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Thermodynamics - Thermal expansion (linear)

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Two cubic samples of the same dimensions (length L) have thermal conductivities [tex]\kappa_1[/tex] and [tex]\kappa_2[/tex]. The slabs are placed adjacently horizontally | 1 | 2 |
    Obtain expressions for the heat flow rate through the composite slab if the temperature difference is applied
    a) from top to bottom
    b) from left to right

    2. Relevant equations

    Heat transfer rate: Q' = [tex]\frac{dQ}{dt}[/tex] = [tex]\kappa \cdot A \cdot \frac{\Delta T}{L}[/tex]

    Thermal expansion: [tex]\kappa = \frac{\Delta L}{L_0 \Delta T}[/tex]

    3. The attempt at a solution
    I'm having trouble even understanding the question. What I can figure out is the above equation to find the expansion length. Then with regard to heat transfer rate, the cross sectional area will be A = 2L² with [tex]\Delta T[/tex] , [tex]\kappa[/tex] and L initially given.

    At first guess I'm think for a), the area is 2A (horizontally slice); and for b) its just A (vertical slice). And then just plugged into the equation for heat transfer rate. But the question is for quite a bit of marks so this seems too easy :P
     
    Last edited: Jul 25, 2010
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  3. Jul 25, 2010 #2

    Mapes

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    The difficulty is in how to handle the two different thermal conductivity values.
    This is not a thermal expansion problem, it's a conservation of energy problem: heat going in must equal heat going out.
    Once you get your final expressions, a useful way to check them is to imagine that [itex]\kappa_1\gg\kappa_2[/itex] (or the other way around). What, intuitively, will happen to the heat flow in the two samples? What do your equations predict?
     
  4. Jul 28, 2010 #3
    One part of the problem is I'm not sure about heat transfer rate. I think it's how fast a certain amount of heat will pass between two points, and that this will increase if the straight distance is increased (like between the ends of a thin rod) or area increased (between a large flat plate). That's what I can get out of the formula, as well as the thermal conductivity of the substance and temperature difference between the points.

    Back to this question, if [tex] /kappa_1 > /kappa_2[/tex] then you will have one block getting much bigger. Perpendicular to their contact surface, this length will increase. But parallel to the surface, you will have unequal surfaces... is it fair to just use the average length then?

    That is, will the heat transfer rate be the same between two identical cubes however they are placed with one side in contact? Like whether they're completely or partially overlapping; when you take the rate between the uneven surfaces.
    (Bad picture)
    |####| ___
    |####||##|
    |####||##|
    |####|
    From left to right, the distance remains the same, but top down it changes...
     
  5. Jul 29, 2010 #4

    Mapes

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    Again, this isn't an expansion problem. Thermal conductivity has nothing to do with thermal expansion. Your second equation above should be

    [tex]\alpha= \frac{\Delta L}{L_0 \Delta T}[/tex]

    where [itex]\alpha[/itex] is the thermal expansion coefficient, a totally different parameter.

    Back to conservation of energy: write your first equation above for each block individually. How do the rates need to be combined when the flux is horizontal? How is it different when the flux is vertical?
     
  6. Jul 29, 2010 #5
    omg! Thanks for clearing that up, I misread the entire question because of one word. Thing is rest of the questions were about expansion so I just confused that one up.

    Right, so then the sizes of the blocks doesn't change. The length is L and 2L for top-down and left-right respectively, and the volume stays 2L³.

    For each block individually, I think it's
    [tex]\frac{dQ}{dt} = \kappa_1 \cdot L^2 \cdot \frac{\Delta T}{L}[tex]
    and similarly for block 2 (simplified)
    [tex]\frac{dQ}{dt} = \kappa_2 \cdot L \cdot \Delta T[tex]

    The dimensions of L are no problem, but I'm still uncertain how heat will distribute itself over 2 different conductivities. In the below equation I don't know if I should multiple, divide or add the 2 values. For left to right, do I just take the average between the two transfer rates for each block?

    Then from top to bottom:
    [tex]\frac{dQ}{dt} = \kappa_1 ?
    kapap_2 \cdot 2L^2 \cdot \frac{\Delta T}{L}[tex]

    Left to right:
    [tex]
    frac{\frac{dQ}{dt} = \kappa_1 \cdot L^2 \cdot \frac{\Delta T}{L}}{\frac{dQ}{dt} = \kappa_2 \cdot L \cdot \Delta T}
    [tex] which just reduces to the ration between conductivites..
     
  7. Jul 29, 2010 #6

    Mapes

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    Can you fix your TeX? :smile:
     
  8. Aug 3, 2010 #7
    Bleah, forgot the / and couldn't edit :p

    For each block individually, I think it's
    [tex]\frac{dQ}{dt} = \kappa_1 \cdot L^2 \cdot \frac{\Delta T}{L}[/tex]
    and similarly for block 2 (simplified)
    [tex]\frac{dQ}{dt} = \kappa_2 \cdot L \cdot \Delta T[/tex]

    The dimensions of L are no problem, but I'm still uncertain how heat will distribute itself over 2 different conductivities. In the below equation I don't know if I should multiple, divide or add the 2 values. For left to right, do I just take the average between the two transfer rates for each block?

    Then from top to bottom:
    [tex]\frac{dQ}{dt} = \kappa_1 \cdot
    kappa_2 \cdot 2L^2 \cdot \frac{\Delta T}{L}[/tex]

    Left to right:
    [tex]
    \frac{dQ}{dt} = \kappa_1 \cdot L^2 \cdot \frac{\Delta T}{L} \frac{dQ}{dt} = \kappa_2 \cdot L \cdot \Delta T}
    [/tex] which just reduces to the ration between conductivites..
     
  9. Aug 3, 2010 #8

    Mapes

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    This equation predicts that no heat will flow if one of the thermal conductivities is zero (that is, if one of the samples is a thermal insulator). But this doesn't sound right; heat could still pass through the other sample. So it looks like you need another combination.

    Be careful - the temperature difference [itex]\Delta T[/itex] is now different for each sample.

    Use conservation of energy: what goes in must come out of each block.
     
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