 We now move on to the next module which is work and heat. So, here we take a close look at thermodynamic work or work as it is understood in thermodynamic analysis and also heat. In engineering thermodynamics work is said to be done by a system. Notice that this is very important work is said to be done by a system if the combined effect of its interaction with the surroundings. Notice that it may the system may interact with the surroundings in many different ways. So, the combined effect of its interactions with the surrounding that is why we use this word combined the surroundings is the raising of a mass. Of course, raising of a mass in a gravitational field means that you are doing some work to increase the potential energy of the mass. So, when I raise a mass in a gravitational field work has to be done by me to increase its potential energy. So, that is how we define work done by a system. So, in this case work is done by the system and the sign convention that we use for work in thermodynamics is that work done by a system is positive because we are always talking about engines. Our thermodynamics as I said earlier started off with development of steam engines. So, the objective for mechanical engineers always has been I give device a certain amount of heat I want this to be converted to useful work. So, heat given to a system is always positive work done by a system is always positive that is the sign convention that we use in engineering thermodynamics. So, in this case when the combined effect of the system is to raise a mass we would say that work done by the system is say 10 kilo joules. Now, negative work is not defined explicitly. Notice that when an external agent does work on the system you may recall that when we discussed a different examples in connection with the thermodynamic system in some instances we saw that the system was doing work in a pushing the atmosphere or raising the mass or raising the piston and so on. In other cases the surroundings were actually doing work to push air inside the cylinder or to push air from a line into a cylinder and so on. In those cases when work is done on the system based on the sign convention we can easily see that the work interaction for the system is negative. Now, we have to be very careful with the language that we use and the signs that we put out here. In other words it is always good practice to say work done by a system is 10 kilo joules and to say work done on a system is 10 kilo joules. It is not good practice to say work done by a system is minus 10 kilo joules. Although the meaning probably can be ascertained it is not good practice to put it that way. The best way to put this is to say work interaction for the system is plus 10 kilo joules or work interaction for the system is minus 10 kilo joules. So, in this way there is no ambiguity and it is clearly understood that the positive number implies that work is done by the system, the negative number implies that work is done on the system. So, it is a good practice to state the work interaction in this way. Rather than saying work done by and work done on although that is probably much more prevalent and more intuitive, you have to be very careful about you know the language and the sign. We can also see that the work done by the system and the work done by the surroundings, the algebraic sum of the work done by the system and surroundings is always equal to 0. So, if the system is doing work in pushing the atmosphere aside, lifting a piston and so on, the system is the work interaction for the system is positive, work interaction for the piston will be negative, work interaction for the mass that is being raised will be negative, work interaction for the atmosphere which is being pushed aside is also negative. So, when I take an algebraic sum of the work done by the system plus all this other work, it will be equal to 0. So, here we actually want to bring out a subtle aspect about a thermodynamic work and the most important idea about thermodynamic work is that it is an interaction of the system with the surroundings, unless there is an interaction, there cannot be a work interaction. So, work is always an interaction of the system with the surroundings. So, here we are looking at a situation where we have a battery which is connected to a motor and the output shaft of the motor turns a pulley which in turn is used to raise a mass. So, the mass let us say goes up. So, the battery supplies current to the motor and the shaft spins and the pulley then winds the rope up so that the mass moves upward. Now, if I look at the battery alone as the system, its interaction with the surrounding is that some amount of electrical current flows out of the out of the system. So, we can easily see that this current can be used to raise a mass in a gravitational field. So, the current is used by I mean through a motor to raise a mass. So, this fits in with the definition that we gave earlier that work is done by a system when its combined effect with the surroundings is the raising of a weight. So, the combined effect of the combined effect of the interaction of this system with the surroundings is the raising of a weight. So, work interaction for this system battery is greater than 0. Now, if I look at the motor alone assuming that the motor is ideal of course, in this case of the battery also we have neglected any heat loss from the battery to the surroundings that can be easily accounted for. So, let us say that there is no heat loss from the battery or the motor to the surroundings. Let us say that the motor is frictionless and there is no dissipation in the windings and it is an ideal motor. In the case that it is an ideal motor then surprisingly you can easily see that the work interaction for the motor will be 0 in case it is ideal because it receives a certain amount of work from the battery. Notice that the battery work interaction for the battery is positive as we already said and the nature of interaction of the battery with the surroundings is that this can be categorized as electrical work. So, if you look at the motor it receives a certain amount of electrical work from the battery and it spins a shaft which means it is putting out a certain amount of shaft work. So, under ideal circumstances when the motor is ideal the amount of shaft work that it puts out will be exactly equal to the amount of electrical work that it is getting numerically equal to the amount of electrical work that it is getting. So, the combined effect of the interaction of the motor with the surroundings will be such that the work interaction is 0 because this work that it is putting out is positive this work that it is receiving is negative. So, the algebraic sum of the two comes out to be 0. So, in the ideal case the work interaction for the motor is 0. In a less than ideal case for instance let us say there is some dissipation in the windings and there is some friction in the shaft and so on in that case the work that it puts out will be less than the electrical work that it receives. So, the shaft work will be less than the electrical work that it receives. So, in this case the shaft work is less than the electrical work in magnitude and so the work interaction for this motor will be negative because it puts out less work than what it receives. So, the net work interaction for the motor is negative in this case. Now, if I isolate the pulley assuming it is a frictionless pulley, in the ideal case the work interaction for this will be 0 just like what it was before. So, in the ideal case this will be 0. So, basically in the ideal case the motor or the pulley merely serves to transmit whatever energy or work it is receiving in either in a different form. So, in this case if the motor is receiving electrical work and it is converting and transmitting this as the form of shaft work. The pulley in the ideal case is receiving shaft work and it is actually converting it, transmitting it as work that can be used to raise the mass. So, that is all the pulley does or that is all the motor does under ideal circumstances. So, the net work interaction for the motor or pulley is 0. And of course, since the mass is being raised in a gravitational field the work interaction for the mass is negative. So, the potential energy of the mass is being increased. So, that means someone else is doing work, the surroundings are doing work to increase the potential energy of the mass. So, we can understand that the work interaction for the mass is negative. So, under ideal circumstances all the electrical work that comes out of the battery will go into raising the energy or potential energy of the mass. Otherwise some of it will be lost here and only the rest of it will go to the mass. What is that? We have also said here that the mass is being raised slowly. That is usually said in order to make sure that you know the elevation of the mass may be evaluated at every instant in time. Remember we said that properties have to be known at every instant in time. Since here the energy of the mass can also be changed through changes in elevation and we need to specify the property elevation with respect to our datum here. So, in order to make sure that the elevation is known at every instant we say it is slowly. Even if it is not raised slowly as long as the elevation is known we can still say that the sum total of the change in kinetic energy and change in potential energy of the mass is actually being taken we is actually being provided or it is a consequence of the work from the battery. We can account for kinetic energy also if it is raised with the finite velocity we can do that also. So, strictly speaking this is not required but it is there to in accordance with what we have said so far that all processes must take place slowly. So, basically what we are saying is if it is raised with a certain velocity then the work that is the work that the surroundings are doing on the mass increases not only the potential energy of the mass but also the kinetic energy of the mass which is which is okay. So, you can see here that depending on where I draw the system boundary the work interactions are different for different system. In fact, if I take this to be my system then notice that the work interaction for the system is 0 the battery is still running motor is still running pulley is still rotating the mass is still being raised but the work interaction for this system is 0 by virtue of the fact that it does not interact with the surroundings at all. Whatever happens happens inside the system when it is not work or heat there is nothing assuming that there is no heat loss to the surroundings it is not heat also. So, you can see that depending on where I draw the system boundary the sign of the work interaction as well as the magnitude of the work interaction both can be very very different. That is a very important point that you need to understand and that is what this and this illustration highlights that both the magnitude and the sign of the work interaction depends on where we draw the system boundary. We now turn to displacement work in the previous slide we sort of referred to different types of work we mentioned that the battery is actually providing electrical work and we saw that here you know this was being raised in a gravitational field there was no network interaction for motor was 0 but it was putting out shaft work. So, you can see that there are different forms of work and we will now try to develop expressions for each one of this form of work starting with displacement work. We have already seen or mentioned displacement work in connection with the definition of a system. So, basically displacement work is that form of work that occurs by virtue of the deformation of the system boundary remember we said that whenever there is deformation of system boundary there is work interaction between the system and the surroundings. So, if the system boundary is you know is being compressed then the surroundings are doing work on the system and if the system boundary expands then we said that the system is doing work on the surroundings. You may recall that in the case of a piston cylinder mechanism the system is doing work because the boundary is expanding system is doing work to raise the piston the mass on top of the piston as well as and push the atmosphere aside. In the case of an evacuated vessel that was being filled from atmosphere we argued that since the system boundary outside the vessel was shrinking that means or that meant that the surroundings are actually pushing the air into the vessel against the resistance of the valve. So that means the work interaction for the system is negative and in case we were filling a vessel with air from a line again there also we argued that since the system boundary in the line was actually shrinking that meant that the air in the line was pushing the system or the air into the vessel against the resistance offered by the valve and so the system the work interaction for the system is negative. So, we will try to now quantify this by deriving an expression. So, we look at a simple situation like this again a piston cylinder mechanism which contains a working substance and so this is our system the one shown in dashed line is our system. So, an external agent initially provides a resistance and the resistance is slowly reduced and the piston moves outward and the system then does work on the external agent may be on the piston also and again on the atmosphere by pushing the atmosphere aside. So, let us take an intermediate instant when instant when the pressure and volume of the system or P and V. So, the piston is now displaced upwards by an incremental distance dx then the work done by the system to accomplish this is given as delta W equal to force times distance move or familiar definition from mechanics. So, P times A times dx and A times dx may be realized or recognized as the change in volume of the system dV is the cross sectional area of the cylinder. So, for the entire process then we may integrate this expression from initial to the final state and so we get the well known familiar integral 1 to 2 P dV as the expression for displacement work. Remember, this is the expression for displacement work and if we actually use P V coordinates remember we said that in this case the number of different ways in which the energy of the system can be changed is 2. This is a simple system what we are looking at here is a simple system. So, the number of ways in which its energy can be total energy can be changed is through work and heat interactions. So, that means 2 properties have to be specified to fix the state of the system. If we choose these 2 properties to be pressure and volume like what we have done here then the nice thing is this integral may be recognized as the area under the process curve. So, if I know the process from 1 to 2 then integral 1 to 2 P dV is the area under the process curve. Of course, we have said or we have discussed the conditions that are required for us to be able to draw a process curve. Remember, we said that the state at every instant must be known which means that it must be a fully registered process only then we can connect all the intermediate states by a process curve like this. So, this actually then brings us to another point that we raised later in the context of partially resisted or unrestrained expansion. If you recall, we made the point that we raised the question whether we use the resisting pressure or the system pressure or the difference in pressures. So, erroneously students have this idea that it should be resisting pressure in some cases, system pressure in some other cases, the difference in pressure in some other cases and so on. Notice that the expression that we have derived here gives us no such choices. It is integral P dV where P is the pressure of the system. Then what about all these other cases? Remember, in the case of a partially resisted process, first of all we cannot partially resisted or unrestrained process, first of all we cannot depict the process in a process diagram at all. So, we will not be able to use this because the pressure at any intermediate instant is not known. So, when we say that at an intermediate instant, let the pressure and volume be P and V that implicitly assumes that pressure and volume are known, which will be the case only if it is a fully resisted process. So, this expression is applicable only for fully resisted processes. So, what it means is that in the case of a fully resisted process, this pressure and the pressure that the surroundings provide on this are almost the same. Remember, we said that there has to be only a small level of disequilibrium. When we lifted small weights one at a time, there was a small level of disequilibrium. That is what we are talking about here. Momentarily, the system pressure is slightly higher than or the surrounding pressure is slightly less than the system pressure. There is a small disequilibrium and so the piston moves up. So, when we say the piston moves up, that is what we mean. So, when the piston moves up by dx, that is what we mean. There is a small disequilibrium. So, there is really no other imbalance. So, the question of which pressure to use does not arise at all. At all instance, that is what we have. So, in fact, we will quantify this in a moment. So, let us summarize what we have said so far. Displacement work cannot be calculated unless the process curve is known. And we have already laid out the conditions that need to be satisfied before the process curve is known. The condition is that it must be a fully registered process. So, the process curve is known only for a fully registered or quasi-equilibrium process. This is the reason why displacement work cannot be calculated for an unrestrained or a partially restrained expansion. We actually cannot calculate it for an unrestrained or partially restrained expansion. But remember if we go back to the example that I showed earlier, let us look at that example. You may recall that even in that case we said that the work interaction was 0. Let us see whether we are contradicting ourselves. So, here we said that in this case the work interaction was equal to 0 because no part of the system boundary is deforming. Notice that we never attempted to calculate the displacement work here using integral PDV. We drew a system boundary and we said that since the system boundary does not deform with time, this work interaction is 0. Displacement work is 0 for this case because displacement work is associated with deformation of the boundary. In this case there is no deformation of the boundary. So, displacement work is 0. We never really tried to apply integral PDV because we had no need to apply integral PDV because all the boundaries are fixed. If on the other hand you had chosen in the case of the right side being evacuated, if on the other hand you had tried to choose something like this as your system, then notice that we will actually try to evaluate integral PDV. Then we say now I am going to use the resisting pressure which is 0 and all that. Notice that in this case integral PDV is not applicable because it is not a fully registered or quasi equilibrium process. So, we are not contradicting ourselves when we say W is 0 in the case of this unrestrained expansion. In this special case, we are able to define a system whose boundaries are fixed. The intermediate states in this case are still not known. So, we have to keep that in mind. So, in principle, displacement work cannot be calculated for an unrestrained or partially restrained expansion because integral PDV is not applicable since the process curve is not known and the process curve is not known because the intermediate states are not known. But in some cases, like what we just saw because the system boundary, the system may be defined in such a way that the system boundaries do not deform, we may say that displacement work is 0. There is no need for us to evaluate integral PDV. So, because it is a fully registered process, I can write this expression for the system pressure P. So, the system pressure P is due to the force exerted by the external agent. So, this is external agent and this is the pressure that is exerted by the atmosphere and this is the pressure due to the piston, the weight of the piston. So, we take the weight of the piston divided by cross sectional area and we get this. So, the pressure at any instant actually when there is mechanical equilibrium, it is a sum of the pressure that is due to the force of the external agent plus the atmospheric pressure plus the pressure due to the weight of the piston. So, that is why there is no question of resisting pressure or system pressure. They are both equal because it is a fully registered process. Now, if I substitute this into the expression for work integral PDV, I can actually get the expression to be like this. This one here is actually the work done against the external agent. So, the work done by the system consists of three components, three pieces. One is the work done against the agent, this is the work done against the atmosphere and pushing the atmosphere aside and this is the work done in raising the height of the piston in elevating the piston from datum or height z2, I am sorry from z1 to z2. Since the piston is being raised, its potential energy increases. So, the work done by the system to do this is actually positive as we as we can see. So, this consists of three components. Now, if I look at the work interaction for the piston, work interaction for the piston has this magnitude, but with the negative sign because its potential energy is being increased by someone else. So, the work interaction is negative, magnitude is the same. So, we can recognize each one of this as the negative of the work interaction against the agent, the atmosphere and the piston. This may be illustrated in the PV diagram like this. So, this rectangle here, notice that this is P atmosphere times V2 minus V1, this is minus W atmosphere. This is again a constant value minus W piston and this is equal to the term that we wrote down earlier. This is equal to MP times G times Z2 minus Z1 and this is equal to P atmosphere, sorry P atmosphere times V2 minus V1 and this is the work done against the agent. So, we can shade different parts of the work interaction and show them very nicely in the PV diagram. And this discussion also shows one of the points that we had raised earlier, but we said we will justify it later. You may recall that in some cases when we had part of the system boundary in the atmosphere or in a line in a line in which air was flowing when the pressure is constant, we said that the exact shape of the system boundary is immaterial. That also becomes clear from this expression. Notice that if the pressure is constant in a certain part of system boundary that is undergoing deformation, then this P actually comes out of the integral and so we have only P times integral 1 to 2 dV. So, that is why the shape of the boundary which is undergoing constant pressure deformation is immaterial. Only the change in volume matters. So, it was some volume before and at some other point the volume is something else. You may recall that we said we will draw in 1000 cc of air. So, that means the initial volume was 1000 cc and the final volume is 0. So, we can evaluate the work interaction that way. The other nice thing about this expression is that not only does it give the magnitude of the work correctly, it also gives it with the proper sign. If I choose the atmosphere as a system, then if I try to calculate work interaction for the atmosphere. So, let us say I use I try to use calculate work interaction for the atmosphere by defining the atmosphere as a system and I try to do integral 1 to 2 P dV for the atmosphere. Since the pressure is constant, I can take the pressure term outside. So, this is P atmosphere times V2 minus V1. Notice that in this case, V2 is the final volume of the atmosphere. V1 is the initial volume of the atmosphere. Since the atmosphere is being pushed aside, V2 for the atmosphere is less than V1. So, V2 for the atmosphere is less than V1. So, this term will automatically come out to be negative. So, if I try to calculate work interaction for the atmosphere alone by identifying the atmosphere as a system, this expression will give the magnitude correctly and the sign also correctly. So, this expression integral P dV can be used quite nicely once the system is defined, it gives both magnitude and sign correctly. So, when a part of the system boundary deforms at constant pressure, the magnitude of the corresponding displacement work is simply the product of pressure and the change in volume. And this is the reason why we said the exact shape of such a boundary is immaterial when the deformation is at constant pressure. The other important aspect, subtle aspect about this is that the pressure P in the integrand is the pressure in that part of the system boundary that is deforming during the process. If we have deformation taking place at different parts of the system boundary. In fact, if you remember, we looked at intake stroke of reciprocating compressor and we saw that different parts of the system boundary in that case were deforming at different pressures. So, in case they are deforming at different pressures, then we have to take the pressure at that part of the system that is undergoing deformation to calculate the work interaction for that part of the system. So, that is not equal to the system pressure because there is a valve in between. So, the part of the system that was outside was an atmospheric pressure, whereas the part of the system that was inside the cylinder is usually at a slightly negative pressure not equal to atmospheric pressure. So, they can be at different pressures because of the presence of the valve and so when we apply this expression, we have to make sure that we take the pressure at that part of the system which is undergoing deformation and then add them all up an algebraic sense to get the combined effect of the system with the surroundings. So, what we do is, so if multiple parts of the system boundary undergo deformation, then the total displacement work is simply the sum of the individual displacement works and we use the subscript B here to denote the boundary. So, in the case of the reciprocating compressor in that example, this will be P atmosphere when we calculate the work interaction for that part which is outside the cylinder and it will be P cylinder when we calculate the work interaction for that part of the system which is inside the cylinder. And then we do the algebraic sum because this expression gives the magnitude along with the correct sign, we can simply do an algebraic sum of all these contributions to get the net or combined work interaction of the system with the surroundings. Here, we again emphasize or try to reiterate the fact that the magnitude and sign of the work interaction depends on where we draw the system boundary. So, here we are looking at the same arrangement but this was our original system. Now, we have drawn our system with the piston also being inside the system. So, when we do it, when we define the system like this, the integral PDV for this would become something like this because now the piston is part of the system pressure exerted by the piston now goes inside. So, notice that this becomes P minus P piston times dV whereas this is P piston plus P atmosphere plus P agent. So, the magnitude of work interaction between these two will be different because the systems are different. So, then the question arises. So, if the magnitude is different, what happened to the difference? Where did it go? So, eventually when we write down first law of thermodynamics, we will take that into account and it will be accounted for properly. So, depending on how you draw the system, the work interaction will be different, heat interaction will also be different. Notice that the total energy content of the system will also be different because of the way the system boundary is. So, any difference in one is taken care of or accounted for in the other part. So, when we finally put all these things together and write down first law, you will see that everything is accounted for properly. But for now, what is important to keep in mind is that work in thermodynamics is an interaction or occurs by virtue of an interaction of the system with the surroundings and the magnitude and sign of the work interaction depends upon the system boundary where we draw the system boundary. So, what we will do in the next few lectures is actually to do a few work examples and illustrate these concepts application of PDV for different situations. We will carry out analysis for different situations, evaluate the work done. Particularly, what we will do is try to see how the integral PDV can be applied or should be applied properly in all these cases.