= Thermal conduction
{wiki}
= Heat conduction
{synonym}
<Thermal conduction> transfers <heat> through microscopic interactions without bulk transport of matter. To predict the heating load through a bar, combine <Fourier's law> with <conservation of energy> and the imposed end <temperatures>. In the <isotropic> model with positive <thermal conductivity> $k$, the conductive <heat flux> points from high to low <temperature>, opposite to the <temperature gradient>. A spoon warming along its handle and <heat> escaping through a wall are familiar examples.
Consider a bar of <length> $L=0.10\,\mathrm m$, constant cross-sectional <area> $A=0.010\,\mathrm{m^2}$ and constant <thermal conductivity> $k=2\,\mathrm{W\,m^{-1}\,K^{-1}}$. Maintain its ends at $T_h=40\,{}^\circ\mathrm C$ and $T_c=20\,{}^\circ\mathrm C$, with $x$ increasing toward the colder end. Assume a <steady state>, one-dimensional <thermal conduction>, no internal sources and no <heat transfer> through the sides. The <temperature> falls by $20\,\mathrm K$ over $0.10\,\mathrm m$, giving
$$
\frac{dT}{dx}=-200\,\mathrm{K\,m^{-1}},\qquad
q=-k\frac{dT}{dx}=400\,\mathrm{W\,m^{-2}},\qquad
\dot Q=Aq=4\,\mathrm W.
$$
The difference between the two readings in <degree Celsius> units is the same $20$-<kelvin> difference. The positive <heat flux> carries <heat> toward increasing $x$.
\Image[/fourier-conduction-steady-bar.png]
{title=Steady <thermal conduction> through a bar: declining <temperature>, constant <heat flux>}
{description=An <isotropic> bar with stipulated constant <thermal conductivity> $k=2\,\mathrm{W\,m^{-1}\,K^{-1}}$, <length> $0.10\,\mathrm m$ and cross-sectional <area> $0.010\,\mathrm{m^2}$. Its ends remain at $40$ and $20\,{}^\circ\mathrm C$, and its sides exchange no <heat>. The lower panel distinguishes $q=400\,\mathrm{W\,m^{-2}}$ from $\dot Q=4\,\mathrm W$.}
The general solution under these assumptions is
$$
T(x)=T_h-\frac{T_h-T_c}{L}x,\qquad
q=\frac{k(T_h-T_c)}{L},\qquad
\dot Q=\frac{kA(T_h-T_c)}{L}.
$$
A proof outline is enough to explain the straight line: <conservation of energy> in any small slice, with no accumulation, sources or side losses, gives $d\dot Q/dx=0$. Constant cross-sectional <area> gives $dq/dx=0$; constant <thermal conductivity> and <Fourier's law> then give $d^2T/dx^2=0$. Two <integrations> and the end <boundary conditions> determine the displayed <affine function>. This is why a visually simple <temperature> profile can encode a continuous flow of <energy>. https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node117.html[E. M. Greitzer, Z. S. Spakovszky and I. A. Waitz, MIT Thermodynamics and Propulsion, section 16.3: steady one-dimensional conduction]{ref}
Doubling the <length> halves the rate, while doubling the cross-sectional <area> doubles it, at fixed end <temperatures> and <thermal conductivity>. This explains why a long narrow support can limit unwanted <heat transfer>, while a short broad path can help remove <heat>. Joining materials leads to <thermal resistance> in series. During initial warming, <internal energy> can accumulate and the <temperature> profile need not be a straight line; the time dependence is described by the <heat equation> and <thermal diffusivity>.
Now join two layers, each of <length> $L_1=L_2=0.10\,\mathrm m$, with constant cross-sectional <area> $A=0.010\,\mathrm{m^2}$ and stipulated constant <thermal conductivities> $k_1=1$ and $k_2=4\,\mathrm{W\,m^{-1}\,K^{-1}}$. Keep the outer ends at $40$ and $20\,{}^\circ\mathrm C$. Assume a <steady state>, one-dimensional <thermal conduction>, no internal sources or side losses, and perfect contact: the <temperature> is continuous at the interface, with no extra contact drop. For each layer, use the area-specific <thermal resistance> $r_i=L_i/k_i$ and $R_i=r_i/A$. The series calculation gives
$$
\begin{aligned}
r_1+r_2=\frac{L_1}{k_1}+\frac{L_2}{k_2}
=0.100+0.025=0.125\,\mathrm{m^2\,K\,W^{-1}},\\
R_1+R_2=\frac{r_1+r_2}{A}=12.5\,\mathrm{K\,W^{-1}},\\
q=\frac{20\,\mathrm K}{0.125\,\mathrm{m^2\,K\,W^{-1}}}
=160\,\mathrm{W\,m^{-2}},\qquad \dot Q=Aq=1.6\,\mathrm W,\\
\Delta T_1=qr_1=16\,\mathrm K,\qquad
\Delta T_2=qr_2=4\,\mathrm K,\qquad
T_{\rm interface}=40\,{}^\circ\mathrm C-16\,\mathrm K=24\,{}^\circ\mathrm C.
\end{aligned}
$$
\Image[/layered-thermal-conduction.png]
{title=Two layers in series: continuous <heat flux>, different <temperature gradients>}
{description=Each layer is $0.10\,\mathrm m$ long with the same cross-sectional <area> $0.010\,\mathrm{m^2}$ and perfect contact. Stipulated constant <thermal conductivities> are $1$ and $4\,\mathrm{W\,m^{-1}\,K^{-1}}$. The <temperature> drops by $16\,\mathrm K$ and $4\,\mathrm K$, reaching $24\,{}^\circ\mathrm C$ at the interface, while $q=160\,\mathrm{W\,m^{-2}}$ and $\dot Q=1.6\,\mathrm W$ throughout.}
The <heat flux> is continuous: otherwise <energy> would accumulate at the interface. Yet <Fourier's law> gives different <temperature gradients>, $dT/dx=-q/k_1=-160\,\mathrm{K\,m^{-1}}$ and $dT/dx=-q/k_2=-40\,\mathrm{K\,m^{-1}}$. Lower <thermal conductivity> requires a steeper <temperature gradient> to carry the same <heat flux>; equal <lengths> therefore give drops in the ratio $4:1$. A kink in the <temperature> profile records the change of material while perfect contact keeps the <temperature> itself continuous.
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