Rate of heat transfer

  • Thread starter Eva Brain
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  • #1
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Homework Statement



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?

Homework Equations



yXap0.jpg


The Attempt at a Solution



Really don't know where to start from.
 

Answers and Replies

  • #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.
 
  • #3
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OK, I have no idea how to even start it? Any help from where to begin?
 
  • #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)
 
  • #5
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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?
 
  • #6
SteamKing
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Yes, that's correct. What kind of temperatures do you have to use with radiation problems?
 
  • #7
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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?
 
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  • #8
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Yes, that's correct. What kind of temperatures do you have to use with radiation problems?
I don't understand Q.?
 
  • #9
SteamKing
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I don't understand Q.?
What temperature scale must you use for radiation problems?
 
  • #10
SteamKing
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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?
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.
 
  • #11
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I have got 0.64. From here where to go now?
 
  • #12
SteamKing
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I have got 0.64. From here where to go now?
For what? It would be better to show your calculations.
 
  • #13
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For what? It would be better to show your calculations.
F_1,2 I have inserted everything into given formula and got the result.
 
  • #14
SteamKing
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Now, you substitute F12 into the equation for the heat transfer rate Q.
 
  • #15
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now, you substitute f12 into the equation for the heat transfer rate q.
Ok. One thing concerns me here and that is thinking about whether or not A is here included as radiation constant?
 
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  • #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.
 
  • #17
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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.
Thank you I have solved it!
 

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