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Steady State Heat Equation in a One-Dimensional Rod

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the equilibrium temperature distribution for a one-dimensional rod composed of two different materials in perfect thermal contact at x=1. For 0<x<1, there is one material (cp=1, K0=1) with a constant source (Q=1), whereas for the other 1<x<2 there are no sources (Q=0, cp=2, K0=2) with u(0) = 0 and u(2) = 0.

    2. Relevant equations
    heat equation: 2rwt0z6.png
    3. The attempt at a solution
    First I did the calculations for the first part of the rod (0<x<1)
    2h30xe8.png
    Then I did the calculations for the second part of the rod (1<x<2)
    212u6v9.png
    Then I tried to set them equal at x=1
    2s9947a.png
    And now I'm stuck. I can't figure out how to find out the values of the constants, and I don't feel confident. I feel like I'm doing something wrong.
     
  2. jcsd
  3. Feb 7, 2012 #2

    vela

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    You need a second boundary condition. Think about the fact that the heat flow out of one region has to equal the heat flow into the other to conserve energy.
     
  4. Feb 7, 2012 #3
    Oh! I get it. Since the bar is in equilibrium, the heat will be flowing away from the heat source into the sinks at the end. That means that the the heat flows should be equal at x=1:
    image.png
    The solving the system of equations yeilds:
    image.png
    And the final solution should be:
    image.jpg
    I checked the boundary conditions and they all seemed correct. Thank you! :)

    P.S. Sorry for posting in the wrong forum. I wasn't sure where this topic went. Thank you for moving it!
     
  5. Feb 7, 2012 #4

    vela

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    Is the heat flow simply du/dx or do you have to throw the Ks in there? I don't know offhand, but it might be something for you to check.

    No problem.
     
  6. Feb 9, 2012 #5
    You must equate fluxes so the conductivity is definitely needed.

    You have continuity of both heat flux and temperature at the interface.
     
  7. Feb 14, 2012 #6
    That makes sense. So instead of [itex]\frac{\partial^2u_1}{\partial x^2} = \frac{\partial^2u_2}{\partial x^2}[/itex], it should have been [itex]\frac{\partial^2u_1}{\partial x^2} = 2\frac{\partial^2u_2}{\partial x^2} [/itex] since I'm setting the fluxes equal.

    Thanks for that extra insight. :smile:
     
  8. Feb 15, 2012 #7
    You must use the first derivative not the second derivative to equate heat fluxes. Eache is multiplied by its respective thermal conductivity.
     
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