 I now look at heat transfer devices. So, what are heat transfer devices? So, boilers. So, you know what happens in a boiler. In a typical boiler, you would have if it is a regular open system, you would have may be water coming in at a certain condition. The condition would usually be some kind of sub pool liquid and at the exit of the boiler, you may have saturated steam or super heated steam. So, all that you really do is you know you try to you know there is a Q interaction, you try to have input some amount of heat. How that heat has been input you know from the surrounding is something which will not normally comes there. You may have burn coal, you may have burn gas or there may be some nuclear reaction which is you know there is some energy output from the reaction. So, we are not considering that we will just say that there is a Q interaction. If you want to consider the whole system and analyze all the remaining that is slightly beyond our course domain to consider what is happening in combustion or what is happening in a nuclear reaction. Similarly, if I look at condensers, all that is really happening is that usually condensers you know may be at the end of a turbine or in air conditioning devices what you want to do is just condense the vapor. So, in a steam turbine you want to condense the steam into water. So, that you can reuse it. So, basically you are just going to extract heat out of the steam. So, that it condenses and becomes water. Similarly, in air conditioning devices or in refrigerators whatever is the refrigerant liquid you want to just remove the heat in the condenser and make it into liquid again. So, these are and then you can look at heat exchangers, heaters, coolers. In a heat exchanger all that is going to happen is there is going to be you can consider you know if there are let us say two fluid one is going to heat the other fluid. So, one fluid is going to you know go down in temperature it is going to have a Q interaction with the other fluid which is going to go up in temperature there is going to be no work interaction between the two either you can consider both the fluid the heating fluid as well as the cooling fluid as two separate systems or you can just analyze it as one single CV. You can realize these are just heat transfer devices. So, the purpose is just to absorb or reject heat and you will realize that this will cause a significant delta H. So, typically you know let us say in boilers not only is the U changing that is the internal energy changing you will realize a specific volume also changes a lot. I mean and one thing we realize is one of the most common default assumptions we will use for such devices in heat transfer is that we will assume that there is negligible change in the pressure or the delta P or the pressure drop is negligible. So, this is one of the default assumptions you will see always, but if you come across a situation as I said earlier where such assumptions do not work you must take the real delta P across the system. But the delta H that is being caused is mostly significantly due to change in U and change in V of such systems there is no work interaction in this system. So, if I draw a schematic of the control volume there is an inlet m dot coming in there is an exit m dot q dot exists and w dot does not exist. So, if you consider the mass conservation equation it is just m dot i is m dot e is m dot if you consider the energy equation you will realize that the w dot term goes away, but this does not mean that the flow work is going away the flow work will exist by default in any open system. So, you will always have a U plus P i and a U plus P e at the inlet and exit and which in turn implies that you have an h e minus h i term always for a for an open system. Of course, unless you say that there is only an inlet and not an exit if there is no exit there will be no h e term and only an h i term, but in a system such as the boiler or a heat exchanger there is a continuous flow there will be a continuous m dot you will have an h e minus h i term always. And for this you will just have to solve the complete second law equation that is the only three terms in that m dot h e minus h i q dot by t n s dot p none of those terms are going to go away because of the flow in and out there is an s dot term because of the q interaction there will be a q dot term and by default in any process that we are considering these are all irreversible processes they will always be an s dot p term which is not going to go away. So, the second law is definitely not going to be simplified, but the first law is going to be simplified. So, this is how you know we are going to write the simplified version now. So, for heat transfer devices the work interaction is 0 that is the external work or W dot s that is going to be 0 q dot is going to be significant because that is what the whole device is meant for to have a heat interaction and the delta h is going to be really significant. What we can do this may not necessarily be true, but what we can do is assume that the change in kinetic energy and the change in potential energy is very close to 0. So, what does this mean it means that you know of course, it does not mean that the kinetic energy by itself is negligible or the potential in energy by itself is negligible it does not mean that. What we are saying is see definitely it is a flow system definitely there is an inlet kinetic energy something is flowing in, but we are going to assume that you know the difference between inlet and exit there is going to be a difference definitely let us say that you know if you are bringing in water in a boiler and bringing out steam this steam may be actually going out faster there may be a difference in the kinetic energy, but the difference if you finally go and check it against the magnitude of q dot and delta h you will realize that the difference is very very or there is going to be a large difference there that is the delta e k that is the change in kinetic energy is going to be very negligible as compared to q dot delta e k that is the change in kinetic energy is going to be very negligible compared to delta h. So, you know as compared to this is basically a magnitude comparison you will realize that you know any variation in kinetic energy that is going to be really insignificant compared to q dot. Similarly, any variation in potential energy will be insignificant as compared to q dot there is definitely going to be a change in the height in a boiler typically you know in a boiler definitely the inlet and exit do not come at the same height, but whatever change is there in the potential energy that is going to be very insignificant compared to q dot delta. So, it does not mean that those two terms are 0 it is just that because their values are insignificant and we unnecessarily do not want to you know solve a complicated equation when some of the terms are you know not significant we will just you know put them as 0 and solve a more simplified equation which gives us more or less correct picture. So, that is all that we are doing by saying that you know these terms are close to 0. So, and as I mentioned earlier the delta p is small in these devices. So, the first law finally, is going to reduce to only this that is earlier we had written for steady state q dot minus w dot is m dot h e minus h i and there was an m dot term with delta e k and there was an m dot term with delta e p. So, the kinetic energy and the potential energy terms we have thrown out the w dot term we have thrown out and we just write it in a very very simple form for all heat transfer devices that q dot is m dot minus m dot multiplied by h e minus h i and as I said for the second law we cannot simplify the equation any further there are only three terms there is definitely an inlet and exit enthalpy sorry entropy difference and since there is a q interaction you cannot throw out the term involving the q interaction and since we are always going to deal with irreversible processes s dot p will always exist. So, there is no simplification as far as the second law is concerned. So, this is what we see for all heat transfer devices. So, after this we will just take up work transfer devices we will analyze then we will go on to see how to solve those equation especially in work transfer devices you know you will realize how that the second law equation becomes slightly simplified because there is no q dot, but we will see the implications of that. So, just to refresh what we did in open system. So, we had taken this control volume along with an inlet plug and an outlet plug and we had considered a closed system initially and we had we can write down our equations for the first law and second law for the closed system and then by appropriately ensuring and taking care of the inlet and exit plug we wrote the first law and second law for the open system that is only the control volume that has been shown in this picture. So, as far as the mass was concerned we just got that the change in mass in the control volume is just the difference in the mass flow rates of what is coming in and going out. For the first law apart from the q dot interaction with the surroundings and the w dot interaction with the surrounding there was a flow associated there was a work associated with the flow in trying to push this mass into the control volume and trying to push this mass outside the control volume or rather you know trying to push this boundary out and this boundary in that is how we were doing it. And as far as the entropy was concerned there was again along with this inlet there was an entropy coming in and entropy going out and due to the q dot interaction there would be an entropy change apart from the entropy production rate that was handled in the second law. So, all these gave us the three equations for conservation of mass the first law and the second law for the open system and finally, we said that if there is steady state we just have three simple equations and I will now shift to the whiteboard here. So, we just wrote m dot i is m dot e is m dot very simple equation as far as steady state mass flow into the control volume goes. Then we had q dot minus w dot s whereas, this is all the other work apart from the flow work and we had m dot h e minus h i plus v e square by 2 minus v i square by 2 plus g z e minus g z i this was first law. And then we had second law we just said that this wrote m dot s e minus s i is summation of all the q dot by t interaction plus the s dot p where by the second law s dot p is always greater than or equal to 0 this is the entropy production rate. So, these were the three sets we got in steady state. So, very simple equations for steady state and we said what is true for now a device which is having only heat transfer. So, you know all heat transfer devices that is boilers condensers and heat exchangers they will have no work interaction. So, let me get the pointer here. So, this term is going to be 0 for all kinds of heat transfer devices we do not involve any kind of work transfer. So, this term will be 0 there will be significant q dot because this is what the device is meant to ensure and there will be a significant m dot h e minus h i or delta h term will be significant and the h the delta h is significant both because the internal energy has increased and because the specific volume changes. We will assume that the delta p in all such interaction is very negligible that very less pressure drops that is what we said. So, most of the difference in h is because the difference in internal energy and the difference in specific volume. We said that these terms they are not 0, but you know when we consider them with the magnet. So, if there will be a difference in the exit and inlet kinetic energies they will not be the same, but this net difference here will be so negligible compared to the huge magnitude we will see in this q dot and h e terms that we will say you know there is no point in considering them in our analysis and we will put them as 0 or you know are equal to 0. Similarly, though there may be small differences in height in the inlet and exit for all our heat transfer devices these this term will also be negligible. Of course, this is for our heat transfer devices in the problem solving we will consider situations which are not of the heat transfer device or the work transfer device where these two terms are not going to be negligible as compared to the other terms in which case we should not neglect them and we will solve them accordingly. For a heat transfer device inlet and exit entropy difference the s e and s i terms will always be there. Since it is a heat transfer device q dot will always be there there will be a q dot by t term and as I said the s dot p term all the processes that we are considering are irreversible by default this s dot p will exist and it will always be greater than or equal to 0. So, the simplified equation in a heat transfer device we will throw out these two terms the v the kinetic energy term and the potential energy term and the work transfer term all that we get is just a relation between q dot and delta h. So, that is the equation that we will get for a heat transfer device. So, let me write that again. So, the mass conservation equation of course that is not changing and we just get q dot m dot h e minus h i. So, that is as simple as it gets. So, this the difference in enthalpy the calculation will depend on what fluid we are using. So, typically in all our applications you know we will either use air in which case you know you will assume air is ideal gas the difference in enthalpy will be very easy in that case h is can be just written as it is only a purely a function of temperature you can just write it as c p delta t and in most other applications like for example, turbines or in refrigeration and air conditioning systems. We will have to read out the delta h from some kind of a table. So, the delta h either from steam tables if you are solving for power plants or if you are solving for some refrigeration system we will have to look up the tables appropriately for that refrigerant and get the difference in delta h, but this is as simple as it can get for a heat transfer device. There is no simplification further for the second law that is all those three terms will exist and we will have to solve accordingly and see if it really makes a difference otherwise there is no simplification at all in the second law all the terms will exist. So, this is how we will simply analyze our heat transfer devices. So, we next go on to work transfer devices. So, work transfer devices these will be turbines. So, typically you know one of the common symbols that we will use for turbine is of this sort. This is the inlet coming in here this is the exit going out here or a compressor in which case I will just do it the other way around here this is the inlet this is the exit or pumps usually we will just draw a circle with P here and show an inlet like this and maybe an exit like this or sometimes we draw an exit like this. So, in these devices again in steady state as far as the mass conservation goes absolutely no difference there is just one m dot is m dot e minus m dot i. So, these are the devices which we will look at. So, it is steady flow situation whatever is coming into the turbine is what is going out same with the compressor and pump the first law is where we will have to look. So, you will realize here that you know mostly we will consider all these devices as adiabatic that is if there is a turbine working it will be well insulated q will be 0. So, I will have to have the q dot term is equal to 0 since it is a work transfer device definitely w dot s is not 0. So, this term is not 0. So, and this w dot is caused by a change in enthalpy and delta h term will be significant. So, delta h is significant. So, again we come back to the potential energy and kinetic energy term definitely you can say that the potential energy term you know the difference between the inlet and exit the height is not. So, great and any difference in the potential energy term is going to be insignificant compared to the w dot or delta h term we will throw it out and you know though in any turbine you know mass is definitely flowing and going out there is a difference in kinetic energy definitely the inlet and exit velocities are not going to the same. But any change in kinetic energy is going to be extremely negligible compared to the work transfer of. So, we will put delta e k and delta e p as insignificant and hence we will just you know put them close to 0 and this is what we will assume for all work transfer So, let me write how it looks simplified. So, I will just get q dot minus w dot m dot h e minus h i plus e square by 2 minus v i square by 2 plus g z e minus z i. So, we have decided adiabatic this is 0 and this is negligible this is negligible and we say minus w dot I should put an s because that is our convention is just. So, this is what we have a very simple equation for all our work transfer devices that is turbines, compressors and pumps. So, extremely simple equation for our steady state again depending on the situation if it is a turbine steam turbine you look up the steam table and get your change in enthalpy read it out of the steam table. If it is a gas turbine you should know that we will just take air as an ideal gas and go ahead with the calculation and look at if you look at the second law there is a simplification possible and that is because m dot s i minus s e now what we have assumed is that all q dot are 0. So, if q dot is 0 then you realize that this term can be thrown out and this term is either only positive or you know equal to 0 if it is equal to 0 there are no irreversibility in the process that is very rare we are analyzing all situations with irreversibility s dot p is bound to be greater than 0. So, you will realize that if this is 0 if this is greater than 0 then s e minus s i is a positive number and we should always have s e greater than s i. So, that will show us whether the procedure is possible. So, if you have a work transfer device which is adiabatic where q dot is 0 and if someone shows you that s e that the entropy at the exit of the fluid is less than entropy at the inlet then you can surely say that it is not possible the process is not possible something is wrong in the calculation or the design. So, by default as long as q dot has been assumed to be 0 you must ensure that your entropy has increased of course, if you are having a turbine which is losing heat you have not insulated it your s e can be less than s i, but then you know that is not a wise design at all whatever heat you are losing you are losing that from the work output as well. So, you usually will well insulate the turbine. So, the only case where s e will be less than s i if you are losing heat to the outside that is not we are not going to consider this situation. So, in all our analysis we must try to ensure that s e is greater than s i this is as long as q dot has been assumed to be 0. So, this is our simplification as far as you know transfer devices. Then we have some other devices for example, nozzles. So, in a nozzle the primary aim is neither to extract work nor do we expect some kind of q dot or heat transfer. So, q dot term is 0 I mean in a nozzle again you know I can draw it if I draw a nozzle typically I will draw some symbol like this there is some m dot going in and m dot going out in steady state. So, if I write down I always start with the full equation q dot minus w dot is m dot I will write h e minus h i plus v e squared by 2 minus v i squared by 2 plus g z e minus z i. So, we said that there is no we are not expecting a work output this is 0 we do not want any work output it is adiabatic this is 0. What we really want is to get high speeds at the exit of the nozzle. So, the difference in kinetic energy is definitely not 0 this is what we really want out of it and this is what we are getting through a change in enthalpy. So, this is also not negligible or it is let me say it is significant and usually we are not going to have any change in height or even if there is going to be a change in height this term is going to be negligible compared to the kind of changes in kinetic energy we are going to see. So, that means for a nozzle only these two terms will exist and all other terms we are either they are 0 or we just are going to neglect them and not consider them in our analysis. So, what you get finally, is that h e plus v e squared by 2 is h i plus v i squared by 2 or usually you know you want only how to calculate v e. So, v e squared by 2 is h i minus h e plus v i squared by 2 is h i minus h e plus v i squared by 2 inlet of the nozzle again this is as we said you know we make a lot of usual assumption we a lot of times we will assume that the inlet of the nozzle has a negligible velocity negligible kinetic energy and this term in many analysis is usually put to 0 and v square by 2 is just given as h i minus h e. If v i is not negligible you just have to consider it in your analysis otherwise this is going to simply reduce to v e square by 2 h i minus h e. Again as far as the second law is concerned you will see that this situation is exactly like the turbine that is q dot is 0 as far as the second law goes. The second law there is no question of having a w dot term. So, what you get is m dot s e minus s i is just equal to s dot p and s dot p has to be greater than or equal to 0. So, the entropy at the exit for the fluid stream will have to be greater in the entropy at the inlet. So, this is assuming that this whole system is adiabatic and you are not losing q dot at all. So, one thing in which people analyze both the nozzle and the turbine is using either a T s or an h s diagram and let me show you how this will look like if we do or how will this look on a h s state space here. So, let me just look at the turbine. So, let us draw an h s curve or a sorry or isobars or on an h s diagram. So, why do we take h s because this is how we can analyze both the delta h and delta s that we see from the first law and second law. It is very easy to picturize the whole thing for a turbine and a nozzle simply by using an h s diagram. The first law will give the delta h will lead to either a work output in a turbine or a change in kinetic energy in a nozzle and we have to ensure that the entropy increases from inlet to exit. So, normally in a heat transfer device there is going to be no change in pressure, but all work transfer devices and nozzles there will be a change in P and in fact that is what is going to drive the entire process. So, if I draw two isobars on an h s diagram I will get them as this P 1 and P 2. Now, you see that on an h s diagram the isobars are actually moving up. If you want to see why this happens then all you have to do is just look at the T d s equations d h is T d s plus v d p. So, you can assume that this is just d h by d s at constant P and this is d h by P at constant s and d h by d s at constant P is what we are seeing this is a isobar the pressure is constant. You see that d h by d s is T it is a positive number the slope is positive and as h increases T increases the slope keeps on changing and becomes higher and higher. So, this is going to have increasing slope an isobar is going to have an increasing slope on an h s diagram. So, I have drawn two isobars here is a lower pressure isobar is a higher pressure isobar and we see a positive slope. So, if I have an inlet state here and if it is a work transfer device like turbine what will happen just is that I will just drop down to the lower enthalpy. So, this is h i this is h e and this difference should cause me the work output or the difference in this will lead to work output. Now, I have drawn in this case I do not know if it is looking exactly vertical in this case I have drawn something where the inlet and exit state are exactly above each other that is at the same entropy. So, this is the special case where the s dot P term the s dot P term is 0 which means the entropy production rate is 0 it is a reversible process and the exit state has the same entropy as the inlet state. So, this is not what you are going to get if you have an entropy production term which is in fact what is going to happen in all turbines you are actually going to reach somewhere here. So, there will be an increase in entropy you will move like this s e here as you see it is towards the right of s i it is an increase in entropy and what you usually will get is something where you go from inlet to exit towards the right as you go to the lower pressure line. Now, I can see that this is let me call this as h e real and this as h e star where h e star is obtained when s e is equal to s i. So, h e star is obtained when the exit entropy is the same as the inlet entropy in that case your work transfer is just m dot times h i minus h e star. Whereas, in the real case your w dot let me call this as w dot star and w dot real may be r here is m dot h i minus h e r this is r is I am just calling r for real here. So, you realize that h e r is going because of the slope of this line h e r is going to be higher than h e star the difference in h e h i and h e is going to be bigger than the difference in h i and h e r. So, this is going to be a smaller number than this. So, in the real case your work output is going to be lesser than the reversible case and you are going to have w dot s real is going to be lesser than w dot. So, because of this there is a term which we call as the adiabatic efficiency of this sorry the isentropic efficiency of this turbine usually denoted by eta and we will say that is the real work upon the so called ideal work when you know the exit entropy is the same as inlet entropy this is what is called as the isentropic efficiency of the turbine. This is the real definition of the actual work upon the so called ideal work if I substitute for you know the difference in enthalpy I will just get this is h i minus h e r upon h i minus h e star. I hope you can see that basically instead of w dot real I have just substituted as h i minus h e real. So, that is the smaller number that is in the numerator that is the actual work divided by the so called ideal work where you know the exit entropy is the same as the inlet entropy. So, this is what is called as isentropic efficiency of the turbine. So, if the turbine is adiabatic one thing you must definitely ensure is that your exit point is not going to lie to the left of this star point. In that case you have created an impossible process you cannot have the exit entropy lesser that we saw from the second law analysis. Of course, if you have a non adiabatic turbine where you are losing heat then you know this is possible probably, but in all our analysis q dot term is going to be 0 in no case your line should move to the left as you go down from p 1 to p 2 that is possible only if you are losing heat. So, you must ensure that in your analysis the point has moved to the right. So, that is what I think you must emphasize and you must tell clear difference to student that this is what is really happening when you are analyzing a turbine. It is a very simple analysis, but you have to tell what it means when we apply the second law and what it means when we apply the first law. So, whereas if I look at the something like a compressor then I just have the opposite or let us say the direction is opposite that I go from a lower pressure to a higher pressure. So, this is the H s diagram I have already shown you that the curves move up here this is let us say I will call this p 2 and this is p 1. So, this is my starting point now I am at this point in this case if I move up isentropically or you know where the exit is at the same entropy. So, at star this is star point S e star is the same as S i then you will see that your work W dot S star is just m dot H i minus H e H i is less than H e this is a negative term and definitely W dot S star is a negative quantity if I write it in this fashion it is work done on the system, but if you want to take only the magnitude you can just calculate it as m dot H e minus H i it is a positive quantity. Whereas, the way we have written using the first law system it will be a negative quantity it is work input into the system. Now, if I do not have if I look at the second law and say that the entropy production rate exist S dot p is there there is no q dot compressor is assumed as adiabatic system. So, you will definitely have S e greater than S i by applying the second law in which case I have to move to a point on the same pressure line to the right. So, I have to move in this direction along the pressure line I may reach somewhere here. So, let me call it actual exit S e and I will actually be moving somewhere on this curve here. So, you are now moving to a point to the right it is at a higher entropy. So, this point I will say S e is greater than S i. So, this is what has happened as I apply the second law. Now, you will realize that in this real case W dot S is m dot H i minus H e sorry in the first case I should put a star H e star W and in the second case it is just W dot S is m dot minus m dot H i minus H e. Now, you will realize that numerically this H e is higher than H e star the difference in H e minus H i is going to be larger than the difference between H e star and H i. So, what you actually end up doing in this case is that you end up putting more work I mean I mean algebraically it is going to be a larger negative number, but what really has happened is you have ended up putting more work into the system and you will realize that this is again due to the irreversible or irreversible process that is occurring you have reached a state with higher entropy. So, you must just realize here that our definition of isentropic efficiency is going to be slightly different than what it was for the turbine. For the turbine we calculated isentropic efficiency or we defined it as what is let us say ideal work sorry actual work upon ideal work the actual work was lesser than the ideal work in that case. In this case if I take the magnitude of the actual work in a compressor the magnitude of the actual work is actually greater than the ideal work. So, this is what has happening in a compressor. So, you know to keep my definition of an efficiency intact where usually we define it such that it is less than 1 we always will for a compression process we will define the efficiency of a compressor as where we calculate or we put the isentropic work or the ideal work in the numerator and I am just taking the magnitude here to make it positive it does not matter even if I do not take the magnitude both of them are negative. So, this is the actual work or I can substitute it in terms of delta H I will get H i minus H e star upon H i minus H e. So, this is what is called as the isentropic efficiency in case of a compressor where you put in work you just realize it that I am putting the ideal work in the numerator and the real work in the denominator and this will be a number less than 1 because the real work will always be greater than the so called ideal work or the isentropic work. So, this is when we are analyzing compression now if I look at the other adiabatic device which is the nozzle in the nozzle we had H e plus V e squared by 2 is H i plus V i squared by 2 and the H s diagram is not going to look different from what it looked for a turbine. There are two isobars here H s P 1 P 2 and we are moving down from one isobar to the lower isobar this is inlet this is exit S e by S i S e is equal to S i corresponds to the isentropic case. So, in this case we will have a V e squared by 2 which let me call as V e star square by 2 and I will call this as E star is equal to H i plus V e star minus H e star plus V i square by 2 the inlet state is the same this is what is going to happen as ideal velocity or ideal kinetic energy that you expect when you are using a nozzle. So, this is you will expect a certain velocity now it is an adiabatic device I apply the second law the q dot term is 0. So, I will get that H e should be greater sorry S e should be greater than S i that is because the entropy production rate S dot P is greater than or equal to 0. So, I should get that S e should be greater than S i in real case. So, again you will realize this looks exactly the same as in a turbine you are going to move towards the right on the S axis that is because you have to show an increase in entropy S e is greater than S i you have to move somewhere to the right of this line and if you are you will the exit pressure we will assume is the same you have taken a nozzle the exit pressure is the same it is the pressure at the exit and you reach a point with higher entropy now in this case. So, in this case you will move in this fashion and again you will notice that it is the same like a turbine where you know the ideal H i minus H e star is going to be greater H i minus H e star is going to be greater than H i minus H e actual. So, this is E actual. So, if this is going to be true then you see in your expression for V e squared here that is just going to be V e squared by 2 is equal to H i minus H e plus V i squared by 2. Now, if I compare this equation let me call this 1 and 2 you will realize that the only difference is in the H i minus H e term here in the ideal case the H i minus H e term is greater than in the real case. So, you will realize that V e squared by 2 is going to be lesser than V e star squared by 2 and the efficiency in such case this is not a work transfer device. So, we have a different definition for efficiency. So, efficiency in all nozzles normally we will call it what is the real V e squared by 2 that we get. So, V e squared by 2 real upon V e squared by 2 ideal. So, this is the you know normal definition for efficiency of a nozzle. Now, sometimes if V i squared by 2 the term is negligible and if you want to throw it out then we can write it only in terms of H i minus H e otherwise this will be H i minus H e plus V i squared by 2 upon H i minus H e star plus V i squared by 2. So, this is the normal definition for efficiency of a nozzle I think it is looking too small on this let me. So, this is the efficiency of a nozzle that we will get if the V i squared by 2 term is negligible then you can just throw it out. We will get it in terms of just the difference in enthalpy and the definition will not be different from the definition of efficiency for a turbine where we were not even considering the kinetic energy term. Here we were considering the kinetic energy term if we decide that the inlet velocity is negligible this is what we get and it should be no different from what it is for a turbine. So, this is what we will get when we have analyzed our nozzle. So, we can analyze some other systems now. So, let me say we have done work transfer devices done heat transfer devices we have done nozzles. So, heat transfer devices would have been boilers. So, typical analysis for a boiler condenser only from the thermodynamic point of view and turbines compressors. So, we can let us say do similar analysis for which is neither a work transfer device nor a heat transfer device nor a nozzle, but may be just a duct you know there are lot of air conditioning ducts in the building. You can do a simple analysis only for a duct again you know the mass conservation equation is the same m dot i is m dot e is m dot and if I look at the heat transfer sorry the first law q dot minus w dot. I have written it even simpler and m dot s e minus s i I just will write this entirely and then you can see which ones you can put to 0. Now, if I take a duct this is how it will be now the duct is not really meant to you know act as a heat transfer device or a work transfer device. So, that way it is similar to the nozzle in putting these two terms as you know 0, but we will need to figure out now there are these three terms and really we need to figure out what is happening with these three terms. Each of them may not be of a huge magnitude, but you know if the duct is moving up and down then there is going to be a change in the potential energy. If there is going to be a change in temperature then there is going to be a change in the you know internal energy. If there is going to be a change in pressure and specific volume the p v term will change and else there will be a delta h term. If there is a change in the velocity at the inlet and exit there will be a delta e k term. So, these three terms we will have to balance out we will see what to do and just like a nozzle or a turbine we have assumed adiabatic this term will go to 0. So, s e will always have to be greater than s i and we should just ensure that if as long as it is an adiabatic duct you have ensured that q dot is 0 you should get that the entropy at the exit of the duct is the same as entropy at the inlet sorry is greater than the entropy at the inlet of the duct. So, that is what you must see. So, one small aside in this is when you write this first law here this equation let me just rewrite it. So, I will just get if I just open up this h i minus h e v i square by 2 v square by 2 etcetera and bring all i terms together inlet terms together I will get h i plus v i square by 2 plus g z i is h e v square by 2 for this I can write as u i plus p i v i plus v i square by 2 plus g z i is u e plus. Now, if I take let us say an ideal gas then u is a function purely of temperature if I take an incompressible fluid also we had assumed that u is purely of temperature and you know it does not depend on pressure. So, if we assume that the temperature has not changed then we can just put u i equal to u e and remove this and I will get an equation which will look and I will write specific volume as 1 by rho and I will get some equation. So, you will recognize that this is some kind of or it is so called the Bernoulli equation, but we have just made it from the energy point of view we have just taken the first law for an open system and very steady steady state one-dimensional analysis where everything is clubbed together in as one stream with constant properties and we get you know so called the Bernoulli equation. So, these were the various analysis that we have done I mean the duct analysis we will do may be lesser, but you must realize that when we do various analysis you should know when to put which term 0. So, for work transfer devices normally we put q dot 0 changes in kinetic energy and potential energy 0 for heat transfer devices it will just be a balance between q dot and h delta h for nozzles it will be only delta h leading to a change in kinetic energy. For other processes you will have to just write down and see what is significant you should not just blindly throw out terms, but depending on the situation you should know which terms to ignore which terms to take into consideration etcetera. So, because we have around 10 minutes let me just look at one example. So, I will just consider you know very first problem in your open thermodynamic system this is problem one and let me just read it clearly. The joule postulated that the temperature of water when it comes down a waterfall rises. So, that is the temperature of water rises when it comes down a waterfall. So, consider the waterfall to be an open thermodynamic system and determine the height through which the water should fall for its temperature to rise by 1 degree centigrade. Assume water to be an incompressible fluid with C p equal to it is given 4.186 and list all the assumptions. So, you know the very first problem it is neither a work transfer or heat transfer device or neither is it you know a nozzles situation. So, we will just write down the equations as we saw them. So, this is first problem. So, let us assume that you know there is some height here water is or let us say water is here and it is falling down and it is getting collected in this lake here. So, this is the height through which the water is falling this is really a delta z that is occurring. So, I will take an open thermodynamic system which will probably begin somewhere here and let me just mark my boundaries till it reaches somewhere in the lower pool here. So, water is entering this system and it is exiting this system in steady state. Let us assume it is steady state waterfall is continuously the same m dot is continuously flowing out and in and out and the same m dot is flowing. So, there is m dot is m dot i is equal to m dot e. So, now if I take write the first law q dot minus w dot m dot delta h. So, we can I mean it is interacting with the surroundings, but we will have to make a few assumptions now. Though it is interacting with the surroundings we can say it is an adiabatic system it is really there is no q interaction we can put this as 0. We are not extracting any work out of this system this is 0 and we have decided that this is an incompressible system this is going to be just a difference in u plus p v terms. Pressure both you know is an incompressible fluid we are not really changing the pressure anywhere the pressure is the same may be the volume will change if there is a change in temperature. So, this may change if there is a change in temperature inlet kinetic energy exit kinetic energy not much difference the way we have drawn the control volume. So, only difference is going to come in this change in the potential energy. So, what we realize is that we have thrown out three terms and we are just going to equate the change in the change in the potential energy to delta h. Even in delta h if you assume that you know the change in volume is not too much with temperature you can even throw out that term and just say the change in delta u is, but in any case we just write for water you know whether it is delta h or delta u just some m c delta t where that c is you know we do not say it is constant pressure c t or constant c v because it includes such effect where we do not expect the volume to change too much. So, we can just say that m c delta t for the water is because of the change in the potential energy. So, we have considered some open system and we have figured out what to put 0, but this what to put 0 depends on how we have considered the system. If we change the system you will realize that some of this terms cannot be 0. So, we have tried to create a system where you know we can conveniently throw out some terms. If we had put the boundary somewhere else you would realize that I cannot throw away some of the terms. So, we are just trying to see how much the temperature is increasing because of a change in potential energy and we have conveniently put the boundary such that our change in kinetic energy is negligible and we are also conveniently assuming that the system is adiabatic. We have no way to calculate if there is a heat transfer to the surrounding that you know the right now we will do. In fact, this was apparently some of the experiment that Joule had himself conducted to see what the change in temperature would be as water comes down a waterfall. Now, most of the other questions after this are going to deal with turbines, compressors, boilers, etcetera and they will start we will start doing them after lunch. So, the first one was just a non none of these devices and you will realize that in most of the problems after lunch we will if it is an air we will calculate delta H using you know C P delta T or something like this. If it is water we will use the steam table. So, in none of our systems right now we are going to talk of refrigerants, but you must realize that if you know it is some condenser using refrigerants maybe we will require some tables with R 11 or something like this. So, at this point with this example we will stop because all the other examples we will except for one or two most of the other examples would be slightly similar. We will also see how to apply the second law and ensure that for most work transfer devices and nozzle system the exit entropy state should be greater than the inlet entropy and we will decide if a particular process is possible or not. Some of the problems will state you know that this is the inlet state, this is the exit state you know if you know this you can definitely calculate delta H and calculate the work transfer out, but the problem will also ask to see if the process itself is possible. In that case we will have to check on the entropy and see whether that process is possible. So, these are all the problems we will do post lunch. Now, I will just stop this session