 So, welcome back. Now, let us see the fourth problem that is question D. Now, this problem is little bit more involved and therefore, please read the language correctly. I have tried to give the language mostly in the mathematical form, so that one can understand the problem in a better way. The problem again goes back to the first law. So, what we say is MA that is the mass of system L. Let us say MA is equal to 3 kg of saturated liquid water at PA that is pressure is equal to 1.5 MPa. So, this is one segment given the mass of saturated liquid that means the position is known with saturated liquid. We know the mass of it, we know the pressure of it and this is mixed with MB that is other fluid now with MB is equal to 2 kg of water maintained at PB is equal to 1.5 MPa and TB is equal to 250 degree centigrade. So, to understand that there are two compartments, one is maintained at given pressure and saturated liquid water that is the condition given. The other of MB is equal to 2 kg which is maintained at pressure of 1.5 MPa and the temperature at 250 degree centigrade. The mixing is adiabatic that means Q is equal to 0 and at constant pressure. So, two conditions are given for the mixing phenomena. Determine the changes in volume delta B12 and energy delta E12 the dryness fraction in the final state X2 and the work done W12 that is during the process from 1 to 2. The thermodynamic state is at first 0.1 or state 1 which goes from this thermodynamic state to thermodynamic state 2 which is maintained at 0.2. The problem says that we have got two fluids maintained at two different thermodynamic conditions given pressure and saturated liquid of 3 kg mass. This is mixed with a state B which is MB at 2 kg, mass at 2 kg of water maintained at some pressure and some temperature. So, we will have to first locate this state A and state B. These two states actually form my thermodynamic state 1. The process of mixing happens which is an adiabatic process and takes place at constant pressure because of which this thermodynamic system goes from state 1 to state 2 where it gets mixed together and what we have to find is during this process when the state changes determine the changes in volume from 1 to 2 delta V1 to 2 energy delta E1 to 2 the dryness fraction X2 at 0.2 and the work done during this process 1 to 2. Alright. So, let us come back to the solution. Let us try to understand how we envisage the situation what is happening here pictorially schematically first and then we will solve this problem. Alright. So, this is my question D. I can visualize this as in a piston cylinder arrangement wherein this is my system which has got two compartments A and B in case small A and small B. Alright. And we can say 1A and 1B. Why 1A and 1B because this is my state 1. Alright. And from here it goes to 2. So, we can say that in this 1A we have got saturated liquid. So, we can say 1A is saturated liquid water at 15 bar and the mass is 3 kg state 1B is 15 bar and at temperature of 250 degree centigrade and the mass is 2 kg in this case. Alright. So, this is what is given about the state 1 which consists of 1A and 1B together. What happens if I mix this let us say we can we can puncture this partition for example or remove this partition then there is a mixing of these gases and because of this mixing of the gases the process happen and this process happens at constant pressure which is maintained at 15 bar. Alright. So, this process is constant pressure adiabatic. The process is from 1 to 2. Okay. Now, let us apply first law. We say Q is equal to delta E plus W. If I neglect delta K E kinetic energy and potential energy and also W others and I assume that there is only expansion work this formulation comes down to delta E is equal to delta U. Alright. So, what we have because this process is adiabatic we say Q is equal to 0 and W expansion is equal to W is equal to P into V2 minus V1. Alright. This is the W now because this is a constant pressure process. So, if I put this back in these equations what I get is therefore delta U plus W expansion is equal to 0 because Q is equal to 0 and therefore I can say now U2 minus U1 plus P into V2 minus V1 is equal to 0. It means that if I club U2 and P V2 which is nothing but H2 minus H1 is equal to 0 or I say that H2 is equal to H1 in that case. Alright. The enthalpy at state 2 is equal to enthalpy at state 1. Let us now see what is enthalpy at state 2? The enthalpy at state 2 is equal to the total mass of M1A and M1B together and corresponding specific enthalpy at state 2. So, I can say now M1A plus M1B which is the total mass at state 2 into H2 which is specific enthalpy at state 2 is equal to M1A H1A plus M1B H1B. Alright. So, this is a state 2 and this is a state 1 in which we have got compartments for A and B. Alright. Here. So, as we understand the state A has 15 bar and saturated liquid. If I go to table 2 and see at 15 bar what is T sat? So, here you can see table 2 and this is 1.5 MPa which is 15 bar and corresponding temperature is 198.28. So, this is the temperature of 1A which is maintained at saturated liquid. So, temperature is C sat at this ratio. So, our 1A is at 198.287 and I can also tell you that the 1B position is 15 bar and 250 degree centigrade. Alright. The pressure is same and it is maintained at 250 degree centigrade. That means the temperature there at 1B is more than T sat. When temperature is more than T sat, we are sure that it is in a super heated vapor region and therefore, one state 1A is saturated liquid which is maintained at 198.287 degree centigrade and the other state which is also at the same pressure but at 250 degree centigrade, it is in a super heated vapor zone. You can see the mixing of these two zones. This mixing of two zones which is a saturated liquid and other one is a super heated vapor and this is a 1A and 1B. Two states together they get mixed because of this a process happens and because of which some work will be done on the system or by the system and that is the problem. We have to basically find out the property changes that happens because of this mixing of two different states which is maintained at 1A and 1B alright for state 1. Alright. So, we find that state A or 1A is at 15 bar and 198.27 degree centigrade and state 1B is at 15 bar and 250 degree centigrade which means that it is a super heated vapor alright. Let us go to the next page. So, we had said that M2 H2, M2 is nothing but M1A plus M1B is equal to M1 H1A plus M1B is M1A H1A plus M1B H1B alright. We got the values from the steam table. We can read that H1A is equal to HF at 15 bar which is what we have read from the steam table. I will not show the steam table again now. I will just write down those values which I feel that you are conversant with these values now. So, HF at 15 bar is equal to 844.56 kilojoule per kg and H1B is equal to H at 15 bar 250 degree centigrade alright and we have found that this is in super heated vapor region. We can get it from table 3 and the H1B for this comes out to be 2923.9 kilojoule per kg. This is what we got from table 3 alright. So, if I substitute these values in this equation 1, so I think substituting we get M2 H2, M2 is 5 kg because 3 kg plus 2 kg together 5 kg into H2 is equal to M1A H1A, M1A is 3 kg into H1A which is 844.56 kilojoule per kg plus 2 kg into 2923.9 kilojoule per kg. From here if I calculate the value of H2 that is the state 2 because what I am going to get from here is enthalpy at state 2 which is also maintained at 15 bar. So, I get H2 is equal to 1676.296 because the process is at constant pressure at 15 bar. I get this value if I compare this H2 at 15 bar with HF at 15 bar and HG at 15 bar I feel that this is what we can see from steam table HF is less than H2 less than HG at 15 bar this is what we can understand from the steam table. If this is true then I know that this is a weight region. When I say weight region that means a 2 phase and when I say 2 phase it has to have some X2 value alright. So, now next important thing is to find out the value of X2 simple formulation for X2 we know that H2 is equal to HF plus XFG at that pressure. So, I can find out the value of H2 is known to me 1676.296 which is just we have calculated HF is known to me at 44.56 X something I want to find out HFG is 1946.4 from where I understand therefore X is equal to 0.4273. Importantly whenever we have got a 2 phase region we have to first find out the value of X then only I can calculate all other values alright. So, we have just calculated the value of X2 at state 2 which is obtained after mixing these two different strings 1A and 1B together. Let us find out the other values also important is to show this schematically on a PV diagram where was 1A where is 1B and when this 1A and 1B got combined where does the state 2 lie alright. So, please this has to be understood because the schematic can actually give you a better picture of the process.