Chapter 13
Momentum Analysis of Flow Systems
Angular Momentum Equation
1345C
The angular momentum equation is obtained by replacing
B
in the Reynolds transport theorem by
the total angular momentum
sys
H
, and
b
by the angular momentum per unit mass
V
r
.
1346C
The angular momentum equation in this case is expressed as
V
m
r
I
where
is the
angular acceleration of the control volume, and
r
is the position vector from the axis of rotation to any
point on the line of action of
F
.
1347C
The angular momentum equation in this case is expressed as
V
m
r
I
where
is the
angular acceleration of the control volume, and
r
is the position vector from the axis of rotation to any
point on the line of action of
F
.
13

28
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View Full DocumentChapter 13
Momentum Analysis of Flow Systems
1348
Water is pumped through a piping section. The moment acting on the elbow for the cases of
downward and upward discharge is to be determined.
Assumptions
1
The flow is steady and uniform.
2
The water is discharged to the atmosphere, and thus the
gage pressure at the outlet is zero.
3
Effects of water falling down during upward discharge is disregarded.
Properties
We take the density of water to be 1000 kg/m
3
.
Analysis
We take the entire pipe as the control volume, and designate the inlet by 1 and the outlet by 2. We
also take the
x
and
y
coordinates as shown. The control volume and the reference frame are fixed.
The conservation of mass equation for this oneinlet oneoutlet steady flow system is
m
m
m
2
1
, and
V
V
V
2
1
since
A
c
= constant. The mass flow rate and the weight of the horizontal
section of the pipe are
kg/s
24
.
45
)
m/s
4
](
4
/
m)
12
.
0
(
)[
kg/m
(1000
2
3
V
c
A
m
N/m
3
.
294
m/s
kg
1
N
1
)
m/s
81
.
9
)(
m
2
)(
kg/m
(15
2
2
mg
W
(
a
)
Downward discharge
: To determine the moment acting on the pipe at point
A
, we need to take the
moment of all forces and momentum flows about that point. This is a steady and uniform flow problem,
and all forces and momentum flows are in the same plane. Therefore, the angular momentum equation in
this case can be expressed as
in
out
V
V
m
r
m
r
M
where
r
is the moment arm, all moments in
the counterclockwise direction are positive, and all in the clockwise direction are negative.
The free body diagram of the pipe section is given in the figure. Noting that the moments of all
forces and momentum flows passing through point
A
are zero, the only force that will yield a moment about
point
A
is the weight
W
of the horizontal pipe section, and the only momentum flow that will yield a
moment is the exit stream (both are negative since both moments are in the clockwise direction). Then the
angular momentum equation about point
A
becomes
2
2
1
V
m
r
W
r
M
A
Solving for
M
A
and substituting,
m
N
70.0
2
2
2
1
m/s
kg
1
N
1
m/s)
kg/s)(4
m)(45.54
(2

N)
m)(294.3
1
(
V
m
r
W
r
M
A
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 Spring '08
 Chung
 Fluid Dynamics, Angular Momentum, Physical quantities, Mass flow rate, Flow Systems

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