More Applications of the Momentum Equation
In this section we will consider the following examples:
1. 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
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:
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.
3. 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. z1 = z2 = z3 and the pressures are all atmospheric = 0. So
By continuity
so
Using this we can calculate the forces in the same way as before.
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.
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 u2 = u3 = u
So we know the discharge in each direction