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Order to create a temperature profile.

  1. Jun 17, 2009 #1
    Hi everyone,

    This is a problem I need to solve in order to create a temperature profile. I have never encountered anything like this, and don't know where to start. Any suggestions?

    [tex]A * \frac{\partial}{\partial x}* \left[ \kappa(T(x)) * \frac{\partial T(x)}{\partial x}\right] = 0 [/tex]

    x : Position
    A : Cross sectional area of the rod
    [tex] T(x) [/tex] : Temperature as a function of position
    [tex] \kappa ( T(x) ) [/tex] : the coefficient of thermal conductivity as a function of Temperature

    I'm trying to solve for [tex] T(x)[/tex].
  2. jcsd
  3. Jun 17, 2009 #2
    Re: Ode

    Asterisks are multiplications?
  4. Jun 17, 2009 #3
    Re: Ode

    Yes, not transforms. That would be insane. =)
    Last edited: Jun 17, 2009
  5. Jun 17, 2009 #4
    Re: Ode

    Hello mherna48,

    Since A is a constant it is divided away with regards to the right hand side. We are left with a derivative equal to 0, giving thus a constant as solution. This means we have (ordinary derivatives because only x is the independent variable and only x):

    [tex]\kappa(T) \cdot \frac{dT}{dx}=\alpha[/tex]

    In which [itex]\alpha[/itex] the first integration constant. Can you proceed from here?

  6. Jun 17, 2009 #5
    Re: Ode

    Thanks for the help coomast.

    I get it to be :

    [tex]T(x) = \frac{\alpha}{\kappa(T(x))} \cdot x + C [/tex]

    Is that right?

    Can I multiply both sides by [tex]\kappa(T(x))[/tex] and square root the right side? Or is that not possible because of the constant?
  7. Jun 17, 2009 #6
    Re: Ode

    Mmmmm, I get the following:

    [tex]\alpha \cdot x + \beta = \int \kappa(T) \cdot dT[/tex]

    In which [itex]\alpha[/itex] and [itex]\beta[/itex] are the integration constants. You can further calculate the integral once you know the dependency of the conductivity with temperature.

    Does this clearify things?

  8. Jun 17, 2009 #7


    User Avatar
    Science Advisor

    Re: Ode

    I am puzzled by this "[itex]\kappa(T(x))[/itex]" What is the reason for the first set of parentheses? Is [itex]\kappa[/itex] a function, so this is [itex]\kappa[/itex] of T(x) or is [itex]\kappa[/itex] simply a constant, so this is [itex]\kappa[/itex] times T (which I would write simply as [itex]\kappa T(x)[/itex]).

    In either case, you can't just tread "[itex]\kappa(T(x))[/itex] as a constant as you seem to be trying to do.

    If it is [itex]\kappa[/itex] times T(x), then [itex]\kappa T \frac{dT}{dx}= \alpha[/itex] can be writen as
    [itex]T dT= \frac{\alpha}{\kappa} dx[/itex]
    and, integrating,
    [itex]\frac{1}{2}T^2= \frac{\alpha}{\kappa}x+ C[/itex]

    If [itex]\kappa[/itex] is a function of T, then
    [itex]\kappa(T)dT= \alpha dx[/itex]
    and, integrating,
    [itex]\int \kappa(T)dT= \alpha x+ C[/itex]
    essentially what coomast gave.

    How you would solve that for T depends heavily on what function [itex]kappa[/itex] is.
  9. Jun 18, 2009 #8
    Re: Ode

    Thanks for the help guys.

    [itex]\kappa(T(x))[/itex] is read as Thermal Conductivity as a function of Temperature, and Temperature is a function of position.

    And yes, I see where I goofed on my calculation. I made a weird mistake and I don't know what I was thinking there. How would you integrate the right side of the equation, or in your case left side:
    [tex] \int \kappa(T(x)) \cdot dT[/tex]

    Is that even possible?
  10. Jun 18, 2009 #9


    User Avatar
    Science Advisor

    Re: Ode

    So what you are saying is that [itex]\kappa(T(x))[/itex] is a function of x. Obviously, how you would integrate that depends on exactly what function that is!
  11. Jun 18, 2009 #10
    Re: Ode

    Hmm. I figured it wouldn't be easy. Maybe I can have Matlab solve it numerically. Thanks for the help everyone.
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