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13-21
Radiation Heat Transfer between Surfaces
13-28C The analysis of radiation exchange between black surfaces is relatively easy because of the absence of reflection. The
4
13-22
13-33 The rate of heat loss from a person by radiation in a large room whose walls are maintained at a uniform temperature is
to be determined for two cases.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque,
12 =F
room which are at a temperature of 295 K is
W78.3=
−=
−=
−])K 295()K 303)[(K W/m1067.5)(m 9.1)(1)(85.0(
)(
444282
4
2
4
1112112 TTAFQ
(b) When the walls are at a temperature of 260 K,
W353=
−=
−=
−])K 260()K 303)[(K W/m1067.5)(m 9.1)(1)(85.0(
)(
444282
4
2
4
1112112 TTAFQ
Qrad
A = 1.9 m2
13-24
13-35 Two coaxial parallel circular disks spaced apart at a distance L. The effect of the distance L on the radiation heat
transfer coefficient between the disks is to be evaluated.
Analysis The problem is solved using EES, and the solution is given below.
"GIVEN"
13-26
13-37 A row of tubes is spaced between two large parallel plates. The net radiation heat flux leaving the bottom plate is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are black. 3 Convection heat transfer is not considered.
Analysis From energy balance, the net radiation heat flux leaving the bottom plate (surface 1) is
4
4
4
6576.0
=
Thus,
−−+−=
−]K)373623)(6576.01(K)283623)(6576.0)[(K W/m1067.5(
)])(1()([
444444428
4
3
4
112
4
2
4
1121TTFTTFq
13-27
13-38 A row of long cylindrical power cables, shielded with PVC insulation, are placed in parallel with a large wall at
high temperature. The surface temperature of the power cables is to be determined whether it is below the operation
temperature specified by the ASTM standard.
2)2]1 2
2tan−1(22−12
12)1 2
4)
Solving for the surface temperature of the power cables yields
⁄=[(373 K)4−200 W/m2
⁄=344 K=𝟕𝟏℃>60℃
13-29
4)]}1 4
⁄
(0.6576)(5.67×10−8W/m2∙K4)[(15W
m∙K)(227−207
0.05 m )−(20 W
m2∙K)(207−20)K
−(0.3424)(5.67×10−8 W
⁄
13-31
13-41E For a square room with specified dimensions and floor, wall and ceiling temperatures, determine the net radiation
heat transfer (a) from floor to walls and (b) from floor to ceiling.
Assumptions 1 All surfaces are assumed black.
Analysis
4
4
13-33
13-43 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 black. 3 Convection heat transfer is not considered. 4
The dryer is well insulated from heat loss to the surrounding.
Properties The latent heat of vaporization for water is hfg = 2257 kJ/kg (Table A-2)
fg
hmQQ
== evap21
Hence,
)(
2
)( 4
1
4
221
4
1
4
2212TTF
DL
TTFAhm fg −=−=
mkg/s 0.0370 =
−
=
−=
−=−=
−444428
3
4
1
4
2
4
1
4
2
4
1
4
221
K )3701000)(K W/m1067.5(
)J/kg 102257(
m) 5.1(
)(
)(
2
2
)(
2
TT
h
D
TT
h
D
TTF
h
D
L
m
fg
fgfg
13-37
13-48 Two parallel black disks are positioned coaxially, where the lower disk is heated electrically. The temperature of the
upper disk is to be determined.
13-38
13-49E A radiation shield is placed between two parallel disks which are maintained at uniform temperatures. The net rate of
radiation heat transfer through the shields is to be determined.
1332 == FF
48.052.01
34 =−=F
. The disk in the middle is surrounded by
black surfaces on both sides. Therefore, heat transfer between
the top surface of the middle disk and its black surroundings
T2 = 650 R,
2 = 1
1 ft
