 We're now going to solve an example problem involving the conservation of energy or the first law of thermodynamics in control volume formulation and what we'll be doing is we'll be looking at the example of an air compressor. Okay, so there we have what our air compressor looks like in terms of the input and the exit properties. What I'm going to do now is I'm going to draw the control volume around the air compressor and I've carefully drawn this so that the flow coming in is normal to the control volume itself as well as the flow leaving. So our compressor goes from 14 psi up to 50 psi, not that much of a pressure rise. We're told it's 20 pound mass per second and 600 horsepower and the thing that they want us to find in this problem is they want us to find the rate of heat transfer. So given that they're asking for the rate of heat transfer, that's a pretty big hint right off the start that this is going to involve using the first law of thermodynamics. So let's set the problem up. We'll start with the assumptions that we're going to make and then we'll start working with the basic equations. So the assumptions, okay, so those are the assumptions that we'll be making in this problem. Steady flow, so it's not changing with time. Uniform properties over the inlet and exit, so that would be temperature as well as the velocity. We'll apply and use the ideal gas equation in this. We're assuming, I mentioned the control surface at one and two is perpendicular to the velocity vector and therefore the shear component disappears. There is no change in potential energy and the inlet kinetic energy is small. If you recall, I think I had v1 was approximately zero. That doesn't mean the mass flow rate is zero, but what that does mean is that the kinetic energy can be neglected on the inlet. So with those assumptions, what we're going to do, we're going to start with the first basic equation and that will be continuity and we will use that in control volume formulation. So that is the continuity equation. We're dealing with the steady flow, so that term disappears. And what we're left with is just the mass flux across the control boundaries, so your input and the exit have to be a little careful with the sign of that. So on inlet control surface one, we had a velocity coming like that, but our area vector would be out in that direction, so that's control surface one. And then for two on exit, the velocity vector was in that direction and our area vector would be there. So the dot product here is a negative and the dot product here is a positive. Okay, so those are equal to the mass flow rate. That's what we get out of continuity. Now the thing to make a comment here, one of the assumptions we made was that v1 was approximately equal to zero. But be careful, that doesn't mean mass flux is zero. All that means is that kinetic energy at one is approximately equal to zero. So that can sometimes be a point of confusion for students. Okay, so continuity next equation that we go to, we're going to go to the energy equation. Our first law of thermodynamics. So let's take a look at that. Okay, so that's the conservation of energy. To begin with, we had mentioned that the velocity coming into and leaving was normal to the control surface. And consequently, the shear work term disappears. We're dealing with the steady flow. And so the time rate of change term disappears. So those two, we can remove right off the bat. Let's rewrite our first law equation. And one thing that I'm going to do here, I mentioned this earlier, the u, u plus p over rho, recall that's u plus pv, specific volume. And that from thermodynamics we know is just enthalpy. So I'll make that substitution as well. Okay, so that's where we get our first law equation, too. Now what we need to do, we need to look at this here. This is the mass flux term. And remember with momentum, we have to be careful with that as well, because the sign depends upon which control surface we're looking at. And so what we'll do now is we will expand that out. Okay, so that's what we get for our conservation of energy. A couple of things that we can do here. First off, we said that the kinetic energy on the inlet was negligible. So that term disappears. Another thing that we can do, we have here rho 1 v1 a1. And then here we have rho 2 v2 a2. We know from continuity that that was just what we called the mass flow rate. And so we can replace that. And recall the minus sign here, that was because we had on control surface one, the vector of velocity coming that way, but the area was actually in that direction. And that's why that's negative. And here we have velocity leaving and we have dA going in that direction. So the dot product is positive. And that's why we have positive in the second term. But with that, let's plug in m dot and we'll rearrange the equations and take a look at what we get. So it looks a lot like what you get in thermodynamics. It is the system formulation or the expression, not for a fixed mass, but for a controlled volume of the first law where you have mass crossing the boundaries. We did say that the potential energy did not change. So that term disappears. We have H2 minus H1. How do we deal with that? Well, if you recall back to thermodynamics, dH is equal to Cp dt. And if we assume constant specific heats, which we probably can because our temperature rise here, we're going from 70 Fahrenheit to 100 Fahrenheit, very, very little change. So with that, we then get H2 minus H1 is equal to Cp times T2 minus T1. So we can make that substitution as well. And with that, we can rewrite the first law or our conservation of energy equation to look something like this. Okay. So looking at our problem, what do we need to get? We have to get a few things here. We have to figure out what the velocity is at two and the temperature difference was given to us. Once we get the velocity at two, we can get the mass flow rate. We know the work coming in. We were told that that was 600 horsepower and then we can get the heat transfer. The biggest trick from here on is the fact that we're using British units. If you're comfortable with British units, it's going to be easy. If you're not, I'm going to slog through and use British units. But another way that you can do it at this point, you convert to SI and go through and solve it that way, which is probably what I would do. But I'm not going to because I'm going to show you how to go through with British units. So here we go. Beginning with continuity, we want to get V2. That will be the mass flow rate divided by rho 2 A2. And the density at two, we can get from the ideal gas equation. So rho is P over RT or P2 is rho RT2, R being the gas constant for air. Okay, so there we got a lot of stuff. We're doing unit conversion in here. That's why it's a little messy. But when you go through and you put in the values, you get the velocity at two is 82.9 feet per second. So that's what we get out of continuity. Now that we know that, what we can do is go back to our first law. Let's look at it here. So this was the formulation. We know velocity. And we have pretty much everything else. We know the mass flow rate. CCP, we can look that up for air. We know the temperature change. We know the work. We can calculate everything. Again, the biggest challenge is going to be dealing with the units. So let's go through and see what we get. Okay, so we have the value of CCP. We have a conversion between horsepower and foot pound for second. So let's go through and plug everything in. So with all of that, we plug it in. We have to do a conversion between foot pound force and BTU here. But what we get is we get Q dot is minus 277 BTU per second. Now, what does that mean, the fact that we get a negative? If you recall, we said that heat transfer into the system and work done on the system are positive. And so in this case, given that it's a negative, that means that it is a heat loss term or heat is losing or leaving our system. And so that would be a case where we have heat loss or heat rejection from our system. So with that, that concludes the example problem of looking at applying the control volume to the first law of thermodynamics or the energy equation.