The Brayton cycle, depicted in the PV diagram of Fig. 20–28, c...

Question

The Brayton cycle, depicted in the PV diagram of Fig. 20–28, can describe a jet engine gas turbine. In process ab the air–fuel mixture undergoes an adiabatic compression. This is followed, in process bc, with an isobaric (constant pressure) heating, by combustion. Process cd is an adiabatic expansion with expulsion of the products to the atmosphere. The return step, da, takes place at constant pressure. If the working gas behaves like an ideal gas, show that the efficiency of the Brayton cycle is
$e = 1 - {(\frac{P_{b}}{P_{a}})}^{\frac{1 - γ}{γ}}$

$e = 1 - {(\frac{P_{b}}{P_{a}})}^{\frac{1 - γ}{γ}}$

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with heat engines. This problem says a spacecraft propulsion system operates using a Brayton cycle. With the following processes going from 1 to 2, we have an a back compression, going from 2 to 3, we have isobaric heating going from 3 to 4, we have an adiabatic expansion and going from 4 to 1, we have isobaric cooling. The gas mixture behaves as an IO gas with the heat capacity ratio of gamma is equal to 1.4 derived expression for the efficiency E of this breakdown cycle. In terms of the pressure ratio P two divided by P one and gamma. Below the question, we're given a pressure volume diagram with the cycle that was described in the problem. We're also given four possible choices. As our answers. For choice A we have the quantity of P two divided by P one raised to the one minus gamma divided by gamma. For choice B, we have one minus one divided by the quantity of P two divided by P one and 20 raised to the one minus gamma divided by gamma. For choice C, we have gamma minus one all divided by gamma. And that's multiplying the quantity of P two divided by P one. And for choice D, we have one minus the quantity of P two divided by P one in quantity raised to the one minus gamma divided by gamma. Now we need to actually calculate the efficiency of this cycle. So recall your definition for efficiency, that's gonna be E is equal to one minus the absolute value of Q out divided by QN where E here is our efficiency, Q out is the amount of heat that's going out of the system. And QN is the amount of heat that's coming into the system. Now, since we have um 280 back processes, we call that an 80 back process has no heat being transferred into or out of the system during that process. That means the heat is actually going in during the isobaric processes. And so the heat that's going to be going into the system is going to go be going into the system during the process, going from 2 to 3, which is the isobaric heating. And this heat's gonna be leaving the system. So it's gonna be a QE out during the process going from 4 to 1, which is the isobaric cooling. So we can use that information to write our heats in terms of the temperatures. So we'll have E is equal to one minus. And here we call your formula or the amount of heat that's required to change a temperatures gas at constant pressure. So that's gonna be n multiplied by CP, multiplied by our change in temperature. And so for out temperature, that's gonna be the process going from 4 to 1. However, we're dealing with the absolute value here. So we actually want this to be a positive quantity. And since I'm cooling, I know that my temperature at 0.4 is going to be greater than this temperature at 0.1. So make sure I get a positive quantity, my temperature difference is going to be T four minus T one. And then that's going to be divided by RQN, which again will use the same formula which is NCP multiplied by our change in temperature. And here, it's just gonna be our normal T three minus T two. Now here and it's gonna be the number of volts of gas NCP is a specific heat capacity for an ideal gas at constant pressure. However, both of those terms cancel. So we needn't worry about them. So just rewriting my formula here, we're gonna have E equal to one minus the quantity of T four minus T one divided by T three minus T two. Now, we ultimately want to write our temperatures in terms of the pressures. That's what we need to write our final answer in. And so recall from the ideal gas law that we're gonna have tea is equal to P multiplied by V divided by Nr, so we can write our temperature in terms of the pressure and volumes. So coming back to our efficiency formula, we'll have E equal to one minus for T four, we're gonna have P four multiplied by V four divided by Nr don't have minus P one multiplied by V one divided by Nr. And that's gonna be divided by, the difference is going to be divided by the quantity of P three multiplied by V three divided by Nr minus P two multiplied by V two divided by Nr, we can simplify things a little bit, the NRS will cancel, but we can also simplify our pressures. We can actually get rid of P three and P four. Because if I look at my diagram here, I know the process going from 2 to 3 is gonna be at a constant pressure. It's an isobaric pro uh process. That means that P three is equal to P two and likewise going from 4 to 1 is an isobaric process. So that means that P four is also going to be P one. So I can write both of those pressures in terms of the other two. So coming back to our equation here for efficiency, we'll have E is equal to one minus. And here I'll substitute in for P four, P one. So I can actually factor it out of that top term. So we'll have P one multiply the quantity of V four minus V one and that's gonna be divided by the quantity. And here, I'll replace P three with P two. And so I can also factor that out of the two quantities in the denominator. They'll have P two multiplying the quantity of V three minus V two. Now, at this point, we need to make use of our adiabatic processes. So we call for an adiabatic process that we can relate the products of the pressure and volumes at the different end points. So for the process of going from 1 to 2, we're gonna have P one multiplied by V one raised to the gamma is equal to P two multiplied by V two raised to the gamma. And so what I'm going to do is solve this for V one. So doing so I'll get V one is equal to V two, multiplying the quantity of P two divided by P one all raised to the one divided by gamma. So I can actually do the same thing or the process going from 3 to 4. And so here I will have V four is equal to P three multiplying the quantity. And here I have P three divided by P four, but P three is equal to P two and P four is equal to P one. So I'm again, just going to get P two divided by P one in quantity raised to the one divided by gamma. So I can substitute in these two expressions for V one and V four. And so when I do that, I'll get E is equal to one minus P one multiplying the quantity or V four will have V three, multiplying the quantity of P two divided by P one in quantity raised to the one divided by gamma, then we will have minus or V one will have V two multiplying the quantity of P two divided by P one in quantity raised to the one divided by gamma. And then all that is divided by the quantity of P two. What plane, the quantity of V three minus V two? So I need to actually simplify our numerator a bit. I'm gonna factor out the quantity of P two divided by P one in quantity raised to the one divided by gamma. And so when I do that, I'll get E is equal to one minus, I'll have P one multiplying the quantity of P two divided by P one in quantity raised to the one divided by gamma. Then that's gonna be multiplying the quantity of V three minus V two. And then that is going to be divided by the quantity of P two. What pain quantity of V three minus two. So the reason why we did this is because we need the volumes to cancel out which they do. So I can simplify our expression still some more. So I'll have E is equal to one minus and here I'll have P two divided by P one in quantity raised to the one divided by gamma. And then that's gonna be multiplied by P one divided by P two. But I can actually write that as P two divided by P one, all raised to the negative one. So since those have these um same base there, I can combine the X components and then I'll get E is equal to one minus the quantity of P two divided by P one in quantity raised to the one minus gamma divided by gamma. So this is the answer that I'm looking for. And now if I come over to my answer choices and compare them to what I got the correct answer choice is actually D so just to recap of what we've done here, we started off with our definition of efficiencies and we wrote our heat in terms of the temperature. We then had to write our temperatures in terms of the pressure. And then we had to relate the volumes and pressures at the different points using our relationships for the AAC processes between volume and pressure. And so once we combine all those concepts together, we were able to calculate an expression for our efficiency in terms of the P two divided by P one and gamma. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Brayton Cycle

Adiabatic Process

Ideal Gas Behavior

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