 So, I am back and let us see what the classification is. All are power type. So, power cycle, power cycle, power cycle, power cycle. When it comes to internal combustion, external combustion, well, car engines we know is internal combustion, steam power plant we know is external combustion. Jet engine is internal combustion and steam locomotive external combustion. Gas cycle, vapor cycle, this is a gas cycle. Steam power plant, vapor cycle. Jet engine, gas cycle, steam locomotive, vapor cycle. Then next one, so we have taken care of this vapor. Now, multi-equipment or single equipment. Car engine is single equipment. The same cylinder piston arrangement, all processes take place. Steam power plant, multi-equipment. We have a boiler, we have a turbine, we have a condenser, we have a pump and we have many other minor components. Jet engine, multi-equipment. We have a compressor, we have a combustion chamber, we have a turbine. Although we do not see them as separate because the whole thing is covered in a very nice aerodynamic casing. Steam locomotive obviously multi-equipment. Finally, whether it is open cycle or closed cycle, this is open cycle because it takes in air and fuel and throws out exhaust gases and we let the nature worry about, let the nature worry about the, you know, reducing the exhaust gas temperature to its ambient value and taking care of any pollutants and other components which are not in the inlet but are present in the exhaust. And you would notice that anything which is internal combustion needs to be an open cycle because today we do not have a technology by which we can design something by which exhaust can be number one cooled. Maybe we can do that. Depressurized if needed, not a difficult thing but take care of all the contaminants. Reduce NOx to N2 and O2 as needed. Carbon dioxide converted back into carbon and oxygen separated or whatever is the fuel particles and oxygen separated. So, if a cycle is a power cycle which is essentially which is internal combustion, we can almost assert very confidently that it will have to be an open cycle. Steam power plant, closed cycle, jet engine, open cycle and steam locomotive is one of those rare things where you have a power cycle which is external combustion, which is a vapor cycle which is multi-equipment but happens to be an open cycle. That is because it does not have a condenser which is a part of the steam power plant. If being a mobile piece of equipment, we do not want to carry that huge bulky condenser on rails. So, what is done is fresh water is put in, it is pumped into the boiler. Boiler generates pressurized steam that drives the steam engine. You do not have a turbine, you have a reciprocating steam engine and at the exhaust of that is thrown out into the atmosphere. And the chugging of a steam engine is essentially the exhaust being used as a driver to take out smoke and flue gases from the combustion chamber in the boiler. So, you will notice that when a steam engine is stationary, there will be no chugging, the smoke will be sort of oozing out on its own from a heat transfer point of view, a natural circulation or a natural convection process. But the moment the engine starts, the wheel starts rotating, notice that the cylinder piston arrangement is directly connected to the wheel. There is no clutch, forget a gearbox in between. So, the moment the engine starts moving, the piston starts reciprocating, the wheel starts moving. At every exhaust, some part of steam is thrown out through the funnel and as it throws out, it creates some sort of a small ejection or a venturi effect. It sucks out some amount of flue gases from the boiler furnace, creating that chug. Every exhaust stroke is one chug of the steam engine. So, actually if you measure it, if you take a recording, you will notice that the chugging frequency and the rpm of the wheel which is connected to the reciprocation frequency, they match each other perfectly or almost perfectly. So, this is a good exercise because now here you have pieces of equipment which are power cycles of various kinds and various combinations. I do not think two of these are exactly the same. Car engine and jet engine, they differ because one is single equipment, one is multi-equipment. Then the steam power plant and steam locomotive differ because one is a closed cycle and one is an open cycle. So, there is at least, you take any one of these four, you will find that there is at least one type of classification which is different. Now, after this we come to the implementation of cycles. So, here the first thing which comes is why not the Carnot cycle. And actually, we cannot analyze this completely unless we encroach on number three performance parameters and one performance parameter which we already know will be added on to by another performance parameter. See the one performance parameter which we know is the efficiency and we know from our theory of thermodynamics is Carnot cycle has the highest efficiency. But remember that Carnot cycle has an highest efficiency only when the situation is like this. Source at a fixed temperature is sink which absorbs it also at a fixed temperature. Now, the sink is ambient environment. So, we can say that well more or less it is at a fixed temperature. However, natural sources which we have like products of combustion or even geothermal energy are quite often not of a fixed temperature. Consequently, if you, we cannot show it on a sketch like this. On a T s diagram as you absorb heat say this is T 1 or T source and this is T sink. T sinks remains unchanged, but as you absorb heat the source temperature itself reduces because it does not have a very large or in principle infinite capacity for absorb providing heat at a fixed temperature. You have flue gases coming out of a combustion chamber. The flue gases are at a high temperature, but the moment you start absorbing heat the temperature reduces. So, depending on the way the heat absorption takes place instead of uniform temperature you will typically have a rising temperature or a temperature reducing temperature. Typically, it is arranged to have a rising temperature. So, this will be the situation and we now agree that this is not exactly a Carnot cycle. Although we may implement it because in principle such a cycle will not be a reversible even though it is reversible it will not be a 2 T heat engine. A 2 T heat engine will have to be of this kind whereas the finite capacity of the heat source will clip the cycle from a corner like this. Similarly, if the sink does not have a uniform temperature it will get clipped in the sink itself. That is one reason why we will see that the Carnot cycle is not implemented in practice or it is difficult to implement in practice. Although we know that it has the highest efficiency, but that highest efficiency is for a fixed source temperature and a fixed sink temperature. Now, that is another characteristic and for which we will have to look at it this way. See if you look up a typical text book, a text book shows an engine like this and although our idea of an engine is this Q 1 heat absorption Q 2 heat rejection, W power output. The efficiency in many text books and we have the efficiency as W by Q 1 or W net by Q 1 if you want to be so specific about it. Many text books show this in the form of what is known as a Sankey diagram. The Sankey diagram is some sort of a flow diagram where the width of a channel pertains to the magnitude. For example, if you want to show traffic a typical highway network or a road network with some sort of a well laid out road network with some sort of a Sankey diagram. Let us say you have a highway connecting two points A and B, but there is a point C and some traffic from A goes to C. We have to determine and say the traffic is only from either A to B or A to C. What would be the width of the road? The width of the road would be equal to the amount of traffic assuming certain idealizations as uniform density and uniform velocity. So from A to B the thickness will be equal to the traffic density from A to B and A to C the thickness density will be the corresponding traffic. Now if the common part of the road that thickness will be that from A to B that from A to C. So part of the highway will be wider say 100 units, but say 80 units go to B. So the road from that junction to B will be only 80 units wide. There will be a small off take to C which will be only 20 units wide. So the Sankey diagram for the engine is quite often shown like this. This is supposed to be the engine. This is where the Q 1 comes in. This is the Q 2 which goes out and this is the W which is delivered and the width of this is if Q 1 is 1 unit, W is 0.3 units. Then this will be 1 unit wide, this will be 1 unit wide, this will be 0.3 units wide and this will be 0.7 units wide. So you get a feeling that the engine is something which converts part of this Q 1 into W and the relative thickness of this, this thickness in the numerator and this thickness in the denominator give the efficiency. So this is numerator, this is denominator and the efficiency would be represented by numerator divided by denominator. This is the Sankey diagram representation. However, this is too simplistic a representation and does not give a complete picture. Let us have another way of looking at it. We know that the engine absorbs the power let me turn it around. Engine absorbs say Q 1 amount of heat. Now inside the engine something happens. Finally, it delivers W, it rejects Q 2, but there are various processes and the only channel is not W. The thing produces certain amount of work which may I may call W plus. Part of it goes out as W, part of it gets recycled into the part of that whole recycled thing in some way gets exhausted as Q 2. So there is a W plus and then there is a W minus. If you look at the cycle, this is a very crude Sankey diagram. If you look at the cycle, we notice that W is made up of two components. Just the way we split the total heat absorption, net heat absorption into two components heat absorbed and heat rejected. So this will be made up of two components say a W plus and subtract that which is W minus. This gives us the W net. You take any power plant. For example, you take an IC engine. There are processes of expansion and there are processes of compression. The process of expansion creates a W plus. Process of compression requires a W minus. What is delivered finally is W plus minus W minus. If you look at the PV diagram of a reciprocating engine, I will not tell you what sort of an engine it is. I will just say that it works between some minimum volume and some maximum volume. Let us say that the cycle looks something like this. Now you have an area and let us say this is zero pressure. This is zero volume. Now you have an area which is shown. Let me show it with dark green. This area is the work done during expansion processes. So this represents W plus. Now let me show something in say orange. So this one is the work required to execute the compression processes. Now let me show something in say red and let me show it horizontally. So this represents the difference between W plus and W minus which is W net. This is something which we do not consider when we look at the efficiency or look at the thermodynamics of cycles. During second law we simply neglected this. We said let W net be the net output from the cycle. But we split Q net into two components which is a Q plus, the Q absorbed and Q minus which is the Q rejected. We define the efficiency as we defined it here, W net by Q 1. Now let us define another this thing. Let me call it simply W r work ratio and let us define this as W minus divided by W plus, magnitudes of both of these. So work required for doing the compression processes and divided by the work which is available through all the expansion processes. This figure will make it clear. Now you will notice that if I have a small orange part but comparatively large green part, then the W net is essentially dictated by W plus and the effect of W minus is small. So we want W r to be low as low as possible. In particular we should say that if W r tends to 1, we are into danger zone. Why? Look at it like this. Let us say that I have an engine which produces 10 kilowatt. Case A, it is possible that W plus is 12 kilowatt. W minus is 2 kilowatt. Giving you a work ratio of 2 by 12 is how much? 2 by 12 is 1 by 6. It is about 0.16. A small volume of W minus is a small work ratio. Consider case B where W plus is, I will take an extreme case, say 1000 kilowatt and W minus is 990 kilowatt. Giving you a work ratio which is almost equal to 1, 990 by 1000, 99 by 100, 0.99. Now so long as things work perfectly, it is okay. But remember that in actual practice, this W plus minus W minus is W net is not really true. The work which is produced during the expansion processes will have to be processed and fed back into the engine to provide that W minus to execute the compression process. For this we will need some mechanism. In case of an IC engine, the mechanism is within the engine. We have a flywheel, we have a crank connecting rod, the throw of the crank and the crankshaft itself. So all this mechanism along with the flywheel will store some part of that W plus. Make it available during the compression process. In case of a large steam plant, W 1 will be fed to the turbine, will be fed to the generator and then the electricity which is generated through may be a set of transformers will be fed to a motor which will run the feed pump, which will consume the power required for pumping or compression. In case of a gas turbine plant, sometimes for almost always the compressor and turbine are on the same shaft. So it is just a question of overcoming some friction in the bearing. But in either case, in any of these cases there is some overhead in converting the required fraction of W plus and making it available as W minus. And let us consider a small fraction. Let us say 1 percent of W 1 is consumed in this and to provide W 2, 1 percent extra is needed. So what happens? 12 kilowatt reduced by 1 percent. So reduce it by 0.12 kilowatt. You get this approximately now say 11.8 kilowatt. W minus instead of 2 kilowatt being provided or let me, let us say 5 percent that will be 10 percent that would make it clear. So 12 minus 1.2 I will have 10.8 kilowatt. Instead of 2, I will now need 2.2 kilowatt. That means my W net available is still 10.8 minus 2.2 that would make it equal to 10.8 kilowatt 8.6 kilowatt. Remember that 10 kilowatt because of this internal friction and other consumption has reduced to 8.6 kilowatt. Now see what happens here? We just do not have that buffer. Even if this reduces by 10 kilowatt, even if this reduces by 10 percent, this becomes 900 kilowatt. And this has to be increased. So this needs to be 990 plus almost 100 kilowatt. So this requires approximately 10.2 kilowatt. This implies that W net is just not available. It will stop working in the small amount of friction. I have used this and we are in trouble. And this just goes to indicate to us, impress on us that a work ratio which is here 1 is very dangerous. Work ratio which is much less than 1 or as low as possible is good. Now come to our Carnot cycle. Our typical view of the Carnot cycle is on the T s diagram. And we know that on the T s diagram, the Carnot cycle is 10. And if we have scaled it up from 0, then we know that the efficiency of the Carnot cycle is represented by Q 1, which is essentially T 1 into the data s. And W net is this. So the efficiency is 0. Looks very well. Unfortunately all text books, when it comes to sketching the Carnot cycle on the P V diagram, sketch it like this. May be for ease of explanation and ease of analysis. A large crooked one isentrope, another isentrope. And they show the four processes. You say 1, 2, 3, 4, 1, 2. The reality is not so. I would recommend that you take a Carnot cycle, make it work on air. And you will find that in reality for a reasonable pressure ratio, reasonable volume ratio, the Carnot cycle is very thin. So the, compared to your W plus, which is the area under the curve 1, 2, 3, which represents W plus, the area under the curve 1, 4, 3, which represents W minus, they are almost equal. Consequently, the network which is available is a very small fraction of the W plus. So for a typical Carnot cycle, you will find that we are nearer to this case than to this case. And this is the reason why we just do not ever attempt to implement the Carnot cycle in real practice. Now if not the Carnot cycle, are there modifications possible for the Carnot cycle, which would get rid of this disadvantage, while keeping the advantage that our T S diagram still looks something similar to this. And our efficiency would still be, if not exactly 1 minus T 2 by T 1, would be very near 1 minus T 2 by T 1. And this under the assumption that we will not have any issue with this. And the answer is yes, there are some attempts to replace Carnot cycle by similar cycles. And that brings us to the two additional cycles, which are of some theoretical importance. One of them is attempted to be implemented in practice with further modification. And this brings us to two cycles, which are known as Ericsson cycle and Sterling cycle. And at this stage, it is necessary for us to include some simple nomenclature with the students find it easy and to understand and remember. See a cycle is usually made up of a number of processes. Although we can sketch a cycle at some bubble on the PV or the T S diagram, in actual practice for ease of implementation, technology dictates that the cycle is made up of a series of processes of some more or less standard kind like constant volume, constant pressure, constant entropy, throttling and all such similar ones. Maybe there are some modifications of these. Consequently, although the cycle in principle is a loop, it is made up of certain identifiable processes, which are rather simpler than a general trace in state space. The Carnot cycle, we say, is made up of two isentropic and two isothermal. So, the first process, if you start from, say, the heat absorption process, first process would be isothermal heat absorption, then isentropic expansion, isothermal heat rejection and isentropic compression. So, we can say that the Carnot cycle is made up of isentropic process followed by an isothermal process followed by an isentropic process followed by an isothermal process followed by an isentropic process. Nothing about which way to write. I can even start with a S process, S T, so be a Carnot cycle. So, this is the heat absorption process, this is the heat rejection process and these two processes have reversible adiabatics or isentropic. So, there is no heat absorption or heat rejection. They just change the temperature of the system. In Ericsson cycle, what we do is, we maintain these two isothermal heat absorption and isothermal heat rejection process, but the two isentropic processes are replaced by either constant volume processes, sorry constant pressure process in case of the Ericsson cycle and constant volume processes in case of the Sterling cycle. So, the Ericsson cycle in our nomenclature in principle becomes T P T P cycle and the Sterling cycle with this nomenclature will become a T V T V cycle. If you show them on the T S diagram, they will look like this. Carnot cycle would be a rectangle. The Ericsson cycle because the isentropes have been replaced by constant pressure lines. So, this is a constant pressure line. This is another constant pressure line and for Sterling cycle, we will have two constant volume lines. This is a constant volume line. This also is a constant volume line. Now, notice one thing immediately that as specified, these will not be two T cycles because in a constant pressure process, if I have to increase the temperature of a gas, then naturally it will have to absorb it at temperatures other than T 1. Similarly, it will have to reject heat at temperatures other than T 2. Same thing is true here as well as here. So, these two processes of constant volume in the Sterling cycle, similarly the two processes of constant pressure in the Ericsson cycle would make these cycles of the non-two T kind. We get rid of that by providing a regenerator or an internal heat exchange device so that, say in the Ericsson cycle, the heat which is required to be rejected from state 2 to 3, the constant pressure cooling process is stored and is made available for heating the fluid from 4 to 1 in the constant pressure heat absorption process. So, whatever is the heat rejected during 2, 3 is made available during 4, 1 and as a result, your Q 1 is still from a uniform constant temperature at T 1 and Q 2 is still at uniform temperature T 2, but there is an internal heat exchange involved. In a similar fashion, the Sterling cycle is also in principle implemented as a two T cycle by providing for internal heat exchange. So, that whatever is the heat rejection during the constant volume cooling process from 2 to 3 is made available for absorbing heat during the constant volume heating process from 4 to 1. As a consequence with this regenerator, an ideal regenerator in either case, the Sterling cycle absorbs heat Q 1 at the fixed temperature T 1 source temperature and rejects heat Q 2 at the sink temperature. Consequently, if we have ideal heat exchangers and we are considering idealized cycle, the eta Ericsson will equal eta Carnot, which will in turn equal eta Sterling if T 1 and T 2 are fixed temperature for all three. Now, Ericsson cycles and Sterling cycles are modifications of the Carnot cycle. They continue to have the Carnot efficiency under the same idealized conditions, but require an ideal heat exchanger and to some extent, they allow us to mitigate the characteristic of a Carnot cycle that the work ratio is very nearly equal to 1. The disadvantage of Ericsson and Sterling cycles is you need an ideal heat exchanger, which is very difficult to have it in practice. And in spite of having an ideal heat exchanger, the work ratio does not significantly decrease below 1. It decreases, but does not significantly decrease below 1. And both these cycles, continue to have the disadvantage that heat absorption and heat rejection Q 1 and Q 2 have to be done at a uniform temperature, which is a bit difficult to achieve in practice, particularly when you use a gas as a working fluid. Between Sterling and Ericsson cycle, the Ericsson cycle is difficult to implement using a cylinder piston arrangement, because we have to execute the constant pressure processes. Whereas, in Sterling cycle, it is a bit less difficult to execute, because in a cylinder piston arrangement, the constant volume process can be reasonably well implemented by just fixing the piston at one place during the execution of that process. And because of that reason that the Sterling cycle is attempted to be implemented as an engine and has been implemented as an engine in a few cases. However, the Sterling cycle or the reverse Sterling cycle is the basis for a large number of our cryogenic machines. Sterling cycles and its modifications are the work horses when it comes to liquid nitrogen plant and liquid helium plant. So, after explaining all this, we should end by saying that the Sterling cycle of the two modifications is a candidate cycle for perhaps future engines at least a few of them, but it is already one of the main contenders for low temperature refrigeration cycle, the reverse version of the Sterling cycle. Now, after this discussion of cycles, let us look at certain performance parameters before we come to the technically significant cycle for power and refrigeration. Now, I am not perfectly sure whether to discuss the performance parameters at this stage, because some of the performance parameters require terminology and some understanding, some exposure of the real cycle. So, it is okay to in an actual teaching to implement the demonstration of performance parameters when you talk about the various cycles, but the disadvantage, a small disadvantage of that is suppose a performance parameter is defined when you explain the Rankine cycle, the students tend to assume that it pertains only to the Rankine cycle. So, one has to be careful saying that although I have defined it while defining or explaining the Rankine cycle, it is and may be applicable to a large number of other cycles. The first and foremost performance parameter is for engines the efficiency. Or what is known as the thermal efficiency and this is defined as the net power output divided by the net heat input, heat supplied or heat absorbed. This is based per unit time, but if you base it on a cycle, this can be W net divided by Q supplied. Either for one cycle or for one kg of working fluid some common basis. When it comes to refrigerators and heat pumps, let us go back to where we sketched the typical refrigerator or the heat pump. Notice that for engines we need the power output, this is what we want to deliver, what we want to or what we have to pay for is the heat absorbed in terms of fuel. So, the efficiency is defined as W divided by heat absorbed. Whereas for a refrigerator what we really need is the refrigeration effect, what we have to pay for is the power to run the refrigerator. So, significant ratio is not W by Q 1, but the refrigeration effect divided by power consumed. For a heat pump on the other side, it will be the heating effect divided by the power consumed. So, compared to efficiency, the refrigerators we have the coefficient of performance which is defined as the refrigeration effect divided by the power supplied. For the heat pumps, the coefficient of performance for heat pumps would be the heating effect divided by the power which is supplied or power which is consumed. If you want instead of supply to use net, that is perfectly okay, but go back to the figure and you should understand what we are defining and why we are defining. One should go back and perhaps right here that your efficiency for any engine is W net or W dot net divided by Q dot absorbed using the nomenclature in this figure. So, COP of a heat pump would be as shown in this figure and COP, this is for COP of a refrigerator, COP of a heat pump would be Q dot heat divided by. The next performance parameter which we talk about is and which we have already talked about is the work ratio. Coming back to the work ratio, this W minus divided by W plus and although it pertains to power cycles, it is also of significance to refrigeration and heat pump cycles. Because although we supply W net, these cycles will have an expander, it will have a compression device. Compression device will be the main consumer of power. If there is an expansion device which produces significant amount of power, we will have to have mechanism to absorb that power and make it available to the compressor. If the expansion device produces a very small amount of power, then that is good for us because we do not have to handle internally large amount of power transverse. It is time now for the tea break. So, we will break for tea.