 Good afternoon to the afternoon session. Let me initially assign some problems for a solution. Then I am around for interaction on these problems and on the morning lecture. And after the tea break, Professor Gaitonde will handle all kinds of general questions too. So, of course, if there are problems in these that I cannot handle, here we available to handle those two. So, let me just assign a few problems to you, which you should do now. So, these are the four problems that I have written in the open system tutorial sheet. Problem number seven is relates to a compressor in a ship, in a ship propulsion system. Problem number eight is regarding a nozzle in a steam turbine. Problem number fourteen is a rigid insulated bottle, which is perfectly evacuated and you let air flow in. And problem number seventeen is with regard to a heat exchanger. So, please do these questions and I think I will be ready to take questions now. Pillai Panvel, is there a question? Over to you. Question number seven, this 5 percent moisture is given. So, sir, I have a doubt say where this moisture is to be considered. Over to you sir. Let me take it over. The question is regarding question number seven. It says at the inlet of the compressor, the steam is at 3.4 bar and 5 percent moisture. So, when you are looking on your steam tables or when you have the steam table. So, you should assume that there is 5 percent of moisture, which is liquid and 95 percent of steam. So, the quality would be 0.95, excess 0.95. So, that is how you must proceed with this. Thank you. NIT Trichy, is there any question from your side? Over. Very good afternoon sir. This question is related to the open system. For example, let me take a heat exchanger or a compressor or a turbine or a boiler. There are basically two analysis like one is thermodynamic analysis, another one is heat transfer analysis. Does these two analysis go in parallel and which analysis is more suitable for determining the performance of the system? Over to you sir. I think this is slightly confusing. I do not know what kind of analysis you are talking of. In the, if you are talking of extreme nitty-gritty as to what is the heat exchanger length, what is the heat exchanger area, we are not going to take those into consideration. So, but if you let us say you are talking in terms of a heat exchanger, we will probably only calculate heat transfer and entropy production. If you think that the entropy production is too much, you may want to design a heat exchanger where the delta T's are very small, but you will soon realize that you are probably having too much area. You and calculation of edge, calculation of all the other things will be in a separate domain which we are not doing. We are only calculating roughly what is the kind of input and output parameters we are having and whether such a process is possible, roughly how much irreversible is it. So, those are the kind of questions we will answer in our analysis. We will not get down to calculating heat transfer coefficients or areas of heat exchanger or lengths of turbine. So, if those nitty-gritties hopefully you will take care of in the heat transfer course. So, we are not getting into those nitty-gritties here. Over. Thank you sir, over and out. Over and out, please handle your problems now. Amal Jyoti, Kerala, any questions from your side? Over. Hello. So, I have two questions. One is that you are shown in the first slide that all the open systems are considered as it is also called control volumes. So, can we consider all the control, all the open systems as the control volumes? Because and second question is that in case of a reciprocating compressor, the volume is constantly changing. Can we call it as a control volume approach? Instead, it is very nice to see. Instead, we can call it simply as an open system approach rather than to call it as a control volume approach. Thank you sir, over to you. So, let me take the question. Yes, open system analysis typically is called control volume approach. So, this is exactly what we do even in fluid mechanics where we are analyzing control volume. Those are nothing but open system. So, anywhere where there is a flow coming in and going out, that is what we are calling as an open system analysis. Now, typically in IC engines and reciprocating compressors, a big chunk of our analysis is done when the walls are closed and the mass does not cross the boundaries. In that case, it becomes a closed system analysis during the part of the analysis and it is not a control volume analysis. I hope that answers your question. Over. Thank you sir, over and out. Yes, S V N I T Surat. Any questions from your side? Related to study flow energy equation applied for duct. You mentioned that the equation which you about duct, it is in terms of pressure and kinetic energy. You also mentioned that it is similar to the Bernoulli's equation. But to the best of my knowledge, when you see the duct, it is of uniform cross-sectional area. Although when there is a pressure drop, it is a frictional pressure drop and certainly it does not result in increase in kinetic energy. Please comment. Over to you. Thank you. So, if you look at the duct there, let me see. I think it is on the previous. I think I have just written this. See, if you are considering some liquid, I mean of course, I did not show any change in area. But you can assume that there may be a change in area. But the other thing is that, you know, there may be change in the density if by chance it happens. Then there can be a change in kinetic energy. But you are right. I mean, in general, if there is a change in area, etcetera, that has been considered. I did not write that down so specifically here. But if there is a change in kinetic energy, it could be because of many reasons. I am just saying that that is not negligible in this case. That is all that was being said. Over. Over and out. Okay. Over and out. Government call a Salem, any questions from your side? There is a question regarding 11th problem. So, this question, the combination of hydraulic pumps and the boiler. Here the question is asked related to the exit velocity from the boiler and the state of the system also asking. So, there is a while solving the statistical energy equation separately from hydraulic pumps as well as the boiler, I hope that some missing data is in this problem exists. Please explain in user. Over to user. In the problem number 11, there is first a pump and then it goes into a boiler. So, what you can do is first consider only the pump as an open system. The inlet state is given, the power is given. So, you can calculate the outlet H of the fluid as it exits the pump. And then that you can consider as the inlet state for the boiler, consider the boiler as a separate open system. And then the Q input into the boiler is given. So, now, once this is known that you can calculate the exit state of the boiler. So, you can consider these as two continuous open systems in a sequence and solve them separately and you should get your answer. Over. Sir, for solving separately for the pump velocity boiler, we cannot proceed further from the pump. I hope there is some missing data, given data. See, I can see very clearly that the inlet to the pump is given as 1 bar 35 degrees C. H at the inlet of the pump can be found out. M dot is given because ray 120 kg per minute is given and the pump power is given as 110 kilowatt. See, if I just apply the first law for open systems, you will just get that W is M dot H e minus H i. H i is known. W is known. M dot is known. So, all you have to do is get H exit and that is all that is needed because that is going to be the H for the inlet of the boiler. So, I do not think there is anything else that is really necessary. Actually, you can even do this as a single control volume and it would not matter. But I think everything to find out the exit state of the pump is already given in there. So, I do not think that should be a problem at all. Over. Sir, what about the velocity where kinetic energy changes in the pump? As already mentioned for turbines, compressors, pumps, we will assume that the net change in kinetic energy from inlet to exit is quite negligible. So, you can safely neglect the changes in kinetic energy for the pump, for the compressor, for the turbine. Over. Sir, the question is they are asking about the exit velocity of the boiler. If you know the inlet velocity of the boiler only, we can able to calculate our exit velocity of the boiler. For knowing velocity of the inlet velocity of the boiler, we should know the data from pump side. See, the mass flow rate of the boiler is given and if m dot is given, m dot is just rho into A into velocity. So, the area is given, the mass flow rate is given, rho will come from the exit state of the boiler because you know what the exit state of the boiler is. You know the pressure and you know the enthalpy. So, you can go to the steam tables and find out the specific volume. So, once rho is known, area is known, all you have to do is calculate velocity because m dot is known, m dot is just rho A V. So, you do not need any other data except the mass flow rate. Over. Thank you very much. Panvel, any question from your side? Over. Regarding question number 7, part B. Sir, is the process possible or impossible? In what context are we have to find out the possibility of the process? Yes, as I said, you know if you just apply the first law, you can just get, you know that delta H, you can correlate delta H and work, but what has been given is directly you have been told what the inlet and exit states are and you have been told to you know yourself ensure isentropic or not isentropic. So, what you should do is just see if the second law is applicable that is whether you need to, whether the second law is being satisfied I should say. So, you just check on the entropy of the inlet and exit states. It is an adiabatic system. So, the entropy at the exit better be more than the entropy at the inlet or at best it should be equal to the entropy at the inlet. So, that is definitely very necessary to check because someone just cannot give you the inlet and exit states and say I will design a turbine like this or a pump like this. You should definitely check whether the exit state entropy is more than the inlet state entropy. So, the limiting case, limiting exit state is of course, you know the going to be the one with entropy equal to the inlet state. So, please check on this fact over. So, may I have any questions from your side? Over. The condition of steam at inlet is wet steam and at outlet it is dry and saturated steam. Compressor can handle two phase mixture? Yes, I mean to some extent yes. So, we are just assuming that in this case the inlet state is given. Of course, we are not talking of you know whether the compressor can really handle such thing or not, but in most cases it can handle especially in sometimes in refrigerator system. It does handle a small amount of moisture over. Over to you sir. Thank you. Amrita Coimbatore, any question from your side? Over. Sir, good afternoon. This is Uday from Amrita Coimbatore. My question is regarding 6.8. There you asked us to find out the exit velocity and exit area. Actually the diameters has not been given there. So, how we have to find the velocity and area in terms of inlet velocity? Yes, if you check the question, the inlet conditions for the nozzle of a steam turbine are 60 bar 350 degree Celsius. The exit conditions are 10 bar and 0.9 dry. So, what you need to do is first this is a nozzle. So, you can go ahead and get the velocity. The steam mass flow rate is given. So, m dot is known. From the nozzle equation, you will get the velocity and since you now know the state of the system 10 bar and 0.9, you can get the density or specific volume and m dot is just rho A v. So, rho is known. Velocity is known. All you have to do is calculate area. So, m dot is given for it. Over. Thank you, sir. NIT, Nakpur, any questions from your side? Over. Sir, my question is steady flow. We derived so many steady flow energy equations for the different mechanical devices were producing, absorbing, non-work producing, non-work absorbing like that. So, in that case, we assume that the flow is steady means our constant mass flow rate is taken place in that particular device. Suppose we have any application like natural convection solar air heater where the flow is taken place because of the natural convection. As the temperature of the plate, absorber plate go on increasing, the flow rate will getting enhanced. So, in that particular case, will that energy equation is suitable? See, if there is no steady flow, that would be tough. But I think you are asking for natural quick. If you are asking for natural convection and if it comes to steady state, then you can take a suitable control volume and try to analyze that and you will realize that those steady state flow situations can be analyzed using this. So, as long as this flow is developing or transient, it is difficult to analyze this. But once it has reached steady state, most of these problems because you can consider, I mean as long as you can calculate, you can consider q dot coming in or going out to the surface. You will have to get some method of calculating m dot and then you can go ahead and solve the problem. Thank you sir. Sir, one more question. Can we have any correlation for calculation of the mass flow rate for the natural convection? What do you say? I think those kind of problems we will deal within the heat transfer course because that is where you will get into what is the rough velocity, what is the rough m dot for natural convection and what are the heat transfer coefficients for natural convection. So, since we have a heat transfer course follow up and that is more relevant to that course, I think that is best answered there and not here, over. Thank you sir, over to you. Amruta, Coimbatore, any questions from your side, over. Sir, this is Uday again from Coimbatore. Still I am not clear with problem 6.8. You explain that it can be found from mass flow rate given. The mass flow rate is given mass flow rate is defined as rho times area times velocity density can be found from the state of the property of specific volume area is not given and velocity is not given there are two unknown variables. So, how can we? So, as I said see this is a problem for a nozzle, it is not a turbine problem. So, knowing the inlet and exit state you know the change in delta H and the delta H is entirely going to result in a change in the velocity. So, one assumption you can definitely make is that the inlet velocity is negligible. So, all the delta H goes in increasing the exit velocity and you will have V e square by 2 is just H e or H i minus H e and that is how you get the exit velocity and from the state as you said you can get the specific volume and since you know the mass flow rate you can get the area. So, the purpose of a nozzle is just to get that exit velocity and since it is a nozzle the very first thing you have to do is actually calculate the exit velocity without anything else and then you can see whether the last thing of course, definitely you need to do is check whether this is possible. I mean since it is assumed to be adiabatic please check whether you know the entropy increases at the exit otherwise someone just cannot directly give you the inlet and exit state and tell you to go ahead with it over. Sir, if that is the case then conservation of mass is not valid because rho A V at entrance must be equal to rho A V at exit. So, if inlet is negligible then conservation of mass is not valid. So, how can we tackle this problem? I think see this is an engineering problem definitely when you are talking of the velocities you should realize that definitely conservation of mass we are assuming to happen and what we are doing is see this is a big nozzle let me draw it here and for the same mass flow rate even here it is around 10 meters per second and if it goes here close to 800 meters per second then you will say why do I need to know the inlet velocity at all because if I square this this is having a negligible effect if I square this here and you realize that there is really no need to bother about the inlet velocity. In fact, in many nozzle problems we just do not bother about the inlet velocity definitely you know you should not assume that you know the flow is being created out of nothing definitely a flow is being created and there is a small flow at the inlet of the nozzle, but we are not really bothered about that because most of the delta it is even if I take this 10 meters per second into consideration it is going to only change in some decimal point at the exit velocity. So, neglecting it is not going to cause any kind of harm. So, that is a very typical analysis that we do for all nozzle problems unless we are talking of nozzle velocity between stages of turbines sometime that the exit of one stage of a turbine the velocity is reasonably high in that case during the design we need to give the inlet state to a nozzle or inlet velocity otherwise in most nozzle problems it is very good assumption to just neglect the inlet velocity component because it is not going to affect the calculation you know significantly at all over. K.K. Varanasiak anything from your side over. Hello sir, I want to know the minus VDP work. Is there any significance of minus VDP work in open system? I think during our analysis you saw that nowhere we have come across this minus VDP term. So, I do not think you should even bother or even try telling your students that there is anything like this it is only something confusing we will not talk about it over. Thank you sir, but in many books it is given like that. So, should we refer to that or we should go by our method only. I can definitely recommend that that section should definitely not be read. So, you must throw those books out if you have them and continue with this because definitely it is not the right way to teach things over. Thank you sir, thank you very much over and out. J.N.T.U. Hyderabad you have raised your hand any questions from your side over. Good afternoon sir. This is regarding the second part of the seventh question. Is the process possible or impossible? Why what is the limiting exit state? Over to you. So, in you should realize whether it is any kind of problem compressor, turbine and even heat transfer device where you know the Q dot. As long see in a turbine and nozzle problem there is no Q dot term. So, you must ensure that S e minus S i comes positive that there is a net entropy production rate or in the best case it is 0 because that is just the reversible process. In a heat transfer device you must ensure that S e minus S i and minus Q dot by T is such that there is a positive entropy production rate. So, these are the things you must ensure to see if the process is possible. The limiting case is obviously when the equation is just satisfied. So, for example, in turbines if S e is equal to S i and the net entropy production rate is 0 that is when the that is the limiting case. Otherwise you should always get something where S dot P is greater than 0 that is M dot into S e minus S i should be a positive number. So, that is definitely what you should get. So, do not just take someone's data on what is the inlet and exit conditions and believe it. I think that is what we have been trying to tell. Definitely you can always calculate W dot using inlet and exit states, but unless you check for the entropy you will not know whether that thing is possible or not. Definitely check if S dot P is going to be greater than or equal to 0 in the best case over. Thank you, sir. Over to you. Over and out. KK, there is no video from your side, but if your audio is working please go ahead with your question. Hello, sir. Can you take an example of charging and discharging of a system or illustration? Only charging of a system or discharging of a system? I hope you do not mean electrical charging or discharging. If it is charging of a cylinder or discharging of a cylinder then that is exactly what problem 14 is which we have assigned to you just now. So, I hope this is what you mean by charging and discharging. Over. Thank you, sir. Over and out. P.H.G. Coimbatore, any question from your side? Over. Sir, in the problem number 7 and problem number 8, so we have to find out whether the process is possible or impossible and you ask that why and what is the limiting exit state and in the problem number 8, same thing, is this process possible or impossible? Why? What is the limiting exit state and exit velocity? Can you guide me for this problem? How to say that process is possible or impossible and what is the limiting state and how to find out that? I am taking over. See, during the class we had drawn an H.S. diagram. Let me draw that again. And well, problem number 7 is a compressor problem and problem number 8 is a nozzle problem. So, in one case, in the compressor problem we go down in pressure P 1, P 2 and in the nozzle problem, sorry, in the compressor problem inlet is at a lower pressure, exit is at a higher pressure and in the nozzle problem inlet is at a higher pressure and exit is at a lower pressure and what we had said clearly in class was that since Q is 0, you must get M dot S e minus S i is entropy production rate which is greater than or equal to 0. Hence, S e should be greater than or equal to S i and we had drawn that this was let us say for example, here i to e, this is the ideal case and in the reality you would go somewhere here. Similarly, in this case you would go somewhere to increasing entropy. Now, what has happened is that we have directly given you the exit state and we have not told you what the entropy is. Now, since you know the state, you should go and find out what the exit state is. See that its entropy is either equal to the inlet entropy or it is more than the exit entropy. The limiting case is always when the exit entropy is same as the inlet entropy that is the reversible case. In all other cases, your exit entropy must be greater than your inlet entropy. So, if the specified problem, let us say the compressor problem, if we have given you the exit state which is lying here to the left as far as entropy is concerned, then this would not have been possible. This process is not possible. So, we would like you to please check what the entropy is at the exit. You cannot blindly take the given data and just do your calculations regarding power input and exit velocity. You must also check in for the entropy condition over and we are getting that answer entropy production in negative value. So, the limiting case is S i is equal to S e. Yes, the limiting case is S i is equal to S e. So, if you are getting a negative value right now, that process is not possible. So, you must figure out the state where S i is equal to S e. That is possible and all other states may be where S e is greater than S i. You must check for those. Those are all possible. Over. Then what about that limiting exit states, sir? We have to take, what state we have to take? What is the limiting exit state? For example, now let us say that your inlet to the compressor is 3.4 bar and 5 percent moisture. Please get the entropy at that state from the steam tables. So, assuming that the compressor is going to deliver at the same exit pressure of 8 bar, please calculate at what condition the entropy would be the same. That would be the limiting case as far as the entropy, as far as the compressor is concerned. As far as the nozzle is concerned, the inlet state is given. So, it is 60 bar 350 degree centigrade that is going to be a superheated state for steam. Please get the entropy from the superheated tables and then again go down at 10 bar. So, exit condition we will always assume that the pressure is specified. So, we will see where on the pressure line and on the isobar line, we will get our exactly S e is equal to S i. That would be our limiting case. So, if S e is greater than S i already, then there is no problem that thing is possible, but the limiting case is when on the 10 bar line you have S e is equal to S i. Over. Thank you, sir. Over and out. Somaya, any questions over to you? Sir, how to solve problem number 9? Yes, I am taking over problem number 9. So, it is given the inlet conditions of a water pump are 1 bar and 25 degree centigrade. The exit pressure is 180 bar. The pump consumes 75 kilowatt of power and pumps 12000 liters of water per hour and this liters per hour that is given at inlet conditions. Determine the temperature of water at the exit of the pump. If we define the ideal pump as the one which does pumping isothermally, what is the efficiency of the pump? So, let us take the system here. It is a pump. So, I will just draw one circle here and what is coming in is given. So, that is 1 bar and 25 degree centigrade and the exit pressure is given. It is 180 bar. Now, you know for a pump, again we had w dot into the pump or if I just want to calculate it. It is m dot h i minus h e. Now, this is the simplified equation that I have. Now, h i is known because I know the inlet conditions. It is 1 bar and 25 degree centigrade. So, if it is sub cooled, as we said, if you apply the incompressible fluid assumption, then even if h is not given in the steam tables, you must just take it as u plus p v. Pressure is known and you must get the u and v for the saturated liquid at 25 degree centigrade. So, that will give you h i and m dot you can calculate from the flow rate. So, the flow rate is 12000 liters per second, not liters per second. Since, you know the flow rate as a volumetric flow rate, at 25 degree centigrade, just get the rho and using the rho and this flow rate, you should get the m dot. So, m dot is known, w dot is already given to you. So, you must, you will automatically get h e. Now, if h e is known and the exit pressure is known. So, one thing is you will have to get h e is u e plus p e v e and in this case, you know what is h e, you know what is p e and you will probably have to calculate what is u e and v e. Now, one method is you can assume some temperature and figure out what the u and v at that temperature is, plug it in here because at that same temperature you will get both u and v, plug it in here and hence get the temperature. So, that is how you must go ahead and try to figure out what the temperature would be at the exit of the pump. Now, if you define the ideal pump as the one which does the pumping isothermally, you will get that, if the temperature does not change, the entropy does not change and it is the same as taking the isentropic efficiency and hence you should get the ideal pump work in that case and then go ahead and calculate the efficiency of the pump over. Sir, what is the significance of flow work in case of open systems? See, in all open systems, you will have at least one place where either mass is coming in or mass is going out and or you may have both just in the most general case that we have considered, we will have may be both of them normally. In fact, nearly in all devices that we are considering, we will have both one inlet and one outlet unless it is a charging discharging situation. Now, in all these cases, wherever mass is flowing in, you will realize that there is some work involved in either pushing in mass into the system or pushing mass out of the system. So, if you take a closed system thing, then you will realize that it involves stretching, pushing the boundaries against either an external pressure or external environment pushing the boundaries into the system. So, both these constitute work, PDV kind of work and hence by default in any kind of system, sorry, an open system analysis, the flow work is definitely going to come because of this and you cannot avoid it. That is one of the reasons the PV term always comes in an open system analysis and since the PV term comes, you can just combine it with a U-term and nearly and you will always see that there is an H-term in any open system analysis over. Anaitin Agpur, any questions from you over? Good afternoon, sir. My question is, can we imagine a quasi-static frictionless adiabatic expansion of a fluid in case of a nozzle so that it can become an isentropic expansion? Over to you, sir. Thank you. See, in fact, in most nozzle analysis, this is what we really do whenever we are calculating exit states of the nozzle. The most common analysis, in fact, this is what we do. We calculate an adiabatic isentropic and hence reversible expansion of the nozzle and that is really our limiting case. So, you can very well imagine this as far as we are concerned. That is the limiting case. It is a possible case and hence you can go ahead and all real cases will be where the entropy will increase at the exit. But they shall be next to impossible, sir. Yes. In fact, all our analysis with reversible adiabatic cases is as good as impossible. In fact, as has been already mentioned quite a few times in the previous lectures, this is just about impossible. You can never get such things even if you try your best. These are for our analysis. It is just giving us the limiting case only. You should know what the limiting case is. This is, but as far as nature is concerned, this is impossible. Everything else which is possible, you must realize that there is going to be an entropy production rate. Thank you. Over to you, sir. Okay. Over and out. See you after 10 half an hour at 4 o'clock. Over.