T ∞, h
t
W
dx
y
L
x
� �
d 2 T
dT
2h
x
+
−
L (T − T ∞ ) = 0
dx 2
dx
kt
2 d 2 θ
dθ
x
+ x
− m 2 Lxθ = 0
(2.56)
dx 2
dx
where
t
A c = w x
L
dA s = 2w dx
θ ≡ T − T ∞
2h
2 ≡
m
kt
A s � 2wL
But we need z 2 (d 2 θ/dz 2 ) + z(dθ/dz) − z 2 θ = 0, for the modified Bessel
function solution
So, z 2 ∼ x, z ∼
√
x
(
)
√
√
√
From the solution form, I 0 2m xL , imply z ∼ x = 2m xL
dz
√ 1
√ 1
2m 2 L
−1/2
= 2m L · · x
= m L √ =
dx
2
x
z
dθ
dθ dz
d 2 θ
d dθ
dz
=
·
;
=
·
;
dx
dz dx
dx 2
dz dx
dx
35
1-D Steady-State Heat Conduction
FIGURE 2.13
A triangular straight fin with variable cross-sectional area.
t
W
dx
y
L
x
� �
d 2 T
dT
2h
x
+
−
L (T − T ∞ ) = 0
dx 2
dx
kt
2 d 2 θ
dθ
x
+ x
− m 2 Lxθ = 0
(2.56)
dx 2
dx
where
t
A c = w x
L
dA s = 2w dx
θ ≡ T − T ∞
2h
2 ≡
m
kt
A s � 2wL
But we need z 2 (d 2 θ/dz 2 ) + z(dθ/dz) − z 2 θ = 0, for the modified Bessel
function solution
So, z 2 ∼ x, z ∼
√
x
(
)
√
√
√
From the solution form, I 0 2m xL , imply z ∼ x = 2m xL
dz
√ 1
√ 1
2m 2 L
−1/2
= 2m L · · x
= m L √ =
dx
2
x
z
dθ
dθ dz
d 2 θ
d dθ
dz
=
·
;
=
·
;
dx
dz dx
dx 2
dz dx
dx
35
1-D Steady-State Heat Conduction
FIGURE 2.13
A triangular straight fin with variable cross-sectional area.