 so let us move on to tutorial 8 basically this is tutorial on entropy starts with say class of inequality entropy outcalculate entropy change then entropy balance for system and so on so first problem, a reversible heat engine receives heat at rate of 2,000 kJ per cycle 1,500 kJ per cycle and 1200 kgs per cycle from 3 reservoirs at 1200 kV, 1000 kV and 800 kV. So, let us draw this reversible heat engine. So, this there are 3 reservoirs one at 1200 kV, another one at 1200 kV, another one at 800 kV and this 2000 kJ per cycle and this gives 1500 kJ per cycle ok. Now, it rejects heat to 2 different sinks, one at 500 kV, another one at 300 kV ok. Now, this let us say qC1 and this is qC2 ok. This is the arrangement. All processes are other than these heat transfer are already static. That means, there is no other heat transfer in the engine with any other device. The overall thermal efficiency of the engine is 60 percent that is 0.6. Find the individual heat transfer to the sinks at 500 and 300 that is qC1 and qC2. So, this is what is asked ok. So, now, this is a, so this reversible heat engine. So, that we have to keep in mind. So, let us say what is the qH, qH equal to 2000 plus 1500 plus 1200 equal to 4700 kJ per cycle ok. Now, work is equal to 4700 that is I will say qH net minus qC net. So, that will be equal to 4700 minus qC1 minus qC2. So, thermal efficiency of the engine equal to 0.6 which is given equal to 4700 minus qC1 minus qC2 which is the work divided by the qH. This is 4700 qH ok. So, that is equation one. But there are two unknowns ok. So, what is second equation? How can we generate? So, you know that cyclic integral of del q by T is equal to 0 for a reversible cycle. This clashes inequality. So, the classic inequality in general states that cyclic integral of del q by T is less than equal to 0, but for reversible cycle it is equal to 0. So, let us apply that. So, what is that 2000 divided by the corresponding temperature 1200 plus 1500 divided by 1000 plus 1200 divided by 800 equal to or minus you have to integrate. So, positive heat transfers positive sign negative heat transfer negative sign I have to put. So, qC1 divided by 500 minus qC2 divided by 300 this sum this is cyclic integration that will be equal to 0. So, that is the second equation which involves qC1 and qC2 ok. So, now these two equations are solved. Solving one and two simultaneously we get qC1 equal to 1200 kilo joule per cycle and qC2 equal to 680 kilo joule per cycle. So, in this case you can see that engine normally engine need not receive heat from only one reservoir. See for example if you take a boiler of a steam power plant for sensible heat it may receive heat from one particular reservoir or combustion chamber and for giving the latent heat it may receive heat from another reservoir or combustion chamber. So, it may be higher temperature and basically for super heating it may require one more reservoir to get the heat or furnace. So, that means multiple heat transfers are possible in engine that is what is illustrated here ok. So, for example the latent heat can be got from say 1000 Kelvin the sensible heat can be got from heat 100 Kelvin reservoir and super heating for saturated vapour to superheated vapour can be got from this as an example I am saying. Similarly there can be multiple things. So, engine basically has several components. Each component is a control volume and together all the components will be taken as a system and when you take as a system only the work and this basically it will also have a work correct. So, work and heat transfers will be involved. Even work you can see multiple work will be there, multiple pumps will be there, multiple stage turbines will be there etcetera. So, we are now talking about a net work here ok. So, this problem illustrates that an engine can work with multiple reservoirs taking multiple amounts of heat and reject multiple amounts of heat to the different reservoirs. This is the first problem. Second one, one k g of air at a pressure of 10, 10 bar and 100 degree centigrade undergoes a reversible polytropic process which may be represented as p v power 1.1 equal to constant. So, please understand that reversible process. Here the p v power 1.1 is constant p is in meter cube per kg specific volume. The final pressure is 2 bar ok. Evaluate the final specific volume, the final temperature and change in entropy. Now, evaluate the work done and the heat transfer at B. Then C, repeat A and B that is evaluate the final specific volume, final temperature and change in entropy and work and heat interactions assuming the process to be adiabatic, but irreversible ok. Please see the difference. First one is reversible process. Reversible process undergone by a system which is having air. So, mass is 1 kg. Then p 1 is 10 bar, t 1 equal to 100 degree centigrade that is 373 Kelvin. Then p 2 equal to 2 bar that is what is given ok. Now, for A, p v power 1.1 equal to constant that is given. Now, let us find we know that equation of state equal to p v equal to r So, for air r is 287 joule per kg Kelvin. So, now we can say v 1 will be equal to r t 1 divided by p 1 which is equal to what? Specific volume 287 into 373 divided by 10 bar. So, 10 into 10 power 5 which is equal to 0.10705 meter cube per kg. So, that is the specific volume 1. Now, I can apply this p 1 v 1 power 1.1 equal to p 2 v 2 power 1.1. So, that means I can I can find v 2 by v 1 equal to p 1 by p 2 power 1 by 1.1 which implies v 2 equal to 0.4625 meter cube per kg. So, this is what it is the initial final states are all fixed. So, now equation of state we can use t 2 we can find t 2 equal to what? p v equal to r t. So, p 2 v 2 divided by r which is equal to 322.23 Kelvin. So, whatever value we want we can we have found that. So, estimate the final specific volume which is 0.4625 final temperature which is 322.23. So, then change in entropy. Basically we are talking about specific entropy here. What is change in entropy? You can also find the total entropy which is equal to m into. So, you may recall the TDS relationship. Since r is ideal gas, delta s can be written as m c v ln t 2 by t 1 plus r ln v 2 by v 1, correct. Now, c v can be got as r by gamma minus 1 which is equal to 717 joule per kg Kelvin, ok. So, now, so we can know we know everything now. So, m is 1, c v is 717 ln, t 2 is 322.23 divided by t 1 is 373 plus 287 r value into ln v 2, v 2 is 0.4625 divided by v 1 is what? v 1 we have calculated here as 0.10705. So, this will be equal to 315 joule per Kelvin. So, part b, part a is solved. Final specific volume, final temperature and change entropy. So, this is the, so this relationship we can find from see TDS equal to du plus pdv. So, this relationship if you write TDS equal to du by t plus pdv by t, then I can write this as c v dt by t plus you know pv equal to rt. So, I will say rt by v I can put, ok. So, rt by tv into dv. So, now this t cancels, so r into dv by v. So, when you integrate this, I get this expression, ok. So, this expression is got from the TDS relationship. There are two TDS relationships. First one is TDS equal to du plus pdv. Second one is TDS equal to dh minus vdp. So, by using the enthalpy thing we can write this. So, from these two we can use anyone and get the value of the change in the entropy. Now, the b part is work and heat interaction. So, what is work? So, b part work interaction equal to what? For pv power n equal to constant, work interaction will be equal to p1 v1 minus p2 v2 divided by n minus 1. So, we can do that. That will be mass is 1, 1 into p1 v1. p1 is basically 10 into 10 power 5 into v1 is 10, 705, ok, minus p2. p2 is given as 2 bar, 2 bar, 2 into 10 power 5 into v2 is 0.4624, which we have calculated, divided by n is 1.1 minus 1. So, that will say work will be equal to 145.7 kilo joules, ok. So, q equal to w plus delta u since we can take delta ke equal to delta pe equal to 0. So, now we can say q will be equal to w is 145.7 plus m into cv into T2 minus T1. So, be careful with those units because this is in kilo joules now, ok. So, it will be 145.7 into 130 plus 1 into cv is 0.717. Now, this is in kilo joules per kg Kelvin, ok. So, that will into T2 is 32.23 minus 307 degree. So, q will be equal to 109.27 kilo joules. So, that is a work in heat answer which is found in this. Then third part, third part is the process is irreversible and adiabatic, ok. The first process was polytropic reversible. Now, it is irreversible and adiabatic. So, now, but it is given that occurring between the same states. So, that means final specific volume, final temperature are the same. Delta S will also be the same because the same states are irreversible. So, no issue. That is what is given in the problem. You can see that repeat A and B assuming the process to be adiabatic and irreversible between the same end states. So, initial state is same, final state also is the same. So, which we have found now 32 equal to 32.23, V2 equal to 0.4625 that is intact. So, that means delta S also will be the same because we cannot calculate delta S otherwise because it is irreversible. So, since we know between the same two states, the delta S we have calculated using a reversible process. For any process occurring between the same end states because entropy is a property delta S will also be the same. So, there is no issue, ok. Now, first law Q minus W equal to delta U. Now, here W we have to calculate because Q is 0 here. So, with that we can calculate W. In this case W will be equal to minus delta U equal to M into Cv into T2 minus T1, but the end states are known to me. So, this will come out to be 36.43 kilo joules. So, this is the answer. So, here the main thing is let us plot a PV diagram basically. You know the 10 bar and this is say 2 bar, ok. Now, some process has occurred. So, now we know the process. So, this is, so let us say this is PV for 1.1 equal to constant, but this is known to me. This is a reversible process. Now, some adiabatic process and that and it can be the same state now. Adiabatic and reversible process that can be represented by dashed line. I do not know the exact path because it is irreversible. But the same end states are known to me, ok. So, you can see this since it is adiabatic Q is 0 and W can be calculated as minus delta U since the states end states are known, the initial and end states are known. I can calculate this work using the internal energy change. What is entropy change? S2 minus S1 that is intact for whatever be the process. I can also another process you can consider. For whatever be the process, since the end states are fixed, the delta S will also be the same. When you calculate, what do you have calculated using the reversible process that will be same for all other process. Because it is a point function, ok. So, that is important of this problem basically. Then third problem, a certain gas has CV of 1.2 kilojoule per kg Kelvin. When it is expanded isentropically, what is isentropically? Delta S equal to 0, isentropic, isoentropic problem, isentropically from a specific volume of, so I will write this. So, V1 equal to 0.0625 meter cube per kg and T1 equal to 540 Kelvin that is given. So, isentropic process, in the isentropic process, T2 is equal to 0.1875 meter cube per kg, sorry, V2, specific volume, 2 is specific volume of 0.1875. So, V2 is 0.1875, its temperature falls by 170. That means, T2 will be equal to 540 minus 170 equal to 370 Kelvin. The isentropic process first, ok. Now, when it is expanded in a adiabatic process, adiabatic process is not isentropic process, ok. Reversibly, so isentropic process should be adiabatic, but not only adiabatic, it is reversibly adiabatic. So, reversibly adiabatic process is called isentropic process. When I say adiabatic process, it may not be reversible, ok. So, that you have to understand. So, here, when it is expanded in an adiabatic process with friction, obviously it is reversible, ok. Then same final specific volume. So, here also V2 equal to 0.1875 meter cube per kg, but now temperature here is basically falls by 30 Kelvin only, that is 540 minus 30 equal to 510 Kelvin. So, these are the states. Initial state, two properties are given to fix the state, two systems. Cv is given as 1.2 kilojoule per kg Kelvin, ok. Now, see this. V1, T1 are given. One process is isentropic process, reversibly adiabatic process. V2 is 0.1875, but temperature has decreased to 370. Adiabatic process which has friction, this is reversible, irreversible process, but the same final specific volume is got, but now temperature is only dropped by 30 Kelvin, that is the final temperature is 510. So, these are the scenarios. Now, what is asked? Find the change in entropy for 1 kg of gas in adiabatic process, ok. So, mass of the gas is 1 kg, that also we can write here, M equal to 1 kg, ok. Now, how will we do this? Change in entropy for change in entropy for isentropic process is 0. So, adiabatic process is reversible, so entropy change we have to calculate. So, how to do this? Ok. For isentropic process undergone by an ideal gas. So, that one of the thing we have to take, ideal gas is P V power gamma equal to constant, ok. P V power gamma equal to constant. How we get this? Because you know that TDS relationship we will take one of them plus P D V. So, now 0 for isentropic process D is equal to 0, correct. So, I can say D U plus P D V equal to M, no, not M. So, C V D T plus R, this is P equal to R T by V, ok, into D V. So, let us also divide by T by T. So, this will be by T and this T goes. So, now this is equal to 0. So, from this I can get the relationship between T and V, T and V. Similarly, using P V equal to R T, I can get relationship between P V, that is what I have got here, ok. So, for isentropic process which is reversibly adiabatic process undergone by an ideal gas, we can say that the process obeys P V power gamma equal to constant, ok. This is not true for steam or R 134A, etcetera. Only for ideal gas which has undergone an isentropic process which is reversibly adiabatic, we can write P V power gamma equal to constant. See, for adiabatic process also we cannot write this. Only when it is reversibly adiabatic I can write this, ok. So, once I know this, I can get the T V, apply this also P V equal to R T. Or straight away from this I can get a relation for T 2 S divided by T 1, ok, where 2 S shows if the state reached through an isentropic process, do you understand? So, when I say 2 is axial state 2, 2 S is the isentropic state 2, ok. So, then I can write that. So, using this I can write this as V 1 by V 2 S power gamma minus 1, ok. Please understand, this is the expression. So, now I can say for isentropic process I can put S here, and S to show that this is constant entropy. Entropy is constant. So, S is constant. So, I can put V 2 S, T 2 S. That is what I have put here. So, once you do this, I can get, now I know T 2 S, what is that? 3 7 T. I know V 2 S, that is 0.1875. T 1 540. Then T 2, sorry, then V 1 0.0625, all are known. So, if I can find, from this I can find gamma equal to 1.344 for this gas, ok. So, the gas knows information is available, only C V is given. Now, from the isentropic process where the end states are known, I apply this equation and get the value of gamma. That is one of the usual information I got from this, ok. Now, go to the next part. What is change in entropy for the adiabatic process? So, delta S equal to M into C V ln T 2 by T 1 plus R ln V 2 by V 1. Now, we are talking about states 2, V 2, T 2, etcetera, because it is an adiabatic process, the actual state 2. So, now, this can be written as delta S can be written as 1 into C V. C V is given as 1.2, correct. Into ln T 2 is now what? 510 divided by 540 plus R. How we calculate R? C V equal to R by gamma minus 1. R equal to gamma minus 1 into C V. Gamma is calculated as 1.344 minus 1 into 1.2. So, that will be 0.4128 kilojoule per kg R value. So, since we know from the isentropic process, gamma value, using that gamma, I can calculate R value. Now, that R value I would use it here. 0.4128 long, 0.1875. This is the final state. This is initial state 625. That is it. So, that will be equal to 0.3849 kilojoule per Kelvin. So, please see this. The information to get properties can also be done using the thermodynamics. For example, in this case, the first process which is isentropic process, the information about the states were used to get the property, gamma. And using that for the irreversible adiabatic process, we have got the entropy change. So, when the states are fixed, you may wonder that in an irreversible process, how can you find the entropy change? I am actually trying to fit a reversible process between the same end states. The end states are given as T1, V1, T2, V2. Using the end states, that is state 1 and state 2, by fitting a reversible process. So, that is what actually we are doing in the TDS relationships. TDS equal to du plus pdv or TDS equal to dh minus vdp. In these two, when we derive it, we have assumed that the process is reversible. So, when you integrate this, we get this. So, that means with the end states, if I can, the state 1 and state 2, if I can fit a reversible process between state 1 and state 2, once the states are known to me, then I can calculate the property change, entropy change. So, that is what we are trying to do here. So, that is very important. So, all the information given the problem are used and the problem is solved. This is the third problem.