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PRESENTER OK Welcome back This is example 3101 from page149 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 Belize a solar hot water heater or an air solar heater And these two plates are25 millimeters apart And some properties of each surface for the plate-- the emissivity epsilon is equal to 015 And the temperature of the plates 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 088 And the temperature of the cover is a little 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 homeward 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 384 from the text And we solve this on per unit area basis because the size of the collector is not given And so we solve it perimeter 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-Boltzmannconstant sigma in this heat exchange equation And then again we use the hotter surface first so that we have good sign convention Temperature of that surface-- 343 minus 323 the temperature of the cooler surface all divided by the emissivity So 1 over 015 plus1 over 088 minus 1 all in the denominator And the Stefan-Boltzmannconstant is 567 times 10to the negative 8 If you've had ahead transfer class you'll remember that number It's pretty easy-- 5 6 7 8 567 times 10 to the negative 8 And when you run those numbers through your calculator you get 246 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'the radiation heat exchange coefficient So here we're going house equation 3101 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-- 246divided 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 endue with is 1232 watts per meter squared Kelvin And that's that That's example 3101 Thank you for listening
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