7-101
V [m/s] t [s] h [W/m2∙K] Re Bi
1 2.906 718.4 4.227 0.003421
2 2.558 816.2 8.453 0.003886
3 2.339 892.6 12.68 0.004251
4 2.179 958.0 16.91 0.004562
5 2.054 1016 21.13 0.004839
6 1.953 1069 25.36 0.005091
7 1.867 1118 29.59 0.005325
8 1.793 1164 33.81 0.005544
9 1.729 1208 38.04 0.005751
10 1.671 1249 42.27 0.005948
20 1.314 1589 84.53 0.007566
30 1.124 1857 126.8 0.008841
40 1.001 2086 169.1 0.009935
50 0.911 2292 211.3 0.010910
60 0.8421 2479 253.6 0.011800
70 0.7869 2653 295.9 0.012630
80 0.7413 2816 338.1 0.013410
90 0.7027 2971 380.4 0.014150
100 0.6696 3118 422.7 0.014850
020 40 60 80 100
500
1000
1500
2000
2500
3000
3500
0.5
1
1.5
2
2.5
3
V [m/s]
h [W/m2·K]
t [s]
t
h
7-102
7-101 A glass spherical tank that is filled with chemicals undergoing exothermic reaction, has a known inner
surface temperature. The outer surface of the tank is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with local
atmospheric pressureat 1 atm. 4 One-dimensional heat conduction through tank wall. 5 The thermal conductivity of the tank
wallis constant.
Properties The thermal conductivity of the tank wall is given to be ktank = 1.1 W/mK.
The properties of air at T∞ = 15°C are k = 0.02476 W/m∙K,ν = 1.470 10−5 m2/s,μ∞ = 1.802 10−5kg/m∙s, and Pr = 0.7323
(Table A-15). The dynamic viscosity of air at the surface Ts,o is to be solved using EES.
Analysis The convection heat transfer coefficient on the outer surface can be determined using the Nusselt number relation
for flow across a sphere. The Reynolds number and the Nusselt number for flow across a sphere are
o
VD
4/1
4.03/25.0 Pr]Re06.0Re4.0[2Nu
s
o
k
hD
The inner and outer radii of the tank are
m5.0=
i
r
and
m 51.0m)01.05.0( =+=
o
r
From Chapter 2 the rate of heat transfer at the tank’s
outer surface can be expressed as
convsph QQ =
))(4(4 ,
2
,,
tank
−=
−
−TTrh
rr
TT
rrk oso
io
osis
oi
)( ,
,,
tank
−=
−
−TTh
rr
TT
r
r
kos
io
osis
o
i
where
h = 70 W/m2 K, Ts,i = 80°C, and T∞ = 15°C.
The problem is solved using EES, and the solution is given below:
“GIVEN”
“h=70 [W/(m^2*K)]””convection heat transfer coefficient”
r_i=0.5 [m] “inner radius”
r_o=r_i+0.010 [m] “outer radius”
T_s_i=80 [C] “inner surface temperature”
T_infinity=15 [C] “ambient temperature”
V=5 [m/s]
“PROPERTIES”
“Air”
Fluid$=’air’
k=Conductivity(Fluid$, T=T_infinity)
Pr=Prandtl(Fluid$, T=T_infinity)
rho=Density(Fluid$, T=T_infinity, P=101.3)
mu_infinity=Viscosity(Fluid$, T=T_infinity)
mu_s=Viscosity(Fluid$, T=T_s_o)
nu=mu_infinity/rho
7-103
h=Nusselt*k/D_o
“SOLVING FOR OUTER SURFACE TEMPERATURE AND k_avg”
7-105
7-103 The components of an electronic system located in a horizontal duct is cooled by air flowing over the duct. The total
power rating of the electronic device is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant
properties.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (65+30)/2 = 47.5C are (Table A-15)
7235.0Pr
/sm 10774.1
C W/m.02717.0
25–
=
=
=
k
For a location at 3000 m altitude where the atmospheric
pressure is 70.12 kPa, only kinematic viscosity of air
will be affected. Thus,
/sm 10563.2)10774.1(
12.70
325.101 255
kPa 66.61@
−− =
=
Analysis The Reynolds number is
4
m) (0.2m/s (200/60)
VD
Air
30C
200 m/min
20 cm
65C
1.5 m
20 cm
7-106
7-104 An ASTM A479 904L stainless steel bar connects two metal plates. Hot air flows across the square bar. A cooling
mechanism removes heat at a rate of 50 W from the bar. The surface temperature of the bar is to be determined to see if it
exceeds the maximum use temperature of 260°C.
Assumptions 1 The flow is steady and incompressible. 2 Uniform surface temperature. 3 Wall effects from the plates on the
bar are negligible.
Properties Using the maximum use temperature for the bar as the surface temperature, the properties of air at the film
temperature of Tf = (Ts + T∞)/2 = (260 + 340)°C/2 = 300°C are (Table A–15): Pr = 0.6935, k = 0.04418 W/m∙K, and ν = 4.765
× 10−5 m2/s
Analysis The Reynolds number is
7-107
7-105 A street sign surface is subjected to radiation and cross flow wind, the surface temperature of the street sign is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The surface temperature is constant. 4 The
street sign is treated a vertical plate in cross flow.
Properties The properties of air (1 atm) at 30°C are given in Table A-15: k = 0.02588 W/m∙K,
= 1.608 10−5 m2/s, and Pr
= 0.7282.
Analysis The Reynolds number is
4
)m 2.0)(m/s 1(
VD
7-109
7-107 A coated sheet is being dried with hot air in cross flow. The effect of air velocity on the convection heat transfer
coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The surface temperature is constant. 4 The
coated sheet is treated a vertical plate in cross flow.
Analysis The problem is solved using EES, and the solution is given below:
“GIVEN”
T_infinity=110 [C]
T_s=90 [C]
L=1 [m]
“PROPERTIES”
Fluid$=’air’
T_film=1/2*(T_s+T_infinity)
k=Conductivity(Fluid$, T=T_film)
Pr=Prandtl(Fluid$, T=T_film)
rho=Density(Fluid$, T=T_film, P=101.3)
mu=Viscosity(Fluid$, T=T_film)
nu=mu/rho
“ANALYSIS”
V [m/s]
7-110
Flow across Tube Banks
7-108C The level of turbulence, and thus the heat transfer coefficient, increases with row number because of the combined
7-110 Air is heated by hot tubes in a tube bank. The average heat transfer coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is constant.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the
assumed mean temperature of 70C and 1 atm based on the problem statement (Table A-15):
k = 0.02881 W/m-K = 1.028 kg/m3
cp=1.007 kJ/kg-K Pr = 0.7177
= 2.05210-5 kg/m-s Prs = Pr@ Ts = 140C = 0.7041
Analysis It is given that D = 0.02 m, SL= ST= 0.06 m, and
D
SL
V=6 m/s
Ts=140C
7-111
7-111 Water is heated by a bundle of resistance heater rods. The number of tube rows is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the rods is constant.
Properties The properties of water at the mean temperature of (15C +65C)/2=40C are (Table A-9):
k = 0.631 W/mK = 992.1 kg/m3
cp=4.179 kJ/kgK Pr = 4.32
= 0.65310-3 kg/ms Prs = Pr@ Ts = 90C = 1.96
Also, the density of water at the inlet temperature of 15C (for use in the mass flow rate calculation at the inlet) is i =999.1
kg/m3.
Analysis It is given that D = 0.01 m, SL= 0.04 m and ST= 0.03 m, and V=
0.8 m/s. Then the maximum velocity and the Reynolds number based on
the maximum velocity become
m/s 20.1m/s) 8.0(
01.003.0
03.0
max =
−
=
−
=V
DS
S
V
T
T
232,18
skg/m 10653.0
m) m/s)(0.01 20.1)(kg/m 1.992(
Re 3
3
max =
== −
DV
D
The average Nusselt number is determined using the proper relation from
Table 7-2 to be
3.269
)96.1/32.4()32.4()232,18(27.0
)Pr(Pr/PrRe27.0Nu
25.036.063.0
25.036.063.0
=
=
=sDD
Assuming that NL> 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become
3.269NuNu ,== DND L
C W/m994,16
m 0.01
C) W/m631.0(3.269 2
,=
== D
kNu
hL
ND
Consider one-row of tubes in the transpose direction (normal to flow), and thus take NT =1. Then the heat transfer surface
area becomes
LLtubes NNDLNA 1257.0 m) m)(4 01.0()1( ===
Then the log mean temperature difference, and the expression for the rate of heat transfer become
C51.45
)]6590/()1590ln[(
)6590()1590(
)]/()ln[(
)()( =
−−
−−−
=
−−
−−−
=
esis
esis
lm TTTT
TTTT
T
LLlmsNNThAQ220,97)C51.45()C)(0.1257 W/m994,16(2===
The mass flow rate of water through a cross-section corresponding to NT =1 and the rate of heat transfer are
kg/s 91.95m/s) )(0.8m 0.03)(4kg/m 1.999(23 === VAm c
W10004.2C)1565(J/kg.C) kg/s)(4179 91.95()( 7
=−=−= iep TTcmQ
Substituting this result into the heat transfer expression above we find the number of tube rows
206=→=→= LLlmsNNThAQ 220,97 W10004.2 7
SL
ST
V=0.8 m/s
Ti=15C
Ts=90C
D
7-112
7-112 Combustion air is heated by condensing steam in a tube bank. The rate of heat transfer to air, the pressure drop of air,
and the rate of condensation of steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the temperature of steam.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the
assumed mean temperature of 35C based on the problem statement (will be checked later) and 1 atm (Table A-15):
k = 0.02625 W/mK = 1.145 kg/m3
cp=1.007 kJ/kgK Pr = 0.7268
= 1.89510-5 kg/ms Prs = Pr@ Ts = 100C = 0.7111
Also, the density of air at the inlet temperature of 20C (for use in the mass flow rate calculation at the inlet) is i = 1.204
kg/m3. The enthalpy of vaporization of water at 100C is hfg= 2257 kJ/kg-K (Table A-9).
Analysis (a) It is given that D = 0.016 m, SL= ST= 0.04 m, and V= 5.2 m/s.
Then the maximum velocity and the Reynolds number based on the
maximum velocity become
m/s 667.8m/s) 2.5(
016.004.0
04.0
max =
−
=
−
=V
DS
S
V
T
T
since
2/)( DSS
TD
+
8379
skg/m 10895.1
m) m/s)(0.016 667.8)(kg/m 145.1(
Re 5
3
max =
== −
DV
D
The average Nusselt number is determined using the proper relation from
Table 7-2 to be
87.70)7111.0/7268.0()7268.0()8379()04.0/04.0(35.0
)Pr(Pr/PrRe)/(35.0Nu
25.036.06.02.0
25.036.06.02.0
==
=sDLTD SS
Since NL =20, which is greater than 16, the average Nusselt number and
heat transfer coefficient for all the tubes in the tube bank become
87.70NuNu ,== DND L
C W/m3.116
m 0.016
C) W/m02625.0(87.70 2
,=
== D
kNu
hL
ND
The total number of tubes is N = NLNT = 2010 = 200. For the given tube length (L = 3 m), the heat transfer surface area
and the mass flow rate of air (evaluated at the inlet) are
2
m .1630 m) m)(3 016.0(200 ===
DLNAs
kg/s 513.7m) m)(3 .04m/s)(10)(0 2.5)(kg/m 204.1()( 3==== LSNVmm TTii
Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become
C) W/m3.116)(m 16.30(
22
−−−=
s
hA
SL
ST
V=5.2 m/s
Ti=20C
Ts=100C
D
7-113
7-113 Combustion air is heated by condensing steam in a tube bank. The rate of heat transfer to air, the pressure drop of air,
and the rate of condensation of steam are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the temperature of steam.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the
assumed mean temperature of 35C based on the problem statement (will be checked later) and 1 atm (Table A-15):
k = 0.02625 W/mK = 1.145 kg/m3
cp=1.007 kJ/kgK Pr = 0.7268
= 1.89510-5 kg/ms Prs = Pr@ Ts = 100C = 0.7111
Also, the density of air at the inlet temperature of 20C (for use in the mass flow rate calculation at the inlet) is i = 1.204
kg/m3. The enthalpy of vaporization of water at 100C is hfg= 2257 kJ/kg-K (Table A-9).
Analysis (a) It is given that D = 0.016 m, SL= ST= 0.06 m, and V= 5.2 m/s.
Then the maximum velocity and the Reynolds number based on the
maximum velocity become
m/s 091.7m/s) 2.5(
016.006.0
06.0
max =
−
=
−
=V
DS
S
V
T
T
6855
skg/m 10895.1
m) m/s)(0.016 091.7)(kg/m 145.1(
Re 5
3
max =
== −
DV
D
The average Nusselt number is determined using the proper relation from
Table 7-2 to be
17.63)7111.0/7268.0()7268.0()6855(27.0
)Pr(Pr/PrRe27.0Nu
25.036.063.0
25.036.063.0
==
=sDD
Since NL =20, which is greater than 16, the average Nusselt number and
heat transfer coefficient for all the tubes in the tube bank become
17.63NuNu ,== DND L
C W/m6.103
m 0.016
C) W/m02625.0(17.63 2
,=
== D
kNu
hL
ND
The total number of tubes is N = NLNT = 2010 = 200. For the given tube length (L = 3 m), the heat transfer surface area
and the mass flow rate of air (evaluated at the inlet) are
2
m .1630 m) m)(3 016.0(200 ===
DLNAs
kg/s 27.11m) m)(3 .06m/s)(10)(0 2.5)(kg/m 204.1()( 3==== LSNVmm TTii
Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become
C) W/m6.103)(m 16.30(
22
−−−=
s
hA
SL
ST
V=5.2 m/s
Ti=20C
Ts=100C
D
7-114
7-114 Water is preheated by exhaust gases in a tube bank. The rate of heat transfer, the pressure drop of exhaust gases, and
the temperature rise of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 For exhaust gases, air properties are used.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the
assumed mean temperature of 250C based on the problem statement (will be checked later) and 1 atm (Table A-15):
k = 0.04104 W/mK = 0.6746 kg/m3
cp=1.033 kJ/kgK Pr = 0.6946
= 2.7610-5 kg/ms Prs = Pr@ Ts = 80C = 0.7154
The density of air at the inlet temperature of 300C (for use in the mass flow rate calculation at the inlet) is
i = 0.6158
kg/m3. The specific heat of water at 80C is 4.197 kJ/kg.C (Table A-9).
Analysis (a) It is given that D = 0.021 m, SL= ST= 0.08 m, and V= 4.5 m/s.
Then the maximum velocity and the Reynolds number based on the
maximum velocity become
m/s 102.6m/s) 5.4(
021.008.0
08.0
max =
−
=
−
=V
DS
S
V
T
T
3132
skg/m 1076.2
m) m/s)(0.021 102.6)(kg/m 6746.0(
Re 5
3
max =
== −
DV
D
The average Nusselt number is determined using the proper relation from
Table 7-2 to be
46.37)7154.0/6946.0()6946.0()3132(27.0
)Pr(Pr/PrRe27.0Nu
25.036.063.0
25.036.063.0
==
=sDD
Since NL =16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become
46.37NuNu ,== DND L
C W/m2.73
m 0.021
C) W/m04104.0(46.37 2
,=
== D
kNu
hL
ND
The total number of tubes is N = NLNT = 168 = 128. For a unit tube length (L = 1 m), the heat transfer surface area and the
mass flow rate of air (evaluated at the inlet) are
2
m .4458 m) m)(1 021.0(128 ===
DLNAs
kg/s 774.1m) m)(1 08m/s)(8)(0. 5.4)(kg/m 6158.0()( 3==== LSNVmm TTii
Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become
C) W/m2.73)(m 445.8(
22
−−−=
s
hA
SL
ST
V=4.5 m/s
Ti=300C
Ts=80C
D
7-118
7-118 Combustion air is preheated by hot water in a tube bank. The rate of heat transfer to air and the pressure drop of air are
to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the temperature of hot
water.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air properties at the
assumed mean temperature of 20C (will be checked later) and 1 atm (Table A-15):
k = 0.02514 W/mK = 1.204 kg/m3
cp =1.007 kJ/kgK Pr = 0.7309
= 1.82510-5 kg/ms Prs = Pr@ Ts = 90C = 0.7132
Also, the density of air at the inlet temperature of 15C (for use in the mass flow rate calculation at the inlet) is i = 1.225
kg/m3.
Analysis It is given that D = 0.022 m, SL = ST = 0.06 m, and V = 4.5
m/s. Then the maximum velocity and the Reynolds number based on
the maximum velocity become
m/s 105.7m/s) 5.4(
022.006.0
06.0
max =
−
=
−
=V
DS
S
V
T
T
since
2/)( DSS
TD
+
313,10
skg/m 10825.1
m) m/s)(0.022 105.7)(kg/m 204.1(
Re 5
3
max =
== −
DV
D
The average Nusselt number is determined using the proper relation from
Table 7-2 to be
49.80)7132.0/7309.0()7309.0()313,10()06.0/06.0(35.0
)Pr(Pr/PrRe)/(35.0Nu
25.036.06.02.0
25.036.06.02.0
==
=sDLTD SS
This Nusselt number is applicable to tube banks with NL > 16. In our case the number of rows is NL = 8, and the
corresponding correction factor from Table 7-3 is F = 0.967. Then the average Nusselt number and heat transfer coefficient
for all the tubes in the tube bank become
84.77)49.80)(967.0(NuNu ,=== DND F
L
C W/m94.88
m 0.022
C) W/m02514.0(84.77 2
,=
== D
kNu
hL
ND
The total number of tubes is N = NL NT = 88 = 64. For a unit tube length (L = 1 m), the heat transfer surface area and the
mass flow rate of air (evaluated at the inlet) are
2
m .4234 m) m)(1 022.0(64 ===
DLNAs
kg/s 646.2m) m)(1 06m/s)(8)(0. 5.4)(kg/m 225.1()( 3==== LSNVmm TTii
Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become
C29.25
C)J/kg kg/s)(1007 (2.646
C) W/m94.88)(m 423.4(
exp)1590(90exp)(
22
=
−−−=
−−−=
p
s
isse cm
hA
TTTT
C73.69
)]29.2590/()1590ln[(
)29.2590()1590(
)]/()ln[(
)()( =
−−
−−−
=
−−
−−−
=
esis
esis
lm TTTT
TTTT
T
W27,430=== )C73.69)(m C)(4.423 W/m94.88(22
lmsThAQ
For this staggered tube bank, the friction coefficient corresponding to ReD = 10,313 and ST/D = 6/2.2 = 2.73 is, from Fig. 7-
27b, f = 0.27. Also, = 1 for the square arrangements. Then the pressure drop across the tube bank becomes
Pa 65.6=
== 2
23
2
max
m/skg 1
N 1
2
m/s) 105.7)(kg/m 204.1(
)1)(27.0(8
2
V
fNP L
Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (15 +25.3)/2 = 20.2C, which is fairly close to the assumed
value of 20C. Therefore, there is no need to repeat calculations.
SL
ST
V=4.5 m/s
Ti=15C
Ts=90C
D
7-119
Review Problems
7-119 Air flows over a plate. Various quantities are to be determined at x = xcr.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Recr = 5105. 3 Air is an ideal gas.
4 The plate is smooth. 5 Edge effects are negligible and the upper surface of the plate is considered.
Properties The properties of air at the film temperature of Tf = (Ts + T)/2 = (65+15)/2 = 40C are (Table A-15)
7255.0Pr , skg/m 10918.1 , KW/m 02662.0k KJ/kg 1007c ,kg/m 127.1 5
p
3===== −
Analysis The critical length of the plate is
m 8364.2
)kg/m m/s)(1.127 3(
s)kg/m 10918.1)(105(
Re
3
55
cr
cr =
==
−
V
x
The calculations at x = 2.84 m are
(a) Hydrodynamic boundary layer thickness, Eq. 7–12a:
0.0197m=
==
5
105
m) 84.2(91.4
Re
91.4
x
x
(b) Local friction coefficient, Eq. 7-12b:
0.00094=== −− 2/152/1
,)105(664.0Re664.0 xxf
C
(c) Average friction coefficient, Eq. 7–14:
0.00188=
== 2/152/1 )105(
33.1
Re
33.1
x
f
C
(d) Total drag force due to friction, Eq. 7-1:
N 0.00086=== 2
m/s) )(3kg/m 127.1(
)m 3.03.0)(00188.0(
2
23
2
2
V
ACF sff
(e) Local convection heat transfer coefficient, Eq. 7-19:
9.210)7255.0()105(332.0PrRe332.0Nu 3/12/153/12/1
xx ===
K W/m1.98 2=
== )9.210(
m .83642
W/m02662.0
Nu x
K
x
k
hx
(f) Average convection heat transfer coefficient, Eq. 7-21:
8.421Nu2)7255.0()105(664.0PrRe664.0Nu x
3/12/153/12/1 ====
K W/m3.96 2==
== x
h
K
x
k
h2)8.421(
m 3.0
W/m02662.0
Nu x
(g) Rate of convective heat transfer, Eq. 7-9:
W17.8=C15))(65m 0.3)(0.3 W/m96.3()( 22 −=−= KTThAQss
Ts = 65C
Air
V = 3 m/s
T = 15C
7-120
7-120 Air is flowing in parallel to a stationary thin flat plate over the top and bottom surfaces: (a) the average friction
coefficient, (b) the average convection heat transfer coefficient, (c) the average convection heat transfer coefficient using the
modified Reynolds analogy are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The edge effects are negligible. 4 The critical
Reynolds number is Recr = 5105.
Properties The properties of air (1 atm) at the film temperature of Tf = (Ts + T∞)/2 = 20°C are given in Table A-15:
k = 0.02514 W/m∙K,
= 1.516 10−5 m2/s, and Pr = 0.7309.
Analysis (a) The Reynolds at the trailing edge of the plate is
5
)m 1)(m/s 2(
VL