 Let us take the next term which is Q-SYS. So, we may write delta S system plus delta S surroundings for any of the systems that we considered earlier and delta S surrounding is nothing but Q surrounding over T surrounding and Q surrounding is minus Q-SYS. So, we may write this like this and finally write Q-SYS equal to this. So, we have decomposed all these three terms. So, what we are going to do now is take WU to the left hand side and develop an expression for WU. So, if you look at these three things, so we basically wrote okay. So, we basically wrote delta E SIS equal to Q SIS minus W SIS and this itself was written as WU plus P0 times V0 minus V1. So, we rearrange and then we get this expression for WU okay. So, this is the expression for WU after we substitute for Q SIS and delta E SIS. So, WU looks like this. So, exergy of the system at state 1 is WU max okay. What is WU max? It is very simple. So, as we can see sigma is greater than 0. So, sigma is always greater than 0 and this is the amount of entropy generated in the universe as a result of internal and external irreversibilities. Both are accounted for that is very important. So, WU will be a maximum when the process is completely reversible no internal or external irreversibilities which means sigma is equal to 0 okay. So, WU max may also be written as WU reversible because all irreversibilities are absent. So, that is just this portion of the expression for WU. So, exergy of the system at state 1 is equal to this that is the maximum work that may be developed or reversible work that can be developed as the system goes from state 1 to 0 okay. So, now the requirements that we listed earlier for instance how to come up with the efficiency for a non-flow process becomes very clear. Once I have this definition of X exergy of the system at state 1 the system can go from state 1 to state 0 in any process. Let us say it goes from state 1 to state 0 in any process there may be some internal irreversibilities there may be some external irreversibilities it does not matter. It goes from state 1 to state 0 and we get a certain amount of work. So, that is let us say WU actual in the process. Now, I know WU max for the same process for the system between the same states. So, I know WU actual I know WU max. So, the efficiency for this process can now is can now be defined simply as WU actual divided by WU max okay. So, that is where the notion of exergy comes in handy when we want to define efficiency for any process while taking into account entropy generated in the universe as a result of the process. We will elaborate this as we go along but it should become clear now why a notion of exergy was necessary in order to develop and another expression for efficiency taking into account irreversibilities generated entropy generated in the universe. Now, this is the exergy of the system as a whole the given mass M specific exergy may be defined like this we divide through by M and we get this to be the specific exergy okay. So, exergy is always greater than 0 and specific exergy is always greater than 0. So, in fact, you may write for instance what is that this actually this entire expression is the amount of work that is developed by the system when it executes a process with internal and external irreversibility. So, this is actually WU actual I am sorry if you will forgive the use of that expression this is WU actual this is maximum possible work minus lost work. So, basically what we are saying is this is the actual work this is the maximum work this is lost work. So, when a system executes a process from say state 1 to state 0 the actual work is the maximum possible work or reversible work minus lost work this was the reason why we paid so much attention to calculating sigma whether it was a flow process or non-flow process sigma is very important because sigma represents lost work. And so, since sigma is a consequence of internal and external irreversibilities both these irreversibilities contribute to lost work that was why we spend so much time in evaluating sigma. So, we can now easily tie up sigma with exergy and then and then come up with new definition for efficiency which is called the second law efficiency understandably because this takes into account irreversibility external and internal. Let us work out a couple of examples let us say that we have a tank of volume 10 meter cube which contains air at 1 ampere and 298 Kelvin. So, this would be something like this. So, this is a we can think of this as a tank containing air at a pressure higher than ambient pressure and at the same temperature as the ambient temperature. So, the question is determine the exergy that is available in this system. So, basically we have a tank which contains initially contains air at 1 ampere 298 Kelvin. So, if the system were allowed to undergo a process from this state to the dead state what is the maximum amount of work that can be generated as a result of this process that is the exergy that is available. So, basically what we would do is we would simply open up one side of the tank and attach a piston to this and as the air expands to the ambient pressure the piston moves up it can be used to lift a weight and what is the maximum work that can be generated as a result of this is what we are looking at. So, notice that once we look at it like this you can see that it is the same as what we saw in this example. So, we can calculate the mass of air in the tank using equation of state. So, the exergy of the air in the tank there is no K e or P e component in this particular case. So, those are those have been set to 0. So, we can easily write this expression u minus u naught is nothing but C v times t minus t naught this may also be rewritten in terms of the using the equation of state and s minus s naught using T d s relationship may be written like this delta s for this process may be written using T d s relationship like this and notice that the initial temperature of the air in the tank is same as the ambient temperature. So, this quantity goes to 0. So, now we may plug in the numbers and we get this to be 14.026 mega joules. You may also recall that we worked out a problem in the previous course where we had the same situation a tank which initially contained air at 1 MPa 298 Kelvin and the air was allowed to expand through a turbine rather than exhausting it to the ambient we allowed the air to expand through a turbine and the turbine generated work as a result of this. So, we allowed it to expand until the pressure in the tank became 100 kPa. So, basically we have done the same thing, but now we are getting a different value from what we had before. There are two reasons for this as this air is allowed to expand notice that its final temperature is less than that of the ambient temperature whereas, in this case we are saying the air initially is a 298 Kelvin it executes a process attains the dead state which is also a 298 Kelvin. And then the process that we considered here was a flow process. So, there the enthalpy is converted to work not the internal energy. So, that is the reason why this number is different from what we have what we saw earlier. So, exergy is always defined with respect to the dead state. So, whatever process it executes it finally attains the dead state which is 100 kPa 298 Kelvin not 100 kPa some other temperature as we had in this situation. That is why this value comes out to be different, but this is consistent with the framework that we are adopting that everything is done with respect to the ambient state or dead state. So, the next example is also something that we have that we saw earlier in the first level course under the topic of finite reservoir. So, here we have a block steel casting which is initially at a temperature of 300 degree Celsius. So, this is allowed to cool to room temperature. So, basically in the in the previous example we said what is the amount of maximum amount of work that can be extracted from this casting. So, we use the ambient as the cold reservoir and the block as the hot reservoir and we ran a reversible engine. So, the block is the hot reservoir produces a certain amount of work and the ambient is the cold reservoir. So, basically we calculated this work that was developed. In fact, that actually is we also mentioned in the previous course that this work would be called the exergy of the block and that is what we are actually going to calculate now. So, in this case we apply the expression for exergy again no ke or pe and you can see that in this case there is no change in the volume, there is no displacement of atmosphere. So, this goes to 0, this goes to 0 and delta S may be evaluated again using a TDS relationship, but now taking into account fact that this is an incompressible or it is a solid actually. So, there is no change in pressure. So, delta S is simply C times natural log T over T naught and if you just rearrange you get this expression finally and again delta U is nothing but specific heat C times delta T, you can you get this expression which is identical to what we obtained in the previous course by treating this as a finite reservoir. What is that? We said the maximum work was possible if you use a reversible engine. The definition for exergy builds that in automatically. When we defined exergy we said that maximum possible work and then we argued that work is maximum possible if internal and external irreversibilities are 0 which means it is executing a perfectly reversible process. So, that is built into the definition of exergy. So, the expression may be used in the straightforward manner to evaluate this. So, what this illustrated is the mechanism by which we would actually develop this work and then we calculated the work by assuming that it is a finite reservoir. Here we can simply apply the expression and calculate the exergy of the block. So, similar to the expression that we developed just now. So, the expression for specific exergy looks like this. We will actually write down without deriving a similar expression for flow exergy. This is for a non-flow process. You must keep in mind that this is for a non-flow process. For a flow process the development is almost is identical and so we will not go through that but we can simply write the expression by noting the following facts. So, in a flow process you know that instead of the specific internal energy you encounter the specific enthalpy and this term will not be present because in the case of flow process there is no displacement of the atmosphere. So, the displacement of atmosphere is encountered only for a non-flow process. So, this term will also be absent. So, which means I can write the specific exergy for a flow process like this where we replace the specific internal energy with H and the V minus V0 term is absent. So, this is of course specific exergy if there is a mass flow rate of m dot then the W dot reversible the power developed maximum power that can be developed would be m dot times psi. So, let us develop some more perspectives on exergy. A lot of interesting perspectives can be developed. Let us go through this sequence and then see what we can come up with. Already we have said that exergy at the dead state is zero at any other state exergy is non-zero and positive that is important you must keep it in mind. Now, first law may be stated like this total energy of an isolated system remains constant although we did not state it in this manner in the previous course, but this is actually a statement of first law total energy of a system remains constant. So, first of an isolated system remains constant. So, application of first law to an isolated system that undergoes a process from 1 to 2 looks like this delta E equal to Q minus W because the system is isolated there is no heat interaction there is no work interaction. So, total energy of the isolated system remains constant. Statement of second law is this entropy of an isolated system increases or remains the same. Again if we write the expression for entropy change of a system it looks like this because the system is isolated there is no heat interaction. So, this is zero and so S2 minus S1 is equal to sigma int no external irreversibility because there is no interaction on the system with the surroundings. So, S2 minus S1 equal to sigma int which is greater than or equal to 0. Now, we may actually make a similar statement regarding exergy. Exergy of an isolated system decreases or remains the same decreases or remains the same. So, basically what we do is we multiply this expression by t naught and then subtracted from the previous expression. So, E2 minus E1 minus t0 times S2 minus S1 is equal to minus t0 times sigma int. So, if I rearrange this left hand side I can write it like this and this is nothing but x2 minus x1 equal to minus t0 times sigma int which is less than or equal to 0. So, notice that exergy of an isolated system remains the same or decreases. So, x2 is the final state one is the initial I am sorry 2 is the final state one is the initial state. So, x2 minus x1 is either negative or if the system executes a reversible process it is 0. So, this is something that we can keep in mind that exergy of an isolated system decreases or remains the same. Now we need to look at other more we need to look at more general situations. So, we defined exergy by saying that I have a system at a given state one and it executes a process and goes to state 0. Now for this process I can calculate I can evaluate w u actual I can evaluate w u max also. Now let us say system goes from state one to state two and state two is not the ambient state how do we and we are supplying heat to this system and it goes from state one to state two. So, that is a more general case we may be supplying work to the system we may be supplying heat to the system or it could be a flow process unsteady flow process where we are also adding mass to the to the device it is no longer a system we can also add mass. So, these are the three possible situations scenarios that we will encounter addition of work addition of heat addition of mass. So, how does the exergy of the system change as a result of this is the more general case. So, when a system does a certain amount of work then it is exergy decreases. So, if you look at this illustration. So, here as the system expands it does a certain amount of work and eventually it reaches the ambient state. So, the exergy of the system was nonzero at the initial state and then exergy decreases as the system executes this process. So, when the system produces positive amount of work then exergy of the system decreases and that is what we have written here when the when the system does an amount of work W its exergy decreases by W. Conversely when an amount of work W is done on the system its exergy increases by W because it is work and exergy is maximum work that is possible work interaction exergy transferred due to work interaction is very easy to to account for when the system does an amount of work it its exergy decreases by that amount when we do a certain amount of work on the system its exergy increases by that amount. When a certain amount of mass having specific exergy fee is added to or removed from a system the exergy of the system increases or decreases by this amount. Of course, you have to be very careful we are using the word system somewhat colloquially here it will no longer be a system if you are adding and removing mass. So, we need to bear that in mind we can still define you know define things in an appropriate fashion so we need not worry about that but if we add or remove a certain mass with specific exergy fee then we are directly increasing or decreasing the exergy again here this is somewhat easy to to calculate. Heat interaction or change of exergy due to heat interaction is subtle and needs to be looked at very very carefully. Let us see why. So, here we have actually illustrated contours of exergy. So, each one of this line that we have each one of this line is a contour value. So, exergy is 0 at the dead state P0 and T0 and then exergy as you can see increases outwards. So, each one of this is a higher value of exergy. Now, we have shown four states A, B, C, D here corresponding to these four states. So, we have shown four states here sort of corresponding to these forces also since P equal to P0 this would lie on the axis basically what we are trying to illustrate are these four situations. So, if had P been equal to P0, A would have been here or B would have been here. So, we are illustrating for a even more general case. So, if you look at A, it is at a temperature T greater than T0, P greater than P0. So, if I add work to this system which is at state A, then its exergy will simply increase and it will move up like this. On the other hand, if I add heat to this system, its temperature will increase and its exergy will also increase and it will move up like this. Same is true with this system also, same is true with this system also. It is at a temperature greater than T0. If I add heat to this system, if I add work to any of this system, their exergy will increase, they will move out radially. If I add heat to this system, it is at a temperature greater than T0, its temperature will increase and its exergy also increases, it moves outward because its temperature also increases. On the other hand, if I take for instance this system whose temperature is less than T0 and if I add heat to that system, then its temperature actually increases which means it moves closer towards the dead state, which means its exergy decreases. If we add heat to the system, which is at a temperature less than T0, its exergy actually decreases. If I add heat to this system again, which is at a temperature less than T0, its temperature will increase and it will move towards the dead state. So, its exergy actually increases as a result of heat addition. Again, let me emphasize for all these four states, addition of work will increase the exergy and the states will move outward. Addition of heat will increase the exergy of state A and B, which are temperatures higher than T0, but actually reduce the exergy of states C and B, which are at a temperature less than T0. Conversely, if I remove heat from this system, let me leave it like this, let me choose a different color here. Conversely, if I remove heat from this system, its temperature decreases and its exergy also decreases. So, the state will move radially inward and the same is true for this also. It will move radially inward. If heat is removed, exergy decreases. For this system, if I remove heat, then its temperature decreases and this system will move radially outward and its exergy will actually increase. So, you have to be very, very careful with transfer of exergy using heat transfer. So, if I reduce the temperature, the temperature decreases further. So, this moves outward and its exergy increases. The same is true in this case also. If I reduce the temperature, if I remove heat from this system, its temperature decreases and the system moves radially outward and its exergy increases. Which is why we need to be very careful when we talk about exergy transfer as a result of heat transfer. It depends on the temperature of the system when it receives the heat or when it rejects the heat. So, we will consider each one of this in turn. So, here we start with the system whose temperature is greater than T0. Now, let us say this system supplies a certain amount of heat Q and this amount of heat that is supplied is received by a reversible engine which then develops a certain amount of what W. This could also be denoted W reversible and we use the ambient as the low temperature reservoir to reject heat. So, we operate a reversible heat engine between this system which is at a temperature greater than T0 and the ambient which is at T0 and the reversible engine produces a certain amount of what W. So, that means when a system which is at a temperature greater than T0 supplies an amount of heat Q, it also supplies equivalently an amount of exergy equal to X which is equal to W. So, when a reservoir supplies an amount reservoir at a temperature greater than T0 supplies an amount of heat Q, it supplies equivalently an amount of exergy X that is shown here. Remember, this is a reversible engine, there are no internal or external irreversibilities. So, whatever exergy is supplied to it simply goes through and this X is equal to W in this case. Now, let us say that this is a system again at a temperature greater than T0, but a certain amount of heat Q is rejected to this system. Then what can we say about the exergy transfer? So, here the system supplied a certain amount of heat Q. So, we said that in this case it is also supplying an amount of exergy X which is equal to W reversible. Now, in this case a certain amount of heat Q is being rejected to this to this system which is at a temperature greater than T0. So, what can we say about exergy transfer is the exergy of the system going to increase or decrease that is what we are looking at. So, in this case exergy of the system decreases what is going to happen in this case. So, when an amount of heat Q is being rejected, we can actually envision reverse engine, reversible engine, reverse engine operating between the ambient and this high temperature reservoir. So, this is a reverse engine which moves heat from the ambient to this reservoir at a higher temperature while taking in an amount of work equal to W. So, we can actually visualize a reverse engine operating like this. So, that means certain amount of work is given to this reverse engine to reject an amount of heat Q to the system or high temperature reservoir. In exergy terms, what this means is we supply an amount of exergy X which is equal to W and this is then transferred to the system. So, when an amount of heat Q is rejected to a reservoir which is a temperature greater than T0, the exergy of the reservoir increases by an amount of X which is equal to W, not Q equal to W which is the work that a reversible engine would have required for accomplishing this. We will develop an expression for this as we go along. So, to summarize, for a reservoir or system which is at a temperature greater than T0, when heat is supplied by the reservoir, its exergy decreases by amount of X equal to this W and when heat is rejected to this reservoir, its exergy increases by an amount X. What we will do in the next lecture is to discuss the situation when this reservoir is at a temperature less than T0. So, what happens to the exergy of the reservoir or system? It is actually a system. So, what happens to the exergy of the system when of a system which is at a temperature less than ambient temperature when A heat is rejected to it, B, it supplies an amount of heat, a certain amount of heat. So, we will look at that in the next lecture.