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Substituting this into Equation 8.29 and rearranging
2
H 2 ∂P
y
u = −
1 −
(8.30)
2μ ∂x
H 2
u max at y = 0
H 2 ∂P
u max = − 2μ ∂x
2
y
u = u max 1 −
(8.31a)
H 2
� H
2
�
2
u max 1 −
y
dy
0
H 2
ρudA
u m =
=
ρA
2H
H
u max y − (y 3 /3H 2 )
0
u max ((2/3)H)
u m =
=
H
H
2
u m = u max
3
Thus,
2
3
y
u = u m 1 −
(8.31b)
2
H 2
b. Friction factor:
− (∂P /∂x) D h
f ≡
2
ρu m /2
ΔPA c + τ w (2w )Δx = 0
ΔP
−2τ w w
−2τ w w
τ w
=
=
= −
ΔX
A c
2wH
H
� �
�
� H
∂u �
3
2y
u m
τ w = μ
= μ
u m
= 3μ
∂y
2
H 2
H
y =H
0
Therefore,
ΔP
−3μu m
= −
Δx
H 2
4A c
4(2wH)
D H =
=
= 4H
P
2w
177
Internal Forced Convection