 So, in the previous course, we it was mentioned that the reversible isothermal and the reversible adiabatic processes are very special processes in thermodynamics. And based on this fact, we also put together the corno cycle out of these two processes alone. The reversible comprises entirely of the reversible isothermal and the reversible adiabatic process. And it may also be recalled that the corno cycle is executed by a fixed quantity of mass. So, basically it is a thermodynamic system which executes a non-flow process. And the work interaction for such a process you may recall would simply be W over m is equal to integral 1 to 2 P dv. And this in PV coordinates, this integral is nothing but the area under the process curve for a system executing a non-flow process. What we are going to do now is develop a similar expression for a control volume in which reversible isothermal or reversible adiabatic process is being executed because we will have occasion to use this as we go along. So, this is a very important derivation. And the most important thing here is that the process that we are considering, both the processes that we are considering are internally reversible. So, reversible isothermal and reversible isentropic, internally reversible. So, in the absence of any internal irreversibilities, we may write the entropy balance equation like this, again for a steady flow situation. So, you may recall that the entropy balance equation looked like this for a steady flow situation. So, in the absence of any internal irreversibilities, this becomes 0. So, we end up with the expression that we wrote down just now. So, this is the expression that we end up with. We may write this as minus m dot integral 1 to 2 ds. Now, steady flow energy equation applied to this control volume reads like this WX dot equal to Q dot plus m dot h1 minus h2 after neglecting KE and PE changes. h1 minus h2 in the same manner as s1 minus s2 may be written as integral 1 to 2 dh. And dh itself, we may recall, may be written like this from the TDS relationship. So, dh itself is equal to TDS minus VDP. So, we may finally write WX dot equal to Q dot minus m dot integral 1 to 2 TDS plus VDP. Now, let us simplify this expression in term for a reversible isentropic process, I am sorry for a reversible adiabatic process and for a reversible isothermal process. So, for a control volume executing an isentropic process, Q dot is equal to 0, ds is also equal to 0. So, this expression simplifies like this. So, for this case Q dot is 0, ds is also 0. So, this expression simplifies to WX dot equal to minus m dot times VDP. So, it simplifies to WX dot over m dot equal to minus integral 1 to 2 VDP. So, for a compression process, you can see that this is the process curve for an isentropic process going from 1 to 2. So, this is a compression process and this one, so let me write it explicitly here. So, this is a compression process going from 1 to 2 is like this and this is an expansion process going from 1 to 2 is like this. So, here we have illustrated an isentropic process for an ideal gas, which is why we have written PV raise to 1.4 equal to constant. So, notice that this area under this curve is equal to minus integral VDP and similarly here. So, this quantity here, so this quantity here WX dot over m dot may be interpreted as specific power, specific power or because the units do not have time dependence on them, units for this quantity, we also refer to it as work interaction. So, specific power or work interaction is just minus 1 to 2 integral VDP and that is nothing but the area that is indicated here for an isentropic process. I am sorry, this is, this area is not correct, so let us just correct it. So, this is the area that we are talking about for an isentropic process. Now, let us just illustrate the same thing for a non-flow process. So, for a system which executes a non-flow process between the same two states, executes a non-flow isentropic process between the same two states, the work interaction for that case W dot over m dot would have been this here. Similarly, so it would have been this area for a steady flow process, the area is as shown here. So, that is very important to keep in mind. Now, let us look at the next situation, reversible isothermal process. So, in case the system executes a reversible isothermal process, then this integral may be simplified. This integral may be simplified quite nicely because temperature is constant, we may take this outside and write it like this. So, we may take the temperature outside and that is a constant and this integral C s, integral C s delta Q dot is the, is nothing but the total heat transferred, net heat transferred to the control volume. So, integral C s delta Q dot is the heat transferred across the entire control surface. So, that is nothing but the net heat transferred to the control volume. So, we may write it as equal to Q dot. So, we get a nice expression like this in this case. We may take the temperature inside the integral because it is a constant. So, W x dot over m dot is equal to Q dot over m dot minus this integral and Q dot over m dot itself may be replaced using this expression. So, then you can see that these two cancel out. So, we end up surprisingly with the same expression as we did for an isentropic flow process. The specific power is equal to minus integral 1 to 2 V dp and the process curve for the reversible isothermal process is shown here using the chain line. So, there is a t equal to constant line. So, let us just see if we can. So, the area that we are talking about in this case is equal to this. And had this been a non-flow process executed by a system between the same states, then the area would have been and similarly here also. The expression is the same, area enclosed is the same in the sense that the area enclosed is given as shown here. Notice that in case of a compression process where we are putting in work or power, the power required for compressing between the same two pressures. Notice that the pressure at 2S and pressure at 1, I mean in each case we are starting from the same initial state and compressing to the same final state here or in this case we are starting from the same initial state and expanding to the same pressure between 2S and 2. So, for both the reversible adiabatic process and reversible isothermal process, we are compressing across the same or expanding across the same pressure in these two illustrations. Notice that the power required is less for the reversible isothermal process for compression process and the power produced by a reversible isothermal process is more in the case of an expansion process. So, this shows that the reversible isothermal process is better when it comes to power absorption for a compression process and power generation for an expansion process. So, it seems to be the ideal process which is somewhat surprising because if you may recall when we defined isentropic efficiency, we said that the isentropic process was the ideal process against which the actual process has to be compared. But this illustration very nicely shows that the reversible isothermal process is ideal in terms of power. So, we will incorporate these sorts of variations from what we studied in the previous course to have better metrics defined for actual processes. So, some of the subtle differences come about because we now have slightly more comprehensive analysis taking into account internal and external reversible. So, what we will do in the next lecture is to start our discussion of exergy. And then the most important thing that will come out of this module is the definition of second law efficiency. And we will compare this with the isentropic efficiency that we defined in the previous course and see how much more general and useful the second law efficiency is and indeed the notion of exergy.