Okay, welcombe back. So when we finished our analysis, we wanted to know, how would the exit temperature from my cooler here how would that temperature scale with fan wind speed? Well, we have all the governing equations on the screen here. So we just need to make sure that this wind speed actually gets into this expression that we use to solve for the exit temperature. So again, we, we take that form of the expression we'd had before, where we had the mass flow rate. And all we need to do is substitute in this expression. Which includes explicitly, the velocity of the air. So we can see the density effect, the area effect, the wind speed, and of course the specific heat. So the temperature scales within this bundle of variables as a function of velocity. So it's a little bit complex. It matters what this number is here as to whether or not we could say, like if we had a very low entrance temperature, we could say well the scale's inversely proportional. But, for the system that we were considering, these, remember, these numbers were the entrance temperature was significant and this temperature increase was about. 20% little bit less then that in terms of the relative to the entrance temperature. So can't quite say it's inversely proportional. There's some waiting associated with that entrance temperature. But it shows you how again, you might set up a spreadsheet and look at the range of velocities that you could achieve with a typical box fan. Cool. Alright, so we've now done a number of different examples for turbo machinery, for compressing air, compressing fluids, flowing things, heating things, cooling things. Now I want to cu, want you to consider one of the hardest systems to analyze. And that's a time varying or a transient system. So, for this example, it's going to seem really simple on the outset when we just consider the system in general. But, when we go to actually analyze a system, we'll see how hard these time varying systems can become to actually gets answers for, get numbers for. So the system we're considering is a car tire that's low on air. So we're going to assume we have a really good air compressor that we're going to use to fill this tire. We're going to assume it has a really large reservoir tank and that airs at ambient conditions 20 degrees C and that it's the reservoir is filled to about 350 psi psi gauge. our tire, which is about ten liters in volume, so this is a typical passenger vehicle tire initially has air at 15 psi and 20 degrees, so ambient conditions again. Now that's a pretty low number. I just this is my aside on. Being a good citizen. You always want to inflate your tires to the appropriate level. Because it really does have a significant effect on your fuel economy. So, we would hopefully never let the air pressure in our tires get that low. And this is kind of an exaggeration, just for this example. so again, be good citizens, keep your air pressure at the appropriate level in your tires because it does have a very significant and direct effect on the fuel economy. Okay, so there's our public service announcement. So our tire's low, it has 15 psi, we want to fill it till we reach the recommended value of 35 psi. So passenger tire, vehicle tires, are about 30, 35 psi if we assume the tire fills very rapidly, so we're going to take our, our, our hose from our compressor, we're going to fill the tire. Let's assume the fill process is very fast. So we can assume that that process is adiabatic. How much air, in mass do we need to fill the tire from 15 psi to 35 psi? And can we determine the temperature of the air in the tire after we've finished filling the tire. So these are linked questions. I'll tell you right now, they're linked. We need we would essentially need to answer both questions simultaneously. So again, we need to define our system. Okay, so we have our tire. Which has a known volume. We know that some initial state one, we have pressure of 15 psi within the tire. And we have an initial temperature of 20 degrees Celsius. We're going to define our system as the air within the tire. I don't want to care, I don't want to include the mass of the tire, so I'm going to put my system as being the volume of air within the tire. We're going to assume that that tire's a nice, beefy, steel-bait, belted radio, because I want to assume the volume is constant, that there isn't any significant volume change. That's going to simplify our analysis a lot, and hopefully the minute I said we're going to assume a constant volume, and no volume change, you're thinking, now I don't need to consider any expansion or compression work that occurs during the process. So, this is a very big assumption. not big assumption, it's an important assumption. I'd argue it's a pretty fair assumption in terms of, the volume isn't going to change much and we're not going to lose too much of our. Energy to expanding the air, expanding the volume of the tire. Okay, so we have our initial state, we've defined our vol, the volume of our system. At the final state we know we're trying to hit a target of 35 psi. we're told that the fill process, itself, is very fast and again we can assume that the process is adiabatic, as a consequence of it being fast. And ultimately, we're trying to find the temperature at the final state. Okay, so we've defined the system, check. Now we need to identify the proper forms of the conservation of mass and conservation of energy. So we look at [UNKNOWN], and remember, always use the most general form and then invoke your simplifications. So we say that for the conservation of energy, we have the change of the mass in the control volume, is balanced by all the mass flow rate into the system and all the mass flow rate out of the system. Well there's nothing coming out. We hope there are no leaks in our tire. so we're going to assume no leaks. And there's no mass leaving the system. Mm. And so what we're left with for the conservation of, of mass, is an expression that looks like this. [SOUND] There's one mass flow rate into the system. That's me filling my air into the tire, right? And we know that there has to be a change in the mass within the control volume. That's what we're trying to achieve. We're trying to put more air. Into the tire. Okay, so now what I have is a differential equation, where as before, when we were considering the continuity expression and the conservation of mass expressions, they were nice and simple constants. Now we have something that looks a little bit harder to work with. So we're going to take that mass expression there. [SOUND] And we're going to see oh, that's simplification, and in order for us to determine the total mass that enters the system, we're going to have to integrate these, this expression as a function of time. So we're going to have to integrate from the initial state to the final state. From state one to state two, okay whenever you see an integral expression, you always have to look at these variables just like we did with expansion and compression work where we had to understand how the pressure would vary as a function of volume. In this case, we have to understand how the mass flow rate varies as a function of time. Now all the language I used in that problem statement, that language was trying to get you to focus on okay. If I have a large reservoir tank, and it's at a pretty high pressure, much higher than the pressure in my tire. In fact it's an order of magnitude higher. What does that mean? What we can assume, I'll tell you what that means. That means that we can assume that the mass flow rate is constant as we fill the tire. And this is something we're going to need, this is information we're going to need next. That the state of the air that is in the tank doesn't change as we fill the tire, so in other words, we can't bleed the tank dry. We can't decrease the amount of material in the tank so much that we affect the state conditions within the tank. That's our assumption. And as long as I have a really high-quality compressor with a large reservoir tank at a very high volume and pressure, that assumption is a good assumption. So, we assume at the mass flow rate is a constant during filling. If you had a little compressor with a little volume. That's not going to be a good assumption, and we'd have to actually have to consider how would the mass flow rate vary as a function of time and solve that integral expression. Okay, so let's keep going with this, so now we've got we've set up the conservation of mass. Now we need to consider, okay. We've got this integral expression for the conservation of mass, which says, this. from state one to state two and we've just said. We're going to assume this mass flow rate in, is a constant, so that let's us say, okay. M dot delta t, which essentially is the mass into the system. And then on the left-hand side of this expression, this is pretty simple. This is just the mass at state two minus the mass at state one is exactly, identically equal to the mass that's been added to the system. And that's very intuitive, and you, many of you may have just written that down for the conservation of mass. You may have just said, hey, look, the final mass minus the initial mass is how much you added to the system. Of course. Except if you did that without going through the thought process of considering the mass flow rate during fill, you skipped a really important step and you skipped something that's very important, in terms of understanding what assumptions were implied. With just righting this down exactly. So again, be very careful and walk through the system and make sure you understand what assumptions, you've invoked. Okay, so that's our conversation of mass. The conservation of energy has to have exactly parallel structure. Okay, so let's look at that. and again, we'll set it up in its most general form, and then we'll simplify it. [SOUND]. Okay so here's the conceptivation of energy most general form. Well we've already said we're going to consider the process fast and therefore adiabatic. We've already said the volume is going to be fixed. We don't have shafts, electrical cords, anything entering in the system so we're not going to have any work in the system so. No expan, an,th, the most important one was the no expansion or compression work. we're going to neglect kinetic and potential energy effects. And of course, we only have mass entering the system. We don't have any mass leaving the system. And on the left hand side here again we're still going to neglect kinetic and potential energy effects. And all of this leads to something that looks pretty elegant and simple. The change in the internal energy and the control volume has to be balanced by the energy introduced by the air entering the system so it's pretty simple. Except again I have a differential so let's go ahead and put that in an integrel form so we have. Okay. So, what we say is that, well, if I assume the mass flow rate was constant in my analysis of the conservation of mass, it has to be constant here, too. And again, because I am assuming I have a large compressor, and the state doesn't change, the state conditions, thermodynamic state conditions, don't change in that air compressor system, I can assume this is, the enthalpy is also constant as a function of time. So I can pull both of these variables outside the integral and it's just a simple expression. So again, We have, the final internal energy minus the initial internal energy is simply the product of the mass into the system times the enthalpy of the air coming into the system. And of course, we have this expression here, for the mass that is enter added to the system, so that we can say. U2 minus U1 is equal to m2 minus m1 times hin. And you're looking at this going, so far, I don't see anything that has an expression for temperature in it. Okay, so how am I going to find the temperature in the of the air in the tire from this expression? Well hopefully you're thinking, okay it's air let's go ahead and invoke some ideal gas assumptions and let's see where that takes us next. Okay, so we're going to go ahead and vote the ideal gas assumption for air and what we see here is that we're going to go, let me just write this down and we'll discuss it. Getting a little crowded here and I apologize for that, but I want to keep all this on that same page so we can see everything together. So what we have here on the left hand side is this expression for the internal energy and it's the product of the mass of the final state and the internal energy of the final state. Mass of the initial state times the internal energy at the initial state. That's a one. Okay. On the right hand side we have the difference between the mass at the final and initial states times the enthalpy at the entrance. Well we know that the ideal gas assumption tells us the difference in internal energy, specific internal energy, is given by. The product of the specific heat at constant volume times the difference in temperature, so we know this. Except, we come down here and say hey, wait a minute, but the mass at state two is very different from the mass at state one. So I can't factor the mass out of this expression. Okay. And on the right-hand side, I have enthalpy which is a product of the specific heat at constant pressure times temperature. So I'm looking at this expression and I can't find a closed form for me to determine temperature. In this expression yet. Let's go through and again kind of think about this some more. We know the ideal gas assumption can give us some information about mass, so for example, we know that the mass at the initial state, can be given by this expression here. Okay so we have the pressure the initial state, we know the volume of the tire, that the specific gas constant. We've gone through that. And there's the initial temperature. Okay fine, m1 can be considered fully defined. So we can go back and we can say, yep I know m1. I can't find m2 using the same type of expression, because of course. M2 is a function of the temperature we're trying to find, T2. Okay, so that's unknown and that's unknown. Okay, enthalpy for an ideal gas is a function of temperature only. We know the temperature of the reservoir, the tank. So I could look this up, this I know. Okay, but I don't know m2, like we said up here. And I don't know u2, because again the internal energy is a function of the temperature. So what I have here is an under defined or under constrained system. I can't solve for the temperature T2. Unless I have the mass at the final state. I can't get the mass at the final state unless I know how much mass I've added to the system. if I had those two pieces of information, I could find the temperature at the final state, or if I could find the temperature through the conservation energy equation. I can go back and find all these masses. So it's under constraint and that's because of the assumptions that we involve in how the film process occurs. So what I want you to think about is that. So I've already shown you this is a really complicated problem. And we need to change the assumptions to make it tractable. So I want you to think about how might we change our assumptions about this fill process to allow us. We need another equation, so we have three unknowns right now, but we don't have enough equations. We actually have four unknowns, and we have three equations. So I need something else in order to make this to actually complete the analysis. So what I want you to do is think about that. How might we change our interpretation of how the tire fill process occurs, in order to allow us to answer this question. And we'll address that next time you come back.
