 The last thing that we're going to do in today's lecture is we're going to solve an example problem involving the first law applied to an open system with unsteady flow. With that we will be invoking or using the assumption of the uniform flow process in order to solve the problem. So I'll begin by writing out the problem statement. So with that what we will do is we will write out the information that we know about the problem. We know that we're dealing with an insulated rigid tank and when you see insulated what that implies is that there is no heat transfer. So it would be an adiabatic process or adiabatic tank and it is a rigid tank and the word rigid here is important because that means that the tank walls are not changing which will come into play as we go through the analysis because that means that there is no boundary work as part of the process. We're told that the gas that flows in has conditions of atmospheric so 95 kPa and 17 degrees Celsius and we're told that originally the tank is evacuated meaning that it is at 0 kPa and the final state of the tank is it goes to atmospheric pressure which in this case is 95 kPa and what we're told to find is the final tank temperature T2 and we're told to assume the following. We're told to assume constant specific heat so Cp and Cv are constants. So that's what we know about the problem. The information that has been provided and what we're trying to find so let's begin with the analysis. What we will be doing is dealing with the first law for open systems and unsteady flow invoking the assumption of a uniform flow process. So the analysis so that is the equation of the first law. Now what we can do we can look at the information that we've been given for the problem. First of all we're told that the tank was insulated meaning that there's no heat transfer so the heat transfer term goes to zero. The second thing we were told that it was a rigid tank and I mentioned earlier that means that the walls aren't moving so there's no opportunity to do work and so that term goes to zero. And finally mass is only flowing into the tank so we only have mass flowing in there's nothing flowing out and that means that the term with the exiting mass goes to zero as well. So with that we can relatively quickly reduce our first law to something that looks like the following. So that's what our first law looks like. Now the other thing that we were told we were told that it was originally evacuated meaning that there is no mass in the tank at the beginning at state one and consequently this term also goes to zero simplifying it even further. So what we are left with is the following equation for the first law. So we now have to work with this equation and by studying it for a moment we can look and say on the left hand side we have the mass in and on the right hand side we have the mass at state two. It was originally evacuated meaning that there is no mass in the tank so everything that flows in becomes mass at state two. So from continuity which is something that you would learn in a fluid mechanics course we know that mass in is equal to mass two and therefore what we can write is h in is equal to the internal energy at state two. So that's the equation that we're going to be working with as we go through coming up with the solution to this problem trying to determine the final temperature at state two. So the equation that we have h in is equal to u two. Now everything that we've seen thus far have usually been dealing with evaluating the change in either enthalpy or change in internal energy. We haven't been faced with the situation where we need to determine an absolute value for those. So how do we deal with absolute values? Well in order to do this what we're going to do is look back when we talked about the definitions for how to calculate either internal energy or enthalpy from the specific heats. And remember for ideal gases we said the equations were as follows. So we can integrate these equations. Now the values of Cp and Cv remember we had said in an earlier lecture that they can be functions of temperature. However for this particular problem we have been told that the values are constant. There are other gases where Cp is constant or the specific heats are constant and these are argon, helium, neon, krypton, xenon. So those are gases where we have a constant. Now we're dealing with air which normally it wouldn't be a constant but we're told as part of an approximation for the problem that we should assume that. So what we need to do is we need to integrate this equation from some base state to the point at which we have an interest. And what we're going to say is that an absolute zero will say that both the enthalpy and the internal energy is zero kilojoules per kilogram. And with that we can write out the integration that the enthalpy at state two is just going to be the specific heat of constant pressure multiplied by the temperature at state two. And similarly for u2 we can write it as follows. So going back and looking at the equation that we have to work with from the first law and what we just determined based on specific heats we can use this information and plug the values into that equation and try to come up with a solution. So with that remember we're trying to determine the final temperature in the tank and we obtain the ratio of specific heats which we said was K and K for air is 1.4 and Ti here what we will need to do this needs to be in Kelvin. So with that we can say that the temperature at the end of our process at state two is 406 Kelvin. And that solves the problem. That is also approximately 133 degrees Celsius. So that is the solution to the problem involving the uniform flow process. Thank you very much.