Question #157213

A 10-FT LONG ROD HAS BOTH OF ITS ENDS INSULATED DURING HEATING OPERATION. BUT EVEN AT ITS INITIAL OPERATION (T=0), THE ROD IS ALREADY LATERALLY INSULATED. FIND U(X,T) IF THE THERMAL DIFFUSIVITY EQUALS 1.5 (X=5.0)



1
Expert's answer
2021-01-28T04:42:33-0500

The heat equation:

ut=k2ux2\frac{\partial u}{\partial t}=k\frac{\partial^2u}{\partial x^2}


We have:

k=1.5k=1.5

uxx=0=uxx=L=0\frac{\partial u}{\partial x}_{x=0}=\frac{\partial u}{\partial x}_{x=L}=0

f(x)=u(x,0)=5f(x)=u(x,0)=5


The solution of heat equation:

u(x,t)=(Acosλx+Bsinλx)ek2λ2tu(x,t)=(Acos\lambda x+Bsin\lambda x)e^{-k^2\lambda^2t}


Applying conditions:

u(x,0)=Acosλx+Bsinλx=5u(x,0)=Acos\lambda x+Bsin\lambda x=5


ux=(BλcosλxAλsinλx)ek2λ2t\frac{\partial u}{\partial x}=(B\lambda cos\lambda x-A\lambda sin\lambda x)e^{-k^2\lambda^2t}


uxx=0=Bλek2λ2t=0    B=0\frac{\partial u}{\partial x}_{x=0}=B\lambda e^{-k^2\lambda^2t}=0\implies B=0


uxx=L=Aλsin(λL)ek2λ2t=0    λ=πn/L\frac{\partial u}{\partial x}_{x=L}=-A\lambda sin(\lambda L)e^{-k^2\lambda^2t}=0\implies \lambda=\pi n/L


u(x,t)=5ek2λ2t=5ek2π2n2t/L2u(x,t)=5e^{-k^2\lambda^2t}=5e^{-k^2\pi^2n^2t/L^2}


u(x,t)=5e2.25π2n2t/100=5e0.22n2tu(x,t)=5e^{-2.25\pi^2n^2t/100}=5e^{-0.22n^2t}


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