 So, now our exercise session for open thermodynamic system, so let us begin with 6.5. We have a turbine, make them sketch a diagram and write as much information as possible on that diagram, so provide enough space, so the moment you say there is a steam turbine, the student should draw a turbine diagram with an inlet and an exit. If you read it, it receives steam at so many kg per hour and develops power of so much kilowatt. Neglecting heat losses determine the change in specific enthalpy of the steam turbine flowing through the turbine if inlet and exit velocities and heights are negligible, if the entrance and exit velocities are 15 meters per second and 320 meters per second respectively and the inlet is 4 meter above the exhaust. It may be a good idea to tell the student that if you look up the steam table, a reasonable problem will have a data edge of easily 100 or a few 100 kilojoules per kilogram. It is also important to impress on them that if air is the working fluid, a typical value of Cp of air is very nearly 1 kilojoule per kilogram Kelvin. So if you have a temperature difference of say 100 degrees C, it is 100 kilojoule per kg delta H, steam will be at least equal to this or usually more than this unless you are doing something small within a stage or something. And now compare this delta H with typically G delta Z, so the gravitational thing. So G is about roughly 10 meters per second per second and your delta Z, what is the typical size of an equipment, if it is a nozzle, if it is 10 centimeters or a few centimeters, if it is a pump fraction of a meter or about a meter, if it is a turbine may be a few meters, a big turbine, if it is a boiler may be 10 meters, 20 meters, a big boiler may be 40, 50 meters but let us take typically 10 meters, so delta Z is 10 meters, big enough. Some boilers will be bigger but most of our equipment will be smaller than 10 meters in size. Assuming that the inlet is at the extreme end in height and exit is at the other extreme end in height, you will be always be higher or I mean lower than 10 meters in delta Z. So what is G delta Z, if you look at it this G delta Z will be of the order of 100 but which unit, this will be 10 joule per kg, so this is of the order of 0.01 kilojoule per kilogram, 100, so 0.1, it is usually negligible when it is of the order of even few tens of kilojoule per kilogram, that is what we are going to have. Then come to your velocity, tell them that higher velocities are attractive in a certain way, for example m dot is rho A B, if you have to pass a certain m dot, a higher velocity means a lower area of cross section, lower area means lower diameter and that means for a given pressure lower thickness of the tube, lower weight of the tube, a cheaper tube but a higher velocity means a higher pressure drop and a higher pressure drop means you are not doing any work for it, it is just like a throttling, we will come to a throttling process. And a higher pressure drop means although the boiler is giving you at certain pressure, you are losing that much pressure when it comes to a turbine, so there is always a give and take between the size of the tube and the pressure drop and typically you say that because of this, velocities are usually restricted to of the order of 10 meters per second. Depending on the fluid, depending on the pressure, temperature and depending on the application, may be in a few cases 20, 30 meters per second will be allowed. In case of some liquids, it may even be lower thing which will be allowed but typically of the order of 10 meters per second. So, assuming that velocity at one end is near 0, the other end is 10 meters per second or even 20 meters per second, you take 20 meters. So, your delta V squared by 2 is going to be 400 minus 0 divided by 2 in joules per kilogram which is 200 joules per kilogram which again is going to be of the order of 0.2 kilojoules per kilogram. Again a very small number. So, this is the way we will justify our assumption that delta E k and delta E p is negligible. But that assumption will not hold for say problem one, we are looking at a waterfall, height of the waterfall comes out to be almost 1000 meters per second or you take a hydroelectric power plant near Mumbai at Koppoli, there is a power plant where the lake Vaavan and Bushi, they are at a height of something like 700 meters. Whereas the tail race, it has a height of something like 30 meters above mean sea level. So, that 600, 700 difference of height is then definitely significant. So, again let us come back to our mainstream exercise 6.0. There is no hint that it is a transient. So, by default we will assume it to m dot is given to be 5,400 kg per hour might as well convert it to how much? 1.5 kg per second. Now, after the initial thing where you confuse them with units etc., there is no need to be so tricky when it comes to units. Once they have shown how to convert units, let them do the unit conversion properly. Let them get eased. There is a W dot, can say W dot S, that is our standard nomenclature. Okay, by the way, why does this S come up here? Why shaft? It may even be PDV work as we say the bladder of a football expanding. Initial idea was this is applicable to systems like pumps, compressors, turbines where the work is provided through a shaft or is extracted through a shaft. That is why this is shaft work. It is possible that it could be any other type of work. So, but the symbolism of W dot S continues. The power developed is 600 kilowatt, so this is 600 kilowatt. Neglecting heat losses, so Q dot is 0. Determine the change in specific enthalpy of steam flowing through the turbine. That means we have to determine h e minus h i. For saying it is a turbine, if he wants to remain in the positive domain and writes h i minus h e, that also is okay. But he should say what he is calculating. h e minus h i is what? And you are given a, entrance and exit velocities are negligible. So, in case a, delta e k, delta e p are small and in case b, inlet and exit velocities are 50 meters per second and 320 meters per second. V i is and inlet is 4 meter above the exit. That is V e minus V i is minus 4 meter inlet is above the exit. Now, I would only allow two types of beginning. If it is a steady state, we can start steady state, start with the steady state equation. If there is no hint that it is a steady state, the student must start with the dm C v by dt and d e C v by dt. Because it is a steady state, we start with the so called steady flow equation. Q dot minus W dot s is m dot into, I will use this, delta h plus delta e k. It is given that this is 0. This value is given, this value is given. So, our expression is only delta h, delta e p. In case a, delta e k, delta e p, delta h and in case b, all that changes is we have calculated delta e k will be g delta z, so delta e p. So, whatever this turns out to be, but remember that this will be in joule per kg approximately 40. So, convert it into the small number which is kilo joule per kg. I always feel that it is a proper idea to make the student write down the units for all intermediate answers. Not only all intermediate answers, I would enforce that the moment you write a number, immediately follow it up with a unit. So, even in a formula, you know g delta z, I would prefer that this be written down as 9.81 meters per second squared multiplied by minus 4 meter. That way it is very clear that it is meter square per second square which is joule per kg. Enforce that, otherwise they will particularly every year, so many students lose marks because when they determine velocity, the fact that data h is in kilo joule per kilogram and needs to be converted to joule per kilogram is forgotten. So, you will instead of 1000 meters per second, you will get only some 30, 40 meters per second. For instead of 100 meters per second, you will only get some 3, 4 meters per second. That factor of square root of 1000 from 30 odd will be there. Again in meter square per second square which is joule per kg, so convert this into kilo joule. Once you obtain that, the only unknown here is data h. Now, 6.7 and 6.8. Let us do 6.8 first and then we can, we have the nozzle of a steam turbine inlet 60 bar 350 degrees C. Exit 10 bar 0.9 dry. Flow rate is given. Determine exit velocity and exit area. Is the process possible or impossible? That apart. We will look at B later. First let us look at A. So, inlet state is given, exit state is given, flow rate is given. We have to determine exit velocity, exit area. So, we sketch. This is a nozzle. Nothing is made. Inlet state is given as 60 bar. Exit state is given as 10 bar 0. Steam flow rate is 10000. Determine exit velocity. What about Q dot? Is it given that it is an adiabatic? So, this will be an assumption. You better decide on your local symbol, the A with a capital A with a circle or small A with a circle or rectangle, so that a student can write down and mark that as an assumption. He does not have to spend time and writing, assuming it to be adiabatic. If he shows it clearly on a figure, good for us. Look at the figure and you know what, what about W dot? 0. It is a nozzle. There is no assumption needed here. It is a nozzle. So, there is no attempt being made to extract the word. So, steady state, so Q dot minus W dot s is m dot into he minus hi plus de squared minus vi square by 2 plus g jd minus steady state energy equation for first law for steady state. And one inlet, one exit, so that form is valid. Q dot has been assumed to be 0. W dot s is 0. What about this g zd by zi, zi minus zi. So, all that happens is we end up with ve squared minus vi squared by 2 equal to hi minus he. Now the inlet state has been given 60 bar 350 degrees C. Exit state has been given as 10 bar x equal to 0.9. So, we can determine everything about the inlet state. We can determine everything about the exit state including enthalpy and entropy. So, we know hi, we know he. To calculate ve, we must know vi. Vi is not known. So, we will make another assumption that vi is much less than and that means we will say that this is negligible compared to vi. And we will make this assumption because there is no mention of vi. And that means we will end up with this simple equation ve squared by 2 equals hi minus he. Only unknown is ve. Only thing remember that by default by steam tables your hi minus he will come in terms of kilo joules per kilogram. And that is not meter square per second square which is required to get ve. So, here you should when you first demonstrated note take care of unit. My trick is to ask the student to calculate and the first one to calculate will give us 130th or 132nd answer because that square root of 1000 will be missing. Now part b is the process possible or impossible. Why? No, but before that in a we have to determine velocity exit. See we have been given m dot which we have not used yet. And m dot is v exit, a exit divided by specific volume exit. This is given. This is calculated. This we have calculated just now. So, a exit can be determined. Velocity is capital V. Small specific volume is small v. There will be no difference. Or if you write it as rho e vea, it does not matter. Calculate rho e. That confusion will always exit. Volume is capital V. Velocity is v. C for velocity is a problem because soon few lectures later the youth see for sonic velocity. But I agree there are people who youth see for velocity and a for sonic velocity. Use your nomenclature. I am not forcing my nomenclature on you. I am using nomenclature which is convenient to me but I agree nomenclature is arbitrary. There are whole textbooks on thermodynamics written where temperature is small t and time is capital T. That is perfectly okay. In the old textbooks, enthalpy was i, entropy was pi. Such textbooks. So, our Mollier diagram is known as I-phi diagram. Sounds very HS or Mollier. I-phi is like I-phi. That is good enough. Now b. Is the process possible or impossible? Notice that it is assumed now to be an adiabatic nozzle. That means s.p must be greater than or at most equal to 0. And that means your se must be greater than or at most equal to si. And now let us sketch our Mollier diagram. We are starting with how much? 60 bar 350 degrees C. This is inlet. And let the exit be 10 bar 0.9 dry. Let me say the 0.9 dry could become here. 0.9 dry could come here. Exit e or exit e. I will write e1, e2 as two options. I do not know. We are not calculated it. This is si. Okay. Obtain si se. Can easily be obtained or computed because everything is known about the states. If se is greater than or equal to si, then it is a possible process. So if the exit is here, it is a possible process. If otherwise, impossible. So if the exit turns out to be here, it is impossible. So that is the question. Is the process possible or impossible? Why is this reasoning? Being adiabatic, if se is greater than or equal to si, it is possible. If se is less than si, it is impossible. What is the limiting exit state and exit velocity? Draw this. Consider this si. This is our e star. This is the limiting exit state is determined as p equals 10 bar. That is p e star is 10 bar. Se star is si. So in terms of p and s, the exit state is fixed. The ideal exit state is fixed. And whatever be the answer, possible or impossible, this remains the limiting case. Now 6.7 is similar except that it is a compressor instead of a turbine. Only difference is the inlet state is given as 5% moisture. That means 95% vapor. That means drainage fraction of 0.95. Because such terms will be there which you may have to explain to the students. Now the steam flow rate is given to be 5 kg per second. Diameters of inlet and exit ducts are 20 centimeters. So this also gives you an opportunity to calculate the inlet and exit velocities and show that the delta ek is pretty small. Again the b part is the similar. Is the process possible or impossible? Only thing is instead of exit pressure being below inlet pressure, exit pressure is above inlet pressure. Now about pumps. Pumps are compressors for. Excuse me sir. So pumps usually handle a liquid and liquids are remember they are almost incompressor. And the problems here 1 bar 25 degrees C 280 bar for the second one 6.10 0.1 bar 25 bar 6.11 1 bar 25 bar. The first one has a pretty high pressure 180 bar. The second one and third one are at 25 bar. So first one maybe you can compute the exit state using that crude compressor table but let us look at problem 2 6.10. I will modify this saying that determine the power required per kg per second of water pump rather than work done per kg of water pump although it means the same thing. Now this is what we have to tell them that for a general pump if we really consider it to be incompressible then the situation will be like this. That is the pump exit state and inlet state. We will make the following assumption steady state a diabetic delta E p delta E k. In that case the first law turns out to be minus W dot s is m dot into h e minus h i. Minus W dot s will be a positive number because the pump will be consuming power. So W dot s in our thermodynamics terminology will be a negative number. We can leave it like this. Now what about h e minus h i? E plus minus U i plus C i V i. Let me say this is equation 1. And then we will say that since Q dot equals 0 ideal pump which is adiabatic would be such that s e equals s i what we say the limiting case s e star is s i. Now if our liquid is incompressible what does s e equal to s i mean incompressible liquid only one property. And if that property is entropy is not changing that means the other properties are not changing. So this means T e equals T i incompressible automatically means V e equals V i or rho e equals rho i if you are so comfortable. And also in particular U e equals U i isentropic because of ideal adiabatic and incompressible model for a liquid leads us to this. Hence 2 becomes in 2 remember U e minus U i will become 0 V e equals V i. So h e minus h i simply become V i or V e same value multiplied by P e minus P i. So h e minus h i simply becomes P e minus P i into V i which is quite often written as P e minus P i by rho the density of the liquid. And hence we get minus W dot s from equation 1 for an ideal pump which pumps an incompressible liquid is m dot into P e minus P i by rho. And since m dot by rho is the volumetric flow rate this becomes P e minus P i into I will write q dot volume which is volumetric flow rate unit meter q which is the standard formula we use in fluid dynamics. Is that clear how that has been obtained in thermodynamics. And you should also show them that if you take an extreme case T s diagram if this is the saturated liquid line suppose quite often we start pumping a liquid when it is the saturated liquid this line is pretty sharp but I am showing it shallow. This is suppose the inlet pressure P i giving you inlet state which quite often is saturated liquid could be subcooled liquid it does not matter here whether it is saturated or not. Exit pressure will be such that this is again not perfectly right because we know that for isentropic this line and this line would almost overlap on each other. So for a T s diagram I am just showing this exit state and I have exaggerated this. This is the ideal exit state e star the actual exit state will be to the right of it may be something like this this is I this is e star e will be somewhere here. But again notice that this is exaggerated the difference between I and e star but if you take the enthalpy entropy diagram then this would not be exaggerated because there will be a significant enthalpy difference. Now using this thing you can solve 6.9, 6.10 and 6.12 now we will spend time on 6.12, 13 and 14 or 6.1 through talk about talks about a venturi meter 6.13 talks about a throttling calorimeter and 6.14 talks about a bottle filling problem 6.12 horizontal venturi meter it is very clear. A venturi meter you should tell them is a duct in which the diameter first reduces and then increases usually to the original diameter and at a steady rate so m dot is 600 kg per minute. Inlet and throat diameters of the venturi are 6 centimeter and 3 centimeter. The throat is the location where the diameter is minimum and we will consider our control volume to be from the inlet plane which is the inlet of the venturi meter to the exit plane which is the throat. So diameter at the exit is given to be 6 centimeters sorry 3 centimeter. Diameter at the inlet is given to be 6 centimeter it is horizontal and there is no transfer of heat or work and no change in internal energy very clearly everything is state in state. Density remains constant at 1000 kg per meter cube. What is the pressure drop between the inlet and throat in bar? Because it is steady state we start with q dot minus w dot equal to m dot into h e minus h i e square minus v i square by 2 plus h e z e minus square. There is no heat transfer given. There is no work transfer given. We do not even have to assume. It is a horizontal venturi meter so this is 0 that also is given. We are also given that there is no change in internal energy from inlet to exit. That means now because of this notice that this m dot goes out as a factor from this equation it will come in again somewhere else. So now we end up with 0 equal to u e plus e e v e minus u i minus e i v i plus v e square minus v i square. It is given that u e and u i are equal. It is also given that the density remains constant at 1000 kg per meter of a fluid mechanics problem but we are showing it that it is thermodynamics that they are Bernoulli equation we are grabbing from over fluid mechanics friends. So this becomes 0 equal to e minus p i divided by rho plus v e square minus v i square. So if you want to determine p minus p i you must determine v e and v i. And how will you determine v e and v i? Again you determine m dot equals rho a i v i equals rho a e v e. I am writing single rho because that is given to be constant 1000 kg per meter. A i and a e are given in terms of diameters and hence you can determine v i and v. And you should tell them that look here the exit diameter is smaller. So area is lower so the velocity will be higher. So the same mass has to flow and the density is the same. So that means exit velocity is higher than inlet velocity. So this term is positive. So this term will have to be negative indicating that the exit pressure will be lower than the inlet pressure. And tell them that by this time perhaps in their chemistry lab they would have done some filtration experiment to click on the process of filtration they would have used a venturi meter connected to the flask in which the filtration takes place, filtrate goes. So why is that done to reduce the pressure so that we get more flow of the filtration process. So next one, 6.20. The throttling calorimeter first why? Tell them that the most common methods, most common parameters which can be measured are pressure and temperature. Easiest, lowest cost over a wide range of pressure and temperature is a wide range of fluids. So this is what we generally do. Whenever you talk of compressed gas it is available in some container at say minus 20 degree C but 100 bar or 30 degree C but 200 bar and things like that. So temperature pressure is the common thing to do. But when it comes to wet steam it cannot be P and T, it could be P or T plus something else. This can be U, V, H, S but you will notice that none of these are easy to measure. It is not easy to measure the specific volume, it is not easy to measure the enthalpy, entropy or internal energy. So what we do is the following. You show them a scheme like this. You say this is where the wet steam is flowing and you measure the pressure, you measure the temperature and you find that they belong to the at or very near the saturation line. So you are not sure whether it is trisaturated, whether it is saturated liquid or whether it is wet steam. So you must have a handle on something else, some other property and that other property we want to have a handle on turns out to be H. And indirectly a very simple measurement using only P and T is possible and that brings us to the throttling calorimeter. Let us say the state here is P1 corresponding to T1 and say a drainage fraction X1, of course enthalpy H1 and all that, state 1. But I am putting T1 in bracket because the moment you know P1, T1 is known, P1 and T1 are on the saturation line. What we do is we take a small sample, we put a sampling tube with some holes on it and we take a small sample out. Then we put it through what we know as a throttle. I will soon tell you what a throttle is. Tell them that a throttle is a device like a small hole which allows a small amount of flow but for allowing that flow it requires a reasonably large pressure drop. So you say that if this pressure is significantly above atmospheric, the pressure drowns things will be atmospheric pressure and a small amount of flow will come. We want a small amount of flow because the major flow should go in the equipment. If we are measuring wetness or drainage fraction of something which is say 1 kg per second, we should not extract half a kg per second, we should extract only a gram or 2 per second, right, that is so we want a small amount of flow. If this is near atmospheric then the pressure drop may be small in which case we will have to use a vacuum pump. Even then finally you tell them that that may not be sufficient, some other tricks may have to be used. But let us say this is large enough and the other side is open to atmosphere but when the pressure reduces to P2 the volume increases, right and if the volume increases the velocity will also increase. We do not want velocity to increase. So we provide a large area of, we have a small flow rate. So we say V1 is small because of small flow rate. We want V2 to be small because we want A2 to be large, we provide a large, a small tube comes but after throttling it diverges into a big channel. And here the pressure is P2 which can be atmospheric if P1 is large enough otherwise it could be connected to a vacuum pump. And out here we measure temperature and we measure pressure and we call that state 2. Now how do you create a throttle? Throttle can be created in various ways. You tell them that you can have a thin tube, a millimeter or less than that in diameter and a long length. Such throttling devices are used in refrigerations, refrigerators, air conditioners etc. Or you could have a plate with a small hole in hole in it or something which they can have a controlled throttle in our household tap. When you open it fully it is not a throttle but you open it slightly which something comes out that is a throttle. Another thing is you compress air in your tire and then you notice that instead of 30 PSI it has gone to 33 PSI. You press that valve and let the air escape. That slight opening of that valve is a throttling process. You are opening it slightly. Only a small amount of flow comes out but the pressure drop is large. So a throttle can be implemented by a number of phase. A slightly open valve or a small orifice. All these things are throttling. So how is the process like? First thing we will assume that the process is steady. Then we will assume that the whole thing is insulated. There is no attempt to do any work. So if I consider my open system to be something like this, draw in appropriate shape. I will again show it here. My open system will be something like this. Inlet 1 throttle exit 2. Q dot 0 it will be reasonably well insulated. W dot s is 0. There is no attempt to remove any work. V exit kept small because of large area that is V2. V inlet V1 also kept small because the m dot is small. We want to take only a small sample. So with steady state we have Q dot minus W dot s equals m dot into delta H plus delta E k plus delta E p. This is well insulated. This is 0. Delta E k is made small. This we assume negligible. We can keep it horizontal but we know that it is going to be rather significant. Only a few joules per kg. So that means this turns out to be delta H equal to 0 or H exit is H inlet. H2 is H1. What good is that? Now we come to the take out the diagram, the HS diagram and show them that on the HS diagram this line saturation line usually droops down like this. Higher pressures are here, lower pressures are here. So if you have a state like this, if this is your state 1 and if this is your pressure for state 2, say atmospheric pressure, throttling means H2 is H1. Do not call it an isenthalpic process because it is a process which is irreversible and in all probability non-causostatic. Suddenly it expands then we give it large area. So we do not worry about what is happening. Show it by a dotted line and also make them understand that there is a large increase in entropy making it a highly reversible process. We are nowhere near an isentropic process. But because of this if state 2 lies in the superheated region then H2 can be determined using the measurements of P22. For that H2 has to be in the superheated region. Once it is in the superheated region you have measured P2 and T2, you can determine H2. You have already measured P1 or T1, one of the two and then H1 and P1 give you the state at the inlet including its dryness fraction and everything else that you want. At this time a student will say that look if the dryness fraction is pretty low then even after throttling it may still remain in the wet zone. So you tell him that look in that case you have to go to still lower pressures by creating appropriate vacuum but then you will say that look if the wetness is so large moisture is pretty large your dryness fraction is pretty low. Even if you go to the triple point pressure you will not get wet zone in superheated. Then you say that look in that case pure throttling calorimeter will not work. You will have to use a throttling separating or separating throttling calorimeter or a heating and throttling calorimeter and if too many students ask you for illustration you say just wait for the next exam I will put a question in it. So is that understood about throttling calorimeter? See when it comes to venturimeter when it comes to throttling calorimeter it is a good idea to tell them what that equipment is and keep this in the exercise sheets rather than in the theory otherwise we will be doing too much of description why are we describing all this all that thing comes up. So put you know derivation regarding pumps derivation regarding throttling calorimeters. Now we will have a derivation about bottle filling and later on about heat exchangers let us do those. Let us have two more problems. Now 6.14 this there is a rigid insulated bottle of volume V0 perfectly evacuated. The stopper is opened and ambient air is allowed to flow in. When the flow stops the stopper is replaced find the finite temperature of air in the bottle. So what we have is a rigid insulated bottle insulated rigid Q is 0. Rigid means W expansion is 0 but it is an open system because the process begins by something flowing in. Let us say M is the mass which has managed to get into the bottle. The volume is V0 initially perfectly evacuated but the environment is at P0 P0 that is the surrounding air ambient air. So initial state M equals 0. Final state some M say M equals M0 let me not call it M0 because not pertains to say M equals M1. Let us say T equals T1 and T equals P1. What can you say about P1? We have opened the bottle air allowed to flow in and moment of flow stop. Flow will stop when P1 equals P0. So this will be equal to P0. Now what about temperature for that and M1 we have to determine using conservation of mass and conservation of energy. So our first equation is going to be our control volume is this. The inner surface of the bottle and the imaginary surface in the neck where we are going to put the stopper. It is an open system one inlet no exit and it is a transient process. So we cannot assume steady state something is changing something is flowing in the flow is stopping all that. So we are going to have dMcv by dt equal to M dot i minus M dot e. There is nothing going out so if I integrate it my this will be final mass M1 minus initial mass 0. I am integrating it over time from opening of the bottle to close of the bottle. What is M dot i that is the amount of mass which has come in. So this is going to be integral M dot i dt over whatever is the time. So this is my M1, trivial equation real first law ecv by dt equals q dot minus w dot s plus M dot i h i plus v i square by 2 plus v z i. There is no M dot e so I will not even write that term. There is a term containing M dot e and the same thing but which I am just not writing. Now let us assume that in ecv, ucv is the important component other components are insignificant. There is a bottle here it is not moving it is not going up and down. What about q dot well insulated? What about w dot s 0? We are making no attempt to extract anywhere. Let us say that the air is travelling only from inside to a small distance. So your g z i would be negligible and your v i square by 2 would also be small. Let us make that assumption. So integrating this out you will get cv in the final state 1 minus what is the ucv in the initial state? 0 will be equal to now this is after integration. h i is the enthalpy of the air going in that is at p naught t naught. Let me call it h naught that is going to be steady. Integral M dot i dt Now what is going to be ucv 1 that is going to be whatever is the mass in the inside the system that is M 1 multiplied by u 1 and since integral M dot i dt is M 1 the right hand side also I can write it as M 1 into h naught is that right? So M 1 cancel out I get u 1 equals h naught. Now till now I have not made any assumption about the fluid which goes in. We are told that it is air and now we will make an assumption that air can be modeled as an ideal gas. Now we will make that assumption till now that assumption was not made. Is it clear? So tomorrow if instead of air I am telling you that it is steam coming in from somewhere up to here it is the same thing beyond this it will change. Now if you assume air to be an ideal gas that means you write this as u 1 equals h naught will be u naught plus p naught v naught and I will also assume constant Cpc. So now I will get u 1 minus u naught is p naught v naught. What is u 1 minus u naught? Cv into t 1 minus t naught under the assumption that air is an ideal gas with constant specific heat. So you will have Cv into t 1 minus t naught equals RT naught because p naught v naught is RT naught. So you will end up with Cv t 1 minus Cv t naught is RT naught. So t 1 will be Cv plus r divided by Cv into t naught which is gamma t naught. The problem is simple you get an algebraic answer because initially there is vacuum. If you say initially there was some pressure the 00 terms will not be there and the answer will not be so direct. Similarly instead of air you say steam goes in then you will have a different answer and that will have to be numerically calculated using steam tables. Here we can continue with algebra because we have assumed the air to be a working fluid with an ideal gas. That is the assumption which gives us such a simple answer. Yes, yes, there is flow must have been stopped because the system has come in equilibrium. No, flow stops because the inner pressure is outer pressure at least from the pressure point of view. But inside it as it comes in there will be some churning when the one that settles down whatever is the temperature final equilibrium temperature is there. Now 6.15 and 6.16 are straight forward problems but they use the isentropic efficiencies of turbines. I think all of you have solved enough problems of that kind. So there should not be any difficulty. It is necessary to explain either 6.17 or 6.18 or both. These are about heat exchangers. So we will tell them something about heat exchangers 6.17 or 6.18. At least one you should do or ask them to do, help them to do heat exchanger. Now what is a heat exchanger? You tell them that a heat exchanger is a device. This is the first device which we will come across in which we have two streams going. They will learn in heat transfer later that there is something like a parallel flow, counter flow, cross flow heat exchanger. Here in thermodynamics we do not have to worry about it. We are only going to talk about energy balance. There are two streams. So two inlet and two exit but they do not mix with each other inside. So consequently we call the stream 1i exit 1e and 2i and exit 2e. And we say that the mass flow rate of 1 is m dot 1. The mass flow rate of the second is m dot 2. So m dot 1 goes in, m dot 1 comes out, m dot 2 goes in, m dot 2 comes out. And you tell them that we will consider that heat exchangers work only in a steady state. Then you tell them that as a total the heat exchanger is well insulated. No work is extracted. So the heat exchanger is overall neither a heat transfer device nor a work transfer device because for the overall heat exchanger as I have shown q dot is 0. W dot s is also 0. So what happens is with steady state if I write the x equation it becomes q dot minus W dot s is for now if there were only one stream I would write m dot 1, delta h1, delta ek1 etc. But there are two streams so I will have to sum it up over two streams. The two streams become m dot 1 into h1e minus h1i plus delta ek1 plus delta ek1 plus m dot 2 h2e minus h2i plus delta ek1 plus delta ek1 plus delta ek1 plus delta ek2 plus delta ek2. This is 0 it is well insulated. This is also 0 we make no attempt to extract any work, any power. And usually these four are negligible. So consequently the first law has applied to any heat exchanger. Can I go to the next page? First law has applied to any heat exchanger when two stream heat exchanger becomes m dot 1 h1e minus h1i plus m dot 2 h2e minus h2i. And now in the problem say 6.17 you will see both streams have a flow rate of 2 kg per second and flow without any loss of pressure that is required. But so m dot 1 is given, m dot 2 is given. Here if you assume it to be an ideal gas that uniform pressure means we do not have to worry about it. One goes from 30 degree C to 80 degree C. So you can determine h1e minus h1i say that is the first stream. The second stream enters the heat exchanger at 150 degree C. So I know h2i. Only unknown remains is h2e which you can calculate. And since it is air and it is moving at constant pressure if you assume Cp is to be uniform as given you can write h1e minus h1i at Cp t1e minus t1i. And here also Cp t2e minus t2i only unknown which remains is t2e. Now there is a question heat transferred between the streams and you must tell them that although we consider the heat exchanger as a whole to be one system which is adiabatic. For detailed study we consider then our first control volume was this the complete heat exchanger. This is the control volume the Hx control volume for which all our equations we have written so far. Now we split it into two control volume for fluid 1. This is fluid 1. I take a control volume which occupies some part of heat exchanger and the first fluid fluid 1 but does not touch any part of fluid 2 that I call the one control volume. Then the remaining part of the heat exchanger which will contain fluid 2 and may be some other parts but does not contain any part of fluid 1 I will call it the second control volume. So 1 and 2 control volumes are partitions of the overall heat exchanger and then what we do is the following. Sir regarding the enthalpy difference one is related with hot fluid another known as cold fluid like that. So I am coming to that one of these will be the hot fluid one of these will be the cold fluid. Then that inlet exit difference will be different for hot fluid and cold fluid. Enthalpy of the hot fluid will be decreasing cold fluid will be. That is what is happening since the sum of these two will be 0 one will be a positive number one will be a negative number. So if H1E is greater than H1I H2E will have to be less than H2I. That is what I meant. So now let me go back to my black colour. Now as you pointed out this term 1 and this term 2 one has to be positive one has to be negative. So the negative term we say it pertains to hot fluid. Enthalpy of the hot fluid we say decreases. The positive term pertains to the cold fluid. The enthalpy of the cold fluid increases. Let us say for example in this case one is the hot fluid and two is the cold fluid example as an illustration. I can number anyone let us say one here is the hot fluid. Then we say the heat interaction from 1 to 2 q dot 1 to 2 is the heat flow rate hot fluid. Look at the direction. This is q dot 1 2 coming out of control volume 1 going into control volume 2. And this is also known in technical terms you can tell them and now you can apply first law individually to these two. You feel like. Our overall equation is still this m dot 1 H1E minus H1I plus m dot 2. Here I have example taken this to be negative, this to be positive. So this we are calling the hot fluid, this we are calling the cold fluid and for the cold fluid we will have if you look up this control volume. Apply first law for cold fluid and you will have q dot 1 2 is m dot 2 H2E minus H2I making the same assumptions as earlier except that it is not adiabatic. It is receiving heat at the rate of q dot 1 2 and fluid 1 is receiving heat at the rate of minus q dot 1 2 which is m dot 1 into H1E minus 1 and 2 I have assumed 1 to be hot, 2 to be cold as an illustration. It could be the other way. Oh entropy production rate it is simple. See the entropy production heat exchanger control volume adiabatic control volume. And if you look at our equation, what is the second law equation? S E 2 minus S I 2. Now this is where the simplification comes in. Both streams flow without any loss of pressure. So for example in 6.17, air on either side, assume ideal gas constant Cp. And since P1, Te1 is Ti1 and Te2 equals Ti2, Sc1 minus Si1 will simply be Cp ln Te1 by Ti1. And a similar expression for Sc2 minus Si2. Calculate the substitute here you know m dot 1 and m dot 2. Remember that in this expression earlier expression H1E minus H1I can be written down as Cp into T1E minus T1I or Te1 minus Ti1. There because it is a temperature difference the temperatures can be substituted in centigrade, Celsius no problem. But in this case these are temperature ratios. Whenever you see temperature ratio or whenever you see temperature by its own like Pv equals RT that temperature must be Kelvin. So if it goes from 120 to 100 do not substitute 100 and 120 degree Celsius. If it is in Celsius convert it into Kelvin and substitute in terms of Kelvin. Now that brings us to the end of today.