Download 04 – The Momentum Equation and its Applications

Momentum Equation and its Applications
Moving fluid exerting force
 The lift force on an aircraft is exerted by the air
moving over the wing.
 A jet of water from a hose exerts a force on
whatever it hits.
 In fluid mechanics the analysis of motion is
performed the same way as in solid mechanics by use of Newton's laws of motion.
 Properties of fluids considered when in motion
F  ma
Used to analyze solid mechanics to
relate applied force to acceleration
In fluid mechanics not clear mass of moving fluid to
use hence the use momentum equation.
 Statement of Newton's Second Law relates the sum
of the forces acting on an element of fluid to its
acceleration or rate of change of momentum.
The Rate of change of momentum of a body is equal
to the resultant force acting on the body, and
takes place in the direction of the force
Consider the streamtube:
Assume flow to be steady and
non-uniform
1
A1
u1
1
2
A2
u2
2
u1t
In time δt a volume of the fluid moves from the inlet a
distance uδt,
Volume entering the stream tube in time δt
 Area  distance  A1u1t
Mass entering stream tube in time δt
 volume  density  1 A1u1t
Momentum of fluid entering stream tube in time δt
 mass  velocity  1 A1u1t u1
Similarly, at the exit, the momentum leaving the
steam tube is
2 A2u2t u2
Calculate force on the fluid
using Newton's 2nd Law
Force = rate of change of momentum
(  2 A2 u2t u2  1 A1u1t u1 )
F
t
 From continuity
Q  A1u1  A2 u2
 Assume fluid of constant density
1  2  
Hence
F  Q (u2  u1 )
or
F  M (u2  u1 )
 Force acts in direction of flow of the fluid.
 Analysis assumed inlet and outlet velocities were in the
same direction - one dimensional system.
Consider two dimensional
u2
2
1
u1
• At inlet the velocity vector, u1, makes an angle,
θ1, with the x-axis.
• At the outlet velocity, u2 makes an angle θ2.
Resolve forces in the directions of the
co-ordinate axes.
The force in the x-direction
 Fx = Rate of change of momentum in x-direction
= Rate of change of mass  change in
velocity in x-direction
 M (u2 cos 2  u1 cos1 )
 M ( u2 x  u1 x )
 Q(u2 cos 2  u1 cos1 )  Q( u2 x  u1 x )
The force in the y-direction
 Fy = Rate of change of momentum in y-direction
= Rate of change of mass  change in
velocity in y-direction
 M (u2 sin 2  u1 sin1 )
 M ( u2 y  u1 y )
 Q(u2 sin 2  u1 sin1 )
 Q( u2 y  u1 y )
The resultant force on the fluid
Fresultant  F  F
2
x
2
y
And the angle which this
force acts at is given by
  tan (
1
Fy
Fx
)
 For three-dimensional (x, y, z) system an extra
force in the z-direction must be calculated.
 This is considered in exactly the same way.
Generally the total force on fluid = rate of change
of momentum through the control volume
F  Q( uout  uin )
 This force is made up of three components:
FR  Force exerted on the fluid by any solid
body touching the control volume
FB  Force exerted on the fluid body (e.g. gravity)
FP 
Force exerted on the fluid by fluid
pressure outside the control volume
Total force, FT, given as the sum of these forces
FT  FR  FB  FP
Force exerted by the fluid on the solid body
touching the control volume is equal and opposite
to FR
So the reaction force, R, is given by
R   FR
Application of the Momentum Equation
Force due to fluid flow round a pipe bend
Force on a nozzle at the outlet of a pipe.
Impact of a jet on a plane surface.
The force due to fluid flow round a pipe bend
Consider a pipe bend with a constant cross section
Lying in the horizontal plane
Turning through an angle of θ°.
Step in Analysis:
1. Draw a control volume
2. Decide on co-ordinate axis system
3. Calculate the total force
4. Calculate the pressure force
5. Calculate the body force
6. Calculate the resultant force
7. Find force on the fluid
Control volume
• The control volume is drawn with faces at the
inlet and outlet of the bend and encompassing
the pipe walls.
Co-ordinate axis system
Co-ordinate axis chosen such that one axis is
pointing in the direction of the inlet velocity.
Calculate the total force
In the x-direction:
F Tx  Q( u2 x  u1 x )
u1 x  u1
In the y-direction:
F Ty  Q( u2 y  u1 y )
u1 y  u1 sin 0  0
u2 x  u2 cos 
u2 y  u2 sin
F Tx  Q(u2 cos   u1 )
F Ty  Qu2 sin
Calculate the pressure force
Fp 
Pressure force at 1 – pressure at 2
F px  p1 A1 cos 0  p2 A2 cos   p1 A1  p2 A2 cos 
F py  p1 A1 sin 0  p2 A2 sin   p2 A2 sin
Calculate the body force
 There are no body forces in the x or y directions.

The only body force is that exerted by gravity a direction we do not need to consider.
Calculate the resultant force
FT x  FRx  FB x  FPx
FT y  FR y  FB y  FP y
FR x  FTx  F p x  FB x
 Q( u2 cos   u1 )  ( p1 A1  p1 A1 cos  )  0
FR y  FT y  F p y  FB y
 Qu2 sin   (  p2 A2 sin  )  0
Resultant force on fluid is given as
Direction of application
Force on bend is the same in
magnitude as force on fluid,
but in opposite direction
Example
• A 45o reducing bend is connected in a pipe line in
on a horizontal floor. The diameters at the inlet
and outlet of the bend are 600 and 300 mm
respectively. Find the force exerted by water on
the bend if the intensity of pressure at the inlet
to bend is 8.829 N/cm2 and rate of flow of water
is 600 lit/sec.
Force on a pipe nozzle
Step in Analysis:
1. Draw a control volume
2. Decide on co-ordinate axis system
3. Calculate the total force
4. Calculate the pressure force
5. Calculate the body force
6. Calculate the resultant force
7. Find the force to resist
Control volume and Co-ordinate axis
1
2
A1
A2
u2
u1
1
One dimensional system
2
Calculate the total force
FT  FTx  Q( u2  u1 )
Continuity
Q  A1u1  A2 u2
 1
1
FTx  Q 
 
 A2 A1 
2
Calculate the body force
The only body force is the weight due to gravity in the
y-direction - but we need not consider this as the only
forces we are considering are in the x-direction.
Calculate the pressure force
Apply Bernoulli equation to calculate the pressure
2
1
2
2
p1 u
p2 u

 z1 

 z2  h f
g 2 g
g 2 g
Frictional losses neglected, hf = 0
Nozzle is horizontal, z1 = z2
Q  1
1 
p1 
 2
2
2  A2 A1 
Pressure outside is atmospheric, p2 = 0
2
Calculate the resultant force
FTx  FRx  F p x  FB x
FRx  FTx  F p x  0
FRx
 1
1  Q
 Q 
  
2
 A2 A1 
2
Force to be resisted
2
 1
1 
 2  2 
 A2 A1 
R   FRx
Example
• A nozzle of diameter 20 mm is fitted to a pipe
of diameter 60 mm. Find the force exerted by
the nozzle on the water which is flowing
through the pipe at the rate of 1.2 m3/min