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throneoo
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Homework Statement
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
Homework Equations
∇2u=r-2 ∂/∂r ( r2 ∂u/∂r ) =D-1∂u/∂t
D>0
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...
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