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This question is inspired by a question from 'Robomaze'. The OP now seems to have lost interest in it but I still think it's an interesting issue.

I've cleaned up the original question and made it more concise.


A pipe (or rod) of total length $2L_2$ has a central section of length $2L_1$ that is heated constantly and uniformly at a rate of $q\,\mathrm{Wm^{-1}}$. The non-heated part loses heat through convection (but not radiation)

What is the temperature profile ($T(x)$) of the pipe at steady state?

centrally heated pipe

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1 Answer 1

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  1. Heated section:

The go-to equation here is Fourier's Heat Equation (with thermal load $q$):

$$\frac{\partial T}{\partial t}=\alpha \frac{\partial^2 T}{\partial x^2}+q$$ Or: $$T_t=\alpha T_{xx}+q\tag{1}$$ 2. Non-heated sections:

These can be treated as convective heat losing fins, the derivation of which can be found here. The resulting differential equation is (also steady state): $$T''-\frac{Ph}{Ak}(T-T_{\infty})=0\tag{2}$$ Let's make a small transformation of dependent variable: $$T-T_{\infty}=u$$


Because we're looking for steady state, so $u_t=0$, $(1)$ becomes: $$\alpha u''(x)+q=0\tag{3}$$

$(3)$ integrates easily: $$u'=-\frac{q}{\alpha}x+c_1$$ $$u=-\frac{q}{2\alpha}x^2+c_1 x+ c_2$$ Also, because the system's geometry is symmetric about $x=0$, that means the maximum of $u$ must also be at $x=0$, so: $$u'(0)=0 \Rightarrow c_1=0$$ $$\Rightarrow u(x)=-\frac{q}{2\alpha}x^2+ c_2$$ $(2)$, after the transformation of $T$ becomes:

$$u''-m^2u=0\tag{4}$$ $$\text{where }m^2=\frac{Ph}{Ak}$$ (For the definition of symbols please consult the link above)

$(4)$ solves to: $$u(x)=c_3e^{mx}+c_4e^{-mx}$$ As a boundary condition, I choose 'no heat loss at the end-points', so: $$u'(L_2)=0$$

Thus: $$mc_3e^{mL_2}-mc_4e^{-mL_2}=0\tag{5}$$

The temperature profile must be differentiable so that the derivatives in $L_1$ must be equal:

$$mc_3e^{mL_1}-mc_4e^{-mL_1}=-\frac{q}{\alpha}L_1\tag{6}$$

$(5)$ and $(6)$ allow to determine $c_3$ and $c_4$.

Finally there can only be one value of $T_1$, so:

$$-\frac{q}{2\alpha}L_1^2+ c_2=c_3e^{mL_1}+c_4e^{-mL_1}\tag{7}$$

With knowledge of $c_3$ and $c_4$, $c_2$ can then be determined.

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