2.1. Heat Transfer Considerations

PRESENTER: Hello, this is example 3.6.1 from page 143 of the Duffie & Beckman text. So this problem is about the Sun, which is a 5,777 Kelvin black body radiator. And we are asked to find the wavelength or lambda, the wavelength at which the maximum monochromatic emissive power occurs. And this part of the question is solved by Wien's displacement law, which states that the maximum wavelength times the temperature of the black body radiator is equal to 2,897.8 micrometers Kelvin. So to divide by T on both sides, and therefore lambda max. This is equal to 2,897.8 divided by 5,777, which is 0.502 micrometers. So that is the wavelength at which maximum power occurs. So building into the second part of the problem, let's just draw what this looks like. This is wavelength, and this is power. The distribution of solar energy looks something like that. And so what it's saying is that at this point that happens, at 0.5 micrometers. So if this is problem A, problem B asks a second question about this same situation. It asks, what is the energy fraction from the source that's in the visible frequency range? So we find these fractional values in table 3.6.1A. And what this is asking, the visible frequency range is from 0.38 micrometers up to 0.78 micrometers. So in this graphic that's given by a range like that. And we're trying to figure out what is going on in between there. So from the table, we want to find what lambda T is. So 0.38 times 5,777 gives us 2,195 and 0.78 times the temperature of the black body radiator is 4,506. So with both of these numbers, you can now look it up in table 3.6.1A and obtain the fraction of energy from 0 up to that wavelength. So in the case of the first one, it's 10% percent. And in the case of the second one it's 56% from that table. So therefore, we can figure out our answer of fvis, the visible frequencies, which is simply 56% minus 10% 0.46, which is 46% of total energy in visible. And again, put the answer in a box so that it's clearer what we've done. And that's that. So thank you for listening. And that's problem 3.6.1.

PRESENTER: OK. Welcome back. This is example 3.10.1 from page 149 of the Duffy and 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. So 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, we 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 the 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, 3 meter squared collector, you would basically multiply this answer by three to obtain your final total heat exchange. So here's the Stefan-Boltzmann constant, sigma, in this heat exchange equation. And then, again, 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 the Stefan-Boltzmann constant is 5.67 times 10 to the negative 8. 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. Cool. 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 can 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.