1. The momentum equation relates the total force on a fluid system to the rate of change of momentum as fluid flows through a control volume. 2. Forces can be resolved into components in different directions for multi-dimensional flows. The total force is equal to the sum of pressure, body, and reaction forces. 3. Examples of applying the momentum equation include calculating forces on a pipe bend, nozzle, jet impact, and curved vane due to changing fluid momentum. Setting up coordinate systems aligned with the flow is important for resolving forces into components.

1. Free and forced vortex, Forces on
pressure conduits, reducers and bends,
stationary and moving blades, torques in
rotating machines.
Dr. Mohsin Siddique
Assistant Professor
NU-FAST Lahore
1
Fluid Mechanics

2. Free and Forced Vortex Flow
2
Vortex flow is defined as flow along curved path.
It is of two types namely; (1). Free vortex flow and (2) forced
vortex flow
If the fluid particles are moving around a curved path with the help
of some external torque the flow is called forced vortex flow.And if
no external force is acquired to rotate the fluid particle, the flow is
called free vortex flow.

3. Forced Vortex Flow (Rotational Flow)
3
It is defined as that type of flow, in which some external torque is
required to rotate the fluid mass.
The fluid mass in this type of flow rotate at constant angular
velocity, ω.The tangential velocity, V, of any fluid particle is given by
V= ω r,
Where, r is radius of fluid particle from the axis of rotation
Examples of forced vortex flow are;
1. A vertical cylinder containing liquid which is rotated about its central axis
with a constant angular velocity ω,
2. Flow of liquid inside impeller of a centrifugal pump
3. Flow of water through runner

4. Forced Vortex Flow (Rotational Flow)
4

5. Free Vortex Flow (Irrotational flow)
5
When no external torque is required to rotate the fluid mass, that
type of flow is called free vortex flow.
Thus the liquid in case of free vortex flow is rotating due to the
rotation which is imparted to the fluid previously.
Example of free vortex flow are
1. Flow of liquid through a hole provided at the bottom of container
2. Flow of liquid around a circular bend in pipe
3. A whirlpool in river
4. Flow of fluid in a centrifugal pump casing

6. Free Vortex Flow (Irrotational flow)
6
The relation between velocity and radius, in free vortex flow is
obtained by putting the value of external torque equal to zero, or
the time rate of change of angular momentum, i.e., moment of
momentum must be zero. Consider a fluid particle of mass “m” at a
radial distance, r, from the axis of rotation, having a tangential
velocity,V, then
Angular momentum=(mass)x(velocity)=mV
Moment of momentum=(momentum)xr=mVr
Rate of change of angular momentum=d(mVr)/dt
For free vortex flow, there is not torque i.e.,
d(mVr)/dt=0
Integrating, we get
mVr=constant orVxr=C1/m=C
Vxr=C

7. Equation of motion for vortex flow
7
Consider a fluid element ABCD
(shown shaded) in figure rotating at
uniform velocity in a horizontal
plane about an axis perpendicular to
the plane of paper and passing
through O.
The forces acting on element are;
(1). Pressure force p∆A on face AB
(II) on face CD
(iii) centrifugal force, mV2/r, acting in
direction away from the center, O
Ar
r
p
p ∆





∆
∂
∂
+

8. Equation of motion for vortex flow
8
Now,
Mass of element=mass density xVolume
Centrifugal force=
Equating the forces in radial directions we
get
Arm ∆∆= ρ
r
V
Ar
2
∆∆ρ
r
V
ArApAr
r
p
p
2
∆∆=∆−∆





∆
∂
∂
+ ρ
r
V
ArAr
r
p 2
∆∆=∆





∆
∂
∂
ρ
r
V
r
p 2
ρ=





∂
∂
Equation gives the pressure variation along the radical direction for a
forced or free vortex flow in horizontal plane

9. Equation of motion for vortex flow
9
The pressure variation in the vertical plane is given by the
hydrostatic law, i.e.,
In above equation, z is measure vertically in the upward direction.
The pressure ,p, varies with respect to r and z or p is the function of
r and z and hence total derivative of p is
Substituting values from above equations we get;
g
z
p
ρ−=





∂
∂
dz
z
p
dr
r
p
dp
∂
∂
+
∂
∂
=
gdzdr
r
V
dp ρρ −=
2
Equation of Motion forVortex Flow

10. Equation of forced vortex flow
10
For forced vortex flow, we have;
Where ω is angular velocity=constt
Substituting the values ofV in equation of motion
of vortex flow
Consider two points 1 and 2 in the fluid having
forced vortex and integrating above equation for
point 1 and point 2, we get
rV ×= ω
gdzdr
r
r
p ρ
ω
ρ −=∂
22
∫∫∫ −=
2
1
2
1
2
2
1
gdzdrrpd ρρω
[ ] [ ]12
2
1
2
2
2
12
2
zzgrrpp −−−=− ρ
ρω
60
2 Nπ
ω =

11. Equation of forced vortex flow
11
If the point 1 and 2 lie on the free surface then,
p1=p2=Patm=0 and hence above equation become;
[ ] [ ]
1122
12
2
1
22
2
2
12
&
2
rVrV
zzgrrpp
ωω
ρωω
ρ
==
−−−=−
Q
[ ] [ ]12
2
1
2
212
2
zzgVVpp −−−=− ρ
ρ
[ ]
[ ] [ ]2
1
2
212
2
1
2
2
2
1
2
0
VV
g
zz
gVV
−=−
−−= ρ
ρ

12. Equation of forced vortex flow
12
If the point 1 lie on axis of rotation then, v1= ω r1=
ω x0=0 and hence above equation becomes;
Thus, Z varies with square of r. Hence, equation is
an equation of parabola.This means the free surface
is paraboloid
[ ] [ ]
[ ] [ ]2
2
22
2
2
212
2
1
2
1
2
1
r
g
V
g
Z
V
g
zz
ω==
=−
[ ] [ ]0212 −=−= zzzZQ

13. Equation of Free Vortex Flow
13
For free vortex flow, we have;
Substituting v for free vortex flow in equation
of motion of vortex flow
Consider two points 1 and 2 at radial distance
r1 and r2 from central axis.The height of
points from the bottom of vessel is z1 and z2.
Integrating above equation for the points 1
and 2 we get
rCV
CconttVr
/=
==
gdzdr
r
C
gdzdr
rr
C
gdzdr
r
V
dp ρρρρρρ −=−=−= 3
2
2
22
∫∫∫ −=
2
1
2
1
3
22
1
gdzdr
r
C
dp ρρ

14. Equation of Free Vortex Flow
14
[ ]122
1
2
2
2
12
2
1
2
1
32
2
1
2
1
3
22
1
11
2
zzg
rr
C
pp
gdzdrrcgdzdr
r
C
dp
−−





−−=−
−=−= ∫∫∫∫∫
−
ρ
ρ
ρρρρ
[ ]
[ ] [ ]12
2
1
2
212
122
1
2
2
2
2
12
2
2
zzgVVpp
zzg
r
C
r
C
pp
−−−−=−
−−





−−=−
ρ
ρ
ρ
ρ
[ ] [ ] [ ]
12
2
1
2
212
12
2
1
2
2122
1
2
2
2
2
12
22
22
zz
g
V
g
V
g
p
g
p
zzVV
g
zz
g
g
r
C
r
C
gg
pp
+−+−=−
−−−−=−−





−−=
−
ρρ
ρ
ρ
ρ
ρ
ρ
g
Vp
z
g
Vp
z
22
2
22
2
2
11
1 ++=++
γγ

15. 15

16. Numerical: Forced vortex flow
16

17. Numerical: Forced Vortex flow
17

18. Numerical: Free Vortex Flow
18

19. Numerical: Free Vortex Flow
19

20. 20
PART II

21. Momentum and Forces in Fluid Flow
21
We have all seen moving fluids exerting forces.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 in the same way as in
solid mechanics - by use of Newton’s laws of motion.
i.e., F = ma which is used in the analysis of solid mechanics to relate applied
force to acceleration.
In fluid mechanics it is not clear what mass of moving fluid we should use
so we use a different form of the equation.
( )
dt
md
ma sV
F ==∑

22. Momentum and Forces in Fluid Flow
22
Newton’s 2nd Law can be written:
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.
The symbols F and V represent vectors and so the change in momentum must be
in the same direction as force.
It is also termed as impulse momentum principle
( )
dt
md sV
F =∑
=
=∑
mV
F Sum of all external forces on a body of fluid or system s
Momentum of fluid body in direction s
( )smddt VF =∑

23. Momentum and Forces in Fluid Flow
23
Let’s start by assuming that we
have steady flow which is non-uniform
flowing in a stream tube.
In time δt a volume of the fluid
moves from the inlet a distance u δt ,
so the volume entering the
streamtube in the time δt is
A streamtube in three and two-dimensions
volume entering the stream tube = area x distance
mass entering stream tube = volume x density
momentum of fluid entering stream tube = mass x velocity
tuA δ11=
tuA δρ 111=
( ) 1111 utuA δρ=
momentum of fluid leaving stream tube ( ) 2222 utuA δρ=

24. Momentum and Forces in Fluid Flow
24
Now, according to Newton’s 2nd Law the force exerted by the fluid
is equal to the rate of change of momentum. So
Force=rate of change of momentum
We know from continuity of incompressible flow, ρ=ρ1= ρ2 &
Q=Q1=Q2
( ) ( ) ( ) ( ) 111222
11112222
1111222211112222
F
F
uQuQ
t
utuA
t
utuA
t
tuuA
t
tuuA
t
tuuAtuuA
ρρ
δ
δρ
δ
δρ
δ
δρ
δ
δρ
δ
δρδρ
−=−=∑
−=
−
=∑
[ ] [ ]1212 uumuuQF −=−= ρ
This analysis assumed that the inlet and outlet velocities were in the
same direction - i.e. a one dimensional system.What happens when
this is not the case?

25. Momentum and Forces in Fluid Flow
25
Consider the two dimensional
system in the figure below:
At the inlet the velocity vector, u1 ,
makes an angle, θ1 , with the x-axis,
while at the outlet u2 make an
angle θ 2.
In this case we consider the forces
by resolving in the directions of the
co-ordinate axes.
The force in the x-direction
Two dimensional flow in a streamtube

26. Momentum and Forces in Fluid Flow
26
The force in the y-direction
The resultant force can be determined by combining Fx and Fy
vectorially as
And the angle at which F acts is given by

27. Momentum and Forces in Fluid Flow
27
For a three-dimensional (x, y, z) system we then have an extra force
to calculate and resolve in the z direction.
This is considered in exactly the same way.
In summary we can say:The total force the fluid = rate of change of
momentum through the control volume

28. Momentum and Forces in Fluid Flow
28
Note that we are working with vectors so F is in the direction of
the velocity.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
So we say that the total force, FT, is given by the sum of these forces:
FT= FR+ FB +FP
The force exerted by the fluid on the solid body touching the
control volume is opposite to FR . So the reaction force, R, is given by
R =-FR

29. Application of the Momentum Equation
29
In common application of the momentum principle, we
use it to find forces that flowing fluid exert on structures
open to the atmosphere like gate and overflow spillways
In the following section, we will consider the application
of momentum principle for the following cases.
1. Force due to the flow of fluid round a pipe bend.
2. Force on a nozzle at the outlet of a pipe.
3. Impact of a jet on a plane surface.
4. Force due to flow round a curved vane.

30. Force due to the flow of fluid round a pipe bend
30
Coordinate system: It is convenient to choose the co-ordinate
axis so that one is pointing in the direction of the inlet velocity.
In the above figure the x-axis points in the direction of the inlet
velocity.
Let’s compute, total force, pressure force, body force and resultant
force
Flow round a pipe bend of
constant cross-section
Control volume

31. Force due to the flow of fluid round a pipe bend
31
1.Total Force:
In x-direction In y-direction
Flow round a pipe bend of
constant cross-section
Control volume

32. Force due to the flow of fluid round a pipe bend
32
2. Pressure force
2
Flow round a pipe bend of
constant cross-section
Control volume

33. Force due to the flow of fluid round a pipe bend
33
3. Body force:
There are no body forces in the x or y directions.The only
body force is that exerted by gravity (which acts into the
paper in this example - a direction we do not need to
consider).
Flow round a pipe bend of
constant cross-section
Control volume

34. Force due to the flow of fluid round a pipe bend
34
Resultant force
Flow round a pipe bend of
constant cross-section
Control volume

35. Force due to the flow of fluid round a pipe bend
35
Resultant force and direction
Finally, the force on bent is same
magnitude but in opposite direction
Flow round a pipe bend of
constant cross-section
Control volume

36. Force on a Pipe Nozzle
36
Force on the nozzle at the outlet
of a pipe. Because the fluid is
contracted at the nozzle forces are
induced in the nozzle.
Anything holding the nozzle (e.g. a
fireman) must be strong enough to
withstand these forces.
Control volume of coordinate
system of nozzle is shown in figure
Control volume of nozzle
Resultant force on nozzle = Total force - Pressure force - Body force

37. Force on a Pipe Nozzle
37
Total Force
Control volume of nozzle
Pressure Force = pressure force at 1 - pressure force at 2
g
VP
Z
g
VP
Z
22
2
2
2
2
1
2
1
1 ++=++
γγ
21 ZZ = 02
=
γ
P






−=





−= 2
1
2
2
2
1
2
2
2
1
11
222 AA
Q
g
V
g
V
P
ρ
γ

38. Force on a Pipe Nozzle
38
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. Control volume of nozzle
Resultant force on nozzle=total force - pressure force - body force

39. Impact of a Jet on a Plane
39
A perpendicular jet hitting
a plane.
Control volume and Co-
ordinate axis
Resultant force of jet = Total force - Pressure force - Body force
Resultant force of jet = Total force+0+0
Total force

40. Impact of a Jet on a Plane
40
A perpendicular jet hitting
a plane.
Control volume and Co-
ordinate axis
Resultant force of jet = Total force - Pressure force - Body force
Hence, the resultant force

41. Force on a curved vane
41
This case is similar to that of a
pipe, but the analysis is simpler
because the pressures are equal -
atmospheric , and both the cross-
section and velocities (in the
direction of flow) remain constant.
The jet, vane and co-ordinate
direction are arranged as in the
figure . Jet deflected by a curved vane
Solve urself I am tired now !!

42. Thank you
Questions….
Feel free to contact:
42