978-0078027680 Chapter 21 Part 5

subject Type Homework Help
subject Pages 13
subject Words 4512
subject Authors John Cimbala, Robert Turner, Yunus Cengel

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21-85
21-125 Two diffuse surfaces A1 and A2 placed at an specified orientation, (a) the expression for the view factor F12 in terms
of A2 and L, and (b) the value of the view factor F12 when A2 = 0.02 m2 and L = 1 m are to be determined.
Assumptions 1 The surfaces A1 and A2 are diffuse. 2 Both A1 and A2 can be approximated as differential surfaces since both
are very small compared to the square of the distance between them.
Analysis (a) The view factor for surfaces A1 and A2 can be determined using the integral
2
2
21
21
2
21
1
21
2
21
1
21
2
21
1
12
coscos
coscos
1
coscos
1
coscos
1
2
2 1
A
r
AA
r
A
dAA
r
A
dAdA
r
A
F
A
A A
 
1
cos2
Lr
(2)
Substituting Eqs. (1) and (2) into the expression for F12, we get
2
2
1
)(cos
A
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21-87
21-127 A cylindrical enclosure is considered. (a) The expression for the view from the side surface to itself F33 in terms of K
and (b) the value of the view factor F33 for L = D are to be determined.
or
1213 1FF
(1)
For coaxial parallel disks, from Table 21-1, with i = 1, j = 2,
1
12/12
2/1
2
2
2
D
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21-89
21-129 Radiation heat transfer occurs between two concentric disks. The view factors and the net rate of radiation heat
transfer for two cases are to be determined.
0.1912
2
1
1
10.0
10.0
,5.0
20.0
10.0 F
L
r
r
L
0.76
)19.0(
m 0314.0
m 1257.0
2
2
12
2
1
21
2
F
A
A
F
(b) The net rate of radiation heat transfer between the surfaces can be determined from
 
 
 
K 573K 1073)K W/m1067.5(
44
428
4
2
4
1
TT
2
1
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21-90
21-130 The base and the dome of a long semi-cylindrical dryer are maintained at uniform temperatures. The drying rate per
unit length experienced by the base surface 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 dryer is
well insulated from heat loss to the surrounding.
2/
2
A
Applying energy balance on the base surface, we have
4
1
4
2
)(
TT
4
1
4
2
))(/(
TTh
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21-91
21-131 Radiation heat transfer occurs between a tube-bank and a wall. The view factors, the net rate of radiation heat
transfer, and the temperature of tube surface are to be determined.
0.658
5.0
2
1
5.0
2
5.0
2
1
5.0
2
1
5.1
3
tan
3
5.1
3
5.1
11
1tan11 D
s
s
D
s
D
Fij
0.419)658.0(
)5.1(
3
ijijij
j
i
jijijijiF
D
s
F
DL
sL
F
A
A
FFAFA
(b) The net rate of radiation heat transfer between the surfaces can be determined from
   
1
1
1
1
1
11
1
4444
j
i
j
j
iji
i
ji
jj
j
ijiii
i
ji
A
A
F
TT
AFAA
TT
q
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21-97
21-137 A cubical furnace with specified top and bottom surface temperatures and specified heat transfer rate at the bottom
surface is considered. The emissivity of the top surface and the net rates of heat transfer between the top and the bottom
surfaces, and between the bottom and the 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.
Analysis We consider the top surface to be surface 1, the base surface to be
surface 2, and the side surface to be surface 3. This system is a three-surface
enclosure. The view factor from the base to the top surface of the cube is from
surfaces is determined by applying the summation rule to be
20.0 ,80.0 ,20.0 323113231221 FFFFFF
We now apply Eq. 21-35 to each surface
 
)()(
1
31132112
1
1
1
4
1
JJFJJFJT
T1 = 700 K
1 = ?
T2 = 950 K
2 = 0.90
T3 = 450 K
3 m
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