 Now, a piston cylinder assembly as in the figure contains 1 kg, 1 kg of saturated steam water mixture with a quality of 0.6 at 100 kilopascals. It is gradually heated until it becomes saturated vapor. So, x2 equal to 1, saturated vapor. At this point, at this instant, a spring, ok, so that is when it is heated, basically it actually moves up and becomes saturated vapor. At this point, a spring with a spring constant of 11.49 kilo Newton per meter just touches the top portion of the piston without exerting any force. So, till this point, it is constant pressure process. P2 is 100 kilopascals, so that is the state 2. The piston has negligible mass, that is m mass of the piston is approximately 0 and its cross sectional area is 0.1 meter square. Now, after the spring just touches the top portion, the heating is continued, heat transfer is continued until the steam attains the pressure. Now, P3 equal to 200 kilopascals, so pressure increases. Why pressure increases is because of the spring is now compressed and that exerts force on the piston. So, the pressure increases, calculate the work and heat interaction in the final temperature of the steam. So, this is the problem. At mass is pressure, so initially only at mass is pressure of 100 kilopascals acts and when the state 1 to state 2 is a constant pressure process, so that is in this. So, let us do the calculation v state 1, so saturated state, saturated liquid vapor mixture at 100 kilopascals and the quality 0.6. So, these are the two things which we know. So, go to the tables now, saturated tables here, so 100 kilopascals is 1 bar, so this is the point here. So, I can take the value of vf vg uf ug from this table that I will do now, so we can say v1 will be equal to vf plus x1 into vg minus vf, so this is equal to from the tables I have taken the value of vf etcetera, 1043 0.001043 is vf value plus 0.6 is the quality into vg. So, vg is 1.694 minus vf is 0.001043, so that will give you v1 as 1.0168 meter cube per kg, so that is the state 1. State 2 is when the pressure just touches the spring constant pressure process, so p2 is 100 kilopascals again, but it has now become saturated vapor, so that is x1 equal to 1 that implies v2 will be equal to 1.694 meter cube per kg and u2 will be equal to 2506 kilojoule per kg. Here also I can calculate u1 ok. So, u1 from this I can calculate u1 as uf plus x1 into ug minus uf which is equal to 417.3, you can take the uf value from tables are 417.3 plus 0.6 is the state 1 x1 into ug is 2506 minus 417.3, so that will be equal to 1670.5 kilojoule per kg. The state 2 u2 equal to what? ug at 100 kilopascals that is 6506. So, now state 1 and state 2 are fixed with properties for evaluated for v1 and v2 u2. So, now for this I can find the work interaction constant pressure process. So, that is m into p into v2 minus v1 which is equal to 1 into 100 into 1.694 minus 1.0168. So, this is equal to 67.72 kilojoule. Now, q1 to 2 equal to w1 to 2 plus m into u2 minus u1 which is equal to 903.2 kilojoules. So, this is when the system is further heated what happens piston rises compressing the spring correct. So, that means, here the pressure increases linearly. So, what is the height the pressure has gone final pressure 3. So, this is the 2 2 3 process is called 2 2 3. So, pressure at the final state is given in the problem as 200 kilopascals here 200 attains the heat. So, until the heat attains the pressure of 200 kilopascals. So, p3 is given as 200 kilopascals that is done. Now, but what is the height which is reached to find the volume correct what height has gone. So, this p3 is equal to p atmosphere plus p spring because the mass of the pressure is negligible. So, there is no pressure in part by the mass. So, what I can do here is 100 kilopascals this is 200 is no 100 plus for the spring I can say k into h divided by area which is equal to 100 plus k is given as 11.49 kilo Newton per meter into h which I do not know divided by area of processing is 0.1. So, using that I will I will get h equal to 0.87 meters. So, the spring is now compressed by 0.87 meters. So, from this I can get the volume what is v3 v3 will be equal to area v3 can be calculated as v2 plus h into area correct. So, that is equal to that will what is v2 we will go back and see what is the value of v2 we have got small v2 as 1.694 into mass is 1 kg. So, capital V2 will be 1.694 meter cube. So, this is 1.694 plus 0.87 into area of cross section is 0.1. So, that will be equal to 1.781 meter cube per kg. So, this is the way you calculate this. I can say v3 specific volume will be v3 by capital v3 by m, but m equal to 1 here. So, this will be 1.781. So, this is actually meter cube with total volume and here this will be ok. So, I got the state 2 fixed as pressure as 2000 say 200 kilo Pascal and volume is 1.781 meter cube per kg ok. Now, this is the situation here what is this state? So, you have to go back to the tables 200 kilo Pascal. So, I go here 200 kilo Pascal I find vg is 0.886, but v2 is 1.781 ok. So, we can say vg at 200 kilo Pascal equal to 0.886 meter cube per kg. v3 is greater than vg which implies state is super heated vapor. So, we have to go for the super heat tables and get the values. Go back here 2 bar table ok. Now, please see the value given is 1.781 exactly we have this. So, 2 bar specific volume is 1.781. So, this is entry straight away no interpolation required because the value which we got for v3 is same as in this entry for 500 degree centigrade. So, u is 3131 kilo joule per kg. So, for v3 equal to 1.781 meter cube per kg at 2 bar ok t equal to 500 degree centigrade and u equal to 3131 kilo joules per kg. So, this is the state. So, let us say u3 t3 500. So, that is the state here. So, now we can find first law q223 equal to w223 plus delta u223 ok. Now, how will you find w? w is basically the work done by the spring and work done by the atmosphere that you have to take into account here ok. So, this work is minus of w atmosphere plus w spring. We can also find a pressure no. So, pressure linearly increases. So, we can take an average pressure in delta v that also we can do or it is p2 plus p3 by 2 because pressure from 2 to 3 increases linearly. So, this is somewhat average pressure into v3 minus v2 that also we can apply. So, we can do this. So, what is p atmosphere? Sorry work atmosphere work atmosphere will be equal to minus p atmosphere into delta v. So, which is equal to minus 8.7 kilo joules. Similarly, pw spring equal to how much? Half minus of k h square which is equal to minus 4.35 kilo joules. So, I know these two values. Now, I can say w223 will be equal to 13.05. Just add these two 13.05 kilo joules plus m into m is 1 into u3 3131 minus u2 2506. So, that will be equal to 638.05 kilo joules. So, these are the heat and voltage. So, total q if you want you can add this q122 plus q223 that will be total heat. Total work is w122 plus w223. So, this is the problem here. So, we will draw the diagram, tv diagram. Now, see this what is the initial state for 400 kPa. And the quality is 0.6. This is state 1. Here x equal to 0.6. State 1 is 6. State 2 is x equal to 1. This is state 2. x equal to 1. So, that is done. Then what happens? The 200 kPa something like this. So, now what is the temperature we have seen? It has reached 500 degrees. So, 500 degrees is somewhere here let us say. So, that is state 3. Now, this linearly increases. So, state 1 to state 2 constant pressure process. State 2 to state 3 there is an increase in the pressure due to spring compression. So, finally, this is for 200 kPa. So, 200 kPa this is 500 degrees centigrade that is the state. So, this is what? 100 kPa it is 99.6 degrees centigrade and this 200 kPa saturated temperature is 120.2 degrees centigrade. So, this is the way you have to show these states. Initial state saturated constant pressure process occurs until saturation where saturated vapor is got. Then the pressure increases linearly and reaches the super industry. So, this is the problem.