 In the last segment, what we did is we came up with the idea of thermal resistances and it is basically a bit of a shortcut technique for calculating conduction problems. So what we're going to do now, we're going to work an example problem involving thermal resistances and heat loss through the wall of a house. Okay, so there is our problem statement. What we have is we have a house wall that consists of two 1.2 centimeter layers of fiber insulating board and then there is 8 centimeter layer of fiberglass pink on the inside between those two walls. So we have a layer, a layer and then inside here we have fiberglass pink and then what we have is we have a layer of brick on the outside. So there would be brick on the outside and then what we're told is that we have convective environments on either side. So we have this layer of brick and then we have two convective environments. One is at minus 20 degrees C, so that's pretty cold and I assume that is on the outside of the brick, so the outside of the house minus 20 degrees C. We have convective environment. We're told the convective heat transfer coefficient is 15 watts per square meter degrees C and then we also have another environment on the inside of the house and so you might have forced air heating or where you could have radiators, but in any event you'll have air circulating inside of the house and the temperature inside the house we're told is 15 degrees C and H is equal to 8, H outside was 15. So that is what we have for this problem and now what we want to do, we want to be able to figure out what is the heat loss, the rate of heat loss through this system. And so you can see we're going convection, conduction, conduction, conduction, conduction and then convection. So we've got a lot of stuff going on, but this is an ideal problem that you can use thermal resistances on and it's a series problem and so what we'll do, we'll work that and we'll begin like all the other problems, let's write out what we know. Okay, so there is all the information that we were given in the problem. The last thing that I'm going to write out, we were told to calculate heat loss per unit area and consequently what I'm going to do is I'm going to assign an area of one meter squared. So that is our unit area and the problem, we're looking for the rate of heat loss or the heat transfer rate, so we're going to assume this is one dimensional and looking at the problem, all of the assumptions that led up to the thermal resistance technique will work and those being we have steady state, one dimensional conduction, there is no internal generation in the wall and the thermal conductivities will assume or not functions of temperature. So with that, what we'll do, we're going to come up with a schematic that has the coordinates and the information in terms of lengths and things on it. So let me write that out. Okay, so there is the schematic for our problem. We have the convective environments on either side and then the wall and just like I said earlier, we will assume the things that are required in order to apply thermal resistances and that was steady state, one dimensional conduction, no internal heat generation and the final thing is K is equal to a constant for all of the materials that we're dealing with. So given that, what we're now going to do, we're going to perform the analysis using thermal resistances and this is a series problem. And so what we'll begin by doing is writing out what our resistance network looks like and we're going to have the number of resistances that we're going to have, there'll be one here, there'll be two there, there'll be three here, four, five and then finally we're going to have six out here. I probably should have used different numbers because that's going to mess this up with the widths. Let me erase that. So for thermal resistances, I will do A, B, C, D, E and F. Each of those is going to have a thermal resistance, although we're going to have a different symbol for it. But anyways, that shows you the number of thermal resistances. So let's take a look at the analysis of this problem. So what we do when we're constructing our network, we begin with the temperature at the one side and that was T infinity i and then we move along and you'll notice we first get to the convective environment. So that has a thermal resistance associated with it. So we put a thermal resistance in and the value of that is going to be RT for thermal resistance and then convection and I'll do an I to denote that that is on the inside. And then we move along looking back at our schematic. The next one that we hit is going to be the wall board, which is on the outside of the wall. So that will have another resistance and that will be due to conduction. So it will be thermal resistance conduction and I will give it the number one because that was the length scale that we had for that one. Then we move into the fiberglass pink insulation and that had length scale two. So that will have its own thermal resistance and that will be RT again due to conduction three. Looking back at our schematic, we then go through another wall board, length L three and so that is going to have another thermal resistance. RT conduction four. Next one, we're going to have the brick, which is on the outside of the wall. And so that is going to have another thermal resistance. Again, it's going to be due to conduction. And that will be five. Oops, sorry, I missed the number here. It should be one, two, three, four. And then finally, what we're going to have is the last one right here. That is convective heat transfer. So we have a thermal resistance to end it here. And that is due to convection and it is convective heat transfer on the outer surface. So I'll put a zero there. And then finally, just like with voltage, you have a voltage drop here. We have a change in temperature and we're going to have a temperature drop. And we then get to T infinity outer. So that is our problem or the scenario that we have. And now heat flux is going to be flowing in this direction. That is Qx and that's what we're after. And the neat thing about thermal resistance is, although we're not going to do it in this problem, once you solve for the heat flux Qx, you can use the equation for thermal resistances in order to figure out intermediate temperatures. And so that's kind of a quick way of computing temperatures at intermediate points. And you'll use the equation for the thermal resistance, basically the equivalent of Ohm's law that we have. And that's how you could evaluate those, which you would do if you're doing electrical circuit analysis as well. You could calculate voltage at intermediate points. Well, here we can calculate temperature. So with that, what I'm going to do, I'm going to write out all of the different thermal resistances. We're going to sum them up because this is a series circuit. We'll put them in the equation and we'll solve for Qx. Okay, so those are all of our thermal resistances. What I'm now going to do, I'm going to sum all of those. And so the total thermal resistance is going to be equal to sum of each individual component because this is a series circuit. And we get 2.5757. And with that, we can now put it into the equivalent of Ohm's law that we have for thermal resistances. So let's do that on the next slide. And what we have then is Qx is equal to, and it's going to be the temperature change from one part of the circuit, from the start of the circuit to the end of the circuit. So looking back at our circuit, we're going from here. Let me do that in red. We're going from there to there. So that's going to be the temperature change over all of the resistances that we've computed. And when you do that, we'll divide by the total of the thermal resistances for all of the individual components. And if we recall, the inner temperature was 15 degrees C. The outer temperature was minus 20. And then that's going to be divided by the sum of all of the resistances, which was 2.5757. And with that, we get Qx is equal to 13.59 watts per meter squared. And so there, what we've done, we've calculated heat loss through this wall using thermal resistances. It's quite a simple technique. It's straightforward. You're just putting these resistance values together. And like I said, if you wanted to go in and evaluate the temperature at one of the intermediate points, you would just use this equation again. But the end points would change. And the resistance values that you're working with would be different. And that would then enable you to determine the temperature at one of the intermediate points within the circuit. So that is an example using thermal resistances provided that the four assumptions apply, that steady state, one-dimensional, no heat generation, and constant thermal conductivity, you can apply this. And you can do this also for cylindrical or spherical coordinate systems. And so it's a pretty quick and efficient method for doing conduction analysis.