 we begin the first session along with me now I have Professor Bhai Chandra Puranik. So, he left a bit early yesterday he had some other meeting. So, now we are ready to take questions. Let me go to NMIT Bangalore 1105 NMIT I think this is the Nikte Minakshi College of Engineering Bangalore. Sir, can you explain three reservoir refrigeration system? I think you are talking about the three reservoir refrigeration system right, is that your question? Yes sir. See our standard thermodynamic understanding of refrigerator would be a cyclic device absorbed from the cold space. The required amount of heat Q1 rejects it to Q2 reject the required amount of heat Q2 to T2 which is supposed to be the ambient or slightly higher than that and we know from our second law of thermodynamics that some work is required to be done. Now the idea is we already have two reservoirs one at one and one at two. Let us assume that the work is obtained by running an engine between a high temperature reservoir T3 and the ambient T2. Suppose I run an engine which let us say absorbs Q3E from some high temperature reservoir this is greater than ambient rejects Q2E to the ambient reservoir at T2 and it is adjusted so that it produces a work W. We can link these two work transferred together and this essentially gives us the idea of a 3T refrigerator. If I consider these two together now then you will end up with a situation which looks like this you have a T3 which is higher than T ambient then you have a T2 which is nearly T ambient and you will have a T1 which is the cold space. Now you will have a cyclic device which will essentially have three interactions. It will absorb Q1 from the reservoir at T1 that is the refrigerated space. It will absorb Q3E at show near I am just combining from the hot reservoir and from the ambient reservoir to the ambient reservoir it will reject Q2 as well as Q2E so Q2 plus Q2E and this is a 3T refrigerator. When you apply second law to this the appropriate relation to apply is the Clausius inequality and the Clausius inequality applied to this would mean the heat absorbed divided by the temperature at which it is absorbed. So Q1 by T1 plus second one is Q3E T3 plus this is heat rejected so I write this as minus Q2 minus Q2E divided by T2. This has to be less than or at most equal to 0. This is the Clausius inequality which should be satisfied for such a refrigerator to work over to you. Now since you understand that this is a combination of an engine and a refrigerator consequently you can have a COP of sort for the refrigerator part you can have an efficiency for the engine part. So this is one situation because in principle there is no work required we do not have the standard COP but even then people define I think it is getting cluttered but let me try to go down. A COP based purely on thermal interactions can be defined as what we need to extract that is Q1 divided by what we need to provide as the driving energy source that is Q3E. This is a thermal COP. Notice that our normal efficiency or the normal coefficient of performance is the ratio between a heat flow rate and power absorption rate but this COP is purely thermal it is the ratio of two different heat flow rates over to you. Heat pump and reversible heat pump sir. No if you want it reversible then the equality should hold and then under that condition you can determine the ideal value of this thermal COP that will depend on T2, T3 and T1. So although our interactions in the definitions are only with T1 and T3 the ideal thermal COP will have T2 also involved in it. From this equation and the first law requirement that Q1 plus Q3E must equal Q2 plus Q2E you can eliminate the one which does not appear here that is Q2 and obtain an expression for the thermal COP in terms of T1, T2, T3. That is a half a page of algebra which you can do yourself over to you. Sir in property relation you have shown the area equivalence in PV diagram and ES diagram sir. Can you hold your mic slightly below I did not hear you properly. Okay sir in the property relation sir. Yes. As per the first law you told that area of PV diagram is equal to area of TS diagram but about the arrow mark how it will be sir or they are in the same arrow mark or something. See the area I think this is one of those major things which all of us should remember is this. First let us have a process in the PV diagram. If it is a PV diagram then a process on the PV diagram naturally represents a quasi-static process and the area under this diagram will be expansion work. If the corresponding process say this is state 1, this is state 2. If the corresponding representation of the same process on the TS diagram is something like this quasi-static process so you will have it as a continuous curve then the area under the curve is integral Tds but that is not equal to dq. This is that is not equal to the heat interaction because second law says that the heat interaction ds is greater than or equal to dq by T. So Tds will be greater than or equal to dq so integral of Tds will be greater than or equal to q that means even if you have a process which is adiabatic that is q equal to 0 it is likely that the area under this curve is positive. If it were a reversible process then naturally for a reversible process this will hold. Now extending this to a cycle if I have a cycle on the PV diagram and its corresponding depiction on the TS diagram the same thing will hold for a cycle this is the work done by the expansion mode through that cycle and this is integral cyclic integral of Tds for the cycle. So this will be by our relation greater than or equal to the total heat absorbed in the cycle assuming both are quasi-static. Once I show a continuous line the only assumption I need to make is quasi-static but if the cycle is reversible then and of course we are considering a fluid that means a simple compressible system then the only reversible mode of work is W expansion in that case this becomes W by second law equality this becomes q and since we know W work done in a cycle is the heat absorbed net heat absorbed in the cycle they have to be equal. Our Maxwell's relations are based on this equality and the mathematical properties of Jacobian over to you. PV diagram for the reciprocating compressor. PV diagram for the reciprocating compressor I think you should look up a text book because all I understand is the PV diagram of the reciprocating compressors can be idealized but under certain conditions it all it also depends on where the valves are set and whether there is a cutoff and all those are interesting diagram you should look at a books and if you want still more interesting diagrams PV diagrams look up the old steam engines book you will have many more interesting in your PV diagrams out there over to you. We have question. Good morning sir. Good morning. Sir can thermodynamics laws be applicable to our biological systems like human arts sir can we consider it as an open system. Yes, first thing is all biological system are necessarily open system because the we consume food we throw out waste stuff we perspire we breathe in and we breathe out. So all biological systems are open thermodynamic system and since thermodynamics is applicable wherever our standard laws of physics and chemistry are applicable. So they are applicable even to biological system having said that biological systems are very very complex systems. The structure is not as simple as just a collection of molecules as in a gas or liquid or a solid. So applying it to biological systems which are never in equilibrium and which are complex system is very difficult but yes thermodynamics is applicable to biological system. We may not have the ability to apply it as properly to biological systems as we can apply it to engines and refrigerators and fluids and maybe even combustion systems. If you really want to check out on this and discuss this there is an old book by I think Schrodinger on thermodynamics of biological systems. It is either one of those two famous people have written it either Heisenberg or Schrodinger. I have seen it in the library of the TIFR in Mumbai. Sir since human heart acts like a pump, can second law of thermodynamics be applicable sir? What is your question? Since human heart acts like a pump sir. Yeah okay the human heart I think in the discussion forum also there was a question. Yes the human heart is a pump but it is a biological pump. So if we the fluid mechanics of the human heart is very well known and using ultrasonic mapping the detailed volumetric efficiencies and all that have been experimentally determined that what they call ejection fraction and all that thing all pathologies do it absolutely routinely. But how is the energy converted from food particles or carbohydrates or whatever it is into that that is not yet fully perfectly assimilated. But yes it is a pump works like a proper pump and hence laws of thermodynamics are applicable to it. But remember that it is a biological system over to you. Can we call the heart is it a PMM1? No there is nothing in our real world which is PMM of any kind 1, 2 or whatever you care to define. The energy input for the heart? The energy input to the heart is the chemicals provided to it through blood and lymph indirectly through our food. It does not have to be a thermal pump or something which consumes normal work input over to you. Physical significance of cutoff ratio in the diesel engine sir. This is something to do with energy conversion we should not worry about it in thermodynamics. Thermodynamics we say give us the cutoff ratio give us the model for combustion process we analyze it further. The cutoff ratio will depend on the type of fuel the way the engine is cooled the way the engine is run. Engine technology beyond thermodynamics over to you. What is the third law of thermodynamics sir? We have not talked about any third law of thermodynamics so I will not say anything about it. Now I think I should now go to some other center the people here are asking me to switch. So thank you for this longest interaction. 1150 Godavath institutions, Atigre Kallapur over to you. Hello, go ahead. Good morning sir. Good morning. Sir in what situation Cp can be equal to Cv for a thermodynamic processor or a gas? First thing is Cp and Cv have nothing to do with a thermodynamic process. Cp and Cv are properties of a fluid and if you go to the exercises thing, if all of you do all the exercises properly you will not ask many of these questions. Go to exercise pr.3 in which you have to show that Cp minus Cv is function of only p, v and t of course in terms of partial derivatives. So anything on the right hand side if it turns out to be 0 there will be no difference between Cp and Cv. You can rewrite the right hand side in terms of the compressibility or the bulk modulus and you can show that if the compressibility is very low or if the bulk modulus is very high then Cp becomes approximately equal to Cv that is Cp minus Cv will turn out to be 0 and if you substitute the ideal gas equation of state on the right hand side you will get Cp minus Cv equal to r over to you. Yes sir, sir when we are considering the effective heat utilization by the insertion of the devices like economizer or any other device then how we are changing the thermodynamic cycle sir. In the sense when there is no economizer what is the thermodynamic cycle and with the insertion of effective heat utilization device that is economizer how will be the change in the thermodynamic cycle. See the on the on over the TS diagram for a Rankine cycle the economizer process is a part of the normal process from pump exit to boiler exit. It is the detail in the minor detail or detail in the way in which the heat is supplied through various areas in the boiler to the feed water which is getting heated and boiled. So our TS diagram does not change whether you have an economizer present or economizer not present over to you. Okay sir means we are not considering economizer into the. No if economizer is there the enthalpy rise in the economizer will be included in the our Q dot supplied. See we write for our boiler Q dot supplied is mass flow rate into H at boiler exit minus H at boiler inlet and this Q dot supplied includes everything that which is supplied in the economizer plus that which is supplied in the evaporator plus that which is supplied in the super heater. If you really look at the detail of the boiler as a thermodynamic device then perhaps we will model the boiler as partly economizer followed by the water balls followed by the drum followed by the super heater and in which case we will write individual energy balances or individual implications of first law and if necessary second law for each of these components. When we do the cycle analysis we do not go into the details of any one of the major components over to you. Sir one more question sir in ranking cycle we require condenser to condense the steam which is coming out from the turbine. Right. Sir can we send the water without the condenser also or to the boiler? Okay. You look at yesterday I mentioned in our cycle classification that ranking cycle is a vapour cycle, multi equipment cycle and although for large power plants it is invariably implemented as a closed cycle. When we have locomotives we do not want to carry that big condenser and it is cooling water or cooling air system around. So we have an open cycle we do not have a turbine there but we have an engine which does not exhaust the steam into a condenser. The steam is exhausted to the ambient. So the disadvantage is yes the cycle will work but then what you are exhausting from the turbine number one will have to come out at ambient pressure or a pressure very near the ambient. So if you condense it you can expand it to a still lower pressure half of ambient or even much lower than half of ambient depending on how well the condenser is designed and operated. So you will lose out on some power of the turbine but the feed pump also will have to now take fresh water which would be under ambient conditions. That is what the that is what is done in a steam locomotive. You have a water tank which is under ambient condition and the pump pumps it into the shell of the steam engine which is the boiler drum essentially over to you. Sir when we are considering the equation for entropy ds equal to dq by t and for a reversible process do not forget ds equals dq by t for a reversible process element. Do not just say ds equal to dq by t that is very dangerous because it is not correct ds is dq by t reversible otherwise if you just want to stop at dq by t you should say ds should be greater than or equal to dq by t over. And that t we are considering as what type of temperature sir atmospheric or final temperature. It is the temperature at the interface of the system at the boundary of the system across which that dq takes place. So see dq is an interaction two systems must be involved a receiving system which is the system under consideration and some donor system the source of that dq. So look at the since dq is an interaction it has to cross the boundary of the system. So find out the location on the boundary where that dq takes place find out what its temperature when the temperature is or was when the dq takes place and that is the temperature which goes in the denominator over to you. Thank you sir. Over to you sir. Over and out. There are a number of hands raised. So what I am going to do now is at any place I think we will restrict to two or at most three questions right. 1006 NIT Tiruchirapairi over to you. Hello sir good morning. The first question is is there any relation between Gibbs function and Helmholtz function and the exergy? They look similar but there is no direct relation because Helmholtz function and Gibbs function they are defined purely in terms of the current state of the system. The state of the environment does not enter into their definitions at all whereas availability and exergy although they look similar in format to the Gibbs function or the Helmholtz function the temperature and the pressure there in availability and exergy is the ambient temperature ambient pressure. So they have nothing to do with Helmholtz function and Gibbs function. Over to you. So what are the significance of isothermal compressibility volume expansion? For example if you take mu J T joule Thompson coefficient if it is 0 the ideal gas expansion during expansion the temperature exchange of temperature is equal to 0 and if it is positive temperature will decrease if it is negative temperature will increase. So likewise any significance is there for isothermal compressibility and volume expansion? See there is nothing isothermal compressibility just tells us how much do you need to increase the pressure to reduce the volume by a given fraction that is at isothermal condition that is without changing the temperature of the system. So there are two compressibilities we define one is as you said isothermal compressibility partial of V with respect to pressure at constant T 1 over V and we put a negative sign to make it a positive number because all stable materials when pressure increases volume decreases. This is isothermal compressibility and the other one you have is isentropic compressibility. This is related to the speed of sound. This is not directly related to the speed of sound that was your question about compressibility what was the other part that you talked about over to you. That is volume expansivity. See the compressibility is considered variation with pressure volume expansivity is the expansion coefficient. So for solids we have a linear expansion coefficient similarly for fluids we have the volumetric expansion coefficient. So there we keep the pressure constant determine the change in volume relative change in volume for a specified change in pressure. So this is you say volume expansivity. We call it the expansion coefficient volumetric expansion coefficient whereas both these things are compressibility. Reciprocal of compressibility is bulk modulus. It is exactly the reciprocal of this. So you will have an isentropic bulk modulus and an isothermal bulk modulus over to you. If we refer let us say that there are n number of equation of state. For calculating thermodynamic properties of new fluid for example refrigerant mixtures all these things. How to choose a proper equation of state. Is there any procedure or we have to develop a new equation of state for that. No you do not have to develop a new equation every time. Actually nowadays I am sure computer programs are available which will fit your data to any type of equations. Usually with the so called virial equation of state or virial form of equation of state where you write PV by RT in the form of a infinite series in one or two variables typically temperature and pressure that type of an equation is used. And I am sure standard statistical packages are available for you to select the right number of terms and the coefficients of those terms. Over to you. Thank you sir. Let me go to some other sentence now. One to one five Bajaj group of institutions Akbarpur over to you. Good morning sir. Good morning. Sir I want to ask that what is the difference between the Rankine cycle and the cycle on which the modern steam power plant works. The basic cycle is the Rankine cycle but it gets modified with two modifications. The first modification is reheat the expansion through the turbine is not in one go. So the high pressure turbine expands it up to some intermediate pressure then it is taken again to the boiler where it is heated again at that intermediate pressure and then it goes through the other turbine or the low pressure turbine where it expands up to the condenser pressure. This is one modification. The second modification is that there is regenerative feed heating typically 5, 6, sometimes 7 rarely feed heaters which improve the efficiency significantly. The modern Rankine plant is basically a Rankine plant with super heat and with regenerative feed heating. There are no other modifications. There are other pieces of equipment in the cycle which take care of control, safety, start up, sudden change in load and such thing. But the basic cycle is the Rankine cycle with super heat and with reheat and regeneration. Over to you. Sir my next question is I have seen on TS diagram of any the isobars in subcooled region are the two consecutive isobars are near to each other. But when we come to the super heat region the same two consecutive isobars have large difference. Earlier I thought that it is a case of change in specific heat at high temperature. No. But if we assume it is an ideal gas then. See the your question is in the subcooled region the various isobars are very near each other on the TS diagram. And the reason for that is very simple is that in the subcooled region the liquid is almost incompressible. And because of that there is hardly any effect of pressure. So you at a certain temperature if you change the pressure the entropy varies very little. You can see it from some detailed steam table. And because of this all the isotherms are all cluttered together. So much so that if you really sketch or plot the TS diagram to scale. You know particularly at low pressures you just cannot make out any difference between the isobars. They almost seem to all cluttered up just to the left of the saturated liquid line. Sir minus question to Dr. Bandarkar. How enthalpy deviation lines done in the psychometric chart and what they are significant. So the question was about enthalpy lines on the psychometric chart. And I think I did explain that the enthalpy you know the simple formula that we use for enthalpy is just C p times the temperature that is the dry bulb temperature. And that is because we have taken the reference at 0 degrees. And plus the specific humidity multiplied by this h f g at 0 which is 2 phi 0 1 plus C p times temperature. So all you have to do really is take a particular edge and vary the T. And what you will realize is that as you vary the T that for a particular edge you will get different values of omega and thus you plot the constant enthalpy lines. Now as far as what is it significant it is only in air conditioning processes you need to know how much heat transfer you have to put into the system or take out of the system if you want to achieve your process. So if you are cooling you will need to remove energy from the system and if you are heating and this is in cold climates you will need to put in energy into the system. And it is a very easy way of doing it just look up the two states that you want. So you want a particular dry bulb temperature and specific humidity or relative humidity mark the points and see the difference in the enthalpy lines passing through those and you know what you should be putting in or taking out that is about it. Thank you. Sir my next question is to Gayatunde sir. Yes go ahead I am here. Dr. Gayatunde sir. Sir you told us earlier that W net in case of Carnot cycle is very low as W positive and W negative are approximately equal to each other. And we know that the efficiency of a cycle is W net upon the total amount of heat input. So sir if W net is approximately equal to 0 then efficiency is also approximately equal to 0 but we say that between two even thermal energy reservoirs. Yeah I know what you are getting at see I did not say W net for the Carnot cycle is negligible. I said W net for the Carnot cycle see W net for any cycle is made up of a W plus and a W minus. So W net is W plus minus W minus. All I have said is the ratio of W plus or W minus to W plus is almost equal to 1 for Carnot cycle. So the efficiency of the Carnot cycle does not change. Suppose you have T equals 1000 K here and T equal to say 500 K here. Then the Carnot cycle working between the two if it takes in say 100 joules from the 1000 K reservoir it will reject 50 joules from the to the 500 K reservoir and it will provide you a W net of 50 joules no doubt about it but this W net will be made up of a W plus of something like of the order of say 1000 joules and W minus will be of the order of 950 joules. Consequently a small amount of friction or any excess any extraneous requirement will kill this difference and the cycle will not be able to deliver any work output W net is pretty small compared to W plus. I did not say W net is pretty small compared to this is which is Q 1. The relation between Q 1 and W net is through the Carnot efficiency which depends only on T 1 and T 2. It is this ratio that I was talking about 1000 to 50. So 50 to 1000 instead of this we would definitely prefer something like 50 joules made up of W plus equal to say 60 joules and W minus it is say 10 joules. This is definitely preferable over to you and this is the last question because I have to go to some other section other center. Sir can the value of gamma be 0 and 1. What is the value of gamma be 0 or 1? Gamma if you consider it ratio of C p and C v I do not think it will go over to 0 because C p and C v are usually of the same magnitude. But yes it can go to 1 if you have very complex molecules in the geysers form. For example if you take saturated water vapor at low pressures gamma turns out to be approximately equal to 1 and for when we say that for incompressible fluids C p equals C v almost exactly. So for that gamma will be 1 over and now I think it is time for me to go to some other center because 1 0 7 7 Federal Institute Ernakulam over to you. Good morning sir. Good morning. My question is related to second law problem SL 12. Sir could you just draw the TV diagram of that process showing the saturation problem? Yes, SL 12. So we have a cylinder piston assembly in which there are two stops. When the piston is at the upper stop it is 3 meter cube. When the piston is at the lower stop it is 1 meter cube. That means the system volume will always be between 1 meter cube and 3 meter cube it cannot go outside this range. The second one is the piston mass and atmospheric pressure are such that the piston floats at 500 kilo Pascal. That means when the system pressure is 500 kilo Pascal the volume is ready to change at and adjust itself to an appropriate value between 1 meter cube and 3 meter cube and remember that thing is 500 kilo Pascal. So the PV diagram will be something like this PV this is 1 meter cube let us say this is 3 meter cube and the pressure at which it will float is in kilo Pascal 500. Now that means the system can take a state at 500 kilo Pascal anywhere on this line. If you try to increase the pressure above 500 it will float up but will stop at 3 meter cube. If you cool it and bring it below 500 it will go along this line. So the process if it is a quasi-static process will have to be on this solid line which is shown here. Now once you understand this it is initially 1 mega Pascal and 500 degree C. So 1 mega Pascal means 1000 kilo Pascal. So it will be pressure somewhere here. So the initial state will be something like this. Details you can calculate because you can determine the volume. Specific volume at 1 mega Pascal and 500 degree C is given. So using that you can determine the mass and then all other properties. It is allowed to cool to 100 degree C by rejecting the heat to the atmosphere at 30 degree C. So at 100 degree C find out from the process where the state will be and you will find that it comes here, it comes here and may be then it goes down here finally giving you a final state here. I am not sure but I think the details of this process are there on the coordinators moodle otherwise I will try to put them up there in a few days. Over to you. Sir could you show the process on the TV diagram along with the saturation line in the TV diagram, temperature volume diagram along with the saturation curve. For that I will have to determine the details and then show it otherwise there is no difficulty in showing it on the TV diagram but on the TV diagram it depends on the process and where the condensation takes place. Here also in TV the link will be between these two but we may not have a horizontal line in between because the horizontal line is because of this constant pressure behavior. Sir I think in this problem at the end of this first constant volume heat rejection the substance is at the wet region. Yes that is possible you will have to determine what state A is which is this intermediate state and what state B is which is this intermediate state. You should not assume that this is a condensation process condensation process may begin somewhere here and may continue up to here or up to here. It is not necessary that this constant pressure process is the condensation process even the condensation can be partly at constant volume and then remaining partly at constant pressure. Yes go ahead. Consider a wet steam. Consider a wet steam having a very low drainage fraction contained in a rigid container which has diatomic wall. A wet steam having a low drainage fraction contained in a rigid container with a diatomic wall. If we remove heat from this wet steam if we remove heat from this wet steam at constant volume that is a process at constant volume actually if we show the process on the TV diagram the drainage fraction is increasing. But during the heat rejection process usually we will expect decrease in the drainage fraction. Sorry it will be better if you mark the state of the steam on the TV diagram with a very low drainage fraction and showing the constant volume heat rejection process. First thing is you have a constant volume chamber fixed volume it has a diatomic wall. So some Q is possible otherwise it is rigid. So your W expansion is 0. So you have 0. I would expect that the total W is 0 because you are not mentioning any stirrer or electrical work. I hope that is the assumption. And your initial state here is liquid plus vapor X near 0. Is that your initial state? Very small drainage fraction over to you. No, okay. Low drainage fraction not a accurately saturated liquid. Is it saturated liquid or is it saturated liquid with some small amount of vapor? Sir small amount of vapor, small amount of vapor like X is equal to 0.1 or like. Okay, say X equal to say 0.1. Now you have it a Q across this and what is the process? The process will have to be a constant volume process. So if you plot it on the PV diagram and let us say this is the volume will become fixed the process can only be like this. So let us say this is your initial state, state 1. And if you are you going to supply heat to it or are you going to reject heat to it? Reject heat, make it reject heat. Is Q greater than 0 or Q less than 0? Q is negative. Q is negative. Okay. Then this is a simple process. This process will be analyzed that delta E is Q minus 0, Q minus W. W is 0 according to our specification. So M into assuming this to be equal to delta U. So this will be M into U2 minus U1 will be equal to Q. So if Q is negative, U2 will be negative. U2 will be less than U1 and then U2 and V2 equals V1 that will give you the final state. Calculate the final state and depending on whether it gets a higher trinus fraction or lower trinus fraction will be determined based on this. My doubt was that actually if the heat is rejected, if the Q is negative, we will expect a condensation but here the dinos... No, no, no. Wait, wait, wait, wait. Again there is an assumption here. When heat is rejected, there is condensation when the pressure is maintained constant. Here you are forcing the volume to be constant. So do not make a general statement that when heat is rejected, there is condensation. That is true only when the pressure is maintained constant. Here you are saying a rigid container so you are maintaining the volume constant. During this process, temperature is also not remaining constant, is it true? Yeah, the temperature will also not change. For example, on this PV diagram, if this is the initial pressure P1, the process may go like this in which case your trinus fraction will increase, then pressure will decrease but the final state is given by this. Rigid container so V2 equals V1 and the second property which determines the final state is U2 which comes out of the first law. Sir, this is a case of phase change with both change in temperature and pressure. Yes, you are right. Over to you. Thank you sir. Thank you very much.