fluid mechanicsvortex flow and impulse momentum

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

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Fluid Mechanics

Free and Forced Vortex Flow

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� 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.

Forced Vortex Flow (Rotational Flow)

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� 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

Forced Vortex Flow (Rotational Flow)

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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

Free Vortex Flow (Irrotational flow)

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� 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 or Vxr=C1/m=C

Vxr=C

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

Arr

pp ∆

∆

∂

∂+


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� Now,

Mass of element=mass density x Volume

Centrifugal force=

� Equating the forces in radial directions we get

Arm ∆∆= ρ

r

VAr

2

∆∆ρ

r

VArApAr

r

pp

2

∆∆=∆−∆

∆

∂

∂+ ρ

r

VArAr

r

p2

∆∆=∆

∆

∂

∂ρ

r

V

r

p2

ρ=

∂

∂

Equation gives the pressure variation along the radical direction for a forced or free vortex flow in horizontal plane


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� 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;

gz

pρ−=

∂

∂

dzz

pdr

r

pdp

∂

∂+

∂

∂=

gdzdrr

Vdp ρρ −=

2

Equation of Motion for Vortex Flow

Equation of forced vortex flow

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� For forced vortex flow, we have;

� Where ω is angular velocity=constt

� Substituting the values of V 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

gdzdrr

rp ρ

ωρ −=∂

22

∫∫∫ −=2

1

2

1

2

2

1

gdzdrrpd ρρω

[ ] [ ]12

2

1

2

2

2

12

2zzgrrpp −−−=− ρ

ρω

60

2 Nπω =


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� 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

ρωωρ

==

−−−=−

Q

[ ] [ ]12

2

1

2

212

2zzgVVpp −−−=− ρ

ρ

[ ]

[ ] [ ]2

1

2

212

2

1

2

2

2

1

20

VVg

zz

gVV

−=−

−−= ρρ


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� 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

rg

Vg

Z

Vg

zz

ω==

=−

[ ] [ ]0212

−=−= zzzZQ

Equation of Free Vortex Flow

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� 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

/=

==

gdzdrr

Cgdzdr

rr

Cgdzdr

r

Vdp ρρρρρρ −=−=−=

3

2

2

22

∫∫∫ −=2

1

2

1

3

22

1

gdzdrr

Cdp ρρ

Equation of Free Vortex Flow

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[ ]122

1

2

2

2

12

2

1

2

1

32

2

1

2

1

3

22

1

11

2zzg

rr

Cpp

gdzdrrcgdzdrr

Cdp

−−

−−=−

−=−= ∫∫∫∫∫−

ρρ

ρρρρ

[ ]

[ ] [ ]12

2

1

2

212

122

1

2

2

2

2

12

2

2

zzgVVpp

zzgr

C

r

Cpp

−−−−=−

−−

−−=−

ρρ

ρρ

[ ] [ ] [ ]

12

2

1

2

212

12

2

1

2

2122

1

2

2

2

2

12

22

22

zzg

V

g

V

g

p

g

p

zzVVg

zzg

g

r

C

r

C

gg

pp

+−+−=−

−−−−=−−

−−=

−

ρρ

ρ

ρ

ρ

ρ

ρ

g

Vpz

g

Vpz

22

2

22

2

2

11

1++=++

γγ

Numerical: Forced vortex flow

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Numerical: Forced Vortex flow

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Numerical: Free Vortex Flow

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Numerical: Free Vortex Flow

19

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PART II

Momentum and Forces in Fluid Flow

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� 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

mdma sV

F ==∑


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� 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 sVF =∑

=

=∑

mV

F Sum of all external forces on a body of fluid or system s

Momentum of fluid body in direction s

( )smddt VF =∑


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� Let’s start by assuming that wehave steady flow which is non-uniformflowing in a stream tube.

� In time δt a volume of the fluidmoves from the inlet a distance u δt ,so the volume entering thestreamtube 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 δρ111

=

( )1111

utuA δρ=

momentum of fluid leaving stream tube ( )2222

utuA δρ=


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� 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= ρ

2&

Q=Q1=Q2

( ) ( ) ( ) ( )111222

11112222

1111222211112222

F

F

uQuQt

utuA

t

utuA

t

tuuA

t

tuuA

t

tuuAtuuA

ρρδ

δρ

δ

δρδ

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?


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� 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


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� The force in the y-direction

� The resultant force can be determined by combining Fx and Fyvectorially as

� And the angle at which F acts is given by


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� 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


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� 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

Application of the Momentum Equation

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� 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.

Force due to the flow of fluid round a pipe bend

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� 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


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� 1. Total Force: In x-direction In y-direction


Control volume


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� 2. Pressure force

� 2


Control volume


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).


Control volume


34

� Resultant force


Control volume


35

� Resultant force and direction

� Finally, the force on bent is same magnitude but in opposite direction


Control volume

Force on a Pipe Nozzle

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� 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


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� Total Force

Control volume of nozzle

� Pressure Force = pressure force at 1 - pressure force at 2

g

VPZ

g

VPZ

22

22

2

2

12

1

1++=++

γγ21

ZZ = 02 =

γ

P

−=

−=

2

1

2

2

21

22

2

1

11

222 AA

Q

g

V

g

VP

ργ


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

Impact of a Jet on a Plane

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A perpendicular jet hitting a plane.

Control volume and Co-ordinate axis

Resultant force of jet = Total force - Pressure force - Body forceResultant force of jet = Total force+0+0

Total force

Impact of a Jet on a Plane

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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

Force on a curved vane

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� 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 !!

Thank you

� Questions….

� Feel free to contact:

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fluid mechanicsvortex flow and impulse momentum

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