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Homework Help: Heat Flow Through Rock Layers Problem

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem is here: 86BTnXI.png

    2. Relevant equations
    q = heat flow
    k = thermal conductivity
    q = -k (dT/dY)

    3. The attempt at a solution
    While this is quite an easy question, I just want to verify that I'm doing it correctly. Would it be correct to begin at the bottom of the rock layer, since heat flow will migrate away from the source, which is at depth? So, for bottom layer of sandstone, q = - 5.3[ (274.266 / (465-412) ] = 27.4 Wm-2, where dT comes from converting the change in temperature (in celcius) to Kelvin.

    Then I move on to the middle shale layer: q = - 1.7 [ (273.609 / (412-402)] = -46.5 W/m2 .

    However, I am confused how the previous heat flow will affect the heat flow above it. Again, it's an easy question, but I'm hung up on something...

    Last edited: Oct 2, 2014
  2. jcsd
  3. Oct 2, 2014 #2
    Those delta T's are way off. Try again.

  4. Oct 2, 2014 #3
    Put the decimal in the wrong spot. Oops. What about that?
  5. Oct 2, 2014 #4
    You are aware that ##\Delta T## in degrees K is exactly the same as ##\Delta T## in degrees C, correct?

  6. Oct 2, 2014 #5


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    Don't worry about that just yet. Calculate the heat flow through each layer. If they are the same or similar then there isn't really a problem.

    What Chet said. Your thinking about delta T is wrong.
  7. Oct 5, 2014 #6
    OK. I am definitely overthinking this problem. I calculated the heat flow from the bottom layer through to the top layer:
    Sandstone: -100 mW/m^2
    Shale: -78.03 mW/m^2
    Sandstone: 74.03 mW/m^2

    Thanks for the help!
  8. Oct 5, 2014 #7
    That's not what I get. I get a positive upward heat flux for all three layers. For the order of the layers shown in the table,

    Sandstone: 74.04 mW/m^2
    Shale: 78.03 mW/m^2
    Sandstone: 111.6 mW/m^2

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