 Exergy can be transferred, added to or removed from a system as a result of its interaction with the surroundings and the interactions with the surroundings may be classified as addition or removal of work, heat and mass. So, we will take a look at what each one of these interaction does in terms of exergy change of the system. So, let us first look at work interaction. So, when certain amount of work is done by the system, let us say an amount of work W is done by the system, its exergy decreases by WU. So, we saw in our earlier example, our earlier illustration, so here the system does certain amount of work as it undergoes an expansion process. So, as it does work, its exergy decreases because it is doing work by raising the mass. So, as the system does work, its exergy decreases by how much ever work it puts out. So, if a system does an amount of work W, its exergy decreases by exactly that same amount and when an amount of work W is done on the system, its exergy increases by the same amount WU. Since exergy is work, work transferred to a system or from the system can be easily converted to or can be directly converted to exergy change. Now, addition or removal of mass to a system or in a device also causes an exergy change. So, let us say that we have certain amount of mass with specific exergy fee. Then if I add that much amount of mass to the system, let us say M to the system, then the exergy of the system increases by an amount equal to M times fee. If that much mass is removed from the system, then the exergy of the system decreases by that much amount. So, this is also relatively straightforward to calculate or compute. Now, when it comes to heat interaction and transfer of exergy through a heat interaction, we have to be somewhat more careful because depending upon the temperature of the system with respect to the ambient temperature, whether the temperature of the system is greater than or less than the ambient temperature, the direction of heat flow and exergy flow can be opposite to each other and that is what we will take a close look at next. Now, here contours of exergy, X equal to constant or plotted on a PT coordinate system. So, notice that this line corresponds to temperature T equal to T naught, the ambient temperature that is this line here and the horizontal line corresponds to P equal to P naught. And the origin of course is the ambient state itself P equal to P naught and T equal to T naught. So, it is equal to 0 at the origin and since exergy is positive, it increases in the outward direction, radially outward direction as shown here, it is in the radially outward direction. Now, let us consider systems are initially at four different states A, B, C and D. Now, if you take system initially at state A, if we add heat to this system, its temperature increases and since it is initially at a temperature greater than T naught, its temperature increases. So, its exergy also increases and a new state is in the radially outward direction because its temperature increases and its exergy also increases. Similarly, if I add heat to a system which is initially at state B, since its temperature is greater than T naught as a result of addition of initially greater than T naught, as a result of addition of heat, its temperature is also likely to increase and it also moves radially away from the origin, so its exergy increases. Now, conversely, if I remove heat from this system or if this system supplies heat to some other device, then its temperature will decrease and its exergy also decreases, so it moves in the radially inward direction and the same is true for this system also which is initially at a temperature greater than T naught. Now, if I look at the systems which are initially at states C and D, now temperature of system C initially is less than T naught, if I add heat to the system then its temperature increases and as a result it moves towards the origin like this, so its exergy actually decreases. If I supply heat to this system then its exergy actually decreases. In contrast to systems which were initially at states A and B, similarly since the system at state D is initially at a temperature less than T naught, if I supply heat to the system its temperature increases and consequently it moves towards the origin, so its exergy decreases. Now, conversely and quite unintuitively, if I remove heat from this system its temperature decreases and the state point moves radially outward and as a result its exergy also increases. Remember exergy increases in the outward direction as shown here, so when the state point moves away from the origin the exergy of that state point is higher, so the exergy increases in this case when I remove heat from this system. Similarly, for the system which is initially at state D, if I remove heat from this system its exergy actually increases, so you can see that the direction of heat transfer and exergy transfer or opposite depending upon whether the temperature is greater than T naught or less than T naught, so that is something that we should keep in mind. Let us explore this in greater detail and see what happens in these four situations, so let us say that initially we have system which is at a temperature T greater than T naught. Now let us say that this, let us say that initially we have something which is at a temperature greater than T naught, I do not want to, we can call it system, no problem, so let us say that we initially have a system which is at a temperature greater than T naught. Let us say that the system supplies a certain amount of heat Q to some other device or system. Now we want to understand, we know that there is a heat transfer, we want to understand what the corresponding exergy transfer is. To understand this we envisage a reversible engine or that operates between the system at temperature greater than T naught and the ambient which is at T naught. So since this reversible engine is supplied with an amount of heat Q, it produces an amount of work W which is equal to Q times 1 minus T naught over T. Now since this is a perfectly reversible engine there are no internal or external irreversibilities, so we are requiring an amount of exergy X equal to W from this engine and since there are no external or internal irreversibilities this means that exergy is not destroyed within this engine. That means that if you are recovering an amount of exergy X equal to W from this engine, it must have been supplied with an amount of exergy X equal to W, that is what is shown here. Let us go through this one more time. This engine produces an amount of work equal to W. So from an exergy perspective, an amount of exergy W is recovered from this engine R, amount of exergy X equal to W is recovered from this engine R. Since it is a reversible engine, there are no internal or external irreversibilities, no exergy is destroyed within this engine. So if you recover an amount of exergy X from this engine, that means that it must have been supplied with an amount of exergy X in the first place. So now if we look at this from the perspective of this system which was initially at temperature greater than T naught, when such a system supplies an amount of heat Q, it is also equivalently supplying an amount of exergy X equal to Q times 1 minus T naught over T. So when a system at temperature greater than the ambient temperature supplies an amount of heat Q, it supplies an amount of exergy X equal to Q times 1 over, I am sorry, Q times 1 minus T naught over T. Now the same system, if it receives an amount of heat Q, then for this situation we can envisage a reversible heat pump that operates between the ambient at temperature T naught and this system which is at a temperature greater than T naught. Now for this heat pump to work, we need to supply an amount of work W and the amount of work W may be evaluated like this. So for this reversible heat pump to reject an amount of heat Q to this system, we can evaluate the amount of work that is required to accomplish that, that is equal to Q times 1 minus T naught over T. So we can envisage a reversible heat pump that operates between the ambient and the system which is at temperature greater than T naught and rejects an amount of heat Q to this system while receiving an amount of work W equal to Q times 1 minus T naught over T. Now from an exergy perspective, when this engine is supplied with an amount of work equal to W, it is also being supplied with an amount of exergy X equal to W. Since it is a perfectly reversible engine, there is no exergy destruction within the engine. This means that whatever exergy is supplied to the engine must be recovered. The same amount of exergy must be recovered and that is what is indicated here. So we may summarize this situation as follows. When a system at temperature greater than T naught receives an amount of heat Q, equivalently it receives an amount of exergy X equal to Q times 1 minus T naught over T. And the direction of heat transfer and exergy transfer are the same in these cases. When the temperature of the system that receives or supplies heat is greater than T naught in the ambient temperature. Now let us look at a system which is at temperature T less than T naught. Now let us say that this system receives an amount of heat Q. Now we envisage a reversible engine R which operates between the ambient which is at T naught and this system which is at a temperature less than T naught. It rejects an amount of heat Q to this system and produces an amount of work W which is equal to Q times T naught over T minus 1. Now from an exergy perspective, since this engine produces a certain amount of work W that means an amount of exergy X equal to W is recovered from the engine. Since the engine is reversible without any internal or external irreversibilities that is no exergy destruction consequently if we recover an amount of exergy X equal to W from this engine it must have been supplied with an amount of exergy X equal to W in the first place that is what is indicated here. So, let us summarize this in words. When a system which is at a temperature less than T naught receives an amount of heat Q it is equivalently supplying an amount of exergy X equal to Q times T naught over T minus 1. Notice that the direction of heat transfer and the direction of exergy transfer are opposite in this case. So, direction of heat transfer and direction of exergy transfer are opposite in this case. Next, same system at a temperature less than T naught let us say that it supplies an amount of heat Q. This is for instance what would happen in the case of a refrigerator. So, when the refrigerant enters the refrigerated compartment heat is picked up from the refrigerated compartment and remember the refrigerated compartment is at a temperature less than the ambient temperature. So, this is for instance what happens in the refrigerator. So, the system at a temperature less than T naught supplies an amount of heat Q. So, I can easily visualize or envision a refrigerator which operates between the ambient T naught and this system which is at a temperature less than T naught while being supplied with an amount of work W which is equal to Q times T naught over T minus 1. So, the refrigerator is supplied with an amount of work equal to W and it transfers an amount of heat Q from the reservoir which is at a temperature T naught and rejects an amount of heat Q naught to the ambient. In terms of exergy since the engine is supplied with an amount of work W we are also supplying the engine with an amount of exergy X equal to W and since the engine is reversible without any internal or external reversibilities there is no exergy destruction which means that whatever exergy is supplied to the engine must be recovered. So, that is what is shown here. So, an amount of exergy X equal to W is recovered from this engine. Now, you can see that the direction of heat transfer and the direction of exergy transfer are opposite to each other in this case also. So, when a reservoir or system at temperature T less than T naught supplies an amount of heat Q it is equivalently receiving an amount of exergy X equal to Q times T naught over T minus 1. So, in these two cases the direction of heat transfer and exergy transfer are opposite. So, when the system is at a temperature less than T naught the direction of heat transfer and exergy transfer are opposite when the system is at a temperature greater than T naught the direction of heat transfer and exergy transfer are the same. This is very, very important that is why we said that it is easy to figure out the amount of exergy transferred and the direction of exergy transfer in the case of work transfer and in the case of mass transfer. Whereas in the case of heat transfer the direction of exergy transfer and heat transfer are dependent upon whether the system that is receiving or supplying the heat is at a temperature greater than or less than the ambient temperature. So, this is very important later on in fact all these situations are important because we will encounter them when we look at thermodynamic cycles for power producing plants as well as power absorbing devices and when we do a second law analysis for these devices this is very, very important being able to figure out whether exergy is transferred to the system or exergy is recovered from the system is very important for doing exergy balance and then evaluating second law efficiency. Now, we will derive an expression for exergy change of a system as a result of both heat and work interactions. Remember if it is a system then mass interactions are not allowed because that will change the mass that is contained within the system. So, we will look at exergy change of a system as a result of heat and work interactions. So, let us start with the system. So, we apply first law to the system. So, delta E equal to Q minus W. So, delta E itself may be written like this and this is equal to Q minus W. Now, the W itself as was mentioned at the beginning of this lecture may be written as a sum of useful work plus the work that is done to displace the atmosphere which is not available as useful work. Now, we may also write for this system based on the principle of increase of entropy we may write delta S system plus delta S surroundings equal to sigma and delta S system is nothing but S2 minus S1. So, delta S system is S2 minus S1 and delta S surroundings is nothing but Q surroundings divided by T surroundings. And we know that our Q reservoir divided by T reservoir in case you know the heat is being supplied from a reservoir. So, if the heat is being exchanged with the surroundings then we may write as Q surroundings divided by T surroundings. In case heat is being supplied or rejected to a reservoir then it will become Q reservoir divided by T reservoir. And we know that Q surroundings or Q reservoir is equal to minus Q system. So, that is where this expression comes from. Now, if we multiply the second expression by T naught which is the ambient temperature and subtract from the first expression we may actually write it like this. So, we end up with an expression that looks like this. So, X2 minus X1 is a change in exergy of the system and that is due to heat interaction with a reservoir or surroundings and useful work that is done. Of course, there is a reduction or destruction of exergy to the amount of T0 times sigma. Now, we may interpret the different terms in this expression as follows. So, the left hand side represents exergy change. So, exergy change is due to exergy transfer minus exergy destroy. So, T0 times sigma is the exergy that is destroyed. So, exergy change is exergy transferred minus exergy destroyed. Exergy transferred itself is made of two terms. One is exergy transferred due to the heat interaction. The other one is exergy transferred due to work interaction. So, if a system does as I mentioned just before this, if a system does a certain amount of work then the exergy of the system changes by WU which is the useful component of the work interaction. So, if a system does an amount of work W then its exergy decreases by WU and conversely if a system does or if a system receives an amount of work W then its exergy increases by WU. So, that is what this term represents. Notice that in this case WU will have the appropriate sign positive if work is done by the system, negative if work is received by the system or done on the system. Similarly for Q also. So, that could be transfer of exergy from high temperature reservoir to the system or rejection of heat to the reservoir at a lower temperature. Now, depending upon the temperature of the reservoir, the sign of the exergy interaction will change as we discussed in detail. Now, if I rearrange this expression then we may write it like this. So, take the WU to the left hand side, we get WU equal to this combination of terms minus T0 times sigma. Notice that if the process is fully reversible meaning sigma is 0 no external or internal irreversibility then sigma is 0, WU is then the maximum value that is possible. So, we may identify the term within this curly bracket as WU reversible. So, to sum up the exergy change of a system due to heat and work interaction looks like this and T0 times sigma is the exergy that is destroyed. So, this is the most general case and the system has both heat as well as work interactions. Now, in the same manner we can develop an expression for rate of change of exergy of a control volume also. As a result of both heat and work interactions with the surroundings for the reservoir. So, for an unsteady flow process with multiple inlet and outlet streams we may write the rate of change of energy total energy within the control volume as q dot minus WX dot plus m dot times this plus minus m dot times this. So, this represents the outgoing stream this represents the incoming stream. So, this is the familiar unsteady energy equation this is the rate of change of entropy within the control volume. And as we can see this represents rate at which entropy is brought into the control volume is the rate at which entropy is leaving from the control volume. This is the rate of change of entropy of the control volume due to the heat interaction with the external reservoir or surroundings and of course, sigma dot is the rate at which entropy is generated during the process both internal as well as external. Now, in the same manner as before if we multiply the second expression by T0 and subtract from the first one we may obtain this expression which is nothing but the rate again expression for the rate of change of exergy within the control volume. So, dx Cv dt is the rate of change of exergy within the control volume it is due to x interactions with the surrounding which is nothing but work interaction and heat interaction plus of course, interaction with reservoir. And then we also have rate at which exergy is brought into the control volume rate at which exergy leaves the control volume and then rate at which exergy is destroyed within the control volume. So, the rate of change of exergy within the control volume is composed of these terms Wx dot term Q dot term and then rate at which exergy is brought in and brought out then rate at which exergy is destroyed. What is that? In combining these two terms we have made use of the fact that Q dot reservoir is equal to minus Q dot. So, we may identify different terms in this expression like this dx Cv dt is the rate of change of exergy and this is the rate at which exergy is supplied to the control volume and this is the rate at which exergy is recovered from the control volume. And these two terms combined together account for or contribute to change of exergy as a result of heat and work interaction with the surroundings or reservoir as the case may be this term represents the rate at which exergy is destroyed. Now, for a steady flow process the time derivative goes to 0 and we may rearrange this the right hand side of this expression just like what we did for a system. So, by taking the Wx dot to the left hand side. So, if you do that we are left with a nice expression like this and this Wx dot reaches a maximum value or attains a maximum value then sigma dot becomes equal to 0 that is a fully reversible process no internal or external reversibility. So, we can see that we can then identify the term within the curly bracket as Wx dot reversible just like we identified Wu reversible you may identify this as Wx dot reversible. Now, we can put all these things together you may recall that we started this module by stating that a different metric is required one apart from energy based metric is required for properly evaluating the performance of devices which are executing non-flow or flow processes. And we pointed out some shortcomings in the energy based definitions and we also pointed out that isentropic efficiency was very limited in its usefulness or utility. And so, we said that we will introduce a new concept or notion called exergy and we have developed that so far. Now, we will put it together and then define second law efficiency for different types of devices work producing device work absorbing device and other devices. And the nice thing about this definition is that it is equally applicable to flow as well as non-flow process no separate definition is required for flow or non-flow process. So, the second law efficiency for a work producing device is defined like this Wu which is a useful work you can say actual useful work although that is understood. So, Wu actual divided by Wu reversible. So, we have a device which produces a certain amount of work it could be let us say piston cylinder mechanism as the flow undergoes expansion it produces a certain amount of work. So, we can evaluate the actual amount of work perhaps from integral PDV and the reversible work you may recall is nothing but the change in exergy of the system. Remember exergy of the system decreases because it is producing work. So, change in exergy in this case we want a positive number. So, we write it as x1 minus x2 and you may also recall that the actual work is less than the ideal or reversible work by the amount of lost work which is nothing but T0 times sigma. So, Wu actual is equal to Wu reversible minus T0 times sigma. So, and this was the lost work. So, if we substitute this into this expression we end up with the following form for the second law efficiency of this particular device. Now, in case it is a steady flow device we will simply understand that Wu means Wx dot and that there is no work done in displacing the atmosphere in this case whereas when we are evaluating Wu actual for non-flow device we need to account for the work done in displacing the atmosphere. This will be demonstrated in the following examples. Now, in case it is a work absorbing device we simply move the reversible work up to the numerator. So, this is nothing but Wu reversible divided by Wu actual and following the same steps as this we may write this as 1 minus T0 sigma divided by x2 minus x1 plus T0 sigma. Now, in case of devices which are neither work producing nor work absorbing a very general form of second law efficiency may be defined like this. It is a 2 equal to exergy recovered divided by exergy supplied. So, we supply a certain amount of exergy to the device and we recover a certain amount of exergy from the device. The difference between the two of course is the exergy that is destroyed within the device and that may be used to define an efficiency for this device. In case of a steady flow device we may write this as sigma across the outlet stream of m dot times psi divided by sigma across the inlet streams for m dot times psi. Of course, we may also understand that exergy recovered is equal to exergy supplied minus exergy destroyed and that is what we have done here T0 times sigma dot is the exergy destroyed. So, we may rewrite this expression in this manner. So, this as you can see covers all the possible cases that we wanted to look at. Flow, non-flow, work producing, work absorbing, neither. So, we can actually calculate the performance metric for any device that we may come up with. That is the utility of second law efficiency. It is far more general and contains more insights because it takes into account entropy that is generated in the universe as a result of internal and external reversibilities. Now, let us see how we calculate second law efficiency for the examples that we worked out in the previous course where we actually did a first law analysis and we determined say for example, entropy generated, work interaction, heat interaction and so on. For the same example, now we will calculate a second law efficiency.