 Welcome back. So, we discuss another exercise. So, here is our problem. It is in fact transient state problem. So, it is slightly different than the other problem that we discussed up till now that were steady state. So, let me read out the question that is at hand. A rigid insulated bottle of volume V naught is perfectly evacuated. The stopper is opened and ambient air at conditions P naught and T naught is allowed to flowing. When the flow stops, the stopper is replaced. Determine the final temperature of air in the bottle. So, you notice at once that the system can be considered an open system with an inlet that is when the stopper is open, air starts coming in. But you realize that there is no exit. So, this system has only one inlet and no exits. And you realize that the mass flow rate is probably driven by the pressure difference. We are not concerned about how the mass flow rate actually changes that we will leave to a fluid mechanism. So, m dot changes as a function of time and as the pressure differential between the outside atmosphere and that in the bottle starts decreasing, m dot probably decreases. When the pressures are the same, you have no longer any mass flow rate. So, that is when at least you have reached some kind of a mechanical or pressure equilibrium. What is the mass contained in the bottle at this time? Well, if we knew the temperature, we would have just used the ideal gas equation and gotten it because we know V naught. So, we could have gotten it. Of course, if it was some other fluid, we would have to know its equation of state. But right now, it seems that we do not know the final temperature. So, how do we proceed? We will use the transient equations of state, but we have actually simplified the problem reasonably because we have assumed an evacuated bottle and you will realize that this simplifies the system reasonably. Let us see how to go about it. So, we draw the bottle here. Maybe it has a small opening and this is where air starts flowing in m dot. It is a function of time. It is not really a constant. So, we will call it as m dot i which changes with time. But we know the inlet conditions. They correspond to P naught and P naught. These are inlet conditions. So, if we write our transient mass flow or mass conservation equation, you realize we had this. So, this is our control volume here which we take along the bottle here in the inlet of the bottle and just where the stopper is here. This is where we put our boundaries. So, this is our control volume. This would be given by m dot i. You realize that there is no m dot e. So, this is 0. If we want to find out how much mass is accumulated, we can integrate this quantity from the initial state to the final state. Of course, we do not know what m dot i as a function of time is, but whatever the mass flow rate here, finally this is mass final minus mass initial is equal to this. This is 0 because we had evacuated the bottle initially. So, we get our mass final here. What about the energy equation? That is the first law for open system. Again, we write something for the transient state d e C v by d t is equal to q dot minus W s dot plus m dot i h i plus v i square by 2 plus g z i minus m dot e h e plus v e square by 2 plus g z e. Notice that these two quantities are not the same. That is because m dot e does not exist and hence we have separated out the terms as it was during our derivation only in steady state. We could have put these two as equal and then we had combined our h i and h e terms together and so on and put the left side as 0, but right now you realize that there is no m dot e. So, this goes to 0. That means this entire term disappears. We will assume that this is well insulated here, the control volume. There is no information given. So, let us just go ahead and assume this is equal to 0. The control volume is not doing any work. Neither are we doing work on it. It is not expanding. This is equal to 0. What about the terms in the bracket here? We realize that z i is not really going to significantly change. Air just enters from somewhere outside and goes into the bottle. There is not too much change in the elevation. So, this can be safely assumed to be close to 0. What we will also assume is that the flow velocity is reasonably small and we will put this term as 0. So, what are we ending up with? We end up with d C v by d t. This is d e C v by d t is equal to m dot i h i. What is h i? It depends on the inlet state which depends on p naught and t naught. So, we can actually find that out because p naught and p naught are known quantities. What about d e C v? Now of course, e is made up of u and k e and p e etcetera. So, we assume that we can only look at this component and these are components of the control volume that do not change because the control volume does not change in height nor is it moving at a particular velocity. So, as such there is really no kinetic energy to talk off here. The system is not moving with any velocity. In general, we will put these two terms as 0 in the sense we will put e is equal to u. Hence, we will just write this as d C v is d t is m dot i h i. So, we have d u C v is m dot h i d t. We integrate it from the initial to final state. You realize that this is u final minus u initial and u initial is 0 because there is nothing in the control volume initially. This is equal to h i its constant. It depends on the outside condition and we have m dot d t which we know is the mass in the control volume finally. I think we had designated it as m f and what is u f? Final mass in the control volume multiplied by specific internal energy. So, we have finally m f u f is m f h i. This h i let us replace it by h naught since the outside conditions were designated as p naught and p naught. We can remove the m f that is the same. So, u final in the bottle is h naught. So, what do we do now? We just write this as u naught plus p naught v naught. So, what we can do now is u f minus u naught is equal to p naught v naught and this is nothing but C v t f minus t naught is equal to p naught v naught and we can write this as r t naught. Once we have this, we can write t final as t naught plus r by C v t naught or we could multiply by this. C v here, C v plus r is C p and then we would have C p by C v t naught which is equal to gamma t naught. So, this is what we get t final in the bottle is equal to gamma times t naught. The final temperature in the bottle will turn out to be gamma times the outside temperature. One can notice that at this point, we had still not assumed anything about the fluid and here because it was air, I have gone ahead and used ideal gas equations and I have assumed that C v does not change with the temperature nor does C p. If it was something like steam which was flowing in, then h naught would have to be determined using steam tables and u f would then you equate it to the h naught and find out at which conditions this is the u f that is obtained at the given p because p is fixed. So, for a given p, you have to find out at what temperature the u is the same as the h coming or the h of the gas which is coming in. If you have any other fluid, you will need separate equations of states or some kind of tables to find this out, but you realize that the temperature in the bottle finally is not going to be equal to the temperature outside. Thank you.