Welcome back. Okay, so we saw that fill, that transient fill process is actually a really complicated problem. We saw that we had three masses that are important in this problem. We have the amount of mass we add to the system, and we had the initial mass and the final mass. We saw with our conservation of energy equation that that had a direct parallel structure to the conservation of mass. And as a consequence we had to solve both equations simultaneously, but we still didn't have enough information to fully solve that problem. So instead, we're going to do something else. We're going to actually say, hey you know what, let's step back, and let's think about what happens when we fill a tire. We know that we initially fill the tire. And again, if we assume that process is fast, that is an adiabatic process. But we also know that after a period of time, that that system will equilibrate to the temperature of the ambient environment. So, if I take this process and I make it a two step process, an initial fill process to some unknown temperature. And then, a cooling process to some known temperature, and known pressure, I can actually make this system tractable for analysis. So, let's set that up, and we'll show you how that works. So again, we're going to consider it a two step process. So, from 1 to 2, the important features are that, that process is adiabatic fill, so that's the adiabatic fill. To some unknown temperature. Okay? And from 2 to 3, that process is a cooling process, so we know that that's going to be q out. Which again, cooling you know that's going to be less than 0. And that's going to occur, we're going to actually set that criteria. That we're going to cool to the target pressure, so where p3 equals 35 psi. And that's pretty much what you do if you filled your tire, right? If you'd filled your tire from the compressor before, you'd sit there, you'd feel it. And you'd wait for a moment for the system to equiliberate, and then you'd measure the pressure. And you'd say, oh okay, you're going to keep filling until you hit the target pressure. Well, that process is actually involves a cooling process. Because the gas is in fact affected by heat transfer during the fill process. So, our criteria from 2 to 3 are that we have a pressure that reaches the target pressure of 35 psi. And it has to hit that target pressure at the ambient conditions of 20 degrees C. Now, I'm giving you numbers, we're not going to crank through the numbers in this example. I want you to do this entirely in the variable form. We don't need to really work. We've gone through the exercise of how do we calculate densities and all sorts of good stuff like that for air before, so we don't need to do that again. What we really want to do is just make sure we understand the governing equations for this system. Okay. So, again, from steps one to two, that hasn't changed, right? From one to two, this is the fill, this is the adiabatic fill to T2. So, we know the conservation of mass. For steps 1 to 2 is exactly the same as what we had before. So, that's m2 minus m1 is equal to mn. Okay. that's a conservation of mass. Okay. The conservation of energy is also exactly the same as what we had before. So, we had the heat trans excuse me, it's adiabatic, so the internal energy, change in internal energy is balanced by the enthalpy associated with the air entering the system. So, we had this expression. Okay. So this is from 1 to 2. Now, from two to three, this is the cooling process. So, this is the cooling to ambient temp, temperature, after filling is complete. This is important because I want to look at the conservation of mass from 2 to 3. Well, remember, always write it in a general form. And we look at this expression we say, well, filling's been complete. So, there is no mass flow rate in or out of the system. So, what you're left with is the mass at state two has to be identically equal to the mass at state three. Okay? So, hopefully that makes sense to everybody. Okay, now we have to consider the conservation of energy from two to three. So, we'll start over here, give ourselves plenty of space. now, I'm going to get a little bit lazy here and take some shortcuts on you. we have just said there is no mass flow rate into or out of the system from states two to three. So, we know that these are going to be, these terms of energy transfer into or out of the system are going to be identically equal to zero too. Again, you always have to have parallel structure, so in my laziness we will eliminate those. right now we won't write all the terms here. Now, here we have a fixed volume. We still have a fixed volume. So, we don't have to worry about any work in this system. So, we know that this is equal to 0, for the same reasons the work transfer was equal to 0 from the fill process. Now, we've already identified that from 2 to 3 is a cooling process, so we can't eliminate this heat transfer term. But we will go ahead and again, neglect kinetic and potential energy effects. So now, from 2 to 3, the governing expression is we'll go ahead and, is going to be the difference in the internal energy from the final state, from the final state to the initial state has to be balanced by the heat transfer from state 2 to 3. and then more specifically here, we know that the internal energy at state 3 is a product of the mass the specific internal energy, and we can write the expression like that. Okay. So now, we've divided this up into a two step process. And the important thing to note is that we have now got a fully constrained final state. because we've said that at state 3, we have the temperature, the pressure, and the volume. Okay, so, a lot of stuff is in front of us right now. So, let's try and balance it all. And let's see if we can maybe put it all on one screen together. Alright, so we're going to put it all together here. We have m1, so the initial and final masses at during the fill process. So, that's m2 minus m1 equals mn. We have m two. Okay, we have mass at state one, given by the idea of gas law. We have, the conservation of energy between, for the second part of the process. And the conservation of mass for the second part of the fill process. Okay. And now, let's go ahead and sort through what we know and what we don't know, and to do that, I'm going to change our colors here. So, we can kind of highlight what we know and what we don't. Okay, so we know the mass flow rate at the initial state, because again, we have the ideal gas relationship here. We know the mass at the final state because we've defined again, that it has to be at ambient temperature and the final fill pressure. So, this is known, this is known, this is known, this is known. Hence, the mass at state one is known. By similar logic, we know that. So, there's just one body in the system, so I'm not going to put a sub script on it, so we remember that. So we know that the final state pressure, the final state temperature, again the volume is known and the gas constant is known, so M3 is known. So, this is known everywhere. This is known everywhere. Okay, we will assume that the specific heat is a function of temperature only, or excuse me the internal energy is a function of temperature only because it's an ideal gas. So, we're going to right now treat any known temperature as being a known internal energy or enthalpy. Okay, so just let me write that down because that's an important thing for us to remember. So, for an ideal gas, the internal energy is a function of temperature only, and the enthalpy, specific enthalpy, is a function of temperature only. So, there are look up tables for this, so in other words. If I told you, the temperature's 298, it's air, you would go look up the specific internal energy, and you could get that value. Conversely, if I told you the internal energy is a thousand kilogils per kilogram, you could go look up, and you knew what substances was. You could go look up what the temperature was, associated with that internal energy. So again, for shortcut purposes, I'm not going to change these expressions from internal energies to specific heats or enthalpies to specific heats. That's what the tables used, the, the ideal gas tables for, for any substance like air, is just a mathematical integration of the specific heats. Okay. So, let's go back to identifying our knowns and unkonowns. Okay, so we've decided m1 is completely known. We know the temperature at t at the initial state within the tire, that's the ambient temperature. So again, that mass is known. We know the temperature at the final state, so that internal energy is known. so if we look at this expression, we will box up what we don't know in just a moment. And then, in the enthalpy here, again, this is the enthalpy associated with the air in the tank, and we know the temperature of the tank. So, this is known. Okay, so we're starting to track things down. So, we know the mass at 3, which means we know the mass at 2, and that was a key connection because I needed the mass at two in order for me to come back to this expression here. And say, well, I know the mass here at two. So, I'm going to come back here and say, oh, I know the mass here at 2. And here at 2. So, this is the key. Because I can make this statement here, I can identify the mass that is at the final state. Because I can get this mass, I can come back here. And I can say, look, in this conservation of energy equation, everything is known except for the specific internal energy, the final state. So, I can solve for u2. Solve, and that is going to be a function of temperature only. And I can then look up the value for that temperature. And that was the intermediate temperature that I was interested in determining. And because I also have m2, I can answer the question of how much mass was added to the system. So then, m2 minus m1, these are both known. Can also be determined at the same time. Once I have that information I have the internal energy at state two. I can now come back and actually determine the heat transfer during the cooling process too. So, can find Q2 to 3 too. Okay. But what was critical was that we really needed to be able to identify the final conditions. As a fixed constraint in order for us to finish the analysis on this problem. So with that, what I want you to do next is I want you to consider how would this analysis change if we had used a small compressor. I've already given you the thermodynamic assumptions associated with that. If you have a small compressor, that means the mass flow rate and the state conditions within the storage tank are changing as a function of time. So, what would you do to complete the analysis if you had to use those constraints of time varying conditions within the fill reservoir. And that's what we'll start with next time.
