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Thermal radiation transfer rate

  1. Apr 24, 2010 #1
    See attached pic

    => Capture.PNG <=
    r=2.2
    h=4.9
    T= 37 Celsius
    Ten= 65 Celsius
    emissivity= 0.81

    For the surface area I used:
    A= 2 pie r^2 + 2 pie r h

    I converted both r and h into meter

    For the thermal radiation transfer I used the following equation

    P= Stefan-boltzmann constant X emessivity X surface area X (T environment^4 - T of the
    cylinder^4)

    My answer is 1.72

    What is my mistake ?
     
    Last edited: Apr 24, 2010
  2. jcsd
  3. Apr 24, 2010 #2

    rock.freak667

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    Well without showing exactly what you did, the only place I can see where you would make a mistake is not converting from Celsius to Kelvin. Did you do that?
     
  4. Apr 24, 2010 #3
    oh sorry. I was planning on attaching an excel file with calculations. Any way here is what I did. Yes I converted from Celsius to kelvin.

    P= 5.6704 x10^08 X 0.81 X 0.009814 (surface area) X (338.15^4 - 310.15^4)
     
  5. Apr 24, 2010 #4

    rock.freak667

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    I don't see anything wrong in your analysis then, why do you have reason to think that is wrong?
     
  6. Apr 25, 2010 #5
    we use wiley plus and when i submited my answer it says it's wrong (there's a 3% tolerance). But I think you can agree that there is nothing wrong with my answer - I calculated it multiple times before posting here.
     
  7. Apr 25, 2010 #6

    Mapes

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    Your S-B constant and area both have a problem with decimal position. How did you calculate the second area? I get a P2/P1 ratio between 2 and 3.
     
  8. Apr 25, 2010 #7

    D H

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    1.72 what? Failing to specify the units that a result is in is a common mistake. Failing to recognize that there are multiple systems of units is another. Note for example that the length of the cylinder is specified as 4.9 centimeters, not just 4.9. Units are very important. "Don't leave home without them."

    My guess: Units.
     
  9. Apr 25, 2010 #8
    @Mapes
    I copied the S-B constant from my book and I verified my area with an online area calculator.

    P2/P1 = 2.75 (I got it right)

    @D H
    1.72W
    I didn't mention the units because they are clearly stated in the screen shot I attached.
     
  10. Apr 25, 2010 #9

    Mapes

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    Got it - was [itex]\sigma=5.67\times 10^{8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}[/tex] a typo? (Should be [itex]5.67\times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}[/tex].) I goofed on the area. And I concur with your final answer.
     
  11. Apr 25, 2010 #10
    Oh sorry about that, I didn't notice it. So there isn't something I missed ? This is frustrating I've been over it a million times and I always get the same answer.

    I asked a friend of mine to get the person he did for him to do mine. He got -1.72 and it is correct. Why is that ?
     
  12. Apr 25, 2010 #11

    Mapes

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    This is the problem with not using units. The whole time I thought that you were getting a ratio (not a power) of 1.72 because you didn't add the "W" after it.

    The answer key is using the convention that net output power is positive. This is an arbitrary choice.
     
  13. Apr 25, 2010 #12
    oh I apologize, I'll avoid doing that in the future.

    Could you elaborate on what you said about the convention?
    All I know that the transfer rate is:
    P= Stefan-boltzmann constant X emessivity X surface area X (T environment^4 - T of the cylinder^4

    Where does a convention come in ?
     
  14. Apr 25, 2010 #13

    D H

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    The convention is which of an outward versus inward net heat transfer is designated as positive versus negative.
     
  15. Apr 25, 2010 #14
    But where does that come in the formula
    P= Stefan-boltzmann constant X emessivity X surface area X (T environment^4 - T of the cylinder^4)

    Edit: I reread the chapter in my book and I found a small part that describes when its positive and when it's negative.

    But there isn't anything in the question that tells us if the boding is absorbing or emitting the radiation.

    A solid cylinder of radius r1 = 2.2 cm, length h1 = 4.9 cm, emissivity 0.81, and temperature 37°C is suspended in an environment of temperature 65°C. (a) What is the cylinder's net thermal radiation transfer rate P1?

    And they're asking about the transfer rate. So why would I assume it's negative
     
    Last edited: Apr 25, 2010
  16. Apr 25, 2010 #15

    Mapes

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    It doesn't appear in the formula. That's why a convention is needed.

    You wouldn't, unless you knew the convention somehow. If you were never taught the convention, and it's not in your book, then the question isn't fair.
     
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