 So, in this lecture we will derive a form of first law of thermodynamics that is suitable for analysis of a control volume. So, we discussed in the beginning of this course the need for using control volumes and the difference between system under control volume. So, I suggest that you know you review the material related to this and make yourself familiar with the situations where a control volume analysis is required. Generally, control volume analysis is required in situations where there is continuous flow of material in and out of a device. So, for these cases it is better to have a form of first law that is very convenient for analysis. It is not a different law, it is just that it is in a form where we can use it in a very convenient manner. Let us see what we mean by this. So, here we have a device and mass flows continuously into the device at the rate of Mi dot and mass flows continuously at the rate of Me dot. So, begin with Mi dot need not be equal to Me dot because we want to develop a very general framework where we can deal with all possible situations. So, for instance if we are trying to let us say fill a vessel from a tank. So, this is an example that we have seen before. So, here is a tank and we are trying to fill the vessel with fluid from the line or here is a tank which we are trying to fill with fluid from the line. Then in this case some mass flows into the vessel. So, our control volume in this case may look something like this. We saw this also before. So, mass flows into the device, but nothing flows out of the device. So, for this case Mi dot is not 0, but Me dot is equal to 0. So, there will be continuous accumulation of mass in the vessel or container until we stop the process and it is also an unsteady process. Now, let us say that you know we have the opposite situation where we actually are trying to empty a vessel. So, initially let us say this vessel contains air or steam at a high pressure and temperature and we are trying to empty the vessel into the atmosphere. In such a case there is continuous outflow of mass from the vessel, but there is no inflow of mass. So, in this case Mi dot is equal to 0, but Me dot is not equal to 0. So, we continue this process until the vessel until the pressure in the vessel falls to a certain value. Typically that is what we would do or until it reaches the ambient pressure beyond which flow cannot be sustained by itself. So, those would be the situation that we would consider for a case like this. And again this is also an unsteady process, but one for which Mi dot is equal to 0, but Me dot is not equal to 0. So, in the general case as I said we want to make sure that we can handle all these cases in our analysis. So, Mi dot need not be equal to Me dot and heat is exchanged by the device with the surroundings at the rate of Q dot kilowatts. Notice that here we no longer say heat is exchanged with the ambient at an amount or equal to some number of kilojoules. We do not use kilojoules anymore because this is a continuous flow device. So, we say that heat is exchanged with the surroundings at a rate of Q dot kilowatts for instance or watts per second basis. Similarly, there is work interaction, external work interaction with the surroundings at the rate of WX dot watts or kilowatts as the case may be. In this case in contrast to a system approach as we said before the control volume is easy to define. We simply take the boundary of the device and define that as a control volume. So, that is very straightforward and easy to do. So, basically what we want from control volume analysis is we want to relate the, so given the inlet properties and the Q dot or WX dot we want to relate the exit properties to these quantities. That is simply what we want or given the inlet and the exit properties and maybe one of these two quantities we want to evaluate the other one. So, that is the sort of analysis we have in mind for a control volume. Notice that this is not a system. So, the control volume is not a system clearly because mass flows in and out and this may not always contain the same amount of mass. But our analysis requires us to relate the incoming quantities and the outgoing quantities in a way where we can evaluate one of this. So, we want to define the control volume and then do the analysis in this way in a straightforward manner. However, we will have to use first law for a system, identify an appropriate system for this, apply first law for a system and then simplify it to suit our requirements for a control volume analysis. For example, this is a system that is appropriate for this analysis. This is how the system works at time instant T. So, at time instant T, we have a certain amount of mass in the device or control volume. Remember, the control volume is just the device itself. So, this is also the control volume. So, at time instant T, we have a certain amount of mass in the device or control volume and a certain amount of mass delta Mi is just about to enter the device. Now, an amount of heat equal to delta Q and an amount of work equal to delta Wx remember or you should note the fact that this is at an instant. So, both these will be in units of joules. So, now there is an exchange of delta Q joules of heat between the system and the surroundings and an exchange of delta Wx joules of work between the system and the surroundings. As a result of which the system changes and becomes something like this. So, this is the system at time T plus delta T. So, this is the control volume here. So, at time T plus delta T, a certain amount of mass is enclosed in the device which need not be the same as the mass at time T. It can be different. So, we have a certain amount of mass in the device or control volume and an amount of mass equal to delta Mi is just about to leave this device at time T plus delta T. But notice that since you are saying this is a system, the amount of mass in the device at time T plus delta T plus delta Mi should be equal to the amount of mass in the device at time T and plus the amount of mass that is going to enter at time T. So, the mass in the system remains the same. So, if you apply first law to this system, we get De equal to De is the change in total energy. This is total energy in the system at T plus delta T minus total energy in the system at T that is equal to delta Q minus delta W. Notice that we are not writing this as delta Wx. Delta Wx is the work interaction between this device and surroundings during this interval, but that is not necessarily equal to the total work interaction of the system because there is displacement work that needs to be accounted for. So, delta W is not equal to delta Wx. Now, the total energy of the system at time T. So, here is a system at time T. So, the total energy of the system at time T is the sum of the total energy contained within the control volume plus the total energy of the mass that is just about to enter the device. So, we can write this as that sum. So, this is the total energy in the control volume plus the total energy that is carried in by the mass that is entering. So, this is lower case E. So, that denotes specific total energy of the incoming stream. So, if I multiply that by the mass of the mass that is about to enter, then I get the total energy that is brought in through that inlet or that is in that part of the system that is a better terminology. So, that is the total energy. Now, the total energy of the system at time T plus delta T is nothing but the total energy of the system or which is enclosed in the device which is nothing but the control volume plus the total energy in this part of the system. So, the total energy in the device or control volume is simply just E C V at T plus delta T and this is specific total energy of the system at the exit. So, multiplying by the mass we get the total energy that is in that part of the system. And of course, you must remember that total energy is nothing but internal energy plus kinetic energy plus potential energy. So, this quantity which is the specific quantity would be the total energy divided by M. So, this would be nothing but specific internal energy plus specific kinetic energy plus specific potential energy. So, that is why this is denoted in denoted using lowercase e. We will take a look at this as we go along. Now, let us turn to the work interaction term delta W. Notice that the work interaction term is the sum of the external work delta W X which could be shaft work or electrical work or any other form of work interaction that this device has in the surroundings. Notice that if you look at this system itself. So, there is deformation in this part of the system boundary. Let me just erase this. So, there is a deformation. So, there is deformation in this part of the system boundary and there is also deformation in this part of the system boundary. In fact, in this part of the system boundary, we have a finite volume at time t and then at time t plus delta t, this becomes 0. And in this part of the system boundary, there is no material at the outlet and at time t plus delta t, we have a certain amount of material at the outlet. So, let us look at this. So, there is a deformation work in these two parts of the system boundary. There is no deformation work in the control volume itself because it has fixed boundaries. You may recall that we mentioned this in our earlier discussion where we said that it is preferable always to have control volumes with fixed boundaries, no deformation. So, the control volume does not deform. There is deformation work in this part of the system boundary and this part of the system boundary while the control surface itself remains fixed. So, delta W is the sum of delta W X which is the work interaction plus displacement work at the inlet plus displacement work at the exit. So, we can easily assume that the pressure at the inlet remains constant during this time interval delta t. That is a reasonable assumption to make that the pressure remains constant in the time interval delta t not throughout. Remember, we are applying this between t plus delta t and t. So, we can easily and safely say that the pressure at the inlet remains at p i during this interval delta t. That is all we are assuming here. So, delta V, the change in volume in this part of the system between time t and t plus delta t may be evaluated like this. It is 0 at t plus delta t and it is equal to this where V i is the specific volume at time t. Similarly, here also we assume the pressure at the exit to remain constant during this time interval delta t because later on we are going to look at examples where the pressure at the inlet or outlet changes continuously. However, this is the instantaneous pressure during this time interval delta t. So, that can be taken to be equal to this. So, if you go ahead and expand this, this is what we get for the total work interaction delta W. Notice that this and this are very special. So, this is the amount of work that is required to push the fluid into and out of the device. So, if you look at this, in this case, the surroundings actually have to do a certain amount of work to push this mass into the device. That is why this term appears with the negative sign in our delta W expression. And if you look at this, the system has to do to push this mass outside. That is why it appears with a positive sign in our expression for delta W. So, notice that this appears with the positive sign and this appears with the negative sign. So, this is the amount of work that is required to push the mass in and out of the device. And hence this term is called the flow work. So, this term is called the flow work for that reason. So, if we combine the above expressions, we end up with something like this which may be rearranged to give. So, I am going to take this and this to the other side. So, if I take these two to the other side, I get something like this. So, I can actually combine this like this and by using our former, by using the definition of EI which we wrote down earlier like this. So, if I combine this specific internal energy term with this PI VI term, I get specific enthalpy. So, this may be written as change in energy during this time interval in the control volume is due to delta Q, heat interaction, work interaction, flow work plus change in enthalpy, change in properties of the fluid between the at the inlet and the exit. If I divide both sides of this expression by delta T and then allow delta T to go to 0, I can do something like this. So, I can divide this by delta T, I can divide this by delta T and divide this by delta T. Note that we have already said that the quantity within the bracket is the instantaneous value. So, it does not change. So, I can divide this by delta T, I can divide this by delta T and then allow delta T to go to 0, then I get this equation DECV DT is equal to Q dot. Notice that now the delta Q has now become Q dot. So, that has now become rate at which heat is supplied or removed from the control volume or device. This is the rate at which work is supplied or removed from the control volume or device. This is the mass flow rate coming in, this is the mass flow rate that is leaving the device or control volume. This equation is known as unsteady flow energy equation, unsteady because we have quantities here which can change with time. So, the total energy contained within the control volume can change with time. Other quantities may also change with time. So, this is a very, very general equation that can be used for doing control volume analysis whether there is a steady flow device or unsteady filling or emptying and so on is a very, very general equation. Now, you must bear in mind that we made a point regarding this being a system which is that the mass that is about to leave plus the mass that is here at time T plus delta T should be equal to the mass that is just about to enter or mass that is in this part of the system plus the mass that is already in the control volume at time T. So, this constraint also needs to be imposed. So, we have to make sure that this constraint is satisfied and that is what we have done here. So, the mass within the control volume at T plus delta T plus the mass in the exit portion of the device must be equal to the mass that is already within the control volume at time T plus the mass which is in the inlet portion of the device. So, then if you rearrange this, take this to the other side and bring this to this side, we may write this again as rate equation which is nothing but rate of change of mass within the control volume is equal to the rate at which mass enters minus the rate at which the mass leaves. So, if the mass entering is more than the mass leaving, then the rate of change of mass within the control volume is positive. So, the mass continues to increase. If the rate at which mass leaves is greater than the rate at which mass enters, then there is overall depletion of mass in the control volume. So, DMCV DT is negative. So, these are the two general equations which can be used for any situation for analyzing a control volume. What we will do next is simplify this to steady flow cases. A majority of applications mechanical engineering that you will encounter or steady flow situation. So, we can simplify this further for steady flow situations. We will work out examples involving steady flow energy equation. Later on, we will also demonstrate the unsteady flow analysis of certain examples. Now, let us say that we have a fan in the room and we turn the fan on in the morning. Initially, the fan, let us say is somewhat cold, we turn the fan on. So, the fan starts spinning and it sort of warms up and after some time, its temperature does not change anymore. So, when you look at the steady state phase, the temperature of the fan does not change anymore. So, if I take the fan as the control volume, the energy, total energy inside the control volume, which is nothing but the internal energy of the fan plus its potential energy plus its kinetic energy. So, its internal energy does not change anymore because its temperature no longer changes. So, there is no change in potential or kinetic energy, it is simply spinning like this, which means that the total energy of the control volume remains constant with time, once it attains a steady state. So, D C V D T is equal to 0. Also, the rate at which mass enters the control volume is the same rate at which mass leaves through the blades around the control volume. So, we can say that this is also, if I take the hub alone as the control volume, then obviously no mass enters or leaves the hub. So, that is always equal to 0. If I draw a bigger control volume, then I can show that the contents of the control volume have attained a steady state and the rate at which mass enters is the same as the rate at which mass leaves the control volume. So, that is the steady state situation. So, once we say that steady state is attained, this should become equal to 0 and this should also become equal to 0. That is very important, note that both time derivatives must become equal to 0 and none of the other quantities in the right hand side of this equation may change. So, that is steady state analysis. So, once these conditions are satisfied, I may drop the time derivative and write this equation like this. Notice that I have made use of the fact here that with D m C v D t equal to 0 m i dot equal to m e dot which may be written as being equal to m dot. That is what I have done here. In case we have multiple inlets and outlets to this control volume. So, for example, I have a control volume like this. So, I have one inlet here, one inlet here, one outlet here and one outlet here, then the steady flow energy equation may be modified to look like this. So, overall, so I have a heat interaction q dot and work interaction W x dot. Then I may write the overall energy equation like this and the mass conservation equation looks like this. This is the steady flow energy equation for multiple inlets and outlets and the overall mass conservation looks like this. So, I may have 1 kg per second entering through this and 2 kg per second entering through this. So, I may have 0.5 kg per second going out here and 2.5 kg per second going out here. So, notice that the mass flow rates may be different in each one of this, but the total mass flow rate 2.5 plus 0.53, 2 plus 1 3 should be satisfied because it is operating at steady state. So, that means the mass conservation equation has to be written like this in the case of multiple inlets and outlets. So, now what we are going to do is we have derived the first law that is suitable for a control volume analysis. What we are going to do now is use this equation to carry out analysis of control volumes. First, we will do steady state analysis, then we will do unsteady analysis.