In this screencast we will work through an example, which we calculate a linear momentum
balance to determine the forces acting on a bend in a piping system. We have water that
flows through a 20 degree reducing bend as shown bellow, at a volumetric flow rate of
0.025 m^3/s. We are going to assume that the flow is frictionless and gravitational effects
are negligible. The pressure at the beginning of the bend point 1 is 2 bar. We want to determine
the x, and y components of the force required to hold the pipe in place. For the sack of
our system We will say that positive y is up, and positive x is to the right. We are
looking for forces, which right away trigger a force balance approach, and we know that
we can use a linear momentum balance to determine the sum of forces acting on our control volume.
So we need to determine our control volume. Now the easiest thing to do is to choose a
control volume of our pipe. So we would draw our control volume along the pipe to include
our points at the surfaces. Now we have flow entering our control surface at point 1, and
our control surface at point 2. So just keep in mind that we are looking at a piping system.
It is not open to atmosphere anywhere. It is just a section of the pipe. Typically when
you see a pipe that ends with this kind of curly exit that indicates that the section
keeps going. So we only have 1 pressure. That at point 1. So some form of energy balance
might help us to determine the pressure at at point 2, because we might need that. So
since it is a force acting on our control volume. So indeed this once of those special
problems that require both an energy and momentum balance together. So the first thing I like
doing is write down all of our known variables and do the conversion up front. So we can
quickly determine the velocities at these points, because the velocity at point 1 or
the average velocity at point 1 is just the volumetric flow rate over the cross sectional
area. Now lets go to our general equations. So from newtons 2nd law of motion. We know
for a control volume that the change in linear momentum in the control volume with time plus
the net rate of change in momentum entering and exiting, which we represent as the volume.
Is equal to the sum of the forces in the x direction, because we have written our x-component
of velocities. So for a steady state processes we have no accumulation so this term drops
out to 0. This is pretty common for these types of problems. So we have 2 control surfaces.
Meaning for the entrance and the exit we have to evaluate this integral. So for control
surface we need 1 at our entrance. We are going to have the integral of the x-component
of the velocity, which is represented here as u. Since it is coming into our system horizontally
we just write V1. This is times the density times the dot product with the normal. So
now this is a common place for people to make mistakes. If we look at control surface The
normal is point out were as our velocity is against the normal. Hence the dot product
is negative. So we multiply by negative V1. Now we are going to assume an average uniform
velocity since we are not told otherwise. Since velocity is not changing over the area.
We just multiply by a 1, and we can get rid of our integral. So this is our control surface
at 1. We do the same for control surface 2. However the x-component of velocity is not
just horizontal if we look back at our piping system we have a y-component and an x-component
exiting this pipe section. So our x-component is V2, cosine of theta. It is to the right,
which we said was a positive direction so we don't have to add a negative sign. We multiply
this by the density, and this times it is leaving our control volume. So the normal
and the direction of flow is in the same direction, hence the dot product is positive. So this
is just V2. Again since nothing is changing as a function of area we could get rid of
the integral and just multiply by A2. So now we have the left side of our equation up here,
and I will set it up as such. So this is equal to the sum of the forces in the X-direction.
So we need to break down this term. Now first we have the resultant force, which I like
to just write as Fr. This is what we are looking for in the problem. Whatever force is left
over to hold that bend in place. We also have the pressure forces, because when we look
at the flow acting on this control volume. We have a pressure acting against control
surface 1, and we have a pressure against control surface 2. So for control surface
1 the pressure is acting against the control volume. So P1 is to the right, which we said
was positive. For the second control surface the pressure is acting against the control
surface, which is in a negative x-direction, but we also have a y component. We have to
break down that vector into just the x component. So P2 will be negative, and it will be a function
of the angle. So this means that now we can add P1, but we are subtracting P2 times the
cosine of theta. So now we have our final equation for our x-direction, and we see that
we have the density. We calculated the velocity, we can calculate the area. We can do the same
for the second points. We know our theta. We know P1, but we don't know P2 and we are
looking for the resultant force (Fr). So we need to determine P2. So how do we determine
the pressure at point 2? Well as we mentioned at the beginning of this video. We can use
an energy balance, and in an energy balance you might be familiar with is the other Bernulli
equation. Which states that the energy at point 1 is equal to the energy at point 2.
Since we are making the viscous term negligible we do not have to worry about that. We can
us the Bernoulli equation, and when we do we can also cancel out our elevation terms
because we are told that the gravitational effects are negligible. Remember that elevation
term is our potential term. Now you see looking for P2 we have our density and our velocities
at points 1 and 2. As well as our pressure and therefore we could solve for P2, and insert
this into our equation. So at this point it is just plug and chug. So lets go and work
through it. Solving for the pressure at point 2. I get 195250 kg/ms^2 or Pa. So now I can
plug P2 into the top equation. Should get something like the following, and I solve
for the resulting force in the x-direction. I get the wrong answer. The reason I get the
wrong answer is because pressure is not a force. We have to make sure to multiply by
the appropriate area, because pressure times area is our force. So I need to rewrite this
equation as the following. Now I have a proper force balance on the right side, and when
we put this all together. Including this negative sign you should get -4838 N. Which is the
same as -4.8 kN. This is primarily due to the pressures at points 1 and 2. Alright so
we are almost done for the x-direction. Going to go though it for the y-direction and we
will be finish. So right away you should see that hey! this is going to be a little easier.
We have no y component at our control surface 1. We only have to worry about control surface
2. So our momentum balance equation should look like the following. Were we have the
sum of the forces in the y-direction equal to the integral at control surface 2. So we
can go ahead and evaluate this. So we get V the y velocity times the velocity vector
dot product with unit normal. Over our differential area. So we evaluate this integral. Our velocity
in the y-direction leaving is downwards. So we have a negative V2 sine theta. This is
going to be times our density, and now again our flow is acting in the same direction as
our normal. So it is just V2 times our area at point 2. So again it is easy to make this
mistake since we are looking at the dot product with the normal. So now the sum of our forces
is our resulting force in the y direction plus P2, sine theta. This is equal to negative
rho, velocity at point 2 squared, sign theta, A2. Don't want to make the same mistake twice.
Make sure that we put in area times our pressure term. So we know P2, we know A2, sine theta,
and we know everything everything on the right side. So we could just plug everything in
and solve our equation. When I plug in all this in and solve for our resultant force
acting in y-direction. I get -681 Kgm/s^2, and our x-direction -4.8kN. So we have 2 negative
forces, which is what we would expect. So when we do this force balance it is the force
acting on the fluid. So when we look at the flow of water coming in and out of this pipe
section. What are we doing to the water? We are transferring the direction of the water
downwards, and to do that we are acting against it. So our resulting force is acting to the
left as well as downwards. So we would expect two negative forces acting against the water.
Which means that the water force acting on this pipe section is the opposite to the right
and up. So if we detach this pipe section and had no forces holding into place what
would happen to it? It would shoot up and to the right. Based on our forces we can see
it would be more to the right. So if we wanted to determine the total force acting on this
pipe section we could combine the 2 vectors to calculate the total force. So our total
resulting force is just going to be equal to the square root of the force in the x-direction,
plus the square of that in the y direction. So when we do this we get a total force of
4.9kN. In what direction do we apply that resultant force, So we apply the resultant
force 8.01 degrees bellow, horizontal to the left. So hopefully this gives you an idea
in both momentum balance and energy balances together to solve these types of problems.