 Second problem, a basic air standard breakdown cycle has a pressure ratio of 10. The air, the R for air is 287 joule per kg Kelvin and the specific heat ratio gamma is 1.4. Air enters the compressor at 290 Kelvin and enters the turbine at 1,120 Kelvin. Assuming the isentropic efficiency of 90 percent for both compressor and turbine, determine the air temperature at the exits of turbine and compressor, net specific work output and thermal efficiency of the cycle. So, now basic breakdown cycle, what are the components in this? When I say cycle, it is a closed cycle, the components are, first there will be a compressor, air compressor which is actually driven by a turbine, a gas turbine. By using the air standard cycle, it is only a, it will only use the air. Then what happens is the air comes in at, it is given is the air comes in at a temperature of 290. So, 290 Kelvin. So, inlet say state one, this is 290 Kelvin and this is compressed to pressure ratio. So, this is exit 2. So, now P2 by P1 equal to 10, that is a pressure ratio. So, you can see that low pressure air at 290 Kelvin enters the air compressor. It is compressed to state 2 where it is, the pressure ratio is P2 by P1 is 10 is given. The values of P1 and P2 are not known, but the pressure ratio is known. Now, I have to add heat to this. So, I will have a heat exchanger where heat is added. So, let us say q dot h, steady state, steady flow, ok. So, some m dot is the flow rate of the air, m dot. Now, temperature, so this is exit of this is actually state 3, state 3. The inlet to the turbine is state 3. So, here temperature T3 is, temperature is 1120 Kelvin, ok. At pressure, this is a constant pressure operation. It operates in constant pressure. So, constant pressure operation happens in the heat exchanger where heat is added. Now, after expanding in the turbine, basically it is adiabatic, compress is also adiabatic and turbine is also adiabatic and it comes out to state 4. Now, I have to close the cycle. So, for that I use one more heat exchanger which is basically operates between low pressure. Here the pressure is P1. So, here P1 equal to P4. Similarly, here in this P2 equal to P3, this also operates at constant pressure. And you can see that in the compressor, pressure is increased from P1 to P2, maintaining a pressure ratio of 10. Similarly, in the turbine, pressure is decreased from P3 to P4, P3 by P4 is also 10. So, that is P3 by P4 also. When I say pressure ratio across turbine and compressor, the pressure ratio is same in this case. The basic cycle, basic cycle and air is the working fluid. So, it is called air standard cycle. Air is the working fluid, heat is added and here in order to cool this to 290 Kelvin, continuously it should be cooled to 290 Kelvin at the pressure of P1, lowest pressure of the cycle. Here, some heat has to be rejected, Q dot C is the heat rejected. Obviously, this turbine runs the air compressor, but also should produce some work, no. So, that is this. I will say, I will say W net because part of the, if Wt is the turbine work, part of the work is supplied to the compressor. So, W dot C. So, W dot T minus W dot C will be equal to net work which is coming out of the turbine which is used to run a generator, etc. So, this is the schematic of a basic Brayton cycle, Brayton cycle. Now, how to fix the states again? Because here two temperatures are given. The inlet to the turbine temperature is given and inlet to the compressor temperature is given. 290 is the inlet to the compressor and the inlet to the turbine is 1,020. These are two temperatures given and the isentropic efficiencies of compressor and turbine are given. Other than that, only the pressure ratio is given, the absolute values of the pressure are not known now. So, how to solve this problem? Here, I can draw the cycle. Let us only take T s diagram which is very important. So, now the isobar. So, for example, this is say P equal to P1 and this is P equal to P2 or I will say P equal to P3 here equal to P4. So, now this is state one, pressure P1, this is 290, this is in Kelvin. Now, the operation of the compressor, compressor increases the pressure of air from P1 to P2. Now, this operation can be isentropic. So, leading to state 2 s or it can be an actual state 2 like this. Now, this compression, this is isentropic compression, this is the actual compression. Okay. Now, heat is added. Now, it goes from this, heat is added. So, in this process Q dot h is added. Now, here obviously work is added, work to the compressor, basically from the turbine itself. Now, this goes to state 3, state 3 here. So, where the temperature is 1120. These are the two temperatures given, 103. Now, in the turbine the operation can be again isentropic because isentropic efficiency is given. So, it cannot be isentropic. But if it is isentropic, it will go to state 4 s, where s3 equal to s4 s, here s1 equal to s2 s. But since isentropic efficiency is given, the actual process leads to state 4, which is a higher entropy. You can see that when there is a irreversibility, the actual state entropy will be higher than the isentropic state entropy, which is the initial state entropy, like that. So, this is the closure of the cycle. Then here, from 4 to 1 heat, here the turbine develops the power and here heat is rejected Q dot c. So, this is the T s diagram of the basic Braydon cycle, ok. So, this is this shows the control volumes involved, four control volumes, compressor, heat exchanger, where heat is added at constant pressure, then gas turbine, which actually does the work. Part of the work is supplied to the compressor, that is W c. And W t minus W c is supplied as network. And in the condenser, sorry, the condenser, instead of condenser, heat exchanger here, where the air is cooled back to 200 Kelvin. So, heat is rejected here. So, this is the Braydon cycle and T s diagram of that, ok. Now, I have to fix the states. Obviously, fixing the state is one of the important thing. I told you, we do not know the absolute pressure, but I can, I do not need to know the absolute pressures also. I do not need to know the ideal pressures, absolute values of them. So, pressure ratio is given, but I need to fix the temperatures, ok, at the actual state, basically 2 and 4. For that, since isentropic efficiencies are given, first I will find the temperatures where, which is attained by an isentropic process. For example, 2 S, from 1 to 2 S, isentropic process, 2 S I can determine. This temperature we find, then apply the isentropic efficiency to get the state 2 temperature. Similarly, I first find the 4 S, assuming an isentropic process for the thurbine, then apply the isentropic efficiency for the thurbine 90 percent to get the actual state. So, that is the thing we have to do. So, ok. Now, air is an ideal gas. So, it obeys Pv equal to RT, ok. P is in Pascals, then V is in meter cube per kg with the specific gas constant given in the problem and this is in Kelvin. And for isentropic process, ok, it obeys Pv power gamma equal to constant, gamma equal to Cp by Cv. So, these two we know already. So, combining these two, I can find the pressure ratio related to the temperature, that is P2 by P1 equal to T2S by T1, T2S divided by T1 power gamma by gamma by gamma minus 1. So, this I can get. When I combine the equation of state Pv equal to RT and Pv for gamma equal to constant, I can get this ratio which also we know. So, similarly, I can write P3 by P4. We know P3 by P4 equal to P2 by P1, correct? Equal to, I can write T3 divided by T4S power again gamma by gamma minus 1. Gamma, gamma is given as 1.4 or is given as 287. So, this is the procedure. First, I will find the isentropic exit state for the compressor which is 2S and for the turbine which is 4S. Then apply the isentropic efficiency for these two, ok. So, let us find these two temperatures first, ok. Now, P2 by P P1 equal to T2S divided by T1 power 1.4 by 0.4 which is equal to which implies 10 power 0.4 by 1.4 equal to T2S divided by 290 which implies T2S equal to what? T2S equal to 560, 560 Kelvin, ok. Now, similarly apply this for the turbine. So, this implies again P3 by P4 is 10. So, 10 power 0.4 by 1.4 equal to T3 by T4S which implies T4S equal to T3 divided by 10 power 0.4 by 1.4 which is equal to 580 Kelvin, ok. So, here you can mark them. So, this T2S is basically 560. It is not to the scale actually. So, this is this is 580, 580 Kelvin, correct. So, these two we get. So, almost same we can say. So, we have to actually mark it properly. So, now for finding the actual states that is T2 and T4, apply the isentropic efficiencies. So, how we will do it? Isentropic efficiency of the compressor equal to what? The isentropic work input to the compressor divided by the actual work input to the compressor because WS is lesser than WA, ok, for compressor. So, this is written as, see for compressor Q dot minus WC dot equal to M dot into H2 minus H1, ok. Now, ignoring KE and PE changes steady state steady flow equation. So, now this is 0. So, that means minus W dot C by M dot equal to W actual, you can say, WC equal to H2 minus H1 or WC equal to H1 minus H2. So, you can supply that, write that here, H1 minus H2S divided by H1 minus H2, ok. So, now isentropic efficiency of compressor equal to 0.9 given in the problem equal to, I will say Cp into T1 minus T2S divided by Cp into T1 minus T2. So, assuming perfect gas, Cp is a constant. So, Cp cancels. So, this is equal to 290 minus 560 divided by 290 minus T2 which implies T2 equal to 590, 590 Kelvin. Similarly, for the turbine Q dot minus WT dot equal to M dot into H 4, this turbine is 3 to 4, correct. So, H4 minus H3 which implies this is 0. So, WT equal to what? WT equal to WT dot by M dot which is equal to H4 minus H3 or WT equal to H3 minus H4. So, isentropic efficiency of the turbine equal to 0.9 given in the problem equal to H3 minus H, here I have to write actual work by a turbine divided by the isentropic work, ok. So, because since WA is less than WS, ok, when there is a power consuming device like a compressor, in the isentropic operation it consumes less power than the actual. So, isentropic work input will be lesser than actual work input. For a power producing device like turbine, the actual work will be lesser than the isentropic work. And the efficiency is always will have the quantity which is greater than the greater to be in the numerator. So, this is equal to H3 minus H4 divided by H3 minus H4S. So, I can say 0.9 equal to T3 minus T4 divided by T3 minus T4S equal to 1120 minus 580 divided by 1120 minus T4E, sorry, this is 580 and this is not known. So, this is T4, ok. So, this implies T4 equal to 634, 634 kilo. So, now actual temperatures are determined. What are they? T2 equal to 590, T4 equal to 634, T1 is 290 and T3 is 1120. So, all the 4 temperatures are determined. Now, I can write the first law, ok. So, specific work done by the turbine equal to WT equal to H3 minus H4 equal to Cp into T3 minus T4, ok. What is Cp? Cp equal to gamma R by gamma minus 1 equal to 1.4 into 287 divided by 1.4 minus 1 equal to 1004.5 joule per kg Kelvin, ok. Similarly, I can write for, ok, first finish this. So, WT will be equal to 1004.5 into T3 is 1120, T4 is 634. So, that will be equal to 488, 488187 joule per kg, ok. Similarly, specific work input to the compressor equal to WC equal to H1 minus H2 equal to Cp into T1 minus T2 which is equal to 1004.5 into 290 minus 590, ok, equal to minus 301350 joule per kg. That is the, so what is the network? Net, specific network is equal to WT plus WC which is equal to 488187 minus 301350 equal to 186837 joule per kg. So, now I want Q which is added. So, go back to this. I have found WT specific by m dot, WC by m dot. Now, Q dot H by m dot, we have to find what is H3 minus H2, ok. QH equal to Q dot H by m dot equal to H3 minus H2 equal to Cp times T3 minus T2 equal to 1004.5 into T3 is 1120 minus T2 is 590, correct. So, that will be equal to 532385 joule per kg. So, what is efficiency? Thermal efficiency equal to W net divided by QH. Please understand, I cannot neglect the, come see in the case of Rankine cycle, the pump work was neglected because it is very small. But here, compressive work cannot be neglected. You can see turbine work is 488187. But the compressive work is more than 3 fourth of that or 3 fourth of that. So, we can neglect that. So, net work, net work output by the heat supply that will be taken into account. So, that will be equal to 18637 divided by 532385 which is equal to 0.3509 or 35.1 percentage. So, that is the efficiency. So, these are the quantities asked. So, here first one is the air temperature of the turbine and compressive existence. Real, that is here T4 and T2. Then, specific work output and for calculating net thermal efficiency, we have to calculate QH also. So, we calculate all the things and this is the answer.