 When we start to adapt the Brayton cycle for thrust production purposes, like in a turbojet, we need to modify a couple of the assumptions that we've been making, the first of which is going to be our assumption that the change in velocity everywhere was zero. Every time that we considered a Brayton cycle that was used for stationary power production, we considered the change in kinetic energy to be negligibly small. But here the whole goal is to produce thrust and that requires a change in velocity. So for us to assume that there was no change in velocity anywhere is going to greatly hurt our ability to produce thrust. Secondly, we are not caring about power output anymore. That is, we are not trying to increase our net power out nor are we evaluating the performance of our engine by considering how much power we're getting per unit of heat input. Instead, the name of the game is thrust. For our purposes, we are only going to be considering engines that are at a constant velocity and they are producing thrust by expelling air at a high velocity and that produces some thrust in the forward direction. Our net thrust is going to involve that amount of force created in the forward direction minus how much force there is as a result of the incoming air hitting the engine and slowing down. So our thrust production is going to be the force at the back end of the engine minus the force on the front of the engine. Furthermore, since we are assuming that the engine is at a constant velocity, we can neglect one of our force terms. That's because remember that we define force as the derivative of momentum with respect to time and momentum is calculated by taking the mass times the velocity, which from the product rule we can write as dm dt times v plus m times dv dt. When our mass doesn't change, we get rid of this term and you're left with mass times acceleration. That's that f equals ma calculation that you've used a million times, but for our purposes we have a mass flow rate, so we have to leave this term in here. However, we don't need acceleration anymore, which means that we can get rid of this term. So we are talking about mass flow rate times velocity and because our force of thrust is going to be the mass flow rate at the exit times velocity at the exit minus the mass flow rate at the inlet times the velocity at the inlet and because if the engine is operating steadily there must be the same mass flow rate of air everywhere, I can write this as mass flow rate of air through the engine times velocity at the exit minus the velocity at the inlet. So that's our force of thrust. I will point out on this equation those velocities are relative to the engine, so we're talking about the air velocity at the exhaust relative to the engine and the air velocity of the incoming air relative to the engine. So what that means is for example if we are analyzing an engine moving at 200 miles per hour through air that is stationary, then the velocity of the incoming air relative to the engine is 200 miles per hour. Even though the air is actually stationary, it appears as though it's moving from the perspective of the engine. Okay, since we're not caring about net power output or rather that we are trying to optimize for something else, we need a new way to evaluate how well our engine is performing and for that we are going to be using propulsive efficiency and propulsive efficiency like any other type of efficiency is going to be what you want or what you're trying to get divided by what you have to put in to make that happen. This is still a power cycle, so what we're putting in is still heat transfer in, but no longer is this net power output. Instead, we are trying to produce propulsive power. Now what is propulsive power? I hear you asking. That's an excellent question. Remember that we calculate work by integrating force with respect to displacement and our force is constant here because the thrust is constant at a constant velocity. We are only considering steady-state operation of engines. This is a very simplified survey of how turbojet engines work. So our force comes out and we're left with force times displacement. And since I'm looking for power, that's going to be force times displacement divided by dt. Again, the force is constant, so I have force times dx dt, which is velocity. Now, which force? The force of thrust. And which velocity? The velocity of our engine. When we multiply the force of thrust by the velocity of our engine, what we get out is propulsive power. And these three equations are how we are going to be measuring the performance of the Brayton cycle when used for turbojet applications. The last modification we need to make to our analysis is that in addition to not caring about the network out, we are actually going to be considering the network out to be zero. And here's the logic. We are saying all of the power produced by the turbine is going into the compressor. So the turbine's only goal in life is to power the compressor. For it to harvest more energy out of the air would reduce how much energy is left for the production of thrust, which is the whole goal. So we are considering the turbine as only powering the compressor. Now, of course, in reality, the turbine's powering other things. There's going to be some power production for the electrical components. We are converting some of that mechanical energy into electrical energy, running electrical systems, battery charging. Furthermore, there are often situations where you're using the shaft work to do something else that requires shaft work, like, for example, running a gas refrigeration system to air condition the aircraft, something like that. But we are considering the turbine as only powering the compressor. So for our purposes, the work out is equal to the work in. The turbine is only powering the compressor. Okay, let's try an example.