 So, I am told we are back online, so what I am going to do is I want all of you to look up your exercise sheets, come to page 19 we have on that page the beginning of the exercise list for cycles. We have from C A 1 to C A 33 and I just browse through and tell you what is involved in each exercise. C A 1 is about the Carnot cycle and its analysis. We have to determine the work ratio, work done per cycle and even the mean effective pressure. You will realize that the volumetric compression ratio is 20 and the pressure at the inlet that is the lowest pressure is 1 bar since the volume reduces by a factor of 20. The temperature rises by a factor of 3 at the point where you have the highest temperature and the lowest volume the pressure will be 20 into 3 that is 60 times very high. So, the pressure ratio for the cycle will be from 1 bar to 60 bar, so ratio of 60 even then you will find the mean effective pressure is pretty low although the highest pressure is as high as 60 bar. Then C A 2 is the same thing to be done for the Sterling and the Ericsson cycle to see in what way they differ from the Carnot cycle. Well 3 and 4 are standard stuff, variation of Carnot efficiency and analysis of some Queer cycles. 5 is characteristic of a typical Brayton cycle, C A 6 is reverse Carnot, so this is a Carnot refrigerator cycle. Then C A 7 is a Joule cycle and 9 are Otto cycle, 10 are 11 are Diesel cycle and up to here I do not think I have any additional idea or additional term involved. But 12 is a Joule cycle that means a mixture of the auto characteristic and the diesel characteristic explain it to them on the PV diagram that part of the combustion is at constant volume or is modelled at constant volume first part, the second part is modelled as at constant pressure and 12 and 13 are on diesel cycle, a Joule cycle again you will notice that 14 is on Otto cycle and petrol engine, 15 is again on diesel engines slightly more involved problems. Now we come to modifications, now I always find it an excellent idea to introduce modifications using exercises, so that the effect is immediately seen. For example 16 is Brayton the reheat modification is introduced here, now 17 is a gas turbine that means essentially Brayton cycle and the modifications introduced in this and compared with each other are reheat, intercooling, regeneration all together, 18 is similar to 17, but now component efficiencies are involved that is in the analysis now we will have to include eta turbine and eta compressor. These isentropic efficiencies are already introduced during the open system analysis, so this is the application of that and 19 is what I called clipped Brayton cycle where the purpose is not to produce power for output, but have a turbine sized exactly to provide the power for the compressor. So the turbine exhaust will be at a pressure higher than the ambient and the high pressure air at the exit of the turbine is expanded through a nozzle and the nozzle exit velocity provides an appropriate amount of thrust. So use the conservation of momentum equation across the jet engine to get the static thrust, this is one problem in the thermodynamics where conservation of momentum will have to be used. Now 20 is Rankine cycle and I think it uses only saturated steam. Now saturated steam cycles for fossil fuel based power plants are rather rare these days, but when it comes to nuclear power plants or solar thermal power plants they are not very uncommon. In fact for nuclear power plants saturated steam cycles are the default cycles used. 21 is the same type of Rankine cycle, but with superheated steam. 22 is based on a Rankine cycle using ammonia, properties of ammonia are given, so I think these are sufficient. 23 is also standard Rankine with some component efficiency, heat turbine is given. 24 is Rankine with reheat, so you can compare the characteristic with that of a cycle without a reheat. Intermediate pressure is given as 8 bar, explain reheat here, here explain intercooling and explain regeneration in braking cycle at the appropriate place and explain reheat in the Rankine cycle here. Then when you come to 25 you will have Rankine with a single stage of regeneration, what you call regenerative heat heating. We are doing a course on thermodynamics, so we do not have to worry about different types of reheaters and various combinations or trains of reheating, for that we go to the steam turbine course which is usually an elective. So just a case with one mixer type of reheater is good enough. Now 26 is, well the word nuclear is immaterial here, you can say it is a steam plant in which dry saturated steam is supplied to the turbine. Now the disadvantage of the dry saturated steam supplied to the turbine is that as the steam expands to a turbine the drainage fraction becomes lower and lower which is not good for the life of the turbine, the turbine blades will get eroded off in no time. So what is done is as it expands at various intermediate stages it is taken out of the turbine put through a separator, the separator will separate it into dry saturated steam and saturated liquid. The dry saturated steam is then sent to the further turbine, the next pressure turbine, intermediate pressure or low pressure whereas the liquid is throttled to the condenser pressure and discharged into the condenser. So this is saturated steam Rankine cycle but with separators between turbines. If you sketch the TS diagram it will look like this, we have saturated steam. Let us say just one separation stage in between, I think in the exercise there are three stages of separation one at 20 bar, one at 5 bar and one at 1 bar. So let us say this is the condenser pressure, this is the boiler pressure, this is the boiler exit state. If I were to expand it completely I would end up with very wet steam at the condenser pressure. So what is done is I will show you just one stage, you expand it in between when you say you have reached a tolerable limit and then that intermediate pressure put it through a separator. If you want to know what a separator is we have already done an exercise come to open systems on page 18, exercise OS 24 contains both a separator and a mixer. So apply first law to the separator, we have all the three states known so the two mass flow rates can be determined, the amount of steam flowing out and amount of liquid flowing out. Let us say this is state 1, first turbine expands it to state 2, then the separator splits it into two part, a state 3 which is dry saturated steam at that intermediate pressure. Let us say P s is the intermediate pressure and another stream at 4 which is saturated liquid at that pressure. Now from 3 you have another turbine bringing it to state 5. So now notice that if this were not there the dryness fraction would have been as low as that for this state. Now we have been able to improve the dryness fraction increasing the efficiency of the second part of the turbine as well as the life. The liquid at P s has a low enthalpy and we are unable to make use of it. So either it is put in a field water heater but in this exercise it is said that it is throttled to the condenser pressure and is mixed with the condenser is completing the cycle. The other part of the cycle I have not shown. 27 is steam regeneration with 2 feed heaters, rank in 2 feed heaters. Coming to the last page, 28 we go back to refrigerator cycles here, refrigeration in air conditioning cycle. This is a joule cycle, there are some component efficiencies also involved. First an ideal cycle then the component efficiencies. Then 29 is the same cycle with the modification that we have an ideal regenerator and then a real regenerator. So at this stage we need to explain to the students how a regenerator works. This is the modification that we talk about. Then 30, 31, 32 and 33 vapor compression refrigeration point. One is with ammonia, another is with R134A, third one is also with R134A and the fourth one 33 is with R22. So we do not have in our steam tables the data of this, but the textbook which you use will have the data at the end in an appendix or you go to the net and you will find a reasonable set of T S diagrams, P H diagrams, H S diagrams as well as tabulated data. Select one and solve these exercises. But again as I said my aim in any course on thermodynamics is not to worry about the final answer, but to really emphasize that the process of obtaining a solution is perhaps more important than the actual numbers which form part of the solution. So that brings us to the end of discussion on cycles and essentially end of the presentation. So we have something like 10 minutes to T time. So I will take a few questions and in the last session between 4 and 5.30 I will be here and hopefully for most of the time professors Bandarkar and Puranik will also be here. So we will take questions. Tomorrow as I have announced earlier 9 to 10.30 would be the general discussion session. 10.30 to 1 will be test 2. 2 to 3.30 would be the final concluding discussion session essentially a feedback session and 4 to 5.30 will be the validatory function. A note to the coordinators, I am told that on moodle a feedback form will be available sometime tomorrow. The participants are expected to fill the feedback form any time from the time it is made available, 2 may be 3 or 4 days later. If they are not at your center after the course, they can go to their homes or their home institutions and fill up the questionnaire from there. The deadline will not be like a test 30 minutes or a few hours. It will be a few days, may be a week at least. That is what I am going to tell them. And if any center wants to have a local validatory function, you may arrange it from 5.30 pm onwards tomorrow. So that brings me to a few questions before I start. 1128 VNIT in Akpur, over to you. Sir, you said in the representation cycle, the throttling process, H1 is equal to H2. So how it is possible, sir? Because in case of that the H2, the temperature and pressure it will drop you. Right, the temperature drops, the first thing, throttling process. I think we have done an exercise on this. The throttling process is a process in an open system. Let us say this is the throttling device, either a capillary or an open wall. If you set up an open system across it, it is a one inlet and one exit. Say this is inlet, this is exit. One inlet and one exit open system, steady state, some m dot flows and the requirement is that P i and P e should be such that there is a significant pressure drop. The areas etcetera are handled such that delta E k and delta E p are negligible. If you look at the enthalpy differences that we look at, this delta E p which would be m, which would be actually g into h, where g is of the order of 10 meters per second per second and h will be of the order of a few meters per second, few meters. So, g h will be of the order of say 40, 50 joules per kilogram which is 0.05 kilo joules per kilogram. Our enthalpy differences are typically tens or hundreds, so sometimes even 1000 kilo joules per kilogram. So delta E p is usually negligible and also except in case of nozzle delta E k is also pretty small compared to delta h. So these are good assumptions. Now with this and the fact that there is no work transfer, there is no heat transfer, just not attempted to extract work and it is reasonably well insulated. So q dot is 0 and so is w dot s. You will notice that the first law of thermodynamics in steady state when you apply to this, you will end up with h e equal to h i. That is what happens in a typical throttling process. And throttling process are commonly applied in refrigeration air conditioning plants of the vapor compression type and they are also used for measuring wetness in steam or moisture of steam using a throttling calorimeter or their its modifications over to you. Why we cannot calculate the efficiency in case of heat pumps? See when it comes to refrigerators and heat pumps, we do not use anything called efficiency at all. We use a parameter known as coefficient of performance and I have defined the coefficient of performance today morning. I think I have it somewhere here in the very beginning. Efficiency is defined only for engines. When you have a refrigerator, coefficient of performance is defined as refrigeration effect divided by power consumed. And if it is a heat pump, our interest is in the heat supply. So we define the coefficient of performance as the rate at which heat is supplied to the hot place divided by the power consumed for running the heat pump over to you. Hello sir. Go ahead. Why the compression ratio in case of diesel cycle is more than that of the Otto cycle. First thing is in case of Otto cycles, you have in actual practice, you have a pre-mixed air-fuel mixture, air-petrol vapor. Now if you compress it by a larger compression ratio, the temperature will rise and it will start, the combustion process will start even before or may start even before the piston reaches the top dead center. We do not want that to happen. Hence the compression ratio is restricted to roughly 10. That is known as pre-ignition and that is not a good thing to happen in an auto engine. Whereas in case of diesel engines, only air is being compressed and we want to reach a temperature high enough where the diesel will get heated, evaporate, mix and then start burning on its own because of the high temperature. Hence we want a compression ratio which is larger and typical compression ratio is around 15, 16, 18. I think 14 to 22 is the range you will find in different diesel engines. Thank you. I think I will take one more center. 1077 Federal Institute of Science and Technology Aernaculum, over to you. Good evening sir. Am I audible sir? Yes, go ahead. Sir, my doubt is regarding the second low problem, SL5. SL5. Sir, you have already solved that problem in the discussion session but my doubt is how to calculate the entropy change in that problem. How to calculate the entropy change in SL5? Remember I will only show the process diagram. Initial pressure is P0. Final pressure we have shown is P0 by 2. Initial volume is initial specific volume is some 2 V0. Final specific volume will be half of that V0. The initial state is P0 T0. This is the initial state 1. The final state will be 2. Occupy is a larger volume and I am showing this just by means of a dotted line. What I have shown here? This dotted line is the isotherm. This is the process representation. The question is how do you calculate S2 minus S1? I think that is the question. Now remember that by definition this equals initial state to final state dQ by T for any reversible process. What the actual process is? We just do not have to worry about. Now set up a reversible process between state 1 and 2. Analyze that in detail and integrate dQ by T over it. Since it is an ideal gas with constant specific heats, actually constant specific heat does not come into picture. We can consider that because the initial state is at a temperature T0, the final state is also at a temperature T0. Let me take a thermal reservoir at temperature T0. Let the system be brought into contact with it across a ideal diathermic wall and I will slowly allow the system to expand from P1 to P2 from state 1 to state 2. As I do that I will execute a reversible isothermal process as shown by Green on this. So this is a reversible process used to evaluate dQ by T0. This is a reversible process used by T. This process has nothing to do with the process which has actually taken place. Sir, can we use the general expressions that we have already derived for the general? Yes, you can use those general expressions which is Cv ln T2 by T1 plus R ln T2 by T1 and all those. Any one of those equations you can use. Sir, for that derivation we have started from a simple compressible system and dQ is equal to dE plus dw reversible. Yes. But here it is not reversible. Then how we can use this equation for this particular? No, no, no by definition. This is the definition of data s. So you have to consider a reversible process for evaluation of data s and using an appropriate reversible process we have derived those general expressions for s2 minus s1 in terms of either temperature ratio or pressure ratio or volume ratio. If you use that general expression you will get here the entropy change to be R into T0 into mass of the system into logarithm of 2 because the final volume and initial volume have a ratio of 2 to each other. Over to you. Sir, I have one more doubt, sir. Yes. I have one more doubt, sir. During the throttling expansion of the vapor compression system entropy is increasing. Yes. Could you please explain that reason for the increase of entropy during that process in the molecular level? See I am not looking at the molecular level but even otherwise it is an irreversible process because fluid is flowing across a finite pressure difference without extracting any work from it. So it is an irreversible process even if I want to I cannot make it go the other way through a throttle valve from a low pressure fluid flowing out as high pressure fluid. So since it is an irreversible process the basic principle of thermodynamics tells us that we have no choice but to have the entropy at the exit higher than the entropy at the inlet. You can go through the property relations and show that it will likely to or it will happen but we do not really have to go into kinetic theory or molecular aspects for this. Over to you. Is it due to the viscosity of the fluid? What did you ask? Due to the viscosity of the fluid or the friction related with the flow. Yes. In a way it is related to friction with the flow. This is a particular example of fluid friction which is comparable to a solid friction between a brake and a brake liner. I think we will stop here now.