 The next example reads like this, reversible heat engine receives 2000 kilojoules of heat from a source at 1000 Kelvin and 3000 kilojoules of heat from another source at 750 Kelvin. So if we were to draw a block diagram for this situation, we would do something like this. So this is one reservoir at 1000 Kelvin, this is another reservoir at 750 Kelvin. So we have a reversible engine. It receives 2000 kilojoules from here and it receives 3000 kilojoules from here. It rejects heat to the ambient at 300 Kelvin. All other processes are adiabatic, determine the work output of the engine and the overall thermal efficiency. So this is how we would have drawn the block diagram for this, there is nothing wrong with this block diagram. Now let us try to actually draw this on a PV diagram. Notice that the problem here appears to be made difficult by the fact that it is receiving heat from two reservoirs which are at different temperatures. Now since it is a reversible engine, we can actually think of it as a Corno engine working with two reservoirs. Remember all reversible engines operating between the same reservoirs have the same efficiency as the Corno engine operating between these two reservoirs. So we can actually visualize this as a Corno engine and try to sketch the PV diagram for the Corno engine like this. So let us say that this is, so we have, so let us say this is the isotherm corresponding to 1000. So let us say this is the isotherm corresponding to 750 and this is the isotherm corresponding to 300. So let us say that we start the Corno cycle from here. It receives 2000 kilojoules from the reservoir at 1000 Kelvin. So that means the process goes something like this. Now it has to get ready to receive 3000 kilojoules from the reservoir at 750 Kelvin and as we said before we now allow, we now remove the Corno engine from the reservoir and move it to an insulated stand and allow the gas to expand until its temperature reaches 750 Kelvin. So which would look something like this. So after 2000 kilojoules has been transferred here, we remove it from the reservoir, keep it on the insulated stand, allow it to expand until it reaches the temperature of the second reservoir. Now it is ready to receive 3000 kilojoules. So that means we go like this. So 3000 kilojoules are transferred. Once 3000 kilojoules of energy is transferred, we then remove the engine from the reservoir, put it on an insulated stand and allow it to expand until its temperature reaches 300 Kelvin which means it is going to look like this. Now we reject unknown amount of heat which is Qc until we go something like this and then it is compressed reversibly and adiabatically to the final state. This is what the cycle will look like. So this is basically a Corno engine receiving heat from two reservoirs. Now notice that I can actually decompose this into two Corno cycles and convert instead of having one engine I can actually convert this into two engines. So if I continue this line like this, this will be one Corno engine, this will be another Corno engine. That is also allowed because they will have the same efficiency, everything else is the same. That is also permitted. In fact that is how we are actually going to do this. So notice that we, so we have two engines one and two, we have split the engine into two which is allowed because they are operating between the same reservoirs. So engine 1 executes 1, 2, 3, X and 6, 1. So engine 1 executes 1, 2, 3, X, 6 and 1 and engine 2 executes 3, 4, 5, X, 3 and the block diagram for this looks like this. So what we drew as engine R in our original illustration is the device inside this box, grey box. So we have actually now filled the grey box and split it into two engines and this is what the layout looks like. Now we can proceed with the analysis. We are asked to calculate the heat rejected to the 300 Kelvin reservoir. That would be nothing but QC1 plus QC2. Now the analysis is very, very straight forward. So for each corner engine we may write this, for each corner engine we may write this and the heat rejected to the coal reservoir, total heat rejected is QC1 plus QC2. With a little bit of algebra substitution of known values we can get QC to be 1800 kilojoules. Now the work produced by the engine is the total heat received minus the heat rejected, QH1 plus QH2 minus QC that is from first law, cyclic integral delta W equal to cyclic integral delta Q. So if you substitute the numbers we get the W to be 3200 kilojoules and the efficiency of the engine to be 64 percent. So this is a very, very important concept that a single corner cycle can be broken into two corner cycles like this and we can then treat each one individually. In fact this forms the basis of a very important concept that we will discuss in the next module which is actually Clausius inequality. So basically the Clausius inequality states that any reversible cycle may be written as the sum of an infinite number of corner cycles. So here we have written it as the sum of two corner cycles, Clausius inequality generalizes that. So we will look at that and then derive that, derive the Clausius inequality in the next module. But for now you can see in a very simple case how that can be true that a single corner cycle can be broken up into two cycles like this and we can then proceed with the analysis in a very straightforward manner. This would have been impossible to analyze further because you know you are stuck there as if you write it in the manner in which we have written the analysis becomes easier. Most important thing is so now you understand that some of the theorems and corollaries that we proved earlier for instance the efficiency of any reverse all reversible cycles operating between the same reservoir is the same as that of a corner engine operating between the same reservoir is very very important because that forms the basis of this decomposition. We need not know what cycle this engine or executes. I can simply say that it is executing or I can simply say that it is equivalent to a corner engine which works like this. As long as I am given the fact that it is a reversible engine I can always assume that it is a corner engine there is nothing wrong with that. So far as I mentioned earlier all the reservoirs that we have looked at reservoirs with which the engines were having heat interactions. So all the reservoirs actually were assumed to be infinite in the sense that we could supply any amount of heat from the reservoir to the engine or we could reject any amount of heat to the reservoir and its temperature will not change. So that was something that we had assumed and we had proceeded with the analysis. In real life of course you realize that there is no such thing as an infinite reservoir. All reservoirs are finite and there will be a change in temperature if you remove heat from the reservoir or there will be an increase or if you remove heat from the reservoir or if you reject heat to the reservoir. If you remove heat then the temperature of the reservoir will decrease and if you reject heat then the temperature of the reservoir will increase. So we will now work out a couple of examples which illustrate how we can extend whatever we have done so far for infinite reservoirs to finite reservoirs. So here we are given that we have a block of mass m which is this one its specific heat capacity is given it is initially at a temperature of T1. Now a heat engine operates between the block and the ambient at a temperature of T0. Determine the maximum amount of work that can be extracted obviously when we want the maximum amount of work we must use or we must employ a reversible engine that is what we have shown here. But what is not given in the problem statement is whether T1 is less than T0 or T1 is greater than T0. In other words is the ambient being used as a low temperature reservoir or a high temperature reservoir. So here we are looking at a situation where T1 the initial temperature is greater than T0. Here we are looking at a situation where the initial temperature T1 is less than T0. Now in both cases the engine will develop work until the temperature of the block becomes equal to the ambient temperature. So here heat is being continuously removed from this reservoir which is finite in size. So its temperature keeps decreasing eventually it will reach the same temperature as the ambient temperature. So the engine will stop developing work. Similarly here heat is being rejected to the block and its temperature will continue to increase until its temperature reaches the ambient temperature at which point the engine will cease to operate. We have to consider both cases and we are asked to calculate the maximum amount of work that can be extracted. So we look at an intermediate instant during the operation of the engine. So during an intermediate instant let the temperature of the block be T. Now an infinitesimal amount of heat delta QH is transferred from the block to the engine. The engine delivers an infinitesimal amount of work delta W and rejects an infinitesimal amount of heat delta QC to the ambient. So this is what is illustrated here. So at an instant this is at a temperature of T and infinitesimal amount of heat delta QH is supplied from this reservoir. The engine produces an amount of work and rejects an infinitesimal amount of heat delta QC to the ambient. Since it is a reversible engine delta QH over T temperature at this instant is equal to delta QC over T naught. Delta QH over T is equal to delta QC over T naught which means delta QC is equal to T naught over T times delta QH. Now if we apply first law to the block during this process, during this process if I apply first law to the block I get dE change in total energy of the block is equal to du because there is no KE or PE change and this is equal to delta Q minus delta W which may be simplified to give delta Q equal to then delta W is equal to 0 because there is no displacement work there is no other form of work so delta W is equal to 0. So from which I may write delta Q equal to MC delta T. Now this is a negative quantity because the temperature of the block is decreasing which makes sense because heat is being supplied by the block to the engine. So if I consider the block as a system, heat is being removed so that means delta Q the sign for Q is negative. Hence heat received by the engine is delta QH equal to minus MC dt. So remember in our expressions for the engine delta QH this is considered to be a positive quantity. We are taking the sign into account when we write expressions of work and so on which is why we have written delta QH as minus MC dt as dt is negative. So then I may write delta QC equal to this which may then be integrated to give me this expression for QC. The total amount of heat supplied by the block from start to finish is MC times T1 minus T0 and the total QC is given by this expression. So the work that is developed by the engine is equal to this for a finite reservoir. So you can see that the concepts that we have developed for infinite reservoirs may also be extended and used in the case of finite reservoirs which are much more realistic because almost all reservoirs in nature are finite reservoirs, their temperatures will change. In fact when you do a course on applied thermodynamics you will probably learn that this expression here the maximum work that we can get from this block is actually called the exergy of the block. This is actually this quantity QH that the block supplies is called as the energy that is contained in the block whereas this work that we can get from this is called the exergy of the block. That is a very, very important concept which you will learn in your second level thermodynamics course or applied thermodynamics course. Now I leave it to you as an interesting exercise to try to draw the cyclic process undergone by the block from beginning to end. I leave it to you as an exercise to try, it is actually quite interesting to try and draw please try it. The second example involving a finite reservoir looks like this. In fact we also actually need to do this for the second case when we have done this only for the first case when T1 is greater than T0. I leave it to you as an exercise to do this for the second case when T1 is less than T0. So here T1 is greater than T0 that we have completed. So I leave it to you as an exercise to do the case when T1 is less than T0 it proceeds along similar lines. So notice that a positive amount of work is developed in this case and in this case even when the temperature of the block is less than that of the ambient. We can still run a heat engine like this which tells you that exergy of the block is positive and its temperature is different from that of the ambient temperature. Here its temperature is higher than that of the ambient here its temperature is lower than that of the ambient in both cases exergy is positive that is that is a very very important concept that you will learn in your next level course alright. So here is the second example. So here instead of a block that we saw before here we have a piston cylinder apparatus which is maintained at a constant pressure. So this is a constant pressure apparatus and here it is explicitly given that this is initially at a temperature T1 which is greater than T0. So some amount of heat is supplied by the constant pressure reservoir which is then used by the reversible engine to calculate I am sorry reversible engine to generate useful amount of work positive work and an amount of heat delta Qc is rejected to the ambient. So this of course is the illustration of this engine at an intermediate instant. So at an intermediate instant this is what the setup looks like because we are asked to calculate the maximum amount of work which is why we have taken it to be a reversible engine. So this problem is very similar to what we saw before except for this reservoir which is now a constant pressure piston cylinder mechanism. So we proceed in the same manner as before because the engine is reversible we can write delta Qc over delta Qc over T0 is equal to delta Qh over T from which we get this expression. In the same manner as before we apply first law to the gas inside the piston cylinder device. So DE is equal to DU no change in KE or PE and this is equal to delta Q minus delta W. The displacement work is non-zero in this case because the piston as work is being supplied the piston moves inwards so this is equal to PDV. So delta Q equal to DU plus PDV which may be written as DH is a change in incremental change in the total enthalpy of the system or the gas inside the piston cylinder device which may be written as mcp dt. And again dt is negative because the temperature is decreasing. So we write delta Qh the heat supply to the engine is minus mcp delta T so that this is a positive quantity. And we proceed in the same manner as before to get Qc equal to this and the work developed by the engine as being equal to this. So once again this quantity here the maximum work that we can develop this is called the exergy of the gas that was inside the cylinder at time t equal to 0. So initially the piston cylinder mechanism contains mkg of a perfect gas at a temperature T1. So the maximum work that it can develop is the exergy of that gas that is what we have written here. So this concludes our examples involving application of second law concept of absolute temperature direct and reverse engines, reversible engines and so on. Most importantly engines operating with infinite reservoirs and engines operating with finite reservoirs. What we will do in the next lecture is start our discussion on entropy.