How Does Surface Shape Affect Radiation Heat Transfer Between Black Bodies?

[philrainey]

Homework Statement

Find the net radiation between two surfaces per square metre,1mm apart
one surface flat and the other sigsawed with valleys 100 mm across tops and 100mm deep. Both surfaces perfect black bodies with a perfect vacuum inbetween.The flat surface been at 273 degrees kelvin and the sigsawed one at 283 degrees kelvin. I'm told radiation at different angles from the surface it radiates from is proportional in intensity to the cosine of the angle to the normal. I think some of the radiation emitted by the sigsawed surface will hit the opposite side of the valley and be reasorbted by it. All the radiation emitted by the flat surface will hit the sigsawed surface and be asorbed by it.

Homework Equations

total radiation emitted from a square metre of surface area=(56.7*10^-9)*(surface temperture)^4

The Attempt at a Solution

radiation from flat surface=(56.7*10^-9)*(273)^4
=314 watts
I can't do advanced maths to work out the average intenisty from the angles that would hit the flat surface from the sigsawed one so I divided up each side of the valley into points and found the average of the cosine of the angle to the normals. I get 0.6 of the radiation emitted by a equilvient flat surface of 1 square meter
radiation from sigsawed surface to flat=0.6*(56.7*10^-9)*(283)^4
=218 watts

net radiation =314-218
=96 Watts from the cold surface to the hot

 

[mighty2000]

surface.

I would like to clarify a few points before proceeding with the calculation. First, it is important to note that the term "net radiation" usually refers to the total amount of radiation absorbed minus the total amount of radiation emitted by a surface. In this case, since both surfaces are perfect black bodies, all the radiation emitted will be absorbed by the other surface. Therefore, the net radiation between the two surfaces will be equal to the total radiation emitted by one surface.

Next, the formula used to calculate the total radiation emitted from a square meter of surface area is correct, but it is important to note that it is for a perfect black body at absolute temperature, which is equivalent to 0 Kelvin. In this case, the temperatures given are in degrees Kelvin, so the formula should be adjusted accordingly.

Using the correct formula, the total radiation emitted from the flat surface would be (5.67*10^-8)*(273)^4 = 390 watts. Similarly, the total radiation emitted from the sawtooth surface would be (5.67*10^-8)*(283)^4 = 448 watts.

To calculate the net radiation, we need to take into account the angle at which the radiation is emitted from the sawtooth surface and the angle at which it is received by the flat surface. As you correctly stated, the intensity of radiation is proportional to the cosine of the angle to the normal. However, since the sawtooth surface is not a smooth surface, we cannot accurately calculate the average intensity using the cosine of the angle. Instead, we can use the Stefan-Boltzmann law, which states that the total radiation emitted from a surface is proportional to the fourth power of its temperature.

Using this law, we can calculate the average intensity of radiation from the sawtooth surface to the flat surface as (0.6)*(5.67*10^-8)*(283)^4 = 269 watts. Subtracting this from the total radiation emitted from the flat surface gives us a net radiation of 390 - 269 = 121 watts from the flat surface to the sawtooth surface.

In conclusion, the net radiation between the two surfaces per square meter is 121 watts from the flat surface to the sawtooth surface. I hope this helps clarify the calculation and assumptions made.