Problem 4.93
[Difficulty: 3]
R
x
y
x
CS
V
2
V
1
p
2
p
1
Given:
Data on adiabatic flow of air
Find:
Force of air on pipe
Solution:
Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equation
p
ρ
R
⋅
T
⋅
=
Assumptions:
1) Steady flow
2) Ideal gas
CV 3) Uniform flow
From continuity
ρ
1
−
V
1
⋅
A
1
⋅
ρ
2
V
2
⋅
A
2
⋅
+
0
=
ρ
1
V
1
⋅
A
⋅
ρ
2
V
2
⋅
A
⋅
=
ρ
1
V
1
⋅
ρ
2
V
2
⋅
=
For x momentum
R
x
p
1
A
⋅
+
p
2
A
⋅
−
V
1
ρ
1
−
V
1
⋅
A
⋅
()
⋅
V
2
ρ
2
V
2
⋅
A
⋅
⋅
+
=
ρ
1
V
1
⋅
A
⋅
V
2
V
1
−
⋅
=
R
x
p
2
p
1
−
A
⋅
ρ
1
V
1
⋅
A
⋅
V
2
V
1
−
⋅
+
=
For the air
ρ
1
P
1
R
air
T
1
⋅
=
ρ
1
200
101
+
1
0
3
×
N
m
2
⋅
kg K
⋅
286.9 N
⋅
m
⋅
×
1
60
273
+
K
⋅
×
=
ρ
1
3.15
kg
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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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