 can set up a few exercises and check it out for yourself. In cycles, I have a number of exercises and many of them are pretty heavy, there are 33 exercises. So, I do not expect you to do these immediately, but in the time that you have today evening and tomorrow morning if at every center there are at least 15-20 people, if they do a division of labor and at least go through each problem and see lay it out, do not do numerical that will take a lot of time, but lay it out and see that they are confident of proceeding with the numerical that will be a job well done. I will now take questions. So, I am first going to JNTU Hyderabad. So, Dr. Brahmara again, good evening and any questions from your side, over to you. Good evening sir. There is one question from the participant Mr. Prasad. I hand out the mic to him. Sir, good afternoon sir. We will consider isentropic process as a ideal process, but in case of a compressor, the work is less than isentropic process. The ideal work is less than in case of compressors, when compared to isothermal process and isentropic process, what is the reason sir? Can you explain it? Thank you, over and out. You said over and out. So, I assume that this is the only question that you had to ask. Anyway, the question asked is the work done in a compressor, when you do the compression isentropically is higher than the work done in the compressor, when you do it isothermally. But in my opinion that the two processes are not comparable, because the end states are different. If your requirement is to compress from say P naught ambient pressure, ambient temperature, compress the gas and if you want to compress gas also at the ambient temperature T naught, that means the initial and final states are isothermal. Now, you have a choice that either you do a direct isent isothermal compression in some way from state initial state to final state, in which case the compressor outlet is the state you want. In the second case, you do first an isentropic compression, that means you will be increasing the temperature of the fluid and then you will be cooling it back to the final state that you want. So, naturally if you have to do cooling, that means you have to extract energy and that means in the isentropic compression you have already added that much extra energy. So, this explains the difference, but you should not say that isentropic compression is more efficient than isentropic compression, because the cases are different. When you do isentropic compression, then if you want to go back to a situation where you want higher pressure, but the same initial temperature, then you have to do cooling. So, that means you are taking a round above truth and if you want that state, you go directly through isothermal compression, then you are directly approaching that state. So, the isothermal compression is a direct truth, it will naturally consume less power than the isentropic compression and cooling, which is a rather indirect truth and hence will consume. Now, let me go to some other center, KJ Somaia Mumbai, Somaia over to you. Hello sir, my question is, we normally talk about constant pressure gas turbine like based on Brayton cycle, but in some textbook I have seen constant volume gas turbine also. So, we normally does not talk about much about this constant volume gas turbine. So, please elaborate on this. Thank you, over to you sir. I would not come across an idea called the constant volume gas turbine because if you have a flowing system, in a flowing system it is very very difficult to create a constant volume process. So, this constant volume gas turbine is something which is may not be an anathema, but something which is odd at the idea of a gas turbine. There may be something called constant volume gas turbine, but I have not come across anything like that. Over. No sir, actually in classification they have mentioned, but they have not given any description about that. So, in the book of Professor Dhamkundwar, I have seen that. Over to you sir. Well, I have not seen that book and if we had just listed it, we can list 10 different things if we does not have to describe it. So, I do not want to comment anything more on that. Over to you. Any other question? Over and out. Vajchan called it sangly, over to you. Hello sir, my question is about Rankin with regeneration cycle. In that case the process 2 to 3 and 6 to 3, these are the mixing processes which should be highly irreversible, but we have shown with a continuous line, is that approximation is good or not? Over to you. Yes, from a purely thermodynamic point of view, showing 2 to 3 and that is the mixing process, you can only show by dotted line saying one stream goes in at 2, another stream goes in at I think 6 and the mixed stream comes out at 3. So, I think you are right, it should be shown only by dotted lines. Over to you. Sir, over and out. PSG Kohimbatur, over to you madam. Good evening sir. I have 2 questions to ask sir. First one is in which engine the Erikson cycle and dual combustion cycle is practiced and the next one is regarding the sterling cycle, that you said that sterling cycle is used for cryogenic refrigeration, whether it is used for liquefaction of gases to produce cryogenic fluids like liquid, oxygen, helium, hydrogen like. Over to you sir. I will answer the second question first. Yes, sterling cycle, if not the basic sterling cycle with minor modifications so that it can be implemented properly is used and is one of the workhorse cycles, that is the mainstream cycles for refrigeration and one of the mainstream cycles for liquefaction of gases. So, whether you have liquid nitrogen production or liquid oxygen production or liquid hydrogen or liquid helium production, a significant load of that work is done by implementing sterling cycles. So, what you have said is right, the sterling cycle is used for liquefaction of gases. The first question was that Erikson cycle and dual combustion cycle. The Erikson cycle by itself is hardly ever implemented in practice, but I showed you and I mentioned that you can have a Brayton cycle with reheat, intercooling and regeneration and I will leave it to you as a exercise to show that if instead of one reheat in which the turbine is split into two parts and you have one reheat in between. If you say that you split the turbine into a number of small parts and have small small reheats in between. Similarly, the compressor can be split into a number of small parts and have small small intercoolers in between and you have a regenerator. If you really implement this, it will be a complex scheme, but if you really implement this, then you are coming somewhere near the Erikson cycle implementation. It becomes very complicated and that is why an Erikson cycle is usually or why it is hardly ever used in practice. About the dual combustion cycle, remember that the constant volume combustion in an auto cycle, the constant pressure combustion in a diesel cycle and the dual combustion cycle where partly you have at constant volume, first you have at constant volume and then you have at constant pressure. These are approximate models for what actually happens in an engine. In an engine whether it is petrol, whether it is diesel or whether it is any other type of IC engine, the combustion does not follow any of these idealizations. Depending on the situation, one can approximate for example, for light fuels like hydrogen, CNG, LPG up to petrol. We can say that the combustion is so fast that you can reasonably approximate as a constant volume process. For diesel whether it is a constant pressure or dual, that is a choice of detail, it is left to you. The actual thing is for look at an actual indicated diagram and you will find the actual thing is neither of the three, neither constant volume, neither constant pressure nor an appropriate mixture of the two. So, over to you. Any more questions? One more question, sir. We are using a representable diagram for representing compressor and turbine, sir. Is there any other specific representation for pump or any thermodynamic devices? Over to you, sir. There is no standard as such. We show an expanding trapezoid for a turbine. We show a reducing trapezoid for a compressor because that is the typical shape of a gas turbine or a steam turbine or a gas compressor. For a pump, usually you show that condenser, you have that sigma type of diagram. That also is a very common industrial practice. For pumps, either you show a circle with an arrow in it or circle in which the inlet pipe goes up to the center and the exit pipe, exhaust pipe comes as a tangent representing some sort of centrifugal pump. But since here we are not saying that it is a centrifugal pump necessarily, circle with an arrow is good enough for a pump. For heat exchangers, etcetera, there is nothing really very special. So long as you create a very simple symbol which is understandable, I think that is good enough. These are anyway schematic block diagrams. These are not actual drawings of the exit pipe. Over to you. Over, sir. Thank you. K. K. Wagnashik, over to you. Hello, sir. My question is on isobaric heating and cooling in open system. You have shown that isobaric heating and cooling happens only in open system. Why it is not in closed system? Isobaric heating and cooling. Over to you, sir. I did not say isobaric heating happens only in the open systems and not in a closed system. We have isobaric heating and cooling examples in many of our exercises in closed systems. I said that it is easy to implement isobaric heating and cooling in an open system. All that you have to do is let the fluid pass through a pipe or a number of pipes, provide reasonable area so that the velocity is low and the pressure drop is small because when a fluid is flowing through a pipe, there will be some pressure drop. So, see to it that the pressure drop is small so that we can still argue that our process is near isobaric, if not exactly isobaric. And then heat it up by burning something on the outside or making another hot fluid flow on the outside. Cooling cool it up by, you know, submerging that pipe in another cold fluid. That is all I said. Over to you. Hello, sir. Another question is when we extract steam at higher pressure, whether it results in reduction in entropy generation? Over to you, sir. I think you are talking about the extraction cycle. I have not done the detailed calculation, but notice that since the overall efficiency increases, I expect the total entropy production of the cycle to go down a bit, but I have not done detailed calculation. Maybe one should do it from that point of view and see exactly how does the overall entropy production behave. Maybe there are some contributions which will go down, some contributions may go up, but my feeling is overall the entropy production is likely to go down. Because if you have the same source and the same sink and if your overall efficiency is going up, at least grossly that means on an overall basis entropy production should be going down. Over to you. So, one more question. Can you please explain us in detail the throttling process which occurs in refrigeration or in expansion of steam? So, I am going to NIT-3C now. May I request every one of you including NIT-3C to restrict yourself to maybe one major question? So, over to you NIT-3C. So, good evening sir. My doubt is about temperature measurement by using constant volume gas thermometer. There is given that relation like theta is equal to 273.16 degree Kelvin into P by Pt by putting reference state as a triple point of water and he is calculating the temperature of the saturated liquid at one atmosphere for different triple point pressure references. He is plotted the theta values and he said that as the triple point pressure reference tends to zero the temperature is arriving at 373.15 degrees. He said means how should he is decreasing? Can you provide me that reference from which you are talking about? Over to you. Sir, from PK Nag, I have read this information sir that is about consistent volume gas thermometer. He said that as limit Pt tends to zero the theta is the approaches to 373.15 degree Kelvin. I think what Professor Nag is talking about is he showing the behavior of various constant volume gas thermometer. Remember that we talked only about the constant volume gas thermometer using ideal gas as a working fluid and when you use various real gases as a working fluid their behavior is not really ideal but their behavior tends to be like an ideal gas when the pressure goes down. So, you work those thermometers at various pressures lower and lower pressures and as you approach lower and lower pressures as you approach zero pressure that means very low pressures the behavior of any gas will be like that of an ideal gas and hence that ratio which is shown there approach the 273.16 Kelvin ratio. I think that should explain it. I do not have Professor Nag's book with me just now but if you really insist I will find it out from it is there in my office somewhere I will extract it and find out and we will discuss more about it. Over to you. Hello sir, one more question regarding work ratio. Sir you defined work ratio as the work consumed by the compressor or pump to that work produced by the turbine but I follow a book Unisanger in that book it is specifically defined for the gas turbine that it is called as the back work ratio. Sir tell me what we should use back work ratio or work ratio? Sir is it same or different? Over to you sir. It is the same you may be calling it as a back work ratio we simply call it work ratio it is a slightly different term it means the same thing. Let me go to NIT Calicut now. NIT Calicut over to you. Hello sir, my question is this cycle has maximum pressure compared to auto diesel and dual cycle. And my second question is why dual cycle is called as limited pressure cycle? Over to you. As I understand the first question was which cycle has the maximum pressure auto diesel or dual. Difficult to ask because in auto cycle the maximum pressure is near the end of combustion but the compression ratio is small. So end of compression the pressure is small but the instantaneous pressure at the end of combustion is pretty high. In case of diesel cycle the compression ratios are large of the order of 16 to 20. So the pressure at the end of compression is high but after that during combustion pressure may increase to some extent. So I think it will my guess is the two things are comparable but diesel is likely to be a bit higher but not much higher. And whether I think you can always find out high performance petrol engine as used in may be F1 racing cars where the maximum pressure may be higher than some other typical diesel cycle type of engine. But I think generally they will be comparable if at all diesel cycle may have a slightly higher pressure. Though I do not know what the second question was can you ask it again please over to you. Hello sir, in the first digital I asked in the reference to diesel engine also auto diesel and diesel cycle I asked it earlier. So compared to diesel and diesel it is having more pressure maximum pressure. See the pressures and parameters of each cycle depend on its application. And for Joule cycles, Brayton cycles the pressure ratios are in the pressures are much lower than what you get in auto and diesel cycles. But otherwise the cycles are not directly comparable to each other. The pressure ratios and maximum pressures are likely to be unrelated to each other. They will depend more on the application and the working fluid. Truba college Indore over to you. What is the need of Brayton cycle when we are Carnot cycle as an theoretical cycle over to you sir. I think I mentioned that that Carnot cycle is a theoretical cycle. It is very good for thermodynamic analysis but it is very bad for implementation. That is because the for a given size the work output will be pretty low. Even if you have everything ideal because the mean effective pressure will be pretty low. Second thing the work ratio is so horribly near one that I doubt that a pure Carnot cycle will ever be operated in practice. And that is why we are looking for alternative cycles and there are good alternatives you know Brayton, Otto, Rankine and all these cycles are very good alternatives to Carnot. Except that on paper they may not have the same efficiency as Carnot for the same temperature limits. The fact is that we are engineers we want to finally have an engine which works and which is not exorbitantly costly. So we will say that look with all due respects to Carnot cycle which we will chew apart in thermodynamics when it comes to actual practice. We will work with Brayton, Rankine and other cycle as needed over to you. So VNIT Nakpur over to you. Good afternoon sir. Our question is with reference to the Mollio diagram. We find that the isentropic expansion of 20 bar and 500 degree Celsius steam up to 0.1 bar and the isentropic expansion of 100 bar and 500 degree Celsius steam up to the same pressure that is 0.1 bar results into almost comparable enthalpy drop 20 bar, 500 degree Celsius into 0.1 bar and 100 bar and 500 C up to 0.1 bar is resulting into comparable enthalpy drop in kilojoule per kg. So why should we go for high pressure turbines? Over to you sir. You are only looking at the enthalpy drop in the turbine. You should also look at the enthalpy rise which you need to provide in the boiler. Calculate the efficiency of the cycle as well as the specific output and then decide what is good for a plant. And finally remember that going to a higher pressure means a thicker equipment because you have to have pressure vessels and pipelines that thick. But if it means more efficiency and more power output that is an attraction. Finally it is the economics of setting up and running the plant which decides and from that point of view today there are one optimal or near optimal turns out to be around 150 bar and roughly 570, 580 degree C and there is another rather costly optimum but that is at the supercritical pressures around the same but sorry temperatures around the same but pressures of roughly 250 bar. Attempts are being made to find out whether something like 350 bar, 750 degree C will be feasible but those are still design attempts we do not have a working plant at those conditions today. Over to you. Point of view who relates work output to the enthalpy drop he would certainly ask when 20 bar, 500 steam is giving the same enthalpy drop as 100 bar and 500 degree C S. Yeah student may ask that but ask him to do the complete cycle analysis and then discuss this further. Over. I think that is the end for us today.