= Steady temperature of a uniformly heated cylindrical wire
{title2=$T(\rho)=T_{\rm air}+\frac{Ha}{2\gamma}+\frac{H(a^2-\rho^2)}{4\kappa}$}
For a long homogeneous circular wire of radius $a$, uniform volumetric heating $H$, constant <thermal conductivity> $\kappa$ and negligible end losses, the steady <heat equation> is $\kappa\rho^{-1}(\rho T')'+H=0$. Regularity at the axis excludes a <logarithm> term, giving $T=T_s+H(a^2-\rho^2)/(4\kappa)$. If the surface obeys <Newton's law of cooling>, $-\kappa T'(a)=\gamma(T_s-T_{\rm air})$, then $T_s=T_{\rm air}+Ha/(2\gamma)$. Here $\gamma$ has units $\mathrm{W\,m^{-2}K^{-1}}$ and $H$ has units $\mathrm{W\,m^{-3}}$. For uniform <electric current> $I$ and constant <electrical conductivity> $\sigma$, <Joule heating> gives $H=I^2/(\sigma\pi^2a^4)$.
Back to article page
