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Chapter 12 Radiation Heat Transfer 
Chapter 12 
RADIATION HEAT TRANSFER 
 
 
View Factors 
 
12-1C The view factor represents the fraction of the radiation leaving surface i that strikes surface j 
directly. The view factor from a surface to itself is non-zero for concave surfaces. 
Fi j→
 
12-2C The pair of view factors and are related to each other by the reciprocity rule 
 where Ai is the area of the surface i and Aj is the area of the surface j. Therefore, 
Fi j→ Fj i→
A F A Fi ij j ji=
 A F A F F A
A
F1 12 2 21 12 2
1
21= ⎯ →⎯ = 
12-3C The summation rule for an enclosure and is expressed as where N is the number of 
surfaces of the enclosure. It states that the sum of the view factors from surface i of an enclosure to all 
surfaces of the enclosure, including to itself must be equal to unity. 
Fi j
j
N
→
=
=∑ 1
1
 The superposition rule is stated as the view factor from a surface i to a surface j is equal to the 
sum of the view factors from surface i to the parts of surface j, F F F1 2 3 1 2 1 3→ → →= +( , ) . 
 
12-4C The cross-string method is applicable to geometries which are very long in one direction relative to 
the other directions. By attaching strings between corners the Crossed-Strings Method is expressed as 
 F
ii j→
= −×
∑∑Crossed strings Uncrossed strings
string on surface 2
 
 12-1
Chapter 12 Radiation Heat Transfer 
12-5 An enclosure consisting of six 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. 
6
5 
4
3
2 
1 
Analysis A seven surface enclosure (N=6) involves N 2 26= = 36 
view factors and we need to determine 15
2
)16(6
2
)1( =−=−NN view 
factors directly. The remaining 36-15 = 21 of the view factors can be 
determined by the application of the reciprocity and summation rules. 
 
 
 
12-6 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. 
2 
1 
4 
5
3
Analysis A five surface enclosure (N=5) involves N 2 25= = 25 
view factors and we need to determine 
N N( ) (5 )− = − =1
2
5 1
2
10 
view factors directly. The remaining 25-10 = 15 of the view 
factors can be determined by the application of the reciprocity and 
summation rules. 
 
 
 
12-7 An enclosure consisting of twelve 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. 2
1
3
9 
11
12
10
4 5
8
6
7
Analysis A twelve surface enclosure (N=12) 
involves view factors and we 
need to determine 
144== 22 12N
N N( ) ( )− = − =1
2
12 12 1
2
66 
view factors directly. The remaining 144-66 = 78 
of the view factors can be determined by the 
application of the reciprocity and summation 
rules. 
 12-2
Chapter 12 Radiation Heat Transfer 
12-8 The view factors between the rectangular surfaces shown in the figure are to be determined. 
Assumptions The surfaces are diffuse emitters and reflectors. 
Analysis From Fig. 12-6, 
 24.0
5.0
2
11
5.0
2
1
31
3
=
⎪⎪⎭
⎪⎪⎬
⎫
==
==
F
W
L
W
L
 
W = 2 m 
(2) L2 = 1 m 
 
L1 = 1 m 
 L3 = 1 m A3 (3) 
A2
 
A1 (1) and 
29.0
1
2
2
5.0
2
1
)21(3
21
3
=
⎪⎪⎭
⎪⎪⎬
⎫
==+
==
+→F
W
LL
W
L
 
We note that A1 = A3. Then the reciprocity and superposition rules gives 
 0.24==⎯→⎯= 3113313131A FFFAF
05.024.029.0 32323231)21(3 =⎯→⎯+=⎯→⎯+=+→ FFFFF 
Finally, 0.05==⎯→⎯= 322332 FFAA
 12-3
Chapter 12 Radiation Heat Transfer 
12-9 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical 
enclosure to its base surface is to be determined. 
Assumptions The surfaces are diffuse emitters and reflectors. 
Analysis We designate the surfaces as follows: 
(2) 
(3) 
(1) 
L 
D 
 Base surface by (1), 
 top surface by (2), and 
 side surface by (3). 
Then from Fig. 12-7 (or Table 12-1 for better accuracy) 
 38.0
1
1
2112
2
22
1
1
1 ==
⎪⎪⎭
⎪⎪⎬
⎫
==
==
FF
r
r
L
r
r
r
r
L
 
1 :rulesummation 131211 =++ FFF 
62.0138.00 1313 =⎯→⎯=++ FF 
 ( ) 0.31==π
π=π
π==⎯→⎯= )62.0(
2
1
22
 :ruley reciprocit 13
11
2
1
13
1
2
1
13
3
1
31313131 Frr
r
F
Lr
r
F
A
A
FFAFA 
 
Discussion This problem can be solved more accurately by using the view factor relation from Table 12-1 
to be 
1
1
2
22
2
1
11
1
===
===
r
r
L
r
R
r
r
L
r
R
 
382.0
1
14334
3
1
111
1
1
5.02
2
2
1
5.02
1
22
2
1
12
2
2
2
1
2
2
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−−=
=++=++=
R
R
SSF
R
R
S
 
618.0382.011 1213 =−=−= FF 
( ) 0.309==π
π=π
π==⎯→⎯= )618.0(
2
1
22
 :ruley reciprocit 13
11
2
1
13
1
2
1
13
3
1
31313131 Frr
r
F
Lr
r
F
A
A
FFAFA
 
 12-4
Chapter 12 Radiation Heat Transfer 
12-10 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base 
is to be determined. 
Assumptions The surfaces are diffuse emitters and reflectors. 
(1) 
(2) 
D 
Analysis We number the surfaces as follows: 
(1): circular base surface 
(2): dome surface 
Surface (1) is flat, and thus F11 0= . 
11 :ruleSummation 121211 =→=+ FFF 
0.5=====⎯→⎯=
2
1
2
4)1(A :ruley reciprocit
2
2
2
1
12
2
1
21212121
D
D
A
A
F
A
A
FFAF π
π
 
 
 
12-11 Two view factors associated with three very long ducts with 
different geometries are to be determined. 
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End 
effects are neglected. 
Analysis (a) Surface (1) is flat, and thus F11 0= . 
1=→=+ 121211 1 :rulesummation FFF 
0.64==
⎟⎠
⎞⎜⎝
⎛==⎯→⎯= ππ
2)1(
2
A :ruley reciprocit 12
2
1
21212121
sD
DsF
A
A
FFAF 
(1) 
(2) 
D 
(b) Noting that surfaces 2 and 3 are symmetrical and thus 
, the summation rule gives F F12 13=
(1) 
(3) (2) 
 a 
 0.5=⎯→⎯=++⎯→⎯=++ 121312131211 101 FFFFFF
Also by using the equation obtained in Example 12-4, 
 F L L L
L
a b b
a
a
a12
1 2 3
12 2 2
1
2
= + − = + − = = = 0.5 
2b
a=⎟⎠
⎞⎜⎝
⎛==⎯→⎯=
2
1A :ruley reciprocit 12
2
1
21212121 b
aF
A
A
FFAF 
L3 = b L4 = b 
 
 L5 L6 
L2 = a 
L1 = a 
(c) Applying the crossed-string method gives 
F F L L L L
L
a b b
a
12 21
5 6 3 4
1
2 2
2
2 2
2
= = + − +
= + − = + −
( ) ( )
a b b
a
2 2
 
 12-5
Chapter 12 Radiation Heat Transfer 
12-12 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. 
Analysis (a) We designate the circular dome surface by (1) and the imaginary flat top surface by (2). 
Noting that (2) is flat, 
(1) 
(2) 
D 
 022 =F
11 :rulesummation 212221 =⎯→⎯=+ FFF 
0.64====⎯→⎯= ππ
2)1(
2A :ruley reciprocit 21
1
2
12212121 D
DF
A
A
FFAF 
(b) We designate the two identical surfaces of length b by (1) and (3), and the imaginary flat top surface by 
(2). Noting that (2) is flat, 
 022 =F
(3) (1) 
a 
b b 
(2) 
5.01 :rulesummation 2321232221 ==⎯→⎯=++ FFFFF (symmetry) 
11 :rule summation )31(2)31(222 =⎯→⎯=+ +→+→ FFF 
2b
a===⎯→⎯
=
+
→+→+
→+++→
)1(
A :ruley reciprocit
)31(
2
)31(2)31(
2)31()31()31(22
A
A
FF
FAF
surr
 
(c) We designate the bottom surface by (1), the side surfaces 
by (2) and (3), and the imaginary top surface by (4). Surface 4 
is flat and is completely surrounded by other surfaces. 
Therefore, and . F44 0= F4 1 2 3 1→ + + =( )
 
 b b 
 (2) (3) 
(1) 
 
a 
 (4) 
2b+a
a===⎯→⎯
=
++
→++→++
→++++++→
)1(
A :ruley reciprocit
)321(
4
)321(4)321(
4)321()321()321(44
A
A
FF
FAF
surr
 
 
 
 
12-13 The view factors from the base of a cube to each of the 
other five surfaces are to be determined. 
(1) 
(3), (4), (5), (6) 
side surfaces 
(2) Assumptions The surfaces are diffuse emitters and reflectors. 
Analysis Noting that , from Fig. 12-6 we read L w L w1 2 1/ /= =
 F12 0 2= .
Because of symmetry, we have 
 F F F F F12 13 14 15 16= = = = = 0.2
 
12-14 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. 
Assumptions The conical side surface is diffuse emitter and reflector. 
h 12-6 (3)
Chapter 12 Radiation Heat Transfer 
Analysis We number different surfaces as 
 the hole located at the center of the base (1) 
 the base of conical enclosure (2) 
 conical side surface (3) 
Surfaces 1 and 2 are flat , and they have no direct view of each other. 
Therefore, 
 F F F F11 22 12 21 0= = = =
11 :rulesummation 13131211 =⎯→⎯=++ FFFF 
2Dh
d2=⎯→⎯=⎯→⎯= 3131
2
313131 2
)1(
4
A :ruley reciprocit FFDhdFAF ππ 
 
 
 
12-15 The four view factors associated with an enclosure formed by two very long concentric cylinders are 
to be determined. 
Assumptions 1 The surfaces are diffuse emitters and reflectors. 2 End effects are neglected. 
Analysis We number different surfaces as 
 the outer surface of the inner cylinder (1) 
 the inner surface of the outer cylinder (2) 
(2) 
 D2 D1
(1) 
No radiation leaving surface 1 strikes itself and thus F11 = 0 
All radiation leaving surface 1 strikes surface 2 and thus F12 = 1 
2
1
D
D===⎯→⎯= )1(A :ruley reciprocit
2
1
12
2
1
21212121 hD
hD
F
A
A
FFAF π
π
 
2
1
D
D1 −=−=⎯→⎯=+ 21222221 11 :rulesummation FFFF 
 12-7
Chapter 12 Radiation Heat Transfer 
12-16 The view factors between the rectangular surfaces shown in the figure are to be determined. 
Assumptions The surfaces are diffuse emitters and reflectors. 
Analysis We designate the different surfaces as follows: 
 shaded part of perpendicular surface by (1), 
 bottom part of perpendicular surface by (3), 
 shaded part of horizontal surface by (2), and 
 front part of horizontal surface by (4). 
(a) From Fig.12-6 
 25.0
3
1
3
1
23
1
2
=
⎪⎪⎭
⎪⎪⎬
⎫
=
=
F
W
L
W
L
 and 32.0
3
1
3
2
)31(2
1
2
=
⎪⎪⎭
⎪⎪⎬
⎫
=
=
+→F
W
L
W
L
 
3 m 
(1) 
 
(3) 
1 m 
 
1 m 
1 m (2) 
1 m (4) 
07.025.032.0 :ruleion superposit 23)31(2212321)31(2 =−=−=⎯→⎯+= +→+→ FFFFFF 
0.07==⎯→⎯=⎯→⎯= 211221212121 :ruley reciprocit FFFAFAAA 
 
(b) From Fig.12-6, 
 15.0
3
2 and 
3
1
3)24(
12 =⎭⎬
⎫== →+FW
L
W
L
 and 22.0
3
2 and 
3
2
)31()24(
12 =⎭⎬
⎫== +→+FW
L
W
L
 
07.015.022.0 :ruleion superposit 1)24(3)24(1)24()31()24( =−=⎯→⎯+= →+→+→++→+ FFFF 
14.0)07.0(
3
6
 :ruley reciprocit
1)24(
1
)24(
)24(1
)24(111)24()24(
===⎯→⎯
=
→+
+
+→
+→→++
F
A
A
F
FAFA
 
3 m 
(1) 
 
(3) 
1 m 
 
 
1 m 
1 m 
1 m
(4) 
 
(2) 
0.07=−=⎯→⎯
+=+→
07.014.0
 :ruleion superposit
14
1214)24(1
F
FFF
 
since = 0.07 (from part a). Note that in part (b) is 
equivalent to in part (a). 
F12 F14
F12
(c) We designate 
 shaded part of top surface by (1), 
 remaining part of top surface by (3), 
 remaining part of bottom surface by (4), and 2 m 
 
2 m
1 m 
(1) 
(4) 
(2) 1 m 
1 m 
1 m (3) 
 shaded part of bottom surface by (2). 
From Fig.12-5, 
20.0
2
2
2
2
)31()42(
1
2
=
⎪⎪⎭
⎪⎪⎬
⎫
=
=
+→+F
D
L
D
L
 and 12.0
2
1
2
2
14
1
2
=
⎪⎪⎭
⎪⎪⎬
⎫
=
=
F
D
L
D
L
 
3)42(1)42()31()42( :ruleion superposit →+→++→+ += FFF 
3)42(1)42( :rulesymmetry →+→+ = FF 
Substituting symmetry rule gives 
 F F
F
( ) ( )
( ) ( ) . .2 4 1 2 4 3
2 4 1 3
2
0 20
2
010+ → + →
+ → += = = = 
20.0)10.0)(4()2( :ruley reciprocit )42(1)42(11)42()42()42(11 =⎯→⎯=⎯→⎯= +→+→→+++→ FFFAFA 
0.08=−=⎯→⎯+=⎯→⎯+=+→ 12.020.012.020.0 : ruleion superposit 12121412)42(1 FFFFF 
 12-8
Chapter 12 Radiation Heat Transfer 
12-17 The view factor between the two infinitely long parallel cylinders located a distance s apart from 
each other is to be determined. 
Assumptions The surfaces are diffuse emitters and reflectors. 
Analysis Using the crossed-strings method, the view factor 
between two cylinders facing each other for s/D > 3 is 
determined to be 
D 
D 
(2) 
(1) 
s 
F
s D s
D
1 2
2 2
2
2 2
2 2
− =
−
×
= + −
∑ ∑Crossed strings Uncrossed strings
String on surface 1
( / )π
 
or 
D
sDs
F π
⎟⎠
⎞⎜⎝
⎛ −+
=−
22
21
2
 
 
 
 
12-18 Three infinitely long cylinders are located parallel to 
each other. The view factor between the cylinder in the middle 
and the surroundings is to be determined. 
Assumptions The cylinder surfaces are diffuse emitters and 
reflectors. 
Analysis The view factor between two cylinder facing each 
other is, from Prob. 12-17, 
D 
D 
(1) 
s 
D 
(surr) 
s 
(2) 
(2) 
 
D
sDs
F π
⎟⎠
⎞⎜⎝
⎛ −+
=−
22
21
2
 
Noting that the radiation leaving cylinder 1 that does 
not strike the cylinder will strike the surroundings, and 
this is also the case for the other half of the cylinder, the 
view factor between the cylinder in the middle and the 
surroundings becomes 
 
D
sDs
FF surr π
⎟⎠
⎞⎜⎝
⎛ −+
−=−= −−
22
211
4
121 
 12-9
Chapter 12 Radiation Heat Transfer 
 
Radiation Heat Transfer Between Surfaces 
 
12-19C The analysis of radiation exchange between black surfaces is relatively easy because of the 
absence of reflection. The rate of radiation heat transfer between two surfaces in this case is expressed as 
 where A)( 42
4
1121 TTFAQ −σ=& 1 is the surface area, F12 is the view factor, and T1 and T2 are the 
temperatures of two surfaces. 
 
12-20C Radiosity is the total radiation energy leaving a surface per unit time and per unit area. Radiosity 
includes the emitted radiation energy as well as reflected energy. Radiosity and emitted energy are equal 
for blackbodies since a blackbody does not reflect any radiation.12-21C Radiation surface resistance is given as R
Ai
i
i i
= −1 εε and it represents the resistance of a surface to 
the emission of radiation. It is zero for black surfaces. The space resistance is the radiation resistance 
between two surfaces and is expressed as R
Ai
i
i i
= −1 εε 
 
12-22C The two methods used in radiation analysis are the matrix and network methods. In matrix method, 
equations 12-34 and 12-35 give N linear algebraic equations for the determination of the N unknown 
radiosities for an N -surface enclosure. Once the radiosities are available, the unknown surface 
temperatures and heat transfer rates can be determined from these equations respectively. This method 
involves the use of matrices especially when there are a large number of surfaces. Therefore this method 
requires some knowledge of linear algebra. 
 The network method involves drawing a surface resistance associated with each surface of an 
enclosure and connecting them with space resistances. Then the radiation problem is solved by treating it 
as an electrical network problem where the radiation heat transfer replaces the current and the radiosity 
replaces the potential. The network method is not practical for enclosures with more than three or four 
surfaces due to the increased complexity of the network. 
 
12-23C Some surfaces encountered in numerous practical heat transfer applications are modeled as being 
adiabatic as the back sides of these surfaces are well insulated and net heat transfer through these surfaces 
is zero. When the convection effects on the front (heat transfer) side of such a surface is negligible and 
steady-state conditions are reached, the surface must lose as much radiation energy as it receives. Such a 
surface is called reradiating surface. In radiation analysis, the surface resistance of a reradiating surface is 
taken to be zero since there is no heat transfer through it. 
 12-10
Chapter 12 Radiation Heat Transfer 
12-24E 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 ε = 0.7 for the bottom surface and 1 for other surfaces. 
Analysis We consider the base surface to be surface 1, the top surface to be surface 2 and the side surfaces 
to be surface 3. The cubical furnace can be considered to be three-surface enclosure with a radiation 
network shown in the figure. The areas and blackbody emissive powers of surfaces are 
 223
22
21 ft 400)ft 10(4 ft 100)ft 10( ===== AAA
 
244284
33
244284
22
244284
11
Btu/h.ft 866,56)R 2400)(R.Btu/h.ft 101714.0(
Btu/h.ft 233,11)R 1600)(R.Btu/h.ft 101714.0(
Btu/h.ft 702)R 800)(R.Btu/h.ft 101714.0( 
=×=σ=
=×=σ=
=×=σ=
−
−
−
TE
TE
TE
b
b
b
The view factor from the base to the top surface of the cube is F12 0 2= . . From 
the summation rule, the view factor from the base or top to the side surfaces is 
 F F F F F11 12 13 13 121 1 1 0 2+ + = ⎯ →⎯ = − = − =. .0 8
since the base surface is flat and thus F11 0= . Then the radiation resistances become 
T1 = 800 R 
ε1 = 0.7 
T2 = 1600 R 
ε2 = 1 
 T3 = 2400 R 
 ε3 = 1 
 
2-
2
131
13
2-
2
121
12
2-
2
11
1
1
ft 0125.0
)8.0)(ft 100(
11
ft 0500.0
)2.0)(ft 100(
11 ft 0043.0
)7.0)(ft 100(
7.011
===
====−=−=
FA
R
FA
R
A
R ε
ε
 
Note that the side and the top surfaces are black, and thus their radiosities are equal to their emissive 
powers. The radiosity of the base surface is determined 
 
E J
R
E J
R
E J
R
b b b1 1
1
2 1
12
3 1
13
0
− + − + − = 
Substituting, 
702
0 0043
11 233
0 500
56 866
0 0125
0 15 0541 1 1 1
− + − + − = ⎯ →⎯ =J J J J
.
,
.
,
.
, W / m2 
(a) The net rate of radiation heat transfer between the base and the side surfaces is 
 Btu/h 103.345 6×=−=−=
2-
2
13
13
31
ft 0125.0
Btu/h.ft )054,15866,56(
R
JE
Q b& 
(b) The net rate of radiation heat transfer between the base and the top surfaces is 
 Btu/h 107.642 4×=−=−=
2-
2
12
21
12
ft 05.0
Btu/h.ft )233,11054,15(
R
EJ
Q b& 
The net rate of radiation heat transfer to the base surface is finally determined from 
 Btu/h 103.269 6×=+−=+= 960,344,3420,7631211 QQQ &&&
Discussion The same result can be found form 
 Btu/h 10338.3
ft 0043.0
Btu/h.ft )702054,15( 6
2-
2
1
11
1 ×=−=−= R
EJ
Q b& 
The small difference is due to round-off error. 
 12-11
Chapter 12 Radiation Heat Transfer 
12-25E 
 "!PROBLEM 12-25E" 
 
"GIVEN" 
a=10 "[ft]" 
"epsilon_1=0.7 parameter to be varied" 
T_1=800 "[R]" 
T_2=1600 "[R]" 
T_3=2400 "[R]" 
sigma=0.1714E-8 "[Btu/h-ft^2-R^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
"Consider the base surface 1, the top surface 2, and the side surface 3" 
E_b1=sigma*T_1^4 
E_b2=sigma*T_2^4 
E_b3=sigma*T_3^4 
A_1=a^2 
A_2=A_1 
A_3=4*a^2 
F_12=0.2 "view factor from the base to the top of a cube" 
F_11+F_12+F_13=1 "summation rule" 
F_11=0 "since the base surface is flat" 
R_1=(1-epsilon_1)/(A_1*epsilon_1) "surface resistance" 
R_12=1/(A_1*F_12) "space resistance" 
R_13=1/(A_1*F_13) "space resistance" 
(E_b1-J_1)/R_1+(E_b2-J_1)/R_12+(E_b3-J_1)/R_13=0 "J_1 : radiosity of base surface" 
"(a)" 
Q_dot_31=(E_b3-J_1)/R_13 
"(b)" 
Q_dot_12=(J_1-E_b2)/R_12 
Q_dot_21=-Q_dot_12 
Q_dot_1=Q_dot_21+Q_dot_31 
 
 
ε1 Q31 [Btu/h] Q12 [Btu/h] Q1 [Btu/h] 
0.1 1.106E+06 636061 470376 
0.15 1.295E+06 589024 705565 
0.2 1.483E+06 541986 940753 
0.25 1.671E+06 494948 1.176E+06 
0.3 1.859E+06 447911 1.411E+06 
0.35 2.047E+06 400873 1.646E+06 
0.4 2.235E+06 353835 1.882E+06 
0.45 2.423E+06 306798 2.117E+06 
0.5 2.612E+06 259760 2.352E+06 
0.55 2.800E+06 212722 2.587E+06 
0.6 2.988E+06 165685 2.822E+06 
0.65 3.176E+06 118647 3.057E+06 
0.7 3.364E+06 71610 3.293E+06 
0.75 3.552E+06 24572 3.528E+06 
0.8 3.741E+06 -22466 3.763E+06 
0.85 3.929E+06 -69503 3.998E+06 
0.9 4.117E+06 -116541 4.233E+06 
 
 
 
 12-12
Chapter 12 Radiation Heat Transfer 
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
ε1
Q
31
 [
B
tu
/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
ε1ε1
Q
12
 [
B
tu
/h
]
 
 
 
 
 
 12-13
Chapter 12 Radiation Heat Transfer 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.0x100
5.0x105
1.0x106
1.5x106
2.0x106
2.5x106
3.0x106
3.5x106
4.0x106
4.5x106
ε1ε1
Q
1 
 [B
tu
/h
]
 
 
 
 
 
 12-14
Chapter 12 Radiation Heat Transfer 
12-26 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. 
T2 = 400 K 
ε2 = 0.9 
T1 = 600 K 
ε1 = 0.5 Assumptions 1 Steady operating conditions exist 2 The surfaces 
are opaque, diffuse, and gray. 3 Convection heat transfer is not 
considered. 
Properties The emissivities ε of the plates are given to be 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 directly from 
2 W/m2795=
−+
−⋅×=
−+
−=
−
1
9.0
1
5.0
1
])K 400()K 600)[(KW/m 1067.5(
111
)( 44428
21
4
2
4
112
εε
σ TT
A
Q
s
&
 
 
 
 
 
 12-15
Chapter 12 Radiation Heat Transfer 
12-27 "!PROBLEM12-27" 
"GIVEN" 
T_1=600 "[K], parameter to be varied" 
T_2=400 "[K]" 
epsilon_1=0.5 "parameter to be varied" 
epsilon_2=0.9 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
q_dot_12=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1) 
 
T1 [K] q12 [W/m2] 
500 991.1 
525 1353 
550 1770 
575 2248 
600 2793 
625 3411 
650 4107 
675 4888 
700 5761 
725 6733 
750 7810 
775 9001 
800 10313 
825 11754 
850 13332 
875 15056 
900 16934 
925 18975 
950 21188 
975 23584 
1000 26170 
 
ε1 q12 [W/m2] 
0.1 583.2 
0.15 870 
0.2 1154 
0.25 1434 
0.3 1712 
0.35 1987 
0.4 2258 
0.45 2527 
0.5 2793 
0.55 3056 
0.6 3317 
0.65 3575 
0.7 3830 
0.75 4082 
0.8 4332 
0.85 4580 
0.9 4825 
 
 
 
 12-16
Chapter 12 Radiation Heat Transfer 
500 600 700 800 900 1000
0
5000
10000
15000
20000
25000
30000
T1 [K]
q 1
2 
 [W
/m
2 ]
 
 
 
 
 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
ε1
q 1
2 
 [W
/m
2 ]
 
 
 
 12-17
Chapter 12 Radiation Heat Transfer 
12-28 The base, top, and side surfaces of a furnace of cylindrical shape are black, and are maintained at 
uniform temperatures. The net rate of radiation heat transfer to or from the top surface is to be determined. 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are black. 3 Convection heat transfer is not 
considered. 
Properties The emissivity of all surfaces are ε = 1 since they are black. 
T1 = 700 K 
ε1 = 1 
r1 = 2 m 
T2 = 1200 K 
ε2 = 1 
r2 = 2 m 
T3 = 500 K 
ε3 = 1 
h =2 m
Analysis We consider the top surface to be surface 1, the base 
surface to be surface 2 and the side surfaces to be surface 3. 
The cylindrical furnace can be considered to be three-surface 
enclosure. We assume that steady-state conditions exist. Since 
all surfaces are black, the radiosities are equal to the emissive 
power of surfaces, and the net rate of radiation heat transfer 
from the top surface can be determined from 
 )()( 43
4
1131
4
2
4
1121 TTFATTFAQ −+−= σσ&
and A r1
2 22 12 57= = =π π ( ) . m m2
The view factor from the base to the top surface of the cylinder is F12 0 38= . (From Figure 12-44). The 
view factor from the base to the side surfaces is determined by applying the summation rule to be 
 F F F F F11 12 13 13 121 1 1 0 38+ + = ⎯ →⎯ = − = − =. .0 62
Substituting, 
kW -762=×−=
×+
×=
−σ+−σ=
 W1062.7
)K 1200-K )(700.K W/m1067.5)(62.0)(m 57.12(
)K 500-K )(700.K W/m1067)(0.38)(5.m (12.57
)()(
5
44428-2
44428-2
4
3
4
1131
4
2
4
1121 TTFATTFAQ&
Discussion The negative sign indicates that net heat transfer is to the top surface. 
 12-18
Chapter 12 Radiation Heat Transfer 
12-29 The base and the dome of a hemispherical furnace are maintained at uniform temperatures. The net 
rate of radiation heat transfer from the dome to 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. 
Analysis The view factor is first determined from 
 
F
F F F
11
11 12 12
0
1 1
=
+ = → =
 (flat surface)
 (summation rule)
Noting that the dome is black, net rate of radiation heat transfer 
from dome to the base surface can be determined from 
 
kW 759.4=
×=
−⋅×−=
−−=−=
−
 W10594.7
])K 1000()K 400)[(K W/m1067.5)(1]( /4)m 5()[7.0( 
)(
5
444282
4
2
4
11211221
π
σε TTFAQQ &&
T1 = 400 K 
ε1 = 0.7 
T2 = 1000 K 
ε2 = 1 
D = 5 m 
The positive sign indicates that the net heat transfer is from the dome to the base surface, as expected. 
 
 
 
12-30 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. 
D2 = 0.5 m 
T2 = 500 K 
ε2 = 0.7 
D1 = 0.2 m 
T1 = 950 K 
ε1 = 1 
 Vacuum 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are opaque, diffuse, and gray. 3 Convection 
heat transfer is not considered. 
Properties The emissivities of surfaces are given to be 
ε1 = 1 and ε2 = 0.7. 
Analysis The net rate of radiation heat transfer between 
the two cylinders per unit length of the cylinders is 
determined from 
 
kW 22.87==
⎟⎠
⎞⎜⎝
⎛−+
−⋅×π=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
ε
ε−+ε
−σ=
−
 W870,22
5
2
7.0
7.01
1
1
])K 500(K) 950)[(K W/m1067.5](m) m)(1 2.0([
11
)( 44428
2
1
2
2
1
4
2
4
11
12
r
r
TTA
Q&
 
 12-19
Chapter 12 Radiation Heat Transfer 
12-31 A long cylindrical rod coated with a new material is 
placed in an evacuated long cylindrical enclosure which is 
maintained at a uniform temperature. The emissivity of the 
coating on the rod is to be determined. 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are opaque, diffuse, and gray. 
Properties The emissivity of the enclosure is given to be ε2 = 
0.95. 
Analysis The emissivity of the coating on the rod is determined 
from 
 ( ) ( )
⎟⎠
⎞⎜⎝
⎛−+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
10
1
95.0
95.011
]K 200K 500)[K W/m1067.5](m) m)(1 01.0([ W8
11
)(
1
44428
2
1
2
2
1
4
2
4
11
12
ε
π
ε
ε
ε
σ
r
r
TTA
Q&
 
D2 = 0.1 m 
T2 = 200 K 
ε2 = 0.95 
D1 = 0.01 m 
T1 = 500 K 
ε1 = ? 
 Vacuum 
which gives 
ε1 = 0.074 
 
 
12-32E 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. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, 
diffuse, and gray. 3 Convection heat transfer is not considered. 
Properties The emissivities of surfaces are given to be ε1 = 0.5 
and ε2 = 0.9. 
Analysis The view factor from the base to the dome is first 
determined from 
 
F
F F F
11
11 12 12
0
1 1
=
+ = → =
 (flat surface)
 (summation rule)
The net rate of radiation heat transfer from dome to the base surface 
can be determined from 
T1 = 550 R 
ε1 = 0.5 
T2 = 1800 R 
ε2 = 0.9 
D = 15 ft 
 
Btu/h 101.311 6×=
⎥⎦
⎤⎢⎣
⎡π
−++−
−⋅×−=
ε
ε−++ε
ε−
−σ−=−=
−
)9.0(
2
ft) 1)(ft 15(
9.01
)1)(ft 15(
1
)5.0)(ft 15(
5.01
]R) 1800()R 550)[(RBtu/h.ft 101714.0(
111
)(
22
44428
22
2
12111
1
4
2
4
1
1221
AFAA
TT
QQ &&
 
The positive sign indicates that the net heat transfer is from the dome to the base surface, as expected. 
 
12-33 Two parallel disks whose back sides are insulated are black, and are maintained at a uniform 
temperature. The net rate of radiation heat transfer from the disks to the environment 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. 
 12-20
Chapter 12 Radiation Heat Transfer 
Properties The emissivities of all surfaces are ε = 1 since they are black. 
Analysis Both disks possess same properties and they are 
black. Noting that environment can also be considered to be 
blackbody, we can treat this geometry as a three surface 
enclosure. We consider the two disks to be surfaces 1 and 2 
and the environment to be surface 3. Then from Figure 12-
7, we read 
 
F F
F
12 21
13
0 26
1 0 26 0 74
= =
= − =
.
. . (
 
 summation rule)
The net rate of radiation heat transfer from the disks into 
the environment then becomes 
( ) ( )
 W5505=
−⋅×=
−=
=+=
− ]K 300K 700)[K W/m1067.5]()m 3.0()[74.0(2 
)(2
2
444282
4
3
411133
1323133
π
σ TTAFQ
QQQQ
&
&&&&
 
Disk 1, T1 = 700 K, ε1 = 1 
Disk 2, T2 = 700 K, ε2 = 1 
0.40 m 
Environment 
T3 =300 K 
ε1 = 1 
D = 0.6 m 
 
12-34 A furnace shaped like a long equilateral-triangular duct is considered. The temperature of 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 End effects are neglected. 
Properties The emissivities of surfaces are given to be ε1 = 0.8 and ε2 
= 0.5. 
Analysis This geometry can be treated as a two surface 
enclosure since two surfaces have identical properties. 
We consider base surface to be surface 1 and other two 
surface to be surface 2. Then the view factor between 
the two becomes . The temperature of the base 
surface is determined from 
F12 1=
 ( ) ( ) K 543T1 =⎯→⎯−++−
−⋅×=
−++−
−=
−
)5.0()m 2(
5.01
)1)(m 1(
1
)8.0)(m 1(
8.01
]K 500)[K W/m1067.5(
 W800
111
)(
222
44
1
428
22
2
12111
1
4
2
4
1
12
T
AFAA
TT
Q
ε
ε
ε
ε
σ&
 
q1 = 800 W/m2
ε1 = 0.8 
T2 = 500 K 
ε2 = 0.5 
b = 2 ft 
Note that .m 2 and m 1 22
2
1 == AA
 
 12-21
Chapter 12 Radiation Heat Transfer 
12-35 "!PROBLEM 12-35" 
"GIVEN" 
a=2 "[m]" 
epsilon_1=0.8 
epsilon_2=0.5 
Q_dot_12=800 "[W], parameter to be varied" 
T_2=500 "[K], parameter to be varied" 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
"Consider the base surface to be surface 1, the side surfaces to be surface 2" 
Q_dot_12=(sigma*(T_1^4-T_2^4))/((1-epsilon_1)/(A_1*epsilon_1)+1/(A_1*F_12)+(1-
epsilon_2)/(A_2*epsilon_2)) 
F_12=1 
A_1=1 "[m^2], since rate of heat supply is given per meter square area" 
A_2=2*A_1 
 
Q12 [W] T1 [K] 
500 528.4 
525 529.7 
550 531 
575 532.2 
600 533.5 
625 534.8 
650 536 
675 537.3 
700 538.5 
725 539.8 
750 541 
775 542.2 
800 543.4 
825 544.6 
850 545.8 
875 547 
900 548.1 
925 549.3 
950 550.5 
975 551.6 
1000 552.8 
 
T2 [K] T1 [K] 
300 425.5 
325 435.1 
350 446.4 
375 459.2 
400 473.6 
425 489.3 
450 506.3 
475 524.4 
500 543.4 
525 563.3 
550 583.8 
575 605 
600 626.7 
625 648.9 
 12-22
Chapter 12 Radiation Heat Transfer 
650 671.4 
675 694.2 
700 717.3 
 
500 600 700 800 900 1000
525
530
535
540
545
550
555
Q12 [W]
T 1
 [
K
]
 
 
 
 
 
300 350 400 450 500 550 600 650 700
400
450
500
550
600
650
700
750
T2 [K]
T 1
 [
K
]
 
 
 12-23
Chapter 12 Radiation Heat Transfer 
12-36 The floor and the ceiling of a cubical furnace are maintained at uniform temperatures. The net rate of 
radiation heat transfer between the floor and the ceiling 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 emissivities of all surfaces are ε = 1 since they are black or reradiating. 
Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the side surfaces to be 
surface 3. The furnace can be considered to be three-surface enclosure with a radiation network shown in 
the figure. We assume that steady-state conditions exist. Since the side surfaces are reradiating, there is no 
heat transfer through them, and the entire heat lost by the ceiling must be gained by the floor. The view 
factor from the ceiling to the floor of the furnace is F12 0 2= . . Then the rate of heat loss from the ceiling 
can be determined from 
 
1
231312
21
1
11
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛
++
−=
RRR
EE
Q bb& 
T2 = 550 K 
ε2 = 1 
T1 = 1100 K 
ε1 = 1 
Reradiating side 
surfacess 
 a = 4 m 
where 
 
244284
22
244284
11
W/m 5188)K 550)(K.W/m 1067.5(
W/m 015,83)K 1100)(K.W/m 1067.5(
=×==
=×==
−
−
TE
TE
b
b
σ
σ
and 
 A A1 2
24 16= = =( ) m 2m
 
R
A F
R R
A F
12
1 12
13 23
1 13
1 1
16 0 2
0 3125
1 1
16 08
0 078125
= = =
= = = =
( )( . )
.
( )( . )
.
 m
 m
 m
 m
2
-2
2
-2
 
Substituting, 
 kW 747=×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−= − W1047.7
)m 078125.0(2
1
m 3125.0
1
 W/m)5188015,83( 5
1
2-2-
2
12Q& 
 12-24
Chapter 12 Radiation Heat Transfer 
12-37 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. 
Properties The emissivities of surfaces are given to be ε1 = 0.1 
and ε2 = 0.8. 
Analysis The net rate of radiation heat transfer between the two 
spheres is 
 ( ) ( )
 W1669=
⎟⎠
⎞⎜⎝
⎛−+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
2
444282
2
2
2
1
2
2
1
4
2
4
11
12
m 4.0
m 15.0
7.0
7.01
5.0
1
]K 400K 700)[K W/m1067.5](m) 3.0([
11
)(
π
ε
ε
ε
σ
r
r
TTA
Q&
 
D2 = 0.8 m 
T2 = 400 K 
ε2 = 0.7 
D1 = 0.3 m 
T1 = 700 K 
ε1 = 0.5 
 
 
Tsurr = 30°C 
T∞ = 30°C 
ε = 0.35 
Radiation heat transfer rate from the outer sphere to the surrounding surfaces are 
W685])K 27330()K 400)[(K W/m1067.5](m) 8.0()[1)(35.0( 
)(
444282
44
22
 =+−⋅×π=
−σε=
−
surrrad TTFAQ& 
The convection heat transfer rate at the outer surface of the cylinder is determined from requirement that 
heat transferred from the inner sphere to the outer sphere must be equal to the heat transfer from the outer 
surface of the outer sphere to the environment by convection and radiation. That is, 
 W9845685166912 =−=−= radconv QQQ &&&
Then the convection heat transfer coefficient becomes 
 
( )[ ] C W/m5.04 2 °⋅=⎯→⎯π= −= ∞ hh TThAQconv K) 303-K (400m) 8.0( W984 222.
&
 12-25
Chapter 12 Radiation Heat Transfer 
12-38 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 ε1 = 0.1 and ε2 = 0.8. 
Analysis We take the sphere to be surface 1 and the surrounding 
cubic enclosure to be surface 2. Noting that F12 1= , for this 
two-surface enclosure, the net rate of radiation heat transfer to 
liquid nitrogen can be determined from 
 
( )
[ ]( )( ) ( )[ ]
W 228=
⎥⎥⎦
⎤
⎢⎢⎣
⎡−+
−⋅×−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−−=−=
−
2
2
444282
2
1
2
2
1
4
2
4
11
1221
m) 6(3
m) 2(
8.0
8.01
1.0
1
K 240K 100K W/m1067.5m) 2(
11
π
π
ε
ε
ε
σ
A
A
TTA
QQ &&
 
Cube, a =3 m 
T2 = 240 K 
ε2 = 0.8 
D1 = 2 m 
T1 = 100 K 
ε1 = 0.1 
 Vacuum 
Liquid 
N2
 
 
 
12-39 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 ε1 
= 0.1 and ε2 = 0.8. D2 = 3 m 
T2 = 240 K 
ε2 = 0.8 
D1 = 2 m 
T1 = 100 K 
ε1 = 0.1 
 Vacuum 
Liquid 
N2
Analysis The net rate of radiation heat transfer to liquid 
nitrogen can be determined from 
 
( ) ( )
 W227=
⎟⎟⎠⎞
⎜⎜⎝
⎛−+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
2
2
444282
2
2
2
1
2
2
1
4
2
4
11
12
m) (1.5
m) 1(
8.0
8.01
1.0
1
]K 100K 240)[K W/m1067.5](m) 2([
11
)(
π
ε
ε
ε
σ
r
r
TTA
Q&
 
 
 
 12-26
Chapter 12 Radiation Heat Transfer 
12-40 "!PROBLEM 12-40" 
 
"GIVEN" 
D=2 "[m]" 
a=3 "[m], parameter to be varied" 
T_1=100 "[K]" 
T_2=240 "[K]" 
epsilon_1=0.1 "parameter to be varied" 
epsilon_2=0.8 "parameter to be varied" 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
"Consider the sphere to be surface 1, the surrounding cubic enclosure to be surface 2" 
Q_dot_12=(A_1*sigma*(T_1^4-T_2^4))/(1/epsilon_1+(1-epsilon_2)/epsilon_2*(A_1/A_2)) 
Q_dot_21=-Q_dot_12 
A_1=pi*D^2 
A_2=6*a^2 
 
a [m] Q21 [W] 
2.5 227.4 
2.625 227.5 
2.75 227.7 
2.875 227.8 
3 227.9 
3.125 228 
3.25 228.1 
3.375 228.2 
3.5 228.3 
3.625 228.4 
3.75 228.4 
3.875 228.5 
4 228.5 
4.125 228.6 
4.25 228.6 
4.375 228.6 
4.5 228.7 
4.625 228.7 
4.75 228.7 
4.875 228.8 
5 228.8 
 
 12-27
Chapter 12 Radiation Heat Transfer 
 
ε1 Q21 [W] 
0.1 227.9 
0.15 340.9 
0.2 453.3 
0.25 565 
0.3 676 
0.35 786.4 
0.4 896.2 
0.45 1005 
0.5 1114 
0.55 1222 
0.6 1329 
0.65 1436 
0.7 1542 
0.75 1648 
0.8 1753 
0.85 1857 
0.9 1961 
 
ε2 Q21 [W] 
0.1 189.6 
0.15 202.6 
0.2 209.7 
0.25 214.3 
0.3 217.5 
0.35 219.8 
0.4 221.5 
0.45 222.9 
0.5 224.1 
0.55 225 
0.6 225.8 
0.65 226.4 
0.7 227 
0.75 227.5 
0.8 227.9 
0.85 228.3 
0.9 228.7 
 12-28
Chapter 12 Radiation Heat Transfer 
 
 
2.5 3 3.5 4 4.5 5
227.2
227.4
227.6
227.8
228
228.2
228.4
228.6
228.8
a [m]
Q
21
 [
W
]
 
 
 
 
 
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
ε1
Q
21
 [
W
]
 
 
 
 
 
 12-29
Chapter 12 Radiation Heat Transfer 
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
ε2
Q
21
 [
W
]
 
 
 
 12-30
Chapter 12 Radiation Heat Transfer 
12-41 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. 
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 ε = 1 for all surfaces since they 
are black or reradiating. 
Steaks, T2 = 278 K, ε2 = 1
Coal bricks, T1 = 1100 K, ε1 = 1 
0.20 m 
Analysis We consider the coal bricks to be surface 1, the steaks to 
be surface 2 and the side surfaces to be surface 3. First we 
determine the view factor between the bricks and the steaks (Table 
12-1), 
 75.0
m 0.20
m 15.0 ====
L
r
RR iji 
 7778.3
0.75
75.0111
2
2
2
2
=+=++=
i
j
R
R
S 
 2864.0
75.0
75.047778.37778.3
2
14
2
1
2/12
2
2/12
2
12 =⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−−=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−−==
i
j
ij R
R
SSFF 
(It can also be determined from Fig. 12-7). 
Then the initial rate of radiation heat transfer from the coal bricks to the stakes becomes 
 
 W1674=
−⋅×=
−=
− ])K 278()K 1100)[(K W/m1067.5](4/m) 3.0()[2864.0(
)(
444282
4
2
4
111212
π
σ TTAFQ&
When the side opening is closed with aluminum foil, the entire heat lost by the coal bricks must be gained 
by the stakes since there will be no heat transfer through a reradiating surface. The grill can be considered 
to be three-surface enclosure. Then the rate of heat loss from the room can be determined from 
 
1
231312
21
1
11
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛
++
−=
RRR
EE
Q bb& 
where 
 W / m K K W / m
 W / m K K W / m
4 2 2
4 2 2
E T
E T
b
b
1 1
8 4 4
2 2
8 4 4
67 10 1100 83 015
67 10 18 273 407
= = × =
= = × + =
−
−
σ
σ
(5. . )( ) ,
(5. . )( )
and A A1 2
20 3 0 07069= = =π( . ) . m
4
 m2 
 
2-
2
131
2313
2-
2
121
12
m 82.19
)2864.01)(m 07069.0(
11
m 39.49
)2864.0)(m 07069.0(
11
=−===
===
FA
RR
FA
R
 
Substituting, W 3757=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−= −1
2-2-
2
12
)m 82.19(2
1
m 39.49
1
 W/m)407015,83(Q& 
 12-31
Chapter 12 Radiation Heat Transfer 
12-42E A room is heated by lectric resistance heaters placed on the ceiling which is maintained at a 
uniform temperature. The rate of heat loss from the room through the floor 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 There is no heat loss through the side surfaces. 
Properties The emissivities are ε = 1 for the ceiling and ε = 0.8 for the floor. The emissivity of insulated 
(or reradiating) surfaces is also 1. 
Analysis The room can be considered to be three-surface enclosure 
with the ceiling surface 1, the floor surface 2 and the side surfaces 
surface 3. We assume steady-state conditions exist. Since the side 
surfaces are reradiating, there is no heat transfer through them, and the 
entire heat lost by the ceiling must be gained by the floor. Then the 
rate of heat loss from the room through its floor can be determined 
from 
T2 = 90°F 
ε2 = 0.8 
Ceiling: 12 ft × 12 ft 
 
T1 = 90°F 
ε1 = 1 
Insulated side 
surfacess 9 ft
 
2
1
231312
21
1
11 R
RRR
EE
Q bb
+⎟⎟⎠
⎞
⎜⎜⎝
⎛
++
−= −& 
where 
244284
22
244284
11
Btu/h.ft 130)R 46065)(R.Btu/h.ft 101714.0(
Btu/h.ft 157)R 46090)(R.Btu/h.ft 101714.0(
=+×=σ=
=+×=σ=
−
−
TE
TE
b
b 
and 
 A A1 2
212 144= = =( ) ft ft2
The view factor from the floor to the ceiling of the room is F12 0 27= . (From Figure 12-42). The view 
factor from the ceiling or the floor to the side surfaces is determined by applying the summation rule to be 
 F F F F F11 12 13 13 121 1 1 0 27+ + = ⎯ →⎯ = − = − =. .0 73
since the ceiling is flat and thus . Then the radiation resistances which appear in the equation above 
become 
F11 0=
 
2-
2
131
2313
2-
2
121
12
2-
2
22
2
2
ft 009513.0
)73.0)(ft 144(
11
ft 02572.0
)27.0)(ft 144(
11
ft 00174.0
)8.0)(ft 144(
8.011
====
===
=−=−=
FA
RR
FA
R
A
R ε
ε
 
Substituting, 
 Btu/h 2130=
+⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
−= −
2-
1
2-2-
2
12
ft 00174.0
)ft 009513.0(2
1
ft 02572.0
1
Btu/h.ft )130157(Q& 
 12-32
Chapter 12 Radiation Heat Transfer 
12-43 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. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 
Convection heat transfer is not considered. 
Properties The emissivities of the horizontal rectangle and the surroundings are ε = 0.75 and ε = 0.85, 
respectively. 
Analysis We consider the horizontal rectangle to be surface 1, the vertical rectangle to be surface 2 and the 
surroundings to be surface 3. This system can be considered to be a three-surface enclosure. The view 
factor from surface 1 to surface 2 is determined from 
 27.0
75.0
6.1
2.1
5.0
6.1
8.0
12
2
1
=
⎪⎪⎭
⎪⎪⎬
⎫
==
==
F
W
L
W
L
 (Fig. 12-6) W = 1.6 m 
(2) L2 = 1.2 m
 L1= 0.8 m A1 (1) 
A2
T3 = 290 K 
ε3 = 0.85 
T2 = 550 K 
ε2 = 1 
T1 =400 K 
ε1 =0.75 
(3)
The surface areas are 
2
2
2
1
m 92.1)m 6.1)(m 2.1(
m 28.1)m 6.1)(m 8.0(
==
==
A
A
 
222
3 m 268.36.12.18.02
8.02.12 =×++××=A 
Note that the surface area of the surroundings is determined assuming that surroundings forms flat surfaces 
at all openings to form an enclosure. Then other view factors are determined to be 
18.0)92.1()27.0)(28.1( 2121212121 =⎯→⎯=⎯→⎯= FFFAFA (reciprocity rule) 
 (summation rule) 73.0127.001 1313131211 =⎯→⎯=++⎯→⎯=++ FFFFF
 (summation rule) 82.01018.01 2323232221 =⎯→⎯=++⎯→⎯=++ FFFFF
29.0)268.3()73.0)(28.1( 3131313131 =⎯→⎯=⎯→⎯= FFFAFA (reciprocity rule) 
48.0)268.3()82.0)(92.1( 3232323232 =⎯→⎯=⎯→⎯= FFFAFA (reciprocity rule) 
We now apply Eq. 9-52b to each surface to determine the radiosities. 
Surface 1: 
[ ]
[ ])(73.0)(27.0
75.0
75.01)K 400)(K.W/m 1067.5(
)()(
1
 
31211
4428
31132112
1
1
1
4
1
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
Surface 2: 2
4428
2
4
2 )K 550)(K.W/m 1067.5( JJT =×⎯→⎯= −σ
Surface 3: 
[ ]
[ ])(48.0)(29.0
85.0
85.01)K 290)(K.W/m 1067.5(
)()(
1
 
31213
4428
23321331
3
3
3
4
3
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
Solving the above equations, we find 
 23
2
2
2
1 W/m 5.811 ,W/m 5188 ,W/m 1587 === JJJ
Then the net rate of radiation heat transfers between the two surfaces and between the horizontal surface 
and the surroundings are determined to be 
 W1245=−−=−−=−= 22211211221 W/m)51881587)(27.0)(m 28.1()( JJFAQQ &&
 W725=−=−= 223113113 W/m)5.8111587)(73.0)(m 28.1()( JJFAQ&
 12-33
Chapter 12 Radiation Heat Transfer 
12-44 Two long parallel cylinders are maintained at specified temperatures. The rates of radiation heat 
transfer between the cylinders and between the hot cylinder and the surroundings are to be determined. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are black. 3 Convection heat transfer is 
not considered. 
Analysis We consider the hot cylinder to be surface 1, cold cylinder to be surface 2, and the surroundings 
to be surface 3. Using the crossed-strings method, the view factor between two cylinders facing each other 
is determined to be 
F s D s
D1 2
2 2
2
2 2
2 2−
= −× =
+ −∑ ∑Crossed strings Uncrossed strings
String on surface 1 ( / )π 
or 099.0
)16.0(
5.016.05.022 2222
21 =π
⎟⎠
⎞⎜⎝
⎛ −+
=π
⎟⎠
⎞⎜⎝
⎛ −+
=− D
sDs
F 
D 
D (2) 
(1) s
T3 = 300 K 
ε3 = 1 
T2 = 275 K 
ε2 = 1 
T1 = 425 K 
ε1 = 1 
(3) 
The view factor between the hot cylinder and the 
surroundings is 
901.0099.011 1213 =−=−= FF (summation rule) 
The rate of radiation heat transfer between the 
cylinders per meter length is 
2m 2513.02/m) 1)(m 16.0(2/ === ππDLA 
 W38.0=−°×=−σ= − 44428242411212 K)275425)(C. W/m1067.5)(099.0)(m 2513.0()( TTAFQ& 
Note that half of the surface area of the cylinder is used, which is the only area that faces the other 
cylinder. The rate of radiation heat transfer between the hot cylinder and the surroundings per meter length 
of the cylinder is 
 21 m 5027.0m) 1)(m 16.0( === ππDLA
 W629.8=−°×=−σ= − 444282434113113 K)300425)(C. W/m1067.5)(901.0)(m 5027.0()( TTFAQ&
 12-34
Chapter 12 Radiation Heat Transfer 
12-45 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 ε = 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(
===
==
ππDLA
A
 
 (summation rule) 1101 12121211 =⎯→⎯=+⎯→⎯=+ FFFF
The temperature of the base surface is determined from 
K 684.8=⎯→⎯−+
−⋅×=
ε
ε−+
−σ=
−
1
22
44
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
11
)(
T
TK
AFA
TT
Q&
 
T1 = ? 
ε1 = 1 
T2 = 650 K 
ε2 = 0.4 
D = 1 m 
 
 
12-46 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 ε = 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 0628.02/)m 2.0(2/
m 0314.04/)m 2.0(4/
===
===
ππ
ππ
DA
DA
 
 (summation rule) 1101 12121211 =⎯→⎯=+⎯→⎯=+ FFFF
The emissivity of the dome is determined from 
 
0.21=ε⎯→⎯
ε
ε−++−
−⋅×−=
ε
ε−++ε
ε−
−σ−=−=
−
2
2
2
2
22
44428
22
2
12111
1
4
2
4
1
1221
)m 0628.0(
1
)1)(m 0314.0(
1
)55.0)(m 0314.0(
55.01
]K) 600(K) 400)[(K W/m1067.5( W50
111
)(
AFAA
TT
QQ &&
 
T1 = 400 K 
ε1 = 0.55 
T2 = 600 K 
ε2 = ? 
D = 0.2 m 
 12-35
Chapter 12 Radiation Heat Transfer 
 
Radiation Shields and The Radiation Effect 
 
12-47C Radiation heat transfer between two surfaces can be reduced greatly by inserting a thin, high 
reflectivity(low emissivity) sheet of material between the two surfaces. Such highly reflective thin plates or 
shells are known as radiation shields. Multilayer radiation shields constructed of about 20 shields per cm. 
thickness separated by evacuated space are commonly used in cryogenic and space applications to 
minimize heat transfer. Radiation shields are also used in temperature measurements of fluids to reduce the 
error caused by the radiation effect. 
 
12-48C The influence of radiation on heat transfer or temperature of a surface is called the radiation effect. 
The radiation exchange between the sensor and the surroundings may cause the thermometer to indicate a 
different reading for the medium temperature. To minimize the radiation effect, the sensor should be coated 
with a material of high reflectivity (low emissivity). 
 
12-49C A person who feels fine in a room at a specified temperature may feel chilly in another room at the 
same temperature as a result of radiation effect if the walls of second room are at a considerably lower 
temperature. For example most people feel comfortable in a room at 22°C if the walls of the room are also 
roughly at that temperature. When the wall temperature drops to 5°C for some reason, the interior 
temperature of the room must be raised to at least 27°C to maintain the same level of comfort. Also, people 
sitting near the windows of a room in winter will feel colder because of the radiation exchange between the 
person and the cold windows. 
 
12-50 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, diffuse, and gray. 3 Convection heat 
transfer is not considered. 
Properties The emissivity of the person is given to be ε1 = 
0.7. 
Analysis (a) Noting that the view factor from the person to 
the walls , the rate of heat loss from that person to 
the walls at a large room whichare at a temperature of 300 
K is 
F12 1=
 W26.9=
−⋅×=
−σε=
− ])K 300()K 303)[(K W/m1067.5)(m 7.1)(1)(85.0( 
)(
444282
4
2
4
1112112 TTAFQ&
 
T2
Qrad
ROOM 
T1 = 30°C 
ε1 = 0.85 
A = 1.7 m2
(b) When the walls are at a temperature of 280 K, 
 W187=
−⋅×=
−σε=
− ])K 280()K 303)[(K W/m1067.5)(m 7.1)(1)(85.0( 
)(
444282
4
2
4
1112112 TTAFQ&
 
 12-36
Chapter 12 Radiation Heat Transfer 
12-51 A thin aluminum sheet is placed between two very large parallel plates that are maintained at 
uniform temperatures. The net rate of radiation heat transfer between the two plates is to be determined for 
the cases of with and without the shield. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 
Convection heat transfer is not considered. 
Properties The emissivities of surfaces are given to be ε1 = 0.5, ε2 = 0.8, and ε3 = 0.15. 
Analysis The net rate of radiation heat transfer with a 
thin aluminum shield per unit area of the plates is 
T2 = 650 K 
ε2 = 0.8 
 
T1 = 900 K 
ε1 = 0.5 
Radiation shield
ε3 = 0.15 
2 W/m1857=
⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −++⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+
−=
−
1
15.0
1
15.0
11
8.0
1
5.0
1
])K 650()K 900)[(K W/m1067.5(
 
111111
)(
44428
2,31,321
4
2
4
1
shield one,12
εεεε
σ TT
Q&
 
The net rate of radiation heat transfer between the plates in the case of no shield is 
2
44428
21
4
2
4
1
shield ,12 W/m035,12
1
8.0
1
5.0
1
])K 650()K 900)[(K W/m1067.5(
111
)( =
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+
−=
−
εε
σ TT
Q no& 
Then the ratio of radiation heat transfer for the two cases becomes 
 
&
& ,
,
,
Q
Q
12
12
1857
12 035
one shield
no shield
 W
 W
= ≅ 1
6
 
 
 
 
 12-37
Chapter 12 Radiation Heat Transfer 
12-52 "!PROBLEM 12-52" 
 
"GIVEN" 
"epsilon_3=0.15 parameter to be varied" 
T_1=900 "[K]" 
T_2=650 "[K]" 
epsilon_1=0.5 
epsilon_2=0.8 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1/epsilon_1+1/epsilon_2-
1)+(1/epsilon_3+1/epsilon_3-1)) 
 
 
 
 
ε3 Q12,1 shield [W/m2] 
0.05 656.5 
0.06 783 
0.07 908.1 
0.08 1032 
0.09 1154 
0.1 1274 
0.11 1394 
0.12 1511 
0.13 1628 
0.14 1743 
0.15 1857 
0.16 1969 
0.17 2081 
0.18 2191 
0.19 2299 
0.2 2407 
0.21 2513 
0.22 2619 
0.23 2723 
0.24 2826 
0.25 2928 
 
 
 
 
 12-38
Chapter 12 Radiation Heat Transfer 
0.05 0.1 0.15 0.2 0.25
500
1000
1500
2000
2500
3000
ε3
Q
12
,1
sh
ie
ld
 [
W
/m
2 ]
 
 
 12-39
Chapter 12 Radiation Heat Transfer 
12-53 Two very large plates are maintained at uniform temperatures. The number of thin aluminum sheets 
that will reduce the net rate of radiation heat transfer between the two plates to one-fifth 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 emissivities of surfaces are given to be ε1 = 0.2, ε2 = 0.2, and ε3 = 0.15. 
Analysis The net rate of radiation heat transfer between the plates in the case of no shield is 
2
44428
21
4
2
4
1
shield ,12
 W/m3720
1
2.0
1
2.0
1
])K 800()K 1000)[(K W/m1067.5(
111
)(
=
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
TT
Q no&
 
The number of sheets that need to be inserted in order to 
reduce the net rate of heat transfer between the two 
plates to onefifth can be determined from 
 
3≅=⎯→⎯
⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε+⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
92.2
1
15.0
1
15.0
11
2.0
1
2.0
1
])K 800()K 1000)[(K W/m1067.5( ) W/m(3720
5
1
111111
)(
shield
shield
44428
2
2,31,3
shield
21
4
2
4
1
shields,12
N
N
N
TT
Q&
 
T2 = 800 K 
ε2 = 0.2 
T1 = 1000 K 
ε1 = 0.2 
Radiation shields 
ε3 = 0.15 
 
 
 
12-54 Five identical thin aluminum sheets are placed between two very large parallel plates which are 
maintained at uniform temperatures. The net rate of radiation heat transfer between the two plates is to be 
determined and compared with that without the shield. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 
Convection heat transfer is not considered. 
Properties The emissivities of surfaces are given to be 
ε1 = ε2 = 0.1 and ε3 = 0.1. 
T2 = 450 K 
ε2 = 0.1 
T1 = 800 K 
ε1 = 0.1 
Radiation shields 
ε3 = 0.1 
Analysis Since the plates and the sheets have the same 
emissivity value, the net rate of radiation heat transfer 
with 5 thin aluminum shield can be determined from 
2 W/m183=
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×
+=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+
−
+=+=
−
1
1.0
1
1.0
1
])K 450()K 800)[(K W/m1067.5(
15
1
111
)(
1
1
1
1
44428
21
4
2
4
1
shield no ,12shield 5,12
εε
σ TT
N
Q
N
Q &&
 
The net rate of radiation heat transfer without the shield is 
 W1098=×=+=⎯→⎯+= W1836)1(1
1
shield 5,12shield no ,12shield no ,12shield 5,12 QNQQN
Q &&&& 
 12-40
Chapter 12 Radiation Heat Transfer 
12-55 "!PROBLEM 12-55" 
 
"GIVEN" 
N=5 "parameter to be varied" 
epsilon_3=0.1 
"epsilon_1=0.1 parameter to be varied" 
epsilon_2=epsilon_1 
T_1=800 "[K]" 
T_2=450 "[K]" 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
Q_dot_12_shields=1/(N+1)*Q_dot_12_NoShield 
Q_dot_12_NoShield=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1) 
 
 
N Q12,shields [W/m2] 
1 550 
2 366.7 
3 275 
4 220 
5 183.3 
6 157.1 
7 137.5 
8 122.2 
9 110 
10 100 
 
 
ε1 Q12,shields [W/m2] 
0.1 183.3 
0.15 282.4 
0.2 387 
0.25 497.6 
0.3 614.7 
0.35 738.9 
0.4 870.8 
0.45 1011 
0.5 1161 
0.55 1321 
0.6 1493 
0.65 1677 
0.7 1876 
0.75 2090 
0.8 2322 
0.85 2575 
0.9 2850 
 
 
 
 12-41
Chapter 12 Radiation Heat Transfer 
1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
500
600
N
Q
12
,s
hi
el
ds
 [
W
/m
2 ]
 
 
 
 
 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0
500
1000
1500
2000
2500
3000
ε1
Q
12
,s
hi
el
ds
 [
W
/m
2 ]
 
 
 12-42
Chapter 12 Radiation Heat Transfer 
12-56E 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. 
T1 = 1200 R, ε1 = 1 
ε3 = 0.15 
T2 = 700 R, ε2 = 1 
1 ft 
1 ft 
T∞ = 540 K 
ε3 = 1 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are black. 3 Convection heat transfer is not 
considered. 
Properties The emissivities of surfaces are given to be 
ε1 = ε2 = 1 and ε3 = 0.15. 
Analysis From Fig. 12-44 we have 52.01332 == FF . 
Then . 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 can expressed as 
48.052.0134 =−=F
]})K 540([48.0])R 1200([(52.0){RBtu/h.ft 101714.0()ft 069.7(15.0
)]([)]([
44
3
44
3
4282
4
2
4
3323
4
1
4
33133
−+−⋅×=
−σε+−σε=
− TT
TTFATTFAQ&
 
Similarly, for the bottom surface of the middle disk, we have 
]})K 540([52.0])R 700([(48.0){RBtu/h.ft 101714.0()ft 069.7(15.0
)]([)]([
44
3
44
3
4282
4
5
4
3353
4
4
4
33433
−+−⋅×=
−σε+−σε=−
− TT
TTFATTFAQ&
Combining the equations above, the rate of heat transfer between the disks through the radiation shield (the 
middle disk) is determined tobe 
 and TBtu/h 866=Q& 3 = 895 K 
 12-43
Chapter 12 Radiation Heat Transfer 
12-57 A radiation shield is placed between two large parallel plates which are maintained at uniform 
temperatures. The emissivity of the radiation shield is to be determined if the radiation heat transfer 
between the plates is reduced to 15% of that without the radiation shield. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 
Convection heat transfer is not considered. 
Properties The emissivities of surfaces are given to be ε1 = 0.6 and ε2 = 0.9. 
Analysis First, the net rate of radiation heat transfer between the two large parallel plates per unit area 
without a shield is 
2
44428
21
4
2
4
1
shield no,12 W/m4877
1
9.0
1
6.0
1
])K 400()K 650)[(K W/m1067.5(
111
)( =
−+
−⋅×=
−+
−=
−
εε
σ TT
Q& 
The radiation heat transfer in the case of one shield is 
 
22
shield no,12shield one,12
W/m 6.731W/m 487715.0
15.0
=×=
×= QQ &&
Then the emissivity of the radiation shield becomes 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −++⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+
−=
−
121
9.0
1
6.0
1
])K 400()K 650)[(K W/m1067.5( W/m731.6
 
111111
)(
 
3
44428
2
2,31,321
4
2
4
1
shield one,12
ε
εεεε
σ TT
Q&
 
T2 = 400 K 
ε2 = 0.9 
 
T1 = 650 K 
ε1 = 0.6 
Radiation shield
ε3
which gives 0.18=3ε 
 
 
 
 12-44
Chapter 12 Radiation Heat Transfer 
12-58 "!PROBLEM 12-58" 
 
"GIVEN" 
T_1=650 "[K]" 
T_2=400 "[K]" 
epsilon_1=0.6 
epsilon_2=0.9 
"PercentReduction=85 [%], parameter to be varied" 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
Q_dot_12_NoShield=(sigma*(T_1^4-T_2^4))/(1/epsilon_1+1/epsilon_2-1) 
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1/epsilon_1+1/epsilon_2-
1)+(1/epsilon_3+1/epsilon_3-1)) 
Q_dot_12_1shield=(1-PercentReduction/100)*Q_dot_12_NoShield 
 
Percent 
Reduction [%] 
ε3
40 0.9153 
45 0.8148 
50 0.72 
55 0.6304 
60 0.5455 
65 0.4649 
70 0.3885 
75 0.3158 
80 0.2466 
85 0.1806 
90 0.1176 
95 0.05751 
 
40 50 60 70 80 90 100
0
0.2
0.4
0.6
0.8
1
PercentReduction [%]
ε 3
 
 
 12-45
Chapter 12 Radiation Heat Transfer 
12-59 A coaxial radiation shield is placed between two coaxial cylinders which are maintained at uniform 
temperatures. The net rate of radiation heat transfer between the two cylinders is to be determined and 
compared with that without the shield. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 
Convection heat transfer is not considered. 
Properties The emissivities of surfaces are given to be ε1 = 0.7, ε2 = 0.4. and ε3 = 0.2. 
Analysis The surface areas of the cylinders and the shield per unit length are 
 
2
33shield
2
22outerpipe,
2
11innerpipe,
m 942.0)m 1)(m 3.0(
m 314.0)m 1)(m 1.0(
m 628.0)m 1)(m 2.0(
====
====
====
ππ
ππ
ππ
LDAA
LDAA
LDAA
The net rate of radiation heat transfer between the two cylinders with a shield per unit length is 
 W703=
−++−++−
−⋅×=
−++−+−++−
−=
−
)4.0)(942.0(
4.01
)1)(628.0(
1
)2.0)(628.0(
2.012
)1)(314.0(
1
)7.0)(314.0(
7.01
])K 500()K 750)[(K W/m1067.5(
111111
)(
44428
22
2
2,332,33
2,3
1,33
1,3
13111
1
4
2
4
1
shield one,12
ε
ε
ε
ε
ε
ε
ε
ε
σ
AFAAAFAA
TT
Q&
 
If there was no shield, D2 = 0.3 m 
T2 = 500 K 
ε2 = 0.4 
D1 = 0.1 m 
T1 = 750 K 
ε1 = 0.7 
Radiation shield 
D3 = 0.2 m 
ε3 = 0.2 
 
 W7465=
⎟⎠
⎞⎜⎝
⎛−+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
3.0
1.0
4.0
4.01
7.0
1
])K 500()K 750)[(K W/m1067.5(
11
)(
44428
2
1
2
2
1
4
2
4
1
shield no,12
D
D
TT
Q
ε
ε
ε
σ&
 
Then their ratio becomes 
 
&
&
,
,
Q
Q
12
12
703
7465
one shield
no shield
 W
 W
= = 0.094 
 
 
 
 
 12-46
Chapter 12 Radiation Heat Transfer 
12-60 "!PROBLEM 12-60" 
 
"GIVEN" 
D_1=0.10 "[m]" 
D_2=0.30 "[m], parameter to be varied" 
D_3=0.20 "[m]" 
epsilon_1=0.7 
epsilon_2=0.4 
epsilon_3=0.2 "parameter to be varied" 
T_1=750 "[K]" 
T_2=500 "[K]" 
sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" 
 
"ANALYSIS" 
L=1 "[m], a unit length of the cylinders is considered" 
A_1=pi*D_1*L 
A_2=pi*D_2*L 
A_3=pi*D_3*L 
F_13=1 
F_32=1 
Q_dot_12_1shield=(sigma*(T_1^4-T_2^4))/((1-
epsilon_1)/(A_1*epsilon_1)+1/(A_1*F_13)+(1-epsilon_3)/(A_3*epsilon_3)+(1-
epsilon_3)/(A_3*epsilon_3)+1/(A_3*F_32)+(1-epsilon_2)/(A_2*epsilon_2)) 
 
D2 [m] Q12,1 shield [W] 
0.25 692.8 
0.275 698.6 
0.3 703.5 
0.325 707.8 
0.35 711.4 
0.375 714.7 
0.4 717.5 
0.425 720 
0.45 722.3 
0.475 724.3 
0.5 726.1 
 
ε3 Q12,1 shield [W] 
0.05 211.1 
0.07 287.8 
0.09 360.7 
0.11 429.9 
0.13 495.9 
0.15 558.7 
0.17 618.6 
0.19 675.9 
0.21 730.6 
0.23 783 
0.25 833.1 
0.27 881.2 
0.29 927.4 
0.31 971.7 
0.33 1014 
0.35 1055 
 
 
 12-47
Chapter 12 Radiation Heat Transfer 
 
0.25 0.3 0.35 0.4 0.45 0.5
690
695
700
705
710
715
720
725
730
D2 [m]
Q
12
,1
sh
ie
ld
 [
W
]
 
 
 
 
 
0.05 0.1 0.15 0.2 0.25 0.3 0.35
200
300
400
500
600
700
800
900
1000
1100
ε3
Q
12
,1
sh
ie
ld
 [
W
]
 
 
 
 12-48
Chapter 12 Radiation Heat Transfer 
 
Radiation Exchange with Absorbing and Emitting Gases 
 
12-61C A nonparticipating medium is completely transparent to thermal radiation, and thus it does not 
emit, absorb, or scatter radiation. A participating medium, on the other hand, emits and absorbs radiation 
throughout its entire volume. 
 
12-62C Spectral transmissivity of a medium of thickness L is the ratio of the intensity of radiation leaving 
the medium to that entering the medium, and is expressed as LL e
I
I λκ
λ
λ
λτ −==
0,
, and τλ =1 - αλ . 
 
12-63C Using Kirchhoff’s law, the spectral emissivity of a medium of thickness L in terms of the spectral 
absorption coefficient is expressed as . Le λκλλ αε −−== 1
 
12-64C Gases emit and absorb radiation at a number of narrow wavelength bands. The emissivity-
wavelength charts of gases typically involve various peaks and dips together with discontinuities, and show 
clearly the band nature of absorption and the strong nongray characteristics. This is in contrast to solids, 
which emit and absorb radiation over the entire spectrum. 
 
 
12-65 An equimolar mixture of CO2 and O2 gases at 500 K and a total pressure of 0.5 atm is considered. 
The emissivity of the gas is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis Volumetric fractions are equal to pressure fractions. Therefore, the partial pressure of CO2 is 
atm 25.0atm) 5.0(5.0
2CO === PyPc 
Then, 
atmft 0.98atmm 30.0m) atm)(1.2 25.0( ⋅=⋅==LPc 
The emissivity of CO2 corresponding to this value at the gas temperature of Tg = 500 K and 1 atm is, from 
Fig. 12-36, 
14.0atm 1 , =ε c 
This is the base emissivity value at 1 atm, and it needs to be corrected for the 0.5 atm total pressure. The 
pressure correction factor is, from Fig. 12-37, 
Cc = 0.90 
Then the effective emissivity of the gas becomes 
0.126=×=ε=ε 14.090.0atm 1 ,ccg C 
 
12-66 The temperature, pressure, and composition of a gas mixture is given. The emissivity of the mixture 
is to be determined. 
Assumptions 1 All the gases in the mixture are ideal gases. 2 The emissivity determined is the mean 
emissivity for radiation emitted to all surfaces of the cubical enclosure. 
 12-49
Chapter 12 RadiationHeat Transfer 
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are 
equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O 
are 
atm 09.0atm) 1(09.0
atm 10.0atm) 1(10.0
2
2
H
CO
===
===
PyP
PyP
Ow
c
 6 m 
Combustion 
gases 
1000 K 
The mean beam length for a cube of side length 6 m for radiation 
emitted to all surfaces is, from Table 12-4, 
 L = 0.66(6 m) = 3.96 m 
Then, 
atmft 1.57atmm 48.0m) atm)(3.96 09.0(
atmft .301atmm 396.0m) atm)(3.96 10.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c 
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1000 K and 
1atm are, from Fig. 12-36, 
17.0atm 1 , =ε c and 26.0atm 1 , =εw 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 1000 K is, from Fig. 12-38, 
039.0 474.0
10.009.0
09.0
 87.257.130.1
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Note that we obtained the average of the emissivity correction factors from the two figures for 800 K and 
1200 K. Then the effective emissivity of the combustion gases becomes 
 0.391=−×+×=εΔ−ε+ε=ε 039.026.0117.01atm 1 ,atm 1 , wwccg CC 
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. 
 12-50
Chapter 12 Radiation Heat Transfer 
12-67 A mixture of CO2 and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical 
container. The rate of radiation heat transfer between the gas and the container walls is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis The mean beam length is, from Table 12-4 
8 m 
8 m Tg = 600 K 
Ts = 450 K 
 L = 0.60D = 0.60(8 m) = 4.8 m 
Then, 
atmft .362atmm 72.0m) atm)(4.8 15.0( ⋅=⋅==LPc 
The emissivity of CO2 corresponding to this value at the gas 
temperature of Tg = 600 K and 1 atm is, from Fig. 12-36, 
24.0atm 1 , =ε c 
For a source temperature of Ts = 450 K, the absorptivity of the gas is again determined using the emissivity 
charts as follows: 
atmft 1.77atmm 54.0
K 600
K 450m) atm)(4.8 15.0( ⋅=⋅==
g
s
c T
T
LP 
The emissivity of CO2 corresponding to this value at a temperature of Ts = 450 K and 1atm are, from Fig. 
12-36, 
14.0atm 1 , =ε c 
The absorptivity of CO2 is determined from 
17.0)14.0(
K 450
K 600)1(
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α c
s
g
cc T
T
C 
The surface area of the cylindrical surface is 
2
22
m 6.301
4
m) 8(
2m) 8(m) 8(
4
2 =π+π=π+π= DDHAs 
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes 
 W101.91 5×=
−⋅×=
α−εσ=
− ])K 450(17.0)K 600(14.0)[K W/m1067.5)(m 6.301(
)(
444282
44
net sgggs TTAQ&
 12-51
Chapter 12 Radiation Heat Transfer 
12-68 A mixture of H2O and N2 gases at 600 K and a total pressure of 1 atm are contained in a cylindrical 
container. The rate of radiation heat transfer between the gas and the container walls is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis The mean beam length is, from Table 12-4 
8 m 
8 m Tg = 600 K 
Ts = 450 K 
 L = 0.60D = 0.60(8 m) = 4.8 m 
Then, 
atmft .362atmm 72.0m) atm)(4.8 15.0( ⋅=⋅==LPw 
The emissivity of H2O corresponding to this value at the gas 
temperature of Tg = 600 K and 1 atm is, from Fig. 12-36, 
36.0atm 1 =εw 
For a source temperature of Ts = 450 K, the absorptivity of the gas is again determined using the emissivity 
charts as follows: 
atmft 1.77atmm 54.0
K 600
K 450m) atm)(4.8 15.0( ⋅=⋅==
g
s
w T
T
LP 
The emissivity of H2O corresponding to this value at a temperature of Ts = 450 K and 1atm are, from Fig. 
12-36, 
34.0atm 1 , =εw 
The absorptivity of H2O is determined from 
39.0)34.0(
K 450
K 600)1(
45.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α w
s
g
ww T
T
C 
The surface area of the cylindrical surface is 
2
22
m 6.301
4
m) 8(
2m) 8(m) 8(
4
2 =π+π=π+π= DDHAs 
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes 
 W105.244 5×=
−⋅×=
α−εσ=
− ])K 450(39.0)K 600(36.0)[K W/m1067.5)(m 6.301(
)(
444282
44
net sgggs TTAQ&
 
 12-52
Chapter 12 Radiation Heat Transfer 
12-69 A mixture of CO2 and N2 gases at 1200 K and a total pressure of 1 atm are contained in a spherical 
furnace. The net rate of radiation heat transfer between the gas mixture and furnace walls is to be 
determined. 
Assumptions All the gases in the mixture are ideal gases. 
Tg = 1200 K 
Ts = 600 K 
2 m 
Analysis The mean beam length is, from Table 12-4 
 L = 0.65D = 0.65(2 m) = 1.3 m 
The mole fraction is equal to pressure fraction. Then, 
atmft .640atmm 195.0m) atm)(1.3 15.0( ⋅=⋅==LPc 
The emissivity of CO2 corresponding to this value at the gas 
temperature of Tg = 1200 K and 1 atm is, from Fig. 12-36, 
14.0atm 1 , =ε c 
For a source temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity 
charts as follows: 
atmft .320atmm 0975.0
K 1200
K 600m) atm)(1.3 15.0( ⋅=⋅==
g
s
c T
T
LP 
The emissivity of CO2 corresponding to this value at a temperature of Ts = 600 K and 1atm are, from Fig. 
12-36, 
092.0atm 1 , =ε c 
The absorptivity of CO2 is determined from 
144.0)092.0(
K 600
K 1200)1(
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α c
s
g
cc T
T
C 
The surface area of the sphere is 
222 m 57.12m) 2( =π=π= DAs 
Then the net rate of radiation heat transfer from the gas mixture to the walls of the furnace becomes 
 W101.936 5×=
−⋅×=
α−εσ=
− ])K 600(144.0)K 1200(14.0)[K W/m1067.5)(m 57.12(
)(
444282
44
net sgggs TTAQ&
 
 12-53
Chapter 12 Radiation Heat Transfer 
12-70 The temperature, pressure, and composition of combustion gases flowing inside long tubes are 
given. The rate of heat transfer from combustion gases to tube wall is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are 
equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O 
are 
 
atm 09.0atm) 1(09.0
atm 06.0atm) 1(06.0
2
2
H
CO
===
===
PyP
PyP
Ow
c
Ts = 600 K 
D = 15 cm 
Combustion 
gases, 1 atm 
Tg = 1500 K 
The mean beam length for an infinite cicrcular 
cylinder is, from Table 12-4, 
 L = 0.95(0.15 m) = 0.1425 m 
Then, 
atmft 0.042atmm 0128.0m) 5atm)(0.142 09.0(
atmft .0280atmm 00855.0m) 5atm)(0.142 06.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c 
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1500 K and 
1atm are, from Fig. 12-36, 
034.0atm 1 , =ε c and 016.0atm 1 , =εw 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 1500 K is, from Fig. 12-38, 
0.0 6.0
06.009.0
09.0
 07.0042.0028.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Then the effective emissivity of the combustion gases becomes 
 0.050.0016.01034.01atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source 
temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity charts as 
follows: 
atmft 0.017atmm 00513.0
K 1500
K 600m) 5atm)(0.142 09.0(atmft 0.011atmm 00342.0
K 1500
K 600m) 5atm)(0.142 06.0(
⋅=⋅==
⋅=⋅==
g
s
w
g
s
c
T
T
LP
T
T
LP
 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm 
are, from Fig. 12-36, 
 and 031.0atm 1 , =ε c 027.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
041.0)027.0(
K 600
K 1500)1(
056.0)031.0(
K 600
K 1500)1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
 12-54
Chapter 12 Radiation Heat Transfer 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 600 K 
instead of Tg = 1500 K. There is no chart for 600 K in the figure, but we can read Δε values at 400 K and 
800 K, and take their average. At Pw/(Pw+ Pc) = 0.6 and PcL +PwL = 0.07 we read Δε = 0.0. Then the 
absorptivity of the combustion gases becomes 
0.0970.0041.0056.0 =−+=αΔ−α+α=α wcg 
The surface area of the pipe per m length of tube is 
2m 4712.0m) 1(m) 15.0( =π=π= DLAs 
Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes 
 W6427=
−⋅×=
α−εσ=
− ])K 600(097.0)K 1500(05.0)[K W/m1067.5)(m 4712.0(
)(
444282
44
net sgggs TTAQ&
 12-55
Chapter 12 Radiation Heat Transfer 
12-71 The temperature, pressure, and composition of combustion gases flowing inside long tubes are 
given. The rate of heat transfer from combustion gases to tube wall is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are 
equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O 
are 
 
atm 09.0atm) 1(09.0
atm 06.0atm) 1(06.0
2
2
H
CO
===
===
PyP
PyP
Ow
c
Ts = 600 K 
D = 15 cm 
Combustion 
gases, 3 atm 
Tg = 1500 K 
The mean beam length for an infinite cicrcular 
cylinder is, from Table 12-4, 
 L = 0.95(0.15 m) = 0.1425 m 
Then, 
atmft 0.042atmm 0128.0m) 5atm)(0.142 09.0(
atmft .0280atmm 00855.0m) 5atm)(0.142 06.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c 
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1500 K and 
1atm are, from Fig. 12-36, 
034.0atm 1 , =ε c and 016.0atm 1 , =εw 
These are base emissivity values at 1 atm, and they need to be corrected for the 3 atm total pressure. Noting 
that (Pw+P)/2 = (0.09+3)/2 = 1.545 atm, the pressure correction factors are, from Fig. 12-37, 
 Cc = 1.5 and Cw = 1.8 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 1500 K is, from Fig. 12-38, 
0.0 6.0
06.009.0
09.0
 07.0042.0028.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Then the effective emissivity of the combustion gases becomes 
 0.0800.0016.08.1034.05.1atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
For a source temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity 
charts as follows: 
atmft 0.017atmm 00513.0
K 1500
K 600m) 5atm)(0.142 09.0(
atmft 0.011atmm 00342.0
K 1500
K 600m) 5atm)(0.142 06.0(
⋅=⋅==
⋅=⋅==
g
s
w
g
s
c
T
T
LP
T
T
LP
 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm 
are, from Fig. 12-36, 
 and 031.0atm 1 , =ε c 027.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
 12-56
Chapter 12 Radiation Heat Transfer 
073.0)027.0(
K 600
K 1500)8.1(
084.0)031.0(
K 600
K 1500)5.1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 600 K 
instead of Tg = 1500 K. There is no chart for 600 K in the figure, but we can read Δε values at 400 K and 
800 K, and take their average. At Pw/(Pw+ Pc) = 0.6 and PcL +PwL = 0.07 we read Δε = 0.0. Then the 
absorptivity of the combustion gases becomes 
0.1570.0073.0084.0 =−+=αΔ−α+α=α wcg 
The surface area of the pipe per m length of tube is 
2m 4712.0m) 1(m) 15.0( =π=π= DLAs 
Then the net rate of radiation heat transfer from the combustion gases to the walls of the furnace becomes 
 W10,276=
−⋅×=
α−εσ=
− ])K 600(157.0)K 1500(08.0)[K W/m1067.5)(m 4712.0(
)(
444282
44
net sgggs TTAQ&
 12-57
Chapter 12 Radiation Heat Transfer 
12-72 The temperature, pressure, and composition of combustion gases flowing inside long tubes are 
given. The rate of heat transfer from combustion gases to tube wall is to be determined. 
Assumptions All the gases in the mixture are ideal gases. 
Analysis The mean beam length for an infinite cicrcular 
cylinder is, from Table 12-4, 
 L = 0.95(0.10 m) = 0.095 m 
Then, 
 
atmft 0.056atmm 0171.0m) atm)(0.095 18.0(
atmft .0370atmm 0114.0m) atm)(0.095 12.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c
The emissivities of CO2 and H2O corresponding to these values at 
the gas temperature of Tg = 800 K and 1atm are, from Fig. 12-36, 
Ts = 500 K 
D = 10 cm 
Combustion 
gases, 1 atm 
Tg = 800 K 
055.0atm 1 , =ε c and 050.0atm 1 , =εw 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 800 K is, from Fig. 12-38, 
0.0 6.0
12.018.0
18.0
 093.0056.0037.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Then the effective emissivity of the combustion gases becomes 
 0.1050.0050.01055.01atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source 
temperature of Ts = 500 K, the absorptivity of the gas is again determined using the emissivity charts as 
follows: 
atmft 0.035atmm 01069.0
K 800
K 500m) atm)(0.095 18.0(
atmft 0.023atmm 007125.0
K 800
K 500m) atm)(0.095 12.0(
⋅=⋅==
⋅=⋅==
g
s
w
g
s
c
T
T
LP
T
T
LP
 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 500 K and 1atm 
are, from Fig. 12-36, 
 and 042.0atm 1 , =ε c 050.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
062.0)050.0(
K 500
K 800)1(
057.0)042.0(
K 500
K 800)1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 500 K 
instead of Tg = 800 K. There is no chart for 500 K in the figure, but we can read Δε values at 400 K and 
800 K, and interpolate. At Pw/(Pw+ Pc) = 0.6 and PcL +PwL = 0.093 we read Δε = 0.0. Then the absorptivity 
of the combustion gases becomes 
0.1190.0062.0057.0 =−+=αΔ−α+α=α wcg 
The surface area of the pipe is 
2m 885.1m) 6(m) 10.0( =π=π= DLAs 
Then the net rate of radiation heat transfer from the combustion gases to the walls of the tube becomes 
 W3802=
−⋅×=
α−εσ=
− ])K 500(119.0)K 800(105.0)[K W/m1067.5)(m 885.1(
)(
444282
44
net sgggs TTAQ&
 
 12-58
Chapter 12 Radiation Heat Transfer 
12-73 The temperature, pressure, and composition of combustion gases flowing inside long tubes are 
given. The rate of heat transfer from combustion gases to tube wall is to be determined. 
Assumptions All thegases in the mixture are ideal gases. 
Analysis The volumetric analysis of a gas mixture gives the mole fractions yi of the components, which are 
equivalent to pressure fractions for an ideal gas mixture. Therefore, the partial pressures of CO2 and H2O 
are 
 
atm 10.0atm) 1(10.0
atm 10.0atm) 1(10.0
2
2
H
CO
===
===
PyP
PyP
Ow
c Ts = 600 K 
20 cm
Combustion 
gases, 1 atm 
Tg = 1200 K 
The mean beam length for this geometry is, from 
Table 12-4, 
 L = 3.6V/As = 1.8D = 1.8(0.20 m) = 0.36 m 
where D is the distance between the plates. Then, 
 atmft .1180atmm 036.0m) atm)(0.36 10.0( ⋅=⋅=== LPLP wc 
The emissivities of CO2 and H2O corresponding to these values at the gas temperature of Tg = 1200 K and 
1atm are, from Fig. 12-36, 
080.0atm 1 , =ε c and 055.0atm 1 , =εw 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 1200 K is, from Fig. 12-38, 
0025.0 5.0
10.010.0
10.0
 236.0118.0118.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Then the effective emissivity of the combustion gases becomes 
 0.13250025.0055.01080.01atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source 
temperature of Ts = 600 K, the absorptivity of the gas is again determined using the emissivity charts as 
follows: 
atmft 0.059atmm 018.0
K 1200
K 600m) atm)(0.36 10.0( ⋅=⋅===
g
s
w
g
s
c T
T
LP
T
T
LP 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 600 K and 1atm 
are, from Fig. 12-36, 
065.0atm 1 , =ε c and 067.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
092.0)067.0(
K 600
K 1200)1(
102.0)065.0(
K 600
K 1200)1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 600 K 
instead of Tg = 1200 K. There is no chart for 600 K in the figure, but we can read Δε values at 400 K and 
800 K, and take their average. At Pw/(Pw+ Pc) = 0.5 and PcL +PwL = 0.236 we read Δε = 0.00125. Then the 
absorptivity of the combustion gases becomes 
0.192800125.0092.0102.0 =−+=αΔ−α+α=α wcg 
Then the net rate of radiation heat transfer from the gas to each plate per unit surface area becomes 
 W101.42 4×=
−⋅×=
−=
− ])K 600(1928.0)K 1200(1325.0)[KW/m 1067.5)(m 1(
)(
444282
44
net sgggs TTAQ αεσ&
 
 12-59
Chapter 12 Radiation Heat Transfer 
 
Special Topic: Heat Transfer from the Human Body 
 
12-74C Yes, roughly one-third of the metabolic heat generated by a person who is resting or doing light 
work is dissipated to the environment by convection, one-third by evaporation, and the remaining one-third 
by radiation. 
 
12-75C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a 
human body increases as (a) the skin temperature increases, (b) the environment temperature decreases, 
and (c) the air motion (and thus the convection heat transfer coefficient) increases. 
 
12-76C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy 
absorbed from a warm surface as liquid water evaporates. The latent heat loss from a human body 
increases as (a) the skin wettedness increases and (b) the relative humidity of the environment decreases. 
The rate of evaporation from the body is related to the rate of latent heat loss by where 
h
& &Q m hfglatent vapor=
fg is the latent heat of vaporization of water at the skin temperature. 
 
12-77C The insulating effect of clothing is expressed in the unit clo with 1 clo = 0.155 m2.°C/W = 0.880 
ft2.°F.h/Btu. Clothing serves as insulation, and thus reduces heat loss from the body by convection, 
radiation, and evaporation by serving as a resistance against heat flow and vapor flow. Clothing decreases 
heat gain from the sun by serving as a radiation shield. 
 
12-78C (a) Heat is lost through the skin by convection, radiation, and evaporation. (b) The body loses both 
sensible heat by convection and latent heat by evaporation from the lungs, but there is no heat transfer in 
the lungs by radiation. 
 
12-79C The operative temperature Toperative is the average of the mean radiant and ambient temperatures 
weighed by their respective convection and radiation heat transfer coefficients, and is expressed as 
 T h T h T
h h
T T
operative
conv ambient rad surr
conv rad
ambient surr= ++ ≅
+
2
 
When the convection and radiation heat transfer coefficients are equal to each other, the operative 
temperature becomes the arithmetic average of the ambient and surrounding surface temperatures. 
 12-60
Chapter 12 Radiation Heat Transfer 
12-80 The convection heat transfer coefficient for a clothed person while walking in still air at a velocity 
of 0.5 to 2 m/s is given by h = 8.6V0.53 where V is in m/s and h is in W/m2.°C. The convection coefficients 
in that range vary from 5.96 W/m2.°C at 0.5 m/s to 12.4 W/m2.°C at 2 m/s. Therefore, at low velocities, the 
radiation and convection heat transfer coefficients are comparable in magnitude. But at high velocities, the 
convection coefficient is much larger than the radiation heat transfer coefficient. 
 
 
0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
V
h
Velocity, 
m/s 
h = 8.6V0.53
W/m2.°C 
0.50 
0.75 
1.00 
1.25 
1.50 
1.75 
2.00 
5.96 
7.38 
8.60 
9.68 
10.66 
11.57 
12.40 
 
 
 
 
 
 12-61
Chapter 12 Radiation Heat Transfer 
12-81 A man wearing summer clothes feels comfortable in a room at 22°C. The room temperature at 
which this man would feel thermally comfortable when unclothed is to be determined. 
Assumptions 1 Steady conditions exist. 2 The latent heat loss from the person remains the same. 3 The heat 
transfer coefficients remain the same. 4 The air in the room is still (there are no winds or running fans). 5 
The surface areas of the clothed and unclothed person are the same. 
Analysis At low air velocities, the convection heat transfer coefficient 
for a standing man is given in Table 12-3 to be 4.0 W/m2.°C. The 
radiation heat transfer coefficient at typical indoor conditions is 4.7 
W/m2.°C. Therefore, the heat transfer coefficient for a standing person 
for combined convection and radiation is 
Troom= 20°C 
Clothed 
person 
Tskin= 33°C
 h h hcombined conv rad
2 W / m . C= + = + = °4 0 4 7 8 7. . .
The thermal resistance of the clothing is given to be 
 Rcloth
2 2 clo m . C / W = 0.109 m . C / W= = × ° °0 7 0 7 0155. . .
Noting that the surface area of an average man is 1.8 m2, the sensible 
heat loss from this person when clothed is determined to be 
 W 104
C.W/m 8.7
1+C/W.m 0.109
C)2033)(m 8.1(
1
)(
2
2
2
combined
cloth
ambientskin
clothedsensible, =
°°
°−=
+
−=
h
R
TTA
Q s& 
From heat transfer point of view, taking the clothes off is equivalent 
to removing the clothing insulation or setting Rcloth = 0. The heat 
transfer in this case can be expressed as 
Troom
Tskin= 33°C
Unclothed 
person 
 
C.W/m 8.7
1
C)33)(m 8.1(
1
)(
2
ambient
2
combined
ambientskin
unclothedsensible,
°
°−=−= T
h
TTA
Q s& 
To maintain thermal comfort after taking the clothes off, the skin 
temperature of the person and the rate of heat transfer from him must 
remain the same. Then settingthe equation above equal to 104 W 
gives 
 Tambient = °26.4 C
Therefore, the air temperature needs to be raised from 22 to 26.4°C to ensure that the person will feel 
comfortable in the room after he takes his clothes off. Note that the effect of clothing on latent heat is 
assumed to be negligible in the solution above. We also assumed the surface area of the clothed and 
unclothed person to be the same for simplicity, and these two effects should counteract each other. 
 12-62
Chapter 12 Radiation Heat Transfer 
12-82E An average person produces 0.50 lbm of moisture while 
taking a shower. The contribution of showers of a family of four to 
the latent heat load of the air-conditioner per day is to be determined. 
Moisture 
0.5 lbm 
Assumptions All the water vapor from the shower is condensed by the 
air-conditioning system. 
Properties The latent heat of vaporization of water is given to be 1050 
Btu/lbm. 
Analysis The amount of moisture produced per day is 
 mvapor = (Moisture produced per person)(No. of persons) 
 = (0.5 lbm/person)(4 persons/day) = 2 lbm/day 
Then the latent heat load due to showers becomes 
 Qlatent = mvaporhfg = (2 lbm/day)(1050 Btu/lbm) = 2100 Btu/day 
 
 
 
12-83 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture 
production in the room are to be determined. 
Assumptions All the moisture from the chickens is condensed by the air-conditioning system. 
Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of 
chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). 
Analysis The total rate of heat generation of the chickens in the breeding room is 
 
 W1020=chickens) (100W/chicken) 2.10(
chickens) of (No. totalgen, totalgen,
=
= qQ &&
100 Chickens 10.2 W 
The latent heat generated by the chicken and the rate of 
moisture production are 
 
kW 0.642=
 W642=chickens) )(100 W/chicken42.6(
chickens) of (No.latent gen,latent gen,
=
= qQ &&
 g/s 0.264 kg/s 000264.0
kJ/kg 2430
kJ/s 642.0
fg
latent gen,
moisture ==== h
Q
m
&
& 
 12-63
Chapter 12 Radiation Heat Transfer 
12-84 Chilled air is to cool a room by removing the heat generated in a large insulated classroom by lights 
and students. The required flow rate of air that needs to be supplied to the room is to be determined. 
Assumptions 1 The moisture produced by the bodies leave the room as vapor without any condensing, and 
thus the classroom has no latent heat load. 2 Heat gain through the walls and the roof is negligible. 
Properties The specific heat of air at room temperature is 1.00 kJ/kg⋅°C (Table A-15). The average rate of 
metabolic heat generation by a person sitting or doing light work is 115 W (70 W sensible, and 45 W 
latent). 
Analysis The rate of sensible heat generation by the 
people in the room and the total rate of sensible internal 
heat generation are 
Chilled 
 air 
Return 
 air 
15°C
25°C
150 Students 
Lights 
4 kW 
 
W 500,144000500,10
W 10,500= persons)150W/person)( 70(
 people)of (No.
lightingsensible gen,sensible total,
sensible gen,sensible gen,
=+=
+=
=
=
QQQ
qQ
&&&
&&
Then the required mass flow rate of chilled air becomes 
kg/s 1.45=°−°⋅=
Δ=
C15)C)(25 kJ/kg(1.0
 kJ/s5.14
sensible total,
air TC
Q
m
p
&
&
 
Discussion The latent heat will be removed by the air-conditioning system as the moisture condenses 
outside the cooling coils. 
 12-64
Chapter 12 Radiation Heat Transfer 
12-85 The average mean radiation temperature during a cold day drops to 18°C. The required rise in the 
indoor air temperature to maintain the same level of comfort in the same clothing is to be determined. 
Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skin temperature 
remains the same. 3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs 
remain constant. 
Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer 
coefficient from the body in still air or air moving with a velocity under 0.2 m/s is hconv = 3.1 W/m2⋅°C 
(Table 12-3). 
Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss by convection, 
radiation, and evaporation, 
 & & & & ( & ) & &Q Q Q Q Q Q Q Qbody, total sensible latent lungs conv rad latent lungs= + + = + + +
Noting that heat transfer from the skin by evaporation and from the lungs remains constant, the sum of the 
convection and radiation heat transfer from the person must remain constant. 
])27318()273[(95.0)(
)()(
])27322()273[(95.0)22(
)()(
44
 newair,
4
newsurr,
4
 newair,newsensible,
44
4
oldsurr,
4
old air,oldsensible,
+−++−=
−+−=
+−++−=
−+−=
ssss
ssss
ssss
ssss
TATThA
TTATThAQ
TAThA
TTATThAQ
σ
σε
σ
σε
&
&
 
Setting the two relations above equal to each other, 
canceling the surface area As, and simplifying gives 
− − + = − − +
− + × × − =−
22 0 95 22 273 0 95 18 273
31 22 0 95 5 67 10 291 295 0
4 4
8 4 4
h hT
T
. ( ) . ( )
. ( ) . . ( )
σ σair, new
air, new
 
22°C 
22°C 
Solving for the new air temperature gives 
 Tair, new = 29.0°C 
Therefore, the air temperature must be raised to 29°C to 
counteract the increase in heat transfer by radiation. 
 12-65
Chapter 12 Radiation Heat Transfer 
12-86 The average mean radiation temperature during a cold day drops to 12°C. The required rise in the 
indoor air temperature to maintain the same level of comfort in the same clothing is to be determined. 
Assumptions 1 Air motion in the room is negligible. 2 The average clothing and exposed skin temperature 
remains the same. 3 The latent heat loss from the body remains constant. 4 Heat transfer through the lungs 
remain constant. 
Properties The emissivity of the person is 0.95 (from Appendix tables). The convection heat transfer 
coefficient from the body in still air or air moving with a velocity under 0.2 m/s is hconv = 3.1 W/m2⋅°C 
(Table 12-3). 
Analysis The total rate of heat transfer from the body is the sum of the rates of heat loss by convection, 
radiation, and evaporation, 
 & & & & ( & ) & &Q Q Q Q Q Q Q Qbody, total sensible latent lungs conv rad latent lungs= + + = + + +
Noting that heat transfer from the skin by evaporation and from the lungs remains constant, the sum of the 
convection and radiation heat transfer from the person must remain constant. 
])27312()273[(95.0)(
)()(
])27322()273[(95.0)22(
)()(
44
 newair,
4
newsurr,
4
 newair,newsensible,
44
4
oldsurr,
4
old air,oldsensible,
+−++−=
−+−=
+−++−=
−+−=
ssss
ssss
ssss
ssss
TATThA
TTATThAQ
TAThA
TTATThAQ
σ
σε
σ
σε
&
&
 
Setting the two relations above equal to each other, 
canceling the surface area As, and simplifying gives 
22°C 
22°C 
0)295285(1067.595.0)22(1.3
)27312(95.0)27322(95.022
448
 newair,
4
 newair,
4
=−××+−
+−−=+−−
−T
hTh σσ
 
Solving for the new air temperature gives 
 Tair, new = 39.0°C 
Therefore, the air temperature must be raised to 39°C to 
counteract the increase in heat transfer by radiation. 
 12-66
Chapter 12 Radiation Heat Transfer 
12-87 A car mechanic is working in a shop heated by radiant heaters in winter. The lowest ambient 
temperature the worker can work in comfortably is to be determined. 
Assumptions 1 The air motion in the room is negligible, and the mechanic is standing. 2 The average 
clothing and exposed skin temperature of the mechanic is 33°C. 
Properties The emissivity and absorptivityof the person is given to be 0.95. The convection heat transfer 
coefficient from a standing body in still air or air moving with a velocity under 0.2 m/s is hconv = 4.0 
W/m2⋅°C (Table 12-3). 
Analysis The equivalent thermal resistance of clothing is 
 Rcloth
2 2 clo m . C / W = 0.1085 m . C / W= = × ° °0 7 0 7 0155. . .
Radiation from the heaters incident on the person and 
the rate of sensible heat generation by the person are 
 
& . & . ( .
& . & . (
Q Q
Q Q
rad, incident rad, total
gen, sensible gen, total
 kW) kW = 500 W
 W) W
= × = =
= × = =
0 05 0 05 10 05
05 05 350 175
Under steady conditions, and energy balance on the 
body can be expressed as 
 
& & &
& & &
E E E
Q Q Q
in out gen
rad from heater conv+rad from body gen, sensible
− + =
− +
0
0=
Radiant 
heater 
or 
 
0W 175)K) 306)[(KW/m 10)(5.67m 8.1(95.0 
)306)(m K)(1.8W/m 0.4(W) 500(95.0
0)()(
4
surr
442-82
surr
22
sensible gen,
4
surr
4
surrconvincident rad,
=+−⋅×−
−⋅−
=+−−−−
T
T
QTTATTAhQ ssss && σεα
Solving the equation above gives 
 Tsurr 266.2 K == °-7.0 C
 Therefore, the mechanic can work comfortably at temperatures as low as -7°C. 
 
 12-67
Chapter 12 Radiation Heat Transfer 
 
Review Problems 
 
12-88 The temperature of air in a duct is measured by a 
thermocouple. The radiation effect on the temperature 
measurement is to be quantified, and the actual air 
temperature is to be determined. 
Assumptions The surfaces are opaque, diffuse, and gray. 
Properties The emissivity of thermocouple is given to be 
ε=0.6. 
Analysis The actual temperature of the air can be 
determined from 
 
K 1111=°⋅
−⋅×+=
−σε+=
−
C W/m60
])K 500()K 850)[(K W/m1067.5)(6.0(K 850
)(
2
44428
44
h
TT
TT wthththf
 
Air, Tf 
Tw = 500 K 
Thermocouple 
Tth = 850 K 
ε = 0.6 
 
12-89 The temperature of hot gases in a duct is measured by a thermocouple. The actual temperature of the 
gas is to be determined, and compared with that without a radiation shield. 
Assumptions The surfaces are opaque, diffuse, and gray. 
Properties The emissivity of the thermocouple is given to be ε =0.7. 
Analysis Assuming the area of the shield to be very close to the sensor 
of the thermometer, the radiation heat transfer from the sensor is 
determined from 
Air, Tf 
Tw = 380 K 
Thermocouple 
Tth = 530 K 
ε1 = 0.7 
 
ε2 = 0.15 
2
44428
21
4
2
4
1
sensor from rad,
 W/m9.257
1
15.0
121
7.0
1
])K 380()K 530)[(K W/m1067.5(
11211
)(
=
⎟⎠
⎞⎜⎝
⎛ −+⎟⎠
⎞⎜⎝
⎛ −
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε
−σ=
−
TT
Q&
 
Then the actual temperature of the gas can be determined from a heat transfer balance to be 
 
K 532=⎯→⎯=−°⋅
=−
=
ff
thf
TT
TTh
qq
22
2
sensor fromconv,sensor toconv,
 W/m9.257)530C( W/m120
 W/m9.257)(
&&
Without the shield the temperature of the gas would be 
K 549.2=°⋅
−⋅×+=
−σε+=
−
C W/m120
])K 380()K 530)[(K W/m1067.5)(7.0(K 530
)(
2
44428
44
h
TT
TT wthththf
 
 12-68
Chapter 12 Radiation Heat Transfer 
12-90E A sealed electronic box is placed in a vacuum chamber. The highest temperature at which the 
surrounding surfaces must be kept if this box is cooled by radiation alone 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 Heat transfer from the bottom surface of the box is negligible. 
Properties The emissivity of the outer surface of the box is ε = 0.95. 
Analysis The total surface area is 
 2ft 67.3)11()12/18(4 =×+××=sA
Then the temperature of the surrounding surfaces is determined to be 
 
F43°==⎯→⎯
−⋅×=×
−=
−
R 503
])R 590)[(RBtu/h.ft 101714.0)(m 67.3)(95.0(Btu/h )41214.3100(
)(
444282
44
surr
surr
surrssrad
T
T
TTAQ σε&
Tsurr 
8 in
100 W 
ε = 0.95 
Ts = 130°F 
12 in 
12 in
 
 
12-91 A double-walled spherical tank is used to store iced water. The air space between the two walls is 
evacuated. The rate of heat transfer to the iced water and the amount of ice that melts a 24-h period are to 
be determined. 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are opaque, diffuse, and gray. 
Properties The emissivities of both surfaces are given to be ε1 
= ε2 = 0.15. 
Analysis (a) Assuming the conduction resistance s of the walls 
to be negligible, the rate of heat transfer to the iced water in the 
tank is determined to be 
 m m2A D1 1
2 22 01 12 69= = =π π( . ) . 
 
 W107.4=
⎟⎠
⎞⎜⎝
⎛−+
+−+⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
ε
ε−+ε
−σ=
−
2
444282
2
2
1
2
2
1
4
1
4
21
12
04.2
01.2
15.0
15.01
15.0
1
])K 2730()K 27320)[(K W/m1067.5)(m 69.12(
11
)(
D
D
TTA
Q&
 
D2 = 2.04 m 
T2 = 20°C 
ε2 = 0.15 
D1 = 2.01 m 
T1 = 0°C 
ε1 = 0.15 
 Vacuum 
Iced 
water 
0°C 
(b) The amount of heat transfer during a 24-hour period is 
 kJ 9275)s 360024)(kJ/s 1074.0( =×=Δ= tQQ &
The amount of ice that melts during this period then becomes 
 kg 27.8===⎯→⎯=
kJ/kg 7.333
kJ 9275
if
if h
QmmhQ 
 12-69
Chapter 12 Radiation Heat Transfer 
12-92 Two concentric spheres which are maintained at uniform temperatures are separated by air at 1 atm 
pressure. The rate of heat transfer between the two spheres by natural convection and radiation is to be 
determined. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an 
ideal gas with constant properties. 
Properties The emissivities of the surfaces are given to be 
ε1 = ε2 = 0.5. The properties of air at 1 atm and the average 
temperature of (T1+T2)/2 = (350+275)/2 = 312.5 K = 
39.5°C are (Table A-15) 
 
1-
25
K 0032.0
K 5.312
1
7256.0Pr
/sm 10697.1
CW/m. 02658.0
==
=
×=
°=
−
β
υ
k
 
Analysis (a) Noting that Di = D1 and Do = D2 , the characteristic length is 
D2 = 25 cm 
T2 = 275 K 
ε2 = 0.5 
D1 = 15 cm 
T1 = 350 K 
ε1 = 0.9 
 Lc =5 cm 
 AIR 
1 atm 
 m 0.05m) 0.15m 25.0(
2
1)(
2
1 =−=−= ioc DDL 
Then 
5
225
3-12
2
3
21 10415.7)7256.0(
)/sm 10697.1(
)m 05.0)(K 275350)(K 003200.0)(m/s 81.9(Pr
)( ×=×
−=−= −υ
β cLTTgRa 
The effective thermal conductivity is 
[ ] [ ] 005900.0m) 25.0(m) 15.0(m) 25.0(m) 15.0( m 05.0)()( 57/5-7/5-455/75/74sph =+=+= −− oioi c DDDD
L
F 
[ ] CW/m. 1315)10415.7)(00590.0(
7256.0861.0
7256.0C)W/m. 02658.0(74.0
)(
Pr861.0
Pr74.0
4/15
4/1
4/1
4/1
eff
°=×⎟⎠
⎞⎜⎝
⎛
+°=
⎟⎠
⎞⎜⎝
⎛
+= RaFkk sph 
Then the rate of heat transfer between the spheres becomes 
 W23.3=−⎥⎦
⎤⎢⎣
⎡°=−⎟⎟⎠
⎞
⎜⎜⎝
⎛= K)275350(
)m 05.0(
)m 25.0)(m 15.0(
)CW/m. 1315.0()(eff ππ oi
c
oi TT
L
DD
kQ& 
 (b) The rate of heat transfer by radiation is determined from 
 W32.3=
⎟⎠
⎞⎜⎝
⎛−+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
ε
ε−+ε
−σ=
=π=π=
−
2
444282
2
2
1
2
2
1
4
1
4
21
12
222
11
25.0
15.0
9.0
9.01
9.0
1
])K 275()K 350)[(K W/m1067.5)(m 0707.0(
11
)(
m 0707.0)m 15.0( 
D
D
TTA
Q
DA
& 
 12-70
Chapter 12 Radiation Heat Transfer 
12-93 A solar collector is considered. The absorber plate and the glass cover are maintained at uniform 
temperatures, and are separated by air. The rate of heat loss from the absorber plate by natural convection 
and radiation is to be determined. 
Assumptions 1 Steady operating conditions exist 2 The 
surfaces are opaque, diffuse, and gray. 3 Air is an ideal gas 
with constant properties. 
Properties The emissivities of surfacesare given to be ε1 = 0.9 
for glass and ε2 = 0.8 for the absorber plate. The properties of 
air at 1 atm and the average temperature of (T1+T2)/2 = 
(80+32)/2 = 56°C are (Table A-15) 
Solar 
radiation 
 θ = 20° 
 Insulation 
Absorber plate 
T1 = 80°C 
ε1 = 0.8 
Glass cover, 
T2 = 32°C 
ε2 = 0.9 
1.5 m 
 L = 3 cm 
 
1-
25
K 003040.0
K)27356(
11
7212.0Pr
/sm 10857.1
CW/m. 02779.0
=+==
=
×=
°=
−
fT
k
β
υ
 
Analysis For θ = , we have horizontal rectangular enclosure. The 
characteristic length in this case is the distance between the two 
glasses L
°0
c = L = 0.03 m Then, 
 4225
3-12
2
3
21 10083.8)7212.0(
)/sm 10857.1(
)m 03.0)(K 3280)(K 00304.0)(m/s 81.9(Pr
)( ×=×
−=−= −υ
β LTTg
Ra 
 2m 5.4m) 3(m) 5.1( ==×= WHAs
[ ] [ ]
747.3
1
18
)20cos()10083.8(
)20cos()10083.8(
)208.1sin(17081
)20cos()10083.8(
1708144.11
1
18
)cosRa(
cosRa
)8.1(sin17081
cosRa
1708144.11Nu
3/14
4
6.1
4
3/16.1
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −×+⎥⎦
⎤⎢⎣
⎡
×
×−⎥⎦
⎤⎢⎣
⎡
×−+=
⎥⎦
⎤⎢⎣
⎡ −+⎥⎦
⎤⎢⎣
⎡ −⎥⎦
⎤⎢⎣
⎡ −+=
++
++ θ
θ
θ
θ
 W750=°−°=−=
m 03.0
C)3280(
)m 5.4)(747.3)(CW/m. 02779.0( 221
L
TT
kNuAQ s& 
Neglecting the end effects, the rate of heat transfer by radiation is determined from 
 W1289=
−+
+−+⋅×=
−+
−=
−
1
9.0
1
8.0
1
])K 27332()K 27380)[(KW/m 1067.5)(m 5.4(
111
)( 444282
21
4
2
4
1
rad
εε
σ TTA
Q s& 
Discussion The rates of heat loss by natural convection for the horizontal and vertical cases would be as 
follows (Note that the Ra number remains the same): 
Horizontal: 
812.31
18
)10083.8(
10083.8
1708144.111
18
Ra
Ra
1708144.11Nu
3/14
4
3/1
=⎥⎦
⎤⎢⎣
⎡ −×+⎥⎦
⎤⎢⎣
⎡
×−+=⎥⎦
⎤⎢⎣
⎡ −+⎥⎦
⎤⎢⎣
⎡ −+=
++++
 
 W1017=°−°=−=
m 03.0
C)3280(
)m 6)(812.3)(CW/m. 02779.0( 221
L
TT
kNuAQ s& 
Vertical: 
 12-71
Chapter 12 Radiation Heat Transfer 
 001.2
m 03.0
m 2)7212.0()10083.8(42.0Pr42.0
3.0
012.04/14
3.0
012.04/1 =⎟⎠
⎞⎜⎝
⎛×=⎟⎠
⎞⎜⎝
⎛=
−−
L
HRaNu 
 W534=°−°=−=
m 03.0
C)3280(
)m 6)(001.2)(CW/m. 02779.0( 221
L
TT
kNuAQ s& 
 12-72
Chapter 12 Radiation Heat Transfer 
12-94E The circulating pump of a solar collector that consists of a horizontal tube and its glass cover fails. 
The equilibrium temperature of the tube is to be determined. 
 
 
 D2 =5 in 
Aluminum tube 
D1 =2.5 in, T1 
ε1 = 0.9 
Air space 
0.5 atm 
Plastic cover, 
ε2 = 0.9, T2 
Water 
T∞ = 75°F 
Tsky = 60°F 
30 Btu/h.ft
Assumptions 1 Steady operating conditions exist. 2 The 
tube and its cover are isothermal. 3 Air is an ideal gas. 4 
The surfaces are opaque, diffuse, and gray for infrared 
radiation. 5 The glass cover is transparent to solar radiation. 
Properties The properties of air should be evaluated at the 
average temperature. But we do not know the exit 
temperature of the air in the duct, and thus we cannot 
determine the bulk fluid and glass cover temperatures at 
this point, and thus we cannot evaluate the average 
temperatures. Therefore, we will assume the glass 
temperature to be 85°F, and use properties at an anticipated 
average temperature of (75+85)/2 =80°F (Table A-15E), 
 
s
k
/ft 101.697/hft 6110.0
FftBtu/h 01481.0
24-2 ×==
°⋅⋅=
υ R 540
11
7290.0Pr
ave
==
=
T
β 
Analysis We have a horizontal cylindrical enclosure filled with air at 0.5 atm pressure. The problem 
involves heat transfer from the aluminum tube to the glass cover and from the outer surface of the glass 
cover to the surrounding ambient air. When steady operation is reached, these two heat transfer rates must 
equal the rate of heat gain. That is, 
Btu/h 30gainsolar ambient-glassglass-tube === QQQ &&& (per foot of tube) 
 
The heat transfer surface area of the glass cover is 
 
 (per foot of tube) 2ft 309.1ft) 1)(ft 12/5()( ==== ππ WDAA oglasso
 
To determine the Rayleigh number, we need to know the surface temperature of the glass, which is not 
available. Therefore, solution will require a trial-and-error approach. Assuming the glass cover temperature 
to be 85°F, the Rayleigh number, the Nusselt number, the convection heat transfer coefficient, and the rate 
of natural convection heat transfer from the glass cover to the ambient air are determined to be 
 
6
224
32
2
3
10092.1)7290.0(
)/sft 10675.1(
)ft 12/5)(R 7585](R) 540/(1)[ft/s 2.32(
Pr
)(
Ra
×=×
−=
−=
−
∞
ν
β oo
Do
DTTg
 
 
( )[ ] ( )[ ]
95.14
7290.0/559.01
)10092.1(387.06.0
Pr/559.01
Ra 387.0
6.0Nu
2
27/816/9
6/16
2
27/816/9
1/6
D
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
×+=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
+= 
 FftBtu/h 5315.0)95.14(
ft 12/5
FftBtu/h 01481.0Nu 2
0
°⋅⋅=°⋅⋅==
D
kho 
 Btu/h 96.6 F)7585)(ft 309.1)(FftBtu/h 5315.0()( 22conv, =°−°⋅⋅=−= ∞TTAhQ oooo&
Also, 
 [ ]
Btu/h 5.30 
R) 535(R) 545()ft 309.1)(RftBtu/h 101714.0)(9.0(
)(
442428
4
sky
4
rad,
=
−⋅⋅×=
−=
−
TTAQ oooo σε&
Then the total rate of heat loss from the glass cover becomes 
 Btu/h 5.375.300.7rad,conv,total, =+=+= ooo QQQ &&&
 12-73
Chapter 12 Radiation Heat Transfer 
which is more than 30 Btu/h. Therefore, the assumed temperature of 85°F for the glass cover is high. 
Repeating the calculations with lower temperatures (including the evaluation of properties), the glass cover 
temperature corresponding to 30 Btu/h is determined to be 81.5°F. 
The temperature of the aluminum tube is determined in a similar manner using the natural 
convection and radiation relations for two horizontal concentric cylinders. The characteristic length in this 
case is the distance between the two cylinders, which is 
 
ft 1.25/12in 25.12/)5.25(2/)( ==−=−= ioc DDL 
Also, 
 (per foot of tube) 2ft 6545.0ft) 1)(ft 12/5.2()( ==== ππ WDAA itubei
 
We start the calculations by assuming the tube temperature to be 118.5°F, and thus an average temperature 
of (81.5+118.5)/2 = 100°F=640 R. Using properties at 100°F, 
 4224
32
2
3
10334.1)726.0(
)/sft 10809.1(
)ft 12/25.1)(R 5.815.118](R) 640/(1)[ft/s 2.32(Pr
)(
Ra ×=×
−=−= −ν
β LTTg oi
L 
The effective thermal conductivity is 
1466.0
]ft) 12/5(ft) 12/5.2[(ft) (1.25/12
)]5.2/5[ln(
)(
)]/[ln(
53/5-3/5-3
4
55/35/33
4
cyc =+=+= −− oic
io
DDL
DD
F 
FftBtu/h 03227.0
)10334.11466.0(
0.7260.861
0.726F)ftBtu/h 01529.0(386.0
)Ra(
Pr861.0
Pr386.0
4/14
4/1
cyc
4/1
eff
°⋅⋅=
××⎟⎠
⎞⎜⎝
⎛
+°⋅⋅=
⎟⎠
⎞⎜⎝
⎛
+= LFkk
 
 
Then the rate of heat transfer between the cylinders by convection becomes 
 Btu/h 8.10F)5.815.118(
ln(5/2.5)
F)ftBtu/h 03227.0(2)(
)/ln(
2 eff
conv , =°−°⋅⋅=−= ππ oi
io
i TTDD
k
Q& 
Also, [ ]
Btu/h 0.25
in 5
in 5.2
9.0
9.01
9.0
1
R) 5.541(R) 5.578()ft 6545.0)(RftBtu/h 101714.0(
11
)( 442428
i
4
o
4
rad, =
⎟⎠
⎞⎜⎝
⎛−+
−⋅⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
o
i
o
o
ii
i
D
D
TTA
Q
ε
ε
ε
σ&
 
Then the total rate of heat loss from the glass cover becomes 
 Btu/h 8.350.258.10rad,conv,total, =+=+= iii QQQ &&&
which is more than 30 Btu/h. Therefore, the assumed temperature of 118.5°F for the tube is high. By trying 
other values, the tube temperature corresponding to 30 Btu/h is determined to be 113.2°F. Therefore, the 
tube will reach an equilibrium temperature of 113.2°F when the pump fails. 
 
 12-74
Chapter 12 Radiation Heat Transfer 
12-95 A double-pane window consists of two sheets of glass separated by an air space. The rates of heat 
transfer through the window by natural convectionand radiation are to be determined. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are 
opaque, diffuse, and gray. 3 Air is an ideal gas with constant specific 
heats. 4 Heat transfer through the window is one-dimensional and the 
edge effects are negligible. 
 5°C 15°C 
 L = 5 cm 
 H = 2 m 
Q& Air 
Properties The emissivities of glass surfaces are given to be ε1 = ε2 = 
0.9. The properties of air at 0.3 atm and the average temperature of 
(T1+T2)/2 = (15+5)/2 = 10°C are (Table A-15) 
 
1-
255
1
K 003534.0
K )27310(
1
7336.0Pr
/sm 10753.4/0.310426.13.0/
CW/m. 02439.0
=+=
=
×=×==
°=
−−
β
υυ atm
k
 
Analysis The characteristic length in this case is the distance between the glasses, m 05.0== LLc
 4225
3-12
2
3
21 10918.1)7336.0(
)/sm 10753.4(
)m 05.0(K)515)(K 003534.0)(m/s 81.9(Pr
)( ×=×
−=−= −υ
β LTTg
Ra 
 539.1
05.0
2)10918.1(197.0197.0
9/1
4/14
9/1
4/1 =⎟⎠
⎞⎜⎝
⎛×=⎟⎠
⎞⎜⎝
⎛=
−−
L
HRaNu 
 2m 6)m 3)(m 2( ==sA
Then the rate of heat transfer by natural convection becomes 
 W45.0=°−°=−=
m 05.0
C)515(
)m 6)(539.1)(CW/m. 02439.0( 221
L
TT
kNuAQ sconv& 
The rate of heat transfer by radiation is determined from 
 ( ) ( )
 W252=
−+
+−+⋅×=
−+
−=
−
1
9.0
1
9.0
1
]K 2735K 27315)[KW/m 1067.5)(m 6(
111
)(
444282
21
4
2
4
1
rad
εε
σ TTA
Q s&
 
Then the rate of total heat transfer becomes 
 W297=+=+= 25245radconvtotal QQQ &&&
Discussion Note that heat transfer through the window is mostly by radiation. 
 12-75
Chapter 12 Radiation Heat Transfer 
12-96 A simple solar collector is built by placing a clear plastic tube around a garden hose. The rate of heat 
loss from the water in the hose by natural convection and radiation is to be determined. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an 
ideal gas with constant specific heats. 
Properties The emissivities of surfaces are given to be ε1 = ε2 = 0.9. The properties of air are at 1 atm and 
the film temperature of (Ts+T∞)/2 = (40+25)/2 = 32.5°C are (Table A-15) 
 
1-
25
K 003273.0
K )2735.32(
1
7275.0Pr
/sm 10632.1
CW/m. 02607.0
=+=
=
×=
°=
−
β
υ
k
 
 
 
 D2 =6 cm 
Garden hose 
D1 =2 cm, T1 
ε1 = 0.9 
Air space 
Plastic cover, 
ε2 = 0.9, T2 =40°C 
Water 
T∞ = 25°C 
Tsky = 15°C 
Analysis Under steady conditions, the heat transfer rate 
from the water in the hose equals to the rate of heat loss 
from the clear plastic tube to the surroundings by natural 
convection and radiation. The characteristic length in this 
case is the diameter of the plastic tube, 
. m 06.02 === DDL plasticc
5
225
3-12
2
3
2 10842.2)7275.0(
)/sm 10632.1(
)m 06.0(K)2540)(K 003273.0)(m/s 81.9(Pr
)( ×=×
−=−= −∞υ
β DTTg
Ra s
 ( )[ ] ( )[ ] 30.107241.0/559.01 )10842.2(387.06.0Pr/559.01 Ra 387.06.0Nu
2
27/816/9
6/15
2
27/816/9
1/6
D =⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
×+=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
+= 
 
2
22
2
2
m 1885.0m) m)(1 06.0(
C.W/m 475.4)30.10(
m 06.0
CW/m. 02607.0
====
°=°==
ππ LDAA
Nu
D
kh
plastic
 
Then the rate of heat transfer from the outer surface by natural convection becomes 
 W12.7=°−°=−= ∞ C)2540)(m 1885.0)(C.W/m 475.4()( 222 TThAQ sconv&
The rate of heat transfer by radiation from the outer surface is determined from 
 
 W26.2=
+−+⋅×=
−=
− ]K) 27315()K 27340)[(KW/m 1067.5)(m 1885.0)(90.0(
)(
444282
44
2rad skys TTAQ σε&
Finally, 
 W 9.382.267.12, =+=losstotalQ&
Discussion Note that heat transfer is mostly by radiation. 
 12-76
Chapter 12 Radiation Heat Transfer 
12-97 A solar collector consists of a horizontal copper tube enclosed in a concentric thin glass tube. The 
annular space between the copper and the glass tubes is filled with air at 1 atm. The rate of heat loss from 
the collector by natural convection and radiation is to be determined. 
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and gray. 3 Air is an 
ideal gas with constant specific heats. 
Properties The emissivities of surfaces are given to be ε1 = 0.85 for the tube surface and ε2 = 0.9 for glass 
cover. The properties of air at 1 atm and the average temperature of (T1+T2)/2 = (60+40)/2 = 50°C are 
(Table A-15) 
 
 
 D2 =9 cm 
Copper tube 
D1 =5 cm, T1 = 60°C
ε1 = 0.85 
Air space 
Plastic cover, 
T2 = 40°C 
ε2 = 0.9 
Water 
 
1-
25
K 003096.0
K )27350(
1
7228.0Pr
/sm 10798.1
CW/m. 02735.0
=+=
=
×=
°=
−
β
υ
k
 
Analysis The characteristic length in this case is 
 m 0.02=m) 0.05-m 09.0(
2
1)(
2
1
12 =−= DDLc 
 850,10)7228.0(
)/sm 10798.1(
)m 02.0(K)4060)(K 003096.0)(m/s 81.9(Pr
)(
225
3-12
2
3
21 =×
−=−= −υ
β cLTTgRa 
The effective thermal conductivity is 
 
[ ] [ ][ ] 1303.0m) 05.0(m) 09.0(m) 02.0( )05.0/09.0ln()( )/ln( 53/5-3/5-3
4
55/35/33
4
cyl =+
=+= −− oic
io
DDL
DD
F 
[ ] CW/m. 05321.0)850,10)(1303.0(
7228.0861.0
7228.0)CW/m. 02735.0(386.0
)Ra(
Pr861.0
Pr386.0
4/1
4/1
4/1
cyl
4/1
eff
°=⎟⎠
⎞⎜⎝
⎛
+°=
⎟⎠
⎞⎜⎝
⎛
+= Fkk 
Then the rate of heat transfer between the cylinders becomes 
 W11.4=°−°=−= C)4060(
)05.0/09.0ln(
)CW/m. 05321.0(2)(
)/ln(
2 eff
conv
ππ
oi
io
TT
DD
k
Q& (Eq. 1) 
The rate of heat transfer by radiation is determined from 
 
 W13.4=
⎟⎠
⎞⎜⎝
⎛−+
+−+⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−+
−=
−
9
5
9.0
9.01
85.0
1
])K 27340()K 27360)[(KW/m 1067.5)(m 1571.0(
11
)(
444282
2
1
2
2
1
4
2
4
11
rad
D
D
TTA
Q
ε
ε
ε
σ&
 
Finally, 
 (per m length) W 8.244.134.11, =+=losstotalQ&
 12-77
Chapter 12 Radiation Heat Transfer 
12-98 A cylindrical furnace with specified top and bottom surface temperatures and specified heat transfer 
rate at the bottom surface is considered. The temperature of the side 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. 
Properties The emissivities of the top, bottom, and side surfaces are 0.70, 0.50, and 0.40, respectively. 
Analysis We consider the top surface to be surface 1, the bottom 
surface to be surface 2, and the side surface to be surface 3. This 
system is a three-surface enclosure. The view factor from surface 1 to 
surface 2 is determined from 
T1 = 500 K 
ε1 = 0.70 
r1 = 0.6 m 
T3 = ? 
ε3 = 0.40 
h = 1.2 m
T2 = 650 K 
ε2 = 0.50 
r2 = 0.6 m 
 
 17.0
5.0
2.1
6.0
2
6.0
2.1
12 =
⎪⎪⎭
⎪⎪⎬
⎫
==
==
F
L
r
r
L
 (Fig. 12-7) 
The surface areas are 
 
2
3
222
21
m 524.4)m 2.1)(m 2.1(
m 131.14/)m 2.1(4/
===
====
ππ
ππ
DLA
DAA
Then other view factors are determined to be 
17.02112 == FF 
83.0117.001 1313131211 =⎯→⎯=++⎯→⎯=++ FFFFF (summation rule), 83.01323 == FF
21.0)524.4()83.0)(131.1( 3131313131 =⎯→⎯=⎯→⎯= FFFAFA (reciprocity rule), 21.03132 == FF 
We now apply Eq. 12-35 to each surface 
Surface 1: 
[ ]
[ ])(83.0)(17.0
70.0
70.01)K 500)(K.W/m 1067.5(
)()(
1
 
31211
4428
31132112
1
1
1
4
1
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
Surface 2: 
[ ]
[ ])(83.0)(17.0
50.0
50.01)K 500)(K.W/m 1067.5(
)()(
1
 
32122
4428
32231221
2
2
2
4
2
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
Surface 3: 
[ ]
[ ])(21.0)(21.0
40.040.01)K.W/m 1067.5(
)()(
1
 
31213
4
3
428
23321331
3
3
3
4
3
JJJJJT
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
We now apply Eq. 12-34 to surface 2 
 [ ] [ ])(83.0)(17.0)m 131.1()()( 321223223122122 JJJJJJFJJFAQ −+−=−+−=& 
Solving the above four equations, we find 
 23
2
2
2
13 W/m 8193 ,W/m 8883 ,W/m 4974 , ==== JJJT K 631
The rate of heat transfer between the bottom and the top surface is 
 W751.6=−=−= 221221221 W/m)49748883)(17.0)(m 131.1()( JJFAQ&
The rate of heat transfer between the bottom and the side surface is 
 W644.0=−=−= 223223223 W/m)81978883)(83.0)(m 131.1()( JJFAQ&
Discussion The sum of these two heat transfer rates are 751.6 + 644 = 1395.6 W, which is practically equal 
to 1400 W heat supply rate from surface 2. This must be satisfied to maintain the surfaces at the specified 
temperatures under steady operation. Note that the difference is due to round-off error. 
 12-78
Chapter 12 Radiation Heat Transfer 
12-99 A cylindrical 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. 
Properties The emissivity of the bottom surface is 0.90. 
T1 = 700 K 
ε1 = ? 
T2 = 950 K 
ε2 = 0.90 
T3 = 450 K 
ε3 = 1 
3 m 
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 . The view factor from the base or the top to the side 
surfaces is determined by applying the summation rule to be 
F12 0 2= .
 F F F F F11 12 13 13 121 1 1 0 2+ + = ⎯ →⎯ = − = − =. .0 8
since the base surface is flat and thus F11 0= . Other view factors are 
 20.0 ,80.0 ,20.0 323113231221 ====== FFFFFF 
We now apply Eq. 9-35 to each surface 
Surface 1: 
[ ]
[ ])(80.0)(20.01)K 700)(K.W/m 1067.5(
)()(
1
 
3121
1
1
1
4428
31132112
1
1
1
4
1
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
ε
ε
εσ
 
 
Surface 2: 
[ ]
[ ])(80.0)(20.0
90.0
90.01)K 950)(K.W/m 1067.5(
)()(
1
 
32122
4428
32231221
2
2
2
4
2
JJJJJ
JJFJJFJT
−+−−+=×
−+−−+=
−
ε
εσ
 
Surface 3: 
3
4428
3
4
3
)K 450)(K.W/m 1067.5(
 
J
JT
=×
=
−
σ
We now apply Eq. 9-34 to surface 2 
 [ ] [ ])(80.0)(20.0)m 9()()( 321223223122122 JJJJJJFJJFAQ −+−=−+−=& 
Solving the above four equations, we find 
 23
2
2
2
11 W/m 2325 ,W/m 985,41 ,W/m 736,11 , ==== JJJ0.44ε
The rate of heat transfer between the bottom and the top surface is 
 2221 m 9)m 3( === AA
 kW 54.4=−=−= 221221221 W/m)736,11985,41)(20.0)(m 9()( JJFAQ&
The rate of heat transfer between the bottom and the side surface is 
 2213 m 36)m 9(44 === AA
 kW 285.6=−=−= 223223223 W/m)2325985,41)(8.0)(m 9()( JJFAQ&
Discussion The sum of these two heat transfer rates are 54.4 + 285.6 = 340 kW, which is equal to 340 kW 
heat supply rate from surface 2. 
 12-79
Chapter 12 Radiation Heat Transfer 
12-100 A thin aluminum sheet is placed between two very large parallel plates that are maintained at 
uniform temperatures. The net rate of radiation heat transfer between the plates and the temperature of the 
radiation shield 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 of surfaces are given to be ε1 = 0.8, ε2 = 0.9, and ε3 = 0.12. 
Analysis The net rate of radiation heat transfer with a 
thin aluminum shield per unit area of the plates is 
T2 = 550 K 
ε2 = 0.9 
T1 = 750 K 
ε1 = 0.8 
Radiation shield
ε3 = 0.12 
2 W/m748.9=
⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε+⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
1
12.0
1
12.0
11
9.0
1
8.0
1
])K 550()K 750)[(K W/m1067.5(
 
111111
)(
44428
2,31,321
4
2
4
1
shield one,12
TT
Q&
 
The equilibrium temperature of the radiation shield is determined from 
K 671.3=⎯→⎯
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
3
4
3
4428
2
31
4
3
4
1
13
1
12.0
1
8.0
1
])K 750)[(K W/m1067.5(
 W/m9.748
111
)(
T
T
TT
Q&
 
 12-80
Chapter 12 Radiation Heat Transfer 
12-101 Two thin radiation shields are placed between two large parallel plates that are maintained at 
uniform temperatures. The net rate of radiation heat transfer between the plates with and without the 
shields, and the temperatures of radiation shields 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 of surfaces are given to be ε1 = 0.6, ε2 = 0.7, ε3 = 0.10, and ε4 = 0.15. 
Analysis The net rate of radiation heat transfer without 
the shields per unit area of the plates is 
2 W/m3288=
−+
−⋅×=
−ε+ε
−σ=
−
1
7.0
1
6.0
1
])K 300()K 600)[(K W/m1067.5(
 
111
)(
44428
21
4
2
4
1
shield no12,
TT
Q&
 
The net rate of radiation heat transfer with two thin 
radiation shields per unit area of the plates is 
 
2 W/m206=
⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε+⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε+⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
−
1
15.0
1
15.0
11
10.0
1
10.0
11
7.0
1
6.0
1
])K 300()K 600)[(K W/m1067.5( 
111111111
)(
44428
443321
4
2
4
1
shieldstwo12,
TT
Q&
 
T2 = 300 K 
ε2 = 0.7 
T1 = 600 K 
ε1 = 0.6 
ε3 = 0.10 
ε4 = 0.15
The equilibrium temperatures of the radiation shields are determined from 
K 549=⎯→⎯
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
3
4
3
4428
2
31
4
3
4
1
13
1
10.0
1
6.0
1
])K 600)[(K W/m1067.5(
 W/m206
111
)(
T
T
TT
Q&
 
 
K 429=⎯→⎯
⎟⎠
⎞⎜⎝
⎛ −+
−⋅×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −ε+ε
−σ=
−
4
44
4
428
2
24
4
2
4
4
42
1
7.0
1
15.0
1
])K 300()[K W/m1067.5(
 W/m206
111
)(
T
T
TT
Q&
 
 12-81
Chapter 12 Radiation Heat Transfer 
12-102 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection and 
radiation and the rate of evaporation of water are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 
Combustion gases are assumed to have the properties of air, which is an ideal gas with constant properties. 
Properties The properties of air at 1200 K = 927°C and 1 atm are (Table A-15) 
/sm 10586.1
C W/m.07574.0
kg/m 2944.0
25-
3
×=υ
°=
=ρ
k 7221.0Pr
CJ/kg. 1173
=
°=pC 
Ts = 105°C
D = 15 cm Analysis (a) The Reynolds number is 
Combustion 
gases, 1 atm 
Ti = 1200 K 
3 m/s 
 373,28
/sm 10586.1
m) m/s)(0.15 (3
Re
25
=×=υ= −
DmV 
which is greater than 10,000. Therefore, the flow is 
turbulent and the entry lengths in this case are roughly 
m 5.1m) 15.0(1010 ==≈≈ DLL th 
which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent 
flow in the entire duct, and determine the Nusselt number from 
 14.76)7221.0()373,28(023.0PrRe023.0 3.08.03.08.0 ====
k
hDNu h 
Heat transfer coefficient is 
 C. W/m45.38)14.76(m 15.0
C W/m.07574.0 2 °=°== Nu
D
kh 
Next we determine the exit temperature of air 
 
 kg/s0.01561=)m 67m/s)(0.017 )(3 kg/m2944.0(
m 0.01767=/4m) 15.0(4/
m 2.827=m) m)(6 15.0(
23
222
2
==
==
==
c
c
VAm
DA
DLA
ρ
ππ
ππ
&
 C107.2)927105(105)( )1173)(01561.0(
)827.2)(45.38(
)/( °=−−=−−= −− eeTTTT pCmhAisse & 
Then the rate of heat transfer by convection becomes 
 W15,010=°−°=−= C)2.107927)(CJ/kg. 3 kg/s)(11701561.0()(conv eip TTCmQ &&
Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinite circular 
cylinder is, from Table 12-4, 
 L = 0.95(0.15 m) = 0.1425 m 
Then, 
atmft 0.075atmm 0228.0m) 5atm)(0.142 16.0(
atmft .0370atmm 0114.0m) 5atm)(0.142 08.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c 
The emissivities of CO2 and H2O corresponding to these values at the average gas temperature of 
Tg=(Tg+Tg)/2 = (927+107.2)/2 = 517.1°C = 790 K and 1atm are, from Fig. 12-36, 
055.0atm 1 , =ε c and 062.0atm 1 , =εw 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 800 K is, from Fig. 12-38, 
 12-82
Chapter 12 Radiation Heat Transfer 
0.0 67.0
08.016.0
16.0
 112.0075.0037.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
Then the effective emissivity of the combustion gases becomes 
 0.1170.0062.01055.01atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
Note that the pressure correction factor is 1 for both gases since the total pressure is 1 atm. For a source 
temperature of Ts = 105°C = 378 K, the absorptivity of the gas is again determined using the emissivity 
charts as follows: 
atmft 0.036atmm 0109.0
K 790
K 378m) 5atm)(0.142 16.0(
atmft 0.018atmm 00545.0
K 790
K 378m) 5atm)(0.142 08.0(
⋅=⋅==
⋅=⋅==
g
s
w
g
s
c
T
T
LP
T
T
LP
 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 378 K and 1atm 
are, from Fig. 12-36, 
 and 037.0atm 1 , =ε c 062.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
0864.0)062.0(
K 378
K 790)1(
0597.0)037.0(
K 378
K 790)1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 378 K 
instead of Tg = 790 K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read 
Δε = 0.0. Then the absorptivity of the combustion gases becomes 
0.1460.00864.00597.0 =−+=αΔ−α+α=α wcg 
The emissivity of the inner surface s of the tubes is 0.9. Then the net rate of radiation heat transfer from the 
combustion gases to the walls of the tube becomes 
 W6486=
−⋅×+=
α−εσ+ε=
− ])K 378(146.0)K 790(117.0)[K W/m1067.5)(m 827.2(
2
19.0
)(
2
1
444282
44
rad sgggs
s TTAQ&
 
(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (Table A-9). Then rate of evaporation of water 
becomes 
 kg/s 0.0644=×
+=+=⎯→⎯=+
J/kg 107.333
 W)6486010,15(
3
radconv
evapevapradconv
fg
fg h
QQ
mhmQQ
&&
&&&& 
 12-83
Chapter 12 Radiation Heat Transfer 
12-103 Combustion gases flow inside a tube in a boiler. The rates of heat transfer by convection and 
radiation and the rate of evaporation of water are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 
Combustion gases are assumed to have the properties of air, which is an ideal gas with constant properties. 
Properties The properties of air at 1200 K = 927°C and 3 atm are (Table A-15) 
/sm 105287.0
/s)/3m 10586.1(
C W/m.07574.0
kg/m 2944.0
25-
25-
3
×=
×=υ
°=
=ρ
k
 
7221.0Pr
CJ/kg. 1173
=
°=pC 
Ts = 105°C
D = 15 cm 
Analysis (a) The Reynolds number is Combustion 
gases, 3 atm 
Ti = 1200 K 
3 m/s 
 114,85
/sm 105287.0
m) m/s)(0.15 (3
Re
25
=×=υ= −
DmV 
which is greater than 10,000. Therefore, the flow is 
turbulent and the entry lengths in this case are roughly 
m 5.1m) 15.0(1010 ==≈≈ DLL th 
which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent 
flow in the entire duct, and determine the Nusselt number from 
 4.183)7221.0()114,85(023.0PrRe023.0 3.08.03.08.0 ====
k
hDNu h 
Heat transfer coefficient is 
 C. W/m59.92)4.183(
m 15.0
C W/m.07574.0 2 °=°== Nu
D
kh 
Next we determine the exit temperature of air 
 
 kg/s0.01561=)m 67m/s)(0.017 )(3 kg/m2944.0(
m 0.01767=/4m) 15.0(4/
m 2.827=m) m)(6 15.0(
23
222
2
==
==
==
c
c
VAm
DA
DLA
ρ
ππ
ππ
&
 C105.0)927105(105)( )1173)(01561.0(
)827.2)(59.92(
)/( °=−−=−−= −− eeTTTT pCmhAisse & 
Then the rate of heat transfer by convection becomes 
 W15,050=°−°=−= C)0.105927)(CJ/kg. kg/s)(1173 01561.0()(conv eip TTCmQ &&
Next, we determine the emissivity of combustion gases. First, the mean beam length for an infinite circular 
cylinder is, from Table 12-4, 
 L = 0.95(0.15 m) = 0.1425 m 
Then, 
atmft 0.075atmm 0228.0m) 5atm)(0.142 16.0(
atmft .0370atmm 0114.0m) 5atm)(0.142 08.0(
⋅=⋅==
⋅=⋅==
LP
LP
w
c
The emissivities of CO2 and H2O corresponding to these values at the average gas temperature of 
Tg=(Tg+Tg)/2 = (927+105)/2 = 516°C = 790 K and 1atm are, from Fig. 12-36, 
055.0atm 1 , =ε c and 062.0atm 1 , =εw 
These are the base emissivity values at 1 atm, and they need to be corrected for the 3 atm total pressure. 
Noting that (Pw+P)/2 = (0.16+3)/2 = 1.58 atm, the pressure correction factors are, from Fig. 12-37, 
 Cc = 1.5 and Cw = 1.8 
Both CO2 and H2O are present in the same mixture, and we need to correct for the overlap of emission 
bands. The emissivity correction factor at T = Tg = 800 K is, from Fig. 12-38, 
0.0 67.0
08.016.0
16.0
 112.0075.0037.0
=εΔ
⎪⎭
⎪⎬
⎫
=+=+
=+=+
cw
w
wc
PP
P
LPLP
 
 12-84
Chapter 12 Radiation Heat Transfer 
Then the effective emissivity of the combustion gases becomes 
 0.1940.0062.08.1055.05.1atm 1 ,atm 1 , =−×+×=εΔ−ε+ε=ε wwccg CC 
For a source temperature of Ts = 105°C = 378 K, the absorptivity of the gas is again determined using the 
emissivity charts as follows: 
atmft 0.036atmm 0109.0
K 790
K 378m) 5atm)(0.142 16.0(
atmft 0.018atmm 00545.0
K 790
K 378m) 5atm)(0.142 08.0(
⋅=⋅==
⋅=⋅==
g
s
w
g
s
c
T
T
LP
T
T
LP
 
The emissivities of CO2 and H2O corresponding to these values at a temperature of Ts = 378 K and 1atm 
are, from Fig. 12-36, 
 and 037.0atm 1 , =ε c 062.0atm 1 , =εw 
Then the absorptivities of CO2 and H2O become 
1555.0)062.0(
K 378
K 790)8.1(
0896.0)037.0(
K 378
K 790)5.1(
45.0
atm 1 ,
45.0
65.0
atm 1 ,
65.0
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
=⎟⎠
⎞⎜⎝
⎛=ε⎟⎟⎠
⎞
⎜⎜⎝
⎛=α
w
s
g
ww
c
s
g
cc
T
T
C
T
T
C
 
Also Δα = Δε, but the emissivity correction factor is to be evaluated from Fig. 12-38 at T = Ts = 378 K 
instead of Tg = 790 K. We use the chart for 400 K. At Pw/(Pw+ Pc) = 0.67 and PcL +PwL = 0.112 we read 
Δε = 0.0. Then the absorptivity of the combustion gases becomes 
0.2450.01555.00896.0 =−+=αΔ−α+α=α wcg 
The emissivity of the inner surfaces of the tubes is 0.9. Then the net rate of radiation heat transfer from the 
combustion gases to the walls of the tube becomes 
 W10,745=
−⋅×+=
α−εσ+ε=
− ])K 378(245.0)K 790(194.0)[K W/m1067.5)(m 827.2(
2
19.0
)(
2
1
444282
44
rad sgggs
s TTAQ&
 
(b) The heat of vaporization of water at 1 atm is 2257 kJ/kg (TableA-9). Then rate of evaporation of water 
becomes 
 kg/s 0.0773=×
+=+=⎯→⎯=+
J/kg 107.333
 W)745,10050,15(
3
radconv
evapevapradconv
fg
fg h
QQ
mhmQQ
&&
&&&& 
 
 
 
12-104 ….. 12-106 Design and Essay Problems 
 
 
KJ 
 12-85
	Heat Chap12-001.doc
	View Factors
	Radiation Heat Transfer Between Surfaces
	Heat Chap12-041.doc
	Radiation Shields and The Radiation Effect
	Radiation Exchange with Absorbing and Emitting Gases
	Heat Chap12-074.doc
	Special Topic: Heat Transfer from the Human Body
	Heat Chap12-088.doc
	Review Problems
	12-104 ….. 12-106 Design and Essay Problems