 Okay, in this segment what we are going to do, we're going to take a look at the approximate solutions for a change in convective boundary condition for the infinite plate. And so we'll look at approximate solutions and the graphical technique. And the graphical technique referred to as being the Heisler chart. So we're looking at the infinite plate. And what we'll begin with, let's see, just remember the infinite plate. This was the geometry. We specified x from the center and then l and l, so the thickness was 2l. But we will begin with the approximate solutions. And we start with the centerline temperature. And this is where we're going to use a lot of the nomenclature that we presented in the last segment, so theta naught star. That's going to be theta naught divided by theta i. And the approximate solution, we get this constant, exponential minus zeta squared. I apologize for my zeta, it's not perfect. And then Fourier, Fourier number. And the Fourier number, if you recall, was alpha t over l squared. And c1 and zeta, where do we get those from? We have to evaluate the bio number first. And the bio number for the slab is hl, l is our characteristic length divided by k, the thermal conductivity. And then knowing that, if a function of bio, you go to tables. And so your textbook should have tables. And from that, you can get zeta and you can get c1. So you would look up at a specific bio number and it would give you c1 and zeta1. Make sure you're reading it for the infinite plate, not for the cylinder or the sphere, because they most likely will also be in the table. So that's how you get the centerline temperature. You use that expression there. And then to get the spatial temperature, again, we use a theta star that we take theta this time, theta naught. And theta denotes temperature at a given spatial location other than the centerline. And in this solution, what we do is we use theta naught star from our previous solution. So you've got to evaluate the centerline temperature first. And then once you've done that, then you can get temperature at a location off of the centerline. And it is expressed in terms of cos and zeta1 times x star, zeta1, that is a function of bio number from the tables. And x star is equal to the spatial location that we're interested in evaluating divided by L, which is the thickness. Now, one thing that you got to be careful about here, this is evaluated in radians. So make sure your calculator is in radians when you evaluate that term, or you will calculate it incorrectly. And then the final thing that we're going to get using the approximate solution is the heat loss, the heat loss from our slab to the surroundings. And again, we're using zeta1, which is a function of the bio number and from tables. And be careful, radians. So don't get that wrong. And then q naught, we talked about q naught with our nomenclature. It refers to the total amount of heat loss if the entire solid was to go from T i to T infinity. So that is using the approximate technique. And using the Heisler charts, I won't show you the Heisler charts yet. We'll do that when we work an example problem in the next lecture. But what I'm going to do is just sketching them so that you get an idea as to what they look like. So using the graphical technique, and these are referred to as being the Heisler charts, they may or may not be in your book depending upon what textbook you're using. But just like before, we start with the center line. The center line is plotted as a function of the Fourier number. And then the curves are going to do interesting things. And these curves, there are a number of them. And they go by one over the bio number. So there would be different curves that come along. I think they all have similar break points. I'll just draw one. So that would be the center line temperature. And just like before, you have to do the center line first. And then you can move on to the spatial. Spatial is a different curve. And here we have theta over theta naught. So you need to evaluate theta naught first before you can go in and evaluate theta, which would be temperature at a given point. And it is plotted as a function of one over the bio number. And the curves, there will be different curves as a function of spatial location. And x over l is how those are plotted. And then finally, you can use charts to get the heat loss. And this is plotted as a function of the Fourier number times the bio numbers squared. And then these charts will be plotted for different bio numbers. So that is heat loss for the infinite plate, giving us the temperature as a function of time at the center line, spatial locations, and then total heat loss, using both the approximate and the Heisler. So it might be a little abstract now. It is, and we'll look at the sphere and the cylinder next. But when we work this in an example problem, it'll become a little clearer. Basically, when would you use the graphical versus the equations? The equations are pretty easy, although for the cylinder, we'll see there are bezel functions. And so you've got to go and look up values in another table, which can be a bit of a pain. The graphs are not as accurate, but they're quicker. And so it's a bit of a mixed call, which is better. But you can use either technique. And you'll see in the example problem, I'll work both techniques. And we get very, very similar results out of them. So anyways, that's the infinite plate. We'll next move on to the cylinder.