 OK, welcome back. This is example 3.10.1 from page 149 of the Duffy Beckman text. And in this problem, we have two parallel plates, an absorber plate, and a cover. So this is a cover absorber system. And this could be like a solar hot water heater or an air solar heater. And these two plates are 25 millimeters apart. And some of the properties of each surface for the plate, the emissivity, or epsilon, is equal to 0.15. And the temperature of the plate is 70 degrees Celsius, which we need to convert to Kelvin. So 70 plus 273 is 343 Kelvin. And the cover, similarly, has an emissivity. The cover is 0.88. And the temperature of the cover is a little bit cooler at 50 degrees Celsius, which converting that to Kelvin to get 323 Kelvin. So in this problem with this system, as defined above, we're asked to find two things. First, we're asked to find the radiation exchange between the two surfaces. And in this case, intuitively, it's helpful to think which direction the heat will flow. That way, you don't have some weird negative sign you're trying to deal with in the end. The heat flows from the hotter surface to the cooler surface. And then second, we're asked to find the radiation heat exchange coefficient under these conditions. So with all of that in mind, let's go ahead and solve the problem. So part A, to solve the amount of radiation heat exchange going on, we're going to use equation 3.8.4 from the text. And we solve this on a per unit area basis because the size of the collector is not given. And so we solve it per meter squared, which then if you had a, say, 3 meter squared collector, you would basically multiply this answer by 3 to obtain your final total heat exchange. So here's the Stefan-Boltzmann constant, sigma in this heat exchange equation. And then again, on the hot, we use the hotter surface first so that we have a good sign convention. Temperature of that surface, 343 minus 323, the temperature of the cooler surface, all divided by the emissivity. So 1 over 0.15 plus 1 over 0.88 minus 1, all in the denominator. And Stefan-Boltzmann constant is 5.67 times 10 to the negative 8, which if you've had a heat transfer class, you'll remember that number. It's pretty easy, 5, 6, 7, 8. 5.67 times 10 to the negative 8. And when you run those numbers through your calculator, you get 24.6 watts per meter squared. So that's from absorber to the cover. And then part B of this problem says under these conditions, what's the radiation heat exchange coefficient? So here, we're going to use equation 3.10.1, which is the definition of H sub R, the heat exchange coefficient. So H sub R equals this heat exchange we calculated in part A, 24.6 divided by the temperature difference, 70 minus 50. And you could do this in Kelvin or Celsius in the denominator. You would get the same number. And so what you end up with is 1.232 watts per meter squared Kelvin. And that's that. That's example 3.10.1. Thank you for listening.