 So, having done this problem now let us do one more problem in order to get complete understanding of how to calculate during a process what parameter changes. So, let us come to the problem number 2 where p 1 again is the same as what was given as 6.5 bar the value of x 1 is given. So, again it tells us that it is a wet steam or a two phase region moment x is given and x is less than 1 we know that it is a two phase mixture. So, x 1 is 0.8 and the point t 2 is 550 degree centigrade. So, what we know is t 2 is 550 degree centigrade and also this process from 1 to 2 is constant pressure process this is very important this tells you that the pressure remains constant. So, on a PV diagram this will be horizontal line earlier we had a vertical line for a PV diagram because it was a constant volume process in this case is a constant pressure process and therefore we will have a horizontal line over here alright. So, let us now solve this problem I will just show in brief in this problem what has been given is pressure what has been also given is x 1 therefore we know that it is a two phase. So, we can easily calculate the properties at point 1 and what has been given for the table for the point 2 is the temperature and corresponding to that we can see what are the values at this temperature and because it is a constant pressure and we can get the details at point number 2 also. So, let us go to table number 1 where we can see what is the state of this point 2. So, as is known to us that the process is constant pressure process that means at point 2 also the pressure is 6.5 bar. So, what we have to see is going to 550 degree centigrade we can see what is the T-sat pressure T-sat at 6.5 bar is and compare that T-sat value to 550 degree centigrade and get the state of point 2. So, first look at the state 1 which is at 6.5 bar which is 0.65. So, again I will have to interpolate this between 0.6 and 0.64 and 0.66 get those value here similarly I will compare this temperature the temperature happens to be around 163 and 164 and we know for sure that 550 degree centigrade is much more than this it implies that we are in a super heated region straight away a very simple calculation can be done here the temperatures are 161 and 162 we our temperature is 550 degree centigrade which implies that we are in a super heated region straight away. So, knowing that the point 1 has got x value that means it is a two phase mixture while the point 2 is in super heated region by understanding its temperature and pressure with this knowledge let us come back to solving this problem. So, question number 2 what we know is P 1 is equal to 6.5 bar what we know is x 1 is equal to 0.8 and for point 2 what we know is P 2 is 6.5 bar and what we also know is T 2 is 550 degree centigrade this is the data. So, after interpolating the values at 0.65 MPa we will calculate the values at for given x value as 0.8. So, we get as v 1 is equal to v f 1 plus x time v f g 1 similarly we have got values of h 1 and s 1. So, v f v 1 as we get are 0.234 meter cube per kg h 1 we get as 2344.4 kilojoule per kg and s 1 we get as 5.77 kilojoule per kg Kelvin and u 1 also we get as 2192.23 kilojoule per kg alright these are the values at state 1 at state 2 we know that T sat at 6.5 bar is around 162 degree centigrade approximately that is what we found and therefore we know that our T 2 is more than T sat at 6.5 bar and therefore this is in superheated region alright and therefore I can get the values of superheated region from now table 3 alright and get the values from table 3 now at 550 degree centigrade again you may require to do interpolation I will not show your table 3 values now here but from here what we get is v 2 is equal to 0.58 meter cube per kg u 2 is equal to 3212.9 kilojoule per kg h 2 is equal to 3591.3 kilojoule per kg and s 2 is equal to 8.1 kilojoule per kg Kelvin this is what is going to be 0.2 importantly I would like to plot this again on a PV diagram so this is a P and V and this is what the state would look like PV diagram so I can draw the pressure at 6.5 bar this is a constant pressure process therefore both the states would lie on this line as we say that at 0.1 we got x 1 is equal to 0.8 so therefore my 0.1 could be somewhere at this point because this is the entire length this is entire from 0 to 1 that is what our dryness fraction is and 0.8 would lie somewhere at this point well 0.2 will in a superheated region outside this dome maybe somewhere at this point this is my process 1 to 2 all right this is what it would look like. So let us come back to the table and put these values at 0.1 and 0.2 at 0.1 x was given as 0.8 the values of v u h and s were calculated by proper interpolation and at 0.2 now where the properties we found from table 3 now where we know that we are in superheated region the different properties of v u h and s are given over here at this point all right thank you very much.