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Heat conduction through sphere

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data

    In general, a sphere with conductivity ##\kappa##, heat capacity per unit volume ##C## and radius ##R## obeys the differential equation at time t:

    [tex] C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r} [/tex]

    Part (a): A sphere with a cavity of radius ##a## generates heat at a rate Q. Heat is lost from the outer surface of the sphere to the surroundings, with surrounding temperature ##T_s##, given by Newton’s Law of Cooling with constant ##\alpha## per unit area. Find the temperature ##T_a## at the surface of the cavity at thermal equilibrium.

    Part (b): A second sphere without a cavity generates heat uniformly at ##q## per unit volume. Like the first sphere, heat is lost to the surroundings at its surface. Find temperature at the center of the sphere ##T_0## at equilibrium.

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    Heat generated in cavity = Heat loss at surface
    [tex]Q = \alpha (T_s - T_R)(4\pi R^2) [/tex]
    [tex]T_R = T_s - \frac{Q}{\alpha (4\pi R^2)} [/tex]

    We will use this to solve for the constants in the differential equation later on.

    At steady state, ##C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r} = 0##.

    [tex]\frac{\partial^2 T}{\partial r^2} + \frac{2}{r}\frac{\partial T}{\partial r} = 0 [/tex]

    Solving, we get:

    [tex]T = A + \frac{b}{r}[/tex]

    We need one more equation with the one for ##T_R## to solve for constants ##A## and ##B##.

    Using ##\int \vec J \cdot d\vec S = k \int \nabla \vec T \cdot d\vec S##:
    [tex]Q = -\kappa \frac{\partial T}{\partial r}(4\pi r^2)[/tex]
    [tex]Q = 4\pi \kappa b[/tex]
    [tex]b = \frac{Q}{4\pi \kappa}[/tex]

    Solving for A:
    [tex]A = T_s - \frac{Q}{4\pi \alpha R^2} - \frac{Q}{4\pi \kappa R} [/tex]

    Together, temperature at surface of cavity is:

    [tex]T_a = T_s - \frac{Q}{4\pi \alpha R^2} + \frac{Q}{4\pi \kappa}\left( \frac{1}{a} - \frac{1}{R} \right) [/tex]

    Part(b)

    I'm not sure how to approach this, as ##T = A + \frac{b}{r}## doesn't work at ##r=0##..

    I have found the temperature at surface though:
    [tex]Q = \frac{4}{3}\pi R^3 q = \alpha (T_s - T_R)(4\pi R^2)[/tex]
    [tex]T_R = T_s - \frac{Rq}{3\alpha}[/tex]
     
    Last edited: Jun 9, 2014
  2. jcsd
  3. Jun 9, 2014 #2

    maajdl

    User Avatar
    Gold Member

    You need to add the source term (q) in heat transfer partial differential equation.
    You then again solve for the steady state solution.
     
  4. Jun 9, 2014 #3
    I don't understand what you mean. And is this for part (a) or part (b)?
     
  5. Jun 9, 2014 #4
    I think you made a sign mistake in the Newton's law of cooling (part a). For part b you need a source term. That means you will solve the non-homogeneous heat conduction differential equation.
     
  6. Jun 10, 2014 #5

    maajdl

    User Avatar
    Gold Member

    This is for part b.

    [tex] C\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial r^2} + \frac{2\kappa}{r}\frac{\partial T}{\partial r} + q [/tex]
     
    Last edited: Jun 10, 2014
  7. Jun 10, 2014 #6
    For part (a), is the sign error here:

    [tex]Q = \alpha (T_s - T_R)(4\pi R^2)[/tex]

    It should be ##T_r - T_s##, since the sphere is hotter. Other than that, is the rest of my working fine?


    For part (b), I read up more on the chapter and I realized there should be a source per-unit-volume term per unit time term.

    Thanks alot!
     
  8. Jun 12, 2014 #7
    I have given this question another go. At steady state, the DE now becomes:

    [tex]\kappa \frac{\partial^2T}{\partial r^2} + 2\frac{\kappa}{r} \frac{\partial T}{\partial r} + q = 0 [/tex]

    Since the homogeneous solution is ##A + \frac{b}{r}##, we now seek the inhomogeneous solution.

    I try ##\beta r^2## where I found ##\beta = -\frac{q}{6k}##.

    So the solution for T is:
    [tex]T = A + \frac{b}{r} - \frac{q}{6k}r^2[/tex]

    But now as ##r \rightarrow \infty##, ##T\rightarrow \infty##..
     
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