Momentum Principle
Chapter Eight
I
n mechanics, the momentum of particle or object is
defined as the product of its mass m and its velocity V,
Momentum  mass  velocity
Momentum  m.V
The particles of a fluid stream will possess, and, whenever the velocity
of the stream is changed in magnitude or direction, there will be a
corresponding change in the momentum of the fluid particles. In
accordance with Newton’s second law, a force is required to produce
this change, which will be proportional to the rate at which the change
of momentum occurs. The force may be provided by contact between
the fluid and a solid boundary or by one part of the fluid stream acting
on another. By Newton’s third law, the fluid will exert an equal and
opposite force on the solid boundary or body of fluid producing the
change of velocity. Such forces are known as dynamic forces, since they
arise from the motion of the fluid and are additional to the static forces
due to pressure in a fluid, which occur even when the fluid is at rest.
The rate of change of momentum 
Mass of fluid per unit time flowing  change of velocity
The rate of change of momentum  .Q.V
(8.1)
Where  is the density of the fluid and Q is the volumetric rate of flow
called discharge. Note that this is the increase of momentum per unit
time in the direction of motion, and according to Newton’s second law
will be caused by a force F, such that:
F  .Q.V
(8.2)
Chapter
Force in any direction  Rate of change of momentum in that direction
8
Chapter Eight
Momentum Principle
This is the resultant force acting on the fluid in the direction of motion.
By Newton’s third law, the fluid will exert an equal and opposite
reaction on it’s surrounding.
The momentum Equation (8.2) was derived for one-dimensional flow in
a straight line, assuming that the incoming and outgoing velocities V1
and V2 were in the same direction.
8.1 Momentum Equation for Two-Dimensional Flow along a Streamline:
Figure (8.1) shows a control volume of fluid, in which V1 makes an
angle  1 with the x-axis while V2 makes a corresponding angle  2.
Vy2
V2

Vx2
P2
2
Control Volume
Rx
y
1
V y1
V1
x

Ry
W
P1
Vx1
Figure 8.1: Momentum equation for two-dimensional flow.
Since both momentum and force are vector quantities, they can resolved
into components in the x and y directions and Equation (8.2) is applied.
Thus if Fx and Fy are the components of the resultant force on the
element of the fluid,
Fx  Rate of change of momentum of fluid in x-direction
Fx
 Mass of fluid  change of velocity
8
Chapter
per unit time
in x-direction
Fx   .Q(V2 . cos 2  V1 . cos 1 )
Fx  Q(V x 2  V x1 )   .Q.V x
165
(8.3)
Momentum Principle
Chapter Eight
Where,
Fx   components of external forces in x-direction
Similarly,
Fy  .Q(V2 . sin 2  V1 . sin1 )
Fy  .Q(V y 2  V y1 )  .Q.V y
(8.4)
Where,
Fy   components of external forces in y-direction
These components can be combined to give the resultant force,
F  Fx2  Fy2
(8.5)
Again, the force exerted by the fluid on the surrounding will be equal
and opposite.
To summarize the above mentioned relations, we can say, in general,
that
Total force exerted on
Rate of change of momentum
the fluid in a control  in the given direction of the fluid
volumein a given direction passing through the control volume
F   .Q(Vout  Vin )
(8.6)
The value of F is positive in the direction in which V is assumed to be
positive.
8.2 Applications of the Momentum Equation
The momentum equation may be used directly to evaluate the force
causing a change of momentum in a fluid. Such applications include
Chapter
The external forces include gravity forces (weight of the control volume
of fluid, which is generally very small and hence neglected), shear
forces, pressure forces, and forces exerted by the solid boundary.
8
Chapter Eight
Momentum Principle
determining forces on pipe bends and junctions, nozzles and hydraulic
machines.
In addition the momentum equation is used to solve problems in which
energy losses occur that cannot be evaluated directly or when the flow is
unsteady. Examples of such problems include local head losses in pipes,
the hydraulic jump and unsteady flow in pipes and channels.
8.2.1 Force Exerted by a Jet Striking a Flat Plate
When a jet of fluid strikes a stationary flat plate at angle  as in Figure
(8.2) it does not rebound, but flows out over the plate in all directions.
In a direction normal to the surface of the plate, the velocity of the
stream will be reduced to zero and the momentum normal to the plate
destroyed. There will, therefore, be a force exerted between the jet and
the plate equal to the rate of change of momentum normal to the plate is
acting on the plate in the direction of motion, with an equal and opposite
reaction by the plate on the jet.
Control surface
u in unit time
A


u
V
V
F
x-direction
x-direction
(a)
(b)
Figure 8.2: Force exerted on a flat plate
8
Chapter
In a direction parallel to the plate, the force exerted will depend on the
shear stress between the fluid and the surface of the plate. For an ideal
fluid, there would be no shear stress and no force parallel to the plate.
The fluid would flow out over the plates so that the total momentum per
second parallel to the plate remained unchanged.
167
Momentum Principle
Chapter Eight
 
A jet of water with a diameter of 60 mm and a velocity of 5 m/s hits a
vertical plate shown in figure. Calculate the force on impact on the jet
on the plate.
Vertical Plate
V 1=5m/s
F
Water Jet 60 mm dia.
V 2=0

Force of the plate on the water in x-direction,
 F  .Q(V2  V1 )
Thus
The discharge Q,
F  .Q(V1  V2 )
(1)
Q  V .A
 (  (0.06) 2 
  0.014m 3 / s
Q  5  
4


But the plate is stationary, i.e., V2=0, so putting the known values into
Eq. (1).
The force of the water jet on the plate is equal to the force of the plate
on the water. They are the same magnitude but in opposite directions.
Chapter
F  1000 0.014  (5  0)  70N
8
Chapter Eight
Momentum Principle
 
A jet of water from a fixed nozzle has a diameter 25 mm strikes a flat
plate at an angle 30º to the normal to the plate. The velocity of the jet is
5 m/s, and the surface of the plate can be assumed to be frictionless.
Calculate the force exerted normal to the plate
a. If the plate is stationary,
b. If the plate is moving with a velocity 2 m/s in the same direction as
the jet.
Control surface
u in unit time
A


u
V
V
F
x-direction
x-direction
(a)
(b)

Force exerted by plate on the fluid in x-direction
 F    Q(Vout  Vin )
Thus
F   .Q(Vin  Vout )
a. If the plate is stationary:
The velocities and discharge in Eq. (1) are next determined,
Chapter
8
Vout  0
Vin  V cos  5 cos30  4.33m / s
169
(1)
Momentum Principle
Chapter Eight
Q  A.V 
  (0.025) 2
4
 5  2.45 103 m3 / s
Now we can solve Eq. (1) for F as follows,
F  1000 (2.45  103 )  4.33  10.63N
i.e., Force exerted by plate on the fluid in x-direction F=10.63N
The force of the jet on the plate is equal to the force of the plate on the
fluid, they are same magnitude but in opposite directions.
b. If the plate moves in same direction as the jet with velocity u:
The velocities and discharge in Eq. (1) are next determined,
Vout  0
Vin  (V  u ) cos  (5  2) cos30  2.6m / s
Q  A(V  u ) 

4
 (0.025) 2  (5  2)  1.47m3 / s
Now we can solve Eq. (1) for F as follows,
F  10001.47  (2.6  0)  3.83N
i.e., Force exerted by plate on the fluid in x direction, F=3.83N
Then the force exerted by the fluid in x-direction will be the reaction to
the force of the plate on the fluid, they are same magnitude but in
opposite directions.
Both velocity and momentum are vector quantities and, therefore, even
if the magnitude of the velocity remains unchanged, a change in
direction of a stream of fluid will give rise to change of momentum. If
the stream is deflected by a curved vane as shown in Figure (8.3),
entering and leaving tangentially without impact, a force will be exerted
between the fluid and the surface of the vane to cause the change of
momentum.
Chapter
8.2.2 Force Due to the Deflection of a Jet by a Fixed Curved Vane:
8
Chapter Eight
Momentum Principle
It is usually convenient to calculate the components of this force parallel
and perpendicular to the direction of the incoming stream by calculating
the rate of change of momentum in these two directions. The
components can then be combined to give the magnitude and direction
of the resultant force, which the vane exerts on the fluid, and the equal
and opposite reaction of the fluid on the vane.
Fy

V1
Fx
y
x
Control surface
V2
Figure 8.3: Force exerted on a curved vane.
 
A jet of water from nozzle is deflected through an angle 60º from its
original direction by a curved vane, which it enters tangentially without
shock with a mean velocity 30 m/s and leaves with a mean velocity 25
m/s. If the discharge from the nozzle is 0.8 kg/s. Calculate the
magnitude and direction of the resultant force on the vane if the vane is
stationary.
y


V 1 = V1x
x
V2x
V2y
Chapter
8


171
V2
Momentum Principle
Chapter Eight

If we draw a control surface around the curved vane, we have
1
Fy
V1= V1x
Fx
Control Surface
2
V2x
V2y
V2
Force exerted on the jet by the van in the x-direction
Fx   .Q (V 2 x  V1x )
(1)
The velocities and the mass of fluid per unit time, .Q, in the forgoing
equation are next determined
V1x  V1  30m / s
V2 x  V2 cos  25 cos 60  12.5m / s
.Q  Mass of fluid per unit time = 0.8 kg/s
Now we can solve Eq. (1) for Fx as follows,
Fx  0.8  (12.5  30)  14.0 N
Force exerted on the jet by the vane in the y-direction,
Fy  .Q(V2 y  V1 y )
(2)
V1 y  0
V2 y  V2 sin  25 sin 60  21.65 m / s
Chapter
The velocities in the Eq. (2) are next determined,
8
Chapter Eight
Momentum Principle
Now we can solve Eq. (2) for Fy as follows:
Fy  .Q V2 . sin 
Fy  0.8   21.65  17.32N
Then the force on the vane will be the reactions to the force of the vane
on the jet, or
R x   Fx  14.0 N
R y   Fy  17.32N
The resultant force, R, exerted by jet on the vane,
R  (14) 2  (17.32) 2  22.27 N
and it will be inclined to the x-direction at an angle
  tan 1 (
Ry
17.32
)  tan 1 (
)  51
Rx
14
22.27 N
O
51
8.2.3 Force Exerted When a Jet is Deflected by a Moving Curved
Vane
8
Chapter
If a jet of fluid is to be deflected by a moving curved vane without
impact at the inlet to the vane, the relation between the direction of the
jet and the tangent to the curve of the vane at inlet must be such that the
relative velocity of the fluid at inlet is tangential to the vane. The force
in the direction of motion of the vane will be equal to the rate of change
of momentum of the fluid in the direction of motion, i.e. the mass
deflected per second multiplied by the change of velocity in that
direction. The force at right angles to the direction of motion will be
equal to the mass deflected per second times the change of velocity at
right angles to the direction of motion.
 
173
Momentum Principle
Chapter Eight
A 3" diameter jet with mean velocity equals to 100 ft/s impinges a
single vane moving at a velocity of 60 ft/s. Compute the force exerted
by the water on the vane.
y
V1=100 ft/s
x


u=60 ft/s
V 2=100 ft/s

If we draw a control surface around the curved vane, we have
Fy
1
V1x = V 1
Fx
60 ft/s
V2
2
V2y
V 2x
Force exerted on the jet by the vane in the x-direction,
Fx   .Q(Vout  Vin )
(1)
Vin  V1x  100 ft / s and Vout  u  Vr
Vr  100  60  40 ft / s
Chapter
The velocities and discharge in Eq. (1) are next determined,
8
Chapter Eight
Momentum Principle
Vout  60  40 cos30  25.36 ft / s
Q  Vr . Anozzle  40 

3
 ( ) 2  1.96 ft 3 / s
4 12
   / g  62.4 / 32.2  1.94slug / ft 3
Now we can solve Eq. (1) for Fx as follows,
Fx  1.94  1.96(25.36  100)  283.82lb f
Force exerted on the jet by the vane in the y-direction,
Fy  .Q(Vout  Vin )
(2)
The velocities and discharge in Eq. (2) are next determined,
Vin  0
And
Vout  Vr sin 30o  40 sin 30  20 ft / s
Now we can solve Eq. (2) for Fy as follows:
Fy  1.94  1.96(20  0)  76 lb f
Then the force on the vane will be the reactions to the force of the vane
on the jet, or
R x   Fx  283.82 lb f
R y   Fy  76.0 lb f
The resultant force, R, exerted by jet on the vane,
Chapter
8
R  (283.82) 2  (76) 2  293.82 lb f
And it will be inclined to the x-direction at an angle,
175
  tan 1 (
Chapter Eight
Ry
Rx
)  tan 1 (
76
)  14.5
283.82
293.82 lbf
O
14.5
Chapter
Momentum Principle
8

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