 So, let us continue with the tutorial. The question B is now again relates to dryness fraction and the problems associated with wet steam. For the following data, a process takes place from state 1 to state 2, alright. So, you got a state 1 defined, state 2 defined. Find the properties H, S, V and T at both these states from point 1 to point 2. What is also given is the initial state P1 that is the pressure is 6.5 bar. That means we have been given initial pressure as 6.5 bar and also corresponding to that if you see problem 1 now, we have given property 1 that is V1 as 0.16 meter cube per kg. That means the initial pressure is 6.5 bar and the V1 is 0.16 meter cube kg and the state 2, one of the properties has been given which is P2 as 2.5 bar and we have also been told that this process from 0.1 to 0.2 is a constant volume process and therefore we know other property also of the 0.2 which is at constant volume. So, V1 is equal to V2. So, indirectly the other property also has been given while in the next problem you can see is constant pressure. So, one has to under it could be constant enthalpy, it could be isentropic process or the entropy can remain constant. So, one has to understand what is the hidden value that has been given in the problem. If it is the constant pressure, constant volume, constant entropy, constant enthalpy any of these properties could be given to you and correspondingly you would have to determine what is the state 2 now. So, the problem therefore is at 0.1 we have got P1 as given 6.5 bar. We have also been given V1 as 0.16. At 0.2 we have been given P2 and again 0.2 we have been given V2 because the process happens at constant volume. What is not known is what is the state 1 where is the state 1 and state 2 lie? Are they single phase? Are they two phase? Are they in super heated region? Are they under the dome? Are they sub cooled region or whatever? This is first we will have to make it clear. So, because we have been given now in this problem pressures have been given P1 is equal to 6.5 bar. Let us now go to table number 2. Look at the values different values that we get at P1 is equal to 6.5 bar and then we solve the problems. So, 6.5 bar is 0.65 MPa. We can go here to the table and we see that we do not have values for 0.65 MPa but we have got values for 0.64 MPa and values for 0.66 MPa. We will have to interpolate between these two values now. So, please write down the values of 0.64 and 0.66 interpolate those values and I will not show you the interpolation now. I will directly write the values that we calculate for 0.65 MPa. Once we know these values we will compare the given other property which was V1 and find out the state of the fluid at state 1. So, coming to the solution of the problem now we say that this question 1 what we know is P1 is equal to 6.5 bar and V1 is equal to 0.16 meter cube per kg. If I write if I divide this page into two part this is state 1 and this could be state 2 and what we know here is P2 is equal to 2.5 bar and also what we know is V2 is equal to 0.16 meter cube per kg because this process of 1 to 2 is constant volume process alright. As I said we will interpolate the values for 0.64 and 0.66 MPa and I will directly write the values for 0.65 MPa here. So, at 6.5 bar I will get Vf is equal to 0.001104 meter cube per kg and I will get Vg is equal to 0.2926 meter cube per kg. What I understand therefore is the V1 value of 0.16 meter cube per kg lie between these two and therefore we say that Vf is less than V1 less than Vg so we are in two phase region. Moment we say two phase region we want to know what is the value of x. So, if I calculate x now I will say for the formulation of V1 is equal to Vf1 plus x time Vfg I can put the values of V1 as 0.16 and correspondingly which I have got here and here I can calculate the value of x1 is equal to 0.545 alright. When I know the value of x I can get the values of h1 which is equal to hf1 plus x1 time hfg1. If I put these values I will get 1815.22 as kilojoule per kg as the value of h. Similarly I will get the values of s1 is equal to 4.56 kilojoule per kg Kelvin and I will get the values of u also as 1711.27 kilojoule per kg alright. So, these are the values which I will get for state 1. Let us now go to state 2 and we know that state 2 we have got P2 is equal to 2.5 bar and at from table 2 I will get the values at P2 is equal to 2.5 bar what I know is V1 is equal to V2 which is 0.16 meter cube per kg at P2 is equal to 2.5 bar I will get Vf2 is equal to 0.00106 meter cube per kg and Vg2 is equal to 0.718 meter cube per kg. So, again we know that Vf2 is less than V2 is less than Vg2 which means that this is also going to be a two phase region. Moment I say two phase region I should know what is the value of x2 alright. It means that importantly that this process entirely happens in the two phase region the point 1 is also in two phase region and the point 1 and 2 both lie in the two phase region only. I would like to calculate s2 again using the formulation for V2 V2 is equal to Vf2 plus x time Vfg V2 is 0.16 Vf2 is 0.00106 plus x time Vfg which is one can find out the difference between these two and I can get from here the value of x2 is equal to 0.22. So, Vfg is 0.717 so x2 happens to be 0.22 corresponding to the x2 now I will get the values of h2, s2 and u2. My values of h2 will be hf2 plus x2 time hfg which is equal to 108.4 value of s2 is equal to 2.81 kilojoule per kg and value of u2 is equal to 978.41 kilojoule per kg and the temperature I can get from the table which happens to be 127.411 degree centigrade which is the saturation temperature at P2 pressure. Importantly I would like to show this process 1 to 2 on a PV diagram the entire process because it happens on because it is in two phase region I would like to look at the 0.1 and look at the 0.2 schematically on a PV diagram and that will actually talk everything about the process itself you know the 0.1 and 2 under a PV diagram we will talk everything about this process. So, let us plot this points 1 and 2 and draw this process 1 to 2 on a PV diagram. So, this is PV typically the PV diagram would look like this. So, the 0.1 this is let us say 6.5 bar and because the process is from 6.5 bar to 2.5 bar this is 2.5 bar and the process is constant volume x is 0.545 around the first case. So, this is my 0.1 and if I come down vertically this is going to be 0.2 alright. So, this is 1 to 2 the process takes place at constant volume and this can be shown on a given PV diagram. So, if I go back to the answer this is my process P 1 is given V 1 is given here P 2 is given as 2.5 bar I can write other values the value of x 1 was 0.545 the value of x 2 is 0.22 which is state 2 here. So, state 1 values are given that we calculated and the value of state 2 are here which is the x changes from 0.45 to 0.22 while u changes are given here h changes are given here and entropy changes are also given here. So, respective values at state 1 and state 2 are shown this process as constant volume and that is why V 1 remains the same in this case.