 questions from NIT Kalikath and Varshan Sangli. Good morning NIT Kalikath, over to you. The question is under what conditions can one have just one reversible process between two states. I am just writing Pt out here but that does not mean much. You can have Pv, Pv, Pvt, Ts. The choice is yours. Actually, since you did not put any restriction on the interactions there will be any number of reversible processes joining the two states. But if you constrain the interactions then there will be some restrictions on the processes. More constraints, higher will be the restriction, lesser will be your choice. For example, if you say you have already said you want reversible processes. Suppose these are isothermal states and if you put a condition that it should also be adiabatic then the isentropic process remains the only process. That is a severe illustration I am providing and depending on the conditions you put you will find more and more restrictions. But it is also possible to put conditions in such a way that you just do not have a single process reversible or otherwise. For example, if you say if it turns out that the entropy at 1 is higher than the entropy at 2 then an adiabatic process will just not be able to take you from 1 to 2 reversible or otherwise. So, it is also possible to create situation in which no process exists otherwise there is nothing unique about a general reversible process between two states and that is why in our definition of entropy we have said that delta s by definition is integral of d q by t for any reversible process joining the initial state to the final state. I think that is what I can tell you just now. Over to you. Let me go to Chen College family. Valchen College, good morning. Over to you. The first question is Van der Waals constant preferably they are calculated in terms of P C and T C. Is it because V C is not easy to measure as compared with P C and T C. And the second question is to what extent it is necessary to define second law efficiency. Over to you sir. The first question was Van der Waals constant A and B. We saw yesterday that A and B can be computed using P C, V C and T C. And we also notice that there are two unknowns and we have three equations. So, one of the equations has to be deleted and it generally turns out that over a wide range of parameters it is P C and T C which are more precisely measurable than V C. And that is the reason why if you have all three values measured then you tend to determine A and B based on P C and T C. But finally, it all depends on a good measurement of P C will tell you P C is this much with an uncertainty so much. Similarly, T C is so much with an uncertainty associated with it and V C is so much with an uncertainty associated with it. Depending on the uncertainties you can now consequent uncertainties can be determined in A and B. And the pair which gives you the least amount of uncertainties in A and B will be the one which you will select. Today uncertainties in P C and T C are an order of magnitude lower than uncertainties in P C and hence we tend to use P C and T C. The second question was about the second law efficiency or what yesterday I called exergetic efficiency. Again I continue that at least in the power plants scheme of things. Second law efficiency is usually not talked about. No power plant is designed considering second law efficiency. Usually design is based on first law efficiencies and isentropic efficiencies. But I again qualified that I am not that much exposed to refrigeration, air conditioning, cryogenics maybe they look at second law efficiency is there. We have a professor Bapati here. So, you are free to target a question at him if you feel like over to you. Sir, another question is that the concept of lost work and exergy destruction. Are these two terms are synonymous or not? See as I explained yesterday that the terms availability, exergy, exergetic efficiency, lost work. See lost work is at least we can straight away come to it by saying W max minus W actual. But when you say loss of exergy then there are energy and energy. These are terms you know anybody can write a paper or write a book, write an article and define them. So, these are not absolutely standard thermodynamic terms. Yesterday we saw that even the definition of availability there are two options and even if you start looking the books and papers you will find on one side of the ocean what is availability and exergy on the other side of ocean it becomes exergy and availability. So, there is absolutely no standardization in this. So, just go do not go on the word find out the context and then you will understand what is being talked about over to you. Thank you sir, over now. It is be a night is Surat now. Good morning Surat, any questions from you over to you. Good morning sir. My question is what is the significance of compressibility factor and in which situation we need to consider it? Over to you sir. The definition of compressibility factor is a bit confusing usual symbol for compressibility factor is z and z is defined as p v by r t for any fluid. That means by definition the compressibility factor for any ideal gas will be 1. So, there are two things compressibility factor is used for. One is comparing compressibility factor for two systems at related pressures temperatures may be related reduced pressures temperatures gives you an idea how good they are how good they can be how well they can be approximated as an ideal gas at least from the p v t point of view in the state space. The second one is it is a dimensionless number. So, you can say that all the equation of state variables p v and t in a dimensionless way can be represented. Then remember that when we come to fluid mechanics and heat transfer we have dimensionless parameters. For example, a flow can be flow inside a duct can be characterized by apart from geometric parameters to flow parameters. One is Reynolds number and another is friction factor. So, the question arises that if you have a number of gases of a similar kind say refrigerants of the methane derivative kinds. Can we have a common dimensionless representation of these equations of state and for that we have what is known in thermodynamics as the principle of corresponding states. In fact, one idea of the principle of corresponding states is something which we have already come across indirectly it is mentioned here. Look at your exercise sheet the F 2 part and this is in with reference to the Van der Waals equation of state. Now the Van der Waals equation of state for a particular fluid which obeys that law will have three parameters a b and r. But then if we write in terms of reduced variables that is p r is p by p c and so on. We have done the exercise in F 2.5 where we have an equation only in terms of p r v r and t r a b and r they get hidden in those reduced pressure, reduced volume and reduced temperature. In thermodynamics we say reduced pressure etc. But we can use the nomenclature of fluid mechanics or heat transfer and say dimensionless pressure, dimensionless volume and dimensionless temperature. So, the equation in F 2.5 is the reduced equation and this says that if you have two Van der Waals fluids at the same reduced temperature and same reduced pressure they should have the same reduced volume. It is like saying that if you have a circular duct with fully developed flow if two situations have the same Reynolds number the two situations should have the same friction factor. Of course, the relative roughness also should be the same. So, that means on a dimensionless plane if the independent variables or control variables are the same the dependent variable should also be the same. That is the principle of corresponding states and using the principle of corresponding states the equation of state for a number of similar fluids can be put together. For example, if you take a large number of refrigerant types most of them will have a this is I think t r. I am not sure exactly at very low pressures it will be like this you will have a and for other pressures you will have this some funny lines like this is for as pressure goes on increasing higher values of p r. So, such a single chart in terms of z can be plotted. The single chart means it is not exactly single chart because the principle of corresponding states is approximately true it is not exactly true for real gases, but with a certain uncertainty you can always predict what the pressure volume temperature relationship for a similar class will be. For example, if you take methane CH4 then that will have similar characteristic on the reduced plane to say CH3Cl and CH2Cl2 and so on. Since many of our refrigerants at least from the chemistry point of view differ from each other only by you know replacing one atom by another atom their principle z plane looks similar and that is our principle of corresponding states in action. Remember that this is approximate and it is not exact. If say all of them were to use the same equation of state they all of them were to be perfectly van der Waals gases then you will have just one z chart. They are not all perfect van der Waals gases. So, the z chart will be fussy over to you. Thank you sir over and out. Good morning NIT 3G over to you. Good morning sir. Sir, I want to know that can we define can we do exergy analysis for the system where the input and output both are in the form of heat like boiler over to you sir. Yes, in principle you can do a combined first and second law analysis which most of us know as exergy analysis for a purely heat transfer device like a boiler and heat exchanger. However the standard exergy analysis out there will give you an idea of lost work and exergy expert would tell you that look if you were to make the boiler thermodynamically ideal or thermodynamically utopian then you would have extracted say 3 kilowatts or 10 kilowatts of power out of it. The boiler engineer will say how do I extract it and what do I do with it? Now what do I do with it is that person said but the you cannot have a boiler modified to extract power from it. So that is why I said yesterday that our standard the common form of exergy analysis although you can apply it to a boiler you do not really make sense of that lost work. So what we can do is remember that in second law the only quantity which is properly defined is the rate of entropy production for an open system in steady state. So it is that which one tries to reduce one does not really say that that much lost work is gained from a boiler. I think but the exergy analysis in the form of combined application of first and second law and looking at the entropy production rate or T naught sp is a good thing to do and can be done in the second law for various equipment over to you. In availability when you are deriving the energy exergy analysis in the availability you say that you put a word called readjusting the environment that is we execute the same process between same initial and final states we execute the problem during the during this we readjust the environment so that additional work obtained is equal to the lost work obtained. Can you please explain me the readjusting the environment is it possible or the user. I think one word was missed in our yesterday's derivation and that is we do not readjust the environment we said that when you have a system say delta E delta V delta S and we consider it doing some work and then we say that look we consider useful work as work done minus P naught delta V where P naught is the pressure of the environment. So that if the system changes by a volume delta V it has to do work against the environment which is equal to P naught delta V then we said heat interactions the heat interactions are split into two classes one is the heat interaction Q naught with the environment which remains at T naught the other interaction is at T and the heat transfer is Q now this is general in the sense that you could have Q 1 at T 1 you could have Q 2 at T 2 etcetera I am not writing the summation just to keep the format simple similarly you could have D Q at a T varying continuous T that can be integrated out. And now you apply first law to this the first law would be delta E is Q minus W so this is Q plus delta V delta V delta V delta V Q naught minus W U plus P naught delta V the second law becomes delta S is Q by T plus Q naught by T naught plus S P with the condition that S P must be greater than or 0 for real life processes S P will always be greater than 0 for thermally ideal processes S P will be 0. So the question is if I want to reduce S P what shall I do now remember that both these are now equations equalities. So if you want to reduce S P you have to change at least one of these three terms if you want to change S P you have to reduce you have to change this is not coming seem properly this is 0 you have to change at least one of these three terms. Now the standard exergy analysis assumes that you do not change these interactions Q you do not change the state end states of the system. So delta S delta P and delta V are not changed all that you change are Q naught. So the idea here is if you want to reduce S P well you have to increase Q naught at least algebraically and when you increase Q naught to make the balance here you have to increase W u. So a minimum of S P will lead to a maximum of W u that is the straight link and when I said that you change something in the environment the only allowed thing is change the heat transfer to or from environment as it is represented by Q naught the environment by itself is not changed it was at the temperature T naught and it remains at T naught only the interaction with the environment as represented by Q naught is allowed to be adjusted so that you can reduce the value of S P ideally to 0. I think that should explain it over to you. Sir one last question sir in the steam table on page number 21 the speed of the sound in dry air at 0 degree centigrade and what one atmosphere it is given but in the case of speed of light in vacuum it is not mentioned what is the pressure and temperature please over to you sir. Good question it is a I think it is a basic question the speed of sound in dry air is it is a speed of sound in a fluid and speed of sound we have not shown but it can be shown that the speed of sound the speed of sound C I think is represented or it can be shown to be partial of P C squared is partial of P with respect to density at constant entropy. This is the derivation for the speed of sound in any fluid you can obtain this is in terms of P rho can be written down as 1 over V. So, this comes in terms of P rho and entropy it is a good exercise to try to reduce a form which uses C P C V P V T and just partial of P with respect to V at constant T or partial of P with respect to rho at constant T that derivation you can do it is a recommended derivation for all of you and that will show you that for any fluid the speed of sound will vary with pressure and temperature. But the speed of light is a different thing speed of sound depends on the pressure waves transferring transmitting through the media whereas the speed of light is the electric movement of electromagnetic waves and even out there you will notice that it is mentioned very properly that it is in vacuum the vacuum does not have a pressure vacuum does not have a temperature a vacuum here and a vacuum anywhere else is the same. And that is why this number 2.998 into 10 raise to 8 meters per second is a fundamental physical constant that is not true for the velocity of sound for any fluid at any pressure or any temperature over to you. Thank you sir over to you V A 19 Akkur over to you. My question is related to property relation P R 11 describe an experiment that will use the relation to determine P 3 without any energy interaction. Yes I got your question I remember that because I think it was asked yesterday this is about P R 11 describe by an experiment that will use this relation to determine CP without any energy measurements. Now just now we have talked about the speed of sound yesterday I gave one answer that if you have a resonating gravity and if you can measure that variation small variations of pressure with temperature those that resonance or those variations of pressure and temperature are at constant entropy. So the left hand side of the equation in P R 11 can be determined but then again like a good professor I will I have given you one exercise in the previous whiteboard sketch. Now I will give you another exercise that rewrite the equation in P R 11 in terms of the velocity of sound and then you will be able to show that perhaps you can measure CP without any energy measurements if you measure the velocity of sound just check that but I will repeat what I said yesterday that you have a resonating acoustic cavity in that cavity the there will be small pressure changes with corresponding temperature changes and that perhaps is one of those few situations in a laboratory where you have almost exact isentropic process. So that can be used over to you. This question is from properties of fluid exercise first law 2 3.1 and e bit there is given 100 bar you have to find out the specific enthalpy volume x n specific entropy and e there is 100 bar and 200 degree centigrade. Sub cold or this question is about this is F 2.1 e this is about steam or water steam whatever and P is given to be P is 100 bar T is 200 degree C whenever you have pressure and temperature we go to the saturation table and first we determine either at 100 bar we calculate P or say T sat at 100 bar if you look up your tables at 100 bar T sat is 311.1 degree C you can also go to the first table but this means that the specified T which is 200 degree C is less than the saturation temperature at the same pressure and that means this is sub cooled liquid and you know our questions even with all you teachers have emphasized one fact to me I think in any textbook on thermodynamics providing this P T diagram of water critical point this is important. What we have is 200 degree C something here 100 bar at 100 bar the saturation temperature is 311.1 degree C the actual temperature given is 200 degree C. So, we are well in the sub cooled zone we are comparing these two now you can also go to table 1 fortunately I think in table 1 also you will see an entry pertaining to 200 degree C. So, you can look at 200 degree C so this was available from I do not think you see it there this is available from table 2 whereas if you take 200 and go to table 1 near the end of page 4 at 200 degree C you will notice that the saturation pressure is 15.54 bar 15.54 in bar. So, this is this information you got from table 2 this information you will get from table 1 and now this state is at a higher pressure than the saturation pressure. So, again it is sub cooled liquid. So, either you notice that the temperature is lower than the saturation temperature at the system pressure or the pressure system pressure is higher than the saturation temperature at the system temperature. But if you sketch this particular what I call the phase diagram pressure temperature diagram that is important and what I have learned during this interaction is that this diagram is very important unfortunately only books in physics and chemistry tend to show it books in engineering do not show it generally. I suppose that will satisfy your answer that will be your answer over to you. One more question I seen in calculations we go for heat balance sheet. So, the heat unaccounted for is it having some relation with the loss work loss that we discussed yesterday over to you sir. No the you write energy balance or energy balance chart they call heat balance chart but remember that there is no such thing as heat balance we have only energy balance properly we have the application of first law of thermodynamics. First thing is there are so many energy interactions across the boundaries of an IC engine it is an open system. So, we have fluids of some composition going in we have fluids at some other compositions coming out. So, we have to measure compositions we have to measure pressure we have to measure temperature plus there are thermal interactions with cooling water there are direct thermal interactions across the surface to the surroundings. And many of these cannot be accurately measured a few of these I am sure are not measured at all. For example, the surface of an IC engine is hot even if it is insulated there will be some heat loss but I do not think we ever measure it we only estimated. And because of this when we substitute everything in the first law the RHS does not become equal to the LHS. So, you end up with certain imbalance and that we say are unaccounted losses but this unaccounted losses we should not just leave it at that either we should be able to do a simple estimate of the unmeasured losses and see that they are of the same order as the unaccounted losses. And also suppose we measure everything our measurements are never perfect there will always be some errors or uncertainties associated with each and every measurement. And because of that even if you try to measure each and every energy flow it is possible that the LHS will not be equal to the RHS but the uncertainty analysis will tell you that the whether the you know the computed difference between LHS and RHS is consistent with the uncertainties of the measurement or not. If it is consistent then it is okay if it is not consistent then something is wrong somewhere either in a measurement or in the identification of a energy transfer link energy transfer stream over to you. So, good morning DS for isolated system is greater than or equal to 0. My question is what processes can or may occur in an isolated system leading to the increase in entropy over to you sir. See in an isolated system anything can happen for example we have illustrations here of if you look at a problem one problem is exercise 2.17 and another problem is exercise in the second law SL 5 so F 2.17 and SL 5 both these are isolated system in case of 2.17 page 7 you have two chambers rigid insulated vessel that itself tells you that if there is no stirrer work electrical work there is no mention of that. That means this is isolated here there are two chambers one containing one kg wet steam and the other one is a liquid steam. Because rigid insulated vessel that itself tells you that if there is no stirrer work electrical work there is no mention of that that means this is isolated here there are two chambers one containing one kg wet steam x equal to 0.4 bar the another containing 0.5 kg dry saturated steam at 2 bar and the partition is allowed to leak and finally you will have one single mass of 1.5 kg at some pressure and some state. So this is a mixing process which is taking place here but since the pressures are different there is some sort of a free expansion also as they mix. In SL 5 you have again an isolated chamber initially in two parts each part of volume V 0 here you have a gas ideal gas at P 0 T 0 this is vacuum and the partition is removed and you have to show that there is a change in entropy. So this is also a free expansion similarly if you think do not be under the impression that it has to be a free expansion consider a simple calorimetric experiment. You have maybe a solid system made up of copper or something some solid and let us say it is isolated. So no change in volume allowed no change in pressure allowed that means no interactions allowed and you say you have here one kg of a solid here you also have one kg of the solid and initially with a thin assume to be a thin partition but that just for imagination all that you can have is a solid at say one at 100 degree C the another at 0 degree C suddenly brought in thermal contact with each other and kept isolated from each other. There will be no work interaction but you have a temperature gradient temperature difference. So there will be a heat flow some simple modeling of this you can even analytically solve the heat flow equation in the heat transfer course but this is a process in an isolated system which naturally it is a natural process we know all three occur naturally so all three will be producing entropy since the systems are isolated the entropy change of the system will be positive. Sir the examples that you have quoted here two different masses are ultimately combining together but what about a system which is left to itself over to you sir. These are examples of systems left to itself if you want another situation you can start with a non equilibrium situation something which is stirred very heavily initially not in a state of equilibrium and left isolated eventually it will come to a state of equilibrium by internal dissipative forces may be viscosity or I mean viscous dissipation or thermal conduction or convection finally it will come to some equilibrium state but here because the initial state is a state of not in equilibrium we are unable to determine exactly what its entropy is all that we can say that because the system is isolated the energy of that initial non equilibrium state will be the energy of the final state and the entropy of the initial non equilibrium state will be lower than the entropy of the final equilibrium state. But what that entropy will be that just by looking at the gross system we will not be able to find out but if you consider that if you assume that you know the temperature variation you know the pressure variation and then you assume that although the initial state was a state of state not in equilibrium but partition into a number of small control volumes or small systems and each system was considered to be reasonably in local equilibrium then you can do a reasonably good estimate of the initial entropy over to you. Once the equilibrium is established there would be no rise further in the entropy never. Yes once an equilibrium is established unless there is a trigger or a stimulus there will be no change from that process it is considered a equilibrium states are considered states in so called stable equilibrium over to you. Over and out. NIT Calicut good morning over to you. Is there any difference between point of gas and real gas? Second question is how do we explain this violation of, tell you statement is a violation of quasi statement of the students. Thank you sir, over to you. The first question was on Van der Waals gas see for any real gas the equation of state is very very complicated. The most important real fluid for us or one of the most important real fluids for us is water and if you go through the website which I think I have mentioned on Moodle just search for IAPWS international association for properties of water and steam and you will find that the number of constants in that equation are of the order of 400 it is a very very complicated equation. So that is for real gases now for real gases which condense like water but for gases like air hydrogen in the state space near ambient pressure and ambient temperature they are far away from their critical point for such gases the deviation from ideal gas behavior is small but even then I will not say that the Van der Waals equation will fit them exactly but it is important because for gases like air hydrogen it gives a better fit than the ideal gas particularly when you go near their critical point does not exactly fit when you try to liquefy them when you try to liquefy them you have to use property tables. But I think when I discuss Van der Waals gas I said the importance of Van der Waals gas is not that it fits any particular gas exactly the importance of Van der Waals equation of state is that it is perhaps the simplest equation of state which gives you the possibility of a critical point and using a b and r you can relate the critical parameters p c v c and p c to the parameters a b and r that is about it. There are more complicated equations which also show the presence of a critical point but perhaps Van der Waals gas Van der Waals equation of state is the simplest equation which shows the critical point that is the only importance of it over to you. Sir, my second question is how do we explain this violation of, shall we explain that is the violation of gas to the students? Thanks. How to use it? Oh I forgot you had asked that question see this is an exercise which is done in many text books and you will find it is there I think in Moran, Shapiro, Nag you take any standard thermodynamics text book Kelvin Planck statement says that something like this you take q 1 from here and produce w equals q 1 is not possible the Clausius statement this is Kelvin Planck and this is Clausius. The Clausius statement says that you have two systems one at a higher temperature one at a lower temperature in fact that is the weakness of Clausius statement we have discussed that earlier. Then a heat flow a cyclic device which provides heat flow from a lower temperature to a higher temperature without any work input is not possible all that you do is assume that one of this is possible and the other is not possible. So, if you assume that Kelvin Planck is true but and Clausius is false then you can combine the two and show that either then you can show that Kelvin Planck is false or you can show that Clausius is true and similarly you can do the following if you assume Kelvin Planck to be false and Clausius to be true then you can show that Kelvin Planck is true or Clausius is false that means you started with certain premises and then you deduce that those premises are incorrect. So, that means there is some inconsistency one of the requirements of any logic and which we have been using in our geometry since our school day is this idea of reductio at absurdum start with some assumptions and show that those assumptions lead to a conclusion that one or more of the assumptions is wrong and that is absurdity reductio at absurdum and this mathematicians have shown us can be used to conclude that one of the assumptions is wrong that means either Kelvin Planck should be false with C false or Kelvin Planck is true C should also be true. So, this only proves that Kelvin Planck and the Clausius statements are equivalent to each other that means both must be true together or both must be false together it does not lead to a proof of either of them it just leads to a proof of their equivalence. So, this is not a proof of the second law the second law remains the fact that the Kelvin Planck statement is true that is the statement of second law and it is a premise there are a basic assumption in thermodynamics over to you. Thank you. Thank you very much sir. Good morning Dr. Brahmara over to you. The question is related to the combined first and second laws which is also referred in some textbooks as exergy analysis or qualitative analysis. Good morning sir. Regarding exergy analysis it is a qualitative analysis it can be applied to the power plants by calculating improving the performance of boiler turbine condenser under pump. Sir is there any scope rather than exergy method to improve the efficiency of the system and apart from that we are able to call it as a work class rather than work class we are able to call it as irreversibility. Thank you sir. The question is about combined first and second laws or availability and exergy analysis the question is it is not just if one should appreciate that in availability and exergy analysis there is nothing new except some definitions and recombination of the first law and the second law of thermodynamics. The basic truth in that is that for any real process the entropy production rate will be positive and from the second law we know that if things are reversible then engines will have the highest efficiency, heat exchangers will have no entropy production. So since it is thermodynamically ideal we tend to believe that that is what is to be achieved and if you do not do exergy analysis or availability analysis and just do things to reduce s dot p that would improve things technically for us but we have also seen that trying to improve that will not always make economic sense because the cost will be very high. So if you appreciate that then you can look at take a relook at your designs and see how to improve them to reduce s dot p thermodynamics just tells you that that if you want to do thermodynamically a better thing reduce the entropy production in your processes that is it you do not really have to do exergy analysis for it over to you. Thank you sir over to you. Good morning PSG Coimbatore over to you. Good morning sir we can understand what is a subcooled liquid and can you please say what is a compressed liquid and at which conditions a liquid can be considered as a compressed liquid over to you sir. Again let me plot the phase diagram for water triple point solid liquid vapor this is the liquid vapor saturation line liquid on this line is saturated liquid any liquid which is in this zone that is at a pressure higher than the saturation pressure or a temperature lower than the saturation temperature this zone including this zone can be called compressed liquid if you draw a dotted line at the so compressed liquid is from this converts like this. However below this dotted line whatever is can also be called subcooled liquid. So a subcooled liquid is always a compressed liquid but a compressed liquid which is at a pressure higher than the PC it is not proper to call it subcooled liquid because the idea of a subcooled liquid is if you keep the pressure same and increase the temperature at some stage you will reach the liquid vapor saturation point that does not happen if you are at the critical pressure or you are above the critical pressure. So I think that should explain to you over to you. Thank you sir over sir.