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Spherically symmetric Heat Equation

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball of radius a, originally at T0, is immersed to boiling water at T1 at t=0. From t≥0, the surface (of the ball) is kept at T1
    Define u(r,t)=R(r)Q(t)=T(r,t)-T1
    ΔT=T0-T1<0

    r,t≥0

    2. Relevant equations
    2u=r-2 ∂/∂r ( r2 ∂u/∂r ) =D-1∂u/∂t
    D>0
    3. The attempt at a solution
    Boundary Condition: u(r=a,t)Ξ0
    Initial Condition: u(r,t=0)=ΔT*Top(0,a) where Top(0,a)=Θ(r)-Θ(r-a) is the top hat function with end-points 0 and a

    This is not a precise formulation as I want u(0,0) to be ΔT and u(a,0)=0. However it conveniently represents u(0≤r<a,0)=ΔT and u(r>a,0)=0

    Let u(r,t)=R(r)Q(t)

    Separation of variables give
    (Rr2)-1d/dr (r2 R')=(D*Q)-1Q'=-μ2 where μ∈R

    The choice of the separation constant is such that the boundary condition can be satisfied

    rR''+2R'+μ2rR=0
    Q'+μ2DQ=0

    The radial equation is simplified by the substitution S=rR, reducing it to S''+μ2S=0

    Thus S=sin(μr) or cos(μr) and Q=exp(-μ2D*t) and u=SQ/r

    Since I expect u to be finite at r=0, S(r=0)=0 is needed to create an indeterminate expression, ruling out the cosine solution.

    S(r=a)=sin(μa)=0⇒μna=nπ ; n∈ℕ

    So the solution has the form
    u(r,t)=r-1∑An sin(μn r)*exp(-μn2Dt)
    u(r,0)=r-1∑An sin(μn r)
    u(r,0)*r=∑An sin(μn r)-----------------------(@)

    Now multiply both sides with sin(μm r) and integrate both sides of (@) w.r.t. r from 0 to a
    I found that An=2ΔT/(π*n/a) * (-1)n+1 ;



    I'd appreciate it if someone could point out where I made a mistake...
     
    Last edited: Mar 14, 2016
  2. jcsd
  3. Mar 19, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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