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Problem 8.86
[2]
Given:
Flow through sudden expansion
Find:
Inlet speed; Volume flow rate
Solution:
Basic equations
p
1
ρ
α
V
1
2
2
⋅
+
gz
1
⋅
+
⎛
⎜
⎝
⎞
⎟
⎠
p
2
ρ
α
V
2
2
2
⋅
+
2
⋅
+
⎛
⎜
⎝
⎞
⎟
⎠
−
h
lm
=
h
lm
K
V
1
2
2
⋅
=
QV
A
⋅
=
Δ
p
ρ
H2O
g
⋅
Δ
h
⋅
=
Assumptions: 1) Steady flow 2) Incompressible flow 3)
α
at 1 and 2 is approximately 1 4) Horizontal
Hence the energy equation becomes
p
1
ρ
V
1
2
2
+
⎛
⎜
⎝
⎞
⎟
⎠
p
2
ρ
V
2
2
2
+
⎛
⎜
⎝
⎞
⎟
⎠
−
K
V
1
2
2
⋅
=
From continuity
V
2
V
1
A
1
A
2
⋅
=
V
1
AR
⋅
=
Hence
p
1
ρ
V
1
2
2
+
⎛
⎜
⎝
⎞
⎟
⎠
p
2
ρ
V
1
2
AR
2
⋅
2
+
⎛
⎜
⎝
⎞
⎟
⎠
−
K
V
1
2
2
⋅
=
Solving for V
1
V
1
2p
2
p
1
−
()
⋅
ρ
1A
R
2
−
K
−
⋅
=
AR
D
1
D
2
⎛
⎜
⎝
⎞
⎟
⎠
2
=
75
225
⎛
⎝
⎞
⎠
2
=
0.111
=
so from Fig. 8.14
K
0.8
=
Also
p
2
p
1
−
ρ
H2O
g
⋅
Δ
h
⋅
=
1000
kg
m
3
⋅
9.81
×
m
s
2
⋅
5
1000
×
m
⋅
Ns
2
⋅
kg m
⋅
×
=
49.1 Pa
⋅
=
Hence
V
1
2
49.1
×
N
m
2
⋅
m
3
1.23 kg
⋅
×
1
1
0.111
2
−
0.8
−
×
kg m
⋅
2
⋅
×
=
V
1
20.6
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This note was uploaded on 05/03/2010 for the course MAE 101b taught by Professor Staff during the Spring '08 term at UCSD.
 Spring '08
 staff

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