 So, now we will begin our next topic and that is open thermodynamic system. We will look at some illustrations of open thermodynamic systems then we will take a specific case for study and derivation so that we have a common structure, common symbolism for derivation. Then we will generalize that and then we will apply it to typical engineering systems. And finally, exercises that is what we, the illustrations are open systems are systems from which mass comes out, mass goes in or both during the process under consideration. So, turbines, compressors, pumps, fans, boilers, condensers, heat exchangers, ducts through which something is flowing, then rooms and buildings unless they are totally sealed off would essentially be any real life system for example car is a very complicated open system fuel goes in exhaust comes out air also is taken in water is periodically supplied human being a very, very complicated open system there is no end to the number of open systems we can think of and this is the unfortunate nomenclature and open system is also known as a control volume. This comes from the earlier nomenclature that what we consider as a closed thermodynamic system was earlier named as a control mass and because of that the open system was given the name control volume but the analogy is not perfect because for a closed system control mass perhaps is okay because the mass remains within the system so it is identifiable as that of the system nothing comes in nothing goes out whereas in control volume the volume of a system is not fixed it may be fixed as in case of a boiler turbine but in case of for example a human being a car tire which gets inflated deflated a water tank in which the level goes up and down the volume is not fixed and when we say that we can have the boundaries of the system any way we feel like we can keep on moving the boundary changing the volume as we feel like so there is nothing special about the volume of a control volume so the although we will generally be using the word open system whenever it comes to subscript because everyone uses CV as a subscript for an open system we will continue to use CV as an open system. Now look at a schematic open system for a typical open system the input and output or inflow and flow could be through ducts or ports all of us know the inlet duct of a turbine exhaust duct of a car similarly we know the inlet port and the transfer port of a two-stroke engine it could be duct or port but it could also be a flow continuously across the boundary because our control surfaces or system boundaries could be imaginary I can put a system boundary half way through this room and air will be flowing all over it left and right even human beings will be crossing that left and right. So we will typically have a discrete model as shown on the left here or a continuous flow model as shown on the right here. Here I have shown two input streams or two inflow streams and three outflow streams you could have any number whereas in a flow field you would have a continuously distributed inflow and outflow and its unsteady flow the inflow and outflow locations could even be keep what we will do is for our purpose of analysis we will use a schematic open system one inlet one exit unfortunately I have made the innards of the control volume or the open system gray that grayness is not very clearly seen there may be the amount of grayness I will change in the final version you see a slightly different shade but that is not really gray right. Now the situation that we consider is that let the state of the control volume be changing with time. So we will say let the volume be v as a function of t mass m as a function of t and all other properties as a function of t we will also assume that the fluids at the inlet and exit are in local equilibrium by local equilibrium we mean that if we take a small sample of the fluid there instantaneously then by doing appropriate experiments on it we can conclude that it has a unique value of temperature pressure and all other properties because the complete system from inlet to outlet we may not find in equilibrium because there may not be a unique value of pressure unique value of temperature and so on. So we do not have to worry about the whole system being in equilibrium what is required is that the inlet and exit or inlets and exit have what we call local equilibrium but that also means that the energy E t entropy S t of the system etc. are formal variables we say let the energy be E t we are not really worried whether we are able to properly evaluate it or not at this stage then we will also assume for algebraic simplicity that the flow situation at inlet and exit is one dimension and that means everything is uniform across the cross section this way I can write mass flow rate as an algebraic equation expression otherwise I will have to integrate it as rho v normal dA integrated over the surface so I will be just writing more complicated expressions but this is not a disadvantage there are more than one inlet exit we will sum them up if it is a continuous flow over a surface we will integrate over it then we will assume that the inlet state is represented by the symbol I subscript I so we will have AI density rho I volume VI energy specific energy EI and so on mass flow rate m dot I similarly the exit state will be denoted by the subscript E so area AE etc etc. And again for algebraic simplicity we will assume that the velocities are normal to the areas so when I write VI it is actually the normal component that I am talking about but again I will use the same VI for the specific kinetic energy so if there is a off normal component include that in the kinetic energy but do not include it in the flow rate that is all that you will have to do. Then we say that the rate of heat transfer to the control is control volume from its surroundings our normal nomenclature is q dot t could be a function of time this would be in watt kilo watt or something like that. Then we will consider the rate at which work is done by the control volume is W dot s s for historical reasons we will come to that later as a function of t it could be time dependent. Now with the proviso that W dot s contains all components of work expansion work for example you are taking the tube of a tire or the bladder of a football you are inflating it it is getting changing in volume so there will be a PDV work but that is included if you are stirring something that is included if some shaft is moving like extracting power from a petrol engine that is included but remember you are making some fluid flow through the system and to push it in and take it out some work is involved that is not included in W dot s that we will take into account separately. So it includes expansion work, stirrer work everything else except the work required for making the fluid flow into and out of the control. Now we consider inlet and exit plugs even here the plugs have been great but so what we assume say at the exit we are considering a time period from t to t plus delta t. So at the exit we say that if the exit plane is at this location at time t at time t plus delta t it would have moved outward slightly that means all the particles which were here at time t have moved at time t plus delta t. The one-dimensional flow and velocity normal is used to shift it in a parallel fashion that is a simplicity we get. Similarly we say that at the inlet plane whatever were the molecules and particles at t plus delta t at t they would have been still away from the inlet plane by some distance. So this is the inlet plug from t to t plus delta t which enters the control volume and this is the exit plug which goes out of the control volume during the period t to t. Now this is the same nomenclature for the inlet plug. The exit plug is at E f at t E prime f prime at t plus delta t. The inlet plug is at C b at t. Now our job is to consider an equivalent closed system because we know how to apply our laws of thermodynamics. First law, second law in the conservation of mass to a closed system. So if I can transpose my behavior of an open system to that of an equivalent closed system I can very easily apply my laws. So what we do is we consider a closed system which occupies the space a, b, c, d, E f, at A at time t that is a, b, c, d, E f, A at time t. So it includes the inlet plug but does not include the exit plug. It occupies the space a, b prime, c prime, d, E prime f. So it includes the exit plug at time t plus delta t but does not include the inlet plug. Now what is the consequence of this? No mass flows across the boundaries of this system during the period because just the plane a, b moves to a prime, b prime. The plane E f moves to E prime, f prime. That is all that happens. So this is a closed system and we apply the conservation of mass to this system then we apply the first law and then we apply the second law. Now here I have tried to show again because the grayness is not there. Maybe here it is there. Can you see the gray part? This is the closed system at time t and this is the closed system at time t plus delta t. Notice that as it goes from t to t plus delta t the inlet plug comes in, exit plug goes out. As if there is a piston sitting at c, b pushing it in and another piston sitting at E f pulling it out to E prime f. So this is at time t, this is at time t plus delta t plus delta t. Is that clear? And now you agree that it is a equivalent closed system which moves from the extent bc to E f at t to b prime c prime to E prime f prime at t plus delta t. So now we are free to apply our laws of conservation of mass, energy and second law of thermodynamics. That is conservation of mass, first law and second law to this equivalent closed system. That is what we are going to do. So now here we have shown the processes and interactions. Well the process is what is happening so you cannot show the process but notice that the state of the system changes from m at t to m at t plus delta t, e at t to e at t plus delta t. So on for all the relevant properties. During that period the heat transferred to the system is q dot delta t. Work done by the system except for pumping or flow in and flow out is w dot s delta t. Something flows out, the work required to be done for that let me call it we. Something flows in the work required to make that thing flow in is wi. So this is our symbolism. Period t to delta t, that is the time level and interactions q dot q interaction, w interaction made up of three components 1, 2 and 3 and change of state. So now we come to conservation of mass, exactly what you were talking about. See mass of the system at t plus delta t is mass of the system at t. I think I should emphasize that these masses energies pertain to the control volume and not to the equivalent open system. And our control volume is always between b prime c prime and e f. So mass of the system at time t plus delta t is mass of the system at t. That is our conservation of mass. But mass of the system at t is mass of the control volume at t plus the inlet plug. And mass of the system at t plus delta t is mass of the control volume at t plus delta t plus mass of the exit plug. And that reminds me that means in this figure for all m, e, v and s I should write subscript c v. That would make things very clear. So we now know the masses of the system at t and t plus delta t. So substitute that in this. Before that I have said that the mass of the exit plug is inlet plug is density at the inlet, area at the inlet, velocity at the inlet into delta t. Similarly, mass of the exit plug is density at exit, area at exit, velocity at exit into delta t. So now we substitute all this into our conservation of mass equation which is this. And we get this expression. Now all that we do is divide throughout by delta t after transposing terms and take the limit as delta t tends to 0. And we get our conservation of mass equation which is dm cv by dt, rate of change of mass of the control volume with time is rho i ai vi minus rho e ai v. Now we do not stop here. What we do is we use the nomenclature that rho i ai vi is rate of inflow of mass, call it m dot i. And rate of outflow of mass rho e ai vi we call it m dot e. And in terms of these we have our standard conservation of mass equation which is dm cv by dt is m dot i minus m dot e. The first law application is a bit more complicated because there are three components of work. So let us try that out. First law for the system. For a closed system delta e is q minus w. So for our equivalent closed system we have to write expression for delta e expression for q and expression for w during the period t to t plus delta t. Now delta e is e of the system at t plus delta t minus e of the system at t. Just like mass, e of the system at t is that of the control volume of t plus exit plug. At t plus delta t, sorry control volume of the system plus inlet plug, at t plus delta t control volume of the mass of the control volume plus exit plug, energy of control volume plus energy of the exit plug. Now energy of the inlet plug is mass of the inlet plug multiplied by specific energy. Similarly energy of the exit plug is mass of the exit plug multiplied by its specific. Substituting all this in the delta e expression, we will get delta e equal to all this big expression which contains mass of the control volume at t, mass of the control volume at t plus delta t, energy in the inlet plug, energy in the exit plug. Then this is the simplified version in terms of m dot e and m dot i. And now next thing is q. q is simply, we know the q dot is the rate of heat transfer, so the amount of heat absorbed will be q dot into delta t. Now we come to work. The work interaction has three components. W dot s is the rate of power output from the system except for the inflow and outflow. So that is W dot s into delta t plus the work required to throw the exit plug out and work required to suck the inlet plug in. Now see here, this is the piston analogy. The work required to push the exit plug out is assuming there is a piston at the exit, pressure would be pe. The displacement would be ve delta t, force would be pea e. So this is either you can consider force into displacement or pe into delta e. In either case this becomes, because writing in terms of m dot, notice rho e a e ve will be m dot e. So this will be m dot e divided by rho e which I am writing as m dot e multiplied by ve. So this becomes pe m dot e delta t. Clear? This is a point which one has to try to explain to the student properly because this is not immediately very clear to many students. Once this is clear, the inlet plug also is treated the same way. The work required to suck in the inlet plug. It is negative here because here the fluid is being pushed out, here the fluid is being pushed in or being sucked in. So the pressure acting on the piston is still outwards. If you remove the piston, things will start leaking out but I am trying to push the piston and the piston is pushing me back. So that is why that negative sign. The system does the work. Control volume or the equivalent system? Actually it is equivalent system. So who does the work is? We are applying it for the equivalent closed system. So the equivalent closed system is doing the work. Later on in the final thing we will be taking limit as delta t tends to 0. So as delta t tends to 0, the region occupied by the control volume and region occupied by the system is the same. So we will say our open system is doing that work. So that is why, notice this is the work done by the system in general, work done for pushing out, work done for pulling in all by our system which I have shown by gray. Now this means that my we plus wi can be written down as m dot e peve delta t minus m dot i pi vi delta t. And now the first law becomes I am combining all these terms. A big expression containing ECV at t plus delta t. This is the energy of the exit plug, energy in the inlet plug. So the left hand side or the first line of this expression is the delta e term. The first term here is the q term and the remaining three terms here are the w term. So this is our complete expression of first law as applied to our equivalent closed system process from time t to time t plus delta t as I had shown earlier by the moving plugs. Now all that we do is transpose and combine terms because you find m dot e and m dot e common here and we keep just the terms containing ECV on the left hand side. And the next thing we do is divide by delta t, take the limit as delta t tends to 0. And now we get what I call is the most general form of the first law for open systems. The conditions here are one inlet, one exit. If there are more than one inlet just sum them up over the inlets more than one exit, sum them up over the exit. If there is no inlet drop that term, if there is no exit drop that term. And notice that if there is neither any inlet nor any exit drop both terms and then you are back to your first law for closed system. Just the time differentiated form. Now this general form we will reduce to a special form. Of course the restriction of local equilibrium will be there but also one dimensional flow and normal flow. So you can relax that by a more complicated expressions if need be. Now we expand EI plus PI VI in a rather restricted sense. We say that what flows is a fluid. A fluid is usually a simple compressible system and hence for us the most significant parts in EI particularly of interest to mechanical engineers would be UI the thermal internal energy, VI squared by 2 the kinetic energy and GZI the gravitational potential energy. That means other components of energy are neglected. If they are there you will have to include them here. This PI VI is being carried over here. Similarly EE plus PE VI, EE is expanded as UE plus VI squared by 2 plus GZ. PE VI is carried over but then we notice that we already have a short form for UI plus PI VI and for UE plus PE VI. So we write them in terms of enthalpies HI and HE respectively. And this is one of the main reason why we have that UI plus PI VI defined as HI. Just a definition of enthalpy. And once you do that you get this is the reasonably general form of the first law for open systems of used to mechanical engineers. But tomorrow you have a dielectric fluid that energy term will have to be included. Tomorrow you have a plasma favorite fluid of Bandarkar. You will have to include all those electrical energy, chemical energy of all the species, magnetic field I thought magnetic energy as required into this EI. So all that happens is EI instead of simply being U or E being U plus V by V squared by 2 plus GZ will be these three components plus any other components which are needed. Now although this is a reasonably general form quite often we will have to simplify this. But before we do the simplification let us see what are the further items that are on our menu. We have to apply second law to the open system. We have to take care of special cases of these laws for open systems. By special cases I mean restricted cases and we will apply these two typical engineering systems and finally we will solve problems because unless we solve problems ideas do not get clear. So what we have done so far? But henceforth we will be using a simpler schematic for showing an open system rather than that big box with inlet plug, exit plug. That was necessary for derivation. That was necessary for understanding. Once it is derived a simple symbolism will do. Our typical open system will just be shown by a small block. I have shown it with a rounded rectangle but it could even be a circle, ellipse whatever is your favorite figure. An inlet stream will be shown and exit stream will be shown. Fortunately for us a large number of our components will have one inlet stream and one exit stream. And we know how to handle two inlet stream, one exit stream, one inlet stream, two exit stream, two inlet, two exits. They happen in heat exchangers, mixers, splitters, separators. All those things we know how to handle. So we will show q dot, we will show w dot s. Since the flow work has already been subsumed using enthalpy we do not have to show the flow work as a separate interaction. That is taken care of by h e here and h i here. We now come to the second law for open systems and you will be wondering that again we will go to that control volume and delta s for the system inlet plug, exit plug. We can derive an expression using the same techniques and I will leave it to you as an exercise to do that. But I will also like you to appreciate something by taking a different route. We will use analogy because I mean you are teachers, you have completed your studies in mechanical engineering, basic mechanical engineering. So you would have realized that say the laws of conduction heat transfer, the laws of convection heat transfer, the laws of diffusion, they are very similar to each other. So many analogy schemes are applied and are very similar. What we are going to do is the following. We will first look at the conservation of mass. For a closed system we have a very simple equation. d m system by d t is 0. For a control volume or open system we have the right hand side instead of 0 we have m dot i minus m dot e. So something which flows in, something which flows out. And the fact that there is no other term indicates that there is no source or sink of mass. But for energy q dot and w dot could be sources or sinks of energy. So this is for our closed system and this is for our open system, conservation of mass. Now let us go to first law which is conservation of energy. For a closed system we have our known d e system by d t is q dot minus w dot. Same thing, delta is q minus w, time differentiated form. And for an open system we have, notice d e system gets replaced by d e c v by d t. q dot remains q dot, we have no fight with it. W dot becomes w dot s, there are two other terms which get hidden here. P i v i m dot i, P e v e m dot e, they are actually part of w dot. So if this is so, what would be in an analogous manner, the second law which I call the entropy principle. For a closed system we have rate of change of entropy of the system with time is sigma or integral q dot by t as explained by Professor van der Kerr to you plus s dot p. And the second law would dictate that s dot p is greater than or at most equal to 0. Is that right? Time differentiated form of the second law for closed system. So when it comes to open system what would happen? This would become d s control volume by d t. This would remain as it is, this would remain as it is plus you will have terms something to do with inflow, something to do with outflow. Notice that in case of mass we had simply inflowing mass, outflowing mass. Out here we had, remember these two terms are actually terms pertaining to w dot. So we have m dot i e i, m dot e e e. So similarly in the second law we will have terms m dot i s i minus m dot e s c. And that is exactly what we are going to have. I would like you as good students of thermodynamics to go through the detailed derivation but be sure you will end up here. So do not take it just because I said by analogy it is so. By analogy we can write it but we can confirm it only when we derive it from first principles. So you should notice the differences as well as the similarities. And if there are differences you should know what are the causes of the differences. For example mass cannot have a source or sink whereas q dot and w dot can change the energy of a system. And mass as it flows in brings with it some energy, as it flows out takes away some energy. In a similar fashion when mass flows in it brings in some entropy, flows out takes away some entropy. That will affect the entropy of the control volume. Again the second law would dictate that s dot p must be greater than or equal to 0. Now many, many of our applications and perhaps for a first course in mechanical engineering for undergraduate students, I think most of our applications would be of a steady state. Not very complicated problems. So a steady state is a very common occurrence. It is not a very common occurrence for aerospace engineers for which the aircraft is never really is in a steady state except while cruising. But for us power plants, pumps, turbines, they tend to run for hours and days together in a similar situation or very slowly changing situation. Such a situation is known as a steady state. So flow rates, states do not vary with time. So many of these parameters are constant and there is no accumulation of mass in the system. There is no accumulation of energy in the system. There is no accumulation of entropy in the system. And this simplifies the situation and such a situation where these three are 0, not at an instant but over a reasonable period of time and q dot, w dot is m dot i and m dot e and inlet and exit states are unchanging over a period of time is known as a steady state. If they have not done a course in fluid mechanics, you should take some time in helping them understand what is meant by a steady state. Now in a steady state, we reduce the equations to their much simpler forms. For example, the conservation of mass becomes m dot i equals m dot e and hence we need not even keep the subscript i and e. We usually replace it by m dot and we say that this is mass flow through the system. That means both at the inlet and exit and steady. So rather than a rate of inflow of mass and rate of outflow of mass, we say rate of flow of mass through the system. First law becomes I am usual habit is to put q dot minus w dot s on one side. On the right hand side, you have this term m dot i multiplied by h e plus e is m dot into there is no i and e. So m dot into this exit properties minus m dot into the inlet properties and usually it is regrouped like this. m dot is a common factor and in the square brackets you have three terms one after another. One is h e minus h i which is usually denoted by delta h. Then this is the delta e k v e square by 2 minus v i square by 2 and this is delta e p g into z e minus z i. In the closed system analysis whenever we talk delta phi for some property phi, it is usually the final state minus initial state. In open systems with one inlet one exit whenever we talk of a delta phi that phi invariably means exit state minus inlet state. So that is usual symbolism otherwise if there is any confusion continue with h e minus h i that is very clear. And now what happens to the second law m dot into s e minus s i becomes sigma or integral as appropriate of q dot by t plus s dot p and of course the real second law dictates that s dot p must be greater than or equal to 0. We will be using these forms very often. Now although these are general forms we have still more simplified forms for special cases and we will look at these one after another. Why do we look at them? Because not all terms in these equations are significant under all situations and we need to make suitable assumptions pretty often. Whenever possible we should check whether these assumptions are valid if some data is available or some reasonable guesses can be made. Now we will now look at some typical classes of devices and look at the default assumptions and the simplified forms that these lead us to. One class of devices is what I call heat transfer devices. Major job is transfer of heat from somewhere to somewhere or something to something and boilers, condensers, heaters, coolers, heat exchangers all these are of this kind. Our air conditioner will contain an evaporator coil that is a heat transfer device. On the other side this is a split unit so outside there will be a condenser coil that also is a heat transfer device. So the heat transfer devices are heat exchangers of some kind but for example a boiler one could say it is a fired heat exchanger. There is no hot stream on other side. There is a combustion which creates a hot system on one side. The purpose is to absorb reject heat and cause a significant amount of delta H of the stream which flows through that is the main purpose. There is no W dot S because we do not have a pump or a turbine in the system. There may be a blower but that is external to the system. So the condenser coil itself does not absorb any work, does not reject any work. So our diagram gets simplified like this. W dot S is 0 inlet exit, Q dot is significant and HE minus HI is significant. So W dot is 0, Q dot and delta H are significant. And usually the kinetic energy that means delta EK and delta PEP represented by V squared by 2 difference and difference in GZ that also is small if not 0. The pressure differences across the streams are also small. So the first law can be reduced to simply Q dot equals m dot into HE minus HI. Mind you this reduction is applicable. This is exact, there is no assumption here because it is a pure heat transfer device. These are significant so they find a place here. These are insignificant so they do not find a place here. If you have a heat transfer device in which this or this is significant you will have to include that here. There is no simplification possible for the second law because the Q dot term is still sitting there. And we generally do not tend to apply or make use of second law unless there is something very special required for heat transfer devices. Now we come to work transfer devices. These are turbines, compressors, pumps, fans, blowers, something which produce a significant amount of power or something contraptions which consume some amount of power. For example, turbines among these are perhaps the only ones maybe I could have written windmills are the only components producing power. Otherwise compressors, pumps, fans, blowers all these will consume some amount of. Purpose, significant W dot S and significant delta H. These two terms will definitely be significant for any work transfer device. Q dot would be small. The adiabatic assumption is quite often valid because we do not want any heat transfer or heat transfer would lead to poor performance or extraneous effects. We would like them to be well insulated or designed in such a way that they are adiabatic. So Q dot is approximately 0 and it is a reasonably good assumption which if Q dot cannot be evaluated or Q dot is not specified this could be a default assumption. And for some reason unlike heat exchangers for particularly for turbines and compressors and pumps, we have certain set symbols like resistors, capacitors, inductors and transistors for the electrical people. A turbine is usually shown by an expanding channel. Compressor is usually shown by a reducing channel and pump, a general pump is shown by a circle with a arrow in it indicating that that is the direction in which the pressure rises and the flow takes place. But sometimes a centrifugal pump will be shown with a tangential outlet. There is nothing standard about it but these are the common symbols. We had a question in the last, right. So work W s should be now comes into the compressor. Yes, but I am using our standard nomenclature that the work done by a system is positive. So by default I will show the arrow of work out of the system. Knowing fully well that for a turbine we expect W dot to be definitely positive. And for compressor and pump it would not be a compressor, it would not be a pump unless W dot s is a significant negative number. But when it comes, because in our equations which we are going to apply, W dot s is the work output. For a compressor it will be negative, we call it work input. But instead of saying 50 kilowatt of work input, when we substituted that we have to substitute minus 50 kilowatt, that is it. This way I am true to the algebra of the equations which we have just written. But having said that when we come to power plants and cycles, we merrily write an outward arrow for the turbine and an inward arrow for the pump or the compressor. But we will have to be conscious of the direction shown when we write the equation relating it to. Now work transfer devices as we have seen are usually adiabatic, quite true assumption, but not always exactly true. And W dot s and delta H are significant. Usually delta E k, delta E p again are pretty small. So if they cannot be calculated or are not specified default assumption is that they could be negligible. Delta p is significant. Usually for all most of turbines, compressors, except for some fans, pumps, delta p is significant. Inlet and outlet pressures are reasonably different. So the first law can be reduced to assuming that they are adiabatic and these other default assumptions are valid. W dot s is m dot into H i minus H e. Notice that for a turbine H e would be significantly below H i giving you W dot s to be a big positive number. For a pump compressor H e will be higher than H i giving you W dot s to be a significantly negative number. The second law is important for these because these are nearly adiabatic or assumed to be adiabatic situations. Then since the q dot by t term can be negligible or can be dropped, m dot into S e minus S i which turns out to be S dot p must be greater than or equal to 0. That means if it is an adiabatic work transfer device, S e must be greater than or equal to S i. That is one characteristic. That brings us to the characteristic of adiabatic work transfer devices and you should spend reasonable amount of time here because if they get confused, they will have this confusion continuing with them when they study turbines, compressors, pumps. So they will have absolutely confused future when it comes to fluid machinery or thermal machinery or refrigeration, air everything. So one should look at their H s diagram. Why H s diagram? Because we have seen that the work transfer is proportional to differences in enthalpy and entropy is going to play a role by the second law. S e is greater than or equal to S i. So H and S are significant properties to worry about and that is the reason now you can go back and show them why the Mollier diagram, usually the only diagram in steam tables is an H s diagram because so long so far we have not considered H s as of any significant. So we look at their H s diagrams and we see for a compressor or a work consuming device and diagram like this and a turbine or a work producing device, a diagram like this. Notice the difference for a compressor P e is higher than P i, H e is higher than H i. For a turbine P e is lower than P i, H e is also lower than H i. It is required that S e be higher than S i or at most equal to it. So the ideal exit state which produces zero entropy would be at the same entropy as the inlet. I have represented it as e star. Similarly for a turbine S e will have to be higher than S i. So the limiting exit state which producing the ideal and limiting exit state which produces no entropy S dot P 0 will have exit entropy equal to inlet entropy. Now a question is asked why is it that you are considering the actual exit state and the ideal exit state at the same pressure P e either here or here. And the answer to that is a ground situation that an open system receives the fluid from some other system before it upstream of it and delivers the fluid to some other system downstream of it. For example, a turbine receives steam from the boiler and provides steam to the condenser for further processing. And whenever we do the design of this, the boiler designer is told that increase the enthalpy from this state, this pressure, your exit pressure should be this because that is what the turbine will accept. The turbine designer is told that the condenser it is at 0.1 bar. So you maintain your exit pressure at 0.1 bar. So whenever you talk of open system because it is linked to other open system, traditionally pressure is the defining boundary variable. So that is why we are free to change the state rather freely within some limits provided the pressure is not disturbed. That is why if I want to improve my turbine, I will reduce its say improve my turbine, I will reduce the entropy production and I will bring S e nearer to S e star. Nothing will go wrong so far as I do not modify my exit pressure. That is why the exit pressure is considered sort of sacrosanct and hence e and e star here as well as e and e star here are considered to be at the same pressure. There is nothing thermodynamic about it. This is just engineering fact. Minor pressure losses, but that we are not going to consider. No, all pressure variations is okay. But when you are considering a real situation to an ideal situation, you cannot tomorrow tell your condenser fellow that look earlier I had a rather poor turbine. I was giving it to you at say 0.94 dryness fraction at 0.1 bar. Now I have a super efficient turbine. I will give it to you instead of 0.94 dryness fraction. I will give it to you at 0.84 dryness fraction. The condenser will be happy, less load for the condenser. Naturally more efficiency for the power plant. But you will say instead of 0.1 bar, I will give it to you at 0.3 bar. Condenser person will shout at you. He said you do whatever you feel like. I want it at 0.1 bar okay. Or 0.1 bar you cannot reduce it to 0.05 bar. You say I have no way I can condense it as 0.05 bar. And if it is 0.5 bar, you see expand it further to 0.1 bar. Why are you wasting that much energy? So all this thing makes the pressure at the interface of two components, something which is sacrosanct. But from a you know ground reality point of view. It is like the borderline between Vagha borderline or Rattari borderline right or checkpoint Charlie in the earlier days. So that is why pressure has a status whenever it comes to a flow. That is why we maintain that S e and S e star could be different but we maintain the pressure the same. So we work with one P e, one P i. Do not change P e. P i anyway is dictated by the exit pressure of the earlier component. So that the engineering fact also has to be impressed on the students. And again make them realize that this figure and this figure are essentially the same. They are not mirror images of each other. Here it goes to the northeast, here it does not go to the southwest. But all that you do, it shift this P i line upwards and you shift the P e line downwards keeping the relative locations of e star and e the same and you will end up with this figure. That is why it is not a mirror image. If at all it is a mirror image, it is a mirror image in the upward downward direction. It is not a rotated image. In whatever be the direction of flow, if it is an adiabatic work transfer device, S e will have to be higher than S e star. Now we come to some detail and some definitions. We now look at, here we looked at adiabatic work transfer devices both of the compressor kind or pressure increasing kind and turbine kind which is the pressure decreasing kind, work absorbing kind and work producing kind. Now we look at a turbine in more detail. Again I have copied the previous pages H s diagram here and what we note is this P e is less than P i for a turbine. Always the exit pressure is lower pressure. The ideal exit entropy is the inlet entropy by second law for an adiabatic turbine. The actual exit entropy is higher than the ideal exit entropy or at most equal to it. So if you look at the enthalpy H e will be higher than H e star. Do you accept that? Why should the constant pressure line be such that at higher entropy you have higher pressure, higher enthalpy? Why shouldn't it be downwards like this? Before lunch when we looked at property relations, we obtain an expression for dh. dh equals T ds plus V dp. This is a constant pressure line. So dp is 0 along this line. So dh is T ds. So dh by ds at constant pressure is temperature. Temperature will always be a positive number. This is the thermodynamic temperature mind you Kelvin. It will always be a positive number. So the slope of the isobar on the H s diagram will be equal to the local temperature. And I am showing here a gas indicating that as you increase the enthalpy the temperature goes up. So the slope goes up. Go to your Mollier diagram. Look at the slope of the isotherms in the superheated zone. They have increasing slope. But look at the slope of the isotherms in the wet vapor zone. They are straight lines. Why? Because as enthalpy changes from saturated liquid to dry saturated vapor along an isobar temperature does not change. In fact most of the enthalpy diagrams will not have isotherms shown in the two-phase zone because that will only clutter the lines out there. They have only isobar shown. So you have to deduce what the isotherm is from the saturation there. So because some student will always ask you, sir why is this going upwards? Why is it not flat? Why is it not drooping downwards? So it is a thermodynamic fact that they have to be sloping upwards because the slope of a constant pressure line on an enthalpy entropy diagram equals the temperature at that point. And the temperature is always a positive number. So we agree that he is greater than or equal to he star. This enthalpy is higher than this enthalpy. Hence he minus he will be lower than or at most equal to hi minus he star. So multiply both sides by m dot. You will get W dot s, the power produced by the turbine less than or equal to W dot s star. And both are positive numbers. We are talking of a turbine. Hence we define the ratio W dot s by W dot s star which will be less than 1 at some sort of an efficiency. And since the basis is an isentropic line, s e star is s i. This is known as the isentropic efficiency of the turbine. I am going a bit fast here because you know the background but spend may be twice thrice the amount of time because this is a very important definition. If they goof it up, they will goof up all throughout the remaining courses. Sir, thermodynamically we can understand or we can but from the physical phenomena point of what actually physically will happen. Why this h e is greater than h e? See I can say why s e is greater than h e star. That is a consequence of the second law. H e. H e is greater than h e star because this has a positive slope. Sir, that is. See just now we said that we have deduced that along an isobar, d h is t d s. So if d s is positive, d h has to be positive. Sir, this if we explain to the students, they may not means be very much comfortable about the t d s relations and maximum. If we want to explain, sir, physically what happens? It is like super situation. That you are coming to a situation where you are trying to extract a physical meaning to entropy. There is a physical meaning to entropy but it is unexpressible. You know, Bandarkar tells the story of a physicist who did write up to Ph.D. everything in physics, worked with entropy left and right without understanding it and then started working with mechanical engineers. And after a few weeks, the first is realization was I did not realize that entropy was so meaningful to you and that you design equipment, turbines, compressors using entropy as a value. For them entropy, some number which increases for adiabatic processes. That is it. For us there is a meaning. Unfortunately, it is a thermodynamic number. The meaning is only in terms of this. And actually absolute values of entropy or even entropy differences. We will realize the significance only when we come to that combined first and second law. Maybe another hour or maybe tomorrow morning. There some significance comes. But finally the answer, what exactly is entropy? Useful thermodynamic property, we should not say anything more than that. Sir, maybe can we explain that the disorder in expansion is more in… No, nothing. It is as orderly as anything else. Only thing there will be, see friction, some amount of viscous friction. You do not call it disorder. It is very difficult to say. We have a, as you grow, as you work with turbines, compressors you develop a feel for entropy. You know, I immediately know when a student does a wrong calculation or reads off the value of entropy. So that way I have a feel for entropy. But do not ask me what entropy is. We will come to that. There are some meanings associated with it. But then, how do you define quality only to be, only in terms of work? Some other word where heat is more important than work, they will put quality as the amount of heat lost that they want to dissipate as much as possible. That may be their situation. Those are interpretations in different ways. So rather than get into 10 different interpretations for 10 different situations, let me say that philosophically that I understand how to use entropy, make excellent use of it. It is an excellent friend. But what exactly it consists of, I do not know. Inside entropy is a black box, I am happy with it. But I can compute numerical values, I can work with it and all my predictions with the help of my good friend entropy come true. That is all I need. So is this clear? Spend enough time in proper understanding and spend enough time in your class to make the students understand. Of course, when you do some calculations the understanding will become more mature. But the basic idea with this figure should be very clear. And the second one for compressors is a bit confusing but the same thing. We again look at it such as diagram. So maybe if you are not using this or I would have preferred to use two bolts, on one bolt I can keep my some of my old transparency. So I would prefer to keep my turbine transparency there and show this transparency. This is for an adiabatic compressor. Looks very similar except that this line which was here has moved up. This line which was here has moved down. Just shift. Maybe you can create an animation doing that. In the previous slide PE was less than PI. Now here PE is greater than PI. That perhaps is the only algebraic difference. Let us see AC star is SI as earlier. AC is greater than or equal to AC star as earlier. HE is greater than HE star as earlier. So HE minus HI is greater than or equal to HE star minus HI almost as earlier. Both of them by m dot. There it was HI minus HE less than or equal to HI minus HI star if you look up. HI minus HE less than or equal to HI minus HI star. All I am writing there is HE minus HI is greater than or equal to HE star minus HI. Or you can even write it like that but notice that W dot S is less than or equal to W dot S star but both are negative. In the earlier case W dot S star could have been 100 kilowatt plus W dot S could have been 80 kilowatt. So 80 kilowatt is less than 100 kilowatt both algebraically and numerical. Here W dot S is less than W dot S star but both are negative. That means if W dot S star it is a compressor. You have to provide work to a power to it. So W dot S star would have been minus 100 kilowatt. W dot S would have been less than that algebraically something like minus 120 or 130 kilowatt. Because both are negative the magnitude of W dot S star is smaller than the magnitude of W dot S. Magnitude of minus 100 is smaller than the magnitude of minus 120. So whenever we define efficiency we will define as lower divided by higher. So in this case we define the isentropic efficiency of the compressor as W dot S star divided by W dot S. I do not have to write magnitudes because both are negative. So I will get a positive number. And second law dictates that that has to be less than or equal to 1 because of this. And this is known as the isentropic efficiency of the compressor. And I have written note the difference but you must spend time emphasizing what the differences and similarities are. But when we go for that reciprocating compressor in that case isothermal is coming as the ideal one. Not for the reciprocating compressor. What is the end state that you need? If my end state required it just increase the pressure with the minimal of power then it is isentropic. But if I want my end state to be higher pressure but the same temperature then rather than do an adiabatic compressor, compression and then an isobaric cooling I should do an isothermal compression. That is the meaning of that ideal is isothermal compression. For example you take a gas turbine compressor. It is never an attempt is made to make it isothermal. In fact it is very well insulated. Why? Because out there the requirement is not I must compress it properly to that pressure. I do not want the temperature to be lower. There isentropic is the best. But if you are looking at a compressor of a refrigeration system remember gas refrigeration system you have compression immediately followed by a cooling. Because you want that higher pressure but that temperature is unnecessary for you. You must cool it because after that either by expansion or throttling when you go to low temperature you will get the reduction in temperature below the refrigerated space and the refrigeration. So now after adiabatic turbine and compressor the next important device which you should tell them is the nozzle. It is an adiabatic device. It is not a work transfer device. It is not a heat transfer device and hence a good assumption for a nozzle is that it is adiabatic. It is a device which converts enthalpy into velocity or kinetic energy. It reduces enthalpy and increases velocity. That is the purpose. Typical symbol for a not a good figure but a typical symbol for a nozzle is a converging passage or a converging diverging passage. And typically you should say that pressure at exit is significantly below pressure at inlet. Enthalpy at exit is significantly below enthalpy at inlet and velocity at exit is significantly higher than velocity at inlet. This is the typical characteristic of a nozzle. There is no attempt to extract work. So w.s is really and 2d0. It is a short thing usually well insulated. So adiabatic is a good assumption for a nozzle. The enthalpy entropy diagram is similar to that of a turbine because enthalpy reduces pressure reduces. And on directly work transfer etc. is not seen. Except that in a turbine the enthalpy reduction is used for power output. Here enthalpy reduction is used for increase in velocity. So actually this is the same diagram I have copied and pasted from the turbine figure. You will find the kinks and small errors will be identical. So you tell them that on the HS diagram a turbine and a nozzle would look very similar. No differences. Now the first law as applied would be H e plus V e squared by 2 is H i plus V i squared by 2. Because you go back to the first law you will notice that delta q dot is 0, w dot S is 0. If we neglect the nozzle it is short. So you neglect that delta g delta z. So the only terms remain is delta H plus delta V squared by 2 is 0. So that means H plus V squared by 2 at H plus V squared by 2 at inlet is H plus V squared by 2 at exit. That is what we have written here. If the students are dumb enough start from there and come to this. Now the ideal exit state is E star as for an adiabatic turbine. So if instead of E we were to go E star our exit state would have been E star and enthalpy H star and velocity V star. So this is the first law for the actual situation. This is the first law for the ideal situation. And now you look here the right hand side is the same because that depends on the inlet case. Out here if H e star is less than or equal to H e that means V star is going to be greater than or equal to V. And hence the isentropic efficiency of the nozzle is defined as the actual kinetic energy at the exit divided by the ideal kinetic energy at the exit. That is the isentropic efficiency of the nozzle. One should bring to their notice that these efficiencies are either in terms of work transfer power or kinetic energy. Work transfer power are also energy transfer rates. It is never in terms of enthalpies or not in terms of velocities in particular. So some people will tend to write this as V e by V e star no because they do not represent energy. Something in the numerator something in the denominator should represent energy or energy change or energy transfer rate. Now after having done that this is a good thing to talk about an adiabatic duct. That means you have a pipeline going. There is no heat transfer. So it may be over a particular length. So you have an adiabatic situation. Since it is a duct, no work is done. You have inlet of the duct, exit of the duct. What would be the first law? Q dot is 0, W dot is 0. So H i plus V i square by 2 plus G z i is H e plus and if U i happens to be U e, the internal energy at inlet happens to be internal energy at exit. For example, it could be an ideal gas coming in and going out at the same temperature or it could be an incompressible liquid again going in and coming out at the same temperature. U i will be equal to U e. In that case this reduces to because from H i U i will go away. We will only have P i V i. I have written it as P i by rho i. And what is this equation? This is the Bernoulli's equation. Now at this state be ready to face a question from a student. I do not know whether in your scheme of things in your college thermodynamics follows fluid mechanics or same semester. So it is possible that by the time you come here you might have learned Bernoulli equation in fluid mechanics. And in fluid mechanics Bernoulli equation is derived as a case of momentum conservation equation. It is an integral form of the Euler equation which is conservation of momentum along a stream line or a stream tube. Out there there is no energy conservation mentioned but here we have shown it as a special case of an energy conservation equation. First law of thermodynamics. So some cranky or interested student will say sir but that was momentum conservation. How come energy conservation reduces to that? Do your homework and answer that question. You should tell them that energy conservation is primary. So you should go and ask your fluid mechanics teacher how come he can derive it from momentum. I hope you do not pick up a fight with your fluid mechanics.