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Problem 4.95
[Difficulty: 3]
R
x
y
x
CS
V
2
V
1
p
2
p
1
V
3
ρ
1
2
Given:
Data on heated flow of gas
Find:
Force of gas on pipe
Solution:
Basic equation: Continuity, and momentum flux in x direction
p
ρ
R
⋅
T
⋅
=
Assumptions:
1) Steady flow
2) Uniform flow
From continuity
ρ
1
−
V
1
⋅
A
1
⋅
ρ
2
V
2
⋅
A
2
⋅
+
m
3
+
0
=
V
2
V
1
ρ
1
ρ
2
⋅
m
3
ρ
2
A
⋅
−
=
where m
3
= 20 kg/s is the mass leaving
through the walls (the software does not
allow a dot)
V
2
170
m
s
⋅
6
2.75
×
20
kg
s
⋅
m
3
2.75 kg
⋅
×
1
0.15 m
2
⋅
×
−
=
V
2
322
m
s
=
For x momentum
R
x
p
1
A
⋅
+
p
2
A
⋅
−
V
1
ρ
1
−
V
1
⋅
A
⋅
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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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