 Yeah, there are three centers who have raised their hands JNTU, Truba and Jabalpur and I will go to them in order just three of these now after that let me continue with that open system derivation. I do not see Dr. Bramara today which is around or which is not but anyway over to you JNTU, any questions? Good morning sir, there is a question regarding the laws of thermodynamics. I hand out the mic to the participant Mr. Harinath. Third law of thermodynamics, over to you sir. Yes, there is a law of called the third law of thermodynamics but there are a few things with that as I think it was discussed earlier. First thing is that mechanical engineers and many many of our engineers do not really need to worry about the third law of thermodynamics and the second issue or more fundamental issue is there is still some doubt in many people's mind and I am a member of that set. Whether third law of thermodynamics is really a fundamental law of thermodynamics because the most common statement of the third law of thermodynamics is that the entropy of a perfect crystal at 0 Kelvin is 0 and the issues here are 0 Kelvin is not a so called defined temperature in thermodynamics. If there is anything special about it that gets derived for example using our first and second laws, we can demonstrate that it will become very very difficult. We require a tremendous amount of energy to bring a finite system to 0 Kelvin that is one issue. The second issue is in any laws of thermodynamics so far we have not used the property of any material. We have not said that first law is for this system, first law is for adiabatic processes, second law is for processes which are possible, impossible. In fact, second law specifically says through Carnot theorem that a reversible 2 T heat engine working between same temperature will have the same efficiency irrespective of whatever is the material fluids etcetera which are used to build that engine. In fact, that is the power of the laws of thermodynamics they do not talk of or they do not depend on any material. Whereas the moment in third law you say perfect crystal that means you are talking of some material it is its structure and all that is one of its weaknesses over. Thank you sir. We have another question on the exercises which we will ask in the afternoon sir. Thank you for the time being over sir. So, over and out from JN to you. Anyway we will meet whenever the need arises. Truba Indore over to you. Good morning sir. Sir in the definition of availability that you mentioned about the various differentials, differential of energy, differential of volume and the differential of entropy. Now, these are the differentials essentially across the system boundaries. Now, in a general case of a thermodynamic system we also have differentials within the boundary that is there could be some pressure differentials, temperature differentials within the boundary and a special case is where there are there is some viscous dissipation. Now, how do we define availability in such a case? Over to you sir. You see already the definitions here on the previous page and you mentioned that we have differentials here of energy, volume and this and you said across the boundary. These are not across the boundary. These are between two states energy, volume, entropy are properties of states. So, they depend only on the state and when you say differences within the boundary of a system that means you are considering a system which is not in equilibrium. Because within the boundary of a system if you have differences in say pressure then that means the system is not in equilibrium. And when you are considering viscous equilibrium that means you are considering that different parts of the system are in relative motion to each other. That itself indicates that the system is not in equilibrium. When we say restrict ourselves to thermodynamics we consider systems in equilibrium. Now, after having said that we have to tackle differences in pressure, we have to tackle variations in temperature, we have to tackle viscosity and variations in velocity which lead to viscous dissipation. So, in that case what we do is remember our very first exercise where we had to determine the force on that dam like piston. And we said that the pressure will be varying from top to bottom. In that case it varied in a reasonably regular way, in a one dimensional way. And we said rather than considering the water as one single system we will consider it as various horizontal slices, thin layers of water one above the other, each one individually in strict thermodynamic equilibrium, each one having an individual pressure. But from one system to next system which will be at a differential height difference of say delta H, the pressure will differ by rho into G into delta H. In a similar fashion when we have to consider a system, a big system with internal variation in pressure, we will split that into a number of small systems, each one of it will be then considered to have a uniform pressure, uniform temperature within it with small differences of pressure and temperature with respect to systems which are its neighbors. In fact this is what we do in fluid mechanics, this is what we do in heat transfer. So when we apply thermodynamics to systems with fluid mechanics also, with fluid movement also and with heat transfer also, we will do this. And there we can talk about viscous dissipation and its relation to entropy production and all that. There is a very interesting book by a doctor or professor Adrian Bejan, B E J A N, it is quite an old book may be 20 years old already. It is I think the title is entropy generation in heat transfer and fluid flow or something like that. And there are a number of papers on this topic but if you want to start looking at fluid flow, heat transfer and the entropy generation effects because of that and doctor Bejan's book is a good place to start, over to you. Sir in the case of this definition of availability that you told, this is applicable for a closed system. But when we talk about this open system, definitely we will have this differential that is the governing thing in open systems. So in that case also this slicing method will do or there is some other approach for that as well, over to you sir. We will soon in a matter of minutes start deriving the open system formulation for availability and now that you mentioned slicing, notice that when we did the derivation for the open systems for convert, all that we have done is converted the equations applicable to a closed system to an open system by considering over a time period t to t plus delta t and equivalent closed system. And we notice that among the assumptions was the fact that one of the assumption was that near the exit there is local equilibrium. So we can define specific energy, specific volume, density, specific enthalpy letter, specific entropy at the exit. At the inlet also we had local equilibrium with specific values of all these properties. So some sort of local slicing we have already done there. All that we have to do is extend this idea of slicing or partitioning all over the system rather than just here we took yesterday or day before yesterday just one big control volume and we considered two small systems one at the inlet, one at the exit. We called them the inlet plug and the exit plug but that itself is the idea of slicing. If you are comfortable with that idea of slicing then we can extend it to even within the system by considering a number of small small open systems or small small control volumes over to you. Sir one final question in the definition of availability you have used this definition to e plus p naught v minus t naught s minus e naught plus p naught v naught. Now why we have not replaced this by h you know enthalpy here? We could also have enthalpy in this definition of availability. Over and out sir. Before I go away from you, you just wait it is good that you are thinking in that direction. You can rewrite this in terms of enthalpy all that if you want to force enthalpy into this equation right it will be an artificial forcing but all that you will have to do is write e as u plus e other and write u as h minus p v. That is an artificial way of doing things. But we know that enthalpy sort of naturally enters the first law of thermodynamics for open systems. So when we do the similar formulation for open systems, the combined loss formulation for open system, you will find that in that formulation enthalpy will end up entering as naturally as it entered the first law of thermodynamics for open systems. So over and out maybe if necessary we will meet again later. I am going to Jabalpur now. It is very nice to see that I can travel in a fraction of a second from one place to another and I think only my reflexes matter even the field of light does not seem to be hindrance here. So over to you Jabalpur. It is related to OS 10. There it is to determine that work done per kg of water pumped and the flow work. Is there any difference between work done per kg of water pumped and flow work? Because to calculate that work we are using minus v dp where v is to be taken constant for water, saturated water. So is there any difference between these two things which is to be determined here? The question is OS 10. So let us define work done per kg as w dot s divided by m dot. Naturally, since we can have only one dot in the denominator this will have meaning only in a steady state because when you have an unsteady state m dot n is likely to be different from m dot e and then if you have to define work done per kg you have to be very specific about whether m dot is i or e if it is unsteady state. Whereas the rate of flow work you can define call it say w dot flow and this by definition in our derivation you would have noticed is m dot e p e v e minus m dot i p i v i. In steady state it will become equal to in steady state w dot flow steady state will then become m dot into p e v e minus p i v i and in case of pumps you will find that w dot flow and w dot s under certain conditions they turn out to be equal. That is an accidental derivation because of all these conditions like incompressible flow isentropic or isothermal which essentially mean the same thing for incompressible flow and steady state. Otherwise there is no direct link between w dot s and w dot flow over to you. Thank you sir over and out. Thank you very much. Now I think we are ready to there is a question from Somaya. I think I will take that up but that is absolutely the last question. Somaya college over to you. Do you have any question? Self-paragmay again attending from Somaya college. Sir I have a very simple question can we say that when there is a equilibrium achieved the dead state is maintained. A system attains dead state when it establishes the equilibrium with the surrounding. Can we say that? This is a simple doubt which I am having because the temperatures are becoming T0 equal to the temperature of the surrounding. So can we say that when equilibrium is established system reaches a dead state? A system is in equilibrium we say a system is in equilibrium when it has a unique set of property. And a system is in dead state when it is in equilibrium and the pressure of the system is the pressure of the surrounding and the temperature of the system is the temperature of the surrounding. By this it is in a state of equilibrium and in equilibrium with the surrounding or the same pressure and temperature of the surroundings that we know as the dead state. Pure state of equilibrium cannot be a dead state among the various states of equilibrium one state is the dead state the one state in which pressure of system is pressure of surroundings and temperature of the system is temperature of the surrounding over to you. Thank you very much. I am now moving over to the tablet and this is where we were when we stopped our derivation and now let us consider an open thermodynamic system. So, application of for the combined application first and second loss system. So, what we are going to do is we will consider a system and now I will not go through the whole system diagram we are now going to do a combined application of the first law to an open thermodynamic system and we will consider our control volume like this and just to keep the algebra simple, I will assume a steady state and do not ask me any questions about how to derive this for the unsteady state. I will only say that absorb this and you should be able to look at this derivation and the closed system derivation and do for yourself the general unsteady state derivation that will be a part of the I declare that as part of the assignment. So, I will not discuss that even if you ask me a question because that is going to be the part of the assignment. Let us say steady state and for simplicity we have one exit and we have one inlet and let us say that it provides a w dot s and it has an interaction q dot 0 with the environment and t naught and other interactions are q at respective t's again as in the case of a closed system this q at t is a generic thing. We will get a term like q by t sum it up or integrate it or do a combination of sum and integration as needed. What about the p naught delta v type of work? The work done in pushing around the atmosphere. Now here we assume that it is a steady state. A steady state means that the delta v of the control volume delta e of the control volume etc would be 0 that means d e c v will be 0 d s c v will be 0 those will be used in the first and second law but similarly d v c v by d t will also be 0 and hence the work or power required to move the atmosphere around will not be there. So, we do not have here an idea of w dot s useful we will live with w dot s itself. So, let us proceed I will again show it simply as inlet exit and this is steady I have a w dot s it is pretty small but it is only for my prompt you have the big figure with you already from the previous page q dot naught at t naught and q at t. First law q dot plus q dot naught minus w dot s equals now steady state. So, I will write m dot now on the right hand side I will have m dot and then we can write h e minus h naught plus v e squared minus v naught squared by 2 plus g z e minus z naught. This is first law let me call this equation one the second equation will be the second law of thermodynamics. We have q dot by t plus q dot naught by t naught on the right hand side we will have equal to m dot s e minus s i and here we will have plus s dot p with the condition s dot p must be greater than or equal to 0 second law and the forms look a bit different compared to the closed system form because we are using the steady state form. If you use the unsteady state form they will look very much like the closed system form because then you will have on one side d e c v by d t for the first law and d s c v by d t for the second law but you are familiar with these forms there is nothing special about them this is just first law and the second law. Now we use the same algebraic procedure which we used earlier we want to eliminate q dot naught between the two equations. So, I multiply equation 2 by t naught and get equation 3. So, I will get q dot t naught by t plus q dot naught minus t s p dot equal to m dot t naught s e minus s i this is equation 3 and then all that we do it subtract equation 3 from equation 1 and then you will get q dot 1 minus t naught by t q dot naught cancels out that was the purpose of going through this algebra from the first equation we have minus w dot s from the second equation we have plus t naught s dot p and from the right hand side we have equal to m dot. Now the terms are usually combined like this you have m dot into t naught s e minus s i here and you have h e minus h i from here. So, you write this as h e minus t naught s e minus h i here I made a small mistake this should be can this should also be i h e minus h i minus t naught s i these two terms remain as it is plus v e squared minus v i squared by 2 plus g z e. Now let me rewrite rephrase this and we should get an equation which gives you this w dot s equal to q dot 1 minus t naught by t minus one term is h e minus t naught s e minus h i minus t naught s i second term v e squared minus v i squared by 2 and third term plus g z e minus i and minus t naught s dot p and now c what do you get you get three terms I will call this as term a I will call this as term b and I will call this as term c and now I will get w dot s max the maximum value the minimum value of s dot p is 0. So, as I reduce s dot p this particular t naught s p will reduce and I will get a higher and higher values of w dot s. So, the maximum values of w maximum value of w dot s I will get is w dot s is these three terms, but only the 1 2 c will be 0. So, w dot s max is q dot into 1 minus t naught by t and this is the power obtained when we absorb heat at q dot rate at a temperature t and the only other interaction that means heat rejection temperature is t naught and this is perfectly corresponding to our term in the closed system. The second term is minus I can write it as this the same I will just write this again this is this is term a this is term b and this is the maximum power obtainable for the process in the open system inlet to exit in steady state of course, steady state is always there when q dot naught or with q dot naught at t naught that means the heat exchange allowed when heat exchange is allowed only with the ambient at t naught. So, this is the maximum power that you can get with q naught at t this is the maximum power you can get from when you go from inlet state to exit state and t naught s dot p is the so called lost power. Now, this is the derivation and now the definition just the way we did a derivation first and then we defined the availability what is the definition here the definition is hidden in these two terms this and notice that we can include these two terms in the definition, but for a large number of open systems these two terms would be usually small and the definition now again not very hard and fast because there are large number of variations the definition is of h minus t naught s this is the term which comes and there was a question I think just before t was that in the availability or exergy analysis why cannot we have h and now we have written the first law for an open system and you will notice that your h is naturally included. So, now we have a definition coming we have terms here like h minus t naught at exit and h minus t naught at inlet. So, this is defined as sometimes it is called a lower case phi or sometime it is called b we will be using the symbol b because this lower case phi quite often at least in handwritten form is confusing with the upper case phi and this is called the specific exergy specific because this is specific enthalpy specific entropy. So, this is specific exergy and this is the definition of exergy and we do not have any use for it because we will always be with mass flow rate we will be using this specific exergy, but if you want the non specific or bulk exergy can be defined as h minus t naught s what is its use I do not know I have never had to use it, but this we use and what is the significance of the exergy the exergy significance of exergy would be if you take minus m dot or let us not put m dot you will say that w dot s max with the open system going from inlet to exit and of course steady state and heat transfer allowed only with the environment this turns out to be minus m dot into delta b where delta b is b e minus b i. So, this tells us that the maximum power you can obtain from an open system in steady state when the only heat interaction allowed is with the environment and no other heat interaction is allowed is the mass flow rate into the reduction minus delta b will be the reduction in specific exergy as it goes from inlet to exit. So, using the specific exergy thing our w dot s can be written down as q dot into 1 minus t naught by t plus m dot into minus delta b and if you see the term here this is minus delta e k and this is minus delta e p this negative sign is taken inside minus delta e k from inlet to exit minus delta e p from inlet to exit and that is where we end our sorry there is a this is w dot x max and w dot s lost is t naught s dot this is where our traditional for the classical availability analysis or exergy analysis ends. Now, what is the problem with the traditional or classical exergy analysis or availability analysis? Let me go back and there is no problem you can set up a lot of exercises a lot of exam papers asking for various definitions and various derivations and various you can set up various numerical problems, but the thing is this is the restriction in traditional or the common analysis of availability and exergy that is when you go from the actual situation to the ideal situation you want to extract more work. So, some other interaction or some change of state should be allowed in the common analysis of the traditional or text bookish analysis of availability and exergy the only variation allowed is with the heat transfer with the environment. The inlet state and the exit state and other interactions are not allowed to be changed the only change is q naught with the environment and also the goodness of the process itself inside if it is non quasi static you can make it quasi static naturally you have to keep on making it more and more quasi static because we are approaching the reversible limit, but now let us look at the situation where this analysis will be difficult or it will not be possible to implement this analysis and the simple situation is this and here we will be looking at the various options that we have only one of the options is the common analysis type of option. Let us consider a turbine it could be a gas turbine it is could be a steam turbine does not matter we will consider a turbine in steady state produces w dot s m dot is the flow inlet state is i exit state is e a usual turbine is insulated a turbine does not take in heat from any other source. So, q dot is also 0 q dot naught is also 0 that means turbine as implemented is an adiabatic and we know that in steady state we will be neglecting delta e k delta e p you can repeat the analysis by including them those two terms will simply be carried over. So, we know w dot s is m dot into h i minus h e and we also know that s dot p should be m dot into s e minus s i and s dot p should be greater than or equal to 0 this much we know and because of this if you draw a say h s diagram or a t s diagram now you see the importance of the h s coordinates h i minus h e represents the enthalpy difference from inlet to exit or decrease in enthalpy from inlet to exit and that is proportional to the power output first law and s e minus s i represents the entropy rise higher the entropy rise irreversible is the process an ideal adiabatic turbine would be one in which the inlet and exit entropy would be the same here the exit entropy is higher than the inlet entropy now this is the actual situation now I want to go to the ideal situation now remember that I have an inlet pressure P i I have an exit pressure P e one ideal situation we have is which we have already considered and using that we have defined the isentropic efficiency of the turbine and in that we have said that let us keep the turbine adiabatic that means we will maintain q dot equal to 0 and q dot not equal to 0 but we will make it reversible means we will make it reversible adiabatic will maintain it adiabatic will also make it reversible that means we will make it isentropic that means the ideal turbine would be would have a state line like this and the ideal exit state would be e star and we have determined the W s star and ratio of W dot s to W dot s star is the isentropic efficiency of the turbine so here we have we have taken this option one where we maintain q dot equal to q dot not equal to 0 but s dot P is made equal to 0 by e going to e star we have changed this exit state from e to e star so that means we have allowed a change of state at exit this is not according to the common analysis of availability and exergy because in the common analysis of availability and exergy the change of state or change of inlet and exit states is not allowed that is one the second thing is while doing so we have said no change in interaction in particular we have said that the turbine was adiabatic it remains adiabatic and as part of that the turbine had no heat exchange with the surroundings and we said that it should continue to have no heat exchange with the surroundings so this also is not according to the common analysis because in the common analysis we consider that to bring s dot P to 0 we are allowed to change this we are not allowing to change that so both these things are not according to common analysis now suppose we were to be suppose I am an expert in availability and exergy and if I am of the opinion that I must have must do everything according to the common analysis that means I should not change the inlet and exit states at all in that case what is the maximum power I will get and how will I get it here in this particular option option one things are clear keep the turbine adiabatic you have it well insulated already leave the insulation like that if at all improve it but the internal working of the turbine you make in such a way that physical friction flow friction non-uniform flow which means non equilibrium states at least locally viscosity effects all these you reduce to the extent possible so that the actual exit state which is at a higher entropy than the inlet state it slowly shifted to a state where the exit entropy equals the inlet entropy but if I take the option 2 which is according to the common availability and exergy analysis then what will happen we will say that although this was my turbine inlet exit I am now going to fix my inlet and exit states the turbine did not have any interaction with a heat source nor did it have any interaction with the surroundings which were at t0 we did not have to worry about t0 but q0 was 0 now I am going to say that I am going to make it better and better but I will keep I and e same but if I want to keep I and e same and if I want to improve W dot s that means I must allow a heat exchange with the environment and my common analysis will tell me that my W dot s max will be m dot into h i minus t naught s i minus h e minus t naught s e I will rewrite this as m dot into h i minus h e plus m dot t naught into notice s i has a negative sign s e as an effective positive sign so s e minus s i notice that this is the actual power currently developed by the turbine first law as applied to the turbine current situation now the question is this I will obtain I was obtaining but how do I obtain this addition to obtain this additional power I must make q dot naught not equal to 0 because this is W dot x actual and that means in the for this situation I must have the ideal turbine working something like this I am just re sketching it inlet unchanged exit unchanged W dot s max but out of W dot s max a part comes as this m dot h i minus h e the other part comes because you are absorbing heat q dot naught from the surroundings which is at t naught and that q dot naught which is equal to this q dot naught needed now how do you absorb it take for example your t naught will be around 300 Kelvin I am assuming 27 degree c people at Nagpur Jaipur etcetera may not agree but let us assume a round figure what will be the temperature on the boundary of the turbine temperature of the boundary of the turbine if it is a steam turbine inlet may be at something like 200 Celsius that means something like 500 Kelvin outlet may be may be at near room temperature so 300 Kelvin but that means because this is at 300 Kelvin and this is out there I must have a reversible 2 t heat engine working out there and I must make my turbine non adiabatic now this is something a turbine operator will not be willing to do so that is why quite often I think that for processes like a turbine the common exergy analysis is something which is to be done but not implemented in practice a good turbine engineer will try to implement in practice this this improves what is known as the isentropic efficiency this would be the purpose of the you know trying to reduce the entropy production to 0 in the second case also we are reducing the entropy production to 0 but not by putting AC equal to Si we are saying entropy production will make 0 by letting AC and Si be where they are but by supplying the requisite amount of heat from the atmosphere or from the ambient by using an appropriate set of reversible 2 t machines okay now there are 15 minutes to lunch there are some very illustrative exercises in the exercise sheet under combined by first and second law in fact CL1 itself is very interesting others are computational but I recommend that you also look at SL10 because there is an aspect of availability and exergy analysis particularly the common availability and exergy analysis which puts all emphasis on power output or work done okay quite often the system is such that there is no work done and that is the situation in SL10 here it says that the only work done is in displacing the environment in fact here it is important for us to have that required energy change volume change and entropy change that is the required change of state okay so here we can look at it as lost heat energy or lost thermal energy and this problem is something which cannot be directly tackled by the common availability and exergy analysis maybe I was not sure whether SL10 should be in the SL part of the exercise sheet or in the combined laws CL part of the exercise sheet I put it in the SL part of the exercise sheet to just emphasize to you that first law and second law can be combined together to obtain either wastage of thermal energy or looking at it from the combined loss point of view as work which could have been obtained but was not obtained okay so we will start solving the exercises after the lunch break but then till lunch break I am available for questions answers discussion over to you through Bah good afternoon sir you just explained about this exergy analysis for open systems I had one query about this option 2 that is the common exergy and availability analysis in that you talked about the heat interaction in that example that you gave of turbine the heat interaction that is heat flowing towards the turbine now usually the turbine people do not have this they try to restrict the heat transfer that is flowing out of the turbine so we cannot usually have the heat flow towards the turbine to increase the maximum work that's why it is kept as zero how do we have this heat transfer taking place from the outside the turbine towards the turbine when the outside temperature is lower than the turbine temperature over to you sir in fact that is the weak point of the option 2 or the common analysis because if you look at a turbine a turbine unless it is in some low temperature refrigeration type of application will have the surface temperature higher than T naught and that's why it will be well insulated now since the ambient is at a lower temperature if you want to enforce the traditional textbook is exergy analysis then from a lower temperature you will have to supply it to a higher temperature by having a reversible engine it is some sort of a heat pump which would absorb some work again in a reversible fashion and that work would add to this and would lead to this going from W dot S to W dot S max and but this is something which is nobody would ever think of doing so that is what I said that if you apply the classical or common exergy analysis you come up with a solution it tells you that from W dot S you can go to W dot S max if you are happy with oh this is only a small difference so I need not do anything then it is okay but if you want to make an attempt at going out here then you are in trouble because you have to try to set up a very very complicated system apart from the fact that whatever is the internal processing that also you have to make reversible so that is the weak point and that is why I do not emphasize the traditional availability and exergy analysis to any extent I just include it because a large number of books and some procedures use that concept and that is why a student is expected to know what is availability what is exergy how to analyze it from the availability and exergy point of view but I think I have already made it a point that there is nothing in it except a few additional buzzwords defined according to the convenience of someone the principles involved are simply the first law of thermodynamics and the second law of thermodynamics and absolutely nothing more over to you and sir what about the relative values of the these work transfers that is the work heat pump and also the work additional work that we have we obtained from the turbine what are the relative values then we could find that this maximum work could be increased secondly sir if we apply the same demonstration for compressor I think we could result in a better because the heat transfer from the compressor is from the compressor towards the environment by supplying some coolant across the compressor so if you could give me the example of compressor again after the break then it will be very handy for us what about the relative values of the work transfer that is required in a reversible heat engine and also the additional work that is obtained from the turbine is there any relation between that over to you sir. Good question in fact we have an actual turbine a turbine with an actual power output of w dot s if you go by the option one you get w dot s star and if you go by option two you will let me say you get w dot s max by the availability and exergy analysis in fact the reason for using star when we came to the definition of isentropic efficiency and not max there is precisely that I did not want you to confuse the w dot s max out there with the w dot s max that you get here so this is actual this is for an adiabatic reversible turbine and this is according to the common exergy analysis and these three values are different and we will take an exercise in one of the exercises there is an air turbine where you can do actual calculations easily and do this and we have defined the ratio of these two as the isentropic efficiency star s and the ratio of this and this is sometimes defined as the exergetic efficiency and your second point is also very good in case of a compressor because a compressor usually will be also hotter and for many other reasons we need to cool it using water or some other fluid maybe instead of that we can use a reversible heat engine there we will discuss that that is a good point but again the temperature difference will be so small that it is very difficult to actually implement an engine there reversible or otherwise so whenever you cool a compressor that higher temperature which the current finds itself in that high temperature coolant or warm coolant can perhaps be used for some other heating purposes may for all you know you could heat it in room heating or incubator heating or laundry or some such purpose the temperature differences will be pretty small to pretty small for anyone to consider designing an engine for it over to you thank you sir over and out nirma over to you good afternoon sir my question about a gas constant is there anything like a gas constant for the gas constant arc for the steam base steam or a superheated steam over to you sir see this is about the gas constant the gas constant for any where any material I should say anything is defined as the universal gas constant divided by the molecular weight okay so for example in the extreme case I say any material so you take graphite pure carbon I think the molecular weight is 12 12 kg per k mole divide the universal gas constant divided by that and you will get the gas constant for graphite graphite is not a gas it is a solid so the gas constant is of no use there so but it will be of use to some fluid and hence the gas constant for any fluid will also be defined like this so gas constant for water is universal gas constant divided by whatever is the molecular weight of water that is 18.086 or something roughly 18 so that gives us the gas constant of water steam or anything it is for a material it is not for the phase of a material of course remember that water is not an ideal gas so after calculating the gas constant r for water do not expect p v equals r t for water at any state water or steam at any state over to you over and out from Irma s g s i t s indoor over to you the utility of exergy and energy and what is the effect of these two parameter on the turbo machine efficiency over to you see the availability analysis or exergy analysis is what we do on pen and paper the turbine does not know anything about this the turbine is a physical system and like any other physical system it will obey the first law of thermodynamics it will obey the second law of thermodynamics it is a real life working system so the thing will be such that you apply properly the conservation of energy across it boundaries it will satisfy the conservation of energy that means it will satisfy first law you apply second law it will demonstrate to you that it is producing enough amount of entropy so it is executing a real life irreversible process that is it beyond that it does not know anything about exergy energy energy entropy nothing it satisfies first law it satisfies second law it is for us to work with all these numbers the turbine does not care about it over next question is what is the effect of exergy and energy on the turbo machine efficiency over to you sir in the afternoon we are going to solve some problems from the exercise sheet cl and there is one of my favorite problems in it in that differences between actual power output ideal power output from the two different methods the isentropic efficiency and exergetic efficiency these things will be clear over over and out now we are going for lunch