 We will now look at few worked examples which demonstrate how to actually apply the expression for displacement work that we derived in the previous lecture, namely W displacement equal to integral PDV. So, we will look at a few examples, demonstrate how to define the system and apply the expression. So, the first worked example reads like this steam main carries steam at a pressure of 1.5 mega Pascal under a temperature of 250 degree Celsius. A rigid vessel that initially contains steam at a pressure of 0.4 MPa is connected to the mains through the valve. The valve is now opened and 0.1 kg of steam from the main slowly flows into the vessel. Notice that this is important slowly flows into the vessel. Determine the work interaction for this process considering the steam finally in the vessel specific volume of the steam in the line may be taken to be 0.142 meter cube per kilogram. You may recall that we had actually looked at such a situation when we discussed the definition of a system and how to define system for analysis for a particular case. So, here we have a line in which steam is flowing. So, this is the line in which steam flows and we have connected a vessel through a valve to this. So, this is the vessel. So, we have connected vessel to the line through a valve and the valve is opened slightly and steam flows slowly into the vessel and if you recall we defined suitable system for this problem to be something like this. So, this was what we defined to be an appropriate system for this problem. So, this is the initial configuration of the system. In the final configuration this part of the system boundary shrinks to 0 volume as all the steam flows inside the vessel. So, this is the initial configuration and the final configuration of the system because this part will be absent and so we can now proceed with the calculation apply integral PDV and calculate the displacement work. So, the part of the system boundary inside the line alone deforms during this process and since 0.1 kg of steam enters the vessel specific volume is also given. So, the volume occupied by this part of the system boundary is 0.1 times 0.142 meter cube per kilogram which is 0.0142 meter cube. So, if you look at the previous slide, the volume occupied by this part of the system boundary is 0.0142 meter cube. So, the final volume of this part of the system boundary is of course, 0 as we have already mentioned. So, this displacement work may be evaluated as integral 1 to 2 P line DV line. How would we get this? This is because only the system boundary in so this comes because the part of the system boundary inside the line alone deforms. So, the displacement work is only due to this part of the system boundary. So, we wrote integral PDV as P line integral DV line. So, P line remains constant so that can be taken outside and so this may be integrated to give V2 minus V1 line and the final volume in this part of the system is 0 and the initial volume is 0.0142 meter cube. So, the work may be evaluated to be work interaction for this system may be evaluated to be minus 21.3. Notice that the negative sign indicates that work is done on this system by the steam in the line. So, the steam in the line pushes this much steam inside the vessel by doing work on this steam. So, that is what the negative line indicates. So, as I said before the nice thing about integral PDV is that not only does it give the magnitude of the work interaction correctly it also gives it with the proper sign. Let us look at the next example. So, here we have a situation where a certain amount of air is enclosed inside a piston cylinder mechanism like this. The pressure initially in the cylinder is 60 kilo Pascal which notice that this is less than the atmospheric pressure of 100 kilo Pascal. So, atmospheric pressure is 100 kilo Pascal on the bottom of the piston and initially the pressure inside the cylinder is 60 kilo Pascal. The cylinder is connected through a valve to rigid vessel which initially contains air at a pressure of 30 kilo Pascal. The lower face of the piston as I have shown here is indicated to is I am sorry exposed to atmosphere at 100 kilo Pascal. Valve is now opened and the piston slowly moves a distance of 12 centimeter from its original position. It is not given whether the piston moves up by 12 centimeters or moves down by 12 centimeters. So, that is something that we have to figure out looking at the physics of the problem. So, we are asked to calculate the work done by each of the following systems. We take the piston to be the system atmosphere and the air and calculate the work interaction. Since the pressure initially in the cylinder is 60 kilo Pascal, pressure here is 100 kilo Pascal. It is quite clear that the piston moves up during the process and the force balance on the piston at any instant not only the initial or the final instant at any instant. Let us say that the piston is located here like this. So, the air inside the cylinder exerts a downward force on this piston and the mass of the piston also exerts a downward force. The atmosphere exerts an upward force on the bottom face of the piston. So, we do a force balance on the piston at any intermediate instant and if we do that then we get the following. P atmosphere times A which is an upward force on the bottom side of the piston has to be equal by P in the cylinder times A which is a downward force and the weight of the piston itself which is also a downward force. So, these forces have to balance at any instant because the process takes place slowly. So, if we rewrite this then we can get an expression for the pressure at any instant in the cylinder in terms of other quantities. Now, notice that the quantities on the right hand side all remain the same. In other words nothing here on the right hand side changes during the process which then tells us that the pressure of the air in the vessel remains constant. So, the pressure I am sorry the pressure of the air in the cylinder remains constant as the process takes place because at any instant the pressure in the cylinder is given by this expression and since nothing in this expression changes during the process we can easily conclude that the pressure in the cylinder remains constant during the process. Since we have given the initial pressure I can substitute the values into into this expression and get the mass of the piston to be 8.155 kilograms. Since the piston moves up during the process the work interaction for the piston may be evaluated as MP mass of the piston times G times change in elevation with respect to a datum and notice that this comes with the negative sign. Since the piston moves up the surroundings or actually by virtue of their interaction or moving the piston up. So, in other words by based on our definition of work if a system by virtue of this interaction results in a mass being moved up then the system is doing positive work. In this case the surroundings by virtue of the interaction with the system is actually moving a mass up in a gravitational field which means that the work interaction for the surroundings is positive and work interaction for the system thus is negative because W system plus W surroundings is equal to 0. So, we get a negative sign for the work for the piston which is 9 point. So, it comes out to be minus 9.6 joules. One can also understand the negative sign in the manner that we have already stated. The potential energy of the piston is being increased by an external agent which means that the external agent is doing work to increase the potential energy which means that the work interaction for the piston is negative. Now, W atmosphere again by applying integral PDV may be written as p atmosphere times V2 minus V1 for the atmosphere. Since piston moves up the atmosphere is actually expanding. So, V2 is greater than V1 and since we know the distance by which the piston moves. So, this is the increase in the volume occupied by the atmosphere and so this comes out to be a positive quantity which means that the atmosphere is actually doing work in pushing the piston as well as the air in the cylinder into the vessel. So, W air again if I take the air as the system if I take the air as the system then I can do the following. I can actually draw a system like this. So, this will be my system what is that only this part of the system boundary, this part of the system boundary deforms all the other parts remain constant. So, when I apply integral PDV for the air I get the expression since the pressure of the air is constant the p can be taken out. So, p air times V2 minus V1 air and since the volume the air in the cylinder decreases during the process we get V2 minus V1 to be negative and so this comes out to the negative sign. Also note that W atmosphere plus W piston plus W air is equal to 0 as it should be. So, in other words we could have actually evaluated W air by using the factor W atmosphere plus W plus piston plus W air equal to 0 but we could also in this case evaluate it like this both are acceptable but the sum should come out to be 0 that is important. So, the next work example involves an arrangement like this. So, we have again a piston cylinder mechanism with the two pistons which are connected by a thin rod of negligible mass and volume and it is open to the atmosphere at the top and bottom. A certain amount of air is enclosed in this space and so initially the piston assembly is in equilibrium. So, the initial pressure of the air is such that the piston assembly is in equilibrium the cross sectional area of the upper piston is 10 centimeter square more than the cross sectional area of the lower piston. The combined mass of the pistons is 5 kg. We now add heat very slowly to the air and this causes the piston assembly to move up by 25 centimeters. Determine the initial pressure of the air, the work interaction for the piston assembly, work interaction for the atmosphere and the work interaction for air the atmospheric pressure is given to be 100 kilo Pascal. So, here we have the atmosphere, here also we have the atmosphere. So, if we do force balance on the piston assembly at any instance. So, let it be at let it be located at some other elevation here. If I do a force balance what is that the pressure of air exerts an upward force on the bottom phase of the top piston and a downward force on the top phase of the bottom piston. Similarly, the atmospheric pressure exerts an upward force here and a downward force here. So, we do force balance on the piston assembly and this is what we get the subscript u denotes upper and the subscript l denotes lower. So, the net force balance looks like this which can then be rearranged to give an expression for the pressure of the air in the cylinder at any instant during the process. So, the pressure at any instant during the process looks like this and once again we notice that all the quantities on the right hand side of this expression remain constant during the process which tells us that the pressure of the air in the cylinder once again remains constant during this process which then simplifies our integral pdv. Initial pressure may be obtained by substituting the values known values into this expression mp is known the area of the upper piston is 10 centimeter square more than the area of the lower piston. So, we use all this information to calculate the initial pressure which comes out to be 149 kilopascal. And the piston assembly in this example also moves up by 25 centimeters which means the work interaction is negative for the piston assembly and it can once again be calculated as mass of the piston times g times change in elevation. So, it comes out to be minus 12.2625 joules. Now, when we want to calculate the work interaction for the atmosphere we again use integral pdv for the atmosphere. Now, we have to be careful because the change in volume notice that the piston assembly moves up by 25 centimeters. So, when I apply integral pdv to this part of the atmosphere. So, I have to apply integral pdv to this part of the atmosphere and then this part of the atmosphere. Notice that in this part of the atmosphere in this part of the atmosphere since the piston moves up the volume change is negative and in this part the volume change is positive because the piston moves up. So, if we go ahead with this. So, if we go ahead then we can actually see that the change in volume in the upper part looks like this p atmosphere Au times change in elevation and the change in volume in the lower part is nothing but Au times this. So, the net displacement work is p atmosphere times Au times that. So, if you go through the arithmetic you get the work interaction for the atmosphere to be minus 25 joules. So, basically what we have done is integral pdv for the atmosphere looks like this you then split it into two parts. So, p atmosphere times v2 minus v1 atmosphere upper plus p atmosphere times v2 minus v1 atmosphere lower and that is what we have done here. So, we can see clearly that the volume increases on the lower side and decreases on the upper side. So, we take that into account and this is what we get finally. Since w air plus w atmosphere plus w piston 0 we can easily evaluate w air to be plus 37.2625 which means that the air is actually doing work to move the piston assembly up and also move the atmosphere aside. Since we have already shown that the pressure of the air in the cylinder is constant you may also evaluate w air by applying integral pdv which is nothing but p air times v2 minus v1 for the air. And once again we note the fact that the volume increases by a certain amount on the upper side and decreases by a certain amount on the lower side for the air and by taking the sign properly for the volume change we can evaluate the work interaction to be 37.2625 which is the same as what we obtained here. So, one has to be careful with calculating the volume change in this problem because the height is given but since the cross sectional area is different for the same displacement the volume change is more on the upper side and less on the lower side because the displacement is the same but the volume change is more on upper side and less on the lower side. So, this needs to be taken into account when we evaluate v2 minus v1. Let us now go to the next example which looks like this again we have a piston cylinder mechanism. So, there is there are a couple of stops in this cylinder and this piston initially rests on the stops. So, this contains a certain gas inside and heat is added to the gas. So, as we add heat the pressure of the gas increases and once the pressure exceeds 115 kilo Pascal then the piston is lifted from the stops and it starts moving up. So, the gas now pushes the piston up initially it lifts it until the pressure becomes 115 kPa the piston remains on the stops. Once it exceeds 115 kPa piston starts moving upward and then it encounters a linear spring and the gas pushes the spring further by 2 centimeters. So, in other words it compresses the spring by 2 centimeters we are asked to calculate the work interaction for the atmosphere piston spring and the gas. All the required data is given in the problem atmospheric pressure is also given. So, since the pressure of 115 kilo Pascal is required to just lift the piston a force balance on the piston gives us that mass of the piston times g divided by its cross sectional area is equal to 115 minus 100 this is p atmosphere times 1000. So, we get the quantity mp times g over Ap to be 15000 from which we can get the cross sectional area of the piston to be 0.00654. Now, it is given in the problem that the piston rises through 12 centimeters initially. So, the 12 centimeter rise is this and then the spring is compressed by 2 centimeters. So, the total distance through which the piston moves is 14 centimeters. So, the work interaction for the atmosphere integral PDV may be evaluated like p atmosphere times V2 minus V1 and since the volume of the atmosphere while the atmosphere is pushed aside the V2 minus V1 is negative for the atmosphere. So, we can calculate this quantity to be minus 91.56 joules and once again similar to both the earlier problems the piston moves up which means the work interaction for the piston is minus m times g times change in elevation which comes out to be minus 13.734 joules. Now, the spring is compressed by 2 centimeters and you all know from your high school physics that when you compress a spring through an amount x then the amount of energy that is stored in the spring is 1 half kx square which means the energy of the spring has been increased by an external agent by an amount of 1 half kx square. So, the work interaction for the external agent is positive and the work interaction for the spring is negative. So, W spring is minus 1 half kx square which gives us minus 20 joules. Since the sum of W gas plus I mean W gas W atmosphere W piston and W spring is equal to 0 we can get W gas to be plus 125.294 joules. So, the gas does work in moving the piston through 14 centimeters pushing the atmosphere aside and also compressing the spring by 2 centimeters. Now, if you look at the process that is undergone by the gas in the cylinder we can appreciate that it is a three step process. So, initially the cylinder is at a pressure of 100 kilopascal. So, now we start adding heat. So, until the pressure increases to 115 kilopascal the piston remains at the same position which means the volume occupied by the gas remains constant. So, that means from 100 kilopascal to 115 kilopascal the gas undergoes a constant volume process. Then as the piston is lifted and until it encounters the spring which means through this movement of 12 centimeters the gas undergoes a constant pressure process because the pressure of the gas is nothing but 115 kilopascal nothing has changed. So, the piston keeps moving up and the gas pressure remains constant at 115 kilopascal. So, once it encounters the spring the pressure on the gas begins to actually change. So, it encounters and it goes through a slightly different process. So, there are three different steps in this. So, first step is one to a constant volume process during which the pressure increases from 100 to 115 kilopascal. Then it is a constant pressure process during which the piston moves up by 12 centimeters and then process 34 where the pressure and volume both increase while the spring is compressed by 2 centimeters. So, we may also actually evaluate the work interaction for the system by saying that it is nothing but integral PDV for all these processes combined that is also possible. So, if you do that W12 will be equal to 0 because it is a constant volume process. W23 is simply 115 times 1000 times the change in volume because it is a constant pressure process. Now, during the third step at any instant a force balance on the piston will give the following. So, if you look at the piston there is the pressure of the gas from below which is exerting a force in the upward direction and then mass of the piston is exerting a force in the downward direction spring is exerting a force in the downward direction pressure the atmosphere is also exerting a force in the downward direction. So, we can write a force balance like this. So, this is the expression for pressure at any instant during the process. Notice that if I write this in terms of volume I can actually write it like this. Notice that in contrast to the previous examples that we saw this step in this step the pressure does not remain constant pressure varies linearly with the volume during the compression step. So, we can now integrate this and then if you go through with the algebra you can show that this actually is equal to you should be able to show that this is equal to 125.294 joules. So, based on the information given it is also left as an exercise to you to work out this value and also sketch the process on a PV diagram and shade the different areas indicating the work interaction for the different components the piston and the atmosphere spring and so on.