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Rate of heat transfer

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    The underside of a smoky layer 7m x 10m is radiating like a flat, isotropic plate at 440°C to the floor of a compartment 1.85m below. The mean emissivity is 0.45 and the floor is homogenous/flat plate at 40°C. What is the rate of heat transfer from the smoky layer to the floor?

    2. Relevant equations

    yXap0.jpg

    3. The attempt at a solution

    Really don't know where to start from.
     
  2. jcsd
  3. Mar 10, 2014 #2

    SteamKing

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    Why don't you start by plugging in the known information from the problem statement into the formulas you were given. At least that will show you have tried to solve the problem rather than immediately giving up.
     
  4. Mar 10, 2014 #3
    OK, I have no idea how to even start it? Any help from where to begin?
     
  5. Mar 10, 2014 #4

    SteamKing

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    In the diagram attached to the OP, which would be the ceiling and which would be the floor? What are the temperatures of the two? (Hint: TH - hot temperature and TC - cold temperature)
     
  6. Mar 10, 2014 #5
    A_1 is the ceiling, A_2 is the floor. A_1 has TH of 440 degrees, and A_2 has TC of 40 degrees?
     
  7. Mar 10, 2014 #6

    SteamKing

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    Yes, that's correct. What kind of temperatures do you have to use with radiation problems?
     
  8. Mar 10, 2014 #7
    I have calculated following X = 3.78 and Y = 5.40

    Having this now I have tried to calculate F_1,2! I get following result 2/3671.16^degrees ( 1.16 + 1135.0^degrees).

    From here I don't know how to compute integeres and degrees? Should I convert this all to pi value, ie 1135.0^degrees is 6.03PI?
     
    Last edited: Mar 10, 2014
  9. Mar 10, 2014 #8
    I don't understand Q.?
     
  10. Mar 10, 2014 #9

    SteamKing

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    What temperature scale must you use for radiation problems?
     
  11. Mar 10, 2014 #10

    SteamKing

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    It's not clear what you've done here. Hint: not all arguments for trig functions have to be expressed in degrees. Here, radians is appropriate. The quantities X and Y are already non-dimensionalized.

    Be sure you set your calculator to 'Radian' mode before doing your calculations. F12 is, I suspect, a factor related to the geometry of the compartment, and probably doesn't have units.
     
  12. Mar 11, 2014 #11
    I have got 0.64. From here where to go now?
     
  13. Mar 11, 2014 #12

    SteamKing

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    For what? It would be better to show your calculations.
     
  14. Mar 11, 2014 #13
    F_1,2 I have inserted everything into given formula and got the result.
     
  15. Mar 11, 2014 #14

    SteamKing

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    Now, you substitute F12 into the equation for the heat transfer rate Q.
     
  16. Mar 11, 2014 #15
    Ok. One thing concerns me here and that is thinking about whether or not A is here included as radiation constant?
     
    Last edited: Mar 11, 2014
  17. Mar 11, 2014 #16

    SteamKing

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    Yes, A is the area of the layer which is radiating the heat. Its dimensions are given in the problem statement.
     
  18. Mar 12, 2014 #17
    Thank you I have solved it!
     
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