 So, the first example that we look at is a certain amount of air, 2 kg of air which is contained in a piston-cylinder assembly which executes a polytropic process with polytropic index N equal to 1.27, air is initially at 100 Kelvin, 300 Kelvin and is compressed to a final pressure of 800 kilopascal. So, we are asked to determine the second law efficiency for this process. So, we have a piston-cylinder mechanism which contains air initially at 100 Kelvin and the air is compressed. At the same time, heat transfer also takes place and we are asked to calculate a second law efficiency for this process. This example was already worked out earlier as I mentioned and the work on heat interactions were calculated to be minus 357.11 kilojoule and minus 116.06 kilojoule and the final temperature was calculated to be 467 Kelvin. Notice that this is a non-flow situation and the displacement work is non-zero in this case. So, the useful work may be evaluated as W sys equal to W u plus p0 times v2 minus v1 from which we may evaluate W useful as minus 217.61 so it is not minus 357.11 minus 357.11 kilojoule is the network interaction for the system. So, this is nothing but the sum of the useful work plus work that is done to displace the atmosphere so the useful work is only minus 217.61 out of the minus 357.11 kilojoules. The entropy generated during this process was determined to be plus 80.56 joule per Kelvin. This was also worked out in the module on entropy in the previous course. So, W u reversible in this case comes out to be minus 217.61 plus 300 times sigma so this 1000 is used to convert the joule here to kilojoules so this comes out to be minus 193.442 so the actual work that is required is minus 217.61 had the process being carried out reversibly the work that is that would have been required is minus 193.442 kilojoules so the second law efficiency W u reversible divided by W u actual comes out to be 89 percent. So, you can see the importance of the conceptor exergy and how it is used to evaluate efficiency for a non-flow process like this. So, in case you have a device and you are thinking of let us say compressing air using this sort of an arrangement you have an idea of what the second law efficiency for this process is. So, or alternatively if you are looking at several different choices or strategies for compressing air from let us say 100 kPa to 800 kPa an analysis like this will allow you to calculate the second law efficiency and figure out which one or select the best design from among the choices that you are looking at. Next example we have we have 5 kg of saturated R134A at minus 15 degree Celsius in a rigid vessel and this is actually stirred by means of a stirrer we draw this nice manner if I can. So, we have R134A inside this and a certain amount of work is transferred to this the R134A vapor is initially at minus 15 degree Celsius. So, you stir the vessel simultaneously heat transferred to the ambient also takes place. So, the ambient is given to be at 30 degree Celsius. So, heat transferred to the ambient also takes place we are asked to calculate the second law efficiency for this process. Notice that although this is a non-flow process since there is no displacement work in this case there is no work done in displacing the atmosphere. So, WU is equal to minus 500 kilojoules this is work that is supplied. So, this is minus 500 kilojoules and the entropy generated for this example was evaluated to be 1.6972 kilojoule per Kelvin. Therefore, WU reversible would be WU plus T0 times sigma this is a lost work. So, minus 500 plus 303 times 1.6962. So, this comes out to be plus 13.9486 plus 13.9486. One thing is that you need to keep in mind both in this example as well as the earlier example is that work is being supplied. So, this was our system. So, what is being supplied to this system? So, if I keep W as a negative number which is what we have done then when I calculate WU reversible notice that I use a plus sign for the lost work because this is negative work. And if I by using a plus sign the reversible work comes out to be less than the actual work as it should. You have to be very careful to keep track of the signs properly. If you do not like to do it this way if you say lost work means lost work you should always have a negative sign then you say that I put in 217.61 in the actual case and in the reversible case I put in less amount of work and the actual value is equal to 193.442. So, the amount of work that I put in the reversible case is less than the actual work by an amount equal to the lost work. So, you have to be very careful with the sign when work is being transferred to the system. So, in this case the system is simply this. So, WU reversible is plus 13.9486. So, you can see how highly irreversible this process is to accomplish a change of state from what we have given here to this final state until the content of the vessel reaches the ambient temperature. So, to go from this initial state to this final state we have put in an amount of work equal to 500 kilojoules heat transfer is also taking place from the system to the surroundings. Had we accomplished the same change in state from the same initial state to the same final state in a reversible manner we could have actually produced an amount of work equal to 13.9486 instead of supplying 500 kilojoules of work. So, the process is very highly irreversible. In fact, it is actually not meaningful to calculate second law efficiency for this process because the sign of the work itself changes. We have supplied minus 500 kilojoules of heat whereas had we carried out the process in a reversible manner we would have recovered or received 13.9486 kilojoules of heat. So, this shows how fundamentally important the notion of exergy and the notion of entropy generated in the universe is. Now, in case you have doubts about this value we can always cross check by doing the following. We can always since w u reversible is x1 minus x2 I can actually evaluate since I know the initial state I know the final state and this is a non-flow process but with displacement work equal to 0 I can actually evaluate the change in exergy of the system as it goes from 1 to 2 and indeed it decreases by an amount 13.947 which means the system would have actually produced an amount of work equal to 13.947. So, this example illustrates the importance of exergy and entropy generated in the universe or lost work. Entropy generated in the universe basically is lost work. Let us look at the next example which is also a non-flow process. So, here we have a piston cylinder mechanism like this. So, we have a piston which initially rests on stops like this. So, heat is supplied to this to the system. So, the system basically is nothing but the water that is in the piston cylinder assembly. So, heat is now supplied from a source at 320 degree Celsius until the temperature of the contents of the cylinder reach 320 degree Celsius. Determine the exergy destroyed and the second law efficiency for this process. So, initially we have the mass is given the volume is also given 250 liters. So, the specific volume may be evaluated. Now, corresponding to a pressure of 1 bar initially it is given that the system is at 1 bar. So, corresponding to a pressure of 1 bar you can easily establish that the initial state is a saturated mixture state. So, X1 is equal to 0.07314. Now, we may evaluate the specific internal energy and the specific entropy from this. Now, we denote the final state not as 2, but as 3. I will explain that in a minute now. So, the final state is superheated because the we know that heat is now supplied until the temperature reaches 300 degree 320 degree Celsius. We also know that a pressure of 3 bar is required to lift the piston which means that the final state the pressure of the contents of the cylinder is 3 bar and the temperature is 320 degree Celsius. So, the water in the cylinder undergoes a constant volume process which means as I supply heat to the cylinder the pressure increases from 1 to 3 bar. So, until the pressure reaches 3 bar the piston is not going to be lifted from the stop. So, it undergoes a constant volume process from 1 to 2 where the pressure is just 3 bar. After that the piston is lifted and it starts moving up. So, it undergoes a constant pressure process from that state to a final state where the temperature is 320 degree Celsius. So, there is an initial constant volume process followed by a constant pressure process. So, we denote the final state as state 3 and that is superheated at 3 bar 320 degree Celsius. So, we can evaluate the required properties specific volume specific internal energy and specific entropy like this. Now, if you apply first law to the system that we have remember this is our system. So, this is the initial system and as the process takes place this part of the system boundary deforms and keeps moving with the piston. Delta E equal to delta U equal to Q minus W there is no change in potential or kinetic energy of this of the system that we have defined. So, it is equal to Q minus W and so, we may expand and write the first law like this and if we substitute the numbers we get the amount of heat to be 5004.558 kilojoules. So, that is the amount of heat that is supplied. Now, the entropy generated during this process sigma is nothing but delta S system plus delta S surroundings. Delta S system is M times final minus initial specific entropy and delta S surroundings is the heat interaction of the surrounding divided by the temperature. In this case the surroundings is nothing but the reservoir at 320 degree Celsius. So, if I plug in the numbers after recognizing that Q source equal to minus Q I get the exergy destroyed T0 times sigma to be 1076.674. In fact, I get the entropy generated to be 3.613 kilojoule Kelvin and we get exergy destroyed to be T0 times sigma to be 1076.674. Now, this is a non-flow process and there is definitely displacement work as you can see from here. So, the piston moves up so, there is definitely displacement work and whenever there is displacement work there is also work that is required to displace the atmosphere. So, we take that into account when we evaluate actual work. So, the actual work, actual useful work is the total work which is this minus the work interaction with the atmosphere. So, the actual useful work comes out to be 312.68 kilojoules and the second law because it is a work producing process the second law efficiency for this process comes out to be like this 22.5 percent. As I can see it is a very, very inefficient process. Why is it so inefficient because we are supplying heat from a source which is at a much higher temperature than the system. So, heat is being supplied across a large temperature difference which is why the entropy generated is very high as a result of this high external irreversibility and the exergy destroyed is also quite high. In fact, the exergy destroyed is actually 3 times the useful work. So, this is not the way you would want to produce 312 kilojoules of work as to there are much better ways of doing that. That is what this example illustrates. The next 3 examples involve steady flow processes. And the last example that we will look at is actually that of a cyclic process. So, we have looked at 3 examples involving non-flow non-cyclic process. Then we will look at 3 examples involving steady flow processes. Then we will close this module by looking at an example of a cyclic process.