 So in the last segment what we did, we talked about the convective rate equation that is Newton's law of cooling. What we're now going to do, we're going to solve an example problem involving convective heat transfer. Let's all begin by writing out the problem statement. Okay, so there is our problem definition or problem statement. What we have is hot air flowing over a flat plate. The air is hotter than the plate, and so that is going to have an impact on the direction in which the heat transfer is taking place. We've been told what the convective heat transfer coefficient is, 75 watts per meter squared degrees C, and the area of the plate is 2 square meters. So we know the free stream temperature, and we know the wall temperature. Those are unknown, so a standard technique of solving any problem, let's begin by writing out what we know, and then what we're looking for. So we're told to find the heat transfer rate. Now one thing to notice, when I wrote out the convective heat transfer coefficient here, notice that I used units of watts per meter squared kelvin, whereas in the problem statement it was watts per meter squared degrees C. Be aware that those are identical. There is no difference from one to the other, and so don't get confused by that. They are the same thing. All right, so what we're going to do, we are going to begin by writing out a schematic. Okay, so we have fluid flowing over a wall. We have no information about the boundary layer, so we'll just draw it as being a free stream temperature T infinity coming over the wall, and we know, however, that the wall temperature is at a lower value. But what I'm going to do, I'm going to begin by just showing this as heat transfer going in that direction, and then we'll see what we get as we go through the process, because that's a standard definition that we've used. Typically we say the wall is hotter than the fluid. Assumptions that we're going to make in solving this. So we're going to assume that we have steady state conditions. What does that imply? That means that the wall temperature is not changing with time, nor is the free stream temperature fluid or the velocity of the fluid coming over it. So we have steady state. Now for analysis, it's pretty simple. What we're going to do, we're going to take Newton's law of cooling that we just saw. We have all of the different values. You'll be pretty much straightforward plugging in the values. So analysis for this. And when you write out Newton's law of cooling, it's always T w minus T infinity plugging in the values. So we get that. When you calculate everything, we get 15,000 watts or 15 kilowatts. And given that we have a 50 minus 150, that's actually a negative. So it's a negative value. And what we need to do is go back and look at the way that we had the drawing. We were showing the heat transfer going from the wall out into the fluid. But given that the fluid is at a hotter temperature or higher temperature than the wall, the fluid is 150 versus the 50 of the wall. Really what we have is a scenario where we have 15 kilowatts and then just add the words into the plate. And what that does is it shows that you know the direction that the heat transfer is taking place. It is going from the fluid into the plate and warming the plate up or heating it up. And so the temperature of the plate eventually would start to rise. So that is an example of how we can do calculations using Newton's law of cooling for convective heat transfer.