13-1
13-2
View Factors
13-1C The view factor
ji
F→
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.
13-4C 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
13-4
13-6 Two coaxial parallel circular disks spaced apart at a distance L. The parameter that would increase the view factor F12
by a factor of 5 is to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis For coaxial parallel disks, from Table 13-1 with ri = rj, we have
2
141
R
D
r
2
1
2
1
)/)(2/1(
1)/(1
11.0 LD
LD +−
+=
→
70273.0
1
=
L
D
2
2
2
2
)/)(2/1(
1)/(1
15.0 LD
LD +−
+=
→
8284.2
2
=
L
D
0.2485=== 8284.2
70273.0
2
11
2
D
L
L
D
L
L
13-7 A row of cylinders is spaced between two large parallel plates. The view factor between the plate and the row of
cylinders is to be determined.
0.8426=
5.3
5
5
Discussion If the spacing between the cylinders is the same as the diameter (s = D), then the view factor would be F12 = 1.
Note that the equation is only valid for s ≥ D.
13-5
13-8 An enclosure consisting of thirteen surfaces is considered.
The number of view factors this geometry involves and the
number of these view factors that can be determined by the
2
1
3
9
8
6
7
13-9 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
5
3
13-10 A semispherical furnace is considered. The view factor from the dome of this furnace to its flat base is to be
(2)
(1): circular base surface
(2): dome surface
Surface (1) is flat, and thus
F
11 0=
.
11 :ruleSummation 121211 =→=+ FFF
0.5=====⎯→⎯= 2
1
2
4
)1(A :ruley reciprocit 2
2
2
1
12
2
1
21212121D
D
A
A
F
A
A
FFAF
1
4
(1)
D
11
4
5
13-6
13-11 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.
h
conical side surface (3)
21122211 ==== FFFF
11 :rulesummation 13131211 =⎯→⎯=++ FFFF
2Dh
d2
=⎯→⎯=⎯→⎯= 3131
2
3131312
)1(
4
A :ruley reciprocit FF
Dhd
FAF
13-12 The four view factors associated with an enclosure formed by two very long concentric cylinders are to be determined.
(2)
the inner surface of the outer cylinder (2)
No radiation leaving surface 1 strikes itself and thus
0=
11
F
2
D
d
D
(2) (1)
(3)
D2 D1
(1)
13-7
13-13 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,
13-8
13–14 A cylindrical enclosure is considered. (a) The expression for the view factor between the base and the side surface F13
in terms of K and (b) the value of the view factor F13 for L = D are to be determined.
131211 =++ FFF
11 =F
or
1213 1FF −=
(1)
For coaxial parallel disks, from Table 13-1, with i = 1, j = 2,
1
12/12
2/1
2
2
2
D
13–12
13–18 A circular cone is positioned on a common axis with a disk and a cylindrical surface oriented coaxially with a disk.
The view factors of the two geometries are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis For the circular cone and disk geometry: The
area for A1, A2, and A3 are
5
2
2/1
2
2
DD
D
2
D
2321 FF =
Applying reciprocity relation between A1 and A2, we have
212121FAFA =
→
5
)/()/( 23
2312211212
F
FAAFAAF ===
For coaxial parallel disks, the view factor F23 is
2
2
2
2
23 =−−=−−=
D
where
2
1
2
21 ====
L
D
RRR
and
642
1
2
1
122
1
2
2=+=+=
+
+=
RR
R
S
0.0767===
5
1716.0
5
23
12
F
F
(circular cone and disk)
For the circular surface and disk geometry: The end surfaces A3 and A4 are treated as hypothetical surfaces. From the
2
2
L
For surface 3:
1716.0
5.02
15.041
12
2
23 =
+−
+=F
where
5.0
2/2/ ====
D
D
L
D
L
r
R
For surface 4:
05573.0
25.02
125.041
12
2
24 =
+−
+=F
where
25.0
2
2/
2
2/
2
====
D
D
L
D
L
r
R
Using Eq. (1), with L = D, we get
0.0290=−=−=−= )05573.01716.0(
4
1
)(
4/
)( 2423
2
2423
1
2
12 FF
DL
D
FF
A
A
F
(circular surface and disk)
13–14
13–21 The view factors between the rectangular surfaces shown in the figure are to be determined.
Assumptions The surfaces are diffuse emitters and reflectors.
(a) From Fig.13-6
26.0
25.0
4
1
25.0
4
1
23
1
2
=
==
==
F
W
L
W
L
and
33.0
25.0
4
1
5.0
4
2
)31(2
1
2
=
==
==
+→
F
W
L
W
L
07.026.033.0 :ruleion superposit 23)31(2212321)31(2 =−=−=⎯→⎯+= +→+→ FFFFFF
0.07==⎯→⎯=⎯→⎯= 211221212121
:ruley reciprocit FFFAFAAA
(b) From Fig.13–6,
16.0
5.0
4
2
0.25
4
1
3)24(
1
2
=
==
==
→+
F
W
L
W
L
and
24.0
5.0
4
2
0.5
4
2
)31()24(
1
2
=
==
==
+→+
F
W
L
W
L
08.016.024.0 :ruleion superposit 1)24(3)24(1)24()31()24( =−=⎯→⎯+= →+→+→++→+ FFFF
16.0)08.0(
4
8
:ruley reciprocit
1)24(
1
)24(
)24(1
)24(111)24()24(
===⎯→⎯
=
→+
+
+→
+→→++
F
A
A
F
FAFA
F
+=
+→
:ruleion superposit
1214)24(1
FFF
12
14
equivalent to
12
F
in part (a).
3 m
shaded part of top surface by (1),
remaining part of top surface by (3),
15.0
67.0
3
2
1
3
3
)31()42(
1
2
=
==
==
+→+
F
D
L
D
L
and
082.0
33.0
3
1
1
3
3
14
1
2
=
==
==
F
D
L
D
L
3)42(1)42()31()42(
:ruleion superposit →+→++→+ += FFF
3)42(1)42(
:rulesymmetry →+→+ =FF
4 m
(1)
(3)
1 m
1 m
(2)
4 m
1 m
(4)
1 m
(1)
(2)
1 m
1 m
1 m
(3)
13–15
075.0
2
15.0
2
)31()42(
3)42(1)42( ==== +→+
→+→+
F
FF
150F07506F3FAFA :ruley reciprocit 421421142424211 .).)(()( )()()()()( =⎯→⎯=⎯→⎯= +→+→→+++→
0.068=−=⎯→⎯+=⎯→⎯+=
+→ 082.015.0082.015.0 : ruleion superposit 12121412)42(1 FFFFF
13–16
13–22 A cylindrical enclosure is considered. The view factor from the side surface of this cylindrical enclosure to its base
surface is to be determined.
top surface by (2), and
(3)
25.0
4
2
22
==
r
r
L
r
1 :rulesummation 131211 =++ FFF
95.0105.00 1313 =⎯→⎯=++ FF
0.119=====⎯→⎯= )95.0(
8
1
8
2
:ruley reciprocit 13
2
1
2
1
13
1
2
1
13
3
1
31313131F
r
r
F
Lr
r
F
A
A
FFAFA
Discussion This problem can be solved more accurately by using the view factor relation from Table 13-1 to be
25.0
4
1
11
1
===
r
r
L
r
R
(2)
(1)
D=2r
13–17
13-23 For a right circular cylinder the view factors F13 and F33 are to be determined.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
5.2
4
10
1
==
r
L
and
4.0
10
4
2==
L
r
From Fig. 13-7 or compute from item 2 of Table 13-1
F12 = 0.12
13–19
13–26 The expressions for the view factors F12 and F21 of two infinitely long parallel plates are to be determined using the
Hottel’s crossed–strings method.
Assumptions 1 The surfaces are diffuse emitters and reflectors.
Analysis From the Hottel’s crossed–strings method, we have
(Crossed strings) (Uncrossed strings)
2 (Length of surface )
ij
Fi
®
S – S
=´
0.3424== 2112 FF
13–20
13-27 Two view factors associated with three very long ducts with different geometries are to be determined.
Analysis (a) Surface (1) is flat, and thus
0
11 =F
.
1=→=+ 121211 1 :rulesummation FFF
0.64==
==⎯→⎯=
2
)1(
2
A :ruley reciprocit 12
2
1
21212121
s
D
Ds
F
A
A
FFAF
(b) Noting that surfaces 2 and 3 are symmetrical and thus
1312 FF =
,
the summation rule gives
0.5=⎯→⎯=++⎯→⎯=++ 121312131211 101 FFFFFF
Also by using the equation obtained in Example 13–4,
0.5===
−+
=
−+
=
2
1
222 1
321
12 a
a
a
bba
L
LLL
F
2b
a
=
==⎯→⎯=
2
1
A :ruley reciprocit 12
2
1
21212121b
a
F
A
A
FFAF
(c) Applying the crossed-string method gives
a
bba 22 −+
=
−+
=
+−+
==
a
bba
L
LLLL
FF
2
22
2
)()(
22
1
4365
2112
(1)
(2)
D
L3 = b L4 = b
L6
L5
L2 = a
L1 = a
(1)
(3) (2)
a
b