�
�
�
2L 1 − L 1 − L 2 + L ac − L 1 − L 4 + L bd
=
2L 1
L ac + L bd − L 2 − L 4
=
2L 1
The concept of the following examples (2) through (4) comes
from [3].
Example 12.2
Determine the view factor between two parallel plates with partial blockages as
shown in Figure 12.6.
)
Length of each crossed string = l 2 + c 2
( ) 2
c
Length of each uncrossed string = 2 b 2 + 2
From Hottel’s crossed-string method, the view factor can be determined as
)
)
�
( ) 2
� 2 ( ) 2
l 2 + c 2 − 2 b 2 + (c/2) 2
c
2b
c
F 1−2 =
= 1 +
−
+
l
l
l
l
Example 12.3
Determine the view factor between two opposite circular tubes as shown in Figure
12.7. From Hottel’s cross-string method, the view factor is
2L 1 − 2L 2
L 1 − L 2
F 1−2 =
=
2A 1
πR
where
L 1 − crossed string abcde
L 2 uncrossed string ef L 2 D 2R
View Factor
245
−
= +
A 1
b
b
2
c
c
l
A 2
FIGURE 12.6
View factor between two parallel plates with partial blockages.