13–43
𝐴1𝐹12 =𝐴2𝐹21 ⟹ 𝐹12=𝐹21 =0.007953 and𝐹23 =1−𝐹21 =0.99205
5.67×10−8W/m2∙K4)1 4
+1−0.95
⁄
13–44
13-54 The radiation heat flux between two infinitely long parallel plates of specified surface temperatures is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered. 4
The surface temperatures are uniform.
Analysis From the Hottel’s crossed–strings method, we have
(Crossed strings) (U ncrossed strings)
S – S
3424.0
2
=
w
The radiation heat flux between the two surfaces is
−=
−444428
4
2
4
11212
)( TTFq
13–46
13–56 Two perpendicular rectangular surfaces with a common edge are maintained at specified temperatures. The net rate of
radiation heat transfers between the two surfaces and between the horizontal surface and the surroundings are to be
determined.
13–48
13-58 Prob. 13-57 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”
12
Q
[W]
T1
[K]
500
525
550
575
600
625
650
675
700
725
750
775
800
825
850
875
900
925
950
975
1000
619.4
620.4
621.3
622.2
623.1
624
624.9
625.8
626.7
627.6
628.5
629.4
630.3
631.2
632
632.9
633.8
634.6
635.5
636.4
637.2
500 600 700 800 900 1000
617.5
621.5
625.5
629.5
633.5
637.5
Q12 [W]
T1 [K]
13–49
T2
[K]
T1
[K]
300
325
350
375
400
425
450
475
500
525
550
575
600
625
650
675
700
436.5
445.5
456
468.1
481.7
496.7
512.9
530.4
548.8
568.1
588.2
609
630.3
652.1
674.3
696.9
719.7
300 350 400 450 500 550 600 650 700
400
450
500
550
600
650
700
750
T2 [K]
T1 [K]
13–50
13-59 A solid sphere is placed in an evacuated equilateral triangular enclosure. The view factor from the enclosure to the
sphere and the emissivity of the enclosure 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 emissivity of sphere is given to be
1 = 0.45.
Analysis (a) We take the sphere to be surface 1 and the surrounding
enclosure to be surface 2. The view factor from surface 2 to surface
1 is determined from reciprocity relation:
0.3023=
=
=
====
===
===
21
21
212121
222
2
2
222
1
)39.10()1)(142.3(
m 39.103m) 449.2(3
4
3
4
m 449.26m) 1(6
m 142.3m) 1(
F
F
FAFA
L
L
A
DL
DA
We note that the tetrahedron has four equal surfaces.
(b) The net rate of radiation heat transfer can be expressed for this two–surface enclosure to yield the emissivity of the
T2 = 420 K
2 = ?
T1 = 600 K
1 = 0.45
1 m
L
L
L
13–51
13–60 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.
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
13–61 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 1414.02/)m 3.0(2/
m 07069.04/)m 3.0(4/
===
===
DA
DA
1 = 0.55
T1 = ?
1 = 1
T2 = 650 K
2 = 0.4
D = 1 m
T2 = 600 K
2 = ?
13–52
13–62E The base and the dome of a long semicylindrical duct are maintained at uniform temperatures. The net rate of
radiation heat transfer from the dome to the base surface is to be determined.
T2 = 1800 R
= 0.9.
Analysis The view factor from the base to the dome is first determined
from
surface)(flat 0
11
=
F
T1 = 550 R
1 = 0.5
D = 15 ft
13–53
13–63 Radiation heat transfer occurs between a sphere and a circular disk. The view factors and the net rate of radiation heat
transfer for the existing and modified cases are to be determined.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is
0.2764=
+−=
+−=
−
−5.0
2
5.0
2
2
12 m 60.0
m 2.1
115.0115.0 h
r
F
0.0691=
=
21
21
)524.4()2764.0)(131.1(
F
F
(b) The net rate of radiation heat transfer between the surfaces can be determined from
( )
( ) ( )
−
−
−
K 473K 873)K W/m1067.5(
44
428
4
2
4
1
TT
13–54
13–64 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
13–57
13–67 A large ASTM B152 copper plate and a large ceramic plate are placed in parallel near each other. The temperature
of the ceramic plate and the net radiation heat flux between the two plates are known. The temperature of the copper plate is
to be determined whether it is below the maximum use temperature of 260°C.
Assumptions 1 Steady state conditions. 2 Uniform surface temperatures on both plates. 3 Surfaces are opaque, gray and
diffuse.
13–59
13-69 A long cylindrical black surface fuel rod is shielded by a concentric surface that has a uniform temperature. The
surface temperature of the fuel rod is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The fuel rod surface is black. 3 The shield is opaque, diffuse, and gray. 4
