 We move on to the sixth problem. An insulated tank A containing 0.4 meter cube, this has a volume of 0.4 meter cube of saturated water vapor, that is, so we can say V equal to 0.4 meter cube and say let us say X1, so XA1 equal to 1 saturated water vapor and P1 equal to 500 kilo Pascal, that is the content in the tank. Connected to initially evacuated, insulated piston cylinder assembly B as shown in the figure here. Nothing is there in this. The mass of the piston here, the mass of the piston is such that the pressure of 200 kilo Pascal is required to lift it. That means if the pressure at the bottom of this cylinder becomes 200 kilo Pascal, then the piston will lift. The, now the valve, this valve is open and the steam flows into the cylinder pushing the, so now what happens? Once the steam starts to flow from this piping to the wall, pressure slowly increases. Once it reaches 200 kilo Pascal, the piston will be lifted up slowly. The process continues until the pressure in the tank falls to 200 kilo Pascal, that is it. Now considering steam as a system, so the steam is now initially was in A. Now it is, so we can see that after steam pressure in A falls to 200 kilo Pascal, so finally, so we will say here 500 kilo Pascal to 200 kilo Pascal, the pressure falls. Then piston in the cylinder B would have reached some height because, so now steam will be there in A as well as B. Steam of course will be there in the pipe and valves, but it is negligible. The steam content in the pipes and the valves are negligible, so we need not take into account of that. So considering steam as the system, determine the A, the mass of the steam initially in the tank. B, the specific enthalpy of the steam at the final state when there is steam in both A and B. And C, a dryness fraction if saturated or the temperature if superheated. So that is the question asked. So here what is given in the state 1 is P1 is, so for 500 kilo Pascal, we have to go to the tables. So 500 kilo Pascal, we have to go to saturation, table 500 kilo Pascal tables here. So this is phi bar, phi bar table, you find that VG is 0.375, VG is 0.375 and here UG is 2561. So we can retrieve this for, because at 500 kilo Pascal or phi bar the quality is 1. That means you can take V as VG and U as UG. So now V1 equal to VG equal to, what is the value? From the tables you have to enter the value 0.375 meter cube per kg. Now you have U1 equal to UG equal to 2561 kilo joule per kg. So this is from the, so corresponding to X equal to 1. So we have taken this value. Now what is the mass? Mass of saturated steam in state 1, that is I will say sorry mA1 will be equal to how much? You know the total volume, V by V1 that is 0.4 meter cube divided by 0.375 meter cube per kg. So that will be equal to 1.0667 kg. So that is the first one, determine the mass of the steam initially in the tank that is 1.0667 kg. Now final pressure is given as, so P2 is given as 200 kilo Pascals that is given. But now the steam is in both A and B, some steam has flown from A and B but we can neglect the steam in the pipelines, small pipelines and the valves. So we will say that steam is present in A and B now, the final state. Now as the piston rises at constant pressure, because there is nothing, no other force. See why there is a pressure of 200 kilo Pascals needed to lift the piston that is there is atmospheric pressure of 100 kilo Pascals plus the piston itself can give a pressure of 100 kilo Pascals. So there will be totally 200 kilo Pascals is required for the piston to be lifted off. So as the pressure in the bottom surface of the piston reaches 200 kilo Pascals, a constant pressure process occurs. So as the piston rises at constant pressure, we can write the first law now, ok. And and the it is given that the insulated piston is insulated, insulated tank that means and A and B are insulated. Do you understand? So we can say q equal to 0, w equal to integral p dv which is occurring at constant pressure for this. So we will see that we have done the mass in initial state as 1.0667 kg. Now p2 is the final pressure is given as 200 kilo Pascals. Now see that the system will be whatever steam which is in A plus whatever is in B. Together we can take as a system for that the pressure in A is 200 kilo Pascals, pressure in B also will be 200 kilo Pascals because as the bottom surface of the piston sees the pressure of 200 kilo Pascals it will gradually lift up at constant pressure. So piston as the piston rises at constant pressure and A and B are insulated we can say q equal to 0 and w equal to p dv equal to p into delta v. This p is 200 kilo Pascals into delta v, ok. So this is the things you have to remember here. Now apply the first law, ok. The first law we apply we can see that q minus w equal to delta u or minus w equal to u2 minus u1. What is minus w? We can say m into p2 into v2 minus v1 equal to m into u2 minus u1 for delta u but m cancels. So now I can rearrange u2, u2 I keep here, p2 v2 I take the other side. So u2 plus p2 v2 equal to u1 which is there, I take the other side. So u1 plus p2 v1, p2 v1. So in the final state p2 equal to 200 kilo Pascals, ok. So now what is u2 plus p2 v2 that is h2. So h2 will be equal to 2561 plus that is u1 plus p2 is 200 kilo Pascals. This is kilo joule per kg. This is kilo Pascals and this will be meter cube per kg. This is v1. So now I substitute I get 2636 for the h2. So what is final state? Final state is h2 equal to 2636 kilo joule per kg and p2 equal to 200 kilo Pascals. So now the enthalpy at the final state is calculated now. That is the second part of the question. So see this b, the specific enthalpy of the steam at the final state that is calculated, h2 is calculated. Now c, calculate the dinos fraction if saturated or temperature if superheated that is c. Now for that I have to go to steam tables. In steam tables 200 kilo Pascals is this, ok, 200 kilo Pascals. You can see that the value of h is maximum h is what? Here 200 kilo Pascals is 2707 hg. At 2 bar that is 200 kilo Pascals the value of hg is 2707. So now at 200 kilo Pascals hg equal to 2707. That means what is this? Since h2 is less than hg, the state is saturated mixture of liquid and vapor. That means I have to find the quality. What is x2? x2 will be equal to h2 minus hf divided by hg minus hf. So now go back to the tables, get the value of this 2 bar. What is the value of hf? hf will be 504.7. So that is 2636 minus 504.7 divided by 2707 minus 504.7. So that is the quality value 0.9677, quality. So if it is, what is asked is if dinos fraction or quality is saturated. Now it is saturated, so we have found this. Now finally the work interaction for the process. What is work interaction? You have to find x2 is got. Now find v2. v2 is vf plus x2 into vg minus vf. So go back to the tables for 2 bar. What is vf? vf is 1.061 into 10 power minus 3 and vg is 0.886. So this is 0.001061 plus quality is 0.9677 into 0.886 minus 0.001061. So that will give you v2 as 0.8574 meter cube per kg. So what is the work done? Work done equal to p2 into v2 minus v1. So that will be equal to 200 into, so m also you can put, m. So 200 into mass is 1.0667 into v2 is 0.8574 minus v1 is what? v1 is 0.375. So that is the value. So it is around 97 kilo joules. So you can see that in this problem, in this problem the valve is open. The valve again mechanical device which will take care of the pressure difference between this. Initially nothing is there. There is a 500 kilo percent pressure here. The valve when it is open this will take care of take care of the pressure difference and slowly allow the steam to flow from A to B. Once the pressure in the bottom becomes 200 kilo Pascal, the piston will be lifted up gradually and the pressure will remain constant. As the steam flows, the pressure decreases from 500 to 200 kilo Pascal, that is the final state. Then after that the flow will not take place because both sides the pressure is same. Now what happens is that is a mass which is flowing from A to B taking the final state of the steam as a system, we can apply the first law. So for that, since it is insulated q is 0 and w will be equal to p into delta v. So that if we apply, we can find the final state specific enthalpy and with that and the pressure which is given, we fix the state and the state is a saturation mixture. So, dynastraction was calculated from that v2 was calculated and we found the value of the work. Next problem, a closed horizontal cylinder is divided into two parts by a frictionless thermally conducting piston held in place by a pin. So, let us draw this to understand better the cylinder horizontal cylinder divided into two parts by frictionless so like this thermally conducting. So, this is thermally conducting held in place by a pin. So, there is a pin which is first put some it may be at some position now say side A contains. So, this is a side A contains 0.01, we will write outside. So, this is 0.01 meter cube of air at 200 kilopascals and 75 degree centigrade. Side B contains 0.3 meter cube of saturated water vapor at 75 degree centigrade of water vapor that means x equal to 1 that is initial condition. So, this is the initial arrangement. The pin is now removed releasing the piston which moves towards the side B. So, this is B. So, when the pin is removed, the piston moves towards side B. Eventually both sides come to an equilibrium. Heat transfer occurs during the process. Please understand that heat transfer can occur from surroundings between A and B also through the piston because that is also thermally conducting. So, heat transfer occurs such that the temperature remains constant. So, here you can see that both temperatures are 75 initially also. Finally, also I will say the temperature remains at 75 degrees in both the chambers. Now, A, what are the final states of water and air? Because state will change now as the piston moves. Determine the sign and magnitude of total heat transfer. So, this is what is asked. So, now let us first understand the states, initial state and find the masses of air and water vapor solution. Mass of the air in A, I will say M A will be equal to what pressure is given. So, 200 kilo Pascal. So, 200 into 10 power 3 Pascal into volume is 0.01 divided by R T. R is 287 given here, 287 into T is 273 plus 75 because you have to use Kelvin. So, that will give you the mass as 0.02 kg. Okay. Now, state in the B. What is this? So, we can say that 75 degrees centigrade X equal to 1. So, what is V? V G. So, I will say V B1 will be equal to V G at 75 degrees centigrade, correct? The specific volume in the B initially. So, that we have to go to the tables. Go to the temperature based, this is pressure based. So, temperature based saturation table and 75 degrees is the temperature we are looking for. What is the value of V G? So, V G is 4.131, 4.131 meter cube per Ag. That is it. And total volume is given as 0.3. So, what is mass of the saturation? Saturated the water vapor in B. That will be V divided by V B, I will say V B1 actually, this is V B1 divided by small V B1, which is equal to 0.3 divided by 4.131, which is equal to 0.7262 kgs. So, we have to note P B also. What is the pressure in the chamber B? Initially, that we can take as the saturation, because it is saturated vapor. So, the pressure should be the saturation pressure at the given temperature of 75 degrees. That is 0.3858. Similarly, I can also take the value of U G for the U. So, 2, 4, 7, 6. So, substituting here, P B will be 0.3858 bar or 38.58 kilo Pascals. And U B1 equal to U G at 75 degrees centigrade equal to 2, 4, 7, 6 kilo joules per kg. So, these are the data from the steam table which you have taken. So, state 1 is clearly fixed, mass is formed, state 2 also is clearly fixed for this. Again, mass of the steam and internal energy etc. has been calculated. Now, temperature remains constant during the process. What happens is the piston moves to the rise right. So, let us see this. Let us for the B we will draw this. Again, we will take a here, we are going to take a P V diagram basically, P V diagram. Now, this is the isotherm of 75 degrees centigrade. Now, we have a state 1. This is for B. State 1 is this. So, this is 38.58 kilo Pascals. That is the saturation pressure at 75 degrees. So, now we have state 1 as the V equal to V G at 75 degrees. That is saturated vapor. This is the critical point. And this is the saturated vapor line and this is saturated liquid line. So, this is the isotherm we have. So, now you see this is the condition at state 1. Now, as the piston moves towards right, what happens it will try to compress, correct. It will try to compress the saturated vapor. When it is compressed and the temperature remains constant. So, that means the this isotherm will be followed. So, it will be compressed and it will go in this line basically. That means pressure will not vary now for B. Finally, when the what happens is the pressure in the A go back. Pressure in the A initially is 200 kilo Pascals. So, as the piston moves towards the right, pressure in the B will not change because it is saturated condition. That is the condensation occurring. That is from the saturated water vapor. The condensation occurs slowly, liquid is formed and pressure and temperature both remain constant due to the phase change. Here, due to the movement of the piston towards right in A, the pressure of the air will decrease from 200 kilo Pascals. And when it reaches the pressure of 38.58 kilo Pascals, piston will not move anymore because afterwards the pressure in the both the chambers will be the same. Because already the pressure in chamber B will be 38.58 till what point? Till it becomes saturated liquid. Whether it reaches this state or I do not know, but you can see that the pressure will remain constant until all the vapor is converted into liquid. So, but between this state itself somewhere you can see that the air pressure will decrease from 200 kilo Pascals to this 38.58 kilo Pascals. Then after that piston will not move. Do you understand? So, keep that in mind. So, we can say that since T remains constant when saturated water vapor is compressed slowly compressed condensers. See this diagram ok condenser and the pressure and temperature remain constant until it becomes saturated liquid. So, we will find a case when it has become after it becomes saturated liquid then further compression occurs. Then it can the pressure can rise. But what happens is within this time if the chamber air pressure in the chamber A reduces to 38.58 the piston will not be able to compress this more. So, keeping that in mind we will find that the P B 2 will be equal to the for B P B 1 equal to 38.58 kilo Pascals. Do you understand? So, that is the thing. So, now we can say P 2. P 2 all common pressure will be this. So, now what will be? So, once you calculate P A 2 also will be equal to P 2 equal to 38.58 kilo Pascals. After this it cannot move. So, that means that what will be the volume in A now volume in A will be equal to use the P V equal to M R T. So, volume in A will be equal to mass. Mass is 0.02 into 287 into temperature remains constant divided by 38.58 into 10 power 3. So, that will be equal to 0.052 meter cube. So, that will be the volume. Initial volume was basically initial volume was given. So, you know the initial volume of this is V A 2 sorry V A 2. So, initial volume was 0.01. So, we can say that V A 1 plus V A 2 will be equal to V B 1 equal to V A 2 plus V B 2. Ok. So, why? Because once the piston moves to the towards the right initially volume in A plus volume in B should be equal to final volume in A and final volume in B added. So, this is the condition. So, we know V A 1 that is 0.01, 0.01 and this is 0.3 equal to now V A 2 is known 0.052 plus V B 2. So, applying this condition I can find V B 2 equal to 0.31 minus 0.052 which is equal to 0.258 meter cube and V B 2 will be equal to what? 0.258 by mass that is meter cube divided by mass in B that is what go back mass in B is 0.07262, 0.07262 kg which is equal to 3.552 meter cube per kg. So, please see that this is the process. So, slowly the from this point slowly the condensation occurs and the pressure is 38.58. It remains constant till all the vapor in the isothermal process, compression process turns into liquid, saturated liquid. Now, then pressure should increase drastically. But what happens is by that time the A, pressure in A also drops to 38.58 from 200. That means, the final pressure should be 38.58. Now, from that I calculated V A 2, final volume in the air chamber that is mass into R into T divided by P. From that I calculated this. Then the volume conservation, initial volume in both chambers should be equal to final volume in both chambers. Using that I calculated the final volume in the steam chamber. From that the specific volume is calculated because specific volume should be used to fix the state of the steam. So, now what is the state? Now, V B 2 is known, final pressure. So, here V B 2 is known. Now, in the same line we are somewhere we will be there. So, what is, we know from the previous that V G 1 is 4.131. V F we will find from the tables. So, here for 75 we can say V F equal to 1.026 into 10 to the power minus 3. Similarly, U F equal to 319.9. We can take these values. So, we can say X 2 for the B will be equal to 3.552 minus 1.026 divided by, this is 4.131. So, what is X 2? X 2 will be 0.86. 0.86. That is it. So, the, now the state 2 will be somewhere here. This is state 2. So, this has happened basically from 1 to 2 it has moved. So, the pressure remains constant. This is, within the time the condensation from saturated vapor to a quality of 0.8 is occurred, pressure has dropped to from the gas, the air chamber from 200 kilopascals to 38.58. So, that is the final state. So, now, what is U 2? U 2 we can calculate as U F plus X 2 into U G minus U F. So, these are, U F and U G can be taken from 75 degrees line. So, that will be equal to, go back to the tables, 313.9, 313.9 plus 0.86 into, this is 2476, 2476 minus 313.9. So, that will be equal to U 2 will be equal to 2173.3 kilojoules per kg. That is it. So, now, we can apply the first law. Now, what is the work involved in the chamber A? Work involved in chamber A will be equal to P A 1 V A 1 into natural logarithm of V A 2 by V A 1. Why? Because isothermal expansion P V equal to constant. Okay, now, applying the values here, it will be equal to 3.297 kilojoules. So, what is work done as the air expands will be absorbed by the steam. So, that will be equal to minus W B. In the steam chamber, it is compressed. That means, as the air expands, the steam is compressed. So, W A plus W B equal to 0 or W A equal to minus W B. So, that we have for A delta U A equal to M C V T A 2 minus T A 1 equal to 0, since T A 2 equal to T A 1. So, that is 0. So, that means, Q for A will be equal to W for A. That is it. So, it is equal to 3.297 kilojoules. So, that is equation one for us. Now, for B, I can say Q B minus W B equal to delta U B equal to M B into U B 2 minus U B 1. So, now, Q B equal to W B equal to minus 3.297, ok, plus M B. M B is 0.07262, 0.07262 into U B 2. U B 2 is this, 2173.3, 2173.3 minus U B 1. U B 1 will be equal to 2476. 2476. So, that will be equal to how much? Minus 25.28 kilojoules, that is Q B. This is 2. So, total heat transfer, Q total equal to Q A plus Q B equal to minus 22 kilojoules. So, this problem, please understand that nothing is insulated. So, the heat transfer can occur from both the chambers to the surroundings plus wind between air and steam chamber also, it can occur because this piston is thermally conducting, ok. So, this is the problem number 7.