 Let us go to the next example. So, here we have a vertical frictionless piston cylinder arrangement. This contains initially 2 kg of water at one bar. So, the piston initially rests on these two stops. And when the pressure inside reaches 3 bar, the piston will begin to move. So, the volume when the piston is on the stops or the volume occupied by the water when the piston is on the stops is given to be 250 liters. Now, we transfer an amount of heat equal to 4500 kilo joules. We are asked to determine the final state of the water and we worked out. And the initial state we have we have the volume which is known. So, V1 equal to 0.25 meter cube. Mass is also known 2 kg. So, which means the specific volume is known at the initial state and the pressure is also known at the initial state. So, with these two pieces of information, we should be able to locate the state initial state. So, the specific volume at the initial state is 0.125 and the pressure is 1 bar. So, you go to the pressure table the steam table and you can see that Vf is equal to this and Vg is equal to this. Since the given specific volume lies between Vf and Vg, the initial state is a saturated mixture. So, we can now show this on a PV diagram like this. So, this is 100 kPa. So, the initial state is a saturated mixture which looks like this. Now, we add heat to the contents of the cylinder. Notice that the volume will remain a constant until the pressure inside reaches 3 bar. Because at that pressure the piston will start to lift. So, the working substance undergoes a constant volume process from 1 bar to 3 bar or 300 kPa. So, let us say that this is 300 kPa. So, the working substance undergoes a constant volume process from 1 bar to 3 bar. Let us denote this state as state 2. So, what we will try to do is determine the amount of heat that is required to go from state 1 to state 2. Notice that at state 2, once again two pieces of information are known to us that is the pressure which is 300 kPa and specific volume V2. Remember V2 is equal to V1. So, using these two pieces of information, we can actually locate the state. So, for process 1-2, if you apply first law, we get delta E equal to delta U again no change in ke or pe. W1-2 is 0 because it is a constant volume process displacement work is 0, there is no other form of work. So, the heat that has to be supplied to go from state 1 to 2 is given by this expression. So, at state 2, as I said before, P2 is 3 bar and V2 is equal to 0.125. So, we go to the pressure table corresponding to 3 bar, we get Vf equal to this, Vg equal to this. Since V2 lies between Vf and Vg, state 2 is also a saturated mixture state. So, when I showed state 2 to be like this, it was an educated guess and that it is a saturated mixture. But now, with the information that is available, we can precisely state that state 2 is a saturated mixture and we can evaluate the dryness fraction using the known volume, a known specific volume like this and that comes out to be 0.2048. And the specific internal energy may also be evaluated at state 2 and that comes out to be 967.15. So, the heat that has to be supplied to go from state 1 to 2 is 794.238 kJ. So, that means of the total 4500 kJ that is supplied, 794.238 goes to raising the state from 1 to 2. Now, we supply the reminder of the heat to reach the final state. Notice that once state 2 is reached and the pressure becomes 300 kilopascal, if I supply heat any further, the piston begins to lift. So, if I look at any intermediate state, let us say this is an intermediate state, if I look at the force balance on the piston at any intermediate state, the forces on the piston are the following. One is pressure force from the contents of the cylinder in the upward direction. The next one is P atmosphere in the downward direction and the pressure due to the weight of the piston which is also acting downwards. However, notice that these two quantities the P atmosphere and the weight pressure due to weight of the piston remain constant during the process, which means that the contents of the cylinder undergo a constant pressure process as we supply as we continue to supply heat, which means that the process looks like this. We do not know where the final state is going to be, whether the final state is a superheated state or whether it is a saturated vapor state or whether it is still a two-phase mixture or not remains to be seen. So, that depends on the amount of heat that we are supplying, but this is the process line. So, that gives us an idea. So, now, we apply first law to process 2-3, we get delta E equal to delta U equal to q 2-3 minus W2-3. Notice that displacement work in this case is non-zero, it is a constant pressure process. So, PDV integral PDV is easy to evaluate, P comes out of the integral. So, we can write this first law in this form and if you rearrange this expression, basically what we are going to do is take this term and combine it with this term here and bring the U2 term to the right-hand side. So, we rearrange it like that, then we get the specific enthalpy H3 equal to U2 plus P2 V2 plus Q2 3 divided by M. P3 is equal to P2. So, I have used the fact that H is equal to U plus PDV. So, now, once again we have two property values at state 3. One is the pressure which is 300 kilopascal, the second one is a specific property which is the specific enthalpy. So, if I plug in the values, I get the specific enthalpy to be 2857. So, we now go to the pressure table and we notice that Hg corresponding to 3 bar is actually less than H3 which means that, so Hg corresponding to 3 bar is less than H3 which means that state 3 has to be a superheated state. So, we go to the superheated table now and we can get the specific volume at state 3 to be equal to 0.7103. Let us just quickly take a look to see how we use the superheated table. So, we have been given the pressure to be 300 kilopascal or 3 bar. So, here we have 3 bar and superheated table, superheated steam 3 bar and we are also given the specific enthalpy to be 2857.53. So, 2857.53. So, we go down the enthalpy column, 2857.53 is going to fall in between these two entries. So, we interpolate and get the final values that we require. Linear interpolation in this case is sufficient or if you have software tools by entering just by entering the pressure and the specific enthalpy, you should be able to get the final values. So, we interpolate to get them specific volume at state 3 to be equal to this. So, the work interaction for entire process 1, 3 is equal to work interaction during 2, 3 because w12 is equal to 0. It is a constant volume process 1, 2. So, we can evaluate the work interaction during the process to be 351.18 that is what we have been asked to find. Let us just check. So, we have been asked to determine the final state. We have determined the final state to be A and superheated state. And we have also evaluated the work done during the process. So, again, when you solve problems like this or do an analysis of problems like this, you must always remember the way in which we developed the material. For a two-phase mixture, first identify the two properties whose values are known. And once you identify them, we locate the state. Once we locate the state, we can then evaluate the property, any other property whose value is required. So, that is the sequence that we follow. Let us go to the next example which is somewhat more involved. So, basically we have insulated piston cylinder arrangement with a metallic partition in between. So, this is a metallic partition, very thin partition does not store any energy. It is rigid, so it remains flat and it does not store any energy, but it is a metallic partition, so that Ta equal to TB all the time. So, the temperature on both sides are always the same. So, that is what the metallic partition does. So, the initial state is given. In compartment A, we initially have steam at 5 bar, 180 degree Celsius. You know that since pressure and temperature are given, that this must be a superheated state because I think it is easy to follow. By now, you should be able to do that. So, the volume is also given. Notice that volume of compartment A is given to be 0.2023 meter cube. And as the process takes place, the volume of compartment A remains the same because the metallic partition is rigid and remains so. So, the working substance in A undergoes a constant volume process. So, that means we can infer from the given information. Compartment B initially has a volume of 0.2 and contains 2 kg of a saturated mixture of liquid water and water vapor. So, then the information that is given is the specific volume is known initially and it is also given to be a saturated mixture. So, we now add heat to this until the pressure becomes equal in both compartments. Bear in mind that the temperature is always equal in both the compartments, but now the pressure is also at the final state the pressure is also become equal. And we are asked to calculate several quantities here A through D and also the amount of heat that is transferred. So, the initial specific volume in compartment B is 0.1. It is initially at 180 degree Celsius because this is initially at 180 degree Celsius. Since the temperatures are the same, this is also at 180 degree Celsius. So, the temperature is given to be 180 degree Celsius. The specific volume is also given. So, you should be able to determine that this is indeed a two-phase mixture and also that the saturation pressure corresponding to this temperature is 10.02 bar. Let us just quickly go to the temperature tables and verify this. So, this is the superheated table. So, we have temperature table now and the temperature is given to be 180 degree Celsius which is this. So, the saturation pressure is 10.02. Notice that the specific volume of the saturated vapor is itself 0.1940. This is 0.001. So, the initial specific volume lies between these two. So, it is a saturated mixture and since it is a saturated mixture, its pressure is equal to the saturation pressure. So, the initial pressure in compartment B is thus equal to the saturation pressure corresponding to 180 degree Celsius which is 10.02 bar. So, we can then retrieve the following values from the steam table and the initial dryness fraction in compartment B may be evaluated like this and 5 bar 180 degree Celsius you should be able to easily show that this is that the state is superheated and we may then retrieve the following values from the superheated table. 5 bar 180 degree Celsius. Let us just see once how this is done. 5 bar 180 degree Celsius. So, this is the location in the table V1 is this and U1 is also equal to this. Now, the mass contained in compartment A may be evaluated from the specific volume and the volume of the compartment itself by using this relationship. So, it comes out to be 0.5 kg. Now, if you look at the pressure in compartment B, this is due to atmospheric pressure plus pressure due to weight of the piston, both of which remain constant during the process which means that the working substance in B undergoes a constant pressure process. Working pressure in I am sorry working substance in compartment A undergoes a constant volume process and the working substance in compartment B undergoes a constant pressure process. So, if we were to sketch the process at least qualitatively on a PV diagram, let us try to do that here. So, if we sketch this qualitatively on a PV diagram, so this is 10.02 bar. So, initially, this is the state in compartment A and this is 5 bar and the initial state in compartment B is over here. So, this is 1 comma A. So, the working substance in B undergoes a constant pressure process which looks like this and the working substance in A undergoes a constant volume process which looks like this until the pressures become equal. So, this will be the final state. So, that is state 2. So, now if you look at the information that is available about the final state, pressure in compartment B remains constant. So, the final pressure is equal to 10.02 bar. Now, the final pressure in A is also equal to 10.02 bar because the final pressures are the same. Now, the specific volume in compartment A remains the same because it undergoes a constant volume process. So, we have two property values at the final state 10 bar and 0.4045. So, final state in A is superheated. This can be easily established and the final temperature comes out to be 607 degree Celsius, again from the superheated table with linear interpolation. Now, since the temperature in both compartments are the same, final temperature in B is also 607 degree Celsius and as this illustration indicates the final state is the same in both compartments. So, there is no separate 2B and 2A. The final state is the same because it lies at the point of intersection of this isobar and this isochore or constant volume line. Now, I take compartment A and B together as my system. So, we take this to be our system and apply first law. Since the partition does not store any energy, I may write delta E equal to delta U which is nothing but the internal energy of the working substance in A and B. There is no storage of internal energy in the partition and there is no KE and PE change also. So, delta E becomes equal to delta U equal to Q minus W. There is no displacement work in compartment A because it is at a constant volume, but there is displacement work in compartment B because it is a constant pressure process. So, we may evaluate the work interaction very simply using this expression and it comes out to be 610.218 kJ and the internal energy change may also be evaluated quite easily and that comes out to be 3576.87. So, the total amount of heat that is transferred comes out to be 4187.088 kJ. So, once again you can see that with the given information, we try to draw a PV diagram or any other coordinates. PV is simple. So, we draw a PV diagram, try to locate the stage and also try to sketch the process at least qualitatively on this diagram, fill in the information as you keep proceeding with the analysis. Initially, we fix the states and try to put in as much information into this diagram as possible with the given information from the problem statement. Then as we go through the analysis, we fill in more and more information so that you get clarity. So, once you have the system identified and then you also enter as much information as possible into the PV diagram, the analysis becomes that much easier. So, the last example that we are going to look at is this one here. So, we have an insulated rigid vessel of volume 500 liters. So, this is the insulated rigid vessel volume 500 liters which is initially evacuated. It is connected to a line in which steam at 20 bar 400 degree flows through a valve. So, here is the line in which steam at 20 bar and 400 degree Celsius flows it is connected through a valve to this line. So, we open the valve and steam is allowed to slowly flow into the vessel until the pressure inside is the same as the line pressure which is 20 bar and then we close the valve at that instant. We are asked to determine the final temperature inside the vessel and the mass that enters. Heat loss may be neglected because it is given to be insulated. KE and PE changes may also be neglected. You may recall that we had already looked at an appropriate system for this case and that system looked like this. So, the system looked like this. So, this is the mass. So, this is the mass that enters from the line which we will label as M line. So, that is the initial system. In the final state, this part of the system boundary diminishes altogether and we have only this part that is left. So, this was discussed in the beginning of the course when we discussed definition of system and different examples for system. So, now, so that is our system. So, if we apply first law to this system, delta E equal to delta U, no KE and PE change equal to Q minus W. Q is 0 because there is no heat loss and the vessel is completely insulated. Now, notice that W is not 0. There is defamation in this part of the system boundary, no defamation in this part of the system boundary, but defamation in this part of the system boundary at constant pressure. So, the displacement work may be evaluated quite easily. The constant pressure is nothing but P line. And as I said, you know the part of the system boundary in the line disappears altogether at the end of the process. So, that means the volume occupied by the system, this part of the system is 0 at the end of the process. So, we end up with something like this and so, notice that U1 is the specific enthalpy of the, notice that this term here delta U, this is actually M times U2 minus U1 where M is the mass of the system. But in this case, since the vessel is initially evacuated, the mass of the system is equal to whatever mass that is going to enter from the line. And U1 is equal to U line, this is the specific internal energy of the mass that is going to enter from the line. So, U1 is equal to U line. Or in fact, you may, if you wish, you may actually rewrite this itself, delta U for the system may be written like this delta U in the, for the system, part of the system in the line plus delta U in the vessel. Since M is equal to 0, since this is initially evacuated, we take that part to be 0. So, basically we end up with something like this. But notice that finally, the steam that was in the line will actually enter and occupy the complete vessel. So, you have to be careful with how you do the analysis. So, we expand this U2 minus U1 like this. And if we rearrange this, take this term to the left-hand side and rearrange, we get something like this. What is that U line plus P line V line is equal to H line. That is the enthalpy of the fluid that is flowing in the line. So, the specific internal energy of the steam that is finally in the vessel is actually equal to the specific enthalpy of the steam that is flowing in the line. So, for the steam flowing in the line, the pressure is given to be 20 bar, temperature is 400 degree Celsius. So, we can get H line to be equal to this from the superheated table. Now, for the steam that is finally in the vessel, again we know two values for two properties, we know that the state is superheated. So, final state of the steam of the working substance in the vessel is superheated. So, we can easily get the final temperature to be 575 degree Celsius. Notice that the steam in the line flows at a temperature of 400 degree Celsius. However, the steam in the vessel is at a higher temperature 575 degree Celsius, because of the fact that the specific internal energy of the steam inside the vessel is equal to the specific enthalpy of the steam that is flowing in the line. So, basically what has happened here is that the work that was done by the steam in the line to push this amount of steam into this is also now included in the specific internal energy of the steam here. So, there is work that is being done and there is also internal energy of the steam. So, the work that is being done is now being added to the internal energy of the steam and that is why the steam here is at a higher temperature than the steam in the line. So, what we will do in the next lecture is start our discussion of first law of thermodynamics for a control volume. So, so far we have looked at first law for a system undergoing cyclic process as well as a non-cyclic process. And we have applied first law to analyze many, many different examples. Some of them involving ideal gases, some of them involving two phase mixtures of refrigerant or steam or water and water vapor. So, for all these problems we could use the system approach quite comfortably and complete the analysis. But there are certain problems for which a modification of the system approach is required to make the analysis spot easier. We will not be deriving a new form of the law, we will simply modify the system approach to come up with an alternative form of first law that can very easily be used for the analysis of these types of problems. We will take it up in the next lecture.