 So we're in part two of the example problem of solving for the reaction force on a reducing elbow, a 90 degree reducing elbow. I'll look back here. This was our problem statement. So what we did in the first part of the problem is we used the conservation of mass to determine v1, the velocity coming into the elbow. Now that we're equipped with v1, we can go ahead and we can apply the conservation of linear momentum, the control volume formulation of it to this elbow. And looking at our free body diagram, what we're after is we're trying to determine the reaction forces we want rx and ry. So what we'll do, we'll start with the x component and then we'll move to the y component of the conservation of linear momentum. So writing out, remember we said that this was a vector equation. I'll write out the x component of the conservation of linear momentum. So we get this. Now, a couple of things happen immediately. First of all, body forces. Body forces in the x direction. There are no body forces in the x direction. The only body force here would be gravity and that does not act in the x direction. So that term is gone. The other one, we were told that the water flows steadily through this elbow. That means that we're not changing anything and consequently we're dealing with steady flow. That term disappears. What we're left with then is the surface forces on in the x direction will be on the left hand side. And the surface forces that we have, we have the reaction force on the control volume. And we also have pressure. And so we were given the gauge pressure. I'm going to subtract off atmospheric. And on the right hand side of the equation, what we have is what is happening across control surface one. So we have to integrate across control surface one. We have velocity u1 coming in. And then that's multiplied by rho and then the dot product of vda. And if you recall, I said in an earlier segment, we've got to be careful. Let's work the sign of that first and then work the sign of u1. So we'll proceed in that manner as we solve this. So writing out the left hand side and then the right hand side, u1. We know that the u direction or u1 is a positive value. But it's multiplied by, if you recall, we talked about this earlier. For control surface one, the area vector will be in that direction. And the velocity is coming in in that way. So the dot product is going to give us a negative. So we have that. Now what I've written here for v1, let me expand out v1 as a vector. And we solved in the first part, u1 was 4. So we can evaluate the magnitude of v1. And that is v1 that I have in the above equation. And that is just the two components and then the square root. So what we end up with, I am going to isolate for rx. So we get this for rx, plugging in the values and calculating. What we find is rx is minus 1.36 kilonewtons. And writing it out as a component with a magnitude with direction. Given that it's minus, that means that it is moving in that direction. And that is to hold the elbow. So that is the x component. Let's take a look at the y component. And again, what we will do, we will use the y component of linear momentum. So we get that. Now, looking at the y component, going back to the free body diagram. Let's even go to the problem statement where it was, it was right here. For the y component, what we were told is P2 is atmospheric pressure. And up here on the other side of the elbow, this is atmospheric as well. So if we look at our free body diagram, we have P atmosphere all around. P atmosphere here, with the exception of control surface 1, we had a higher pressure there. That was P1. And then down here, it's P atmosphere. Consequently, there is no force in the y direction due to pressure. And the net result of that is the first term here. We can say P atmosphere balances. Now, body force, there is a body force in the y direction. And that is due to the mass of the fluid and the gravity vector. But they did not give us any details about the volume. So unknown volume, therefore, we're going to neglect that. We're not given information and so we can't solve for that. So those two go away. And finally, we also have steady flow, like we said earlier. So that disappears as well. And with that, what we're left with on the left hand side, oops, sorry. I was a little too quick here. That is there. But the pressure force balances, but we still have the reaction force. If we look at our free body diagram right here, we still need to account for the reaction force. So I shouldn't have thrown that out so quickly. And what we're left with on the left hand side is going to be R y, but you'll notice there's no pressure term there. And then that is going to be equal to an integral across control surface 2. And for that, we have v2, and then it's multiplied by rho v dot da. Again, let's be careful with this dot product and let's take a look at it. So we have da is the vector pointing out, and we were told or we know that v2 is going in that direction. And consequently, the dot product of that is going to be a positive. So what we then get for R y is v2 times positive. And then we have what is in here, which is essentially resolving the components of the dot product. Let's look at the vector for v2. We have u2 in the i, which was 0 plus v2 in the j, which we were told. So we can say the magnitude of v2 is the magnitude of that vector, which is going to be u2 squared plus v2 squared square root. And that is how we will get the magnitude of that, which we will use on the next slide. So v2, we can write it out as a vector with 0 in the i minus 16j. Our little v2 in our equation is minus 16, so let's look back. This is where we have to be careful. You have to carry that minus sign here. Remember I told you about that earlier? We know the dot product was positive, so we resolved that, but we've got to be careful about the sign of v2. And this tells us that v2 is going to be a negative. Therefore we write R y is minus 16 multiplied by the density of water as 1000. We have the vector component there, the magnitude of the vector, that's 16 squared plus 0, and then the area. Doing all of that, we get R y is minus 640 Newtons. It's a negative. What is that telling us? It's telling us that that is 640 Newtons in the down direction. So what we can write is the force required to hold the elbow. We can say it is fx is 1.36 kilo Newtons in the left, and fy is 640 Newtons down, and we can then resolve that vector, and the angle would be 25.2 degrees, and the magnitude is 1,503 Newtons, or 1.5 kilo Newtons. So that is the solution to the problem of the reducing elbow. One final note that I will make is if you look back at our free body diagram, now let's see where was it right here. We were computing the force on the control volume, and I should make a note there. So what we can say is that although we were calculating force on the control volume, that really is the same force that would be required to hold the elbow stationary, because the entire elbow is enclosed by the control volume. So that is an application of control volume analysis using the conservation of linear momentum, and what we were able to do was solve a problem. We started with continuity, and then we went to linear momentum. But that gives you an idea as to how you can apply this.