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Thus,
(3μu m /H 2 )(4H)
24μu m H
f =
=
2
2
ρu m /2
ρu m H 2
24μ
f = u m Hρ
But,
ρu m D H
4u m Hρ
Re Dh =
=
μ
μ
Finally,
96
f =
(8.32)
Re Dh
c. Energy equation:
∂T
∂T
∂ 2 T
∂ 2 T
ρC p u
+ υ
= k
+
+ q ˙ + Φ
∂x
∂y
∂x 2
∂y 2
Φ = 0 (because of low-speed flow)
q ˙ = 0 (no internal heat generation)
dT
d 2 T
= const ⇒
= 0
dx
dx 2
The governing equation becomes
∂ 2 T
u ∂T
=
(8.33)
∂y 2
α ∂x
With solution (after substitution of Equation 8.31b)
3 u m y 2
y 4
dT
T =
−
+ c 1 y + c 2
2 α
2
12H 2 dx
Boundary conditions:
y = 0, dT /dy = 0 → c 1 = 0
y = b, T = T s
5 u m H 2 dT
c 2 = T s − 8 α
dx
2
4
3 u m
y
y
5 dT
T =
H 2
−
−
+ T s
(8.34)
2 α
2H 2
12H 4
12 dx
178
Analytical Heat Transfer
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�
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Thus,
(3μu m /H 2 )(4H)
24μu m H
f =
=
2
2
ρu m /2
ρu m H 2
24μ
f = u m Hρ
But,
ρu m D H
4u m Hρ
Re Dh =
=
μ
μ
Finally,
96
f =
(8.32)
Re Dh
c. Energy equation:
∂T
∂T
∂ 2 T
∂ 2 T
ρC p u
+ υ
= k
+
+ q ˙ + Φ
∂x
∂y
∂x 2
∂y 2
Φ = 0 (because of low-speed flow)
q ˙ = 0 (no internal heat generation)
dT
d 2 T
= const ⇒
= 0
dx
dx 2
The governing equation becomes
∂ 2 T
u ∂T
=
(8.33)
∂y 2
α ∂x
With solution (after substitution of Equation 8.31b)
3 u m y 2
y 4
dT
T =
−
+ c 1 y + c 2
2 α
2
12H 2 dx
Boundary conditions:
y = 0, dT /dy = 0 → c 1 = 0
y = b, T = T s
5 u m H 2 dT
c 2 = T s − 8 α
dx
2
4
3 u m
y
y
5 dT
T =
H 2
−
−
+ T s
(8.34)
2 α
2H 2
12H 4
12 dx
178
Analytical Heat Transfer