9-21
9-32E A hot plate with an insulated back is considered. The rate of heat loss by natural convection is to be determined for
different orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant
properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (130+75)/2 = 102.5F are (Table A-15)
1–
23
R 001778.0
R)4605.102(
11
7256.0Pr
/sft 101823.0
FBtu/h.ft. 01535.0
=
+
==
=
=
=
−
f
T
k
Analysis (a) When the plate is vertical, the characteristic length is the height of the
plate.
ft. 2== LLc
Then,
8
223
3-12
2
3
10503.5)7256.0(
)/sft 101823.0(
)ft 2)(R 75130)(R 001778.0)(ft/s 2.32(
Pr
)( =
−
=
−
=−
LTTg
Ra s
6.102
7256.0
492.0
1
)10503.5(387.0
825.0
Pr
492.0
1
Ra387.0
825.0
2
27/8
16/9
6/18
2
27/8
16/9
6/1
=
+
+=
+
+=Nu
222
2
ft 4)ft 2(
F.Btu/h.ft 7869.0)6.102(
FBtu/h.ft. 01535.0
===
=
==
LA
Nu
k
h
s
and
Btu/h 173.1=−=−= C)75130)(ft 4)(F.Btu/h.ft 7869.0()( 22
TThAQss
(b) When the plate is horizontal with hot surface facing up, the characteristic length is determined from
ft 2
2
L
A
Q
Insulation
Air
T = 75F
Plate
Ts = 130F
L = 2 ft
9-22
9-33E Prob. 9–32E is reconsidered. The rate of natural convection heat transfer for different orientations of the plate as
a function of the plate temperature is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
L=2 [ft]
T_infinity=75 [F]
T_s=130 [F]
“PROPERTIES”
Fluid$=’air’
k=Conductivity(Fluid$, T=T_film)
Pr=Prandtl(Fluid$, T=T_film)
rho=Density(Fluid$, T=T_film, P=14.7)
mu=Viscosity(Fluid$, T=T_film)*Convert(lbm/ft-h, lbm/ft–s)
nu=mu/rho
beta=1/(T_film+460)
T_film=1/2*(T_s+T_infinity)
g=32.2 [ft/s^2]
“ANALYSIS”
“(a), plate is vertical”
delta_a=L
Ra_a=(g*beta*(T_s-T_infinity)*delta_a^3)/nu^2*Pr
Nusselt_a=0.59*Ra_a^0.25
h_a=k/delta_a*Nusselt_a
A=L^2
Q_dot_a=h_a*A*(T_s-T_infinity)
“(b), plate is horizontal with hot surface facing up”
delta_b=A/p
p=4*L
Ra_b=(g*beta*(T_s-T_infinity)*delta_b^3)/nu^2*Pr
Nusselt_b=0.54*Ra_b^0.25
h_b=k/delta_b*Nusselt_b
Q_dot_b=h_b*A*(T_s-T_infinity)
“(c), plate is horizontal with hot surface facing down”
delta_c=delta_b
Ra_c=Ra_b
Nusselt_c=0.27*Ra_c^0.25
h_c=k/delta_c*Nusselt_c
Q_dot_c=h_c*A*(T_s-T_infinity)
9-23
Ts
[F]
a
Q
[Btu/h]
b
Q
[Btu/h]
c
Q
[Btu/h]
80
7.714
9.985
4.993
85
18.32
23.72
11.86
90
30.38
39.32
19.66
95
43.47
56.26
28.13
100
57.37
74.26
37.13
105
71.97
93.15
46.58
110
87.15
112.8
56.4
115
102.8
133.1
66.56
120
119
154
77.02
125
135.6
175.5
87.75
130
152.5
197.4
98.72
135
169.9
219.9
109.9
140
187.5
242.7
121.3
145
205.4
265.9
132.9
150
223.7
289.5
144.7
155
242.1
313.4
156.7
160
260.9
337.7
168.8
165
279.9
362.2
181.1
170
299.1
387.1
193.5
175
318.5
412.2
206.1
180
338.1
437.6
218.8
80 100 120 140 160 180
0
50
100
150
200
250
300
350
400
450
500
Ts [F]
Q [Btu/h]
Qa
Qb
Qc
9-26
9-35 A vertical plate with length L is placed in a quiescent air, and the expressions, having the form
n
L
CRa=Nu
, for the
average heat transfer coefficient are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties.
Properties The properties of air at Tf = 20°C are k = 0.02514 W/m∙K,
= 1.516 × 10−5 m2/s, Pr = 0.7309 (from Table A-15).
Also, β = 1/Tf = 0.003413 K-1.
Analysis The Rayleigh number (Lc = L) is
38
3-12
3
)K 003413.0)(m/s 81.9(
)(
TL
LTTg s
−
9-27
9-36 A circuit board containing square chips is mounted on a vertical wall in a room. The surface temperature of the chips 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 The heat transfer from the back
side of the circuit board is negligible.
Properties Based on the problem statement, the properties of air at 1 atm and the
anticipated film temperature of (Ts+T)/2 = (35+25)/2 = 30C are (Table A-15)
1–
25
K 0033.0
K)27330(
11
7282.0Pr
/sm 10608.1
C W/m.02588.0
=
+
==
=
=
=
−
f
T
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and
thus the Nusselt number depends on the surface temperature which is unknown. We start the solution process by “guessing”
the surface temperature to be 35C for the evaluation of the properties and h. We will check the accuracy of this guess later
and repeat the calculations if necessary. The characteristic length in this case is the height of the board,
m. 5.0== LLc
Then,
8
3-12
3
)m 5.0)(K 2535)(K 0033.0)(m/s 81.9(
)( =
−
−
LTTg
Air
T = 25C
Tsurr = 25C
PCB, Ts
= 0.7
1210.18 W
L = 50 cm
9-29
9-38 A printed circuit board (PCB) is placed in a room. The average temperature of the hot surface of the board is to be
determined for different orientations.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with
constant properties. 3 The local atmospheric pressure is 1 atm. 3 The heat loss
from the back surface of the board is negligible.
Properties We evaluate air properties at a film temperature of (Ts+T)/2 = 32.5C and
1 atm based on the problem statement. Then, for an air temperature of T = 20C, the
corresponding surface temperature is Ts = 45C. The properties of air at 1 atm and
32.5C are (Table A-15)
1–
25
K 003273.0
K)2735.32(
11
7275.0Pr
/sm 10631.1
C W/m.02607.0
=
+
==
=
=
=
−
f
T
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and
thus the Nusselt number depends on the surface temperature which is unknown.
(a) Vertical PCB . We start the solution process by “guessing” the surface temperature to be 45C for the evaluation of the
properties and h. We will check the accuracy of this guess later and repeat the calculations if necessary. The characteristic
length in this case is the height of the PCB,
m. 2.0== LLc
Then,
7
3-12
3
)m 2.0)(K 2045)(K 003273.0)(m/s 81.9(
)( =
−
−
LTTg
Insulation
Air
T = 20C
PCB, Ts
8 W
L = 0.2 m
9-30
)()(
44
−+−=
surrssss
TTATThAQ
9-32
T
[F]
Ts,a
[C]
Ts,b
[C]
Ts,c
[C]
5
32.54
28.93
38.29
7
34.34
30.79
39.97
9
36.14
32.65
41.66
11
37.95
34.51
43.35
13
39.75
36.36
45.04
15
41.55
38.22
46.73
17
43.35
40.07
48.42
19
45.15
41.92
50.12
21
46.95
43.78
51.81
23
48.75
45.63
53.51
25
50.55
47.48
55.21
27
52.35
49.33
56.91
29
54.16
51.19
58.62
31
55.96
53.04
60.32
33
57.76
54.89
62.03
35
59.56
56.74
63.74
510 15 20 25 30 35
25
30
35
40
45
50
55
60
65
T [C]
Ts [C]
Ts,a
Ts,b
Ts,c
9-35
9-42 Absorber plates whose back side is heavily insulated is placed horizontally outdoors. Solar radiation is incident on the
plate. The equilibrium temperature of the plate is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas
with constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the anticipated film
temperature of (Ts+T)/2 = (115+25)/2 = 70C based on the problem
statement are (Table A-15)
1–
25
K 002915.0
K)27370(
11
7177.0Pr
/sm 10995.1
C W/m.02881.0
=
+
==
=
=
=
−
f
T
k
Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh number and
thus the Nusselt number depends on the surface temperature which is unknown. We start the solution process by “guessing”
the surface temperature to be 115C for the evaluation of the properties and h. We will check the accuracy of this guess later
and repeat the calculations if necessary. The characteristic length in this case is
m 24.0
)m 8.0m 2.1(2
)m 8.0)(m 2.1( =
+
== p
A
Ls
c
Then,
7
225
3-12
2
3
10414.6)7177.0(
)/sm 10995.1(
)m 24.0)(K 25115)(K 002915.0)(m/s 81.9(
Pr
)( =
−
=
−
=−
cs LTTg
Ra
The Nusselt number relation for the horizontal hot surface, facing up is
04.60)10414.6(15.015.0 3/173/1 === RaNu
C. W/m208.7)04.60(
m 24.0
C W/m.02881.0 2=
== Nu
L
k
h
c
2
m 96.0)m 2.1)(m 8.0( ==
s
A
In steady operation, the heat gain by the plate by absorption of solar radiation must be equal to the heat loss by natural
convection and radiation. Therefore,
W1.501)m 96.0)( W/m600)(87.0( 22 === s
AqQ
])K 27310()273)[(1067.5)(m 96.0)(09.0(
C)25)(m 96.0)(C. W/m208.7( W501.1
)()(
4482
22
44
+−++
−=
−+−=
−
s
s
skyssss
T
T
TTATThAQ
Its solution is
C7.89 =
s
T
which is not very close to the assumed value of 115C. We repeat the calculations at a new anticipated surface temperature of
95C. The properties are to be evaluated at the film temperature of (95+25)/2=60C.
25
/sm 10896.1
C W/m.02808.0
=
=
−
k
Insulation
Air
T = 25C
Absorber plate
s = 0.87
= 0.09
600 W/m2
L = 1.2 m
9-36
])K 27310()273)[(1067.5)(m 96.0)(09.0(
C)25)(m 96.0)(C. W/m759.6( W501.1
)()(
4482
22
44
+−++
−=
−+−=
−
s
s
skyssss
T
T
TTATThAQ
C93.5=
s
T
This is close to the assumed surface temperature of 95C. Therefore, there is no need to repeat the calculations.
If the absorber plate is made of ordinary aluminum which has a solar absorptivity of 0.28 and an emissivity of 0.07,
the rate of solar gain becomes
9-39
9-45 Stainless steel bolts (ASTM A437 B4B) are used to secure two horizontal metal plates together. A ceramic plate is
used for preventing the bolts in the metal plates from cooling below the minimum suitable temperature of −30°C. The upper
surface of the ceramic plate is subjected to natural convection with cold hydrogen gas at −60°C and thermal radiation
exchange with the surrounding. The minimum thickness of the plate is to be determined.
Assumptions 1 Steady state conditions. 2 H2 is an ideal gas. 3 The local atmospheric pressure is 1 atm. 4 Conduction through
the plate is one-dimensional. 5 Contact resistance at the interface is negligible.
Properties The properties of H2 gas at the film temperature of Tf = (T + Ts )/2 =(−60−40)/2 = −50°C are (Table A–16) Pr =
0.6562, k = 0.1404 W/m∙K, ν = 6.624 × 10−5 m2/s, β = 1/Tf = 1/ (−50 + 273 K) = 0.004484 K−1
Analysis The characteristic length of the plate is
