Okay, welcome back. Now, let's finish our analysis of our microprocessor. So, remember, we had some general information that we had to interpret, and we wanted to then determine what the temperature of the air would be. The cooling air after it had been directed across the microprocessor. So last time we set up our systems, so let's go ahead and redraw that on this figure. We have our little microprocessor and we know that it's dissipating heat so we know that that's heat out of the system. We're going to put that on a rate basis. And we defined our system not as the microprocessor, but as the volume of air above the microprocessor. And we'll just assume a direction of flow here and we had our fan. And what we had was a lot of information that was somewhat general that we were going to have to assign or interpret based on what we thought were best estimate. So for example we define state one as the error starting across the surface of the microprocessor and then we define state two of the air as it exits this zone of cooling above the micro processor. So we had to define the system. And we said well we know we were told that it was an ambient system. So we're using ambient air, so we'll just assume 298 Kelvin or 20 degrees Celsius. It's gotta be atmosphere. We're not going to contain this in any sort of pressurized or evacuated system. At the entrance and we know that the pressure is going to be atmospheric at the exit as well. And obviously the question we're trying to answer is this question of what is the temperature. So last time we invoked a few assumptions too we assumed that our system was steady state. And we solved the continuity equation, the conservation of mass, that told us that we had no accumulation of mass within the system. And so the mass flow rate was a constant. And that was equal to the density of the air, times the area, times the velocity. Now, remember, we can use these conditions at either the entrance or the exit to define the mass flow rate. Now, we were looking for the temperature, and so we know that we're not going to get any temperature information from the mass flow rate of the exit, so instead, we're going to look to the conservation of energy to tell us the temperature of the system. And we solved, again, a using a steady-state assumption. We assume that there was no work transfer in the system, which is a good assumption. There's no electrical energy, there's no shafts, no expansion or compression. And again, those are all assumptions that are linked to our definition of a system. So you want to choose our system wisely. For example, we choose our fictional volume as having a fixed volume, right? This little air. rectangle here the, the volume here. We assume has fixed volume and that way we don't have to worry about expansion or compression work. So again we want to be thoughtful in how we find the system. And we can revise that system based on our solutions to the conservation of mass or the conservation of energy. That's a four show, we're going to cover that in the next example. Okay. so we solve the conservation of energy. We neglected kinetic and potential energy effects, and that gave us the simplified form of the conservation of energy. We took our solution for the mass flow rate and, used, substituted that into the conservation of energy equation. And then we invoked an assumption of an ideal gas for air to allow us to substitute from the enthalpy to the difference in temperature times the specific heat at constant pressure. And remember, that c p comes from the fact that we have enthalpies. Right, that's the definition of enthalpy, and the specific heat at constant pressure. Okay, all of that, and we realized we needed to evaluate the mass flow rate, and when I left you, you were supposed to determine the density that we were going to substitute in for the mass flow rate. Now, if we look at this expression, here. We're told what the air speed is entering the system. So we have the velocity at state one. This area here we have to be thoughtful about. And we have to decide, okay, what's the appropriate area? And this is the area we're referring to here. This area of this, of our control volume. The entrance to our control volume. So. That really requires some understanding of the heat transfer processes and I'll give you some background on that. so first of all this dimension is going to be whatever the characteristic dimension of the CPU is. So depending upon whether it's a square configuration or rectangle, but we're probably talking about something around the order of a centimeter or two in terms of this dimension. Now, in terms of the dimension above the surface of the CPU, the heat transfer occurs entirely within, what's referred to as the thermal boundary layer. So, we would need to have some estimate of what the, that distance is for the thermal boundary layer. And for this problem, we're going to just assume a good number for that area, and I believe. We are going to treat that as, yes. So if I assume that this dimension here is about two centimeters, I assume this dimension here is about one centimeter. And again, it's a little bit unfair of a task for you right now. Because it does require some understanding of the physics associated with heat transfer. But the way you can think of it right now is. How close can you get to the CPU before the air is affected by that heat transfer? So, essentially you can think of it as, maybe a good analogy would be if you water flowing over let's say a hot surface. And if you are sticking your finger in the water, how far can you go in that water before you start feeling the temperature increase associated with that surface. Now, don't do this at home. Be safe, but, that's a good way for us to get the characteristic dimension. So, we have our area, it's one centimeter by two centimeter. Because we want this to be a fixed volume, we're going to assume that swept area is the same at the entrance and the exit. So I have this area here at A1. And I just need to know what the density is at state one. So then we ask ourselves do we have enough information to fully define the state? Yes we do because we have pressure and temperature. Alright. That's state one. And so that's where we left off. Again, a lot of work to try and get to the point of saying hey I need this specific gas constant. Why do I need the specific gas constant? Because I need the density. Okay. So we're going to determine the density at state one and that's simply given by the ideal gas relation. Okay. Where we have the pressure at state one and the temperature at state one, this gas constant is the specific gas constant for air, so we can even be explicit and add a subscript to denote that's for air. The specific gas constant is related to the universal gas constant through the molecular weight of air. So, that was your task, was to go find the molecular weight of air and calculate the specific gas constant for air. Which, again, requires that you know the universal gas constant and the molecular weight of air. So the universal gas constant has many, many units and you just have to pick whatever set of units are appropriate for your system since our problem is setup in SI units those are the units that we're going to use. So the universal gas constant is given by 82.0568. Cubic centimeters, atmospheres, mole, Kelvin. The molecular weight of air is represented by 28.97 grams per mole. Recall that molecular weight is always grams per mole or kilograms per kilomole. And remember that air is actually a mixture of gases, it's the one kind of violation to our rule of we won't study mixtures. We treat it as essentially a simple, single composition substance and if you realize that air is mostly nitrogen, it's like 70, 75% nitrogen, 25% oxygen, close, but not quite that. you would expect it to have a molecular weight that looks really close to nitrogen, since it's, you know, almost three quarters nitrogen, little more than that. And indeed, nitrogen has a molecular weight of 28, oxygen has a molecular weight of 32, so we can see the slight bias towards a slightly higher molecular weight. But for routine calculations or just introductory calculations, we can just use the molecular weight of air, assuming it's a simple single substance. Single composition material with the molecular weight of 28.97 grams per mole. So now we just watch our units and we see that my, remember the over bar denotes thermal basis and because we're going from mole to mass space we don't have an over bar on a specific gas constant for air and we do that calculation and we find the specific constant for air is 2.832 cubic centimeter, atmosphere, gram, Kelvin. Okay, now we're going to take that information and we're going to calculate the gas density. So remember, we had this expression, so we just take the specific gas constant we just determined and plug that in. To that expression and we get a number. And I just want to go through this in detail. And I'm sure you can do these calculations just fine. But I want you to see how careful you need to be with the units. To make sure that you keep a consistent set of units and don't introduce an error into the system. And then our temperature at state one. So this is the pressure at state one, is one atmosphere. The temperature at state one is 298 Kelvin. again, we want to be careful about units, so we're going to move from cubic centimeters to cubic meters. Sorry, that's a 100 there. going to move from grams to kilograms. And if we do that, we can see that we have nice cancellation of all of these units and it's going to give me the proper units for density as 1.185 kilograms. Per meter cubed. now, there are look up tables for the densities, density of air that you can find out, in atmospheric pressure in particular, on the internet. So you could just Google, what's the density of air at 298 and you're going to get a number. You can do that for a pretty good range, a pretty large range of temperatures. I also told you to go out and find the specific heat for air. Same sort of, using the same sort of method. Which is to just determine what the specific heat is by Googling specific heat for air atmospheric. Or what we call STP condition, standard temperature and pressure conditions. and you can find that information in lots of different places. So the specific heat, as you you know, as you determined, is. Pretty trivial at these conditions it's a value of essentially one. So 1.005 if we want to get some significant figures in there. And again with units and SI units of kilo Joules per kilogram Kelvin now. Moving from this part of the equation to this part of the equation, remember, we invoke the ideal gas assumption, the ideal gas model. Okay, we're going to check, I'm going to show you some more information on how to check if an ideal gas approximation is correct. And atmospheric conditions for air, it is. You [UNKNOWN] remember, we told, we went through the exercise of what are the types of conditions where we'd start to worry about ideal gas approximations. High pressures, cold temperatures. So we're not too worried in this system. So in another example or two I'll show you how to go ahead and check that rigorously. but what I want to focus on right now, this is my aside on the specific heats as from, moving from here to here assumes a constant specific heat as a function of temperature. Okay, this number is assumed to be constant as a function of temperature. In reality, the specific heat as a function of temperature actually varies. So if we have a diatomic, and it's very related to the physical structure, the molecular structure of the substance we're talking about. For diatomics like so that's molecules that consist of two atoms. Like oxygen, like nitrogen, we expect a roll off behavior, where you reach a point where the specific heat is approximately constant. So there you say, hey, wait a minute, you just told me we're assuming a constant specific heat, but look, this curve is a very strong function of temperature. So how can we make that approximation? There are two ways you can make that approximation, and it's a good approximation. One is you say hey look, my problem considers the temperature region where the specific heat really is approximately constant. So you can say I'm at a sufficiently high temperature, where this is a good approximation. Alternatively, you can be in these regions here, where you do see some variability to the specific heat and you have two choices. You can either integrate that expression, so determine what the specific heat is as a function of temperature and then this region it's linear, so you just integrate the line, you integrate the expression of specific heat as a function of temperature. And. The simplest method is to say, well, if this is my initial case, and this, this initial state, and this is my final state, is to just assume an average that's somewhere, an arithmetic average that's somewhere between the two end points. So there are lots of ways we can make this approximation pretty legitimate, pretty close to being a constant specific heat. And again for air it really is constant and particularly for the temperature region that.we're looking at. So that's a good assumption. So again hopefully you went out, you found this number and you actually should have found, hopefully, a table that's, gave you the specific heats as a function of temperature and you went and looked at well, the specific heat from 298 to let's say 398. Just guessing what we think the exit temperature would be doesn't vary much varies in like the forth significant figure. So hopefully, you came back with a number that was essentially one in case. So again, that was a little bit of an aside. So let's finish it up. Let's rap up the rest of our analysis here. So, we need to find the mass flow rate. So we did if you crank through, I'm sorry and did the rest of that calculation, so we found the density of air. Let's take that density, plug it into this expression. Let's assume, again, the area is 1 centimeter by 2 centimeters, and that's an approximation. we're given the wind speed, so we plug all those in and we get a mass flow rate that's equal to 2.13 times 10 to the minus 4 kilograms per second. We take this mass that mass flow rate plug it in to our conservation of energy equation here. Again we looked up our value for the specific heat at the entrance condition. The entrance temperature is known, that's 298 Kelvin. And, again, we're trying to dissipate heat of about 10 watts, so, all of these variables are known. The specific heat, mass flow rate, excuse me, the heat transfer rate, the mass flow rate, the specific heat, and the entrance temperature. So now we just need to solve for T2. In an algebraic form that looks like this, and we're going to want this expression here in just a couple of seconds. And we go ahead and we plug in all those numbers and what we get is, the first part of this expression is 46.6 Kelvin and we add that to the entrance temperature of 298 Kelvin, and what we get is that the exit temperature is 345 Kelvin. Or if you prefer, that's 71.7 degrees Celsius, or if you prefer even further that we do that in Fahrenheit, that's about 161 degrees Fahrenheit, so pretty warm. That's a lot of temperature increase for that volume flow rate of air. So remember we have a tiny little CPU dissipating a lot of power. This is a huge issue in pretty much any power system or micro electronics where trying to miniaturize the system. Dissipating heat in laptops, in cell phones. Are huge challenges. So these numbers are pretty accurate in terms of the amount of heat we need to dissipate. And in terms of the amount of air speed that's available. And remember, your little cell phone doesn't have a fan. Those are the flows that we get associated with convective flow. So we put a fan on your desktop, and we get that answer. But a lot of these systems rely on passive cooling, which is just whatever air induction is introduced by the ambient conditions. So you can see the challenges associated with keeping these systems operating below a threshold temperature. So, what I want you to what I want you to think about next is what happens to the exit temperature as a function of the fan wind speed. So in other words, if we doubled the fan speed, the air speed from the fan I should say, what would happen to the temperature? So if you're a design engineer, these are the type of criteria that you would have to consider. Which is okay, I can't let the system get above this temperature. What's the wind speed I need to achieve that? And that's going to dictate how you, how much you need to power your, your fan. So think about that and we'll answer that question next.
