21–61
21–98 Prob. 21–97 is reconsidered. The effects of the rate of the heat transfer at the base surface and the temperature of
the side surfaces on the temperature of the base surface are to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
a=2 [m]
1000
552.8
21–62
T2
[K]
T1
[K]
300
325
350
375
400
425
450
475
500
525
550
575
600
625
650
675
700
425.5
435.1
446.4
459.2
473.6
489.3
506.3
524.4
543.4
563.3
583.8
605
626.7
648.9
671.4
694.2
717.3
300 350 400 450 500 550 600 650 700
400
450
500
550
600
650
700
750
T2 [K]
T1 [K]
21–63
21–99 A long semi-cylindrical duct with specified temperature on the side surface is considered. The temperature of the base
surface for a specified heat transfer rate is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
Properties The emissivity of the side surface is = 0.4.
Analysis We consider the base surface to be surface 1, the side surface to be surface 2. This system is a two-surface
enclosure, and we consider a unit length of the duct. The surface areas and the view factor are determined as
2
2
2
1
m 571.12/m) 1)(m 0.1(2/
m 0.1)m 0.1)(m 0.1(
DLA
A
1101 12121211 FFFF
(summation rule)
The temperature of the base surface is determined from
K 684.8
1
22
4
4
1
428
22
2
121
4
2
4
1
12
)4.0)(m 571.1(
4.01
)1)(m 0.1(
1
]K) 650()[ W/m1067.5(
W1200
1
1
)(
T
TK
AFA
TT
Q
21-100 A hemisphere with specified base and dome temperatures and heat transfer rate is considered. The emissivity of the
dome is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
Properties The emissivity of the base surface is = 0.55.
Analysis We consider the base surface to be surface 1, the dome surface to be surface 2. This system is a two-surface
enclosure. The surface areas and the view factor are determined as
222
2
222
1
m 06283.02/)m 20.0(2/
m 03142.04/)m 20.0(4/
DA
DA
1101 12121211 FFFF
T1 = ?
1 = 1
T2 = 650 K
2 = 0.4
D = 1 m
T2 = 600 K
2 = ?
21–64
21-101 Two very large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between
the two plates is to be determined.
0.5 and 0.9.
Analysis The net rate of radiation heat transfer between the
two surfaces per unit area of the plates is determined
1
9.0
1
5.0
1
1
11
21
A
s
T2 = 400 K
2 = 0.9
21–67
21-104 Liquid nitrogen is stored in a spherical tank this is enclosed by a concentric spherical surface at 273 K. The rate of
vaporization for the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Heat transfer by radiation
only.
Properties The emissivity of the two surfaces is given
21–68
21-105 Two very long concentric cylinders are maintained at uniform temperatures. The net rate of radiation heat transfer
between the two cylinders is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray. 3 Convection heat transfer is not considered.
kW 29.81
W810,29
5
5.3
55.0
55.01
1
1
])K 500(K) 950)[(K W/m1067.5](m) m)(1 35.0([
1
1
)(
44428
2
1
2
2
1
4
2
4
11
12
r
r
TTA
Q
21-106 Two phase gas-liquid oxygen is stored in a spherical tank this is enclosed by a concentric spherical surface at 273 K.
The heat transfer rate at the spherical tank surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Heat transfer by radiation
only.
D2 = 0.5 m
D1 = 0.35 m
Vacuum
21–69
21-107 Two concentric spheres are maintained at uniform temperatures. The net rate of radiation heat transfer between the
two spheres and the convection heat transfer coefficient at the outer surface are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are
opaque, diffuse, and gray.
spheres is
W1270
2
44
4282
2
2
2
1
2
2
1
4
2
4
11
12
m 2.0
m 15.0
7.0
7.01
5.0
1
K500 K700KW/m 1067.5m) 3.0(
1
1
r
r
TTA
Q
Radiation heat transfer rate from the outer sphere to the surrounding surfaces are
)(
444282
44
22
surrrad TTFAQ
2 = 0.7
1 = 0.5
Tsurr =30C
= 0.35
21–70
21-108 A spherical tank filled with liquid nitrogen is kept in an evacuated cubic enclosure. The net rate of radiation heat
transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be 1 = 0.1 and 2 = 0.8.
W228
2
2
m) 6(3
m) 2(
8.0
8.01
1.0
1
21-109 A spherical tank filled with liquid nitrogen is kept in an evacuated spherical enclosure. The net rate of radiation heat
transfer to the liquid nitrogen is to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered. 4 The thermal resistance of the tank is negligible.
Properties The emissivities of surfaces are given to be 1 = 0.1 and
Vacuum
21–72
1
21
Q
[W]
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
227.9
340.9
453.3
565
676
786.4
896.2
1005
1114
1222
1329
1436
1542
1648
1753
1857
1961
2
21
Q
[W]
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
189.6
202.6
209.7
214.3
217.5
219.8
221.5
222.9
224.1
225
225.8
226.4
227
227.5
227.9
228.3
228.7
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
200
400
600
800
1000
1200
1400
1600
1800
2000
1
Q21 [W]
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
185
190
195
200
205
210
215
220
225
230
2
Q21 [W]
21–75
21-113 A circular grill is considered. The bottom of the grill is covered with hot coal bricks, while the wire mesh on top of
the grill is covered with steaks. The initial rate of radiation heat transfer from coal bricks to the steaks is to be determined for
two cases.
21–76
21-114E Top and side surfaces of a cubical furnace are black, and are maintained at uniform temperatures. Net radiation heat
transfer rate to the base from the top and side surfaces are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
not considered.
Properties The emissivities are given to be
= 0.7 for the bottom surface and 1 for other surfaces.
T2 = 1600 R
summation rule, the view factor from the base or top to the side surfaces is
8.02.0111 1213131211 FFFFF
T1 = 800 R
1 = 0.7
21–78
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1.0x106
1.5x106
2.0x106
2.5x106
3.0x106
3.5x106
4.0x106
4.5x106
1
Q31 [Btu/h]
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-200000
-100000
0
100000
200000
300000
400000
500000
600000
700000
1
Q12 [Btu/h]
0.0x100
5.0x105
1.0x106
1.5x106
2.0x106
2.5x106
3.0x106
3.5x106
4.0x106
4.5x106
1
Q1 [Btu/h]
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. 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.
21–116 A hole is drilled in a spherical cavity. The maximum rate of radiation energy streaming through the hole is to be
determined.
21–117 The wavelengths at maximum emission of radiation for both daylight and incandescent light are to be determined.
Assumptions 1 The sun and the incandescent light filament behave as black bodies.
Analysis The wavelength at maximum emission of radiation can be determined using the Wien’s displacement law:
Km 82897)( power max
.T
21–80
21-118 The fraction of the incident solar radiation that is absorbed by the human skin is to be determined.
273232.0
1
0
f
616797.0273232.0890029.0
21
f
109971.0890029.01
2
f
0.6916
)109971.0(0.1)616797.0(5.0)273232.0(0.1)(0.1)(5.0)(0.1 2211
0
fff
21-119 The variation of emissivity of an opaque surface at a specified temperature with wavelength is given. The average
emissivity of the surface and its emissive power are to be determined.
Analysis The average emissivity of the surface can be determined from
21 and
corresponding to
TT 21 and
. These functions are determined
from Table 21-2 to be
890029.0mK 9000=K) m)(1500 6(
273232.0mK 3000=K) m)(1500 2(
2
1
2
1
λ
λ
fT
fT
and
0.5243 )890029.01)(0.0()273232.0890029.0)(85.0()273232.0)(0.0(
, m
2
6
0.85