
1040.6
55.01)730.0(
1
1
55.01
25.0
6
525.04.0
1.0
4
4
25.0*
25.025.04.01.0






























 










































 












I
Pr
Pr
F
F
x
x
F
h
h
g
l
l
g
p
p
p
l
tp


 
or K W/m2810
2  )K W/m2995(9369.09369.0 2ltp hh 
Discussion The non-boiling two-phase heat transfer coefficient (htp) is about 7% lower than the liquid phase heat transfer 
coefficient (hl). 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-84 
 
10-94 A two-phase flow of air and silicone (Dow Corning 200
®
 Fluid, 5 cs) is being transported in an 11.7-mm diameter 
horizontal tube, where condensation occurs on the tube outer surface. The overall convection heat transfer coefficient is to be 
determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow inside tube is non-boiling and does not involve phase 
change. 3 Fluid properties are constant. 4 The thermal resistance of the tube wall is negligible. 5 Isothermal tube surface. 6 
Film condensation occurs on the tube outer surface. 
Properties Inside the tube: The properties of liquid silicone (liquid) are given to be μl = 45.7×10
-4
 kg/m·s, ρl = 913 kg/m
3
, kl = 
0.117 W/m·K, σ = 19.7×10
-3
 N/m and Prl = 64. The propertiesof air (gas) are given to be μg = 18.4×10
-6
 kg/m·s, ρg = 1.19 
kg/m
3
, Prg = 0.71. 
Outside the tube: The properties of water at the saturation temperature of 40°C are hfg = 240710
3
 J/kg and v = 0.0512 kg/m
3
 
(Table A-9). The properties of liquid water at the film temperature of Tf = (Tsat + Ts)/2 = (40 + 20)/2 = 30°C are (Table A-9), 
 
5.42=Pr
K W/m615.0
KJ/kg 4178
/sm10801.0/
skg/m10798.0
kg/m 0.996
,
,
26
,
3
,
3
,
f
fl
fpl
llfl
fl
fl
k
c










 
Analysis Inside the tube: The flow pattern factor (FP) can be calculated using 
 
9898.0
)kg/m 19.1kg/m 913()m 0117.0)(m/s 81.9(
)m/s 34.9m/s 5.13)(kg/m 19.1(
tan
2
)011.0()011.01(
)(
)(
tan
2
)1(
2
332
23
1
2
2
1


















































gl
lgg
P
gD
VV
F
 
The inclination factor (I
 *
) for horizontal tube (θ = 0°) is calculated to be 
 1sin
)(
1
2


 

 gD
I
gi
 
The liquid phase heat transfer coefficient is calculated using the Seder and Tate (1936) equation: 
 
K W/m3246
m 0117.0
K W/m117.0
8.39
7.45
)64()21718(027.0
PrRe027.0
2
14.0
3/18.0
14.0
3/18.0






 






















D
k
h l
s
l
ll


 
where the in situ liquid Reynolds number is 
 21718
)m 0117.0)(skg/m 107.45(011.01
)kg/s 907.0(4
1
4
Re
4





 D
m
l
l
l

 
Using the general two-phase heat transfer correlation, Eq. (10-38), the two-phase heat transfer coefficient (htp) is estimated to 
be 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-85 
 
 
 
K W/m3337
104.18
107.45
64
71.0
9898.0
9898.01
1008.21
1008.2
55.01)9898.0)(K W/m3246(
1
1
55.01
2
25.0
6
425.04.0
1.0
5
5
2
25.0*
25.025.04.01.0






























 










































 












tpi
g
l
l
g
P
P
Pltp
hh
I
Pr
Pr
F
F
x
x
Fhh


 
Outside the tube: The modified latent heat of vaporization is 
 J/kg 102464=0)K2K)(40J/kg 4182(0.68+J/kg 102407)(68.0
33
sat,
*  sfplfgfg TTchh 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
K W/m9584
m) K(0.0117)2040(s)kg/m 10798.0(
)K W/m615.0)(J/kg 102464)(kg/m 05.0996)(kg/m 996)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat,
3
,
*
,,
horizontal
























DTT
khg
hh
sfl
flfgvflfl
o


 
Noting that the thermal resistance of the tube is negligible, the overall heat transfer coefficient becomes 
 K W/m2475
2 




9584/13337/1
1
/1/1
1
oi hh
U 
Discussion The condensation heat transfer coefficient is almost 3 times higher than the non-boiling heat transfer coefficient. 
This is expected, as heat transfer coefficient involving phase change is much higher than that without phase change. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-86 
 
10-95 Air-water mixture is flowing in a 5° inclined tube with diameter of 25.4 mm, and the mixture superficial gas and liquid 
velocities are 1 m/s and 2 m/s, respectively. The two-phase heat transfer coefficient (htp) is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The properties of water (liquid) at bulk mean temperature Tb = (Ti + Te)/2 = 45°C are, from Table A-9, μl = 
0.596×10
-3
 kg/m·s, ρl = 990.1 kg/m
3
, kl = 0.637 W/m·K, and Prl = 3.91. The properties of air (gas) at bulk mean temperature 
Tb = 45°C are, from Table A-15, μg = 1.941×10
-5
 kg/m·s, ρg = 1.109 kg/m
3
, and Prg = 0.7241. Also, at Ts = 80°C we get μs = 
0.355×10
-3
 kg/m·s from Table A-9. 
Analysis From the superficial gas and liquid velocities, and void fraction, the gas and liquid velocities can be calculated as 
 m/s 030.3
33.0
m/s 1


sg
g
V
V m/s 985.2
33.01
m/s 2
1






sl
l
V
V 
The gas and liquid mass flow rates are calculated as 
 kg/s 10619.5
4
m) 02540(
)m/s 1)(kg/m 109.1(
4
4
2
3
2

.D
VAVm sggcsggg  
 kg/s 003.1
4
m) 02540(
)m/s 2)(kg/m 1.990(
4
2
3
2

.D
VAVm sllcslll  
Using the gas and liquid mass flow rates, the quality is determined to be 
 4
4
4
10599.5
10619.5003.1
10619.5 








gl
g
mm
m
x


 
The flow pattern factor (Fp) can be calculated using 
 
670.0
)kg/m 109.1kg/m 1.990()m 0254.0)(m/s 81.9(
)m/s 985.2m/s 03.3)(kg/m 109.1(
tan
2
)33.0()33.01(
)(
)(
tan
2
)1(
2
332
23
1
2
2
1


















































gl
lgg
p
gD
VV
F
 
The inclination factor (I
 *
) for θ = 5° is calculated to be 
 023.95sin
N/m 068.0
)m 0254.0)(m/s 81.9)(kg/m 109.1kg/m 1.990(
1sin
)(
1
22332




 

 gD
I
gl
 
The liquid phase heat transfer coefficient is calculated using: 
 
K W/m11754
skg/m10355.0
skg/m10596.0
m 0254.0
K W/m637.0
)91.3()103100(027.0
PrRe027.0
2
14.0
3
3
3/15/4
14.0
3/15/4
















 


















s
ll
lll
D
k
h


 
where the in situ liquid Reynolds number is 
 103100
)m 0254.0)(skg/m 10596.0(33.01
)kg/s 003.1(4
1
4
Re
3





 D
m
l
l
l

 
Thus, using the general two-phase heat transfer correlation, the value for htp is estimated to be 
 
 
021.1)023.9(
941.1
6.59
91.3
7241.0
670.0
670.01
10599.51
10599.5
55.01)670.0(
1
1
55.01
25.0
25.025.04.01.0
4
4
25.0*
25.025.04.01.0


























 










































 










I
Pr
Pr
F
F
x
x
F
h
h
g
l
l
g
p
p
p
l
tp


 
or K W/m12,000
2  )K W/m11754(021.1021.1 2ltp hh 
Discussion The inclination factor is I
*
 = 1 when the tube is at horizontal position, since sin(0°) = 0. The two-phase heat 
transfer coefficient for horizontal tube would be htp = 10300 W/m
2
·K, which is about 14% lower than that of 5° inclined tube. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-87 
 
10-96 Mixture of petroleum and natural gas is being transported in a pipeline that is located in a terrain that caused it to have 
an average inclination angle of 10°. The two-phase heat transfer coefficient is to be determined. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The properties of petroleum (liquid) are given to be μl = 297.5×10
-4
 kg/m·s, μs = 238×10
-4
 kg/m·s, ρl = 853 kg/m
3
, 
kl = 0.163 W/m·K, σ = 0.020 N/m, and Prl = 405. The properties of natural gas are given to be μg = 9.225×10
-6
 kg/m·s, ρg = 
9.0 kg/m
3
, and Prg = 0.80. 
Analysis From the gas and liquid mass flow rates, the superficial gas and liquid velocities can be calculated: 
 m/s 748.0
)m 102.0()kg/m 0.9(
kg/s) 055.0(44
232

 D
m
A
m
V
g
g
g
g
sg
 
 m/s 296.2
)m 102.0()kg/m 853(
)kg/s 16(44
232

 D
m
A
m
V
l
l
l
l
sl

 
Using the superficial velocities and void fraction, the gas and liquid velocities are found to be 
 m/s 400.3
22.0
m/s 748.0


sg
g
V
V m/s 944.2
22.01
m/s 296.2
1






sl
l
V
V 
Using the gas and liquid mass flow rates, the quality is determined to be 
 310426.3
055.016
055.0 




gl
g
mm
m
x


 
The flow pattern factor (Fp) can be calculated using 
 
7802.0
)kg/m 0.9kg/m 853)(m 102.0)(m/s 81.9(
)m/s 944.2m/s 40.3)(kg/m 0.9(
tan
2
)22.0()22.01(
)(
)(
tan
2
)1(
2
332
23
1
2
2
1


















































gl
lgg
p
gD
VV
F
 
The liquid phase heat transfer coefficient is calculated using: 
 
K W/m1.419
skg/m10238
skg/m105.297
m 102.0
K W/m163.0
)405()7601(027.0
PrRe027.0
2
14.0
4
4
3/15/4
14.0
3/15/4
















 


















s
ll
lll
D
k
h


 
where the in situ liquid Reynolds number is 
 7601
)m 102.0)(skg/m 105.297(22.01
)kg/s 16(4
1
4
Re
4





 D
m
l
l
l

 
The inclination factor (I
 *
) for θ = 10° is calculated to be 
 9.74810sin
N/m .0200
)m 102.0)(m/s 81.9)(kg/m 0.9kg/m 853(
1sin
)(
1
22332




 

 gD
I
gl
 
Thus, using the general two-phase heat transfer correlation, the value for htp is estimated to be 
 
 
999.1)9.748(
225.9
29750
405
80.0
7802.0
7802.01
10426.31
10426.3
55.01)7802.0(
1
1
55.01
25.0
25.025.04.01.0
3
3
25.0*
25.025.04.01.0


























 










































 










I
Pr
Pr
F
F
x
x
F
h
h
g
l
l
g
p
p
p
l
tp


 
or K W/m838
2  )K W/m1.419(999.1999.1 2ltp hh 
Discussion Since Vg > Vl, there will be slippage between the gas and liquid phases. When Vg ≠ Vl, slippage between the gas 
and liquid phases exists. When Vg = Vl, slippage between the gas and liquid phases is negligible, and the flow is called 
homogeneous two-phase flow. 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-88 
 
10-97 Air-water mixture is flowing in a tube with diameter of 25.4 mm, and the mixture superficial gas and liquid velocities 
are 1 m/s and 2 m/s, respectively. The two-phase heat transfer coefficient (htp) for (a) horizontal tubee (θ = 0°) and (b) vertical 
tube (θ = 90°), are to be determined and compared. 
Assumptions 1 Steady operating condition exists. 2 Two-phase flow is non-boiling and it does not involve phase change. 3 
Fluid properties are constant. 
Properties The properties of water (liquid) at bulk mean temperature Tb = (Ti + Te)/2 = 45°C are, from Table A-9, μl = 
0.596×10
-3
 kg/m·s, ρl = 990.1 kg/m
3
, kl = 0.637 W/m·K, and Prl = 3.91. The properties of air (gas) at bulk mean temperature 
Tb = 45°C are, from Table A-15, μg = 1.941×10
-5
 kg/m·s, ρg = 1.109 kg/m
3
, and Prg = 0.7241. Also, at Ts = 80°C we get μs = 
0.355×10
-3
 kg/m·s from Table A-9. 
Analysis From the superficial gas and liquid velocities, and void fraction, the gas and liquid velocities can be calculated as 
 m/s 030.3
33.0
m/s 1


sg
g
V
V 
 m/s 985.2
33.01
m/s 2
1






sl
l
V
V 
The gas and liquid mass flow rates are calculated as 
 kg/s 10619.5
4
m) 02540(
)m/s 1)(kg/m 109.1(
4
4
2
3
2

.D
VAVm sggcsggg  
 kg/s 003.1
4
m) 02540(
)m/s 2)(kg/m 1.990(
4
2
3
2

.D
VAVm sllcslll  
Using the gas and liquid mass flow rates, the quality is determined to be 
 4
4
4
10599.5
10619.5003.1
10619.5 








gl
g
mm
m
x


 
The flow pattern factor (Fp) can be calculated using 
 
670.0
)kg/m 109.1kg/m 1.990()m 0254.0)(m/s 81.9(
)m/s 985.2m/s 03.3)(kg/m 109.1(
tan
2
)33.0()33.01(
)(
)(
tan
2
)1(
2
332
23
1
2
2
1


















































gl
lgg
p
gD
VV
F
 
The liquid phase heat transfer coefficient is calculated using: 
 
K W/m11754
skg/m10355.0
skg/m10596.0
m 0254.0
K W/m637.0
)91.3()103100(027.0
PrRe027.0
2
14.0
3
3
3/15/4
14.0
3/15/4
















 


















s
ll
lll
D
k
h


 
where the in situ liquid Reynolds number is 
 103100
)m 0254.0)(skg/m 10596.0(33.01
)kg/s 003.1(4
1
4
Re
3





 D
m
l
l
l

 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-89 
 
(a) The inclination factor (I
 *
) for horizontal tube (θ = 0°) is 1I . Thus, using the general two-phase heat transfer 
correlation, the value for htp is estimated to be 
 
 
8727.0
941.1
6.59
91.3
7241.0
670.0
670.01
10599.51
10599.5
55.01)670.0(
1
1
55.01
25.025.04.01.0
4
4
25.0*
25.025.04.01.0
horiz ,


























 










































 










I
Pr
Pr
F
F
x
x
F
h
h
g
l
l
g
p
p
p
l
tp


 
or 
 K W/m10,300 2  )K W/m11754(8727.087271.0 2horiz , ltp hh 
(b) The inclination factor (I
 *
) for vertical tube (θ = 90°) is calculated to be 
 
05.93
90sin
N/m 068.0
)m 0254.0)(m/s 81.9)(kg/m 109.1kg/m 1.990(
1
sin
)(
1
2233
2





 

 gD
I
gl
 
Thus, using the general two-phase heat transfer correlation, the value for htp is estimated to be 
 
 
 
30.1
05.93
941.1
6.59
91.3
7241.0
670.0
670.01
10599.51
10599.5
55.01)670.0(
1
1
55.01
25.0
25.025.04.01.0
4
4
25.0*
25.025.04.01.0
 vert,


























 










































 










I
Pr
Pr
F
F
x
x
F
h
h
g
l
l
g
p
p
p
l
tp


 
or 
 K W/m15,300 2  )K W/m11754(30.130.1 2 vert, ltp hh 
Discussion The two-phase heat transfer coefficient of vertical pipe is about 49% higher than that of horizontal pipe: 
 49.1
300,10
300,15
horiz ,
 vert,

tp
tp
h
h
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-90 
 
10-98 Air-water mixture flows through a 0.0254 m stainless steel pipe at specified flow condition. Using the concept of 
Reynolds analogy, the two phase convective heat transfer coefficient is to be determined. 
Assumptions 1 Steady state operating conditions exist. 2 Two phase flow is non-boiling in nature and does not undergo any 
phase change. 3 Fluid properties are constant. 
Properties The properties of water and air are calculated at a system temperature and pressure of 25
o
C and 201 kPa using 
EES. The thermo physical properties of water and air are, 
 kg/m.s 101.84 ,kg/m 2.35
N/m 90710 andK W/m 5950 k , 266 kg/m.s, 108.9 ,kg/m 997.1
53
l
43




gg
lll ...Pr


 
Analysis We first need to determine superficial Reynolds number and the actual velocities for each phase. 
8537
s)(kg/m108.9
0.0254(m)0.3(m/s))997.1(kg/m
4-
3




l
sll
sl
DV
Re

74612
s)(kg/m101.84
0.0254(m)23(m/s))2.35(kg/m
5-
3




g
sgg
sg
DV
Re


 
The actual phase velocities are calculated from the known values of superficial velocities and void fraction. 
m/s 2.14
0.861
0.3(m/s)
1






sl
l
V
V and m/s 26.74
0.86
23(m/s)


sg
g
V
V 
The mass flow rate of each phase is, 
 kg/s 0.15)(m0254.0
4
0.3(m/s))(kg/m1.997
4
2232 

 DVm slll 
kg/s 0.027)(m0254.0
4
23(m/s))(kg/m35.2
4
2232 

 DVm sggg 
Thus the total mass flow rate is, kg/s 177.0027.015.0  gl mmm  
Since this is a vertical upward flow of air water we can use the Reynolds analogy given by Equation (10-40). 
p
l
n
l
tplm
p
l
tp
ρ
ρ
m
m
F
h
h

















 
The single phase heat transfer coefficient is calculated as, 
14.03154 )/()/(PrRe027.0 slllll Dkh  
The in-situ Reynolds number required in single phase heat transfer equation is calcutaed as, 
 22579
0.86-10.0254(m)(kg/m.s)108.9
0.15(kg/s)4
1
4
Re
4-






 D
m
l
l
l

 
KW/m 3878
10664
1098
0.0254(m)
K)(W/m5950
266225790270 2
140
4
4
3154 














.
//
l
.
..
..h 
783.0
)2.35)(kg/m-(997.10.0254(m)9.81
2.14)-)(26.742.35(kg/m
tan
2
86.0)86.01(
)(
)(
tan
2
)1(
2
3
23
1
2
2
1












































p
p
gl
lgg
p
F
F
gD
VV
F





 
The two phase density required in Reynolds analogy equation is calculated as, 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-91 
 
333 kg/m 141.6)2.35(kg/m0.86)1(kg/m99786011  .).()( gltp  
Single phase pressure drop for turbulent pipe flow (with superficial Reynolds number of Resl = 8537) is calculated by first 
calculating the single phase friction factor from either the Mood chart (Fig. A-20) or the Colebrook equation (Eq.8-74) as 
follows, 
 0323.0
8537
51.2
7.3
4.25/002.0
log2
Re
51.2
7.3
/
log2
1



















llsll ff
D
f

 
Where from Table 8-3 for stainless steel pipe, the roughness is  = 0.002. 
Single phase pressure drop is then calculated as, 
 Pa/m 06.57
0.0254(m)2
)(m0.3)(kg/m1.9970323.0
2
)/(
2232
, 



D
Vf
dLdP sllllf

 
Thus the two-phase friction multiplier is, 
876
)57.06(Pa/m
2700(Pa/m)
.
)dL/dP(
)dL/dP(
l,f
tp,,f
l  
Thus the two-phase heat transfer coefficient calculated using Reynolds analogy (Eq. 10-40) is, 
922
876
)997.1(kg/m
)141.6(kg/m
1770
150
7830 20
50
3
3
50
.
h
h
.
.
.
.
ρ
ρ
m
m
F
h
h
l
tp
.
.
.p
l
n
l
tplm
p
l
tp


































 
Thus the two phase heat transfer coefficient is: htp = 11324 W/m
2
·K 
Discussion The use of Reynolds analogy strongly depends on the correct estimation of the two-phase pressure drop. Most of 
the correlations available for two-phase pressure drop fail to correctly estimate the two-phase pressure drop and hence based 
on the calculated value of two-phase pressure drop, the Reynolds analogy may not predict the two- phase heat transfer 
coefficient correctly. 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-92 
 
Review Problems 
 
10-99 Water is boiled at Tsat = 120C in a mechanically polished stainless steel pressure cooker whose inner surface 
temperature is maintained at Ts = 130C. The time it will take for the tank to empty is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 120C are (Tables 10-1 and A-9) 
44.1Pr
CJ/kg 4244N/m 0550.0
skg/m 10232.0kg/m 121.1
J/kg 102203kg/m 4.943
33
33





l
pl
lv
fgl
c
h



 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically 
polished stainless steel surface (Table 10-3). Note that we expressed the 
properties in units specified under Eq. 10-2 in connection with their 
definitions in order to avoid unit manipulations. 
Analysis The excess temperature in this case is C10120130sat  TTT s which is relatively low (less than 30C). 
Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m400,228
44.1)102203(0130.0
)120130(4244
0550.0
1.121)-9.8(943.4
)10)(220310232.0(
Pr
)()(

























 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The rate of heat transfer is 
 W7174) W/m400,228(m) 20.0(
4
1 22
nucleate  qAQ 
 
The rate of evaporation is 
 kg/s 003256.0
kJ/kg 102203
 W7174
3evap



fgh
Q
m

 
Noting that the tank is half-filled, the mass of the water and the time it will take for the tank to empty are 
   kg 446.4m) 30.0(4/m) 20.0()kg/m 4.943(
2
1
2
1 23   Vlm 
 min 22.8 s 1365
kg/s 003256.0
kg 446.4
evap
evap
m
m
mt

 
Heating 
 
130C 
120C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-93 
 
10-100 Water is boiled at sea level (1 atm pressure) and thus at a saturation (or boiling) temperature of Tsat = 100C in a 
mechanically polished AISI 304 stainless steel pan placed on top of a 3-kW electric burner. Only 60% of the heat (1.8 kW) 
generated is transferred to the water. The inner surface temperature of the pan and the temperature difference across the 
bottom of the pan are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling (this assumption will be checked later). 4 Heat transfer through the bottom of the pan is one-dimensional. 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
Kc
h
pl
l
fg




J/kg 4217
m/skg 10282.0
J/kg 102257
3
3
 
 
Also, ksteel = 14.9 W/mK (Table A-3), sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless 
steel surface (Table 10-3). Note that we expressed the properties in units specified under Eq. 10-2 in connection with their 
definitions in order to avoid unit manipulations. 
 Analysis The rate of heat transfer to the water and the heat flux are 
 
22
222
 W/m25.46=)m 69 W)/(0.0701800(/
m 07069.04/m) 30.0(4/
 W1800=kW 8.1kW 360.0



s
s
AQq
DA
Q


 
Then temperature difference across the bottom of the pan is determined directly from the steady one-dimensional heat 
conduction relation to be 
 C3.10 




Kk
Lq
T
L
T
kq
 W/m9.14
m) )(0.006 W/m460,25(
 
2
steel
steel

 
 The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be 
used to determine the surface temperature when the heat flux is given. 
Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
75.1)102257(0130.0)100(4217
0589.0
0.60)9.81(957.9
)10)(225710282.0(460,25















 
  s
T
 
It gives 
 C105.7sT 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
Electric burner, 3 kW 
P = 1 atm 
100C Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-94 
 
10-101 Water is boiled at 84.5 kPa pressure and thus at a saturation (or boiling) temperature of Tsat = 95C in a mechanically 
polished AISI 304 stainless steel pan placed on top of a 3-kW electric burner. Only 60% of the heat (1.8 kW) generated is 
transferred to the water. The inner surface temperature of the pan and the temperature difference across the bottom of the pan 
are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the boiler are negligible. 3 The boiling regime is 
nucleate boiling (this assumption will be checked later). 4 Heat transfer through the bottom of the pan is one-dimensional. 
Properties The properties of water at the saturation temperature of 95C are (Tables 10-1 and A-9) 
85.1Pr
N/m 0599.0
kg/m 50.0
kg/m 5.961
3
3




l
v
l



 
Kc
h
pl
l
fg




J/kg 4212
m/skg 10297.0
J/kg 102270
3
3
 
 
Also, ksteel = 14.9 W/mK (Table A-3), sfC 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless 
steel surface (Table 10-3). Note that we expressed the properties in units specified under Eq. 10-2 in connection with their 
definitions in order to avoid unit manipulations. 
Analysis The rate of heat transfer to the water and the heat flux are 
 
222
222
kW/m 25.46= W/m25,460=)m 69 W)/(0.0701800(/
m 07069.04/m) 30.0(4/
 W1800=kW 8.1kW 360.0



s
s
AQq
DA
Q


 
Then temperature difference across the bottom of the pan is determined directly from the steady one-dimensional heat 
conduction relation to be 
C3.10 




Kk
Lq
T
L
T
kq
 W/m9.14
m) )(0.006 W/m460,25(
 
2
steel
steel

 
The Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. 
Assuming nucleate boiling, the temperature of the inner surface of the pan is determined from Rohsenow relation to be 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
3
3
1/2
33
85.1)102270(0130.0
)95(4212
0599.0
0.50)9.81(961.5
)10)(227010297.0(460,25















 
  s
T
 
It gives 
 C100.9sT 
which is in the nucleate boiling range (5 to 30C above surface temperature). Therefore, the nucleate boiling assumption is 
valid. 
Electric burner, 3 kW 
P = 84.5 kPa 
95C Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-95 
 
10-102 Water is boiled at 1 atm pressure and thus at a saturation temperature of Tsat = 100C by a nickel electric heater 
whose diameter is 2 mm. The highest temperature at which this heater can operate without burnout is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water are negligible. 
Properties The properties of water at the saturation temperature of 100C are (Tables 10-1 and A-9) 
 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
Kc
h
pl
l
fg




J/kg 4217
m/skg 10282.0
J/kg 102257
3
3
 
Also, sfC 0.0060 and n = 1.0 for the boiling of water on a nickel 
surface (Table 10-3). 
Analysis The maximum rate of heat transfer without the burnout is simply the critical heat flux. For a horizontal heating wire, 
the coefficient Ccr is determined from Table 10-4 to be 
 
151.0)399.0(12.0*12.0
1.2 < 399.0
0589.0
60.09.957(8.9
)001.0(
)(
*
25.025.0
2/12/1






 





 

LC
g
LL
cr
vl


 
Then the maximum or critical heat flux is determined from 
2
4/123
4/12
max
 W/m1,280,000
)]60.09.957()6.0(8.90589.0)[102257(151.0
)]([


 vlvfgcr ghCq 
 
Rohsenow relation which gives the nucleate boiling heat flux for a specified surface temperature can also be used to 
determine the surface temperature when the heat flux is given. Substituting the maximum heat flux into Rohsenow relation 
together with other properties gives 
 
3
sat,
2/1
nucleate
Pr
)()(







 





 

n
lfgsf
slpvl
fgl
hC
TTcg
hq


 
 
3
3
1/2
33
75.1)102257(0060.0
)100(4217
0589.0
0.60)-9.8(957.9
)10)(225710282.0(000,280,1
















  s
T
 
It gives the maximum temperature to be: 
C109.6sT 
Ts= ? 
Heating wire 
1 atm 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-96 
 
10-103 Water is boiled at Tsat = 100C by a chemically etched stainless steel electric heater whose surface temperature is 
maintained at Ts = 115C. The rate of heat transfer to the water, the rate of evaporation of water, and the maximum rate of 
evaporation are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100C are 
(Tables 10-1 and A-9) 
75.1Pr
N/m 0589.0
kg/m 60.0
kg/m 9.957
3
3




l
v
l



 
CJ/kg 4217
m/skg 10282.0
J/kg 102257
3
3




pl
l
fg
c
h
 
Also, sfC 0.0130 and n = 1.0 for the boiling of water on a chemically etched stainless steel surface (Table 10-3). Note that 
we expressed the properties in units specified under Eq. 10-2 in connection with their definitions in order to avoid unit 
manipulations. 
Analysis (a) The excess temperature in this case is C15100115sat  TTT s which is relatively low (less than 30C). 
Therefore, nucleate boiling will occur. The heat flux in this case can be determined from Rohsenow relation to be 
 
2
3
3
1/2
33
3
sat,
2/1
nucleate
 W/m900,474
75.1)102257(0130.0
)100115(4217
0589.0
0.60)9.8(957.9
)10)(225710282.0(
Pr
)()(
















 








 





 


n
lfgsf
slpvl
fgl
hC
TTcg
hq



 
The surface area of the bottom of the heater is 2m 005027.0m) m)(0.8 002.0(  DLAs . 
Then the rate of heat transfer during nucleate boiling becomes 
 W2387 ) W/m900,474)(m 005027.0( 22nucleateboiling qAQ s 
 
The rate of evaporation of water is determined from 
 kg/h 3.81 = kg/s 101.058 3


J/kg 102257
J/s 2387
3
boiling
nevaporatio
fgh
Q
m

 
(b) For a horizontal heating wire, the coefficient Ccr is determined from Table 10-4 to be 
 
151.0)399.0(12.0*12.0
1.2 < 399.0
0589.0
60.09.957(8.9
)001.0(
)(
*
25.025.0
2/12/1






 





 

LC
g
LL
cr
vl


 
Then the maximum or critical heat flux is determined from 
2
kW/m 1280

2
4/1234/12
max
 W/m1,280,000
)]60.09.957()6.0(8.90589.0)[102257(151.0)]([ vlvfgcr ghCq 
 
Water, 100C 
115C 
Steam 
100C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission.10-97 
 
10-104 The initial boiling heat transfer coefficient and the total heat transfer coefficient, when a heated steel rod was 
submerged in a water bath, are to be determined. 
Assumptions 1 Steady operating condition exists. 2 The steel rod has uniform initial surface temperature. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16, 
 ρv = 0.3831 kg/m
3
 cpv = 1997 J/kg·K 
 μv = 2.045 × 10
−5
 kg/m·s kv = 0.04345 W/m·K 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 400°C, which is much larger than 30°C for water from Fig. 
10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 
24
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10476.6
)400(
)400)(02.0)(10045.2(
)]400)(1997(4.0102257)[3831.09.957)(3831.0()04345.0(81.9
62.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
Using the Newton’s law of cooling, the boiling heat transfer coefficient is 
 )( satfilmfilm TThq s  → 
sat
film
film
TT
q
h
s 


 
 K W/m162 2 



K )100500(
 W/m10476.6 24
filmh 
The radiation heat transfer coefficient can be determined using 
 )()( satrad
4
sat
4
rad TThTTq ss   → 
sat
4
sat
4
rad
)(
TT
TT
h
s
s




 
 K W/m08.43
K )100500(
K )373773)(K W/m1067.5)(9.0()( 2
444428
sat
4
sat
4
rad 







TT
TT
h
s
s 
Then, the total heat transfer coefficient can be determined using 
 radfilmtotal
4
3
qqq   → )(
4
3
)()( satradsatfilmsattotal TThTThTTh sss  
or 
 
K W/m194
2 


)K W/m08.43(
4
3
K W/m162
4
3
22
radfilmtotal hhh
 
Discussion The boiling heat transfer coefficient (hfilm) is 3.76 times the radiation heat transfer coefficient (hrad). 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-98 
 
10-105 The boiling heat transfer coefficient and the total heat transfer coefficient for water being boiled by a cylindrical metal 
rod are to be determined. 
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = 300°C are, from Table A-16, 
 ρv = 0.3831 kg/m
3
 
cpv = 1997 J/kg·K 
 μv = 2.045 × 10
−5
 kg/m·s 
kv = 0.04345 W/m·K 
Analysis The excess temperature in this case is ΔT = 
Ts − Tsat = 400°C, which is much larger than 30°C for 
water from Fig 10-6. Therefore, film boiling will 
occur. The film boiling heat flux in this case can be 
determined from 
25
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10152.1
)400(
)400)(002.0)(10045.2(
)]400)(1997(4.0102257)[3831.09.957)(3831.0()04345.0(81.9
62.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
Using the Newton’s law of cooling, the boiling heat transfer coefficient is 
 )( satfilmfilm TThq s  → 
sat
film
film
TT
q
h
s 


 
 K W/m288 2 



K )100500(
 W/m10152.1 25
filmh 
The radiation heat transfer coefficient can be determined using 
 )()( satrad
4
sat
4
rad TThTTq ss   → 
sat
4
sat
4
rad
)(
TT
TT
h
s
s




 
 K W/m93.23
K )100500(
K )373773)(K W/m1067.5)(5.0()( 2
444428
sat
4
sat
4
rad 







TT
TT
h
s
s 
Then, the total heat transfer coefficient can be determined using 
 radfilmtotal
4
3
qqq   → )(
4
3
)()( satradsatfilmsattotal TThTThTTh sss  
or 
 
K W/m306
2 


)K W/m93.23(
4
3
K W/m288
4
3
22
radfilmtotal hhh
 
Discussion The boiling heat transfer coefficient (hfilm) is about 12 times the radiation heat transfer coefficient (hrad). 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-99 
 
10-106 Water is boiled at Tsat = 100C by a spherical platinum heating element immersed in water. The surface temperature is 
Ts = 350C. The boiling heat transfer coefficient is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible. 
Properties The properties of water at the saturation temperature of 100C are (Table A-9) 
3
3
kg/m 9.957
J/kg 102257


l
fgh

 
The properties of water vapor at (350+100)/2 = 225C are (Table A-16) 
C W/m03581.0
CJ/kg 1951
skg/m 10749.1
kg/m 444.0
5
3





v
pv
v
v
k
c


 
Analysis The film boiling occurs since the temperature difference between the surface and the fluid. The heat flux in this case 
can be determined from 
 
 
 
2
4/1
5
33
sat
4/1
sat
sat
3
film
 W/m399,27
)100350(
)100350)(15.0)(10749.1(
)100350)(1951(4.0102257)444.09.957)(444.0()03581.0)(81.9(
67.0
)(
)(
)(4.0)(
67.0


























TT
TTD
TTchgk
q s
sv
spvfgvlvv



 
The boiling heat transfer coefficient is 
 C W/m110 2 




C)100350(
 W/m399,27
)(
2
sat
film
satfilm
TT
q
hTThq
s
s

 
 
 
350C 
100C 
Water 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-100 
 
10-107 The initial boiling heat transfer coefficient and the total heat transfer coefficient, when heated steel ball bearings are 
submerged in a water bath, are to be determined. 
Assumptions 1 Steady operating condition exists. 2 The steel ball bearings have uniform initial surface temperature. 
Properties The properties of water at the saturation temperature of 100°C are hfg = 2257 kJ/kg (Table A-2) and ρl = 957.9 
kg/m
3
 (Table A-9). The properties of vapor at the film temperature of Tf = (Tsat + Ts)/2 = 400°C are, from Table A-16, 
 ρv = 0.3262 kg/m
3
 cpv = 2066 J/kg·K 
 μv = 2.446 × 10
−5
 kg/m·s kv = 0.05467 W/m·K 
Analysis The excess temperature in this case is ΔT = Ts − Tsat = 600°C, which is much larger than 30°C for water in Fig. 10-6. 
Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from 
25
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
 W/m10052.1
)600(
)600)(02.0)(10446.2(
)]600)(2066(4.0102257)[3262.09.957)(3262.0()05467.0(81.9
67.0
)(
)(
)](4.0)[(

























TT
TTD
TTchgk
Cq s
sv
spvfgvlvv



 
Using the Newton’s law of cooling, the boiling heat transfer coefficient is 
 )( satfilmfilm TThq s  → 
sat
film
film
TT
q
h
s 


 
 K W/m175 2 



K )100700(
 W/m10052.1 25
filmh 
The radiation heat transfer coefficient can be determined using 
 )()( satrad
4
sat
4
rad TThTTq ss   → 
sat
4
sat
4
rad
)(
TT
TT
h
s
s




 
 K W/m15.62
K )100700(
K )373973)(K W/m1067.5)(75.0()( 2
444428
sat
4
sat
4
rad 







TT
TT
h
s
s 
Then, the total heat transfer coefficient can be determined using 
 radfilmtotal
4
3
qqq   → )(
4
3
)()( satradsatfilmsattotal TThTThTTh sss  
or 
 
K W/m222
2 


)K W/m15.62(4
3
K W/m175
4
3
22
radfilmtotal hhh
 
Discussion The boiling heat transfer coefficient (hfilm) is 2.82 times the radiation heat transfer coefficient (hrad). 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-101 
 
10-108E Steam at a saturation temperature of Tsat = 100F condenses on a vertical plate which is maintained at 80F. The rate 
of heat transfer to the plate and the rate of condensation of steam per ft width of the plate are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The plate is isothermal. 3 The condensate flow is wavy-laminar over the 
entire plate (this assumption will be verified). 4 The density of vapor is much smaller than the density of liquid, lv   . 
Properties The properties of water at the saturation temperature of 100F are hfg = 1037 Btu/lbm and v = 0.00286 lbm/ft
3
. 
The properties of liquid water at the film temperature of  2/)( sat sf TTT (100 + 80)/2 = 90F are (Table A-9E), 
 
FftBtu/h 358.0
FBtu/lbm 999.0
/hft 02965.0/
hlbm/ft 842.1slbm/ft 10117.5
lbm/ft 12.62
2
4
3






l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
Btu/lbm 1051=
F)80F)(100Btu/lbm 999.0(0.68+Btu/lbm 1037
)(68.0 sat
*

 splfgfg TTchh
 
Assuming wavy-laminar flow, the Reynolds number is determined from 
 
145
h) 1(
s) 3600(
)h/ft 02965.0(
ft/s 2.32
)Btu/lbm 1051)(hlbm/ft 842.1(
F)80100(F)ftBtu/h 358.0(ft) 4(70.3
81.4
)(70.3
81.4ReRe
82.0
3/1
2
2
22
2
820.0
3/1
2*
sat
wavyvertical,





































lfgl
sl g
h
TTLk

 
which is between 30 and 1800, and thus our assumption of wavy laminar flow is verified. Then the condensation heat transfer 
coefficient is determined from 
 
FftBtu/h 875
h) 1(
s) 3600(
)h/ft 02965.0(
ft/s 2.32
2.5)145(08.1
F)ftBtu/h 358.0(145
2.5Re08.1
Re
2
3/1
2
2
22
2
22.1
3/1
222.1wavyvertical,






















l
l gkhh

 
The heat transfer surface area of the plate is 
 2ft 4ft) ft)(1 4(  LWAs 
Then the rate of heat transfer during this condensation process becomes 
 Btu/h 70,000 F)80100)(ft 4)(FftBtu/h 875()( 22sat ss TThAQ
 
The rate of condensation of steam is determined from 
 lbm/h 66.6
Btu/lbm 1051
Btu/h 000,70
*oncondensati
fgh
Q
m

 
4 ft 
80F 
Condensate 
Steam 
100F 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-102 
 
10-109 Saturated ammonia at a saturation temperature of Tsat = 25C condenses on the outer surface of vertical tube which is 
maintained at 15C by circulating cooling water. The rate of heat transfer to the coolant and the rate of condensation of 
ammonia are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 3 The tube can be treated as a vertical plate. 4 
The condensate flow is turbulent over the entire tube (this assumption will be verified). 5 The density of vapor is much 
smaller than the density of liquid, lv   . 
Properties The properties of ammonia at the saturation temperature of 25C are hfg = 116610
3
 J/kg and v = 7.809 kg/m
3
. 
The properties of liquid ammonia at the film temperature of  2/)( sat sf TTT (25 + 15)/2 = 20C are (Table A-11), 
 
463.1Pr
C W/m4927.0
CJ/kg 4745
/sm102489.0/
skg/m 10519.1
kg/m 2.610
26
4-
3







l
l
pl
lll
l
l
k
c



 
Analysis (a) The modified latent heat of vaporization is 
 
J/kg 101198=C)15C(25J/kg 47450.68+J/kg 101166
)(68.0
33
sat
*

 splfgfg TTchh
 
Assuming turbulent flow, the Reynolds number is determined from 
 
2142
253)463.1(151
)s/m 102489.0(
81.9
J/kg) 1098kg/m.s)(11 10519.1(
)1525()463.1(4927.020690.0
253Pr151
)(Pr0690.0
ReRe
3/4
5.0
3/1
22634
5.0
3/4
5.0
3/1
2*
5.0
turbvertical,








































lfgl
ssatl
v
g
h
TTLk

 
which is greater than 1800, and thus our assumption of turbulent flow is verified. Then the condensation heat transfer 
coefficient is determined from 
 
C W/m4873
)/sm 102489.0(
m/s 81.9
)2532142(463.1588750
C) W/m4927.0(2142
)253(RePr588750
Re
2
3/1
226
2
75.05.0
3/1
275.05.0turbulentvertical,
























l
l gkhh

 
The heat transfer surface area of the tube is 2m 2011.0m) m)(2 032.0(  DLAs . Then the rate of heat transfer during 
this condensation process becomes 
 W9800 C)1525)(m 2011.0)(C W/m4873()( 22sat ss TThAQ
 
(b) The rate of condensation of ammonia is determined from 
 kg/s 108.180
3-


J/kg 101198
J/s 9800
3*oncondensati
fgh
Q
m

 
Discussion Combining equations llL hk / and Lhh )3/4( , the thickness of the liquid film at the bottom of the tube is 
determined to be 
 mm 0.135m 100.135=
 C) W/m4873(3
C) W/m4927.0(4
3
4 3-
2




h
kl
L 
The assumption that the tube diameter is large relative to the thickness of the liquid film at the bottom of the tube is verified 
since the thickness of the liquid film is 0.135 mm, which is much smaller than the diameter of the tube (3.2 cm). Also, the 
assumption of turbulent flow is verified since Reynolds number is greater than 1800. 
Ammonia 
25C 
Condensate Ltube = 2 m 
D =3.2 cm 
15C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-103 
 
10-110 Saturated refrigerant-134a vapor condenses on the outside of a horizontal tube maintained at a specified temperature. 
The rate of condensation of the refrigerant is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 
Properties The properties of refrigerant-134a at the saturation temperature of 35C are hfg = 168.210
3
 J/kg and v = 43.41 
kg/m
3
. The properties of liquid R-134a at the film temperature of  2/)( sat sf TTT (35 + 25)/2 = 30C are (Table A-10), 
 
C W/m.0808.0
CJ/kg. 1448
kg/m.s 10888.1
kg/m 1188
4
3





l
pl
l
l
k
c


 
Analysis The modified latent heat of vaporization is 
 
J/kg 10178.0=C)25C(35J/kg 14480.68+J/kg 102.168
)(68.0
33
sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
C W/m1880
m) C(0.015)2535(s)kg/m 10888.1(
)C W/m0808.0)(J/kg 100.178)(kg/m 41.43)(1188kg/m 1188)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
4
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
The heat transfer surface area of the pipe is 
 2m 3299.0m) m)(7 015.0(  DLAs 
Then the rate of heat transfer during this condensation process becomes 
 W6202C)2535)(m 3299.0)(C W/m1880()( 22sat  ss TThAQ
 
The rate of condensation of steam is determined from 
 kg/min 2.09

 kg/s 0.03484
J/kg 10178.0
J/s 6202
3*oncondensati
fgh
Q
m

 
R-134a 
35C 
Condensate 
25C 
Dtube = 1.5 cm 
Ltube = 7 m 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-10410-111 Saturated refrigerant-134a vapor condenses on the outside of a horizontal tube maintained at a specified temperature. 
The rate of condensation of the refrigerant is to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 
Properties The properties of refrigerant-134a at the saturation temperature of 35C are hfg = 168.210
3
 J/kg and v = 43.41 
kg/m
3
. The properties of liquid R-134a at the film temperature of  2/)( sat sf TTT (35 + 25)/2 = 30C are (Table A-10), 
 
C W/m.0808.0
CJ/kg. 1448
/sm101590.0/
kg/m.s 10888.1
kg/m 1188
26
4
3







l
pl
lll
l
l
k
c



 
Analysis The modified latent heat of vaporization is 
 
J/kg 10178.0=C)25C(35J/kg 14480.68+J/kg 102.168
)(68.0
33
sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
C W/m1581
m) C(0.03)2535(s)kg/m 10888.1(
)C W/m0808.0)(J/kg 100.178)(kg/m 41.43)(1188kg/m 1188)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
4
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
The heat transfer surface area of the pipe is 
 2m 6597.0m) m)(7 03.0(  DLAs 
Then the rate of heat transfer during this condensation process becomes 
 W430,10C)2535)(m 6597.0)(C W/m1581()( 22sat  ss TThAQ
 
The rate of condensation of steam is determined from 
 kg/min 3.52

 kg/s 0.05859
J/kg 10178.0
J/s 430,10
3*oncondensati
fgh
Q
m

 
R-134a 
35C 
Condensate 
25C 
Dtube = 3 cm 
Ltube = 7 m 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-105 
 
10-112 Steam at a saturation temperature of Tsat = 40C condenses on the outside of a thin horizontal tube. Heat is transferred 
to the cooling water that enters the tube at 25C and exits at 35C. The rate of condensation of steam, the average overall heat 
transfer coefficient, and the tube length are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube can be taken to be isothermal at the bulk mean fluid 
temperature in the evaluation of the condensation heat transfer coefficient. 3 Liquid flow through the tube is fully developed. 
4 The thickness and the thermal resistance of the tube is negligible. 
Properties The properties of water at the saturation temperature of 
40C are hfg = 240710
3
 J/kg and v = 0.05 kg/m
3
. The properties of 
liquid water at the film temperature of  2/)( sat sf TTT (40+30)/2 
= 35C and at the bulk fluid temperature of  2/)( outin TTTb (25 
+ 35)/2 = 30C are (Table A-9), 
4.83=Pr
C W/m623.0
CJ/kg 4178
/sm10724.0/
skg/m10720.0
kg/m 0.994
26
3
3








l
pl
lll
l
l
k
c



 :C35At 
 
5.42=Pr
C W/m615.0
CJ/kg 4178
/sm10801.0/
skg/m10798.0
kg/m 0.996
26
3
3








l
pl
lll
l
l
k
c



 :C30At 
 
Analysis The mass flow rate of water and the rate of heat transfer to the water are 
 
 W58,830=C)25C)(35J/kg kg/s)(4178 408.1()(
kg/s 408.1]4/m) 03.0(m/s)[ )(2kg/m 996( 23water


inoutp
c
TTcmQ
VAm

 
 
The modified latent heat of vaporization is 
 J/kg 102435=C0)3C(40J/kg 41780.68+J/kg 102407)(68.0 33sat
*  splfgfg TTchh 
The heat transfer coefficient for condensation on a single horizontal tube is 
C W/m775,11
m) C(0.03)3040(s)kg/m 10720.0(
)C W/m623.0)(J/kg 102435)(kg/m 05.0994)(kg/m 994)(m/s 81.9(
729.0
)(
)(
729.0
2
4/1
3
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll
o


 
The average heat transfer coefficient for flow inside the tube is determined as follows: 
 
C W/m7357
m 0.03
358.9)(C) W/m615.0(Nu
9.358)42.5()906,74(023.0PrRe023.0Nu
906,74
100.801
m) m/s)(0.03 2(
Re
2
4.08.04.08.0
6-
avg







D
k
h
DV
i

 
Noting that the thermal resistance of the tube is negligible, the overall heat transfer coefficient becomes 
 C. W/m4528
2 




775,11/17357/1
1
/1/1
1
oi hh
U 
The logarithmic mean temperature difference is: 
C102.9
)5/15ln(
515
)/ln(
lm 





oi
ei
TT
TT
T 
The tube length is determined from 
m 15.1




C)102.9)(m 03.0()C W/m4528(
 W830,58
)(
 
2
lm
lm
 TDU
Q
LTUAQ s

 
Note that the flow is turbulent, and thus the entry length in this case is 10D = 0.3 m is much shorter than the total tube length. 
This verifies our assumption of fully developed flow. 
Steam 
40C 
Condensate 
25C 
Cooling 
water 
35C 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-106 
 
10-113 Saturated steam condenses on a suspended silver sphere which is initially at 25C. The time needed for the 
temperature of the sphere to rise to 50C and the amount of steam condenses are to be determined. 
Assumptions 1 The temperature of the sphere changes uniformly and thus the lumped system analysis is applicable. 2 The 
average condensation heat transfer coefficient evaluated for the average temperature can be used for the entire process. 3 
Constant properties at room temperature can be used for the silver ball. 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.60 kg/m
3
. The 
properties of the silver ball at room temperature and the properties of liquid water at the average film temperature of 
 2/)( ,sat avgsf TTT (100 + 37.5)/2 = 69C  70C are (Tables A-3 and A-9), 
C W/m429
CJ/kg 235
/sm10174
kg/m 500,10
26
3





l
p
k
c


:BallSilver 
 
C W/m663.0
CJ/kg 4190
skg/m10404.0
kg/m 5.977
3
3





l
pl
l
l
k
c


: WaterLiquid
 
Analysis The modified latent heat of vaporization is 
 
J/kg 102435=C)5.37C(100J/kg 41900.68+J/kg 102257
)(68.0
33
sat
*

 splfgfg TTchh
 
Noting that the tube is horizontal, the condensation heat transfer coefficient is determined from 
 
C W/m9916
m) C(0.012)5.37100(s)kg/m 10404.0(
)C W/m663.0)(J/kg 102435)(kg/m 60.05.977)(kg/m 5.977)(m/s 8.9(
815.0
)(
)(
815.0
2
4/1
3
33332
4/1
sat
3*
sph
























DTT
khg
hh
sl
lfgvll


 
The characteristic length and the Biot number for the lumped system analysis is (see Chap. 4) 
 
1.0 0462.0
)C W/m429(
)m 002.0)(C W/m9916(
m 002.0
6
m 012.0
6
6/
2
2
3





k
hL
Bi
D
D
D
A
L
c
c

V
 
The lumped system analysis is applicable since Bi < 0.1. Then the time needed for the temperature of the sphere to rise from 
25 to 50C is determined to be 
 
s 0.202











 tee
TT
TtT
Lc
h
c
hA
b
tbt
i
cpp
s
 
10025
10050)(
s 009.2
m) C)(0.002J/kg 235)(kg/m (10,500
C W/m9916
009.2
1-
3
2
 V
 
The total heat transfer to the ball and the amount of steam that condenses become 
kg/s 102.29
5-




J/kg 102435
J/s 81.55
J 81.55C)2550)(CJ/kg 235)(kg 009500.0(])([
kg 009500.0
6
m) 012.0(
)kg/m 500,10(
6
3*oncondensati
sphere
3
3
3
sphere
fg
ip
h
Q
m
TtTmcQ
D
m



V
 
Silver 
sphere Steam 
100C 
Ti = 25C 
1.2 cm 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-107 
 
10-114Steam at a saturation temperature of Tsat = 100C condenses on a suspended silver sphere which is initially at 25C. 
The time needed for the temperature of the sphere to rise to 50C and the amount of steam condenses during this process are 
to be determined. 
Assumptions 1 The temperature of the sphere changes uniformly and thus the lumped system analysis is applicable. 2 The 
average condensation heat transfer coefficient evaluated for the average temperature can be used for the entire process. 3 
Constant properties at room temperature can be used for the silver ball. 
Properties The properties of water at the saturation temperature of 100C are hfg = 225710
3
 J/kg and v = 0.60 kg/m
3
. The 
properties of the silver ball at room temperature and the properties of liquid water at the average film temperature of 
 2/)( ,sat avgsf TTT (100 + 37.5)/2 = 69C  70C 70C are (Tables A-3 and A-9), 
C W/m429
CJ/kg 235
/sm10174
kg/m 500,10
26
3





l
p
k
c


:BallSilver 
 
C W/m663.0
CJ/kg 4190
skg/m10404.0
kg/m 5.977
3
3





l
pl
l
l
k
c


: WaterLiquid
 
Analysis The modified latent heat of vaporization is 
 
J/kg 102435=C)5.37C(100J/kg 41900.68+J/kg 102257
)(68.0
33
sat
*

 splfgfg TTchh
 
Noting that the tube is horizontal, the condensation heat transfer coefficient is determined from 
 
C W/m7886
m) C(0.03)5.37100(s)kg/m 10404.0(
)C W/m663.0)(J/kg 102435)(kg/m 60.05.977)(kg/m 5.977)(m/s 8.9(
815.0
)(
)(
815.0
2
4/1
3
33332
4/1
sat
3*
sph
























DTT
khg
hh
sl
lfgvll


 
The characteristic length and the Biot number for the lumped system analysis is (see Chap. 4) 
 
1.0 092.0
)C W/m429(
)m 005.0)(C W/m7886(
m 005.0
6
m 03.0
6
6/
2
2
3





k
hL
Bi
D
D
D
A
L
c
s
c

V
 
The lumped system analysis is applicable since Bi < 0.1. Then the time needed for the temperature of the sphere to rise from 
25 to 50C is determined to be 
 
s 0.634











 tee
TT
TtT
Lc
h
c
hA
b
tbt
i
cpp
s
 
10025
10050)(
s 6392.0
m) C)(0.005J/kg 235)(kg/m (10,500
C W/m7886
6392.0
1-
3
2
 V
 
The total heat transfer to the ball and the amount of steam that condenses become 
kg/s 103.57
4-




J/kg 102435
J/s 5.869
J 5.869C)2550)(CJ/kg 235)(kg 148.0(])([
kg 148.0
6
m) 03.0(
)kg/m 500,10(
6
3*oncondensati
sphere
3
3
3
sphere
fg
ip
h
Q
m
TtTmcQ
D
m



V
 
Silver 
sphere Steam 
100C 
Ti = 25C 
 3 cm 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-108 
 
10-115 There is film condensation on the outer surfaces of 8 horizontal tubes arranged in a horizontal or vertical tier. The 
ratio of the condensation rate for the cases of the tubes being arranged in a horizontal tier versus in a vertical tier is to be 
determined. 
Assumptions Steady operating conditions exist. 
Analysis The heat transfer coefficients for the two cases are related 
to the heat transfer coefficient on a single horizontal tube by 
Horizontal tier: 
 tube1 ,horizontal tubesN of tier horizontal hh  
Vertical tier: 
4/1
 tube1 ,horizontal
 tubesN of tier vertical
N
h
h  
Therefore, 
 
1.68=8=
/
Ratio
1/4
4/1
4/1
 tube1 ,horizontal
 tube1 ,horizontal
 tubesN of tier vertical
 tubesN of tier horizontal
 tubesN of tier vertical
 tubesN of tier horizontal
N
Nh
h
h
h
m
m






 
Horizontal tier 
Vertical tier 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-109 
 
10-116E Saturated steam at a saturation pressure of 0.95 psia and thus at a saturation temperature of Tsat = 100F (Table A-
9E) condenses on the outer surfaces of 144 horizontal tubes which are maintained at 80F by circulating cooling water and 
arranged in a 12  12 square array. The rate of heat transfer to the cooling water and the rate of condensation of steam are to 
be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are 
isothermal. 
Properties The properties of water at the saturation temperature of 
100F are hfg = 1037 Btu/lbm and v = 0.00286 lbm/ft
3
. The 
properties of liquid water at the film temperature of 
 2/)( sat sf TTT (100 + 80)/2 = 90F are (Table A-9E), 
 
FftBtu/h 358.0
FBtu/lbm 999.0
/hft 02965.0/
hlbm/ft 842.1slbm/ft 10117.5
lbm/ft 12.62
2
4
3






l
pl
lll
l
l
k
c



 
Analysis (a) The modified latent heat of vaporization is 
 
Btu/lbm 1051=
F)80F)(100Btu/lbm 999.0(0.68+Btu/lbm 1037
)(68.0 sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
FftBtu/h 1562
ft) F(1.2/12)80100)(hlbm/ft 842.1](s) 3600h/ 1[(
)FftBtu/h 358.0)(Btu/lbm 1051)(lbm/ft 00286.012.62)(lbm/ft 12.62)(ft/s 2.32(
729.0
)(
)(
729.0
2
4/1
2
3332
4/1
sat
3*
horiz























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 4-tube high vertical tier becomes 
 FftBtu/h 2.839F)ftBtu/h 1562(
12
11 22
4/1 tube1 horiz,4/1 tubesN horiz,
 h
N
h 
The surface area for all 144 tubes is 
 2total ft 678.6= ft) ft)(15 12/2.1(144  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
 Btu/h 11,390,000 F)80100)(ft 6.678)(FBtu/h.ft 2.839()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 lbm/h 10,837
Btu/lbm 1051
Btu/h 000,390,11
*oncondensati
fgh
Q
m

 
n = 144 tubes 80F 
L = 15 ft 
P = 0.95 psia 
Cooling 
water 
Saturated 
steam 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-110 
 
10-117E Saturated steam at a saturation pressure of 0.95 psia and thus at a saturation temperature of Tsat = 100F (Table A-
9E) condenses on the outer surfaces of 144 horizontal tubes which are maintained at 80F by circulating cooling water and 
arranged in a 12  12 square array. The rate of heat transfer to the cooling water and the rate of condensation of steam are to 
be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are 
isothermal. 
Properties The properties of water at the saturation temperature of 
100F are hfg = 1037 Btu/lbm and v = 0.00286 lbm/ft
3
. The 
properties of liquid water at the film temperature of 
 2/)( sat sf TTT (100 + 80)/2 = 90F are (Table A-9E), 
 
FftBtu/h 358.0
FBtu/lbm 999.0
/hft 02965.0/
hlbm/ft 842.1slbm/ft 10117.5
lbm/ft 12.62
2
4
3






l
pl
lll
l
l
k
c



 
Analysis (a) The modified latent heat of vaporization is 
 
Btu/lbm 1051=
F)80F)(100Btu/lbm 999.0(0.68+Btu/lbm 1037
)(68.0 sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
FftBtu/h 1375
ft) F(2.0/12)80100)(hlbm/ft 842.1](s) 3600h/ 1[(
)FftBtu/h 358.0)(Btu/lbm 1051)(lbm/ft 00286.012.62)(lbm/ft 12.62)(ft/s 2.32(
729.0
)(
)(
729.0
2
4/1
2
3332
4/1
sat
3*
horiz























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 4-tube high vertical tier becomes 
 FftBtu/h 8.738F)ftBtu/h 1375(
12
11 22
4/1 tube1 horiz,4/1 tubesNhoriz,
 h
N
h 
The surface area for all 144 tubes is 
 2total ft 1131= ft) ft)(15 12/2(144  DLNAs 
Then the rate of heat transfer during this condensation process becomes 
 Btu/h 16,712,000 F)80100)(ft 1131)(FBtu/h.ft 8.738()( 22sat ss TThAQ
 
(b) The rate of condensation of steam is determined from 
 lbm/h 15,900
Btu/lbm 1051
Btu/h 000,712,16
*oncondensati
fgh
Q
m

 
n = 144 tubes 80F 
L = 15 ft 
P = 0.95 psia 
Cooling 
water 
Saturated 
steam 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-111 
 
10-118 Ammonia is liquefied in a horizontal condenser at 37C by a coolant at 20C. The average value of overall heat 
transfer coefficient and the tube length are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. 3 The thermal resistance of the tube walls is 
negligible. 
Properties The properties of ammonia at the saturation temperature of 310 K (37C) are hfg = 111310
3
 J/kg and v = 11.09 
kg/m
3
 (Table A-11). We assume a tube outer surface temperature of 31C. The properties of liquid ammonia at the film 
temperature of  2/)( sat sf TTT (37 + 31)/2 = 34C are (Table A-11) 
 
C W/m4602.0
CJ/kg 4867
skg/m10303.1
kg/m 0.589
4
3





l
pl
l
l
k
c


 
The thermal conductivity of copper is 401 W/mC (Table A-3). 
Analysis (a) The modified latent heat of vaporization is 
 
J/kg 101133=C)31C(37J/kg 48670.68+J/kg 101113
)(68.0
33
sat
*

 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
C W/m7693
m) C(0.038)3137(s)kg/m 10303.1(
)C W/m4602.0)(J/kg 101133)(kg/m 09.11)(589.0kg/m 0.589)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
4
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
Noting that there are two 2-pipe high, two 3-pipe high, and one 4-pipe high vertical tiers in the tube-layout, the average heat 
transfer coefficient is to be determined as follows 
 
C W/m5908
14
544045845664694
413222
413222
C W/m5440C) W/m7693(
4
11
C W/m5845C) W/m7693(
3
11
C W/m6469C) W/m7693(
2
11
2321
22
4/1 tube1 horiz,4/11
22
4/1 tube1 horiz,4/12
22
4/1 tube1 horiz,4/11









hhh
h
h
N
h
h
N
h
h
N
h
o
 
Let us check if the assumed value for the rube temperature was reasonable 
 
C1.31)37()038.0()5908()20()030.0()4000( tubetubetube 

TTLTL
TAhTAh oooiii

 
which is very close to the assumed value of 31C. Therefore, the assumption was good. The overall heat transfer coefficient 
based on the outer surface is determined from 
C W/m2012 2 
















 11
5908
1
)401(2
)0.3/8.3ln(038.0
4000030.0
038.01
2
)/ln(
o
ioo
ii
o
o
hk
DDD
hD
D
U 
(b) The rate of heat transfer is 
 W10833.2J/kg) 10kg/s)(1133 3600/900(
53*
oncondensati  fghmQ 
 
Then the tube length may be determined from 
 
m 4.96

LL
TAUQ oo
)2037(m) (0.038C)(14) W/m2012( W10833.2 25 

 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-112 
 
10-119 Saturated ammonia vapor at a saturation temperature of Tsat = 25C condenses on the outer surfaces of a tube bank in 
which cooling water flows. The rate of condensation of ammonia, the overall heat transfer coefficient, and the tube length are 
to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. 3 The thermal resistance of the tube walls is 
negligible. 
Properties The properties of ammonia at the saturation temperature of 25C are hfg = 116610
3
 J/kg and v = 7.809 kg/m
3
 
(Table A-11). We assume that the tube temperature is 20C. Then, the properties of liquid ammonia at the film temperature of 
 2/)( sat sf TTT (25 + 20)/2 = 22.5C are (Table A-11) 
 
C W/m4869.0
CJ/kg 4765
skg/m10479.1
kg/m 5.606
4
3





l
pl
l
l
k
c


 
The water properties at the average temperature of (14+17)/2 = 15.5C are (Table A-9) 
 
7.98Pr
C W/m590.0
skg/m10124.1
CJ/kg 4185
kg/m 0.999
3
3






k
c p


 
Analysis (a) The modified latent heat of vaporization is 
 
J/kg 101182=
C)20C(25J/kg 47650.68+J/kg 101166
)(68.0
3
3
sat
*


 splfgfg TTchh
 
The heat transfer coefficient for condensation on a single horizontal tube is 
 
C W/m9280
m) C(0.025)2025(s)kg/m 10479.1(
)C W/m4869.0)(J/kg 101182)(kg/m 809.7)(606.5kg/m 5.606)(m/s 8.9(
729.0
)(
)(
729.0
2
4/1
4
33332
4/1
sat
3*
horizontal
























DTT
khg
hh
sl
lfgvll


 
Then the average heat transfer coefficient for a 4-pipe high vertical tier becomes 
 C W/m6562C) W/m9280(
4
11 22
4/1 tube1 horiz,4/1 tubesN horiz,
 h
N
hho 
The rate of heat transfer in the condenser is 
 
 W10970.1)1417)(CJ/kg 4185)(kg/s 69.15()(
kg/s 69.15)m/s 2()m 025.0)(25.0()kg/m 999(1616
5
inout
23


TTcmQ
Am
p
c

  V
 
Then the rate of condensation becomes 
 kg/s 0.167



J/kg 101182
 W10970.1
3
5
*cond
fgh
Q
m

 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-113 
 
(b) For the calculation of the heat transfer coefficient on the inner surfaces of the tubes, we first determine the Reynolds 
number 
 440,44
skg/m 101.124
)kg/m m)(999.0 m/s)(0.025 2(
Re
3-
3




VD 
which is greater than 10,000. Therefore, the flow is turbulent. Assuming fully developed flow, the Nusselt number and the 
heat transfer coefficient are determined to be 
 9.275)98.7()440,44(023.0PrRe023.0 4.08.04.08.0 Nu 
 C W/m6511)9.275(
m 0.025
C) W/m590.0( 2 

 Nu
D
k
hi 
Let us check if the assumed value for the rube temperature was reasonable 
 
C3.20
)25)(6562()5.15)(6511(
tube
tubetube



T
TT
ThTh ooii
 
which is sufficiently close to the assumed value of 20C. Disregarding thermal resistance of the tube walls, the overall heat 
transfer coefficient is determined from 
 C W/m3268 2 














 11
6562
1
6511
111
oi hh
U 
(c) The tube length may be determined from 
 
m 5.05








L
L
TUAQ
)1714(
2
1
25m) (0.025C)(16) W/m3268( W10970.1 25 

 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-114 
 
10-120 Saturated steam at 270.1 kPa pressure and thus at a saturation temperature of Tsat = 130C (Table A-9) condenses 
inside a horizontal tube which is maintained at 110C. The average heat transfer coefficient and the rate of condensation of 
steam are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The tube is isothermal. 3 The vapor velocity is low so that Revapor < 
35,000. 
Properties The properties of water at the saturation temperature of 130C are hfg = 217410
3
 J/kg and v = 1.50 kg/m
3
. The 
properties of liquid water at the film temperature of  2/)( sat sf TTT (130 + 110)/2 = 120C are (Table A-9), 
 
C W/m.683.0
CJ/kg. 4244
/sm10246.0/
kg/m.s10232.0
kg/m 4.943
26
3
3







l
pllll
l
l
k
c



 
Analysis The condensation heat transfer coefficient is determined from 
C W/m8413
2 




































4/1
3
3
3332
4/1
sat
sat
3
internal
C110)C)(130J/kg 4244(
8
3
+J/kg 102174 
m) C(0.025110)s)(130kg/m 10232.0(
C) W/m683.0)(kg/m )50.14.943)(kg/m 4.943)(m/s 8.9(
555.0
)(
8
3
)(
)(
555.0 splfg
sl
lvll TTch
DTT
kg
hh


 
The heat transfer surface area of the pipe is 
 2m 7854.0m) m)(10 025.0(  DLAs 
Then the rate of heat transfer during this condensation process becomes 
 W151,132C)110130)(m 7854.0)(C W/m8413()( 22sat  ss TThAQ
 
The rate of condensation of steam is determined from 
 kg/s 0.0608


J/kg 102174
J/s 151,132
3oncondensati
fgh
Q
m

 
 
Steam 
270.1 kPa 
Condensate 
110C 
Dtube = 2.5 cm 
Ltube = 10 m 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-115 
 
Fundamentals of Engineering (FE) Exam Problems 
 
 
10-121 When boiling a saturated liquid, one must be careful while increasing the heat flux to avoid “burnout.” Burnout 
occurs when the boiling transitions from _____ boiling. 
(a) convection to nucleate (b) convection to film (c) film to nucleate 
(d) nucleate to film (e) none of them 
 
Answer (d) nucleate to film 
 
 
 
 
10-122 Heat transfer coefficients for a vapor condensing on a surface can be increased by promoting 
(a) film condensation (b) dropwise condensation (c) rolling action (d) none of them 
 
Answer (b) dropwise condensation 
 
 
 
 
10-123 At a distance x down a vertical, isothermal flat plate on which a saturated vapor is condensing in a continuous film, 
the thickness of the liquid condensate layer is δ. The heat transfer coefficient at this location on the plate is given by 
(a) /lk (b) fh (c) fgh (d) gh (e) none of them 
Answer (a) /lk 
 
 
 
 
10-124 When a saturated vapor condenses on a vertical, isothermal flat plate in a continuous film, the rate of heat transfer is 
proportional to 
(a) 4/1sat )( TTs  (b) 
2/1
sat )( TTs  (c) 
4/3
sat )( TTs  (d) )( satTTs  (e) 
3/2
sat )( TTs  
 
Answer (c) 4/3sat )( TTs  
 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-116 
 
10-125 Saturated water vapor is condensing on a 0.5 m
2
 vertical flat plate in a continuous film with an average heat transfer 
coefficient of 7 kW/m
2
K. The temperature of the water is 80
o
C (hfg = 2309 kJ/kg) and the temperature of the plate is 60
o
C. 
The rate at which condensate is being formed is 
(a) 0.0303 kg/s (b) 0.07 kg/s (c) 0.15 kg/s (d) 0.24 kg/s (e) 0.28 kg/s 
 
Answer (a) 0.0303 kg/s 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
hfg=2309 [kJ/kg] 
dT=20 [C] 
A=0.5 [m^2] 
h=7 [kJ/m^2-K-s] 
mdot=h*A*dT/hfg 
 
 
 
 
 
10-126 Steam condenses at 50ºC on a 0.8-m-high and 2.4-m-wide vertical plate that is maintained at 30ºC. The condensation 
heat transfer coefficient is 
(a) 3975 W/m
2
ºC (b) 5150 W/m
2
ºC (c) 8060 W/m
2
ºC (d) 11,300 W/m
2
ºC (e) 14,810 W/m
2
ºC 
(For water, use l = 992.1 kg/m
3
, l = 0.65310
-3
 kg/ms, kl = 0.631 W/m°C, cpl = 4179 J/kg°C, hfg @ Tsat = 2383 kJ/kg) 
 
Answer (b) 5150 W/m
2
ºC 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
T_sat=50 [C] 
T_s=30 [C] 
L=0.8 [m] 
w=2.4 [m] 
h_fg=2383E3 [J/kg] "at 50 C from Table A-9" 
"The properties of water at (50+30)/2=40 C are (Table A-9)" 
rho_l=992.1 [kg/m^3] 
mu_l=0.653E-3 [kg/m-s] 
nu_l=mu_l/rho_l 
c_p_l=4179 [J/kg-C] 
k_l=0.631 [W/m-C] 
g=9.81 [m/s^2] 
h_fg_star=h_fg+0.68*c_p_l*(T_sat-T_s) 
Re=(4.81+(3.70*L*k_l*(T_sat-T_s))/(mu_l*h_fg_star)*(g/nu_l^2)^(1/3))^0.820 
"Re is between 30 and 1800, and therefore there is wavy laminar flow" 
h=(Re*k_l)/(1.08*Re^1.22-5.2)*(g/nu_l^2)^(1/3) 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-117 
 
10-127 An air conditioner condenser in an automobile consists of 2 m
2
 of tubular heat exchange area whose surface 
temperature is 30
o
C. Saturated refrigerant 134a vapor at 50
o
C (hfg = 152 kJ/kg) condenses on these tubes. What heat transfer 
coefficient must exist between the tube surface and condensing vapor to produce 1.5 kg/min of condensate? 
(a) 95 W/m
2
K (b) 640 W/m
2
K (c) 727 W/m
2
K (d) 799 W/m
2
K (e) 960 W/m
2
K 
 
Answer (a) 95 W/m
2
K 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
hfg=152000 [J/kg] 
dT=20 [C] 
A=2 [m^2] 
mdot=(1.5/60) [kg/s] 
Q=mdot*hfg 
Q=h*A*dT 
 
 
 
 
 
10-128 Saturated water vapor at 40C is to be condensed as it flows through a tube at a rate of 0.2 kg/s. The condensate 
leaves the tube as a saturated liquid at 40C. The rate of heat transfer from the tube is 
(a) 34 kJ/s (b) 268 kJ/s (c) 453 kJ/s (d) 481 kJ/s (e) 515 kJ/s 
 
Answer (d) 481 kJ/s 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
T1=40 [C] 
m_dot=0.2 [kg/s] 
h_f=ENTHALPY(Steam_IAPWS,T=T1,x=0) 
h_g=ENTHALPY(Steam_IAPWS,T=T1,x=1) 
h_fg=h_g-h_f 
Q_dot=m_dot*h_fg 
 
"Wrong Solutions:" 
W1_Q=m_dot*h_f "Using hf" 
W2_Q=m_dot*h_g "Using hg" 
W3_Q=h_fg "not using mass flow rate" 
W4_Q=m_dot*(h_f+h_g) "Adding hf and hg" 
 
 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-118 
 
10-129 Steam condenses at 50ºC on the outer surface of a horizontal tube with an outer diameter of 6 cm. The outer surface of 
the tube is maintained at 30ºC. The condensation heat transfer coefficient is 
(a) 5493 W/m
2
ºC (b) 5921 W/m
2
ºC (c) 6796 W/m
2
ºC (d) 7040 W/m
2
ºC (e) 7350 W/m
2
ºC 
(For water, use l = 992.1 kg/m
3
, l = 0.65310
-3
 kg/ms, kl = 0.631 W/m°C, cpl = 4179 J/kg°C, hfg @ Tsat = 2383 kJ/kg) 
 
Answer (c) 6796 W/m
2
ºC 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
screen. 
T_sat=50 [C] 
T_s=30 [C] 
D=0.06 [m] 
h_fg=2383E3 [J/kg] "at 50 C from Table A-9" 
rho_v=0.0831 [kg/m^3] 
"The properties of water at (50+30)/2=40 C are (Table A-9)" 
rho_l=992.1 [kg/m^3] 
mu_l=0.653E-3 [kg/m-s] 
c_p_l=4179 [J/kg-C] 
k_l=0.631 [W/m-C] 
g=9.81 [m/s^2] 
h_fg_star=h_fg+0.68*c_p_l*(T_sat-T_s) 
h=0.729*((g*rho_l*(rho_l-rho_v)*h_fg_star*k_l^3)/(mu_l*(T_sat-T_s)*D))^0.25 
 
 
PROPRIETARY MATERIAL. © 2015 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course 
preparation. If you are a student using this Manual, you are using it without permission. 
10-119 
 
10-130 Steam condenses at 50ºC on a tube bank consisting of 20 tubes arranged in a rectangular array of 4 tubes high and 5 
tubes wide. Each tube has a diameter of 6 cm and a length of 3 m and the outer surfaces of the tubes are maintained at 30ºC. 
The rate of condensation of steam is 
(a) 0.054 kg/s (b) 0.076 kg/s (c) 0.315 kg/s (d) 0.284 kg/s (e) 0.446 kg/s 
(For water, use l = 992.1 kg/m
3
, l = 0.65310-3
 kg/ms, kl = 0.631 W/m°C, cpl = 4179 J/kg°C, hfg @ Tsat = 2383 kJ/kg) 
 
Answer (c) 0.315 kg/s 
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES 
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T_sat=50 [C] 
T_s=30 [C] 
D=0.06 [m] 
L=3 [m] 
N=4 
N_total=5*N 
 
h_fg=2383E3 [J/kg] "at 50 C from Table A-9" 
rho_v=0.0831 [kg/m^3] "at 50 C from Table A-9" 
"The properties of water at (50+30)/2=40 C are (Table A-9)" 
rho_l=992.1 [kg/m^3] 
mu_l=0.653E-3 [kg/m-s] 
c_p_l=4179 [J/kg-C] 
k_l=0.631 [W/m-C] 
g=9.81 [m/s^2] 
h_fg_star=h_fg+0.68*c_p_l*(T_sat-T_s) 
h_1tube=0.729*((g*rho_l*(rho_l-rho_v)*h_fg_star*k_l^3)/(mu_l*(T_sat-T_s)*D*N))^0.25 
h_Ntubes=1/N^0.25*h_1tube 
A_s=N_total*pi*D*L 
Q_dot=h_Ntubes*A_s*(T_sat-T_s) 
m_dot_cond=Q_dot/h_fg_star 
 
 
 
 
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