 Welcome back. So now we pick a problem where we also want to apply the second law. So up till now we were only applying the first law for open systems. Now we will take up something where we will apply the second law. So let us consider this problem here. A ship propulsion system incorporates a compressor which receives steam at 3.4 bar with 5 percent moisture. It delivers a dry and saturated at 8 bar. Steam flow rate is 5 kilograms per second. The compression is adiabatic. Diameters of inlet and exit ducts are 20 centimeter. The mechanical efficiency of the machine is 92 percent. What do we have to find out? A, determine the power required to drive the compressor and B is the process possible or impossible. Why? What is the limiting exit state? And we have to assume that the exit pressure is fixed. So how do we approach it? We could always just go ahead and blindly apply the first law for open systems and right now we will do that foolishly. So that we will come up with some answer and then we will realize that it would have been better if we had actually looked at the second law too. But let us go ahead with the first law for open systems right now. So what do we write? We write q dot minus w dot s is equal to m dot h e minus h i plus v e squared by 2 minus v i squared by 2 plus g z e minus z i. And we realize that it is mentioned that the process is adiabatic. So we say this should be 0. We need to find this out. Whatever we get out of this is the actual work input that we will get. It will be a negative number since it is a compressor. Now this is the work which the compressor will take in and we realize that there is something called or something given as the mechanical efficiency. That means there is an efficiency of transferring the power from the element which is driving it that is really some kind of a motor and there is an efficiency of transferring this power to the compressor. And hence the amount of power that is required will be more than this and we will use the efficiency to calculate that. What else do we have? We have no information regarding z e and z i. So we will say okay this is assumption this should be equal to 0. What do we have regarding v e and v i? We realize that the areas have been given and hence if we use m dot is rho e A e v e is equal to rho i A i v i in steady state. We should be able to get v e and v i because if we know the inlet and exit states then rho e rho i or specific volume or specific volume at inlet or exit are known and since m dot is given and A e and A i are given we will be able to calculate this and substitute it here. What about h e and h i? The states have been given and hence we can calculate it and this also is given. So let us go about doing this, what information do we have? We have at the inlet 3.4 bar 5 percent moisture. This means that the dryness fraction is 95 percent or 0.95. So x is 0.95 that means the vapor is 95 percent and the moisture is 5 percent that is what 5 percent moisture would mean. So x or the quality is 0.95. So we can go ahead and calculate. So we take up the steam tables that have been provided. 3.4 bar is 0.34 megapascal. We can have h f at this which is 579.91. We can have h g at this looking at the table 2730.6 and h f g is equal to 2150.7. All these are in kilojoule per kg. So what do we do next? We can say yes because I know x I can calculate h at the inlet. I just have to multiply x to h f g and add it to h f and we realize that we will get it roughly as 2623.08 kilojoule per kg. Similarly, I can get si which I am not going to use right now and we will get it as 6.6881 kilojoule per kg kelvin. What about h e? It is given 8 bar and dry saturated. So we can just look up the table directly get h g at 8 bar because it is dry saturated and we get that as 2768.3 kilojoule per kg. On a similar note, we can go ahead and calculate specific volume at the inlet and specific volume at the exit. At the exit, v e is directly given as 0.24034 meter cube per kg from the steam tables. Whereas, v i again we use the same interpolation method and go ahead and complete the calculation. I am not doing it here. It is 0.51176 meter cube per kg and hence we can write m dot is equal to velocity at inlet area at inlet upon specific volume at inlet is the same as v e a e upon specific volume at the exit. Now a i is equal to a e and it is just pi d square by 4. We are assuming a diameter here is given is pi multiplied by 0.2. We are calculating it in meters. So 20 centimeters is 0.2 meters. So we use it in meters and we go ahead and get the area here. We substitute the area here. We can substitute this. So we can get v i is m dot multiplied by specific volume and divided by a i. We will get that as 81.44 meters per second. As you can see, this is not so high but it is not negligible either and v e by a similar procedure will be calculated as 38.25 meters per second and we realize that this is not as large as v i but we hold on to it and now we come to our first law. We had already put the terms with q dot n g z e minus z i as 0 and we would have gotten minus w dot s is m dot multiplied by h e minus h i h i plus v e square by 2 minus v i square by 2. We have this as 5 kg per second. We have the values for h e, h i, v e, v i. We substitute all of this and we get that minus w dot s is equal to 713.17 kilowatt. So, please notice h e and h i, we would have gotten in kilo joule per kg. If we had used v e and v i, we would have got v e square by 2 and v i square by 2 in joules per kilogram and we will have to divide this by 1000 to get kilo joule per kg and that is what has been substituted here and I have not completed those steps. But you please do that and add it to h e minus h i and one should get the result that the answer is minus w dot s is 713.17 kilowatt or w dot s is equal to minus 713.17 kilowatt. This is a negative quantity because heat is sorry the work is done on the system and this is what the compressor would need to go ahead with its compression. But since there is a mechanical efficiency, the actual work needed would be more than that and it would be w dot s upon the mechanical efficiency which I am putting as eta m. So, we can go ahead and finish this calculation and we will get some number. Now, we need to look at the second law now. So, if I draw on the h s diagram, we have our isobars here. This is h, this is s, we start at a lower isobar, this is a compression process, this is p i, this is p e, this is inlet state and we had s i and this is e star. Now, since q dot is equal to 0, this is an adiabatic process, we would have the second law for steady state as m dot s e minus s i is equal to q dot by t plus s dot p where this term has to be greater than or equal to 0. This is given as 0 because it is an adiabatic process and since this is greater than 0, s e necessarily must be greater than or equal to s i. Now, s e equal to s i is the isentropic process. Now, s i we have already calculated, this was equal to 6.6881 kilojoule per kg Kelvin. If we go ahead and calculate s e at 8 bar, dry saturated, we realize that this is equal to 6.6616 kilojoule per kg Kelvin, which means that this is lesser than s i and we have actually come here and we are expecting that this is the process. Now, since q dot is equal to 0, such a process is not possible and we should have checked this initially and not done all our calculations regarding the first law because though the first law allows us to calculate whatever we want using it, we have to see whether we are restricted by the second law and we realize that at the fixed exit pressure of 8 bar, we would have actually reached this state isentropically. This is the limiting state and we realize that since this is dry saturated, this is actually in the superheated zone. Of course, we could have reached any other superheated part of 8 bar like this and this would have been perfectly allowed. In fact, to get to this state, we would have to lose heat and this would have to be a negative quantity and it would be of course difficult, we would have to calculate or integrate q dot by T and at different temperatures, different q dot is escaping and so on and it would involve some loss of heat if you ever wanted to move to a lesser entropy and since this is an adiabatic process, this process is simply not possible and we should have actually checked on this before. So, we can leave it as an exercise that we will consider the limiting case where Se is equal to Si in which case we can find out what the exit state is and then go ahead and complete the calculation and we will get a different W dot S. So, I would like you to go ahead and do this right now. Thank you.