 So, I can calculate other property changes delta u is equal to u 2 minus u 1 which is equal to m 2 u 2 minus m 1 a u 1 a plus m 1 b u 1 b ok. This is my second equation I can get the value of u 2 is equal to u f plus x u f g or delta u and again putting the respective values together we will get u 1 a is equal to u f at 15 bar and u 1 b is equal to u at 15 bar and 250 degree centigrade alright. Putting these values I can get 15 bar and 250 degree centigrade from which I can get u 1 b. So, I can write as u 1 b is equal to 2695.9 kilojoule per kg alright. I can do further calculations get the values from table I know u f is equal to 842.83 kilojoule per kg and u g is equal to 2593.4 kilojoule per kg this is what values I will get from table 2. So, from this table if I put this back in this equation I will get u 2 is equal to 842.83 that is u f plus 0.4273 u g minus u f putting these respective values over here I get u 2 is equal to 1590.85 kilojoule per kg alright. Having got these values now I can get delta u using this equation now I can put these respective values over here and I apply the same way as I did earlier I can now compute delta u is equal to 33.96 kilojoules. Similarly, I do with the specific volume change also and I get delta v is equal to minus 0.0228 meter cube alright. So, there is a volume change that happens from v 1 to v 2 alright which is very important to understand there is a reduction in volume that happens because of this mixing process. Let us now understand the expansion work w expansion work which is p into delta v pressure is 15 into 10 to the power 5 Newton per meter square into delta v is minus 0.0228 meter cube from here what I get w expansion is equal to minus sign 342.00 joules which is minus 342 kilojoules. So, this is the work done on the system because it has got negative sign. So, w expansion is 34.2 kilojoules. What we can do also is to show the schematic of this on a p v diagram if I want to show on a p v diagram now right this is my state 1 a this is line of 15 bar alright. So, this is 1 a which is maintained at saturated liquid and at 15 bar. So, this is point 1 a 1 b we have found that it is a superheated region. So, somewhere around this point 15 bar 250 degree centigrade 15 bar saturated liquid and we also found out that when these two streams came together this stream came from here this stream came from here both of them came and this mixing resulted in a wet steam which has got x as 0.4273. So, somewhere around here was state 2 this is how we can show that because of mixing of 1 a and 1 b 0.2 came as a state 2 because of this process. So, if I see the piston arrangement this was 1 a and 1 b and as soon as the state 1 a got mixed with state 1 b the state 2 resulted in decrease in volume and because of which this piston came somewhere at this point and which is why we got w expansion work as negative which is work done on the system alright. So, because of this saturated liquid and superheated steam got together the piston actually came inside which is what we say well done on the system. Answer to the entire question now is we got delta V12 as a result of this mixing we had reduction in volume minus 0.0228 meter cube we had delta E12 approximated as delta U12 which is 33.96 kilojoule w expansion work is negative minus 34.2 kilojoule and we found that at 0.2 which is a wet steam we got a 0.4273 this is a dryness fraction of the wet steam at state 2 what we got as a result of this mixing phenomena this is a very little bit involved problem where we understood that two streams of different nature got together of one A and one B because of which property changes occurred and because of which some work got done on the system. I think this four problems actually will give you a good idea about what kind of problems that need to be solved using water properties, water phase diagrams and again integrating all that knowledge to basically understand the first law of thermodynamics applied to practical systems. Thank you very much.