This preview shows page 1. Sign up to view the full content.
Problem 4.60
[Difficulty: 3]
Given:
Data on flow at inlet and outlet of pipe
Find:
Ratio of outlet to inlet momentum flux
Solution:
∫
⋅
=
A
x
dA
V
u
r
ρ
mf
Basic equation: Momentum flux in x direction at a section
Assumptions:
1) Steady flow
2) Incompressible flow
Evaluating at 1 and 2
mf
x1
U
ρ
⋅
U
−
π
⋅
R
2
⋅
()
⋅
=
mf
x1
ρπ
⋅
U
2
⋅
R
2
⋅
=
Hence
mf
x2
0
R
r
ρ
u
2
⋅
2
⋅
π
⋅
r
⋅
⌠
⎮
⌡
d
=
2
ρ
⋅
π
⋅
u
max
2
⋅
0
R
r
r1
r
R
⎛
⎝
⎞
⎠
2
−
⎡
⎢
⎣
⎤
⎥
⎦
2
⋅
⌠
⎮
⎮
⎮
⌡
d
⋅
=
2
ρ
⋅
π
⋅
u
max
2
⋅
0
R
y
r2
r
3
R
2
⋅
−
r
5
R
4
+
⎛
⎜
⎝
⎞
⎠
⌠
⎮
⎮
⎮
⌡
d
⋅
=
mf
x2
2
ρ
⋅
π
⋅
u
max
2
⋅
R
2
2
R
2
2
−
R
2
6
+
⎛
⎜
⎝
⎞
⎠
⋅
=
⋅
u
max
2
⋅
R
2
3
⋅
=
Then the ratio of momentum fluxes is
mf
x2
mf
x1
1
3
ρ
⋅
π
⋅
u
max
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details