 Okay, we're now going to look at an example problem involving control volume application of conservation of linear momentum and what I'll begin by doing is writing out the problem statement. Okay, so there's the problem statement. What we have is a reducing elbow and essentially what this says we have liquid. What is the liquid? The liquid I believe was water. Where does it say? It says right here. Okay, so we got water. Water coming in. It's flowing in in this direction here. Uniform flow and then it is flowing out. It's a reducing elbow. That means that the diameter on the exit or the area of the exit is going to be smaller and consequently we have the liquid going at a higher velocity coming out. The other thing that's happening is we have a high pressure at one. So P1 is above atmospheric and this is P2 down here. We're told that P2 is equal to atmospheric. So essentially it's kind of like a jet that goes through a 90 degree change. What we are looking for is the force required to hold the elbow in place because due to the fact that the fluid is changing and it's accelerating and going and we have a high speed jet coming out we will have a reaction force on it. So let's take a look at the solution to this problem. So the first thing that we want to do is we want to be able to define the control volume and this is a little bit of an art to this. I have to be careful with the way that you define a control volume. Depending on how you do it you could actually make the problem more complex than it needs to be and that will come with practice. Not much of a reduction there but there should be a reduction. So this is control surface 1, control surface 2 and we were told we have v1 coming in here, v2 down there and what I'm going to do I'm going to define the control volume that looks something like this. So we're crossing the interfaces where we have mass crossing the boundary which is good because we want to be able to do that and then what I'm going to say is that there's some reaction force here on our control volume we'll have reaction force in the x and reaction force in the y and I'll define the coordinate system again and that is what is holding the elbow in place as the fluid is coming through. So in order to solve this we're not going to jump right into the conservation of linear momentum. The first thing we're going to do is we're going to use the continuity equation or conservation of mass. So let's do the conservation of mass. Now why might we be doing that? Well let's take a look at the problem statement that we have here. We're given v2 but we are not given v1. We need to figure out what the velocity on inlet into this elbow is. So that's why we're going to apply the conservation of mass first of all and we'll find v1 through that process. So looking back at the conservation of mass for a control volume now we were told that the water flows steadily that means that it's steady flow so this term is gone. So what we're left with is an integral over the two control surfaces where you have mass crossing the boundary and that is surface 1 and surface 2. Now for control surface 1 if you recall we had that we were told that the velocity was going that way but the vector out of control surface 1 is going to be in that direction and the velocity was in that direction. So through the dot product we know that this is going to be a negative and for control surface 2 it was that way the area vector is going to be outward in that direction and the velocity vector is going out like that and consequently this will be a positive. So remember we got to be careful with that when we go through doing control volume analysis. So what we can do is rewrite the conservation of mass equation in the following way. The first term is going to be negative and I could cancel out the density at this point but I haven't done that yet. Okay so we get that plugging in the values let me cancel density out because we're not going to keep it it's incompressible. We get v1a1 equals v2a2 we can then isolate for v1 and we'll take the magnitude of v2 and multiply by the ratio of the areos and what we end up with is v1 is four meters per second we can express that as a vector so v1 is then equal to plus four i and that is meters per second. So that's the first part and once we have that we can now move on to the conservation of linear momentum and I'm going to do that in the next segment and what we'll be doing there is we'll be applying the conservation of linear momentum to solve for rx and ry.