The best way to answer this question is to go back to the conservation of energy and just look at the, answer the question mathematically. So what would heat loses mean to our turbine? So let's again, we'll start with the, ooh, we'll start with most general form of conservation of energy. So we start with, and then we'll simplify. We're going to start with the storage term, the heat transfer term, the work transfer term. We have the mass flow into the system, and all the energies associated with that. The mass transfers out of the system. And remember we assumed we actually justified that we could neglect kinetic energy. And potential energy is even smaller, so we're going to neglect changes in kinetic and potential energy. We're going to assume it's a steady-state device. And so what we did last time was we actually solved for this portion of the expression. We had the mass flow rate times the enthalpy difference. But now we have this heat transfer term that we're not going to cancel because we don't want to consider this adiabatic. Well this is simply the adiabatic work that we just determined. And this heat transfer term, because it's going to be a heat loss, and it's going to be heat transfer out of the system, that's less than zero. This term is greater than zero. So we know that the work transfer associated with a turbine that experiences heat losses is less than the work transfer, for an adiabatic turbine. So it makes good sense. If we expend some of our energy in heat transfer, we don't have that energy available in work transfer. So again, the heat loses are a negative impact on the power out of the turbine. Okay, so we're going to take a look at one more example before we move on to some additional concepts. I want you to consider in this system, a microprocessor. Let's take a desktop computer or a laptop computer. And on your desktop computer, those CPUs or those microprocessors are cooled using a fan. So we're using ambient air to cool the microchip, and let's assume the microchip dissipates about ten watts of energy. And that's pretty consistent. It will generate, you know, the heat that's generated by those microprocessors can be anywhere from one to tens of watts of power. so what we want to know is, if we have to cool that chip, we have to remove that 10 watts of heat transfer from the chip, using air that's directed, flowed over, we flow over the CPU what's going to be the temperature out of the computer if the air speed into the system is 0.9 meters per second? So, just as we back up for a moment and give you some context. power dissipation in CPUs is a huge issue. This is very limiting in terms of failure and performance of your computers. So I'm sure as you're aware, as your computers get smaller and smaller, the amount of energy that they have to dissipate by heat is getting higher and higher on a flux basis. In other words, a lot more energy is coming out of a lot smaller locations. You have a lot less surface area to work with. Okay, so what I want you to do, this very open ended on purpose, I want you to make the system tractable for you to be able to conduct the analysis. So first step, just like we talked about before, the four steps define the system. Well, we have a CPU, right with its little contacts here. And we know that we are using air to cool that CPU. Okay. so this is a generic system. We're asked what the temperature is out of this heat exchanger type system. So in this case, my heat exchanger is the heat that's being dissipated from the microchip. So this is heat out of the chip and into the air. Those are the two systems I have that I could consider and if I want to know what the temperature of the air is then I have to consider the air as the system. And I think that might be very intuitive and obvious, but it isn't always. But again, if you're asking for thermodynamic information it has to be associated with the, the fluid that you're talk, the system that you consider has to be associated with the fluid that is relevant for that thermodynamic data. So if you want to know the air temperature, the air really has to be the system you consider. So we've got this little box that's above the CPU, that's the air that, where the airflow occurs in contact with the CPU. And there's a lot of good heat transfer questions that are associated with this too, so it kind of shows you how heat transfer and thermodynamics are very, very linked. So, how big should this box be? Right. How far up should it extend? Well, you know that to a certain extent, imagine if you have a little box heater, a little flat film heater, that a certain area above that above that heater, you're not going to feel, the heat transfer from that heater isn't going to be observable in that air. So, that's probably your upper bound. Where the temperature is unaffected is where the upper bound of the control volume would be. And then what's the width? Well that'd nominally be the width of the film, or the width in this case of the microprocessor. So this is if we were to look at it in an isometric view, it'd be something like this. Where again, our control volume is really this rectangle that's associated with the air above the surface of the microprocessor. So that's the system that I'm concerned with. It has a temperature at the inlet. We're told its ambient air so we're going to consider that to be 25 degrees. That's a good ambient air middle of the road ambient air temperature. So, 298 Kelvin or, 25 degrees Celsius. We're asked what the exit state temperature is, so that's what we're trying to find. We know that's an unknown. And we're given the heat transfer is 10 watts on a rate basis. And we're also given that the speed of the air in, into this de facto heat exchanger is 0.9 millis, 0.9, excuse me, metres per second. And that's pretty box, if you have a box fan, kind of a typical fan that you might pick up at a hardware store, that's a pretty, it's on the low end of a box fan. But it's, you know, so you're talking about a typical, not industrial grade fan type speed, so low end on the fan type speed. So you could probably increase the heat transfer by increasing the velocity of the air by maybe an order of magnitude just using kind of conventional fans. Okay, so we've defined the system. What are the appropriate forms of the conservation of mass and conservation of energy? For conservation of mass, the first question you ask yourself, does mass cross the system boundary? Sure, right? We can see the mass flow into the system and the mass of the air flow out of the system. were going to assume the system is steady state. And then we can say that the mass, there's one mass flow rate, and it's a constant into and out of the system. So, what's the parallel form of the conservation of energy? Again, let's use the most general form here. And I promise you we won't write it down in this gory detail much more. But again, it's good for us to develop good habits. N minus all the out terms, and as you might exam, as you might expect, you know, there are no changes in the potential energies so those are not going to be relevant. But the wind speeds we're talking about here, the air speeds that we're talking about, are small so you might expect the kinetic energy effects are going to be small. again its steady just like in this conservation of mass. We better have that simplified. We know there's heat transfer. There's no work transfer within this control surface. There's electrical work into the chip, but that's not our system. That electrical work is being used in part to create this heat transfer, so there's no work transfer, because there's no shafts within the air, you know, there's no electrical work that crosses the air system boundary, no expansion, no compression. We're assuming a fixed control volume region. So, what we're left with is a nice, simple system that says, the heat transfer is simply the mass transfer times the enthalpy at the exit minus the enthalpy at the entrance. Okay. Do we have a fully-defined entrance state? Well, we said the ambient temperature is about 290 Kelvin. We can also state that at the entrance state the ambient pressure is about one atmosphere. Okay. At the exit state we know that we also have its, its, you know, our desktop is on our desk or our laptop is on our lap, which means that pressure at the exit is going to be one atmosphere too. Now often people will have a kind of knee jerk reponse to say it's a constant pressure process. That's over stepping your interpretation. We don't know that that's a constant pressure process. We just know that at the entrance and at the exit we're going to end up on the same isobar. Okay. So we have a fully-defined entrance state. We think it's air, so you know it's not condensed. You know it's not in the, in the saturation region, because it's air and it's ambient conditions. So pressure and temperature are enough to fully define this state. At the exit condition you're trying to find temperature, and you have an assumption about the pressure which, which is a pretty good assumption. Okay, so that's not enough to fully define the exit state. Okay, so we go back and we look over here and we say but I have the heat transfer rate. So I can use the heat transfer rate to find the enthalpy at the exit. Okay, but that's not the temperature. We're looking for temperature. I used the conservation of mass. I used the conservation of energy. Those are my two big guns, those are my best tools. What's left? I need another equation to connect enthalpy to temperature. And at this point, you should all be having the aha moment. But it's air. It's a gas. So I'm going to invoke that convenient state relation, the ideal gas model, which says there's a direct relationship between temperature of the gas and the enthalpy of the gas. So if we assume air is an ideal gas, well then we, with constant specific heats, then we know that this heat transfer can be described as simply the product of the specific heat, the constant pressure specific heat, times the difference in the two temperatures at the inlet and outlet. Some people will say, I use CP because it's a constant pressure process. Uh-huh, I just told you it wasn't a constant pressure process, it just happens to be the entrance and exit states are at the same pressure. You're using a specific heat at, at constant pressure, this is what, you know, this is how we refer to it, the specific heat at constant pressure, because this is the enthalpy. If this had been internal energy, you would have used CV, but because it's enthalpy, you'll use CP. Okay? So make sure you understand why we invoked the specific heat at constant pressure. Okay, phew! So we found the proper forms of the conservation mass and the conservation energy. We sort of found the missing state information, what we did is essentially use the state relations between enthalpy and temperature, to get us to the point where we say, this is known, the temperature at the inlet. we have the speed of the gas entering the system. So we think that's how we're going to get the mass flow rate. We know the heat transfer, thats been given to us, that's the ten watts, the heat transfer rate. So we think we can find T2. We look at these variables and in general we say, the specific heat at constant pressure is a thermodynamic property. Okay? It's a constant, it's a constant, specific heat, so we can define it at the entrance state conditions. So in principle this sounds pretty straightforward. That there's not a whole lot left for us to do. In practice, it's much more complicated. Okay. Primarily because we don't have those convenient lookup tables in terms of enthalpies. So what we need to do is be smart about the state information that we do need. So we need the specific heat at constant pressure. That is generally tabulated as a function of temperature, so one of the questions, one of your assignments, is to go find that information. At 298 Kelvin what is the specific heat for air? Okay, and the mass flow rate we know is given by this expression here where we need the density of air. So the specific heats pretty easy. You're going to Google specific heat air standard temperature and pressure and you're going to get a number. And that's very, you'll find it a, a million different places. The density of air is typically not tabulated. Instead what we tabulate, well we know, instead we say, hey wait we're smart enough to know that the ideal gas law says that the pressure of the gas is equal to the product of the density times the specific gas constant, times the temperature. This is the specific gas constant. Recall that the specific gas constant is related to the universal gas constant through the molecular weight. So we know that R for air, the specific gas constant, is equal to the universal gas constant divided by the molecular weight of air. And I will tell you the molecular weight of air, even though air is a mixture of oxygen and nitrogen and all sorts of good stuff we generally assign a molecular weight of simple, we generally treat air as if it's a simple mixture. A simple fluid, not a mixture, excuse me, is what I was trying to say. So we say the molecular weight for air is equal to 28.97 grams per mole and again we consider it a single composition fluid. Okay, so your project, before you, you look at the next segment, is for you to determine the density of air at the entrance conditions to our question here. And I want you to go through and be very careful about your units. I've given you molecular weight of air. You need to go find an appropriate form of the units for universal gas constant, and chug through that units conversion. Plug in the appropriate pressure and temperature at the entrance state, to get the density. And then if you're really feeling gung-ho, I want you to, to make an approximation about the area, the flow area that would be appropriate for a CPU device. And evaluate the mass flow rate for the system. And we'll see how close your numbers come, to the numbers I calculated next time. Thank you.
