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CHAPTER 4
4.1 For Configuration 11, Appendix D, find Fd1−2by (a) area integration, and (b) contour integration. Compare
the effort involved.
Solution
(a) From equation (4.23)
Scos θ1=ˆn1·s12 =ˆ
k·s12 =z,
Scos θ2=−ˆn2·s12 =−ˆı ·s12 =c,
and, from equation (4.21)
S2=c2+y2+z2.
b
c
a
A2
dA1
y
z
θ
2
θ
1
n1
n2
Thus
cos θ1cos θ2
cz dz dy
Fd1−2=1
2π"tan−1b
c−c
√a2+c2tan−1b
√a2+c2#
(b) For equation (4.26), we need
=−dy ˆfor z=a,0≤y≤b,
=−dz ˆ
kfor y=0,0≤z≤a.
Therefore
(s12 ׈n1)·ds2=c dy for z=0,0≤y≤b,
Fd1−2=1
2πI(s12 ׈n1)·ds2
S2
b
Z
c dy
a
Z
b
Z
c dy
a
Z
68 RADIATIVE HEAT TRANSFER
4.2 Using the results of Problem 4.1, find F1−2for Configuration 33 in Appendix D.
CHAPTER 4 69
4.3 Find F1−2for Configuration 32, Appendix D, by area integration.
70 RADIATIVE HEAT TRANSFER
4.4
h
A2
dA1
a
r
Evaluate Fd1−2for Configuration 13 in Appendix D by (a) area integration, and (b)]
contour integration. Compare the effort involved.
Integrals such as the one in ψ2are listed in common integration tables, at least for a first-power denominator,
i.e.,
dψ
dψ
Fd1−2=2h2ZR2
0
(h2+a2+r2
2)r2dr2
[(h2+a2+r2
2)2−4a2r2
2]3/2
=h2ZR2
2
(h2+a2+η)dη
Fd1−2=h22C(2η+D)−2(Dη+2C2)
q√X
0
,
where
=1
4a2
2a2η−2a2(h2+a2)
p(h2+a2)2+2(h2−a2)η+η2
0
CHAPTER 4 71
2−h2−a2
s2=x2ˆı +y2ˆ=R2(ˆı cos ψ2+ˆsin ψ2)
ds2=R2(ˆcos ψ2−ˆı sin ψ2)dψ2.
Thus
Fd1−2=1
(s12 ׈n1)·ds2
=R2
2πZ2π
0
h2+a2+R2
2+2aR2sin ψ2
dψ2
h2+a2+R2
Fd1−2=1
2
1+R2
2−h2−a2
q(h2+a2)2+2(h2−a2)R2
72 RADIATIVE HEAT TRANSFER
4.5 Using the result from the Problem 4.4, calculate F1−2for Configuration 40, Appendix D.
CHAPTER 4 73
4.6 Find the view factor Fd1−2for Configuration 11 in Appendix D, with dA1tilted towards A2by an angle φ.
(4.23),
c
i
s12 =−cˆı +yˆ+zˆ
k,ˆn1=−sin φˆı +cos φˆ
k,ˆn2=ˆı,
S2=c2+y2+z2,
Scos θ1=ˆn1·s12 =csin φ+zcos φ,
Scos θ2=−ˆn2·s12 =c,
and
c(csin φ+zcos φ)dy dz
πZa
0
2(c2+z2)(c2+b2+z2)+1
2(c2+z2)3/2tan−1b
c2+z2#dz.
It does not appear possible to carry out the z-integration!
Using contour integration:
s12 ׈n1=ysin φˆ
k−zsin φˆ+ccos φˆ+ycos φˆı
From equation (4.26)
Fd1−2=1
2π
b
Z
y=0
ccos φdy
c2+y2+
a
Z
z=0
bsin φdz
b2+c2+z2−
b
Z
y=0
(ccos φ−asin φ)dz
a2+c2+y2
2π"cos φtan−1b
c+bsin φ
74 RADIATIVE HEAT TRANSFER
4.7 Find Fd1−2for the surfaces shown in
the figure, using (a) area integration,
(b) view factor algebra and Configu-
ration 11 in Appendix D.
a
b
A2
y
x
CHAPTER 4 75
4.8 For the infinite half-cylinder depicted in
the figure, find F1−2.
r1
with
a2
1+b2
1=a2
2+b2
2=4R2,
ai=2Rcos αi=2RR+ri
ai
.
Thus
ai=p2R(R+ri),bi=p2R(R−ri),
and
F1−2=√2
4"r1−r1
R−r1+r1
R+r1+r2
R−r1−r2
R#.
76 RADIATIVE HEAT TRANSFER
4.9 Find Fd1−2for the surfaces shown in the
figure.
Solution
a
c
d
A2b
A5
Fd1−2=φ a+d
c,e
c!−φ d
c,e
c!−φ a+d
c,e−b
c!,+φ d
c,e−b
c!
where
φ(X,Y)=1
2π(X
√1+X2tan−1Y
√1+X2
+Y
√1+Y2tan−1X
√1+Y2).
Note that (e−b) may be positive (e>b) or negative (b>e).
CHAPTER 4 77
4.10
R
α
α
Find the view factor of the spherical ring shown in the figure,
to itself F1−1, using the inside-sphere method.
78 RADIATIVE HEAT TRANSFER
4.11 Determine the view factor for Configuration 51 in Appendix D, using (a) other, more basic view factors given
in Appendix D, (b) the crossed strings rule.
CHAPTER 4 79
4.12
2
A1
r2 = r3r4
r1
A2
A3
A4
θ
Acap
To reduce heat transfer between two infinite concentric cylin-
ders a third cylinder is placed between them as shown in the
figure. The center cylinder has an opening of half-angle θ.
Calculate F4−2.
80 RADIATIVE HEAT TRANSFER
4.13
A1
A2
A4
R3R
R
2RA3
Consider the two long concentric cylinders as shown in the figure.
Between the two cylinders is a long, thin flat plate as also indicated.
Determine F4−2.
CHAPTER 4 81
4.14
a
A2
b
ϕ
xbxa
RbRa
x
Calculate the view factor F1−2for surfaces on a cone as shown
in the figure.
Solution
From view factor algebra
2X−qX2−4R2
X=(xa−xb)2cot2ϕ
2+x2
a+x2
b
x2
a
,
and
dx
2
x2
a−x2
b1−1
2X+1
2qX2−4x2
with Xas defined above.
82 RADIATIVE HEAT TRANSFER
4.15
D
A2
h= 2 D
A1
s
s
A3
A4
Determine the view factor F1−2for the configuration shown in the figure, if
(a) the bodies are two-dimensional (i.e., infinitely long perpendicular to the paper);
(b) the bodies are axisymmetric (cones).
