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21-22
21-43 The variation of emissivity of a surface with wavelength is given. The average emissivity and absorptivity of the
0.154
(at 5800 K)
21-44 The variation of absorptivity of a surface with wavelength is given. The average absorptivity, reflectivity, and
emissivity of the surface are to be determined at given temperatures.
Analysis For T = 2500 K:
633747.0mK 5000=K) m)(2500 2( 1
1 λ
fT
The average absorptivity of this surface is
0.383
)633747.01)(7.0()633747.0)(2.0(
)1()( 11 21
ffT
Then the reflectivity of this surface becomes
0.617 383.0111
Using Kirchhoff’s law,
, the average emissivity of this surface
0.7
0.2
21-26
21-49 An opaque plate is being heated uniformly at the bottom and the top surface is exposed to natural convection and
irradiated uniformly. The radiosity of the plate is to be determined.
Assumptions 1 The plate has a uniform temperature. 2 Heat loss from plate’s side surface is negligible. 3 The surroundings
are treated as an isothermal surface, Tsurr = T∞. 4 Kirchhoff’s law is applicable.
33.0
67.01
The radiosity of the plate is given as
EJ
Substituting Eq. (2) into (1) gives
TThJq
EJ
)( elec EqTTh
2
W/m639.6
33.01
)K 353()K W/m1067.5)(67.0() W/m1000)(33.0(K )780()K W/m7)(33.0(
1
)(
442822
4
elec
ss TqTTh
J
21-28
The View Factor
factor from a surface to itself is non-zero for concave surfaces.
21-54C The cross-string method is applicable to geometries which are very long in one direction relative to the other
i
21-30
21-56 Two coaxial parallel circular disks spaced apart at a distance L. The parameter that would increase the view factor F12
by a factor of 5 is to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
2
2
2
2
12 )/)(2/1(
1)/(1
1
)]2/([2
1)]2/([41
1LD
LD
LD
LD
F
2
1
2
1
)/)(2/1(
1)/(1
11.0 LD
LD
70273.0
1
L
D
2
2
2
2
)/)(2/1(
1)/(1
15.0 LD
LD
8284.2
2
L
D
0.2485 8284.2
70273.0
2
11
2
D
L
L
D
L
L
21-57 A row of cylinders is spaced between two large parallel plates. The view factor between the plate and the row of
cylinders is to be determined.
0.8426
5.3
5
5
21-31
21-58 An enclosure consisting of five surfaces is considered. The number of
view factors this geometry involves and the number of these view factors that
can be determined by the application of the reciprocity and summation rules
are to be determined.
21-59 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base is to be
(1): circular base surface
Surface (1) is flat, and thus
F
11 0
.
11 :ruleSummation 121211 FFF
0.5 2
1
2
4
)1(A :ruley reciprocit 2
2
2
1
12
2
1
21212121D
D
A
A
F
A
A
FFAF
21-60 The view factor from the conical side surface to a hole located at the center of the base of a conical enclosure is to be
determined.
conical side surface (3)
Surfaces 1 and 2 are flat, and they have no direct view of each other. Therefore,
2
1
D
h
(3)
21-32
21-61 The four view factors associated with an enclosure formed by two very long concentric cylinders are to be determined.
the inner surface of the outer cylinder (2)
No radiation leaving surface 1 strikes itself and thus
0
11
F
2
D
21-62 View factors from the very long grooves shown in the figure to the surroundings are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected.
D2 D1
(1)
21-33
21-63 A cylindrical enclosure is considered. (a) The expression for the view factor between the base and the side surface F13
in terms of K and (b) the value of the view factor F13 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
21-36
21-66 The view factors from the base of a cube to each of the other five surfaces are to be determined.
Because of symmetry, we have
21-67 The view factors between the rectangular surfaces shown in the figure are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
25.0
33.0
3
1
33.0
3
1
31
1
3
F
W
L
W
L
and
32.0
67.0
3
2
33.0
3
1
)21(3
21
3
F
W
LL
W
L
07.025.032.0 32323231)21(3
FFFFF
W = 3 m
(2)
L2 = 1 m
L1 = 1 m
L3 = 1 m
A3 (3)
A2
A1
(1)
(1)
(3), (4), (5), (6)
side surfaces
21-37
21-68 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base
surface is to be determined.
side surface by (3).
05.0
25.0
4
4
4
2112
2
22
1
1
1
FF
r
r
L
r
r
r
r
L
1 :rulesummation 131211 FFF
95.0105.00 1313 FF
0.119 )95.0(
8
1
8
2
:ruley reciprocit 13
2
1
2
1
13
1
2
1
13
3
1
31313131F
r
r
F
Lr
r
F
A
A
FFAFA
(3)
(1)
L=2D=4r
D=2r
21-39
21-70 The expression for the view factor F12 of two infinitely long parallel plates is to be determined using the Hottel’s
crossed-strings method.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis From the Hottel’s crossed-strings method, we have
)strings Uncrossed()strings Crossed(
0.204
w
2
Discussion The Hottel’s crossed-string method is applicable only to surfaces that are very long, such that they can be
considered to be two-dimensional and radiation interaction through the end surfaces is negligible.
21-71 The view factor between the two infinitely long parallel cylinders located a distance s apart from each other is to be
determined.
)2/(2
22
1 surfaceon String2
strings Uncrossedstrings Crossed
22
21
D
sDs
F
or
D
sDs
F
22
21
2
D
(2)
(1)
s
