 In this lecture we're going to be taking a look at solutions to the heat diffusion equation. We'll be looking at solutions that provide us with temperature distributions in two dimensions. So these are two D solutions to the heat diffusion equation. Now in order to proceed with the analysis that we're going to do we have to make a number of assumptions. The first one it's the obvious one I just said we're assuming that we're dealing with two-dimensional heat transfer so temperature will only be a function of x and y. We will assume that we have steady state conditions and that will simplify the heat diffusion equation for us. And then finally what we're going to do we will assume that there is no internal generation within the two-dimensional object that we're looking at. So let's begin by looking at the heat diffusion equation and we'll make some simplifying assumptions. Okay so with the approximations that we're making first of all the derivative with respect to the z-coordinate system will go away. There is no generation so that goes away and we're dealing with steady state consequently transient terms disappear as well. And so what we're left with is the following from the heat diffusion equation. Okay so even though that is simplified significantly that is still a partial differential equation. And consequently there are not a great a large number of analytic solutions that exist to that equation. We'll be taking a look at a couple of solutions in this lecture segment. So there are a number of limited analytical solutions but there are not that many. So if you go and open up pretty much any undergraduate textbook in heat transfer you will find the solution for the temperature distribution in a square plate. That's what we'll be doing here. I won't be going through and giving us the solution. You can look at any book and you'll find that. If you go to graduate heat transfer textbooks they'll have more solutions to different types of geometries. But the one that we'll look at here is probably one of the simplest ones. So let me sketch out the coordinate system that we'll be looking at for this problem. Okay so there is the flat plate that we're looking at. The plate extent is in here. And what we're after we're trying to determine what is the temperature as a function of X and Y within that plate. We're given the dimensions H and W. And the boundary conditions each side of the plate has a different temperature. This, this and this are all at the same temperature and the top of the plate is at a different temperature. And so that is what then gives us a temperature distribution within this plate. So if you look at any book you'll find two common boundary conditions or two common solutions to different boundary conditions. So let me write out the boundary conditions now. So boundary condition number one. And this is the boundary condition that leads to a nice analytical solution for the problem. But it's not a very practical boundary condition as we'll see in a moment. And what boundary condition number one states is the temperature at the top of the plate. Let's go back and look at the image. The temperature up here T2. So we're looking at the temperature distribution along the top of the plate. We're saying that it is a sine function. How you would get a sine function for the temperature distribution. You could shine a laser. A laser has a Gaussian function which looks like a sine function. But that would probably be a tough thing to be able to actually fabricate experimentally. So you can see that this is a little bit of a hypothetical solution. One that the mathematicians dreamed up and it enabled them to solve the equations in a nice clean way. So what we're finding the width of the plate and goes from zero to W. And the temperature at zero or at the base is going to be T1. Okay. So not a perfect sine wave but you get the idea. Essentially we're saying that the boundary condition at the top is a sine wave. Again kind of a in terms of reality not really the kind of temperature distribution that you would find. But one that makes the math a little cleaner and easier. Boundary condition number two is actually one that's more realistic. But it's a little more difficult to solve as we'll see. And what boundary condition two states is temperature two at the top of the plate is simply temperature two. It's a different temperature from the other three edges of the plate. And so that's boundary condition number two. Now as I mentioned I'm not going to go into the details of how you solve this. You use a technique called separation of variables. But I will give you the solution. And like I said if you go into any undergraduate heat transfer book chances are you'll find the solution to this problem in it. So let's take a look at solution two boundary condition number one. So that's what we get with boundary condition number one. And it's not too complex of a function. We do have hyperbolic signs in there. But let's take a look at what we get to when we use boundary condition number two. And that will make this solution look a lot better. Okay so that is solution using boundary condition number two. And you can see it is a little bit more complex mainly because it's an infinite series. And so it's very difficult to evaluate by hand. You'd want to use a computer for that. And consequently we can see that boundary condition number one results in a lot easier solution because that's one that we can calculate. It's not an infinite series. It's just plugging values in. But what we're going to do now we're going to take a look at these solutions. And so we will begin by looking at the solution using boundary condition number one to begin with. And then we will look at the solution using boundary condition number two. So let's go and take a look at a plot using boundary condition number one. So what we have here, this is the temperature distribution in our plate with the sinusoid boundary condition. So if you recall we said that this side of the plate was T1 100 degrees C. We said this was T1 100. Sorry it's kind of hard to read that. Let me move that a little bit. We said that the lower surface here was also T1 100 degrees C. And then up at the top here, this is where we had that sine distribution. T2 was 100 plus 10. And I've assumed the dimensions of this are one unit, whatever the unit might be. I guess it'd be meters with what we're doing here. But that was the profile. And if you recall, we had something like this. And so that was our sine wave distribution. And we go from 100 degrees C up to 110. And when we look at the contour plot, it looks quite nice. We see that the green, and here we have our temperature scale. So we can look at it. The green is around 100 degrees C, which we would expect for all of these walls because they're all at around 100 degrees C. And then with the sine temperature distribution, we would expect that in the center we're going to move up to about 110. And sure enough, if we look at the red color here, that's close to what we're going to. So that solution comes out to be quite nice. Now let's take a look at the second solution that we had looking back. And this is the one with the more realistic boundary condition. And for that boundary condition, we said, if we look back at our plate, for that boundary condition, we said that T2 was just some temperature. It was not a sine distribution. And consequently, it resulted in this infinite series, which was much more complex. So let's go take a look at what that solution looks like. And this now is the solution to that boundary condition, the one where we had 10, and what I've assigned is 10 degrees C up here, and then 100 degrees C here, 100 degrees C here, and 100 degrees C here. And you look at it and you think, well, that looks not bad. That actually looks pretty good. It kind of looks like the one that we just looked at. Nice, nice smooth curves. But what I should say is that this is only for one term in the infinite series. And in order to evaluate it, you have to have many, many, many more terms. So this is not an accurate solution that we're looking at here. It just gives you an idea as to what it looks like to begin with. And so I'm just going to sketch these. All of these temperatures are the same. And then this one up here, you know what I should do? I should do it green because it's green. That one is at 10 degrees C, so it's at a very, very different temperature. But what we can do, we can look at the convergence of this series. And so by convergence, that means let's increase n, and we'll go up. And so we'll go to 2, to 3, to 4, and looking back at our infinite series, which is here, what I'm referring to is the number of summation terms. So we started off with n equals one. Now let's look at 2, 3, 4. And what we're going to do, we're going to walk all the way up to 200 terms. And so let's take a look at that now and see what it looks like when we go through this process. So I've put it into video and we're beginning here with one term. And there you can see what we were looking at earlier. That's with 2, 3, 4. You see as you get more and more terms in the solution, we get some weird things going on at the top. And that's what they call ringing because what's happening there, these hyperbolic sine functions, they're starting to approximate a step change, a step discontinuity, which is very difficult to approximate. So there's 40 terms, 50. It's starting to look pretty good there. 70, 80, 90, 100, all the way up to 200. So there we can see the final image has 200 terms. And that one is actually getting to be close to the actual temperature distribution that you would find in this flat plate. I'm not entirely sure what you would need in order to have a fully converged solution, but we're getting pretty close once we have these 200 terms. So anyways that gives you an idea as to the fact that if you do want to calculate the temperature distribution in a flat plate, it's 2D. It really is not that trivial of a solution technique, especially if you're going to apply boundary conditions that are more like what we find in reality. And so that's a demonstration of the heat diffusion equation for two different boundary conditions for the simplest thing that we can look at. And that is of a flat plate.