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So I was given this boundary value problem: A thin rod insulated along the length of $10$ cm with ends kept one at $50$ C and the other in contact with a fluid bath at $150$ C. The initial distribution is given by $f(x)$ where $0 \leq x \leq 10$. I need to find the steady-state solution $U(x)$.

So far from this I have derived the following B.V.P:

$u_t = k u_{xx}$

  • $u(x,0)=f(x)$
  • $u(0,t)=50$
  • $u(10,t)=150$
  • $u_x(10,t)=-h(u(10,t)-150)$ (Newton's Law of Cooling where $h>0$ is just a constant)

Then for my steady-state I let $u_t=0$ $\implies$ $u_{xx}=0$ $\implies$ $U(x)=Ax+B$

Now from the first B.C. I got that $B=50$

I got stuck in trying to find $A$. My first guess would have been that it would be $\frac{T(10)-T(0)}{10}$ as a typical straight line slope however in the solution it is given as $\frac{100h}{1+10h}$. From the Cooling Law B.C. I naturally got $U_x(10)=A=u_x(10,t)_{t\rightarrow\infty}=-h(150-150)=0$, which doesn't make much sense.

I don't know how to approach this, and now I am also doubting whether my B.C. are even correct. Any hints (assuming they are)?

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  • $\begingroup$ There seems to be an 'A' missing from your version of Newtons' Law of Cooling. Is that what the question is asking for? $\endgroup$
    – JMLCarter
    Commented Jun 18, 2018 at 0:43
  • $\begingroup$ @JMLCarter if you are referring to the cross-sectional area of the rod I think all the constants are collected in $h$, or are you referring to $A$ the constant of integration of the steady-state solution? $\endgroup$
    – Jepsilon
    Commented Jun 18, 2018 at 0:46
  • $\begingroup$ The heat transfer coefficient h does not incorporate the the heat transfer surface area A. What's h in this case? en.wikipedia.org/wiki/Newton%27s_law_of_cooling $\endgroup$
    – JMLCarter
    Commented Jun 18, 2018 at 0:54
  • $\begingroup$ @JMLCarter it is just described as a "constant of proportionality" $\endgroup$
    – Jepsilon
    Commented Jun 18, 2018 at 0:56
  • $\begingroup$ You haven't captured the thermal gradient in your expression of Newton's Law of Cooling. From the cool end of the rod to the hot end. $\endgroup$
    – JMLCarter
    Commented Jun 18, 2018 at 1:04

1 Answer 1

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$$u_x(10)=A=-h(10A+50-150)$$ So, $$A=\frac{100h}{1+10h}$$ So $$u(10)=50+10\frac{100h}{1+10h}=\frac{50+1500h}{1+10h}$$ So the temperature at x = 10 is not 150.

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