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Rectangular Plate BVP problem

  1. Feb 3, 2009 #1
    THe following exercise deal with the steady state distribution of the temperature in either 2-dimensional plates or 3-dimensional regions.
    Problem: A 10X20 rectangular plate with boundary conditions . at the lower side where there is poor insulation the normal derivative of the temperature is equal to 0.5 times the temperature

    The rectangular 10 X 20 plate is as follows

    bottom part : poor insulation
    left side: Insulation
    right side: Kept at 10 degrees
    top side: Kept at 20 degrees

    Write the BVP

    Solution: What i found was the expression for poor insulation but i think this is more for one space variable, [tex]u_{x}(0,t)[/tex] + au(0,t) = 0
    i wrote some BVP:
    [tex]u_{x}(0,y) = 0[/tex]
    u(x,0) = poor insulation ( i dont know how to do this part)
    u(10,y) = 10
    u(x,20) = 20

    Is this even right?
    any help would be appreciated
  2. jcsd
  3. Feb 4, 2009 #2
    Hello jc2009,
    The following is correct:
    u(10,y) = 10
    u(x,20) = 20
    The left hand side is insulated, this means that no heat passes through this boundary and thus the heat flux is zero. This is given by the law of Fourier, or:
    [tex]\frac{\partial u}{\partial x}=0[/tex]
    The bottom is now easily found as:
    [tex]\frac{\partial u}{\partial y}=\frac{1}{2}\cdot u[/tex]
    Keep in mind the meaning of the normal for the two boundaries.
    The PDE of the BVP problem is:
    [tex]\frac{\partial u}{\partial t}=k\cdot \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right][/tex]
    Now it should be possible to write down the complete BVP.
    Hope this helps
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