 Welcome back. So, we were discussing an adiabatic turbine and an adiabatic compressor. We looked at the first law for both of them and we looked at the reaches diagram. Now, we look at an another adiabatic open system and this is a nozzle. A nozzle is not work transfer and what is expected from a nozzle is that we increase the velocity of the fluid and this is a device which is used mostly in turbines and even elsewhere. So, what happens is if we write down the first law for a nozzle in full, let us just write down the steady state first law which is q dot minus w dot s is equal to m dot h e minus h i plus v e square minus v i square by 2 plus g z e minus z i. The nozzle is expected to be an adiabatic device. So, we say this is nearly equal to 0, the q dot. We are not extracting or inputting any work into the system. So, this is equal to 0. Typically, we do not expect any change in the potential energy of such a system and this is reasonably good assumption to put this as equal to 0. So, what finally remains is that there is a significant delta h between the inlet and exit of the nozzle and as a result we get an increase in the velocity of the fluid at the exit. So, we have finally that h e minus h i plus v e square by 2 minus v i square by 2 is equal to 0. It can also be written as h e plus v e square by 2 is equal to h i plus v i square by 2. Now, we just depict the nozzle as it is typically shown in an open system manner. We do not draw trapezoid here, but the fact is that we go from a higher pressure to a lower pressure. The trapezoid we have already used for a turbine, but this is the inlet. So, we just draw i here. We go from a higher pressure and a higher enthalpy to the exit here and this is normally how we show it here. What we really get out is v e. So, we can assume negligible inlet velocity. Even if it is there, it is much smaller than v e. So, this is really what we are interested in v e and typically this is what is written. So, for this p e is much lesser than p i. That is the exit pressure is less than the inlet pressure. h e is less than h i and v e is definitely much greater than v i. We have assumed of course that w dot s is equal to 0 and it is reasonably adiabatic q dot is 0. We have also said delta p e is nearly 0. So, we realize that in this case what has happened is as far as the h s diagram goes, it would be the same as that of a turbine because there is a significant drop in h which instead of the work output has resulted in an increase in the velocity and the second law is going to be the same as that for a turbine in the sense that we will have the exit entropy s e minus s i should be equal to s dot p which is a quantity greater than 0. The s e would be greater than s i and hence one can draw the h s diagram as follows. This is h and s this is the inlet isobar p i, this is the inlet state, this is the exit isobar p e and ideally we would have come from i to e star. So, we would have got an s e star is equal to s i. We had p i which is greater than p e in this case and we had h i greater than h e in this case and if the process is not ideal and if there is entropy production one expects to move in this fashion towards increasing entropy. This is an adiabatic case and we would have got an e here such that s e is greater than s i or s e I should say greater than or equal to here s e is definitely greater than or equal to s e star which means that the first law we had written h e plus v e squared by 2 which is equal to h i plus v i squared by 2. We can of course write even for the ideal process the same h e star plus v star squared by 2 is equal to h i plus v i squared by 2. This is the first law it does not change, but since we see that just as in a turbine there is a lesser delta h now actual compared to the ideal delta h and since h e star is a quantity which is less than or equal to h e and hence this implies that v e star is a quantity greater than or equal to v e. We have v e star greater than v e this is what we expect the nozzle is used to accelerate the fluid to a higher velocity and what we get or what we are expected to get ideally that is v e star is higher than v e which is what we get actually. So, because the nozzle has not performed to what we expect because of irreversibilities and because it is not ideal we have v e star that is the ideal velocity obtained at the exit is greater than the actual velocity at the exit. Now, because of this we can of course define an efficiency for a nozzle similarly as what we had done for a turbine and compressor and that is what we can write down here. We will call it the isentropic efficiency of the nozzle since we are basing it exactly on the same principle we say that what we would have got ideally is v e star by 2 this is the ideal expected output from the nozzle what we get actually this is ideal this is actual the actual is v e square by 2 v e square by 2 is a smaller quantity and just as in a turbine we will define isentropic efficiency of the nozzle which we can write as v e square by 2 that is what we get actually upon v e square by 2 or v e star by 2 this is the ideal kinetic energy output from the nozzle and this number as you can see because v e star square by 2 is greater than or equal to the actual v square by 2 this number will be less than or equal to 1 as expected and this is what we call as the isentropic efficiency of the nozzle. Thank you.