 Welcome back. So, after dealing with the problem involving the second law in the last snippet, let us deal with a similar problem not with a compressor, but with a nozzle. So, let me just read out the problem. The inlet conditions for the nozzle of a steam turbine are 60 bar 350 degree centigrade. The exit conditions are 10 bar 0.9 dry. What do we have to find out? A, if the steam flow rate is 10000 kilogram per hour, determine the exit velocity and area and B is the process possible or impossible. Why? What is the limiting exit state and exit velocity? Assume that the exit pressure is fixed. Now, how do we approach this? The last time around, we had figured out that the process was not possible and gone ahead and done a lot of calculations using the first law. Now, let us first find out, since it is easier to find out whether the process is possible, we will go ahead and do that first. Now, you will notice that there is nothing mentioned regarding whether the process is adiabatic or not, that is whether there is heat transfer or not. But right now, since there is nothing given, we can only assume that q dot is 0. And if q dot is equal to 0, we know that because s dot p has to be positive, the exit entropy has to be greater than the inlet entropy. So, let us go ahead and do those calculations. So, we will do it systematically, draw the diagrams, figure out whether something is possible or not. This is our nozzle and we had inlet 60 bar 350 degrees centigrade and we had exit at 10 bar and 0.9 dry, which means the dryness fraction is 0.9 and m dot is given as 10000 kilogram per hour. So, because we know the inlet state 60 bar 350 degrees centigrade, it is in the superheated zone. We look up the superheated tables at 6 megapascal because that is what is 60 bar 350 degrees centigrade and we get that h i should be equal to 3043.9 kilojoule per kg. And similarly, s i is equal to 6.3357 kilojoule per kg kelvin. What about exit? So, this is at inlet. What about exit? We know that it is 10 bar, which means this is 1 megapascal and it is 0.9 dry. So, we can find h easily. This would be 0.9 multiplied by h fg at 1 megapascal plus hf at 1 megapascal and we would get this as 2575.66 kilojoule per kg. So, this is h e. We would go ahead and do a similar exercise, 0.9 multiplied by s fg plus s f and we realize that we get this as 6.1404 kilojoule per kg kelvin. Now, we draw our H s diagram. This is our inlet to the nozzle. So, let me write this p i, this is p e, this is i and we would have expected an ideal process to go like this if it was adiabatic and we are assuming q dot equal to 0 because nothing is mentioned about it. So, in that case, we expect that s e should be greater than s i or the process should have been in this fashion. So, this should have been e star and this should have been e. But what do we notice? We notice that s i is 6.3357 and s e is 6.1404 both of these kilojoule per kg kelvin and s e is actually less than s i which means that we are expecting that the process would go in this fashion and this is e given. So, this is just not possible because we have assumed q dot is equal to 0 which would have meant that s e should have been greater than s i and we find that s e is actually less than s i which means this is not possible. What would have been possible is if exit e was equal to e star or greater than e star. So, let us go ahead with our calculations assuming that we get s e equal to s i that is our ideal exit state. So, in this case go ahead assuming s e equal to s i which means e equal to e star. So, if this is so, we realize that let me draw this diagram again. This is what we are getting. So, we know what s is here. So, let us just find out that s e is also equal to 6.3357 kilojoule per kg kelvin. So, this means we would need to find our dryness fraction here x. So, we will just go ahead and find it. We will use 6.3357 and subtract the sf at this point which is 2.1381. We will look at the tables for 1 mega Pascal and divide by sfg which is 4.4470 and this will turn out to be 0.949. So, that means that e star would have been a more drier state than what was given in the problem. And in fact, if the process is actually irreversible, it will be even more drier and we should not be surprised if it was completely dry. So, what is h? Now it is very simple. We use this x multiplied by hfg which is the same hfg as before and add hf. So, both of these are at 1 mega Pascal and we get this as 2664.1 kilojoule per kg. And this should be a quantity that is greater than the previous quantity. So, this is he. We also would like to calculate the specific volume that is because we have been asked to find out the velocity and the area. We get ve here which is the specific volume. We go ahead and do a similar procedure. I will not do it here. It is 0.18352 meter cube per kg. So, now we find out m dot. We convert it into kgs per second instead of the kilogram per r units that have been given. So, I just have to divide by 3600 and I will get 2.7778 kilogram per second. Now we can write our first law. We have q dot minus w dot s is equal to m dot he minus hi plus ve squared by 2 minus vi squared by 2 plus g ze minus zi. We realize that we have assumed this is adiabatic. So, this is 0. This is a nozzle. No work is expected. This is 0. And nothing has been mentioned about the heights. It is a good assumption to go ahead and assume that this is 0. In this calculation, we realize that we do not need the mass flow rate now because all we get is he minus hi should be equal to vi squared by 2 minus ve squared by 2. Now, we realize that we do not know the inlet velocity, but in such nozzles, it is a good assumption that it is a reasonably negligible velocity compared to the exit velocity. In fact, the nozzle is meant to accelerate the fluid to a higher velocity. So, we say that this is nearly equal to 0, which means that we get ve squared by 2 is equal to hi minus he. So, all we have to do now, calculate this. So, ve squared is nothing but 2 multiplied by hi minus he. And we see that we have to convert this into joule per kg. So, we first substitute the values. Hi was 3043.9 and he. Now, we have calculated as 2664.1. And we multiply this by 1000. So, that our units are correct. We need it in joule per kg. And we get now ve should be equal to 871.55 meters per second. This is a pretty large velocity. And now, since we have m dot should be equal to a e ve by specific volume at the exit. We know m dot, this is 2.7778. We know this, this is 871.55. We know this, this is 0.18352. We substitute these values and get the exit area as 5.849 multiplied by 10 to the power minus 4 meter square. So, we have made a few assumptions here, but those are reasonably logical for the nozzle. We have assumed it is adiabatic. There is no change in inlet and exit heights and the inlet velocity is 0. Of course, we first figured out whether the problem as given was possible. We discovered it was not possible. We figured out what was the limiting state and went ahead by solving for that limiting state. So, these are the answers for the limiting state. They are not for the problem as given. We are trying to find out what the velocity at the exit is and the area at the exit is for the limiting state. Thank you.