 Welcome back. So, now we come to another problem involving turbines and this problem involves isentropic efficiency. It is a reasonably simple problem, but it involves the entropy production rate. So, let us just go ahead and read the problem. An adiabatic steam turbine handles 10 kilograms per second of steam. The inlet state is 10 bar and dry saturated. The exit pressure is 1 bar. The isentropic efficiency of the turbine is 0.8 and we have to determine the exit state, the power output and the entropy production rate. Of course, you notice that it is an adiabatic turbine and hence q dot is equal to 0 because it is given as such. So, let us go ahead and draw the turbine. This is i and what do we have? It is 1 megapascal. This is because it is 10 bar. The h i is equal to 2777.1 kilojoule per kg. The inlet entropy is equal to 6.5850 kilojoule per kg Kelvin. So, these are quantities that we have directly picked up from the steam table by looking at 1 megapascal and it is saturated dry state. So, now let us draw the HS diagram just to have a feel of what is happening. This is p i and this is p e. So, we can draw the inlet state and the so-called exit state which is ideal. So, once we get the exit ideal state, let us see how we will use the isentropic efficiency. So, this is i and this straight down with the same entropy is e star as per our momentum. So, here at the exit, we do not know the real state except what the pressure is. The pressure is given as 1 bar which is 0.1 megapascal, but we will have to go ahead and calculate what the enthalpy and what the dryness fraction is and how do we proceed? We first figure out what e star is. So, as far as e star is concerned, it is s e star is the same as s i because this is the so called ideal process where the entropy at the exit is the same as the entropy at the inlet. This is not we get actually, but we have to go ahead and do this calculation. Once we get this, we can get x star that is the dryness fraction assuming this. How do we get it? We use the entropy at s e star which is 6.5850 and subtract s f at 1 bar which is 0.1 megapascal. From the tables, we find it is 1.3028 and divided by s f g which turns out to be 6.0561 and hence we get this as 0.8722. Once we get x star, we can get h e star that is the exit enthalpy assuming the ideal process and we know how to get it which is just x times h f g plus h f where h f g and h f are calculated at 1 bar. I would not do the calculation here and we can get this as 2386.4 kilo joule per kg. Now, we know this is not the real exit state. We use the definition of the isentropic efficiency and we know what it is. It is just h i minus h e upon h i minus h e star. So, we know that the real state would have been here e and this difference here delta h real let us say would have been lesser than the h i minus h e star. So, we get h i minus h e is the isentropic efficiency multiplied by h i minus h e star. So, we know what this is h i minus h e star because we know both h i and h e star and we know that this is been given as 0.8. So, we can go ahead and calculate that h i minus h e should be equal to 312.56 which means that the real h e should be 2464.54 kilo joule per kg and this number should be greater than the ideal h e star. So, it is greater. Now, what do we do? We calculate the real exit state and the real entropy at the exit. So, let us just proceed to be that. So, x at the exit is equal to 2464.54. So, we use h e here minus h e star. So, we can get h e star. So, h e is equal to 417.5. We use h f at 0.1 mega Pascal divided by 2257.4 which is h f g. So, we write down the answer for it. I will just do it directly. It is 0.9068 and we realize that this is drier than the state which we would have achieved ideally and of course, the process of even more irreversible you would have probably got an even drier state. So, we know what the real exit state is now. Now, it is reasonably easy to get s e because all we have to do is use x multiplied by s f g plus s f where s f g and s f are calculated at 0.1 mega Pascal and we get that as 6.7945 kilo joule per kg kelvin. So, we get that as 6.7945 it is easy. Now, we write the first law steady state q dot minus w dot s is equal to m dot h e minus h i plus v e square by 2 minus v i square by 2 plus g z e minus z i. Now, notice that nothing has been said about the inlet and exit velocities and the difference in heights and we have already discussed this much earlier when we were solving our second problem that it is a reasonably good assumption to put these quantities as 0 because they are reasonably negligible compared to the delta h s that we get. So, we put this as nearly equal to 0 nearly equal to 0 and we have been given it is an adiabatic turbine there is no harm in then putting q dot is equal to 0. Then we get w dot s is nothing but m dot h i minus h e we have transpose the negative term here and it is reasonably simple 10 kilograms per second multiplied by 312.56 kilo joule per kg this is equal to 3125.6 kilo joule per second. Of course, this unit is also kilowatt. What about the entropy production rate? We realize that the second law when we wrote it it would have been m dot s e minus s i is equal to q dot by t plus s dot p where s dot p is greater than equal to 0 it is adiabatic this is 0. So, what we have s dot p is nothing but m dot s e minus s i and this is 10 kilograms per second and multiplied by the difference between s e and s i. So, s e is now 6.7945 as per our calculation and s i is 6.5850 and the units for these are kilo joule per kg Kelvin. So, we can remove the kg kilo joule per second is nothing but kilo watt we multiply this and get s dot p is 2.095 kilo watt per kilo. So, this is our entropy production rate this is because the turbine is not really isentropic it is not going to its ideal state there are irreversibilities in the turbine it comes out as a dryer state and it also means that the exit entropy is higher than inlet entropy and we can easily calculate the entropy production rate. So, right now the s dot p is not really useful but we at least go ahead and calculate it what really is important is we know the exit state and we can go ahead and finish our calculation as to what the real work output is which is what we have done. Thank you.