Problem 4.101
[Difficulty: 3]
CS
x
c
y
2
h
d
Given:
Data on flow in 2D channel
Find:
Maximum velocity; Pressure drop
Solution:
Basic equations: Continuity, and momentum flux in x direction
Assumptions:
1) Steady flow
2) Neglect friction
Given data
w2
5
m
m
⋅
=
h5
0
m
m
⋅
=
Q
0.025
m
3
s
⋅
=
ρ
750
kg
m
3
⋅
=
From continuity
QU
1
2
⋅
h
⋅
w
⋅
=
U
1
Q
2w
⋅
h
⋅
=
U
1
10.0
m
s
=
Also
ρ
−
U
1
⋅
A
1
⋅
A
ρ
u
2
⋅
⌠
⎮
⎮
⌡
d
+
0
=
U
1
2
⋅
h
⋅
w
⋅
w
h
−
h
y
u
max
1
y
2
h
2
−
⎛
⎜
⎝
⎞
⎠
⋅
⌠
⎮
⎮
⎮
⌡
d
⋅
=
wu
max
⋅
hh
−
()
−
[]
h
3
h
3
−
⎛
⎝
⎞
⎠
−
⎡
⎣
⎤
⎦
−
⎡
⎣
⎤
⎦
⋅
=
max
⋅
4
3
⋅
h
⋅
=
Hence
u
max
3
2
U
1
⋅
=
u
max
15
m
s
=
For x momentum
p
1
A
⋅
p
2
A
⋅
−
V
1
ρ
1
−
V
1
⋅
A
⋅
⋅
A
2
ρ
2
u
2
⋅
u
2
⋅
⌠
⎮
⎮
⌡
d
+
=
Note that there is no R
x
(no friction)
p
1
p
2
−
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics, Flux

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