 In this, one kg of water, next question, one kg of water at 30 degrees is contained in a leak proof frictionless piston cylinder operators as shown in the figure. The piston has a cross sectional area, so AP equal to 0.065 meter square and mass of the piston is 40 kilograms and initially rest on the stops as shown in the figure and enclosed volume V1 is 0.1 meter cube. Heat is now transferred to the water until the temperature reaches, so T2 is 200 degrees centigrade. This is the data given. So initial temperature T1 is 30 degrees centigrade. Ok, mass is 1 kg. This is the data given. So what is asked? Reduce the temperature of the water when the piston rises from the stops. Ok, now what is the initial condition? P1, sorry T1 equal to 30 degrees centigrade and V1 equal to 0.1 meter cube, M equal to 1 kg. From this I can find small V1 equal to V1 by M which is equal to 0.1 meter cube per kg. Ok, so now two values are there. T and V are there. From that we will try to fix the state. So go back here, temperature based tables, saturation tables. So here 30 degrees centigrade, 30 degrees centigrade is here and I can see that the temperature 30 degrees centigrade, the pressure is 0.0425 bar. Ok, now what is the VG value? 32.9. This is 0.0010045. Ok, so now the state is saturated because VF equal to 0.0010049, ok, sorry 45 meter cube per kg. This is at 30 degrees and VF equal to 32.9 meter cube per kg. These are the data. So now we can find that VF is less than V1 is less than VG, this is VG, sorry so that means the state 1 is, state 1 is saturated mixture. So we can find quality X1 will be equal to V1 minus VF divided by VG minus VF which is equal to 0.003, 0.003. So from that I can find the internal energy U1, ok. So U1 equal to UF that is 125.8 plus X into UG, 2416 minus 125.8. So this is UF at 30 degrees and this is the X1 and this is UG at 30 degrees, ok then this is UF at 30 degrees. So from this I can get the value of U1 as 132.67 kilo joules per kg. Ok, so now this is the initial state is fixed now. Now I have to find what is the pressure here now? Pressure will be 4.25 kilo Pascal which is equal to P sat at 30 degrees. This we saw from the steam table, correct? So this pressure will not be enough to move the piston. So what is the pressure? Pressure required to move the piston, ok. So that I will say as P atmosphere plus M P G divided by AP. So now P atmosphere is given here in this problem as 94 kPa, mass of the piston is 40 and AP is 0.065 and G is 9.81. So substitute these values 94 kPa, so 94000 Pascal plus 40 into 9.81 divided by AP is 0.065. So that will be equal to 100 kPa. So now initially the pressure is only 4.25 kPa. When the pressure reaches 100 kPa, then only the piston will lift. So you have to add heat, constant volume process will take place from 4.25 kPa to 100 kPa, then the piston will lift. So here I will say 30 degrees, line is somewhere here, 30 degrees. So corresponding pressure, so this is 30 degrees I will say, corresponding pressure can be drawn like this. So this is the P sat, this is 4.25 kPa. So now we were in some stage here. This is state 1. Quality is 0.003. We calculated the quality also, correct? 0.003. So this is state 1 for me. Now a constant pressure process should occur till we move to state, say let us say when piston just lifts, let that be state 2. So that means P2 equal to 100 kPa. Volume does not change. So V2 will be equal to same as V1, V1 equal to what? So now V1 is 0.1. 0.1 meter cube per kg. So now go to the steam tables again. Here go to the pressure based tables now, 100 kPa, 100 kPa 1 bar, so this line. 100 kPa you can see that Vf is 0.001043 and 1.694. So Vf equal to 0.001043 and Vg equal to 1.696, meter cube per kg, 694, sorry, 694 meter cube per kg. So now we can say again the state is still saturated only. So for example, it may be like this, some saturated state somewhere, ok. So now I will find the quality again. X2 will be equal to what? V2 minus Vf divided by Vg minus Vf. So substituting this I will get 0.05842. So you can see this, the quality has reduced to the state 2 now, ok. So now this is what? This is 100 kPa. So that means this temperature will be, since it is saturated, this temperature will be for 100 kPa, 99.62, that will be the temperature here, 99.62 degrees centigrade. So constant volume process has occurred. So at this point the piston will lift off because pressure has reached 100 kPa. Do you understand? So what is the temperature? T of the system, T2 is equal to 99.62. So that is the temperature when the piston will just lift off. That is the question asked. What is the temperature of the water when the piston first rises? That is the T saturation at the 100 kPa, ok. Now finally, the temperature is just 200 degrees centigrade. Do you understand? Now what happens when the piston is now free to move, ok? That means P remains constant at 100 kPa because there is nothing, there is no spring or anything. So the piston just moves at constant pressure. But now the temperature should, so I will draw this, this is actually 100 kPa. Now the temperature, final temperature is given as T2 equal to 200 degrees centigrade, T2, sorry, I will say P3. T3 is 200 degree and P3 here, P3 will be equal to 100 kPa. So now it goes like this, becomes saturated, then it superids to 200 degrees centigrade. So this is the process. So you can see 1 to 2 constant volume, then constant pressure process occurs and it goes to superheated condition. So final condition is superheated steam, superheated steam at 100 kPa, 200 degrees centigrade, ok. Now we have to find the, what is asked is, yeah, this is amount of heat transfer work done by the water, both you have to find amount of heat transfer. For that first state 2, state 3 should be fixed. State 3 is what? P3 equal to 100 kPa same as P2 and T equal to 200 degrees centigrade. So go to the tables, superheated tables for this is the 100 kPa or 1 bar, temperature is 200. So find the V, V is 2.172 and U is 2658. So V equal to 2.172 meter cube per kg and U equal to the U3 V3. Now this is 2658 kilojoules per kg, that is it. Now what is work done in the constant pressure process? Work done in constant pressure process equal to m into P2 or P3 whatever into P3 minus V2, ok. So that will be equal to what? 1 into 100 kilo Pascal into V3 is 2.172 and V2 is 0.1. So that will be equal to 207.2 kilojoules, that is part 3. Then part 4 is heat, first law. So that will be equal to 207.2 plus m, m is 1 into U3 minus. Now you can take the total change. Even in the constant volume process, there will be change in internal energy. You can see this diagram. 1 to 2 is occurring. There will be difference in the here temperature increases. So due to that internal energy increases, similarly here the volume increases and again volume and temperature both increases in this. So you get the total change in internal energy is final state minus initial state here. So U3 minus U1. And this is the diagram with reasonable saturation lines for representing the process undergone. So that will be equal to 207.2 plus 1 into 2658 which is U3 minus U1 is 132.67 which is equal to 2732.73 kilojoules. So that is the answer, ok. So this is about this problem where you can see that first a constant volume process occurs, then the piston can move off from the stops. And then now constant pressure process occurs. Initially saturation condition occurs, but the quality initially is higher. 0.003, say lower, 0.003. Now quality improves to 0.05842, then it becomes liberated. So this is the answer. So I will stop here.