 So the last two cycles that we're going to look at considering problem-solving tips will be the Sterling and the Brayton. So let's start with the Sterling and I'll write out the PV diagram. So that's the PV diagram for the Sterling. Now a couple of things that you can note in terms of solving these problems. First of all the specific volume between four and one is the same. Another thing is from two to three you have the same thing where the specific volume is the same. You can write the specific volume using the ideal gas equation. And finally we can write out an expression. I didn't draw the regenerator on here, but we actually have heat transfer in our region. So we would assume that to be equal to the heat transfer from four through one, from four to one, and we can make the substitution. Here we have the change in internal energy between one and four. And so what we can use is we can use the expression of that CV delta T. And depending upon the working fluid, if we're using helium, one of the things about helium, if you remember when we were talking about property data and specific heats, for helium the specific heat is not a function of temperature and consequently we can use the specific heat that would be in the back of the book and be independent upon the temperature change that you may have through that process. And so that would be an exact relationship then for the heat transfer between four and one. So those are some of the things that you can use if you're faced with a problem dealing with the Sterling. The last thing you want to do, let's take a look at the Brayton cycle or the gas turbine cycle. And again, just like before, what I'm going to do, I'll write out the TS and the PV diagrams. So those are the diagrams for the Brayton cycle. Now what we'll do is we will go through and look at whatever information may be presented and then different steps that you can take for solving. So quite often you will be given the pressure ratio for the engine. And if you recall, that is the pressure rise that you'll have in your compressor. And you might sometimes be given atmospheric temperature. You might be given the temperature after the combustor T3. So if you're trying to go from state one to state two, now taking a look back at our TS diagram, this is an isentropic process. So we're looking at state one to state two. It's isentropic. And consequently we can use our relative pressure that you would have in the tables in the back of the book. Because recall the relative pressure and relative specific volume only work if it's an isentropic process. So that's how you could work your way through to fine temperature two. And then looking at our process again, the other thing to notice is that when we're going through the combustion process or the heat addition, we are saying that this is constant pressure, a constant pressure process. And so we can write P2 is equal to P3. And in a like manner, we can say that the heat rejection process is also at constant pressure and therefore P4 equals P1. Now the next thing we're going to do is we're going to look at going from state three to state four. So let's go back to our diagram. Three through four is going from here, from here down to there. Again if you know the temperature of the gas is coming out of your combustor, you can then go and look at the relative pressure in the back of the book of your textbook. And you can use the pressure ratio of the engine. That would then give you the relative pressure at state four, which then would give you the temperature at four, which would be the temperature at the exit of the compressor, or sorry, of the turbine, not the compressor. Now the last thing we're going to do, we're going to invoke or use the first law for different steps or parts of our process. And we'll begin with the compression process. And so that's how we can write compression. There is no heat transferred during compression, usually. Unless you have an intercooler, but we're not talking about an intercooler here. And so with that, you can write out that your work compression in equals H2 minus H1. And you might wonder what happened to the minus sign. And I remember our definition of work as being positive when it is coming out of the system. If you're doing work into the system, it's negative. And consequently, we can flip the sign around. So that's where that comes from. And then for the turbine, similarly, the first law we can write out. Again, there is no heat transfer in the turbine, so that disappears. Now this is work coming out. And we get that for the work from the turbine. So those are some of the equations. The last thing I want to say, and you may have wondered when I was drawing the diagram why I put this state four prime here. That would be the state that you would be getting work out of the turbine in, for example, a turbojet. In a turbojet, what's happening is the work from the turbine, which we would have from here, would be going directly into the compressor. And the remaining enthalpy in the fluid from there to there is going into producing the velocity of the jet or of the exhaust, which is providing a form of propulsion for the engine. So in the case of a turbojet, so that's what you do in the case of a turbojet. You're saying the work of the compressor in is equal to the work coming out of the turbine. And then going from state four prime down to, let's see, what did we have here? State four. That is what is used to accelerate the fluid through the nozzle of your jet engine. And consequently, that concludes today's lecture. We looked at problem solving tips for the auto, the diesel, the sterling, as well as the Brayton. So thank you very much. We'll see you next time.