# Application of the Momentum Equation

In this section we will consider the following examples:

- Force due to the flow of fluid round a pipe bend.
- Force on a nozzle at the outlet of a pipe.
- Impact of a jet on a plane surface.
- Force due to flow round a curved vane.
- A curved vane on a Pelton wheel turbine.
- Impact of a jet on An angled plane surface.

## 1. The force due the flow around a pipe bend

Consider a pipe bend with a constant cross section lying in the
__horizontal plane__ and turning through an angle of .

Why do we want to know the forces here? Because the fluid changes
direction, a force (very large in the case of water supply pipes,)
will act in the bend. If the bend is not fixed it will move and
eventually break at the joints. We need to know how much force
a support (thrust block) must withstand.

Step in Analysis:

- Draw a control volume
- Decide on co-ordinate axis system
- Calculate the
**total**force - Calculate the
**pressure**force - Calculate the
**body**force - Calculate the
**resultant**force

1 __Control Volume__

The control volume is draw in the above figure, with faces at
the inlet and outlet of the bend and encompassing the pipe walls.

2 __Co-ordinate axis 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.

3 __Calculate the total force__

In the x-direction:

In the y-direction:

4 __Calculate the pressure force__

5 __Calculate the 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).

6 __Calculate the resultant force__

And the resultant force **on the fluid** is given by

And the direction of application is

the force **on the bend** is the same magnitude but in the
opposite direction

## 2. Force on a pipe nozzle

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.

The analysis takes the same procedure as above:

- Draw a control volume
- Decide on co-ordinate axis system
- Calculate the
**total**force - Calculate the
**pressure**force - Calculate the
**body**force - Calculate the
**resultant**force

1 & 2 __Control volume and Co-ordinate axis are shown in
the figure below.__

Notice how this is a one dimensional system which greatly simplifies
matters.

3 __Calculate the total force__

By continuity, , so

4 __Calculate the pressure force__

We use the Bernoulli equation to calculate the pressure

Is friction losses are neglected,

the nozzle is horizontal,

and the pressure outside is atmospheric,,

and with continuity gives

5 __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.

6 __Calculate the resultant force__

So the fireman must be able to resist the force of

## 3. Impact of a Jet on a Plane

We will first consider a jet hitting a flat plate (a plane) at an angle of 90, as shown in the figure below.

We want to find the reaction force of the plate i.e. the force the plate will have to apply to stay in the same position.

The analysis take the same procedure as above:

- Draw a control volume
- Decide on co-ordinate axis system
- Calculate the
**total**force - Calculate the
**pressure**force - Calculate the
**body**force - Calculate the
**resultant**force

1 & 2 __Control volume and Co-ordinate axis are shown in
the figure below.__

3 __Calculate the total force __

As the system is symmetrical the forces in the y-direction cancel i.e.

4 __Calculate the pressure force.__

The pressure force is zero as the pressure at both the inlet and
the outlets to the control volume are atmospheric.

5 __Calculate the body force__

As the control volume is small we can ignore the body force due
to the weight of gravity.

6 __Calculate the resultant force__

Exerted **on the fluid.**

The force **on the plane** is the same magnitude but in the
opposite direction

## 4. Force on a curved vane

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 below.

1 & 2 __Control volume and Co-ordinate axis are shown in
the figure above__.

3 __Calculate the total force in the x direction__

but , so

and in the y-direction

4 __Calculate the pressure force.__

Again, the pressure force is zero as the pressure at both the
inlet and the outlets to the control volume are atmospheric.

5 __Calculate the body force__

No body forces in the x-direction, = 0.

In the y-direction the body force acting is the weight of the fluid. If V is the volume of the fluid on he vane then,

(This is often small is the jet volume is small and sometimes ignored in analysis.)

6 __Calculate the resultant force__

And the resultant force **on the fluid** is given by

And the direction of application is

exerted on the fluid.

The force **on the vane** is the same magnitude but in the
opposite direction

## 5. Pelton wheel blade

The above analysis of impact of jets on vanes can be extended
and applied to analysis of turbine blades. One particularly clear
demonstration of this is with the blade of a turbine called the
*pelton wheel*. The arrangement of a pelton wheel is shown
in the figure below. A narrow jet (usually of water) is fired
at blades which stick out around the periphery of a large metal
disk. The shape of each of these blade is such that as the jet
hits the blade it splits in two (see figure below) with half the
water diverted to one side and the other to the other. This splitting
of the jet is beneficial to the turbine mounting - it causes equal
and opposite forces (hence a sum of zero) on the bearings.

A closer view of the blade and control volume used for analysis can be seen in the figure below.

Analysis again take the following steps:

- Draw a control volume
- Decide on co-ordinate axis system
- Calculate the
**total**force - Calculate the
**pressure**force - Calculate the
**body**force - Calculate the
**resultant**force

1 & 2 __Control volume and Co-ordinate axis are shown in
the figure below.
__

3 __Calculate the total force in the x direction__

and in the y-direction it is symmetrical, so

4 __Calculate the pressure force.__

The pressure force is zero as the pressure at both the inlet and
the outlets to the control volume are atmospheric.

5 __Calculate the body force__

We are only considering the horizontal plane in which there are
no body forces.

6 __Calculate the resultant force__

exerted on the fluid.

The force **on the blade** is the same magnitude but in the
opposite direction

So the blade moved in the x-direction.

In a real situation the blade is moving. The analysis can be extended to include this by including the amount of momentum entering the control volume over the time the blade remains there. This will be covered in the level 2 module next year.

## 6. Force due to a jet hitting an inclined plane

We have seen above the forces involved when a jet hits a plane
at right angles. If the plane is tilted to an angle the analysis
becomes a little more involved. This is demonstrated below.

(Note that for simplicity gravity and friction will be neglected
from this analysis.)

We want to find the reaction force normal to the plate so we choose the axis system as above so that is normal to the plane. The diagram may be rotated to align it with these axes and help comprehension, as shown below

We do not know the velocities of flow in each direction. To find these we can apply Bernoulli equation

The height differences are negligible i.e. *z _{1 }=
z_{2 }= z_{3}* and the pressures are all atmospheric
= 0. So

*u*

_{1 }= u_{2 }= u_{3 }= uBy continuity

*Q*

_{1}= Q_{2}+ Q_{3 }

*u*

_{1}A_{1}= u_{2}A_{2}+ u_{3}A_{3}so

*A*

_{1}= A_{2}+ A_{3 }

*Q*

_{1}= A_{1}u

*Q*

_{2}= A_{2}u

*Q*

_{3}= (A_{1}- A_{2})uUsing this we can calculate the forces in the same way as before.

- Calculate the
**total**force - Calculate the
**pressure**force - Calculate the
**body**force - Calculate the
**resultant**force

1 __Calculate the total force in the x-direction__.

Remember that the co-ordinate system is normal to the plate.

but as the jets are parallel to the plate with no component in the x-direction.

, so

2. __Calculate the pressure force__

All zero as the pressure is everywhere atmospheric.

__Calculate the__**body**force

As the control volume is small, hence the weight of fluid is small,
we can ignore the body forces.

4. __Calculate the resultant force__

exerted **on the fluid**.

The force **on the plate** is the same magnitude but in the
opposite direction

We can find out how much discharge goes along in each direction on the plate. Along the plate, in the y-direction, the total force must be zero, .

Also in the y-direction: , so

As forces parallel to the plate are zero,

From above

and from above we have so

as *u _{2 }= u_{3 }= u*

So we know the discharge in each direction