 Good morning everyone. We are on our final day and as informed yesterday, what happened? Two hands have already been raised from Calicut and Panvel. So, let us go there. Okay, first to NIT Calicut. Good morning Calicut. Over to you. There is the tutorial sheet page number 3. Page number 3 tutorial sheet. Problem number 1.6. The stress is proportional to same and the stress is equal to m by a. That is equal to y d delta L by L and delta L is delta L by L into a into L by y and I am substituting in the integral of dL. But how can I substitute dL, delta L for the stress strain equation what I am having in the very incremental change in the length? Over to you sir. Thank you. I think this is a straightforward problem. I do not have the solution with me. This is a straightforward problem. Unfortunately, I do not have the solution with me today. I took away all my stuff from this all yesterday. Just send me an email or put it on Moodle and do not forget to write your name and the place you are from below the Moodle. Because I have stopped answering and commenting on Moodle comments which are sort of anonymous. So, do that and I will provide you the detailed solution. Maybe upload it on Moodle itself. Just give me a few days. Over to you. Thank you. And then another question on the second law regarding the improvement of the corner cycle. Efficiency given by 1 minus T2 by T1. This is the effective way to increase the efficiency of the engine. Since the T2 is the operation of the environment, T2 is fixed and we can only vary the T1 for the source temperature. So, that is my understanding sir. Is this right or is there anything else you should tell the students? Over to you. Thank you. The question is about the Carnot or similar cycles where the, see on the, this is just the temperature scale. This is T environment and suppose we have a Carnot type machine working between T1 and T2. Now T2 cannot be less than T environment whereas T1 can be as high as possible. So, the efficiency of Carnot which is 1 minus T2 by T1. So, you are right to improve efficiency since T2 cannot be lowered below T environment. And making it approach T environment very close will become very costly. The only way to improve is to increase T1. And if you look at today, if you burn fossil fuels, T1 can in principle be as high as roughly, you know for good fossil fuels, the adiabatic flame temperature will be of the order of 2000, more than 2000 Kelvin. But that means around 1700, 1800 Celsius. But we do not have material today which will withstand that temperature continuously for a long period of time. Consequently, of course we do not have Carnot machines, but other machines which come anywhere near it. For example, gas turbine cycle, Rankine cycle, the attempt has always been to increase the T1, the highest temperature in the cycle. The problem is not from the point of view of thermodynamics. The problem is from the point of view of designing or getting materials and appropriate design which those materials which will last for a long time under those conditions. Over to you. Thank you very much sir. There is another question, sorry for page number two on the tutorial sheet sir, tutorial sheet. I have done this problem. This problem I have got H2, that is HEE Exit has 120.9 kOhm per kg and HF has 105. But the pump temperature, for pump temperature we need to do the double heater positions in this problem. Am I going in the right direction? Please answer. The question pertains to exercise OS 9. We have a water pump in lead conditions are 1 bar 25 degrees C. Exit pressure is 180 bar and the pump power consumption is given. So, we have a pump standard nomenclature I will use W dot S. In lead is 1 bar 25 degrees C. Exit is 180 bar. The one of the conditions is determine the temperature of water at the exit of the pump. So, you will have to get your right H and pressure is given. You will have to do double interpolation no doubt you have to do double interpolation here or actually I have with me and maybe your library also has a much more detailed steam tables. In that case if you have an ISO bar in that available at 180 bar you will be getting away with a single interpolation. We have to do double interpolation because in our steam tables the 180 bar pressure is not listed. I think in our steam tables we have pressure only at 150 bar and 200 bar that is why double interpolation is necessary. Over to you. Good morning. Pillai College, Panvel. Over to you. Question. First question is regarding the mean temperature of heat addition in case of the ranking cycle. Can you please explain it sir? And the second question is regarding the exercise number PR14. Thank you sir. Over to you. As I understand the first question is some explanation of the mean temperature of heat addition. It is asked pertaining to the ranking cycle and the second question is PR14. Let me talk about the heat addition. Remember that we realized the following. On a TS diagram if we have a quasi static process then the area under that curve that is integral Tds is greater than required to the heat absorb. The equality signs holds only for a reversible heat addition. Now if you assume that the process is reversible then Q will become equal to integral Tds from say the initial state 1 to the final state 2. And then we can write this as T mean into S2 minus S1 that is sort of a mean height providing you the same area. This is the way quite often T mean is calculated. This is for a simple closed system. The corresponding relation for an open system will turn out to be H2 minus H1 is T mean S2 minus S1. And this is the way you can calculate the mean temperature of a Rankine cycle or mean temperature of heat addition not only in a Rankine cycle you can determine the mean temperature of heat addition in any process. So, under certain assumption you can even determine the mean temperature of heat rejection this way except that the process will be from higher entropy to a lower entropy generally. And the area will be negative but since delta S will also be negative the T mean will always turn out to be positive. Apart from this there is no other thermodynamic significance involved in this over to you. And sir this T mean for this Rankine cycle or any other vapor cycle how it affects the efficiency of the cycle thermal efficiency of the cycle over to you sir. Under certain assumptions you will be able to show certain assumption that all processes are reversible quasi-static you will be able to show that the if you calculate T mean of heat addition and call it Tm1, T mean of heat rejection and call it Tm2 and you can show that the efficiency will be 1 minus Tm2 by Tm1 but that is only under the case of under the assumption that all processes are not only quasi-static but also reversible. For example, if you solve or formulate a Brayton cycle problem or a Rankine cycle problem with isentropic efficiencies of compressors, turbines, pumps whatever are the work transfer devices to be 1 that is they are really isentropic then this formula will hold. But if you just take a Rankine cycle and just make the turbine non-isentropic provides an isentropic efficiency of something like 0.8 or 0.9 in that case this formula will not work there will be a deviation. This formula is likely to give you a different efficiency than the actual one. Now let me go to exercise Br14. Now you will notice that of the four properties which you have defined of course the basic thermodynamic property is u but the other defined properties are h, a and g out of which we have realized as we have solved problem that u is very good that is the basic thermodynamic property but it is absolutely it is the driver property for closed system h turns out to be the so called driver's seat property for open system. G we do not have much of an application for G but a and g will now come slowly into their own. The specific molar Gibbs function is the chemical potential and wherever you have processes taking place at the same pressure and same temperature particularly when chemical equilibrium is involved. G is a very important property nothing great about it except that when chemical equilibrium is involved the property combination u plus p v minus t s will turn up so often that we find it useful to have a short form and that short form is G. Now a we have seen that the natural variables for a are t and v not t and p in this case and here entropy turns out to be a natural variable and entropy is not a primary variable which you can directly measure. So when it comes to direct measurement since G is dependent on t p, a is dependent on t v they become attractive properties but here G has a small disadvantage that whenever you have two or more phases together t and p cannot really be independent variables. But for a t and v will always be independent variable for a simple system in any part of the state space even during the situations where the two phases are together. So now if you look up our derivation we have a defined as u minus this is definition u minus t s and we have already derived that s is from this first derivation was d a was minus s d t minus p d v and that is why we say that t and v are the natural independent variables for a and that is why we took a as a this problem talks about a as a function of t v then entropy or specific entropy turns out to be partial of a with respect to t at constant v and p turns out to be partial of a with respect to v at constant t. So the isothermal derivative with respect to volume gives you pressure with a negative sign and the isometric or constant volume derivative with temperature with a negative sign gives you entropy. Now look at it like this that means if I have a specified as a function of t and v all over the state space then naturally t and v are independent variable. So t v will be known from here s and p are known the moment you get all these four variables at any state then a equals u minus t s. So this gives you u equals a plus t s. So apart from these four properties now you have u available to you and since p and v is also available now you can calculate h equals u plus p. So all these properties are known and since the derivative of u with respect to t at constant v it is c v you can determine c v and by other relations you can determine c p you can determine all other variables properties for example isothermal expansion coefficient then isentropic expansion coefficient compressibility all those things you can determine. The question that arises is that is the last part including any change of phase for saturation line. Now it is not very direct how to determine this it is rather indirect and I am setting up my assignment in which one exercise will be on this based on the van der Waals gas. So let me show you how it is done we have these derivations. So at any stage we can determine not only h but I should complete this you are also able to determine g which is h minus t s. So all these t and v are independent but moment you are given t and v and this function a as a function of t and v you can determine entropy pressure u h and g not only at that t and v but by doing analytical differentiation or numerical differentiation over that range you can also determine their derivative as needed. Now what happens when there is a change of phase remember that we have a simple system what many text books call a pure substance come to PR 1 3 you should be able to show that if there are two phases here the illustration is for the gas and the liquid phase or liquid and vapor phase f and g if you do this calculations you will be noticing that g f almost exactly equals g g if you do the calculations for water at 1 bar 10 bar 100 bar at different pressures and this should raise our curiosity that may be g f equals g g and which you can show for a simple system because we have our d g is something into d t plus something into d p and if you take for our system say the t s diagram or p v diagram whatever is your choice and you take the same pressure same temperature saturation you take the f point and g point integrate this from f to g and what happens is integral f to g of d g some confusion here because this g here is Gibbs function whereas this g here is this point g if you want you can write this as a and b does not matter this will turn out to be 0 because over this path f to g d t is 0 as well as d p is 0 f to g is an iso bar as well as an isotherm. So, this turns out to be 0 and this implies that g f is g g now what is the difference between f and g at f and g you have same temperature, same pressure, same Gibbs function, but all other properties are different particularly v f and v g are different and remember that for our a t v the values of a will also be different, but remember that for a given temperature if there is a two phase zone for this given temperature you will find that there are two distinct values of v f and v g for which the Gibbs function will be the same and we will use that characteristic what is done is the following given a t v at shown earlier by differentiation obtain an expression either an expression or a computational procedure whatever that leads to pressure as a function of t v and Gibbs function as a function of t v then after having done you select a t say t naught some selected t fixed then of course to hunt out we know that for water v f suppose you have low temperature v f and v g are very widely spread. So, you choose a range of v reasonably wide and wide enough so that you expect both the saturated liquid specific volume and the dry saturated vapor specific volume if it exists to be included in this range. Now over this range that means your t zero is a fixed at some value and over this range obtain these p t zero v g t zero v now plot something like this p against g or g against p whichever way what is on the y axis what is on the x axis does not matter and you will first start with a tabulation that t naught is fixed you will use different values of v v 1 v 2 v 3 and so on you will get values of p v 1 p 2 p 3 and so on and g 1 g 2 g 3 and so on. We have seen that if there is a two phase point you must have two different values of v f and v g which gives you at the same temperature in any way temperature we have fixed we are exploring an isotherm. So, these two different values of v f and v g for a fixed temperature should give you the same value of p and same value of g and that means it is possible that your curve with something like this it does not loop over then that means in the range no change of phase, but if your curve loops over looks something like this that means as you change v your pressure increases your g increases, but after sometime it turns back and there is a point and say you approach this point like this you have one value of v say v a and you approach this point again at some other value of v say v b then if such a point exists then this p of that particular point happens to be the saturation pressure corresponding to t naught and of the two v a the two values v a and v b one is saturated phase one another is saturated saturation of phase two the lower if you have go lower to higher the lower and you are you know exploring in the liquid vapour zone the lower one will be v a that will pertain to the saturated liquid state the higher one v b will pertain to the dry saturated vapour state the lower one v a will pertain to saturated liquid state and the higher one will pertain to dry saturated vapour state I think I cluttered it up let me draw it again may be I should draw it on a new bigger page pressure against Gibbs function or Gibbs function against pressure t naught it selected you go from v over a range and determine p and g as v varies and if you get curves like this no looping over that means this is say t one no change of phase and if you select t naught equals t two you will sometimes get a loop like this or sometimes you will get a loop like this in this case suppose this is t two this is t three let us say t two t two means and you get different points on this loop you get different points on this loop as you vary v from lower to higher values and you have a common point here and the pressure at the common point is the saturation pressure at t two similarly out here this will be p sat at t two and so on and the Gibbs function is naturally the g f and g g at t two this is the Gibbs function g f and g g at t three and take this t two t two loop this point will be obtained using v a it will also be obtained using v b that means there are two values of v a and v b which give you the same pressure and the same Gibbs function and those two values of v a and v b these are saturation volume for the two phases at t two similarly here there will be one obtained v a one another point the same point will be obtained at v b and then the values of v a and v b out here are the saturation a specific volume for the two phases at t three where the saturation temperature saturation pressure is given by this so again let me explain using a as a function of t and v by appropriate differentiation and using property relations obtain pressure as a function of t and v Gibbs function as a function of t and v then select a particular temperature t naught here I have shown three isotherms t naught equals t one t naught equals t two t naught equals t three and for a selected temperature t decide on a range of v in which you expect the liquid vapor saturation to exist if you take water and if you look at your steam tables you get a specific volume of water as low as 0.001 so in the range of water you should take the specific volume as low as 0.0008 to start your exploration and the perhaps the highest value for specific volume turns out to be at the triple point 206.2 tri saturated vapor you take something like 220 or 230 meter so in that range you start exploring and then it is possible that if the temperature is high enough you will get simply a line like this no looping there may not even be a maximum and minimum whereas as you go to lower and lower temperatures say for example at this temperature t two then you plot you will get a loop like this and this loop means there are two distinct values of v say v a and v b for both of these values you will get the same value of pressure p and the same value of gives function g and by more detailed calculation and interpolation there you will be able to 0 in with reasonable accuracy of the required accuracy to the saturation pressure and saturation gives function more important for us is the saturation pressure is the saturation pressure at the selected temperature and the two values v a and v b will correspond to the lower one will correspond to the saturated liquid phase and higher one will correspond to the dry saturated vapor phase. I will not force you to do this for water substance because that will mean writing a program of few thousand lines in length because you are handling 400 parameters and you cannot do that program in a matter of 49 so I will take a Vander Waals like gas set up a very simple problem in which you should be able to do it. I think that is enough detail about this over to you K. K. Wagnashik good morning over to you. Hello sir good morning Satish Surya Unishi from K. K. Wagnashik sir my question is related to a reciprocating air compressor in our thermodynamics course that is on pre air delivery I have come across a term called pre air delivery just tell something about over to you sir. Many of these terms are technical terms used in industry and particularly when you have a fluid which is compressive. See when you talk of a pump which pumps liquid you say it pumps so many liters per hour and we know that for a liquid over a reasonably narrow range of temperature and pressure the density is not really an issue remains more or less constant. So when you ask when you say that a water pump capable of pumping say 50 liters per minute or 600 liters per hour or something like that you do not ask at what density whereas when you take a gas the question arises that what pressure and temperature at which you will measure the density for some reason industry always works with meter cube per hour and CFM and something like that it does not talk about kg per second whereas good engineers know that it is mass which is conserved volume will not be conserved but industry still insists on CFM and CMM and something like that. So naturally the question arises at what pressure and temperature physicist and chemist and sometimes engineers always also talk about under STP standard temperature and pressure and the standard temperature and pressure is defined as may be 300 Kelvin and 1 atmosphere or 25 degree C and 1 atmosphere in the proper context it is defined. I think I am not sure what free air delivery is but I think free air means pressure at ambient temperature and pressure. So when compressor says that it provides a free air delivery of so much I think that means the flow rate is equivalent to that meter cube per hour or whatever it is multiplied by the density at the ambient pressure and temperature. I am not sure but it will be something like that you can check it out the issue arises only because we have volume specified or volumetric flow rate specified but for our thermodynamic calculations we will need mass flow rate. So we will need density to be specified indirectly the density is specified by saying that it is at STP. So quite often you will see particularly in chemical engineering flow rates are mentioned in meter cube per minute or meter cube per hour but they will use a capital N before that Nm cube per hour that means normal meter cube per hour and that indicates that the meter cube per hour the meter cube or density is to be calculated at normal temperature and pressure over to you. Thank you sir, over and out. Good morning Velour, over to you. Sir, do CV exist for solids and liquids? The question is does CV exist for solids and liquids? May I bring your attention to exercise PR 3? We have derived a relation between CP and CV for a fluid. Now here we have variation of volume with temperature at constant pressure. This is something like our expansion coefficient but this dp by dt at constant volume we do not really have an expression for it but this can be converted like this. You can write using the chain relation dp by dt at constant volume will be with a negative sign dp by dv at constant t multiplied by dv by dt at constant p and hence because of this you have to write dp by dt at constant p square over cp minus cv will turn out to be now notice I will write this negative sign I will write this t then I have a dv by dt at constant p dv by dt at constant p. So, I will have dv by dt at constant p square and this I will write it as partial of v divided by p at constant t and if you see this I will divide this by v I will divide this by v and I will take the negative sign here. Now this is isothermal compressibility the rate at which volume changes with pressure at constant temperature in a relative way because we have one over v. If you take a liquid or a solid the value of this will be such that you will end up with cp minus cv almost 0. So, cp minus cv will turn out to be almost 0 for liquids and solids and that is why instead of cp and cv you tend to use only the word c, but because cp is a very common thing we continue using the word cp check this derivation out I have done it in a hurry. So, may have made a mistake check this whole thing out over to you. Thank you sir over and out. NIT Calicut good morning over to you. Explain about the psychometric chart and its properties. The question is you want an explanation about psychometric charts and its properties. I am not going to do it for the simple reason that we have restricted ourselves to a thermodynamics where we did not consider mixtures. So, the topics not included in our basic course because of lack of time and which you cannot if you include whatever has been taught so far to an undergraduate course 3 hours a week then why I think what we have learnt is more than sufficient, but if you have a bigger course. So, I mentioned yesterday that we do not even get time for doing cycle analysis, but I included cycle analysis because at many places in some way or the other it is included may be by deleting some part of the course. So, if you say cycles, but then you also have mixtures this will not be simple system and in mixtures you have two parts one non-reacting mixtures which leads to one illustration is air plus water almost always in its vapor form mixtures which leads to what is known as psychometry and the other one is reacting mixtures as say fuel plus air and liquid and so on this leads to a basic study of combustion. So, these aspects are not included in our scheme of things for this particular workshop of course there are other aspects for example, we have included cycles, but something called compressible flow the basic compressible flow that also can be included see in psychometry you have equilibrium between two different components, but no reaction that means air remains air water remains water in psychometry, but in reacting mixture fuel and air do not remain fuel and air. So, you have oxygen from air getting consumed and some part of fuel getting consumed and you end up with non-air components like carbon dioxide carbon monoxide sulphur dioxide and what have you in when it comes to compressible flow along with all over laws of thermodynamics you need the law of conservation of momentum and of course, one can always argue that the moment you include conservation of momentum you are not really restricting yourself to the domain of thermodynamics you are encroaching on the domain of fluid dynamics. So, may be that is true that is why compressible flow will sometimes be taught in the thermodynamics type of courses applied thermodynamics type of courses or sometimes it will be taught in the fluid dynamics type of course over time. The question is why is dual cycle called as limited pressure cycle well the answer is I do not really know, but one thing one can guess is that if you consider dual cycle as a see dual cycle has a part in combustion which is partly at constant volume and partly at constant pressure. Suppose this is the end of compression then in a dual cycle you have the heat addition partly at constant volume and then partly at constant pressure. If you consider it to be a modification of the diesel cycle then you will say that it is a modification with considered an initial rapid rise in pressure and then an constant pressure process, but you can consider it to be a modification of the petrol cycle and in petrol cycle you could say that since every all heat is absorbed at constant volume the pressure at the end of combustion or end of heating will be pretty high and then you can say that look I am limiting it to this pressure and when it comes to this pressure I am moving the piston to the right or increasing the volume to keep the pressure constant. May be that is the reason I guess it is called a limited pressure cycle, but it is only a question of nomenclature there is no thermodynamics involved in it over to you. Good morning Dr. Brahmara and other side Jayanthi you over to you. Good morning sir. We cannot believe we have come to the last day of the workshop and as far as I am concerned I did not find any difference between the one I had attended live with you teaching to us directly and teaching online here to the participants it was as good as it was just like the same. Coming to the we have two questions here sir. The first question is on can you please suggest some books some good books on compressible flow and applied thermodynamics which we call here as thermal engineering. First thank you very much for your comment for you may be it is very similar, but I am I have found it well not difficult but a bit odd that in front of me there are only equipment and technicians hardly any students and on this day I think at this particular instant of time there is not a single student not a single teaching assistant they are all busy with setting up the test for you and checking that things go well. As a teacher I think you are also a teacher so you will appreciate this that our teaching style and our teaching enjoyment depends on the feedback we get from our students they may not say anything immediately, but you know their facial expression their attentiveness their body language all this tells us whether they are attentive whether they are understanding whether they are interested you can get this as a group feedback and at least for the first few rows you also get individual feedbacks because the detail body language one can make out. Now when it comes to books unfortunately I took all the books with me home yesterday the two applied thermodynamics books the two mechanical engineering thermodynamic books Moran and Shapiro as well as Sontak, Borgnak, Van Vilen these have chapters which include compressible flow and which include basic thermal engineering. Professor Achyudhan's book had something but Sears, Salinger and Zimansky have nothing they have essentially been written from the physics point of view so you look up Moran and Shapiro or Sontak you will find at least for an extended course in thermodynamics or a basic course in applied thermodynamics you will find a reasonable amount of material, but if you consider a proper applied thermodynamics course then I think you do not have a really single book doing justice to that you will have to refer to different books may be one for psychrometry refrigeration one for IC engines one for gas turbines and one for steam turbines. There are some books they have now become a bit old you know earlier there was a book by published in 70s late 60s or 70s by Kadambi and Manohar Prasad they were professors in IIT Kanpur at that time for some reason no second edition third edition came up that is the situation today may be the coordinators or some interested participants can come together and write a book may be e-book which we can put up on some site and the advantage of an e-book is although it is not a traditional commercial proposition the audience would be very very wide so if it is good it will really become quote unquote popular in that sense and the second advantage of an e-book is you can almost continuously update it you know you find an error in a matter of minutes or hours it can be corrected you want to modify something add something just create it and link it up to it may be we should start with a web book on thermodynamics and then others can you know link up for compressible fluid flow to it combustion to it refrigeration to it and then we will have a may be in 10-15 years we will have a website on thermal engineering for mechanical engineers over to you Thank you sir participants would like to know about the portraits of the scientists that contributed to thermodynamics which you said you will place in model and that is second question on throttling sir I will hand out the mic to the participant The DS diagram expansion process 3 to 4 is an incline I will win some improper increases inside the system that means some heat is entering into the system so the increase the constant enthalpy process is inclined in the DS diagram The question is about the throttling process on the in the vapor compression cycle I sketched yesterday the vapor compression cycle on two planes first one was the T s plane in which let us say this is the I am clipping the top part of it let us say this is the lower pressure line this is the upper pressure line then we start off with saturated saturated vapor compress it then cool it and condense it I will I am being warned here that I am reusing page 10 but I will draw the detailed diagram again on the next page I am just revising what I have done so from 1 compress it to 2 say ideally isentropic compression then cool first and then condense it to 3 and then we said that we do not want to have a turbine or an expander from 3 to 4 but you have to reduce the pressure from the higher pressure to a lower pressure so we use a throttling process now the throttling process under certain condition turns out to be a process in which the final enthalpy is the same as the initial enthalpy and it is an irreversible process adiabatic so the entropy increases entropy does not increase in this case because of addition of heat and then the evaporation process the T s diagram is important because this adiabatic entropy increasing process is very clearly seen if you plot it on the ph diagram that is not very clearly seen the diagram will be like this and traditionally even you will find in many textbooks the diagram shows this as a straight line continuous line which is not true and this is then shown as a constant entropy process now the detailing of this process it is not really a continuous process because we do not know what is happening in between if you look up the throttling exercise from your exercise sheet I think let me tell you which exercise I am referring to this is there is a throttling calorimeter exercise in the open system OS 13 refer to OS 13 also the throttling process was explained in that the throttling process is a process in which a fluid is made to go through a contraption it could be either a porous plug or it could be a small hole just a plate with a small hole as in case of a needle valve or it could even be a small capillary that means a duct with small diameter d and a length which is pretty long a few meters a typical capillary will be the other day I have noticed the specification of a capillary a 0.8 mm diameter and length of something like 2.2 meters that depends on how much pressure and how much flow you want to have this was for a small refrigeration now here what happens is the whole thing is well insulated so if you have a throttling process let me show it like this show this as a throttle showing that a large delta p is obtained but there is no heat flow there is no attempt to extract work and you provide sufficient area at the inlet and exit to make delta ek negligible and you can neglect delta ep and if you take a control volume like this across the throttling device with inlet our thing say inlet at 3 and exit at 4 and all you have to assume it is steady state delta ek delta ep negligible they can be made so no q dot and no w dot you will end up with the condition that h 3 equals h 4 and since it is a real life process you will find that s 3 is less than s 4 or s 4 is greater than s 3 if you use the property relation you will have d h equals what is look up the property relation you will find that d h is equal to t d s plus v d p so here we have d h 0 we have d p negative and since t and v are always positive that means your d s will have to be positive and entropy will have to increase in an isenthalpic process when pressure reduces and that means that also means that in an isenthalpic process the pressure has to reduce if it has to be a real life process over to you again here it properly but I think what is requested is exergy analysis of fuels or exergy analysis of fuel cells in either case that is a distinct topic by itself so I am not going to do it in this course over to you thank you sir over to you over and out thank you very much