9-61
9-66 A cylindrical propane tank is exposed to calm ambient air. The propane is slowly vaporized due to a crack developed at
the top of the tank. The time it will take for the tank to empty is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas
with constant properties. 3 The local atmospheric pressure is 1 atm. 4
Radiation heat transfer is negligible.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (-42+25)/2 = -8.5C are (Table A-15)
1–
25
K 003781.0
K)2735.8(
11
7383.0Pr
/sm 10265.1
C W/m.02299.0
=
+−
==
=
=
=
−
f
T
k
Analysis The tank gains heat through its cylindrical surface as well as its circular end surfaces. For convenience, we take the
heat transfer coefficient at the end surfaces of the tank to be the same as that of its side surface. (The alternative is to treat the
end surfaces as a vertical plate, but this will double the amount of calculations without providing much improvement in
accuracy since the area of the end surfaces is much smaller and it is circular in shape rather than being rectangular). The
characteristic length in this case is the outer diameter of the tank,
m. 5.1== DLc
Then,
10
225
3-12
2
3
10869.3)7383.0(
)/sm 10265.1(
)m 5.1](K )42(25)[(K 003781.0)(m/s 81.9(
Pr
)( =
−−
=
−
=−
DTTg
Ra s
( )
 
( )
 
1.374
7383.0/559.01
)10869.3(387.0
6.0
Pr/559.01
387.0
6.0
2
27/8
16/9
6/110
2
27/8
16/9
6/1
=
+
+=
+
+= Ra
Nu
222
2
m 38.224/m) 5.1(2)m 4)(m 5.1(4/2
C. W/m733.5)1.374(
m 5.1
C W/m.02299.0
=+=+=
=
==
DDLA
Nu
D
k
h
s
and
W8598C)]42(25)[(m 38.22)(C. W/m733.5()( 22 =−−=−= ss TThAQ
The total mass and the rate of evaporation of propane are
D = 1.5 m
L = 4 m
Propane tank
0
Ts = –42C
Air
T = 25C
9-62
9-67 Hot water flows in a horizontal pipe with a known inner surface temperature. The pipe outer surface is exposed to cool
air. The outer surface temperature of the pipe is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 The pipe thermal conductivity is
constant. 4 Radiation heat transfer is negligible. 5 Local atmospheric pressure is 1 atm. 6 The film temperature is 40°C (this
assumption will be verified).
Properties The properties of air at the assumed Tf = 40°C are k = 0.02662 W/m∙K, ν = 1.702 × 10−5 m2/s, Pr = 0.7255 (Table
A-15), and β = 1/Tf = 1/ 313 K.
Analysis With the assumption that Tf = 40°C, the pipe outer surface temperature is estimated as
9-64
9-69 Hot engine flows in a horizontal pipe with a known inner surface temperature. The pipe outer surface is covered with a
layer of insulation. The outer surface temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 Thermal conductivities of pipe and
insulation are constant. 4 Contact resistance is negligible. 5 Radiation heat transfer is negligible. 6 Local atmospheric
pressure is 1 atm.
Properties We first assume the film temperature is Tf = 50°C. Then, the properties of air at Tf = 50°C are k = 0.02735 W/m∙K,
ν = 1.798 × 10−5 m2/s, Pr = 0.7228 (Table A-15), and β = 1/Tf = 1/323 K .
The thermal conductivities of the pipe and the insulation are kpipe = 15 W.m∙K and kins = 0.15 W.m∙K, respectively.
Analysis With the assumption that Tf = 50°C, the outer surface temperature is estimated as
Discussion The results from the iterations are as follows:
Iter Ts,o [°C] Ra Nu h [W/m2∙K]
1 90 1.863 × 106 17.38 6.791
2 74.15 1.672 × 106 16.86 6.447
3 74.80 1.681 × 106 16.88 6.462
4 74.77 1.681 × 106 16.88 6.462
As Ts,o changes through the iterations, so does the film temperature used for evaluating the properties.
9-65
9-70 Reconsider Prob. 9-69. Hot engine flows in a horizontal pipe with a known inner surface temperature. The pipe
outer surface is covered with a layer of insulation. The effect of the insulation layer thickness on the outer surface
temperature is to be evaluated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
L=2 [m] “Length exposed to natural convection”
D_i=0.05 [m] “Inner pipe diameter”
t_pipe=5e-3 [m] “Pipe wall thickness”
T_infinity=10 [C] “Cool air temperature”
T_s_i=90 [C] “Inner surface temperature”
“PROPERTIES”
g=9.81 [m/s^2]
Fluid$=’air’
k=Conductivity(Fluid$, T=T_film)
Pr=Prandtl(Fluid$, T=T_film)
rho=Density(Fluid$, T=T_film, P=101.3)
Ra=(g*beta*(T_s_o-T_infinity)*D_o^3)/nu^2*Pr
Nusselt=(0.6+0.387*Ra^(1/6)/((1+(0.559/Pr)^(9/16))^(8/27)))^2
h=k/D_o*Nusselt
“Heat conduction through cylindrical layers”
D_interface=D_i+t_pipe*2
D_o=D_interface+t_ins*2
R_pipe=ln(D_interface/D_i)/(2*pi#*L*k_pipe)
R_ins=ln(D_o/D_interface)/(2*pi#*L*k_ins)
R_total=R_pipe+R_ins
“Heat balance”
Q_dot=(T_s_i-T_s_o)/R_total “Heat conduction through cylindrical layers”
Q_dot=h*A_s*(T_s_o-T_infinity) “Heat loss by natural convection”
A_s=pi#*D_o*L “Surface area exposed to natural convection”
tins [m] Ts,o [°C]
0.005 74.77
0.006 72.41
0.007 70.21
0.008 68.16
0.009 66.24
0.010 64.45
0.011 62.76
0.012 61.18
0.013 59.68
0.014 58.27
0.015 56.94
Discussion As the insulation layer thickness increases, the heat loss through the pipe is reduced. Therefore the outer surface
temperature decreases with increasing thickness of insulation.
0.005 0.007 0.009 0.011 0.013 0.015
50
55
60
65
70
75
80
tins [m]
Ts,o [°C]
9-66
9-71 A hot fluid flowing as a fully-developed laminar flow inside a horizontal pipe with constant surface temperature. The
pipe outer surface temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 Radiation heat transfer is negligible.
4 Local atmospheric pressure is 1 atm. 5 The film temperature is 50°C (this assumption will be verified).
Properties We first assume the film temperature is Tf = 50°C. Then, the properties of air at Tf = 50°C and 1 atm are k =
0.02735 W/m∙K, ν = 1.798 × 10−5 m2/s, Pr = 0.7228 (Table A-15), and β = 1/Tf = 1/323 K .
Analysis With the assumption that Tf = 50°C, the outer surface temperature is estimated as
C902
,=−=
TTT fos
The Rayleigh number is
)m 045.0(K )1090()K 27350)(m/s 81.9(
)(
312
3
,=
−+
−
−
oos DTTg
9-67
9-72 A hot liquid flowing inside a horizontal pipe with a known mass flow rate and temperature difference of the pipe inlet
and outlet. The pipe outer surface temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 Local atmospheric pressure is 1 atm.
4 The film temperature is 35°C (this assumption will be verified). 5 The Tsurr is the same as the air temperature.
Properties We first assume the film temperature is Tf = 35°C. Then, the properties of air at Tf = 35°C are k = 0.02625 W/m∙K,
ν = 1.655 × 10−5 m2/s, Pr = 0.7268 (Table A-15), and β = 1/Tf = 1/ 308 K.
The emissivity of the black oxidized copper pipe outer surface is ε = 0.78 (Table A-18)
Analysis With the assumption that Tf = 35°C, the outer surface temperature is estimated as
9-68
9-73 A hot liquid flowing inside a horizontal pipe with a known mass flow rate and temperature difference of the pipe
inlet and outlet. The pipe outer surface temperature is to be determined whether or not it is safe from thermal burn hazards.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 Local atmospheric pressure is 1 atm.
4 The film temperature is 35°C (this assumption will be verified). 5 The Tsurr is the same as the air temperature.
Properties The properties of air at Tf = 25°C and 1 atm pressure are k = 0.02551 W/m∙K, ν = 1.562 × 10−5 m2/s, Pr = 0.7296
(Table A-15), and β = 1/Tf = 1/298 K . The emissivity of the black painted surface is given as ε = 0.88.
Analysis With the assumption that Tf = 25°C, the outer surface temperature is estimated as
9-71
9-76 Water in a tank is to be heated by a spherical heater. The heating time is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The temperature of the outer surface of the sphere is constant.
Properties Using the average temperature for water (15+45)/2=30C as the fluid temperature, the properties of water at the
film temperature of (Ts+T)/2 = (85+30)/2 = 57.5C are (Table A-9)
1–3
26
K 10501.0
12.3Pr
/sm 10493.0
C W/m.6515.0
−
−
=
=
=
=
k
Also, the properties of water at 30C are (Table A-15)
CJ/kg. 4178 and kg/m 996 3== p
c
Analysis The characteristic length in this case is Lc = D = 0.06 m. Then,
8
3-132
3
)m 06.0)(K 3085)(K 10501.0)(m/s 81.9(
)(
−
−
−
DTTg s
Water
T,avg = 30C
Resistance
heater
Ts = 85C
D = 6 cm
D = 6 cm
9-75
9-80 A metal spherical tank is filled with hot liquid and the inner surface temperature is known. The tank is covered with
a layer of insulation and outer surface temperature is to be determined whether it is safe for thermal burn prevention.
Assumptions 1 Steady operating conditions exist. 2 Surface temperatures are constant. 3 The tank wall thermal conductivity
is constant. 4 Local atmospheric pressure is 1 atm. 5 The Tsurr is the same as the air temperature. 6 The film temperature is
30°C (this assumption will be verified).
Properties The properties of air at the assumed Tf = 30°C and 1 atm pressure are k = 0.02588 W/m∙K, ν = 1.608 × 10−5 m2/s,
Pr = 0.7282 (Table A-15), and β = 1/Tf = 1/303 K. The thermal conductivity of the tank is given as ktank = 15 W/m∙K.
The thermal conductivity and the emissivity of the insulation are given as kins = 0.15 W/m∙K and ε = 0.35, respectively.
Analysis With the assumption that Tf = 30°C, the outer surface temperature is estimated as
9-76
Natural Convection from Finned Surfaces and PCBs
9-83C Removing some of the fins on the heat sink will decrease heat transfer surface area, but will increase heat transfer
9-84 An aluminum heat sink of rectangular profile oriented vertically is used to cool a power transistor. The average natural
convection heat transfer coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Radiation heat transfer
from the sink is negligible. 4 The entire sink is at the base temperature.
Analysis The total surface area of the heat sink is
2
2
m 01463.0)m 0762.0)(m 0048.0)(2()m 0152.0)(m 0762.0)(6)(2(2
=+==
fins
nLbA
−
−
C)22120)(m 021465.0(
)(
TTA
stotal
stotal
b =1.52 cm
9.68 cm
Power
transistor
Heat sink
9-77
9-85 Aluminum heat sinks of rectangular profile oriented vertically are used to cool a power transistor. A shroud is placed
very close to the tips of fins. The average natural convection heat transfer coefficient is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Radiation heat transfer
from the sink is negligible. 4 The entire sink is at the base temperature.
Analysis The total surface area of the shrouded heat sink is
2
2
m 006835.0)m 0762.0)(m 0317.0()m 0762.0)(m 0145.0)(4(
m 013898.0)m 0152.0)(m 0762.0)(6)(2(2
=+=
===
unfinned
fins
A
nLbA
−
−
C)22108)(m 035486.0(
)(
TTA
stotal
stotal
b =1.52 cm
9.68 cm
Power
transistor
Heat sink
Shroud
9-78
9-86 A heat sink with equally spaced rectangular fins is to be used to cool a hot surface. The optimum fin spacing and the
rate of heat transfer from the heat sink are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air
is an ideal gas with constant properties. 3 The
atmospheric pressure at that location is 1 atm.
Properties The properties of air at 1 atm and 1 atm and
the film temperature of (Ts+T)/2 = (85+20)/2 = 52.5C
are (Table A–15)
1–
25
K 003072.0
K)2735.52(
11
7222.0Pr
/sm 10823.1
C W/m.02753.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length in this case is the
height of the surface Lc = L = 0.18 m. Then,
7
3-12
3
)m 18.0)(K 2085)(K 003072.0)(m/s 81.9(
)( =
−
−
LTTg
W = 15 cm
L = 18 cm
H
S
85C
T
= 20C
9-79
9-87E A heat sink with equally spaced rectangular fins is to be used to cool a hot surface. The optimum fin spacing and the
rate of heat transfer from the heat sink are to be determined.
Assumptions 1 Steady operating conditions exist. 2
Air is an ideal gas with constant properties. 3 The
atmospheric pressure at that location is 1 atm. 4 The
thickness t of the fins is very small relative to the fin
spacing S so that Eqs. 9-32 and 9-33 for optimum fin
spacing are applicable.
Properties The properties of air at 1 atm and 1 atm
and the film temperature of (Ts+T)/2 =
(180+78)/2=129F are (Table A-15E)
1–
23
R 001698.0
R )460129(
11
7217.0Pr
/sft 101975.0
FBtu/h.ft. 01597.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length in this case is the fin height,
in. 8== LLc
Then,
7
223
3-12
2
3
21 10058.3)7217.0(
)/sft 101975.0(
)ft 12/8)(R 78180)(R 001698.0)(ft/s 2.32(
Pr
)( =
−
=
−
=−
LTTg
Ra
The optimum fin spacing is
in 0.292==
== ft 02433.0
)10058.3(
ft 12/8
714.2714.2 4/174/1
Ra
L
S
The heat transfer coefficient for this optimum spacing case is
The number of fins and the total heat transfer surface area is
fins 16
08.02916.0
6=
+
=
+
=tS
w
n
2
ft 2.226=
ft) 12/2.1(ft) (0.08/12162ft) 12/8(ft) (0.08/1216ft) ft)(1.2/12 12/8(162
22
++=
++= ntHntLnLHAs
Then the rate of natural convection heat transfer becomes
Btu/h 194.7=−=−= F)78180)(ft 226.2)(F.Btu/h.ft 8578.0()( 22
TThAQss
Discussion If the fin thickness is disregarded, the number of fins and the rate of heat transfer become
fins 21
2916.0
6=
+
=s
w
ts
w
n
2
ft 2.8=ft) ft)(1.2/12 12/8(2122 == nLHAs
Btu/h 245=−=−= F)78180)(ft 8.2)(F.Btu/h.ft 8578.0()( 22
TThAQss
Therefore, the fin tip area is significant in this case.
W = 6 in
L = 8 in
H = 1.2 in
S
180F
T = 78F