 Good morning. So, let us continue with the tutorial number 8 on entropy, the sixth problem. Here 1 kg saturated vapor of R134A is contained in an insulated frictionless piston cylinder device at a pressure of 300 kilopascals. It is slowly and reversibly compressed to a pressure of 700 kilopascals. So, first process is slowly and reversibly compressed. Now, during this process basically it is insulated. Then one of the insulation is removed and it is gradually cooled at a constant pressure to 20 degree centigrade. Determine the work and heat interactions and entropy change. The property tables for R134A is also given. The first table is the saturation table and the second one is the superheated table at 700 kilopascals. So, mass is 1 kg, initial pressure is 300 kilopascals and it is given that saturated vapor that means, x 1 equal to 1. So, that means, you can go to tables because it is saturated state. So, you can go to saturation table and corresponding to pressure of 300 kilopascals, we can take the values of V1 as Vg at 300 kilopascals which is equal to 0.06778 meter cube per kg. Similarly, U1 will be Ug at 300 kilopascals which is equal to 230.55 kilojoule per kg. Now, it is slowly and reversibly compressed and it is adiabatic. So, reversibly means, so, reversible and adiabatic process is obviously, an isentropic process. So, here delta S will be equal to S2 minus S1 will be equal to 0. So, we will also take the value of entropy at state 1 which is Sg at 300 kilopascals. So, from the table, it is 0.9312 kilojoule per kg Kelvin. Now, state 2, state 2 is S2 equal to S1, one of the property because reversibly and adiabatically compressed. So, that is basically isentropic process. So, S2 will be equal to S1 0.9312 kilojoule per kg Kelvin and the pressure is given. It is compressed to a pressure of 700 kilopascals. So, that is P2 equal to 700 kilopascals. So, now the state 2, two properties for the state 2 is known. One is the pressure, another one is the entropy. So, now see here. First we will go to the saturation table. For corresponding to 700 kilopascals pressure here, you see the entropy value. Entropy value Sg is 0.9201, but the entropy value at the state 2 is equal to state 1 which is 0.9312, 0.9312. So, we can find that S2 is greater than Sg at 700 kilopascals. You understand, no? So, 700 kilopascals, Sg is 0.9201, S2 is 0.9312 is greater than that. So, the state 2 should be superheated. So, now go to the superheated tables for 700 kilopascals. I find the entropy value of 0.9314. In fact, I have given the tables here where the value is closer to what we get in the state 2 entropy. So, we can just take this. There is not much difference. It is only 0.0002. So, we can take this as the closest entry in the superheated tables for 700 kilopascals. So, we can say to a fixed V2 as 0.03 meter cube per kg and U2 as 247.5 kilojoule per kg. So, all the properties for state 1 to 2, 1 and 2 are fixed. Then R134A is cooled at constant pressure that is 700 kilopascals to 20 degrees centigrade. So, what is state 3? It is fixed. State 3 is P3 equal to 700 kilopascals and T3 equal to 20 degrees centigrade. Okay, go back. You can see this 700 kilopascals. The temperature in the second column is nothing but the saturation temperature for the given pressure. That means the saturation temperature for a pressure of 700 kilopascals is equal to 26.7 degrees centigrade. But for the state 3, the temperature is 20 degrees centigrade. Okay, so this means the state 3 is subcooled liquid. Okay, since T3 is less than T sat at 700 kilopascals, that is 26.7 degrees centigrade, this state is subcooled liquid state, subcooled liquid. So, we know that for subcooled liquid state, we should use temperature-based saturation tables at 20 degrees centigrade. So, go back here. Here 20 degree table is there, saturation table. Here you can see that the value of VF and UF you have to take and obviously SF also you have to take. So, here VF for 20 degree is 0.816 into 10 power minus 3. So, V3 equal to VF at 20 degrees centigrade equal to 0.816 into 10 power minus 3 meter cube per kg. Then U3 equal to UF at 20 degrees centigrade which is equal to 78.85 kilo joule per kg. Similarly, S3 equal to SF at 20 degrees centigrade which is equal to 0.3006 kilo joule per kg Kelvin. So, properties, liquid properties at all the three states are now taken. Okay, now for process 1 to 2, reversibly adiabatic process. So, Q1 to 2 equal to 0, first law Q1 to 2 minus W1 to 2 equal to U2 minus U1. So, this is 0. So, we can say W1 to 2 equal to U1 minus U2 equal to M into U1 minus specific energy change. So, that will be equal to here 1 kg. So, mass is 1 kg and U1 is 230.55 U2 here U2 is 247.5. So, 1 into U1 is 230.55 minus 247.5. So, this will be equal to minus 16.95 kilo joules. So, that is the work interaction for process 1 to 2. Then for process 2 to 3 where there is a constant pressure cooling, correct. So, here Q2 to 3 minus W2 to 3 equal to U3 minus U2. Of course, we are neglecting the changes in the potential and kinetic energy in both processes. So, now this is the thing. So, here what is W2 to 3? It is a constant pressure cooling that means 700 into M into 700 into V3 minus V2. So, that is 1 into 700. 700 is the pressure P2 equal to P3. So, that 700 into what is V3? V3 is here 0.816 into 10 power minus 3 minus V2 is 0.03. So, this will be equal to minus 20.43 kilo joules. So, that means Q2 to 3 will be equal to minus 20.43 plus M into U3 minus U2. So, substituting the values, I get Q2 to 3 as 189.08 kilo joules. Now entropy change. Entropy change for the total state 1 to state 3 will do. So, entropy change delta S equal to S sorry S M into S3 minus S1 is equal to 1 into S3 is 0.3006 minus S1 is 0.9312 equal to minus 0.6306 kilojoule per K Kelvin Kg Kelvin. So, this will be Kelvin actually. So, now you see this here the first process is an isentropic process and second process is a cooling process. So, entropy decreases. So, that is fine. So, this is about the problem. Here first one is an isentropic process clearly described as an insulated container where there is a slow and reversible compression happening. That is the first process. The second process there is a heat rejection. So, enthalpy sorry the entropy decreases because of the heat transfer out of the system. Okay seventh problem an insulated cylinder as shown in the figure is divided into two compartments each of volume 0.025 meter cube. So, let us say the left side is A and compartment B is the right side compartment. The left chamber is filled with air at 100 kilo Pascal and 25 degree centigrade. So, we can note the volumes first. So, VA1 is 0.25 meter cube equal to VB1. Okay now PA1 equal to 100 kilo Pascal under TA1 equal to 25 degree centigrade. Right chamber has water vapor. The left chamber is connected to a valve. So, the left chamber is connected to a valve to a line in which air is flowing at a pressure of 600 kilo Pascal under 25 degrees centigrade. The right chamber contains saturated water vapor that means XB1 equal to 1. The valve is opened allowing the air to flow from the air line to the left chamber until pressure reaches the line pressure that means PA2 equal to 600 kilo Pascal. At this point the valve is closed once the pressure reaches 600 kilo Pascal in A valve is closed for air R equal to 287 and gamma equal to 1.4. Determine the mass of the air initially in the left chamber, the mass of steam in the right chamber, final temperature in each chamber and mass that enters in the left chamber due to the opening of the valve. Okay, so this is the problem here. So, let us do this P for air we can say P V equal to M into 287 into T, T in Kelvin. So, we can apply for the state 1 that is 100 kilo Pascal. So, 100 into 10 power 3 Pascal into volume is 0.025 equal to mass, mass you have to find. So, I will say mass in A into 287 into temperature is 25 degrees that is 273 plus 25. So, from this I can get the mass of the air in the chamber A initially. So, I will say MA1 because some mass is coming in also correct when the valve is open. So, it is MA1 that will be equal to 0.0292 kg. Okay, that is the initial mass in the chamber A. Okay, now for steam, what should be the pressure of the steam? It is given as saturated water vapor. So, what should be the pressure of the steam? The piston is free to move. Okay, so, insulated frictionless piston is free to move. So, if the steam's pressure is different than the pressure of the air chamber, then the piston will not be in equilibrium. So, that means Pb1 should be equal to Pa1 equal to 100 kilo Pascal. So, now we are looking for a state where the pressure is 100 kilo Pascal and the quality is saturated vapor. Okay, so, go to steam tables. From steam tables you can get the values. So, you will see that at 100 kilo Pascal Vb1, the specific volume of the steam in the chamber B is equal to Vg at 100 kilo Pascal. So, that is we can take from the tables. So, Vb1 will be equal to 1.694 meter cube per kg. This is the Vg at 100 kilo Pascal. So, now we can find the mass of the steam which is equal to Vb divided by Vb1 which is equal to 0.025. This is the total volume meter cube and this will be meter cube per kg. Okay, now that will be 0.025 divided by 1.694 which is equal to 0.01475 kg. So, we can easily see that for the piston, see it is insulated frictionless piston. It is insulated. So, there may be no heat transfer. Basically the cylinder itself is insulated. So, for steam or air that will be not be any heat transfer basically. Between air and steam also there will be no heat transfer that is fine, but the pressure has to be same in both the chambers at the equilibrium point. Okay, so for example, if there is a pressure gradient, okay from air pressure is higher than steam pressure then the piston will move to the right or else it will move to the left. So, as the pressure is same then only the piston will stay at a particular state that is equilibrium state where the properties are defined. So, in this case specifically that is the concept we have used and so this is nothing but what Vg at 100 kilopascals. So, from that I have got the mass. Okay, now what is the process happening in steam? In the steam as the valve is open the air passes to the chamber A. Now this makes the pressure to increase in chamber A due to which the piston moves to the right. As the piston moves to the right, so this air comes in the piston moves to the right, the steam undergoes an compression. It is slow process when the valve is slowly open slowly air enters and the piston is frictionless. That means there is nothing which will cause an irreversibility in the process. So, smooth transition of piston takes place as it goes from the initial point towards the right side compressing the steam. So, piston as it moves to the right it actually causes an adiabatic reversible compression for the steam. Okay, so we can note that steam undergoes reversible that is due to the frictionless piston adiabatic compression. Okay, so adiabatic because the cylinder itself is insulated and also the piston is insulated. So, due to which completely chamber B is adiabatic and slow compression takes place with because of the frictionless piston it is reversible also. So, the process is therefore isentropic that is s2 will be equal to s1. Okay, so again from steam tables s1 will be equal to sg at 100 kilopascals. So, what is that? That will be equal to 7.359 kilo joule per kg Kelvin. Now, this will be equal to s2. Okay, what is the final pressure? Final pressure for chamber A is 600 kilopascals. Okay, now as the air enters the chamber A slowly the piston goes towards the right. Once the pressure in the chamber A reaches 600 kilopascals obviously the pressure in the steam chamber will also be 600 kilopascals. Temperature may be different in both. Please understand that. So, pressure has to be the mechanical equilibrium is attained. So, we can say p2 will be equal to 600 kilopascals. Okay, now the state 2 for the steam is fixed. State 2 is what? p2 equal to 600 kilopascals and s2 equal to 7.359 kilo joule per kg Kelvin. So, now two properties are enough to fix the state. Correct. So, what is this state? So, first we will see that go to saturation tables and see for 600 kilopascals. What is the value of sg? So, we find that sg at 600 kilopascals is basically less than s2. So, that means what? s2 is already greater than sg. So, the state is the state 2 is superheated vapor. So, you have to go to superheated from superheated vapor table for 600 kilopascals. Okay, we know that for superheated state lot of tables are there corresponding to a given pressure. So, go to the pressure corresponding to 600 kilopascals by interpolation. Now, the second property was entropy being 7.359. So, we can use this as the primary column and interpolate. For other by interpolating, we will get the other properties, required properties. So, Ts2 is found as 296.5 degrees centigrade and Vs, sorry V2, C2 and V2. V2 equal to 0.4312 meter cube per kg. So, we can find one of the properties. U2 also you can find. Okay, U2 also you can find. So, now this is the state 2 fixed for this. Once the specific volume is known, mass of the steam is known to me already. So, what is the final volume of the steam? That is, I will say V, V2. What is the final volume of the chamber B? That is the volume, total volume of steam is equal to 0.4312, that is V2 into MS, that is 0.01475. So, which is equal to 0.00636 meter cube VB2. Okay, now please see that initially VB1 was 0.025. Now compression has happened. So, volume is decreased to 0.00636. That means the volume of A will increase. So, we have this. Initial volume of A plus initial volume of B should be equal to initial final volume of A plus final volume of B, because in the same chamber total volume should be conserved. So, you know this, this is 0.025 plus 0.025 equal to, now VA2 should be calculated from this equation. So, VB2 we know the value. So, which implies VA2, the total volume of A in the final stage is what? That will be 0.04364 meter cube. So, you can see that the volume has increased from 0.025 to 0.043. So, now state 2 for A is also fixed, because you know volume pressure is 600 kilo bus is given. So, you know these two. From that we can calculate the required values. Planning for example, temperature if you want we can use the equation of state and calculate temperature. So, before that we will see what is happening for air. There is a control volume. There are two things happening basically. Here there is a mass transfer from the air line to the chamber A, but volume also is increasing for A. So, this is a hybrid control volume. In system mass remains constant, there will be no mass transfer, volume can change. In a control volume mass can come in or mass can go out, but volume will not change. In this particular case for A you will see both are changing that is mass is coming in. So, mass of the air initially was found to be 0.0292 kg. Now it is increasing because of the incoming mass from the air line plus the volume also is changing. So, this is called an hybrid control volume in which both mass and volume changes. So, we have to write the equation for this. So, first we will finish this theme which is easier. First law is written as q minus w equal to delta u obviously, insulated and it is an isentropic process. So, q is equal to 0. So, we can say minus w equal to delta u. Delta u is what? m into u2 minus u1. So, from the super editables u2 can be got as 22795.4 kilo joule per kg. u1 is previously got here from this initial state itself. Ok. Here yes we have said that this state is fixed basically. So, from that u1 will be equal to ug at 100 kilo Pascal's 100 kilo Pascal's. So, from that we can get 2506 kilo joule per kg for u1. So, you know both the values now. So, I can say this is mass is 0.1 sorry 0.01475 into u2 2 this is by interpolation ok 2795.4 minus u1 is 2506. So, once you fix the state with two properties the other properties can be easily got. So, this will be equal to 4.268 kilo joules that means, w equal to minus delta u equal to minus 4.268 kilo joules. So, the work involved work involved in compressing the steam is minus that is what it is compression. So, negative steam for the steam it is negative work 4.268 kilo joules. Now, what is the I will say work for steam specifically. So, what is work for air which is compressing steam air is compressing steam. So, this work for the steam has come from the air chamber ok. So, that will that is I can write the only environment for the steam is air or for the air is steam basically. So, in case of work transfer. So, I can say w r equal to minus w s which is equal to 4.268 kilo joules. So, the work involved in this steam this is due to the volume change of the air correct. So, that is in got it we have got from the applying the first law for the steam chamber ok. Now, let us proceed and write the first law for air for air q minus say let us integrate this. First I will say a small amount of q small amount of delta w equal to we can say d e c v minus sorry plus mass which is going out into h e minus like for control volume we can write, but q is 0 insulated. So, this also I can write as d m e mass which is a small amount of mass which is coming in into h e minus sorry going out into h e minus a small amount of mass which is coming in d m i into h i ok. Here obviously, the kinetic energy and potential energy changes are neglected that means, this I can write as u c v. So, I can write minus delta w equal to d u c v converted as m c v u c v ok. This is also 0 there is nothing going out only something is coming in. So, minus d m i into h i. Now, mass conservation mass conservation d m c v by d t equal to m. So, d t will not the differentially will write. So, d m c v equal to d m i minus d m e. So, this is 0. So, I can say d m i can be written as d m c v itself. So, this is written as d m c v u c v minus instead of d m i I can put d m c v itself d m c v into h i. So, now, we have to see what is h i? h i is the enthalpy of the incoming air from the air line. So, what is this? Air line has a fixed pressure and temperature. If you go back and see the diagram 600 kilopascals 25 degree centigrade. So, it has a enthalpy and the constant property air flow takes place in the air line. So, the air which is coming in from the air line into the chamber will have the same enthalpy. So, I can say h i will be equal to h of the line. Do you understand? So, that is the thing here. So, now, this is what the expression is. How to do this? Now, integrate this. Integrating we can say w minus w equal to this is a total work in involving this process is equal to m 2 u 2 minus m 1 u 1. Understand that m 2 is the mass of the air at the final state and u 2 is the internal energy at the final state. So, d here for example, take this when you want to integrate this from state 1 to state 2 d m d m c v u c v ok. Now, this can be written as m c v u c v 1 to 2 which is equal to m 2 u 2 minus m 1 u 1. So, that is what? So, first term. Second term h i is a constant because this is constant because what the flow of air has a constant property when it flows to the air that comes in. So, the line enthalpy will not change. So, the incoming enthalpy also will be constant. So, this need not be taken into account for the integral. So, I can say this minus h line into this if you integrate d m i it will be m 2 minus m 1 that is it correct. So, what is m 2? m 2 is the final mass in the chamber A. So, this will be the integrated equation. So, what is W here? W is already got W equal to 4.268 kilo joules correct 4.268 kilo joules because that is the work involved in compressing this steam and that is given by the A to B air chamber to the steam chamber ok. Now, this can be written as 4268 joules. So, let us write this minus 4268 equal to m 2 I do not know. What is the final temperature of this? That also I do not know basically. Please understand that u 2 can be written as C v t 2 u 1 can be written as C v t 1 t 2 comma t 1 in Kelvin ok. Now, C v itself can be calculated as what? R by gamma minus 1 equal to 287 by 0.4 equal to 717.5 joule per kg Kelvin ok. Similarly, h line h line can be calculated as C p into t line t line in Kelvin. So, now, C p equal to R C p equal to gamma R by gamma minus 1 which is equal to 1004.5 joule per kg Kelvin ok that is it. So, now h t line is what? t line is equal to 25 degree centigrade correct here 25 degree centigrade. So, that will be equal to 298 Kelvin. So, now, I do not know T 1 I know, but T 2 I do not know ok. So, I have to write this m 2, m 2 is what? So, we know that p 2 v 2 equal to m 2 or T 2. So, now, I do not know T 2 because here u 2 will have C v T 2 correct. So, now, let us try to eliminate m 2 here. So, which implies m 2 you will be equal to what? p 2 v 2 divided by R T 2. I know p 2, p 2 is 600 into 10 power 3 600 kilo Pascal into v 2 is what? v 2 I have calculated already. That is the R chamber here v a 2 is the v 2 0.04634 4364 0.04364 divided by 287 into T 2 ok. Now, so now, you can substitute this m 1 I know u 1 also I know. So, for m 2 you substitute something in terms of T 2 itself do you understand. So, I can write minus w equal to for m 2 I can write this as 600 into 10 power 3 into 0.04364 divided by 287 into T 2 into C v is 717.5 into T 2 actually T 2 cancels in the first term minus m 1 m 1 I know correct. What is m 1 I have already done here from this 0.0292 kg 0.0292 into u 1 u 1 is 717.5 into T 1 is what is T 1 25 degrees centigrade that is 298 298. So, this is the first term minus. So, this I know. So, T 2 is cancelled. So, first two terms I can evaluate the numbers second term h line h line is 1004.5 into 298 this is also known to me, but here m 2 again I can substitute this I will just for brief I will write P 2 v 2 by T R T 2 ok minus m 1 is 0.0292. So, now T 2 is only unknown in this equation we have eliminated m 2 in terms of T 2. So, I get so because w is known to me 4268 correct. So, I can eliminate eliminate m 2 in terms of T 2 and write calculate T 2 itself as 378.25 Kelvin or 105.25 degrees centigrade that is it. So, you can see in this problem that is a high bridge control volume first of all. So, in the control volume where both mass is changing as well as the volume is changing that is a displacement work which is involved. So, this work is not any other form of work in control volume basically we can write w x that is the work involved due to some other say for example, turbine work compressor work ok shaft work electrical work anything, but not displacement work because volume is not going to change. But in this case since the volume is changing the displacement work comes into play. So, that displacement work was calculated for the steam chamber. Since the immediate neighbor of the or the only neighbor of the steam chamber is the air chamber this work for the steam has to come from the air chamber. So, that is very important. So, the key here is calculating the work involved for the steam chamber and equating that to the negative of the work involved for the air chamber. So, that is very key. So, in that way we have found the displacement work involved in the air chamber. Now, control volume equation is applied here and integrating the control volume equation we got two unknowns in the equation one is U2 written in terms of T2. So, M2 and T2 are the unknowns, but equation of state is the second equation which we used to eliminate M2 or writing M2 in terms of T2 itself and making this energy equation to have only one unknown T2. So, once the T2 is evaluated we can solve the problem. So, this is what is asked. So, you can see the problem final temperature in each chamber and mass that enters. Mass that enters is very simple. So, once you know the final mass now T2 is known V2 is known. So, you can find the final mass that is it. So, M A2 sorry M A2 minus M A1 will be equal to 0.21208 kg because you know now volume is known final volume A2 V A2 is known P A2 is known that is 600 kilo vascals that is known P A2 and now T A2 also is known this is T A2 basically correct. So, all the three are known. So, M2 can be found M A2 can be found. So, once M A2 is found then M A1 is already known. So, subtract these two you get the mass which has come in final mass is more than the initial mass. So, this is a very interesting problem in which first one important thing what was identified is that is a slow compression due to frictionlessness of the piston. This compression is slow and it can be taken as adiabatic and reversible compression. So, the process undergone by this team is adiabatic reversible process which was taken as the isentropic process. So, the second state for this team was 6 dash S2 equal to S1 one of the properties. But understanding that always there will be a mechanical equilibrium between air and the steam chamber the pressure final pressure of the air which is given as 600 kilo vascals will be the pressure of the steam chamber as well. So, two properties we have got one is the entropy another one is the pressure and from these two properties we have got the other properties like v2 u2 etcetera and applied the first law for the steam from which we got the work done which is nothing but minus of delta u and that work done on the steam will be done by the air so that we got. So, you can clearly see that work done by the air is positive. So, now applying the control volume energy equation for the air side and recognizing that the work term involved here is the displacement work not the any other form of work substituting that work here we got the temperature in the air column in the final state from that we can find the mass ok this is the solution.