 Welcome back. So, now we will discuss first law with the open system. So, by looking at the last snippet, you must have already realized that what we are doing is taking the control volume and adding the inlet and exit plugs and analyzing a closed system. And that is what we will continue doing. We will analyze a closed system. So, let me draw our open system and the inlet and exit plugs again. And this time I will include far more terms because once we come to open systems, we have to think about the heat transfer and work transfer rates. So, let me just draw an open system. So, this is our control volume here which is not shaded. And the plug on the left is our inlet plug. We have already mentioned it. I will not write it again. And the plug on the right is our exit plug and this is how we considered our system earlier. Now, this time we will also show the heat transfer and work transfer into the control volume. So, let me show this is q dot into delta t. So, q dot is the heat transfer rate and q dot delta t is the overall heat transfer that takes in a time delta t. And that is what we require to analyze our process. So, we are trying to analyze the process that takes place from time t to t plus delta t. So, in this time the system sees a heat transfer equal to q dot delta t as long as delta t is small, the q dot over that small time period can be considered constant and hence the overall heat transfer rate is q dot delta t. Similarly, now we will write an expression for the work terms. So, by our regular time convention, I will just draw the work as an arrow pointing outside like this. So, what we have drawn here are the three different work components. One is w dot s delta t. We have already mentioned that this is the regular work with the system or the control volume does on the surroundings or the surroundings do on the control volume which includes expansion work, stirrer work, electrical work and all other work that you can think of except the work done in pushing in the fluid at the inlet and pushing out the fluid at the exit. Those are denoted here by w inlet on the left and w exit on the right. So, this is the overall interactions of the system with the surrounding and the system overall has energy e at time t and energy e at time t plus delta t and the difference between these two we will call it as delta e. So, this when I write system here, this is the closed system I am considering. Similarly, we will have energy e for the control volume at time t and energy e for the control volume at time t plus delta t. So, these are the energy that we consider. These are the work transfer and heat transfer that we consider and now we are ready to write our equation and we will write our first law for the closed system and then we will see how by including all the terms that we can think of will come up with the first law for the open system. So, we now write the first law for the closed system delta e is q minus w. So, what is delta e? We have already mentioned it delta e. So, let me be clear this is the closed system. So, for the closed system and delta e for the closed system is just e system e plus delta t minus e of the system. Now, what is e of the system at time t and what is e of the system at time plus delta t? The energy of the system at time t is nothing but the energy of the control volume at time t plus the energy of the inlet mass plus because that is the mass at time t and it is the same mass at time t plus delta t, but the volume occupied is different. So, e of the system at time t sorry I have written system here, here I have written this is the same subscript. So, e system at time t is nothing but e control volume at time t plus energy of the inlet plug. What is the energy of the inlet plug? It is nothing but the mass contained in the inlet plug multiplied by its specific energy. So, we had a specific energy that is energy per unit mass we use a lower case letter for it small e. So, it is m dot i delta t which is the total mass in the inlet plug multiplied by e i and we have said over a small time delta t the energy of or the specific energy of this inlet plug is fixed and we were able to define it that is what we had mentioned initially that there is locally some kind of an equilibrium and we know exactly what the energy or the entropy or density at the inlet and exit. So, this is what we have done we have now written down an expression for energy of the inlet plug. So, this can of course, also be written as rho i a i e i delta t. We can of course, have energy of the exit plug similarly is rho e a e v e delta t into e e where e e is the specific energy at the exit side. So, overall we will have delta e of the system is equal to e c v e plus delta t plus e exit plug minus e c v at t plus e e delta t. Because this is e system at t plus delta t and this is e plus delta t. So, now let me just rewrite this whole expression. So, what we have here is delta e for the closed system. So, as I said when I use system it is the closed system. This is equal to energy of the control volume at time t plus delta t minus energy of the control volume at time t minus the energy of the inlet plug and added to the energy of the exit plug. So, this is just rewriting our previous equation. So, now we will consider expressions for the heat transfer and the work transfer. So, in our expression q during the process that is a process which goes from t to t plus delta t is nothing but q dot times delta t. What is w dot? We have said that this made up of three components. One is w dot s and then there is the work at the inlet and exit. So, in this expression w e is the work at the exit and w i is work at the inlet. And we will now write down expressions for w e and w i. So, let me start with an expression for w i. This can be written as just an expression as f times delta x or p times delta v. Basically, it can be looked upon as some kind of an expansion work or a work where a force is moving a distance delta x. So, this is how we write it. What is w i? It is nothing but the force at the inlet which is p i into a i multiplied by the distance level which is nothing but v i times delta t. So, this is the work done at the inlet. So, you can consider this as some force and this as some distance or you can consider this as pressure and this as some delta v. So, this is how you can consider the work at the inlet. And what we will do is we will realize that this is really work done on the system. We will put a negative sign here and what we can have is w e is p e a e v e delta t. Now, the next step that we do is substitute for a i v i by using our expression for m dot and we realize that we had m dot i is nothing but rho i a i v i. Hence, a i v i is nothing but m dot i upon rho i is nothing but m dot i into specific volume at the inlet. So, w i is just minus p i m dot i v i delta. So, we will continue further with our equations now. So, now we have an expression for the inlet and exit works. So, we already had derived in the previous sheet an expression for the inlet work, which was in terms of the mass flow rate at the inlet multiplied by the pressure multiplied by the specific volume into delta t. Similarly, we can derive an expression for the exit work in terms of the mass flow rate at the exit multiplied by the pressure at the exit multiplied by the specific volume and delta t. So, this overall is our expression for w e and w i. Overall, now we can get back to our first law for the closed system, where we have an expression for every term that is the energy the delta e for the system and the w for the system and the q for the system. And hence, I will write the following expression. On the left side, we have an expression for delta e, which is this we had derived it earlier. This is nothing but q and this is nothing but w, where we have three components of w. One is w dot s and delta t, one is work at the exit and one is work at the inlet. So, we have in essence written an expression for the first law for the closed system. And now, we will just transpose the terms here and ensure that we write the first law for the control volume. So, what we have done is we have transpose the terms and taken all terms considering the energy of the control volume on the left side. This is e c v t plus delta t minus e c v at time t. And on the right side, we are writing the heat transfer and work transfer, the regular ones. And we have added the terms which involve the inlet for the energy at the inlet plug along with the work done term at the inlet plug. And similarly, for the incoming energy and the work done at the exit. So, we will now come up with an expression for e i and e e. So, we are considering a reasonably simple system where we can consider that the energy is made up of the internal energy and the kinetic energy and the potential energy. Right now, we are considering only these three terms. Of course, if there are other terms chemical energy or some kind of electronic energy, those will have to be considered. But that is just an addition and hence we can just write e as u plus kinetic energy plus the potential energy and hence I will just write an expression for that. So, we have an expression for e i u plus e i e i plus e i square by 2 plus g z i. And similarly for e e. So, we can write our combined term which includes the incoming energy and the work done at the inlet as e i plus p i e i as follows. We notice now that there is a combination of u plus p v and we already know that we have defined this as enthalpy and here it is of course specific enthalpy. So, it is h i plus v i square by 2 plus g z i. Similarly, I will just write e e plus p e v e as h e plus v e square by 2 plus g z e. So, overall now we will have the following expression. So, we have taken delta t on the left side and we know now what to do as delta t tends to 0. The left hand side can be taken as a derivative of the energy of the control volume as a function of time. So, we will write that in the next expression. So, this is our final expression for the first law of an open system. Thank you.