Chapter 5
5.4d Conservation of Angular Momentum
V
r
m
O
dr
R

R
O
Figure 5.4-8 A point mass m and a distributed mass M
are rotating at a uniform angular velocity .
The linear momentum of a mass m moving in the x direction with a velocity Vx is mVx. The
angular momentum (L) of a point mass m rotating with an angular velocity  rad/s in an arc
having a radius of curvature R is mVR = mR2. The angular momentum has dimension of
“length times momentum,” and is therefore called the “moment of momentum.” If the mass
is not a point but a rigid distributed mass (M) rotating at a uniform angular velocity, the
angular momentum of the differential mass dM is given by
dL = r2dM
The total angular momentum of the mass M is obtained by integration over the entire mass
L =
 r
2
dM =   r 2 dM = I
M
(5.4-5)
M
where the moment of inertia I is defined as I =
 r dM
2
M
As an example, we want to determine the moment of inertia for the flywheel of width W,
radius R, and mass M, whose cross section is shown in Figure 5.4-8.
dM = 2rdrW
I=
 r dM
2
M
=
r
M
2
2rdrW = 2W 
R
0
R2
R2
r dr = WR 2 = M 2
3
2
For a fixed mass, the conservation of linear momentum is equivalent to Newton’s second
law:
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


d (mV )

dV
 F = m a = m dt = dt
Let  be the torque acting on the system, the conservation of angular momentum is written
as
 
=
 F R =
d ( I )
d
=I
dt
dt
For a flow system, angular momentum may enter or leave the system by convection. As an
example, consider the impeller of a centrifugal pump, whose cross section is shown in Figure
5.4-9.

r2
r1
V1
V2
Figure 5.4-9 Cross section of pump impeller.
The impeller rotates with angular velocity , and its rotation causes the fluid to be thrown
radially outwards between the vanes by centrifugal action. The fluid enters at a radial
position r1 and leaves at a radial position r2; the corresponding tangential velocity V1 and V2
denote the inlet and exit liquid velocities relative to a stationary observer. Making an angular
momentum balance on the impeller yields
d ( I )
 r1V1)in  ( m
 r2V2)out + 
= (m
dt
For steady-state system, there is no accumulation of angular momentum, and the torque
required to rotate the impeller is
 (r22  r12)
 = m
The corresponding power required to drive the pump is the product of the angular velocity
and the torque
Power = 
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Example 5.4-53 ---------------------------------------------------------------------------------Consider an ideal garden sprinkler shown in Figure 5.4-10. The central bearing is well
lubricated, so the arms are free to rotate about the central pivot. Each of the two nozzles has a
cross-sectional area of 5 mm2, and each arm is 20 cm long.
2
0
c
m
2
0
c
m
u

Figure 5.4-10 An ideal garden sprinkler.
u
If the water supply rate to the sprinkler is 0.0001 m3/s, determine:
(a) The velocity u(m/s) of the water jets relative to the nozzles.
(b) The angular velocity of rotation, , of the arms, in both rad/s and rps.
(c) The applied torque required preventing the arms from rotating.
Solution -----------------------------------------------------------------------------------------(a) The velocity u(m/s) of the water jets relative to the nozzles
Q
10 4 m 3 / s
u=
=
= 10 m/s
2A
2  5  10 6 m 2
(b) The angular velocity of rotation, , of the arms, in both rad/s and rps.
Making an angular momentum balance on the steady state sprinkler yields
 r1V1)in  ( m
 r2V2)out + 
0 = (m
 [(r2V2)out  (r1V1)in]
 = m
Water enters at a radial position r1 and leaves at a radial position r2; the corresponding
tangential velocity V1 and V2 denote the inlet and exit water velocities relative to a stationary
observer. We assume that (r2V2)out >> (r1V1)in, therefore
 (r2V2)out
  m
Since the arm is free to rotate, there is no applied torque or  = 0 and V2 = 0. The water
discharge velocity is zero as seen by a stationary observer. Let unoz be the velocity of the
nozzle, then
V2 = u  unoz = 0
 unoz = u = 10 m/s
The angular velocity is then
3
Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 103
5-33
 =
unoz 10
=
= 50 rad/s
0 .2
r2
 =
50
= 7.96 rps
2
(c) The applied torque required preventing the arms from rotating.
If the arm is not rotating then unoz = 0
V2 = u  unoz = 10  0 = 10 m/s
 (r2V2)out = (0.0001 m3/s)(1000 kg/m3)(0.2 m)(10 m/s) = 0.2 Nm
  m
5.4e Momentum Balance on Moving Systems
We will consider a system that is moving with constant velocity while streams carrying
momentum and energy may flow into and out of the system. The absolute stream velocity in
the x direction Vx is related to the system velocity Vsx and the stream velocity relative to the
system Vrx by the relation
Vx = Vsx + Vrx
The analysis for a system moving with a constant velocity will be simplified if a momentum
balance is applied to a control volume moving with the system velocity.
Example 5.4-6 ---------------------------------------------------------------------------------Figure 5.4-11 shows a plan of a jet of water impinging against a cone that is held stationary
by a force F opposing the jet, which divides into several radially outwards streams, each
leaving at an angle of 30 degree with respect to the horizontal. The velocity of the water jet is
18 m/s and its diameter is 8.0 cm.
V2
Water
V1
60o
F
V2
Figure 5.4-11 Jet impinging against a cone.
Determine the force need to:
(a) Hold the cone stationary.
(b) Move the cone away from the jet at 5 m/s.
Solution -----------------------------------------------------------------------------------------5-34
(a) Force required holding the cone stationary
Applying the x-momentum balance on the control volume shown with the dash line
F + A1V1V1x  A2V2V2x = 0
We will assume that the area of flow at the inlet is the same as that at the outlet: A1 = A2,
therefore V1 = V2. For steady state system
m = A1V1 = A2V2
Therefore
F = A1V1(V1x  V2x) = A1V1(V1  V1cos 30o) = A1V12(1  cos 30o)
F = (1000)(0.042)(182)(1  cos /6) = 218.2 N
(b) Force required moving the cone away from the jet at 5 m/s.
We now choose the control volume that moves with the cone so that the velocity of the water
jet relative to the control volume is
V1r = V1  Vs = 18  5 = 13 m/s
The x-momentum balance is now written with the relative velocities
F + A1V1rV1xr  A2V2rV2xr = 0
 = A1V1r = A2V2r
Since A1 = A2, therefore V1r = V2r and m
The force required for this case is then
F = A1V1r(V1xr  V2xr) = A1V1r(V1r  V1rcos 30o) = A1V1r2(1  cos /6)
F = (1000)(0.042)(132)(1  cos /6) = 113.8 N
Example 5.4-74 ---------------------------------------------------------------------------------As shown in Figure 5.4-12, a boat of mass M = 1,000 lbm is propelled on a lake by a pump
that takes in water and ejects it, at a constant velocity of V2r = 30 ft/s relative to the boat,
through a pipe of cross-sectional area A = 0.2 ft2. The resistance force F of the water is
proportional to the square of the boat velocity Vs, which has a maximum value of 20 ft/s.
What is the acceleration of the boat when its velocity is Vs = 10 ft/s?
4
Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 98
5-35
+
V2r
V2r
Vs
M
M
V1r
F
F
Figure 5.4-12 Jet-propelled boat.
Solution -----------------------------------------------------------------------------------------Choose the control volume C.V. moving with the boat then the velocity of water entering the
C.V. is
V1r =  Vs
The mass flow rate of water entering or leaving the C.V. is
 = A2V2r = (62.4)(0.2)(30) = 374.4 lbm/s
m
The x-momentum balance is now written with the relative velocities
d ( MV s )
dV
 V1r  m
 V2r + kVs2
= M s = Ma = m
dt
dt
At maximum velocity Vs = 20 ft/s, there is no acceleration, therefore
 (V2r  V1r)  k =
kVs2 = m
k=
m
(V2r  V1r)
Vs2
374.4(30  20) 374.4
=
20 2
40
The acceleration at Vs = 10 ft/s is given by
a=
k
m
(V1r  V2r) +
Vs2
M
M
a=
374.4
374.4
(10  30) +
(102) =  6.55 ft/s2
1000
40  1000
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