 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 W. 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 W. 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 a 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, X 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 at 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 its initial 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 this 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 this system supplies 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 recovering 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 T0, 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 T0 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 minus T0 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 T0 and this system, which is at a temperature greater than T0. 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 T0 over T. So, we can envisage a reversible heat pump that operates between the ambient and this system, which is at temperature greater than T0 and rejects an amount of heat Q to this system while receiving an amount of work W equal to Q times 1 minus T0 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 T0 receives an amount of heat Q, equivalently it receives an amount of exergy X equal to Q times 1 minus T0 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 T0, the ambient temperature. Now, let us look at a system which is at temperature T less than T0. 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 a T0 and this system which is at a temperature less than T0. It rejects an amount of heat Q to this system and produces an amount of work W which is equal to Q times T0 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, there 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 T0 receives an amount of heat Q, it is equivalently supplying an amount of exergy X equal to Q times T0 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 T0. 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 T0 supplies an amount of heat Q. So, I can easily visualize or envision a refrigerator which operates between the ambient T0 and this system which is at a temperature less than T0 while being supplied with an amount of work W which is equal to Q times T0 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 T0 and rejects an amount of heat Q0 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 irreversibilities, 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 T0 supplies an amount of heat Q, it is equivalently receiving an amount of exergy X equal to Q times T0 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 T0, the direction of heat transfer and exergy transfer are opposite. When the system is at a temperature greater than T0, 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 transfer 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 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 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 our work producing device is defined like this WU which is a useful work. We 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 the 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 irreversibilities. 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.