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Thermal radiations

  1. Jan 10, 2008 #1
    in thermal radiations what is the difference between effective surface area and surface area of emitting body ,and how to calculate effective surface area of ang body????????
     
  2. jcsd
  3. Jan 11, 2008 #2
    What is the context of the question? You can assume that the MBR has an effective surface area of the entire universe, but that the surface area of the emitting body would be the size of the universe when the radiation was first released.
     
  4. Jan 11, 2008 #3
    consider a case of a hollow cylinder with one side open to outside &
    external area is covered by insulating material, temperature is T
    assuming perfect BB how to calculate heat lost by it in 1 sec,

    E=[tex]\sigma[/tex]*T[tex]^{4}[/tex]*?

    length is l, radius is r.


    as i think ?=[tex]\pi[/tex]*r[tex]^{2}[/tex]
     
  5. Jan 11, 2008 #4

    GT1

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    Your body loses heat by convection and radiation.
    For convection you need to estimate H and then you can calculate how much heat you lose by convection.
    your body also loses heat by radiation, but because you are dealing with cylinder , some of the radiation is reflected by the inner sides of the cylinder and only some of the radiation escape outside of the body. you need to know what fraction of the energy escapes to the outside (look for shape factor on any heat transfer book).
    your heat transfer area by convection and radiation is the inner surface area of the cylinder. A=PI*D*L
     
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