 problem number 8. 1 kg of air initially at 4 bar 550 Kelvin, 1 kg of air is in the, again a similar type of cylinder where the movable piston is there. Frictionless piston initially held by pin is also present. So, one side is air at 4 bar 550. The second side is 1 kg of wet steam. Wet steam means it may be a mixture of liquid and vapor at 1 bar is there. That is it. The frictionless piston that separates air and steam is thermally conducting like in the previous problem and is also free to move. The steam initially occupies 1 meter cube. So, we can write for this, ok. So, A side for A. So, this is I will call this side as A and this side as B like in the previous problem. For A, P A1 equal to 4 bar, T A1 equal to 550 Kelvin, M A, M A equal to 1 kg. Ok, that is all. Here, for B, M B equal to 1 kg, 1 bar pressure is given. So, P B1 equal to 1 bar and also is given. The steam initially occupies 1 meter cube on the right side. That means B B1 equal to 1 meter cube. That is it. So, these are the initial conditions given. The piston is now allowed to move by removing the pin. The piston moves slowly, ok. See, there is a pressure difference between A and B. A at 4 bar, B at 1 bar. So, that the piston actually moves to the right. So, A undergoes a polytropic process P v power n equal to constant as the piston moves. Ok, now the piston cannot go to equilibrium. It has to move under, there is a stops here, stops. So, piston finally comes to the stop due to the stopper ring which is present here. The, you can understand that it cannot move. That means that still the pressure between these two chambers can be different. In the previous problem, there was no stop present. So, the piston will go and rest at a particular point where the pressure in both chambers will be same. Now, here it is not so. It cannot, need not be same. Ok, now what is given is, when the piston goes to this position in this stop, like this, ok, what happens? The wet steam becomes saturated steam at the pressure of 2 bar. That is P B 2 equal to 2 bar and X 2 equal to 1. At this state, temperature of air is equal to that of the steam. That means, I will say T B 2 here, T B, sorry, T A 2 will be equal to T B 2. I do not know this, we can calculate this. But what is this? This is nothing but T saturated, saturation temperature at 2 bar. So, that will be here, equal to T sat at 2 bar. So, these are the information given to us. Why this is possible? Because, basically, there is a heat transfer between A and B, so due to which the temperature finally becomes equilibrium, but pressure need not be at equilibrium. So, determine the initial dinos fraction of this steam, given these three. So, from this I can calculate V B 1 also, as 1 by 1 equal to 1 meter cube per kg. Now, the state is fixed at 1 bar pressure, specific volume is 1 meter cube per kg. From that I can fix and do it. So, initial dinos fraction, final temperature of air, that is already found, final temperature of air will be equal to T sat at 2 bar, ok. So, it is given, the condition is given like this. So, V B 2 will be T sat at 2 bar, so that will be the final temperature. And third is the value of the polytropic index, that is P V equal to P power n equal to constant, what is the value of n? Then the work transfer for the air and heat transfer during the process, ok. So, these are the things asked. Solution for A, side A, I can find volume, initial volume as 1, 1 kg into 287 into 550 Kelvin divided by 4 bar equal to 0.394625 meter cube, ok. Now, for B, we have already seen that P B 1 equal to 1 bar, V B 1 equal to 1 meter cube per kg, that is going to fix the state. So, now, what you do is, at 1 bar, you go to this table here, pressure based table, 1 bar pressure. So, this is the one, 1 bar pressure. So, what is the value of V F 0.001043 and V G is 1.694. Now, we can see that V 1 was 1 meter cube per kg, that means it is a saturated mixture. So, we can note this. Similarly, I can note the U also, 417.3 and 2506. So, go here, U first V F equal to 0.001043 meter cube per kg and V G equal to 1.694 meter cube per kg and we can find X 1 as V B 1 minus V F divided by V G minus V F which is equal to 1 minus 0.001043 divided by 1.694 minus 0.001043 equal to 0.59. So, initial quality is 0.59. Then, we can find U 1 equal to U F plus X 1 into U G minus U F which is equal to U F is 400 17.3 plus 0.59 into 2 506 is the U G value minus 400 17.3. So, that will be equal to 2 sorry 1 6 4 9 0.63 kilojoule per kg. That is the initial state is 6 for the steam. Okay now, at state 2 for B, P B 2 equal to 2 bar and it is given that X equal to 1, that is X 2 equal to 1. That means V equal to what? V B 2 will say V B 2 equal to 0.886 for 2 bar which is what? This is actually I will say this is V G at 2 bar which is equal to plus X equal to 1 0.886 meter cube per kg that will be V B 2. So, what is V capital V B 2 that will be equal to M in M B into V B 2 it is 1 kg. So, it is again same 0.886 meter cube that is the value. Now, apply this V A 1 plus V B 1 will be equal to V A 2 plus V B 2 that should be obeyed. So, implies V A 2 minus V A 1 will be equal to V B 1 minus V B 2 equal to 1 minus 0.886 equal to 0.114 meter cube that means V B 2 can be found correct. So, V A 1 is known so, you can find V B 2 sorry V A 2 you can find V A 2 as V A 1 is known here 0.394625 plus 0.114 equal to 0.508625 meter cube that is it. So, now, you can find the pressure. So, before that I know what is T A 2 equal to T B 2 equal to T sat at 2 bar pressure. So, go to the stream tables. So, here 2 bar what is saturation temperature 120.2. So, equal to 120.2 degrees centigrade. So, R converted into Kelvin 393.2 Kelvin. So, that is this. Now, find P ok P will be equal to because the pressure can be different. Now, we cannot say pressure in the left hand side it will be 2 bar now. So, P A 2 will be equal to 1 mass into 287 R into final temperature 393.2 divided by the final volume 0.508625 which is equal to 221.86987 you can say 87 kilopascals. So, that is so, you can see that now the still there is a pressure difference between A and B. A is at 221.87 B will be at 200 kilopascals because of the stops. Pressure difference can be there, but temperature is different sorry same in both. So, now P V power n equal to constant is given. So, apply that. So, we can also say here P 1 for A for R. So, now we can use that P 1 P A 1 ok V A 1 power n equal to P A 2 V A 2 power n substituting 400 initial, initial volume is 0.394625 power n equal to final is 221.87 kilopascals into 0.508625 power n. So, from this I can find the value of n equal to 2.3224. So, what is the work done by the air is equal to P A 1 V A 1 minus P A 2 V A 2 divided by n minus 1 which is equal to 34.03 kilojoules delta U of air will be equal to m A into C V into T A 2 minus T A 1 which is equal to 1 into 0.7175 V C V value into this mass ok and here this will be 1 that is 393.2 Kelvin minus initial temperature was higher. So, this is negative minus 156.8 kilojoules where this is in kilojoule per kg Kelvin ok. Now, what is Q of air? Q of air will be just add this W of air plus delta U air so which is equal to minus 78.474 kilojoules done. Now, for steam work involved in the steam work then on the steam is by the air. So, I can say W of air minus of that. So, that will be equal to minus 34.03 kilojoules. So, Q of steam will be equal to W steam plus m B into U B 2 minus U B 1 which is equal to 846.54 kilojoules. So, that is the value. So, in this problem basically the pressure is not same in the chamber because there is a stop after the position moves due to the pressure difference towards the right compressing the steam after a particular value after the particular movement it actually stopped by the stops. So, that after that it cannot move. So, pressure difference prevails between these two, but temperature is same that is given in the problem. That is understandable because this piston is thermally conducting. Okay. Now, the next problem we have frictionless thermally conducting piston separates air and water. Again this is like the same. So, at the horizontal cylinder two compartments, piston is there air is in the left chamber, water in the right chamber. Okay. Initial volumes of A and B that is V A 1 equal to V B 1 equal to 0.5 meter cube and the initial pressure in B that is P B 1 equal to 700 kilo Pascal's. So, please see there is no pin and doll here. Okay, that we have to understand here. There is no pin and doll. So, obviously what happens if P B 1 is 700 kilo Pascal's, P A 1 also should be same pressure. So, I can easily say this is there. Because otherwise at every instant the pressure in both the chambers should be same because when the piston is in equilibrium. So, a state one where the equilibrium condition at state one when the pressure for the B is 700 kilo Pascal that should be the pressure for the A. Okay. Now mass of the liquid in B, mass of liquid in B equal to 0.9 kg. So, basically it has water. That is water means it may be in any phase, liquid and vapor in this case. Okay. So, now specifically the liquid mass is given as 0.9 kg. This is not a total mass. Okay. So, M B can be written as M of the liquid in B plus M of vapor in B. Now, this is only given of the total mass. Now heat is transferred to both air and water. Heat is transferred to both air and water until the final pressure and temperature are 2 mega Pascal's and 700 degree that is P A2 equal to P B2 equal to 2 mega Pascal's equal to 2 into 10 power 6 Pascal's and T A2 equal to T B2 equal to 700 degrees centigrade. This is given. So, now when this is in equilibrium, what will be the condition? So, what is to be determined? Mass in A, mass in B, total heat transfer during the process. These are the three things asked. Solution. So, since the thermally conducting piston is there, okay, we can say that T A1 will be equal to T B1. Finally, also you can see that T A, T A2 equal to T B2 and pressure also will be at every time instant will be same. Say, initially 700 degree Pascal's, finally 2 mega Pascal's. So, this is the thing we have to understand here in this case. The mass of the liquid in B is initially given here. So, we have to find this mass also or total mass then find V B. So, these are the things. So, first of all, let us take A. So, for A, pressure is 700 degree Pascal's. What is M A? M A will be equal to 700 into 10 power 3 into volume is given, 0.5 PV by M. I do not know, M I have taken here, R 287. For air you can take R equal to 287 given in the problem into temperature. Temperature is what, T A1 equal to T B1. So, what is T A1? So, we have to understand this. 700 kilo Pascal's. Okay. So, now first you have to find, go to the tables for the 700 kilo Pascal's here. That is here, 7 bar. This, we will find the values of V F and V G. V F equal to 0.001108 and this is 0.273 V G. So, at 700 kilo Pascal's from saturated tables, I can say V F equal to 0.0011, I will see it again, 1108, 1108 meter cube per kg and V G will be equal to 0.273, 0.273 meter cube per kg. So, now the state is actually what, state is saturated. So, what will be T B1? Saturated state. So, T B1 will be equal to here, 165. That is T sat at 700 kilo Pascal's, 165 degree centigrade. Why? It is clearly given that the mass of the liquid is given. That means, some mass of vapor should also be there. So, that means, there is a saturated mixture of liquid and vapor. So, temperature should be at this saturation temperature at the given pressure, 165. So, here I can fix this at 165 and here I substitute it, convert it into Kelvin. So, that will be the mass of A. What is mass of A now? 2.7843 kg. Mass of A is calculated, first one. Mass in B you have to do. So, here I will continue with the first A, P A2 is 2 mega Pascal's. T A2 is also given, 700 degree centigrade, 700 degree centigrade equal to 973 Kelvin. So, now I can find V A2. V A2 will be equal to what? V A2 equal to, what is the mass we have calculated? 2.7843 into 287 into 973 divided by 2 into 10 power 6, which is equal to 0.38875 meter cube. Okay. Now, why I am not calculating mass in B? Because I cannot. Why? You can see that I know that this is a saturated mixture because specifically mass of liquid in B initially is given. That means, mass of liquid is given, then there will be mass of vapor. So, I can only estimate that there is a mixture of liquid and vapor. So, I can say it is at saturated mixture condition. So, the temperature should be 165 because that is the saturation temperature at 700 kilo Pascal, that I know. But other than that, there is no information for me to fix the state here. Let us, I do not know what is, see, temperature and pressure at, for a saturated mixture will not be able to fix the state because I need quality or any other property, but it is not there. So, I cannot proceed with this. But since the final state is given for air, I can at least find the final volume. Okay. Now, from this, I can find the final volume of the steam. Okay. That is what I am trying to do here. What is that? Again, the VA1 plus VB1 equal to VA2 plus VB2. The volume in both chambers initially and finally should be the same. So, VA1 is 0.5 plus this is also 0.5 and this I have calculated now 38875 plus VB2 implies VB2 will be equal to 0.61125 meter cube. So, now state 2 for steam now we will take for steam at state 2 P equal to PB2 equal to 2 mega pascals and TB2 equal to 700 degree centigrade VB2 I know VB2 equal to 0.66 or 61125 meter cube. So, these three I know. So, now I will go to the tables 2 mega pascals. What is 2 mega pascals? 2 into 10 power 6 or 20 into 10 power 5 that is 20 bar. This is 2 mega pascals 20 bar. So, this is given in bar. Okay. So, this table is given in bar. So, 20 bar table I find what is the final thing. Temperature is 2 for temperature at 20 bar saturation temperature is 212.4. But what is the temperature given? That is 700 degrees centigrade. So, the state should be superheated. So, since T sat at 2 mega pascals is less than TB2 the state 2 is saturated superheated superheated vapor. So, I have to go to the superheated vapor tables for 20 bar. Go down superheated tables, go to the pressure of 20 bar now here 16 at 20 bar. 20 bar temperature is also given. 20 bar temperature is this. So, now I can take the value of V as 0.223 and U as 3471. So, I can say from superheated tables for 20 bar or 2 mega pascals U equal to, sorry V equal to first I will say 0.223 meter cube per kg and U equal to 3471 that is U2 I will say U2 here 3471 kilo joules per kg this you can retrieve. Now I can find the mass in B because I know both total volume and specific volume. So, I will say VB2 here also I can say UB2. So, now what is this mass will be capital VB2 total volume divided by the specific volume which is equal to 0.61125 divided by 0.223. So, that will be the mass because in the system here system final mass finally or initially mass will be the same. So, finding the mass at the final state because I can easily fix the final state now I have found the mass using the final state that will be 2.741 kg. So, now I know the mass from that I can find the initial. So, what is asked is mass in B now I have found during this then I can find the. So, initial state I need to get the heat transfer correct. For initial state I have to get the quality of the initial state for that I have to fix the state itself. So, I will do that now. So, for V, state one, state one is what 700 kilopascals and volume, capital volume is that is capital V is 0.5 and mass is 2.741 kg. So, now I can find the initial specific volume as 0.5 divided by 2.741, initial specific volume that will be equal to okay I will calculate that this is what the initial value. So, from this I can find the x1, x1 is what here for 700 kilopascals Vf is 0.001108 meter cube by kg Vg is 0.273. So, I can say x1 will be equal to Vb1 minus 0.001108 divided by here Vg equal to 0.273. So, 0.273 minus 0.001108. So, that will give you the value of x1, okay. Now, please see that partially I know what is the mass of the liquid in the B at state one that is given as 0.9 kg. Total mass is 2.741. So, what is the mass of vapor? So, I will say mass of vapor in the B initially will be equal to 2.741 minus 0.9. So, that will be equal to 1.841 kgs, okay. Now, what is quality? State away I can find the quality and you not find this. I can also do like this, but state away quality will be equal to m Vb1 divided by mb. This will be equal to this. So, finally I will get quality as 0.67.65. From this I can find the state one initial specific internal energy Ub1 has 696.4, this is Uf plus x1 0.67165 into 2573 minus 696.4. This is Ug, this is Uf, this is x1. So, that will be equal to 1956.82 kilojoules per kg. So, this is the value. Now, apply the law. So, easy to take A plus B as the system. That means W will be equal to 0 since delta V equal to 0, delta V. Okay. Now, I can find if I have a first law is W0 equal to delta U, delta U is written as delta U in A plus delta U in B, correct. So, now this is equal to for A it is ma Cv ta2 minus ta1 plus mB into, this is Uub2 minus Ub1. Okay. Now, what is this? 2.7843 into Cv is 0.7175 into ta2 is 973 minus 438, 165, 165 degrees centigrade plus mB. mB is now calculated as 2.741 into Ub2. What is Ub2 in the previous slide? Ub2 is 3471, 3471 minus, here Ub1 you have now calculated as 1956.82. So, this will give you the value of Q, total Q as 5219.156 kilojoules. This is the problem. So, it is an interesting problem where the thermal equilibrium prevails at every state. Pressure also is same because of the piston being free to move as well as thermally conducting. So, with this you can solve the problem. The only main thing is initial state of the water cannot be fixed because it is a saturated mixture. The mass of the liquid alone is given. So, the final state from the final state mass is gone and from the mass initial state is fixed. So, that is the important problem.