(2)

L 1 + L 4 − L bd

F 1−4 =
(12.8)
2L 1
(3)

L ac + L bd − (L 2 + L 4 )

F 1−3 =
(12.9)
2L 1
(1) F 1−2 + F 1−ac = 1
F 1−2 = 1 − F 1−ac
L ac
= 1 −
F ac−1
L 1
L ac
= 1 −
(1 − F ac−2 )
L 1
L ac
L ac L 2
= 1 −
+
·
· F 2−ac
L 1
L 1 L ac
L ac
L 2
= 1 −
+
· (1 − F 2−1 )
L 1
L 1
L ac
L 2
L 2 L 1
= 1 −
+
−
·
· F 1−2
L 1
L 1
L 1 L 2
L ac
L 2
= 1 −
+
− F 1−2
L 1
L 1
L 1 + L 2 − L ac
∴ F 1−2 =
2L 1
(2) Similarly, F 1−4 = ((L 1 + L 4 − L bd )/2L 1 )
(3) F 1−2 + F 1−3 + F 1−4 = 1
∴ F 1−3 = 1 − F 1−2 − F 1−4
L 1 + L 2 − L ac
L 1 + L 4 − L bd
= 1 −
−
2L 1
2L 1
244
Analytical Heat Transfer
1
2
3
4
a
b
c
d
FIGURE 12.5
Concept of Hotell’s cross-string method for 2-D geometry.