 Welcome back. So, we take up another problem in open systems. Let me read out the problem. Steam enters the nozzle of a steam turbine with a velocity of 5 meters per second at a pressure of 40 bar and the temperature is 600 degrees centigrade. The pressure and temperature at the nozzle exit were measured at 1 bar 200 degrees centigrade. So, you have to determine the exit velocity, the entropy production rate, if the flow rate of steam is 1.5 kilograms per second and the isentropic efficiency of the nozzle. So, let us begin. So, you see that part A is reasonably straight forward. One just has to write the open system equation for the first law and simplify it and all the simplifications that we made during our discussion for a nozzle are applicable. So, what we do is we assume that the nozzle is adiabatic. So, Q dot is 0. Of course, work transfer out is 0. We say that there is negligible change in the potential energy. So, finally, just a balance between the difference in edge and the difference in the kinetic energy. So, let us write this down again and see how it goes. We write the first law for open systems and steady state that is our assumption. So, we have Q dot minus W dot S is equal to m dot h e minus h i plus difference in k e plus difference in potential energy. This is assumed 0. It is adiabatic. There is no work transfer. You assume that there is negligible change in the potential energy and now the m dot is of no significance in this equation. You would get h e plus V e squared by 2 is h i plus V i squared by 2. Of course, one will notice that V i is the inlet velocity is given and it is not really significant. One could have as well neglected it. In fact, I would like you to do that exercise. We can take it here and h i can be read out from the steam tables because the conditions are given. It is 40 bar or 4 megapascals and 600 degree centigrade. So, this can be read out. This is 3674.9 kilojoule per kg. What about h e? That is at 1 bar or this is 0.1 megapascal and 200 degree centigrade. So, even this can be read out. So, you realize that you have this quantity and this quantity the h V i squared by 2 you could as well neglect or take it as it is and you should remember to convert kilojoule per kg for h in joule per kg. So, you have all quantities except V e squared by 2 and that can be now calculated. You can calculate V e squared by 2 by either assuming V i is 0 or the very small velocity that has been given. You will only get a negligible difference at probably the first or second decimal place. So, this is something that you must do. We will take up the next part which is regarding the entropy production rate and you realize that this is a steady state system with q dot assumed to be 0 and all you would actually get is that m dot s e minus s i is equal to s dot p. So, s e can be read out from the steam tables. This would be 7.8356 kilojoule per kg Kelvin s i can be read out from the steam table. This is 7.3705 kilojoule per kg Kelvin m dot is given. This is 1.5 kilograms per second and hence you will get s dot p. You notice that s e is more than s i which is as it should be because if it was lesser than this process would not have been possible for an adiabatic open system and we can now easily subtract multiply by 1.5 and we would get s dot p which is the entropy production rate. Now, look at the part c of this problem. You have been asked to find the isentropic efficiency. We had already defined this for a nozzle and just for the sake of repeating let me just draw it again here. So, you have h s here. You have two isobars. This is your inlet isobar. This is your exit isobar. This is your inlet state and you would have reached some point isentropically here. Instead of that, you have reached a point at this is e star. You have reached e which is at higher entropy. So, what do we get? We get for the isentropic state s e star is equal to s i. Using this s e star we now find out what must be h e star and you will realize that if I do this I get at 0.1 mega Pascal. The state is still superheated. This is I should write it as 7.3705 kilo joule per kg kelvin and you will realize that if you look at the tables that have been given to you s 100 degrees C at 0.1 mega Pascal is equal to 7.3610 and s 105 degree centigrade is 7.3885 both these are kilo joule per kg kelvin and our state 7.37 falls in between. You realize that the temperature is somewhere between 100 and 105 degree centigrade. If you interpolate it, it will come out to be 101.73 degree centigrade and what we do now is we just go ahead and interpolate for h 2 and we realize that h e star that is assuming that the exit was isentropic. We can calculate h e star to be an interpolation between 100 and 105 degree centigrade that is at 101.73 and we get roughly 2679.36 kilo joule per kg. So, we apply the first law for open systems assuming this and we get v e star by 2 and what we do then this of course can be found out by h i plus v i square by 2 minus h e star and once we get v e square by 2 the isentropic efficiency for the nozzle is just the actual v e square by 2 upon the v e square by 2 that one would have obtained if the nozzle was operating isentropically. So, this is what one should get. So, we would have gotten a value for v e star by 2 use that that is the maximum v e star by 2 you would have got if the nozzle was isentropic use the actual v e square by 2 that divided by the ideal v e square by 2 would give you the isentropic efficiency. Thank you.