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Try this calculation

  1. Dec 15, 2008 #1
    I have a much better shape that is much easier to calculate.Try calculating this one the net radiation between two spheres one within the other a perfect vaccuum inbetween.outer Big sphere is 0.13m diameter at 1500 kelvin. the little sphere is .01m diameter at 1550 kelvin. what is the net radiation and in which direction.

    so the little sphere area is0.000314metres square at 1550 kelvin=(56.7*10^-9)*(1550^4)*0.000314=102.7 watts.
    Total radiation from the big sphere area of0.053066 metres square=(56.7*10^-9)*(1500^4)*.053066=
    15232 watts radiating from all the surface of the colder big sphere
    but only 7.7% of this radiation hits the little sphere as the rest misses it and hits the big sphere again on opposite side.15232*.077=1172watts so the net radiation is 1172-102.7=1069.3 watts from the colder big sphere to the smaller hotter sphere . I'm led to beleive the intenisity of radiation is proportional to the cosine of the angle to the normal to the surface it is radiating from.
  2. jcsd
  3. Dec 15, 2008 #2
    by the way they are both perfect black bodies
  4. Dec 16, 2008 #3
    my calculation is wrong can someone please look at my calculation and logic to show me where I am wrong?

    big sphere 0.13m diameter
    little sphere 0.01m diameter
    taking a point on the big sphere and working out the % of total radiation from that point that hits the little sphere.First the angle from the normal to edge of little sphere
    make x the angle from the surface that just hits the inner sphere
    radiation intensity proportional to cosine to the normal of the radiation or the sine of x
    antidifferential of sinx is -cosx
    enter angle x=-cos1.4932381
    =- 0.0774905
    At the normal 90 degrees=-cos1.5706
    Area under sin wave between normal and =0--0.0774905
    Area under angle that hits small sphere =0.0774905*2
    % of total radiation that hits little sphere=0.1549869/2
    2 been the area under a sin wave for 180 degrees?

    big sphere temp=1500 Kelvin
    little sphere temp=1550 Kelvin
    area of little sphere=0.000314 metres square
    area of big sphere =0.053066 metres square

    heat radiated by little sphere all that hits big sphere=(56.7*10^-9)*(1550^4)*0.000314
    =102.76 Watts
    total heat radiated by surface of big sphere=(56.7*10^-9)*(1500^4)*0.053066
    =15232.26364 Watts

    radiated heat from big sphere to little sphere=15232*0.07749345
    =1180.3 Watts
    net radiation =1180.3-102.76
    =1077.6 Watts????????It is wrong
  5. Dec 16, 2008 #4


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    Staff: Mentor

    So you're asking the same question you did two days ago and hoping for a different answer this time? You already acknowledged you don't know how to calculate the effect of geometry, so you know the 7.7% is wrong and I already pointed out that you are misusing the Stefan-Boltzman Law and gave you the correct form of the equation. If you want to know how to derive the law mathematically (as opposed to guessing about how it works), just ask that question - maybe someone can help with that.
  6. Dec 17, 2008 #5
    O.K I'll do that
  7. Dec 19, 2008 #6
    It was the shape of the sigsawed surface radiating that was just way too hard for me to calculate the effect of geometry, the sphere in a sphere is much much simplier and I want to find out where the calculation is wrong . The Stefan-Boltzman Law is P=area * emissivity*absolute temperture^4 I don't know how the formular in the website http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html is produced or what it is based on, I'm sure it doe not consider the surroundings a perfect sphere. I'm asking for help as to where the logic I used in the calculation is wrong or where the number crunching is wrong.
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