 So, in this lecture, we are going to develop expressions for the change in entropy of a fluid across a control volume and develop an expression for the rate of entropy generation in a control volume due to internal irreversibilities. And we will also develop an expression for rate of entropy generation in the universe as a result of this process. Now the starting point for this development is something that we discussed in the previous course. So, in the previous course, we derived this expression for the change in entropy of a system as it undergoes a process. So, basically delta S which is the change in entropy of the system is the sum of two quantities. One or the first quantity is integral 1 to 2 delta q over t. So, this is the entropy transfer term. And the second one is sigma int which is the entropy generation due to internal irreversibilities that is very important. Remember, we are talking about entropy change of a system as it undergoes a process from state 1 to state 2. So, that means the entropy generation is only due to internal irreversibilities. So, basically what this expression says is that entropy change of a system is the sum of entropy change due to transfer of entropy from the surroundings and entropy generation due to internal irreversibilities. Now entropy transfer can be greater than equal to or less than 0 depending upon the direction of heat transfer. So, if we supply heat to the system, then we are supplying entropy, this will cause the entropy of the system to increase. And if you are removing heat from the system, then of course, this will cause the entropy of the system to decrease. If the system is let us say adiabatic, then q is 0 and of course, so there is no entropy transfer to the system. Now, entropy generation is always positive or equal to 0. If the process is free of internal irreversibilities, then sigma int is equal to 0. Otherwise, as a result of internal irreversibilities such as friction, dissipation, mixing and so on, sigma int is positive. So, the net change in entropy of the system is due to entropy transfer and entropy generation due to internal irreversibilities. Now, this expression as it stands is not useful for calculating entropy change of a system, numerical value for entropy change of a system because there is no expression for calculating sigma int. Sigma int is a notion of internal irreversibility. The only thing we know about sigma int is that it is positive or equal to 0. Even this may be very difficult to compute in actual scenarios. So, the usefulness of this expression lies in being able to tell the change in the magnitude, not magnitude, the change in entropy of the system. In other words, from this expression, we will be able to tell whether the entropy of the system is going to increase, decrease or remain the same. Therein lies the usefulness of this expression. The actual calculation of entropy change of a system as we showed in the previous course has to be done using either tedious relations or ideal gases or using tabulated values of entropy for steam and R134A. It was also demonstrated in the previous course that the change in entropy of the universe is equal to change in entropy of the system plus change in entropy of the surroundings because the universe comprises of the system plus the surroundings. So, the change in entropy of the universe is due to change in entropy of the system plus change in entropy of the surroundings and this is equal to sigma. Sigma represents the total entropy generated in the universe due to both internal as well as externally irreversibilities. Note that this does not have a subscript int or ext which would indicate internal or external irreversibilities. This is the entropy generated due to internal and external irreversibilities as the system executes a process. And as we can imagine just like sigma int, sigma is also greater than or equal to 0. It is equal to 0 if there are no internal or external irreversibilities and it is positive in case any one of these irreversibilities is present. I urge you to consult the appropriate lecture in the previous course because there is a detailed discussion of how we actually show this to be greater than or equal to for all possible combinations of internal and external irreversibilities. Now, this quantity as we can imagine is a very, very important quantity because in some sense it actually is a very good performance metric of the process. For instance, if a system executes a process which results in no change in entropy of the universe then one must assume that that is the best possible process. On the other hand, the higher the amount of entropy generated in the universe due to irreversibilities, the less desirable the processes. So, this is an indication or is a performance metric of processes. And in fact, this will be the basis for us to develop an efficiency for any process in the general case which is what we are going to do in the next module in this course. So, we will define what is called a second law efficiency with this idea in mind that if the entropy generated in the universe is 0, it is an ideal process. Otherwise, the more the entropy generated, the less desirable the processes. So, these two are very important. So, what we will do now is try to develop similar expressions for a control volume, one like this for entropy change across a control volume, one like this for the rate of entropy generation in the universe due to a steady flow process, remember or due to a flow process. Remember, since we are dealing with the flow process, we no longer say just entropy generation, we have to talk about rate of entropy generation. So, sigma dot int or sigma dot is what we are interested in and we will try to develop expressions similar to this for the case of a control volume or a flow process also. The starting point for this is the same illustration that we used in the previous course when we derived an expression for the first law applied to a control volume, when we derived the unsteady flow energy equation. So, we have a device which looks like this. So, this is the rectangular box is a device, it has for the sake of simplicity and without any loss of generality, one inlet and one outlet. Fluid enters the device at the rate of mi dot say kg per second and leaves at the rate of me dot me dot kg per second mi dot need not be equal to me dot in the general case. In a particular case, for example, for a steady flow process, we may take mi dot to be equal to me dot, but for the general development, we do not assume mi dot to be equal to me dot. Now, heat is exchanged with the device at the rate of q dot, let us say watts and external work or power is supplied or removed from the device at the rate of Wx dot watts. Now, as we did before, we simply take the boundary of the device to be our control volume. That is the simplest possible definition and that is what we will do. So, this is our control volume. So, what we want to do now is develop an expression for entropy change across the control volume. So, to this end, what we do is in fact, basically what we want to do is start with this expression because this is entropy change for a system. This is entropy change for a system. So, we will start with this expression and then modify it so that we end up with an expression that involves this control volume. Now, for that purpose, we need to identify an appropriate system. So, let us say that we freeze the device at a time instant t. So, at time instant t, we freeze the device and as you can see at this instant, some amount of mass is present inside the control volume and amount of mass equal to delta Mi is just about to enter the control volume. So, we already have some mass inside the control volume and an amount of mass equal to delta Mi is about to enter the control volume. Now, we supply delta Q to the system, an amount of heat. There is a heat interaction to the magnitude of delta Q to the device and there is a work interaction to the magnitude of delta Wx to the device. So, now, we allow the flow process to take place and at time t plus delta t, the system looks like this. Again, there is a certain amount of mass inside the device and now an amount of mass equal to delta Mi is just about to leave the device. Earlier, we had a certain amount of mass in the device and an amount of mass delta Mi was just about to enter the device. Now, we have an amount of mass, certain amount in the device different from what it was before and an amount delta Mi is just about to leave the device. Notice that the system that we are looking at is shown in gray and that always contains the same amount of mass. So, this is the system at time t, this is the system at time t plus delta t and it contains the same amount of mass between time t and time t plus delta t. Now, we apply this expression, this expression for the system that we have identified. So, delta S for the system. Now, remember, we are talking about an infinitesimal time interval, infinitesimally small time interval. So, we use dS instead of delta S. So, the change in entropy of the system dS between time t plus delta t and time t is equal to S at t plus delta t for the whole system minus S at t. Now, at time t plus delta t, the system is comprised of two parts. One is whatever is inside the device or control volume. Remember, we identified the device to be our control volume. So, whatever is inside the device or control volume plus this small bit which is at the outlet. So, we may expand this and write this as equal to this. So, this is nothing but entropy of the mass that is contained in the control volume at t plus delta t plus entropy of the mass which is just about to leave that is nothing but Se times delta Me. But this is the specific entropy of the mass that is about to leave the control volume. So, we take the specific entropy, multiply that by the mass and we get I am sorry multiply that by the mass and we get the entropy of the mass that is about to leave the control volume. So, this is S t plus delta t and in the same manner this is S t. So, this is the entropy of the mass that is within the control volume at time t and at time t there is a part here which is about to enter the control volume. So, we write the entropy of that part of the system as Si times delta Mi. Now, we also know from this expression, remember this is an expression for entropy change for a finite process. So, if we want entropy change for an infinitesimally short duration process then we may write this in differential form like this. So, we say dS equal to delta Q over T plus delta sigma Int. So, this is for a process that takes place between two at distinctly different states 1 and 2 and this is entropy change of a system for an infinitesimal change in the state of the system. So, this is what we would write. Now, for the system at hand, so this is the system that we are talking about remember SCV is what we I am sorry. So, what we are looking at is change infinitesimal change in entropy of the system and if you look at the system here notice that the heat interaction occurs over the entire surface of the device. So, normally we will simply write dS equal to delta Q over T plus delta sigma Int. Now, in this case because the heat transfer is taking place over the entire device and the device need not be at constant temperature between inlet and outlet. So, this part of the device may be at some temperature, this part may be at some other temperature. So, the amount of heat transferred here will be different from the amount of heat transferred here. So, the net heat interaction of the device with the surroundings then has to be evaluated as a surface integral like this. So, this is the net heat transfer across the device between time T plus delta T and time T. Normally for a system we would have simply left this as delta Q over T because the entire system would have been at the same temperature. But here because we are looking at a device where the temperature may change from inlet to outlet we have to evaluate this across the surface of the device. So, this is the control surface which is the surface of the control volume and T B is the temperature at each part of the control surface. So, basically what we are saying here is, so temperature here is T B some value. Say let us say T B it could be say at some location I this could be T B at some location J and so on. So, the temperature varies across the surface of the control volume. So, we evaluate this as an integral and delta sigma int of course is left as it is because that is only internal irreversibilities inside the device. So, this is the most important thing that you should keep in mind that we need to evaluate this as an integral across the entire control surface. So, now putting everything together we end up with an expression like this and if we divide both sides by delta T and allow delta T to become to approach 0, then we finally end up with an expression like this DSCV DT. This is the rate of change of entropy within the control volume is equal to Mi dot Si rate at which entropy is brought inside the control volume minus Me dot Se rate at which entropy exists the control volume plus this term which is rate at which is entropy is transferred to the control volume and sigma dot int rate at which entropy is generated within the control volume as a result of internal irreversibilities. So, basically what this is is that this is an entropy balance equation. Now in case we are looking at steady flow devices then this term would go to 0 and Mi dot would be equal to Me dot. So, we may simplify this and then write this like this for a steady flow situation. Mi M dot times Si minus Se plus integral across control surface delta Q dot over TB is equal to sigma dot int that is rate of entropy generation due to internal irreversibilities. Now notice that this expression again is not very easy to evaluate because we do not have a means of calculating sigma int. We do not have a closed form expression for calculating sigma int. This integral will also be difficult to calculate in many, many practical situations unless delta Q dot is 0 meaning the device is adiabatic or the system is isothermal in which case the integral simplifies and we may take the T outside and just evaluate this as the total heat divided by temperature. So, it simplifies if it is either adiabatic or if it is isothermal. Otherwise this expression again is not very easy to use. However, it has its utility value and that is what we will see now through a couple of examples.