Problem 8.127

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Problem 8.127 [Difficulty: 2]
Given: Flow through rectangular duct
Find: Pressure drop
Solution:
Basic equations
p1
ρ
α
V1
2
2
⋅+ g z1⋅+
⎛⎜⎜⎝
⎞
⎠
p2
ρ
α
V2
2
2
⋅+ g z2⋅+
⎛⎜⎜⎝
⎞
⎠− hlT= hlT hl hlm+= f
L
D
⋅ V
2
2
⋅
Minor
f
Le
D
⋅ V
2
2
⋅
⎛⎜⎝
⎞
⎠∑+=
Dh
4 a⋅ b⋅
2 a b+( )⋅=
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data Q 1750 cfm⋅= L 1000 ft⋅= b 2.5 ft⋅= a 0.75 ft⋅=
At 50oF, from Table A.9 ρ 0.00242
slug
ft3
⋅= μ 3.69 10 7−⋅ lbf s⋅
ft2
⋅= ρw 1.94
slug
ft3
⋅=
Hence V
Q
a b⋅= V 15.6
ft
s
⋅= and Dh
4 a⋅ b⋅
2 a b+( )⋅= Dh 1.15 ft⋅=
Re
ρ V⋅ Dh⋅
μ
= Re 1.18 105×=
For a smooth duct 1
f
2− log 2.51
Re f⋅
⎛⎜⎝
⎞
⎠⋅=
so f 0.017=
Hence ∆p f
L
Dh
⋅ ρ⋅ V
2
2
⋅= ∆p 0.031 psi⋅=
or, in in water h
∆p
ρw g⋅
= h 0.848 in⋅=