 So, welcome back. Let us go to the tutorial 2 now and let us have some more illustrative example based on steam tables. The question is, find a state of a closed system containing water substance of given mass as mentioned below, which you can see in the next table. Specify the zone in which the point lies and property V, U, H and S as appropriate. Also comment on the approximation value if applicable. So, we have just now seen table 3 now and let us try to use table 3 as much as we could in the given problem. The question is, you got some mass given now that means the mass also is very important parameter. Some two properties are given. For example, here pressure is 100 bar, temperature is 250 degree centigrade. In other case pressure is given, some state is given. In the third case temperature is given and some enthalpy is given. So, minimum two variables are known to you about the system and therefore what we are expected now to get the values of V, U, H and S. So, first and foremost important is to understand where the state lies, where the thermodynamic state lies, system lies and accordingly we will decide whether I should go for table 1 or table 2 or table 3. Also important is to note that these are not, some mass could be not only 1 kg sometimes it could be 2 kg or 3 kg also and most of the properties are given as per kg. In that case, you will have to multiplied by the mass associated with those particular problems. Let us now look at the first problem where the mass is 1 kg, property P pressure is 100 bar that means 10 MPa and property 2 is T is equal to 250 degree centigrade. So, let us first try to see in which zone it lies. So, question number 1, m is equal to 1 kg, P is equal to 100 bar and T is equal to 250 degree centigrade. Let us go to table 1 and let us see what is the P sat value for 250 is alright. So, let us refer to table 1 and find out corresponding P sat value and then we can compare that with P alright. So, let us go to steam table, table 1 and let us see what is P sat 250. So, here we are temperature is 250 degree centigrade and corresponding P sat P is 3.9762 which is P sat value alright. So, let us note down this value and compare it with the P that we have been given. So, P is equal to 3.976 MPa which is obviously our P which is 10 MPa is more than P sat T alright. So, P given is more than P sat T because P given is 10 MPa and therefore, we can say that the region is sub cooled region, sub cooled or compressed region, compressed liquid region or zone. Now, once I know that we are in sub cooled region, I know that I have to refer to table 3. Now, I would like to go to table 3, look at the point of pressure and temperature and get the respective values at that particular point. So, now I will go to table 3, go to the value of 10 MPa and 250 degree centigrade and get the corresponding values. So, can we go to table 3 now? So, here we are, we are at table 3 10 MPa and we want to see the value at 250 degree centigrade. So, if we come down, this is 250 degree centigrade and these are the values for V, U, H and S respectively. So, we can just note this value that V is equal to 0.00124115 meter cube per kg, U is also given, H is also given and S is also given and we can write these values in the answer sheet that we have. So, can we now write these values after locating the point, the region we can now see, we can write the values in our answer sheet. This is the answer sheet and at T is equal to 100 bar, T is equal to 250 degree centigrade, I can write these values as what we have seen. Now importantly, having written these values, the problem is actually solved for us, but I have just explained to you that this can have approximation also and we have just seen what approximation that it exists that we can take the values at PSAT also. And the corresponding PSAT because we are at 100 bar, which is much higher than the PSAT value, can we now see if I take the values at PSAT, which is what you see here in this table, PSAT T is 39.762 bar. So, just for the sake of it and just for the comparison purpose, I would like to get the values from the table, at pressure is equal to 39.762, temperature is 250 degree centigrade and let us get those values and let us just compare the values at this saturation condition and values at sub cooled condition in this case. So, let us go to this particular table now. So, let us go to table number one and find out those properties. So, here we are table number one at 250 degree centigrade, PSAT is 39.762 MPa, as we know corresponding to that, we can get the values of Vf, Uf, then Hf and Sf. The value of Hf however is that 250 and 3.9762. If I want to get the values at 100 bar, I will use the formulation for H is equal to Uf plus P time Vf, isn't it? As we have already seen, I will do these corrections and I will not compute all those values right now. Let us find out the enthalpy at 100 bar by knowing the formula that H at 100 bar is equal to Uf that means saturation value at this point plus P times Vf, put that in formulation and get these values and let us compare it in the table that we just saw. So, let us come back to the table now here and at P is equal to 39.762 bar at P is equal to 250 degree centigrade, I get these values. So, you can see that from here that V, this is Vf value and this is actually V and you can see that there is 1 2 4 has become 1 2 5, there is hardly any change. As far as U is concerned, there is some change, percentage wise it will be less than 1 percent, enthalpy is considered, there is some change, S is concerned, there is hardly any change. If I see percentage change, you can see that the percentage change is very very small, very very small 0.85, 0.69, 0.69 and 0.51. This is a very small percentage change which has been evolved evaluated, this has been evaluated at the approximate condition that we have evaluated the properties at saturation condition, isn't it? At saturation temperature 250 degree centigrade, at those points we have considered this, this is Vf, Uf, H evaluated at 100 bar, but taking the values of at Vf and Uf and Sf also at this point, surely because of the fact that we know that subcooled liquid is incompressible or liquid is basically incompressible, alright.