 So we just took a look at Fourier's law and we said that that was the equation that enables us to calculate heat transfer when we have conduction. What we're going to do now, we're going to take a look at an example problem applying Fourier's law for one-dimensional heat transfer. Okay, so there's our problem statement. We're given a brick wall and we're told that one side of it is at 20 degrees C. The other side is at 5 degrees C. It's 30 centimeters or 12 inches thick. We're given the thermal conductivity and we're asked to solve for the rate of heat transfer going through that wall. So let's write out, and we should do this whenever we're solving problems, begin by writing out what we know, so our knowns. We were given the dimensions of the wall and consequently we can then determine the area. The thickness of the wall, we were told the thermal conductivity. Remember from Fourier's law, k was the thermal conductivity so we use k, middle k for that, and then we were told the temperature difference on either side of the wall. What are we after? We're after the rate of heat transfer and that is q. And then finally what we'll do, I'll do this on the next slide, we're going to write out a schematic. You should always do schematics when you're solving engineering problems. It kind of helps you conceptualize and understand what is going on. And it's also a way to communicate for somebody else who's looking at your analysis. So x is the horizontal distance, this is our brick wall, and we were told the inner and outer temperatures. So what we're going to assume is that that is measured by a thermal couple or some other mechanism right on the wall. We were told the width of the wall was 30 centimeters. Thermal conductivity was k. And what we're after, we're trying to determine this rate of heat transfer coming through the wall that will have q. So one of the things you have to be very careful with Fourier's law and applying it is making sure that you have the right coordinate system with your temperatures. And so here we will say x is equal to x i and then x is equal to x outer. And you'll see why in a moment when we put the values into Fourier's law. And I'm going to assume a couple of things in solving this problem. One is that we will assume that we have steady state conditions. So what does that mean? Well we're going to talk about steady state and then we're going to talk about transient conduction later on in the course. But steady state conditions means that the temperatures are not changing. So it's not like you have somebody taking a blow torch to the left hand side and raising this temperature up to 80 degrees C. So we're assuming that things are remaining at stable conditions and that the temperature is not changing. Second thing, we're assuming that we have one-dimensional conduction through the wall. Even though this wall is finite, we're going to assume that we're looking in the middle region of the wall. If we were to go to the edges of the wall, then you would get more than 1D. You'd have 2D, maybe 3D conduction. And the third thing that we're going to assume is that we have constant thermal conductivity both throughout the wall as well as constant thermal conductivity that does not change with temperature. So those are the three things that we're going to assume. Now in order to do the analysis of this problem, what I'm going to do is I'm going to use Fourier's law or Fourier's equation that we saw in the last segment. And that stated qx is minus kadt by dx. And we're going to substitute in some of the values to begin with. First of all, we're going to start with the temperature differential, dt by dx or the gradient, I should say. And for this, this is where you have to be careful to make sure that your coordinate system and the temperatures that you're using are consistent. And so in evaluating this, we take ti minus t outers, so the change in temperature from the inner to the outer divided by the change in distance from the inner to the outer surface. And one thing I should say here is notice that this can also be written in another way. I could have flipped it around t outer minus t inner and then x outer minus x inner. And by having a consistent sign convention for the coordinates, we would get the same answer. But we're not going to do that here. We're going to use this form of it. But just as long as you're consistent with the way that you're applying your coordinate system and your temperatures, you should be fine. So let's enter in our values minus 0.69 and I'm going to explicitly put all of the units. I usually won't do this, but I will for this equation or for this problem, given that we're just starting out, but you'll get used to it as we go through the course. And so we can move a little quicker later on. And then temperature five degrees C. And what I'll do is I'm going to do this with the outer minus the inner. So the outer temperature is five degrees C minus 20 degrees C. So I was wrong. I'm not using that one. I'm using the second one. I apologize. So we're actually using that outer minus inner. And then we're going to divide that by let's look back at our schematic. So we're going to take x outer minus x inner. And that is just 30 centimeters. So expressing that in meters. And we have that. So plug this into your calculator and you find Qx for this brick wall is 1035 watts or 1035 joules per second. So there we go. That's our first heat transfer calculation that we have done in this course. We've estimated the heat loss from a wall knowing the temperature conditions on either side of that wall. So if you were to do this for a house, for example, you could then figure out what is the heat loss from the house. And that starts giving you interesting things for engineering analysis. So anyways, that is the first problem dealing with conduction, what we're going to do in the next lecture, we're going to move into convection. And then we will look at radiation. And that will take us through the three different physical mechanisms of heat transfer.