Problem 6.16
[Difficulty: 3]
Given:
Flow in a pipe with variable area
Find:
Expression for pressure gradient and pressure; Plot them; exit pressure
Solution:
Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity
Basic equations
QV
A
⋅
=
Given data
ρ
1.75
slug
ft
3
⋅
=
p
i
35 psi
⋅
=
A
i
15 in
2
⋅
=
A
e
2.5 in
2
⋅
=
L1
0
f
t
⋅
=
u
i
5
ft
s
⋅
=
For this 1D flow
Qu
i
A
i
⋅
=
uA
⋅
=
AA
i
A
i
A
e
−
()
L
x
⋅
−
=
so
ux
() u
i
A
i
A
⋅
=
u
i
A
i
A
i
A
i
A
e
−
L
x
⋅
⎡
⎢
⎣
⎤
⎥
⎦
−
⋅
=
a
x
u
x
u
∂
∂
⋅
v
y
u
∂
∂
⋅
+
=
u
i
A
i
A
i
A
i
A
e
−
L
x
⋅
⎡
⎢
⎣
⎤
⎥
⎦
−
⋅
x
u
i
A
i
A
i
A
i
A
e
−
L
x
⋅
⎡
⎢
⎣
⎤
⎥
⎦
−
⋅
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
∂
∂
⋅
=
A
i
2
L
2
⋅
u
i
2
⋅
A
e
A
i
−
⋅
A
i
L
⋅
A
e
x
⋅
+
A
i
x
⋅
−
3
=
For the pressure
x
p
∂
∂
ρ
−
a
x
⋅
ρ
g
x
⋅
−
=
ρ
A
i
2
⋅
L
2
⋅
u
i
2
⋅
A
e
A
i
−
⋅
A
i
L
⋅
A
e
x
⋅
+
A
i
x
⋅
−
3
−
=
and
dp
x
p
∂
∂
dx
⋅
=
pp
i
−
0
x
x
x
p
∂
∂
⌠
⎮
⎮
⌡
d
=
0
x
x
ρ
A
i
2
⋅
L
2
⋅
u
i
2
⋅
A
e
A
i
−
⋅
A
i
L
⋅
A
e
x
⋅
+
A
i
x
⋅
−
3
−
⌠
⎮
⎮
⎮
⎮
⌡
d
=
This is a tricky integral, so instead consider the following:
x
p
∂
∂
ρ
−
a
x
⋅
=
ρ
−
u
⋅
x
u
∂
∂
⋅
=
1
2
−
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 Fall '07
 Lear
 Fluid Mechanics, Trigraph

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