 So, fourth problem steam enters a turbine at 100 bar. So, let us first draw a turbine at 100 bar 600 degrees centigrade and undergoes a steady adiabatic expansion process. So, this is adiabatic and expands. The exit the steam exits the turbine at a pressure of 0.1 bar that are these are infligences and it does some work. So, and a steady operation, steady operation which is given steady. What will be the possible maximum specific work output of the turbine and the corresponding exit state of the steam? That is the first part, ok. Second part is, so let us say this is inlet I will say 1 and this is exit this is 2. So, we are worried about the steady operation. It is like what is the change between the properties at state 1 and state 2 or inlet and exit. So, first one is the maximum possible maximum specific work output. The second is if the isentropic efficiency of the turbine is 0.9, what is the exit state? Specific enthalpy of the steam and the specific entropy change during the process, ok. Now, solution P 1 equal to 100 bar, T 1 equal to 600 degrees centigrade, P 2 equal to 0.1 bar. Now, fix the state. So, what are the things I need? It is a steady state steady flow device, steady state steady flow device. So, this means I need enthalpy value basically for the first law and the S value entropy value, ok. Now, please understand that for possible maximum work output, ok the expansion process should be reversible. So, that means thus for reversible adiabatic expansion delta S will be equal to 0. So, that means the state 2 I will say S because this is state 2 S which is due to an isentropic process delta S equal to 0. So, this will be equal to S 1. So, I need properties H and S. Why I need H? Because specific work output of turbine W T will be equal to what? H 1 minus H 2. Yes, I will say, ok. I will say this is maximum. That will be H 1 minus H 2. Yes. So, I need enthalpy and entropy, etcetera. So, how to do this? So, now let us take from the steam tables. From the steam table for 100 bar 600 degrees centigrade, the state is superheated vapour. So, from superheated table, property table at 100 bar or 400 bar at 600 degrees centigrade, H 1 will be obtained as 3625 kilojoule per kg. Similarly, S 1 will be got as 6.903 kilojoule per kg Kelvin, ok. Now, we will continue. So, S 2 S equal to S 1 equal to 6.903 kilojoule per kg Kelvin, ok. Now, P 2 is given as 0.1 bar. So, now, we can see that state 2, yes, that is the, when we will also plot this to understand better, T S diagram will be used for this. So, here this is the, so 100 bar pressure, let us take 600, the state 1. Now, state 2 is 0.100 bar, this is state 1. And the state 2 now, state 2, yes, will be somewhat here. So, here let us say this is 2, because it basically comes like this, S 2 S equal to S 1 in this point. So, now, this is 2 S, this is 0.1 bar pressure, ok, and this is the 0.1 bar. So, this is the expansion, isentropic expansion shown by solid line, ok, straight line, vertically down line, ok. Now, please see that I have to calculate the state 2 now. From saturation table for 0.1 bar, ok, I can take the values of S F equal to 0.649 kilojoule per kg Kelvin, S G equal to 8.15 kilojoule per kg Kelvin. So, now, I find that S F is less than S 2 S is less than S G. So, the state 2 S is saturated mixture. So, now, what I can do, I will find the quality. X 2 S is equal to 6.903 minus 0.649 divided by 8.15 minus 0.649, which is equal to 0.8337. I will find the exit state specific enthalpy H 2 S equal to H F plus X 2 S into H G minus H F, which is at the 0.1 bar. So, from 0.1 bar saturation table, I will take H F and H G value and find this. So, this H 2 S will come out as 2187.01. So, that means the specific work in isentropic process will be equal to H 1 minus H 2 S, which is equal to 3625 minus 2187.1, which will be equal to 1437.8989, which is not 1, ok. So, this is kilojoules per kg. So, this will be also kilojoules per kg. So, this is the maximum possible work, specific work done by the turbine, which is possible only when the process is isentropic process, ok. There is no change in the entropy S 2 S equal to S 1. So, that is the first part. Second part will be isentropic efficiency equal to 0.9. It is given. So, now what is this? So, now the actual state 2 will be like this. I cannot draw a solid line, I will draw a dashed line. So, this is the actual state 2. It may not be a dash. This is 2 S, ok. Now this is 1 bar pressure line. Now the actual state will be like this here. So, this will be 2 actual state 2, ok. So, you can see that the entropy now increases. S 1 is equal to S 2 S equal to S 1, but S 2 is greater than S 1. The entropy due to irreversibility will increase, ok. You know that. So, adiabatic process, the entropy will increase only due to irreversibility. So, that is what is this. So, I can write entropy isentropic efficiency will be equal to H 1 minus H 2 divided by H 1 minus H 2 S, which is actual work, specific work divided by the isentropic specific work. So, that is equal to 0.9, which implies the actual specific work will be equal to 0.9 times the isentropic specific work which will be equal to 1294.19 kilojoule per kg. Now you can see that the actual work put is lesser than the isentropic because of this irreversibility, ok. Now what is this? Basically H 1 minus H 2. So, I can find H 2 from this as 2330.809 kilojoule per kg. So, now I what is the state 2? Actual state 2, state 2 is P 2 equal to 0.1 bar and H 2 equal to 2330.809 kilojoule per kg, ok. This is the state 2. Now we can see that the, from the saturation, we have to see that H G at P 2 is greater than H 2. So, the state is saturated, that is mixture, saturated mixture. So, we can find X 2, actually X 2 equal to what? H 2 minus H f divided by H G minus H f. So, which is equal to 0.89378. So, now you find that the quality when there is a isentropic process, the quality was 0.8337. Now the quality has increased to 0.89378 because the entropy has increased, no? So, the quality has increased. This is benefit of the irreversible process. When you have higher quality at the exit of the turbine, the turbine life will increase. So, that is the benefit. If you have isentropic process, the quality is lower, the quality has improved here, but work has reduced. You can work in isentropic process is 1437 because reduced to 1294, ok. So, this is this. Then I find since X 2, X 2 is got, X 2 is got, S 2 can be found as S f plus X 2 into S G minus S f which is equal to 7.353 kilojoule per kg Kelvin, ok. So, now what is delta S? S 2 minus S 1, specific entropy change will be equal to 7 point final, 7.353 minus initial is 6.903. So, which is equal to 0.45 kilojoule per kg Kelvin. So, this is the difference. So, we have found all the things which are asked. First one is basically when what is the maximum possible work, specific work that is possible only when the reversible process occurs in the expansion. So, adiabatic expansion reversibly occurring where is isentropic. So, delta S equal to 0. When in that case, we can find the state 2 S as S 2 S equal to S 1 and P 2 equal to 0.1. From that, we find the quality X 2 S, then enthalpy, then with the enthalpy change we can find the work output. Now, when the isentropic efficiency is given, then we can calculate the actual work from which we can calculate the actual exists enthalpy, specific enthalpy. From that enthalpy value and the pressure value, we can fix the state. That is P 2 equal to 0.1 bar, S 2 equal to 2330.809 kilojoule per kg. Now, with that enthalpy, we can find what is the state. In this case, it is a saturated mixture with the quality of 0.89378. The quality is improved now. From that, I can calculate the S. And once final state S is known, the specific enthalpy change S 2 minus S 1 can be calculated. So, this is the answer for this. Then the third problem, fifth problem is steam enters an insulated nozzle. So, nozzle insulated nozzle against steam, insulated nozzle at 8 bar, 8 bar under 200 degrees centigrade with negligible velocity. So, I will say V is state 1 or entry state and exit state. So, V 1 equal to 0 and exits at 2 bar, 2 bar. For steady operation, calculate the exit velocity. So, steady operation, steady state, steady flow. Calculate the exit velocity and exit state of the steam. Isentropic efficiency is given as 0.95. That is the thing. So, isentropic efficiency is given. So, that means, first I have to assume the process, the expansion process as isentropic. Then I calculate the exit velocity. So, let V 2 S be the velocity when the process is isentropic. Now, and I can say V 2 be the actual exit velocity with considering the isentropic efficiency. So, the state, exit state pressure is 2 bar in both the cases. Now, how to do this? So, now, again you can say P 1 equal to 8 bar, P 2 equal to 2 bar, T 1 equal to 200 degrees centigrade. Okay. Now, from this P 1 and T 1, I can find H 1 from the stables. So, state 1, first of all, state 1 is superheated stable state. So, that means, H 1 from the tables I can take H 1 as 2839 kilojoule per kg and entropy also I want, correct. So, entropy at state 1 will be equal to 6.816 kilojoule per kg Kelvin. So, these are the values. Now, for isentropic process P 2 equal to 2 bar and S 2 S equal to S 1 equal to 6.816 kilojoule per kg Kelvin. So, with this I can fix the state 2. So, from saturation tables for 2 bar, I will get the values of S f equal to 1.53 kilojoule per kg Kelvin and S g equal to 7.127 kilojoule per kg Kelvin. So, now I find that S f is less than S 2 S is less than S g. So, the state 2 S is saturated mixture, okay, saturated mixture. So, I will find the X, X 2 S equal to 6.816 minus 1.53 divided by 7.127 minus 1.53 is equal to 0.9444. So, from this I can find the enthalpy this as H f plus X 2 S into H g minus H f. These are all taken at 2 bar. So, this is equal to 504.7 plus 0.9444 into 27007 minus 504.7 which is equal to 2584, 2584.63 kilojoule per kg, okay. So, this is fixed now. So, we can apply the first law. First law for the nozzle, for nozzle u dot minus w x dot equal to m dot into H 2 minus H 1, okay. Now, please see this. No, sorry, this is kinetic energy I have to add, this here for this final state kinetic energy S. So, I have told it, let V 2 S be the velocity and the process is isentropic. So, for that I can add this. It is also H 2 S, okay, minus H 1, that is it. So, this is this. But please see, it is insulated. So, this is 0. Then there is no work involved in this nozzle. So, that means I can write this equation. So, 0 equal to this. H 2 plus V 2 S by 2 equal to H 1. Or please see that H 1 and H 2 are in kilojoule per kg. So, that means V 2 S square by 2 should be written as V 2 S by 2000 that you have to remember, okay. So, now I can say V 2 S square equal to 2000 into H 1 minus H 2 S, okay. That is the equation. So, from this I can get V 2 S itself as 713.26 meter per second. So, from zero velocity, it is accelerated to 1113.26 meter per second. Now adiabatic efficiency or isentropic efficiency of the nozzle is given as 0.95 which is equal to what? This equal to V actual kinetic energy, kinetic energy due to the actual velocity V 2 A divided by the kinetic energy due to the isentropic expansion, okay. So, that is the isentropic efficiency, definition of this. Because the process is isentropic that will give you highest to higher velocity. But any irreversibility present, any irreversibility present during expansion like friction, etcetera will reduce the velocity. So, V 2 A will be less than V 2 S. So, the kinetic energy due to the actual process will be less than the kinetic energy due to the isentropic process. So, this implies V 2 A is equal to 695.2 meter per second. You can calculate this. So, this is the velocity. So, you can see that due to irreversibility, but not much. Isentropic efficiency is 0.95. So, there is a reduction in the corresponding reduction in the velocities between the isentropic process and the actual process. So, now we have to calculate using this. I can write for actual process, okay, H 2 plus V 2 A, V 2 A square divided by 2000 equal to H 1. So, from this I can get H 2 equal to, I will put H 2 A also I can put H 2 A equal to what from this equation. So, we know V 2 A now, H 1 is known. So, from that I can get this is 2597.35 kilo joule per kJ. Okay, now state 2 A is 6 dash P 2 equal to 2 bar. The pressure does not change, but the enthalpy now is used. H equal to H 2 A equal to 2597.35 kilo joule per kg. So, now I find that H 2 A is less than H G at 2 bar. So, which implies state is saturated mixture. So, that means X 2 A can be found as H 2 A minus H F divided by H G minus H F. These are all got at 2 bar. So, which is equal to 2597.35 minus 504.7 divided by 2707 minus 504.7 which is equal to 0.95. That is the final state. So, that is what is asked here, correct. Exit velocity and exit state. Exit state is the if it is a saturated mixture then you have to specify the quality as the exit state or if it is superheated then you have to give the temperature or any other property. So, this is the problem. So, isentropic efficiency in these two problems basically the isentropic efficiency is used. Isentropic efficiency is given like this. The actual process is shown in the dashed line in the here, the actual process. The isentropic process is a solid vertical line. So, if you see that the isentropic process produces higher work output for a work producing device or higher kinetic energy in nozzle and so on. Similarly, the irreversibility reduces the work output and reduces the kinetic energy. So, we cannot precisely measure this. By experiments, we can specify what is called the isentropic efficiency. Based upon that, you consider an ideal case of isentropic expansion, get the work output or kinetic energy etcetera for the isentropic process, then use the isentropic efficiency to calculate the actual work output or actual kinetic energy and so on. So, these are the two things which we have seen now.