 So, after understanding the concept of dryness fraction, what is the dryness fraction and how to calculate? Let us now have a tutorial on dryness fraction, for which this tutorial is there on dryness fraction. The question is, find a state of a system, find a state of a closed system containing water substance of given mass as mentioned below. Apply the zone in which the point lies and properties V, U, H and S as appropriate. Also comment on its quality, which is a dryness fraction if applicable. So, basically depending on whatever has been given, for example, your question number 1 is mass 1 kg, property has been given as 0.6, while property 2 has been given as 10 bar. That means for a given pressure and given dryness fraction, find out where this state lies, at the same time find out the properties V, U, H and S at those properties, at that particular thermodynamic state. So, first problem is mass has been given, the fact that the X has been given, we know that it is wet steam and the property has been given as 10 bar. So, I know pressure therefore, what would I do? I will first go to the steam table and see table number 2, where the pressure is the base. So, let us come to this is my question, the first question is mass is given, one property has 10 bar and X has been given as 0.6. So, these two parameters have been given, we will have to find out the values of V, U, H and S. So, let us go to table number 2 and locate pressure is equal to 10 bar and find out the properties at that particular location. Can we see table number 2? So, 10 bar is 1 MPa, this is the table on this table, this is 1 MPa alright. So, at 1 MPa, write down all the values first, what has been given is X is equal to 0.6. So, we will require both the parameter Hf and Hg, Sf and Sg, Uf and Ug alright as given over here. So, Vf, Vg, Uf, Ug, Hf, Hg, Sf, Hg all these parameters are required, please note those parameters and let us go to the solving of the problem now. So, question 1, what is given to us is X is equal to 0.6 and P is equal to 10 bar or 1 MPa. So, from table 2, we could get that Vf is equal to 0.0011273 meter cube per kg and Vg is equal to 0.19436 meter cube per kg. So, simple is I know Vfg is equal to Vg minus Vf which is equal to 0.1932 meter cube per kg alright. Now, I will have a simple formula V is equal to Vf plus X time Vfg alright, I will just write the value of Vf and Vg, X is equal to 0.6, so we say Vf is equal to 0.0012723 plus 0.6 into 0.1932, this is equal to 0.117066 meter cube per kg alright. Similarly, this is the value of V that we got, similarly I can do S is equal to Sf plus X time Sfg, H is equal to Hf plus X time Hfg and U is equal to Uf plus X time Ufg, I will get all these values from table 2 for Uf and Ufg and get that S is equal to 4.8 kilo joule per kg Kelvin, H is equal to 1971.28 kilo joule per kg and U is equal to 1854.176 kilo joule per kg and the mass that was given to us was 1 kg. So, I will just multiply this parameters by 1 kg and then get the value of S as 4.8 kilo joule per Kelvin, H is equal to 1971.28 kilo joule and U is equal to 1854.176 kilo joules alright, so this will be my answer to this question. So, I can come back here now and put those properties and the results that we got alright. So, these are the results now, V is equal to so many meter cube, U is equal to 1854, H is equal to 1971 and S is equal to 4.806 alright. So, importantly is when X is given I know that the two phase and therefore, we have to go to table 1 or table 2, get Vf and Vfg, Uf and Ufg, F and Hfg, Sf and Sfg and get those values.