 We are back here and now as you noticed I have Professor Bhandarkar on my right hand side and Professor Poranik on my left hand side. So we are ready to take questions, first I am going to 1, 2, 5, 7, HCTM, 1, 2, 5, 7, HCTM, Kaital, Haryana, over to you. There is one question number, RAC. All the pH, all the psychrometric chart are drawn on a corresponding to the atmospheric pressure and to find the WBT, what method should be adopted? Similar in the case of PS1, how to find the WBT? There is one standard equation known as the carrier equation, in that we need saturation temperature corresponding to saturation pressure, but to find the value of WBT in all the cases when pressure is not atmospheric, over to you sir. I think in the carrier equation you notice that there is a PT term which is the total pressure, so if you just substitute the total pressure there that would be good enough and you will get, you just notice that the temperature has to be given in degrees Celsius, whereas all the pressure units have to be in consistent units. So if you give bar, then all the pressure should be in bar and you will get your answer also in bar. Thank you. So how to find this PV dash? Okay, so the question you are asking is that how to find PV dash? PV dash is nothing but the saturation pressure at the WBT. So if you are given the PV, you are actually supposed to get a value of WBT and the P saturation corresponding to that WBT will be available from the steam table. So as I said earlier if you use that equation you have to do trial and error, but PV dash is directly the saturation pressure corresponding to the WBT and that has nothing to do with the total pressure, you just have to look up your steam tables for it. Thank you. I have it out, how to find WBT? Neither WBT is known, we know the total pressure, we know the WBT, how to find the WBT? See even at one atmosphere, okay the question is that how will you find the WBT if the total pressure and the DBT is known? The answer is if only the total pressure and DBT are known, no one you cannot find a WBT because there are many WBTs possible which depend on PV the saturation the vapor pressure of water vapor in the atmosphere. So either you have to know PV or you have to be given specific humidity with which you will find PV or you have to be given relative humidity with again with which you will find PV. So in some or the other form you have to find PV either by using dew point temperature or relative humidity or specific humidity or PV should be given itself. Without that you cannot find the WBT because there is no unique WBT if a given total pressure and DBT are known. The third quantity regarding how much water vapor is there in the atmosphere has to be known. Thank you. It means WX is known to find the WB. Thank you sir, over and over. Thank you. 1113 HD Party College, Indapur, over to you. Good evening sir. Good evening. Hello. Good evening sir. Sir, my question is for the same capacity of the plant whether we can select external combustion engine or internal combustion engine for the same capacity or for the same power output. Yes, your question is how do you select an external combustion engine or an internal combustion engine? First thing is this is a major policy decision and the advantage of an external combustion engine is that the combustion takes place in a separate furnace or a separate combustion chamber. The working fluid does not get into contact with the combustion process at all. So if you have what are known as unclean fuels which produce lots of pollutants and lots of residues then naturally you would prefer external combustion because the working fluid in the cycle just does not come to know what had changed in the outside world. So I would prefer as far as possible for such fuels external combustion because internal combustion means you have to see to it that the not only air but the fuel and the air fuel mixture and the combustion products are compatible with the materials which are used. So the material used has to be compatible with this. So that is what I think I hinted in the morning that if you have internal combustion engine then engineering is that much more difficult because the materials have to be selected properly and the life of an internal combustion engine is usually lower than that of an external combustion engine mainly because the internal combustion engine various components come in direct contact with the gases undergoing combustion and also the exhaust products which are quite often corrosive in nature over to you. Sir one more question is there any loss of efficiency in case of external combustion engine as compared to IC engine? The question is whether there is a loss in efficiency. Loss in efficiency see the external combustion cycles and the internal combustion cycles are different but if you say that suppose I have a Brayton cycle and I implement it using air and internal combustion with a given fuel and implement another cycle with air but the heat exchange through a heat exchanger so convert it into an external thing. Yes the efficiency for a given air temperature that is combustion temperature will be higher for the internal combustion cycle lower for the external combustion cycle because the external combustion cycle will have to run at a lower temperature because of the data T required for heat transfer over to you. Thank you sir over and out sir. Thank you. One two zero five Mahatma Gandhi missions college Noida Uttar Pradesh over to you. Yes sir. Yes sir. I have three very small questions in case of heat engines you define the second law efficiency of the engine likewise can we not define the second law COP of the refrigerator? Your question was about the second law efficiency. I think in the definition of our efficiencies of engines we have not defined anything called the second law efficiency. We have just defined efficiency as a ratio of W dot net divided by Q dot supplied. So I do not think that is either based on first law or second law because it is just a ratio it is not a first law efficiency or second law efficiency over to you. Because in some books you know we are using that term in many books we are using that term second law efficiency of the engine likewise there is a second law COP of the refrigerator. I like the COP of the refrigerator ratio they are also defining that they are also defining that refrigerating efficiency. Okay but in proper thermodynamics these when it comes to engines and refrigerators the terms first law efficiency and second law efficiency are not standard definitions. So when some books uses it I will have to be very careful I will have to see what the definitions they have provided are and if there are any exercises I will have to solve those exercises using the definitions provided in that book over to you. The Otto cycle and diesel cycle that you have discussed are the air standard Otto cycle or diesel cycle? Yes the Otto cycle the diesel cycle and the Brayton cycle which I have described are what are what I have described is the standard representation and the standard analysis of these cycles. The word air standard is an old thing we do not have to really use the word air because the analysis assumes the working fluid not really an air but any gas which is which is assumed as ideal gas with constant specific heats will do. So the I do not really need to use the old terminology air standard cycle or air standard analysis I think the standard analysis is good enough because the same analysis is applicable in case of Brayton cycle if it works with say helium. So we would not then call it helium standard efficiency we will say it is a standard efficiency if helium is the working fluid over to you. Okay the next question you use the word potential then you derive the that expression for availability of flow system. So this potential is which potential whether you would like to call as a work potential or thermodynamic potential or thermal potential? No I have not defined any potential there all I have said is that the availability notice that if there is no heat transfer to any system other than the ambient then the maximum work that can be obtained is the decrease in that availability that is what we have shown. So at that stage I simply said that the availability can be considered at some potential the decrease in which gives you the maximum amount of work which is possible. Since what is extractable is work you can say it is work potential but I have not defined it as a potential I have said it can be considered at something similar to a potential because the idea in physics of a potential is a decrease in potential is the interaction typically work interaction over to you. Thank you thank you over to you. Thank you over and out 1054 national college of engineering Tirunel valley over to you. Hello sir. Good evening sir. Good evening. Sir I have a doubt in cycles sir is there any difference in standard analysis for diesel cycle and Brayton cycles because there are two cycles are under in the same classification of cycles sir. No the question is about comparison of diesel cycle and Brayton cycle at cycles themselves they are different. So although the procedure for standard analysis is similar the results will be different because the diesel cycle has one constant pressure process and one constant volume process apart from two isentropic processes whereas Brayton cycle has both processes constant pressure apart from two processes which are isentropic over to you. Hello professor. Yes. That yesterday's test the isentropic process may not be reversible one answer we have received and what are all the restricted case this is possible sir. There seems to be some confusion an isentropic process only means a ds is 0 and I think I have isentropic adiabatic and reversible in my at the end of my second law I have spent some time on comparing this and I wrote the following expression ds is dq by t plus dsp. First thing is ds is a change in property so the left hand side of this equation is an exact differential dq by t is a process dependent thing in exact differential in general because dq is an exact differential and since dq is in general exact differential ds will also be an exact differential because you cannot have two exact differential being equated to an exact differential. Now the second law says that dsp has to be greater than 0 or at best it can be equal to 0 dq we know a system can absorb it system can reject it so dq by t can be greater than 0 for an adiabatic process it can be 0 for some other process it can be less than 0 and ds is a change in property so it can be equal to 0 it can be greater than 0 it can be less than 0 we have the steam tables you select two states depending on what is initial and what is final you can have any ds any sign for ds. Now notice that isentropic only means that ds is 0 isentropic does not mean that dq is 0 or dsp is 0 for example you can have dsp positive and dq negative making the right hand side equal to 0 and balancing this equation so this goes to show that isentropic means isentropic it does not automatically mean that the process is adiabatic it also shows that isentropic process does not automatically mean that the process is reversible I think that should explain your query over to you. Sir one more question yes that auto cycle and diesel cycle basic doubt after end of the reaching the point number 4 last two processes expansion process when the exhaust valve gets open 4 directly goes to 5 then it is coming to 6 why 4 directly goes to 6 sir. See this is an idealized analysis so that is why we tend to split the cycle into ideal sub processes that is why during induction we did not take care or we assume no pressure drop as the air comes in for example similarly we have assume the compression and expansion processes in auto cycle and part of the second expansion process in the diesel cycle to be isentropic when it comes to 4 the pressure is higher than ambient and if at that stage two things will happen first thing is exhaust valve will open in actual practice it will not open suddenly it will open slowly start opening even before it the piston reaches TDC but in the ideal cycle we assume that it opens suddenly so that our expansion process can be smoothly extended up to the bottom dead center point 4 and then when it is open suddenly all that happens is if the internal pressure is higher than ambient sufficient air will rush out of the exhaust port to drop the pressure to 1 so that is the almost constant volume 4 to 1 pressure reduction process and then of course the piston moves from BDC to TDC and the idealization of this exhaust process where the fluid is pushed out is that we assume it to be at constant pressure neglecting any rise in pressure because of the flow resistance in the exhaust valve the exhaust pipe silencer convertor and all that over to you thank you professor over you and over and out thank you 1089 KLE college of engineering and technology Chikodi over to you good afternoon sir yes go ahead it is question related to the idealization of the cycle sir yes how this can you repeat this idealization concepts see when the cycle is based on a gas as a working fluid it is possible for us to idealize things and obtain or try to obtain algebraic relationships between interactions and then simple formula or not so complicated formula for efficiency and some other parameters this idealization requires two things we have to idealize the working fluid which we do by assuming it to be an ideal gas with constant specific heats so that the equation of state is simply PV equals RT and delta u can be written down as Cv delta T delta H can be written down as Cp delta T the second idealization which we have to do is about the processes in real life no process will be isentropic no process will really be adiabatic and nothing will be reversible similarly because the piston is continuously moving we will have hardly any constant volume process and hardly any constant pressure process so the idealization requires that we neglect pressure drop through ducts that is one of the idealizations we do second thing is compression and expansion processes which are adiabatic are assumed to be which are approximately adiabatic are assumed to be not only adiabatic but also reversible and hence isentropic the third one are the processes in which heat transfer takes place these are the combustion processes and the part of the exhaust process in IC engines in the combustion process depending on how fast the combustion takes place for example it takes place quickly in auto cycle petrol engine so we tend to model it or we do model it a constant volume combustion it is only a model and an approximation to provide us simple results algebraically similarly to begin with we model the combustion process in diesel cycle as a constant pressure process knowing fully well that in real life it is nowhere near constant process so the next level of approximation for a diesel cycle is a dual cycle for which you can do the analysis you will get a more complicated formula over to you. So my next question is to Puranik sir go ahead easier sir you have told regarding the purely convergent nozzle it is been told that when the pb by p0 is equal to 1 there will be no flow and p by p0 is equal to 1 so the back pressure reduces such that pb by p0 will increase or will be more than the critical pressure ratio so the my question is so when the back pressure reduces the area will increase as the area increases the velocity flow will decrease as the velocity flow will decrease the Mach number has to be less than 1 but in the second case the pb is the pb reduces to critical pressure and it is been shown that the Mach pressure is exactly equal to 1 so why it is like that. Your question is on the operation of purely convergent nozzle so there seems to be some confusion as far as what you understand by the back pressure and the role of the back pressure is so let me first clarify that we have a nozzle that is fixed in shape so at any particular actual location the area of cross section is fixed the largest area of cross section is at the inlet and it monotonically decreases to become the smallest area at the exit and that is why we call this purely convergent nozzle now the back pressure is actually the pressure of the reservoir into which this nozzle is discharging so the back pressure is actually external to the nozzle absolutely so what ends up happening is that if the back pressure is equal to your upstream stagnation pressure there is absolutely no driving potential for the flow and therefore the flow is 0 however as you start reducing the back pressure then slowly there is a delta p available for the flow to establish and to take place through the nozzle and as you increase the delta p by continuously decreasing the back pressure the flow the mass flow rate and the velocity and the Mach number etc will all keep on increasing this will happen until the back pressure is so low that the exit pressure is going to be equal to the critical pressure corresponding to whatever stagnation pressure you have at which point the Mach number in the exit plane becomes equal to 1 so what you need to remember is that the back pressure is external to the nozzle and in the operation characteristics as we describe those it is assumed that we have some control over this back pressure and we are in a position to change this back pressure according to our will hopefully once you understand that your your query would be clarified thank you so I have one more question for you sir in your slide page number 38 it is told that if m is equal to 1 dv by v is finite only if dA by A is equal to 0 but actually the area velocity relation is dv by v is equal to the ratio of dA by A to the m square minus 1 so if m is 1 and dA by dA is 0 dv by dv will be infinite why it is given that if m is equal to 1 dv by v is finite only if dA by A is 0 equal to 0 yeah so your question is related to the interpretation of the area velocity relation when it comes to Mach number of 1 so as you as you quoted the area velocity relation is dv over v is equal to dA over A divided by m squared minus 1 what I wanted to convey here is that if m is equal to 1 is a situation that is going to occur in your in your nozzle then because m is equal to 1 that denominator m squared minus 1 will become 0 however physically speaking something divided by 0 which would imply that things are going to be infinite is not going to be acceptable because the left hand side relates to the velocity which is a physical quantity so just because m is equal to 1 is going to be reached somewhere there is no reason to believe on any basic physical principle that the velocity can really shoot to infinity so this paradox is resolved essentially by by saying that if m is equal to 1 is going to be reached in order for dv over v which is a physical quantity to be finite we must come up to the conclusion that dA over A also has to be equal to 0 that the reason is because then the right hand side will become 0 over 0 sort of a situation which typically has a finite limit from calculus so it is a combination of a calculus concept and a physical interpretation associated with not having a physical quantity like velocity suddenly for no reason going to infinity hopefully these two concepts combined together is going to explain your answer thank you your question I am sorry thank you sir in case of sub cooled water region we are approximating that is a incompressible fluid is there a concept behind that sub cooled in case of sub cooled water sir yeah I think the question is addressed to me the question is question pertains to properties of sub cooled water and what I have presented is only an approximation to be used in case two things happen number one you do not have the data in your steam tables which would be the case if you are using the Mathur method steam tables and you are at reasonably low pressure and number two the pressure range that you are talking about is not very high and I mentioned a mentioned as an example that this is a reasonably ok approximation up to a pressure of roughly 50 bar nothing special about it now up to 50 bar the water sub cooled water or compressed liquid water can be shown to be reasonably incompressible not exactly because pressure does affect the density you can assume it to be reasonably incompressible and if it is reasonably incompressible for such a fluid you can show such a liquid you can show because fluid is almost always considered compressible whereas liquid you may consider incompressible. The three main thermodynamic properties volume internal energy and entropy will depend only on one parameter and that would be temperature or if you take the quartet temperature volume internal energy and entropy select one and all other things would become fixed volume goes out of contention because volume is invariant so that means there is a direct one to one relationship between temperature and internal energy and entropy. So if you want to determine the properties of the sub cooled liquid go to a stage where the pressure is different and but the temperature is same and one such state which is usually tabulated is the saturated liquid state. I have provided links to a more detailed steam table the NIST steam table it is available for free download I recommend you download it you need not print it but in the sub cooled liquid zone you will notice how good this approximation is because they have tabulated compressed water and superheated steam. Unfortunately they have not tabulated the internal energy you will have to calculate the internal energy from the values of enthalpy and specific volume and pressure but you can look at the entropy values and notice that they are reasonably independent of pressure remember it is an approximation so do not expect this to be true in an exact sense over to you. Please repeat the concept of dead state what we explained in the availability concept. There is nothing special about the concept of dead state the dead state is defined only as of course it was yesterday when a system is brought by some means to a pressure and temperature which is the ambient pressure and ambient temperature that is supposed to be a dead state of the system because if a system is at the ambient pressure and ambient temperature unless you do some interaction with the system like pressurizing it or heating it it is provide some sort of an interaction it will not move away from that state that is why it is called a dead state. So the dead state of a system means look at the ambient pressure ambient temperature and bring the system down to that pressure and that temperature and that is the dead state of the system over to you. Please repeat that the super saturations are yesterday. I will not repeat it because I do not think it is proper to discuss that in detail in the thermodynamics course. I recommend you look up the books by Moore and Seawarding you will get enough information there on super saturation particularly that in a nozzle and yesterday's raw videos I am told have already been uploaded the slides are also up there. So if you want to go back to that portion where I mentioned something about super saturation you can refer to that over to you. Thank you sir over and on. Thank you 1071 Amrita Viswavidyapetam Coimtore over to you. Sir good evening sir. So my question is about to problem number SL3 so I have tried I am not getting any idea so kind like you idea to solve SL3 sir. Exercise SL3 okay you have a thermally inflated cylinder closed at both ends that means our overall system is an isolated system it seems to be an isolated system. Thermally insulated cylinder closed at both ends that means Q is 0 and W expansion is 0 fitted with a leak proof friction less diathermic piston which divides the cylinder into two parts initially the piston is clamped in the center okay so the piston is in the center and it is diathermic it is clamped in the center on one side you have 1 liter of air at 300 K and 3 bar so you have air 1 liter 300 K 3 bar on the other side you have 1 liter of air at 300 K and 1 bar this is also air this is 1 liter this is 300 K and this is 1 bar the piston is diathermic so 300 K on either side is okay it will maintain that it needs to be clamped because pressure on one side is 3 bar pressure on the other side is 1 bar so when the piston is released and reaches equilibrium in pressure and temperature at some new position so because of this if I release the piston notice what will happen there is a higher pressure on one side 3 bar there is a lower pressure on the other side so the piston will start accelerating towards the 1 bar side there will be some back and forth jerky movement it is not given to be quasi static and we have to compute the final pressure and temperature and change in entropy what irreversible process has taken place you will notice that there is no hint of any electrical work or any stirrer work so we can consider that the total work is 0 so that means with Q equal to 0 W equal to 0 it is an isolated system so the first law will simply tell us delta E is 0 assuming that this is a stationary system on a table or in a lab so there will be delta E K and delta E P to be 0 so we can say that first law finally reduces to 0 equals delta U because we can assume delta E K delta E P also to be negligible or essentially 0 we have to determine the final state in the final state remember assume the final state will be made up of a uniform temperature because the piston is diathermic so let T2 be the final temperature the we also know that the final pressure also know that the final pressure will have to be the same on either side so because the piston is frictionless so let P2 be the pressure on either side we also know the final volume the final volume will be 1 liter on the left hand side plus 1 liter not 1 liter on the left hand side and 1 liter on the right hand side the final total volume V2 of say this is the A part of the system this is the B part of the system V2A plus V2B is going to be 2 liter that is we know but we do not know what is T2 and we do not know what is T2 we can apply the following relations the equations which we have is one equation is this the second equation is this delta u is 0 the third equation is going to be the equation of state for part A that means P2VA2 by T2 will be P1A which is 3 bar V1A which is 1 liter V1A which is divided by T1 which is 300 K the fourth equation would be same thing for the second part so P2VB2 divided by P2 will be pressure of 1 initial state B part volume of 1 B part divided by T1 I can write T1 because it is the same thing on either side and these four equations will give you the four unknowns T2 P2 V2A and V2B so you have four equations four unknown solve them remember what the four equations are one is the fixed geometry so the volume remains unchanged VA VA and VB totally 2 liter second equation is first law third equation is the equation of state applied between the initial and final state of part A the second is the equation of state the fourth one is the equation of state applied to part B between its initial state and final state and once you know the initial and final states of the two subsystems involved A and B you can determine the entropy change of A as well as the entropy change of B sum it up and you will get the entropy change of the total isolated system over to you. Thank you another is the significance of classius-clapeyron equation over to you sir. The significance of the clausius-clapeyron equation is the following if you take the clausius-clapeyron equation tells us how the saturation pressure varies with temperature this turns out to be if for example if you apply it to the liquid vapor equilibrium this will be hfg divided by Vfg and this comes out by applying in the two phase zone say on the T s diagram let us say you take at some point in the T s diagram you determine the value of partial of p with respect to t at constant V and you also determine partial of p with respect to t at constant s and for this obtain a relation using the Maxwell's relation obtain a relation using the Maxwell's relation and you will notice that because in the two phase zone an isotherm is also an isobar you will find that both these things are equal and that is obvious so because in the two phase zone pressure is a function only of temperature so whether you keep the volume constant or the entropy constant does not matter and from that you will be able to show that dp by dt in the saturation zone turns out to be the way entropy varies with specific volume either at constant pressure or at constant temperature and since in the two phase zone constant pressure also means constant temperature and s and v at constant p or t would only be a function of the dryness fraction x one additional algebraic step will show that this is sfg divided by Vfg this is one way of deriving the Clausius-Clapeyron equation another way of deriving is in the exercise sheet for example after second law if you go to the property relations part I think the derivation of the Clausius-Clapeyron equation is Pr 17 but it can also be derived using the technique provided in Pr 18 using a small Carnot cycle between two temperatures t and t minus dt also you will find that various text books have various other tricks of using the Clausius-Clapeyron equation coming to the second part of your question that how what is the application of this let me again come to the dt diagram of water this is the triple point and I said that the solid liquid line bends towards the left indicating that as you increase pressure the boiling point reduces whereas as you increase pressure sorry the melting point reduces but as you increase pressure the boiling point increases and now here you will notice that if I apply my dp by dt to the saturation of the solid liquid kind then this means that I must have s on the right hand side s liquid minus s solid divided by V liquid minus V solid now s liquid minus s solid can be shown to be equal to 1. 1 over T H solid liquid this is the latent heat of fusion this is the melting point and this now turns out to be 1 over rho liquid minus 1 over rho vapor now this is a positive number this is a positive number but we know that ice floats on water that means the density of the solid this is liquid and solid so density of the solid is lower than the density of the liquid and hence reciprocal of the density of solid will be higher than reciprocal of the density of the liquid hence the denominator will be a negative number since everything else is positive this turns out to be a negative number that means if you increase the pressure the saturation temperature will reduce any solid liquid combination where the solid floats on liquid as in case of water will have this characteristic that the melting point will decrease as the pressure increases whereas wherever you have a solid liquid combination of a material where the solid sinks in the liquid you will find that the melting point increases as the pressure increases apply this to the liquid vapor equilibrium at this point and you will notice that because the vapor always floats over the liquid a bubble always rises so the density of the vapor is lower than the density of the liquid and hence the as the pressure increases temperature also increases this is a typical example where thermodynamics puts a restriction between the relations of properties but not on the properties themselves thermodynamics will never tell us that property of one this property of this substance at this particular temperature pressure has to have this value but it will give us some funny relations saying that for example this is the relation between latent heat temperature and the two densities and that is related to how the saturation pressure varies with saturation temperature or saturation temperature varies with saturation pressure that is about it over to you. Thank you very much sir for nice explanation over and out sir. 1112 SUV college Tirupati Chittur over to you Dr. Munshed. Yes go ahead. Sir I have one question regarding the exergy regarding exergy go ahead. Yes sir what happens to the exergy when a system performs a cyclic process under the cyclic process? I could not hear your question can you repeat it? Sir when a system undergoes a cyclic process what happens to the exergy of that system? See if we define the exergy I suppose we have defined exergy yesterday as the flow exergy as H minus T0S. So it is a funny property which depends on the state also depends on the environment temperature. So when a system executes a cycle naturally it comes back to the same enthalpy and same temperature the same entropy and hence the exergy change of a system after executing a cycle will be 0 over to you. One more question sir. Hello sir in yesterday examination there was a question what are exists at atmospheric condition is given as super saturated steam how it is I did not understand that one. What you mean is water exists in normally in air in super heated steam you need an explanation. Over to professor Bandarkar, psychromic tree expert. See water you should realize that this is in vapor form it is a gaseous form and at the saturation vapor pressure the temperature of the water is above the saturation temperature for that vapor pressure and hence it is super heated form of the substance. So that is why it is in a super heated form. In fact you can say it is no different from any of the gaseous that is existing like nitrogen and oxygen all of them are existing in super heated form. Thank you. Sir one more question sir. Hello. Yes go ahead. Sir why that liquid vapor saturation line is ending at critical point or for that will there be any change in that line in two different ways? The question is why is the liquid vapor saturation line ending at the critical point. Thermodynamics does not have any answer to this but all I can say is that the structure of the three phases is such that solid as a distinct phase standard solid at some crystalline locally crystalline structure. The gaseous on the other hand have a absolutely random unstructured behavior leading to kinetic theory and all that. Liquids are somewhat in between but being fluids they are more nearer to gases than to solids and particularly as you go to higher and higher temperatures automatically means higher and higher temperatures but for a particular fluid as you go to higher and higher temperatures the intermolecular forces in the liquids reduce whereas the kinetic behavior, kinetic energy of liquid particles increases and at certain stages the kinetic energy increases and the kinetic energy becomes comparable and equal to that of a vapor whereas the intermolecular forces become negligibly small when that happens that balancing point is known as the critical point means at critical point this balance takes place removing the distinction between a liquid and a vapor. If you want more details you will have to look up research papers and papers on the statistical thermodynamics and kinetic theory of liquids over to you. There is one more question sir. Hello sir. Go ahead. Sir in some of the books I have seen that the loop in a PV diagram as available work and whatever that is given as unavailable work. No that is not really true. Your question is a loop in a PV diagram for a cycle. We have seen in the morning or yesterday I think I showed it as what is this is the network output maybe I should go back to that particular slide. Now notice we have a loop in the PV diagram and what is shown by horizontal red lines this is the net output and the area under the curve which is doubly hashed like this is actually the work done by processes which are essentially the compression type of processes to complete the loop. Nothing about unavailable work do not bring in availability and unavailability in such simple things. It will only confuse our availability analysis for that. This is just work done in expansion part of the cycle and work done in the compression part of the cycle. The difference would be the net work delivered to the outside world. Thank you sir. Over and out.