Heat Flowing from a Chip to Its Heatsink

A copper bar links a hot chip at one end to a cooler heatsink at the other. Starting at room temperature, the bar warms from the chip end and settles into a straight-line temperature profile, the equilibrium. What's left over decays exactly like the heat equation with zero boundaries.

How it works

Nonzero boundary conditions and the equilibrium temperature

In one sentence

For $u_t = k\,u_{xx}$ on $0 < x < L$ with $u(0,t) = T_1$ and $u(L,t) = T_2$ , write $u = u_E + v$ with $u_E(x) = T_1 + (T_2 - T_1)x/L$ . Then $v$ solves the heat equation with $v(0,t) = v(L,t) = 0$ and $v(x,0) = f(x) - u_E(x)$ , so $v = \sum B_n e^{-k(n\pi/L)^2t}\sin(n\pi x/L)$ and $u \to u_E$ as $t \to \infty$ .

General formula

u(x, t) = {\htmlData{sym=0}{u_E(x)}} + {\htmlData{sym=1}{v}}(x, t), \qquad {\htmlData{sym=0}{u_E(x)}} = {\htmlData{sym=2}{T_1}} + ({\htmlData{sym=3}{T_2}} - {\htmlData{sym=2}{T_1}})\frac{x}{\htmlData{sym=4}{{L}}}, \qquad {\htmlData{sym=1}{v}} = \sum_{n=1}^{\infty} {\htmlData{sym=6}{B_n}} e^{-{\htmlData{sym=5}{k}}(n\pi/{\htmlData{sym=4}{L}})^2 t}\sin\frac{n\pi x}{\htmlData{sym=4}{{L}}}

Split the problem in two: where the bar ends up (a straight line) plus how far it still has to go (a transient that fades).

What the symbols mean

$u_E(x)$
Equilibrium temperature
The straight-line profile the bar settles into.
$v$
Transient
How far the bar still is from equilibrium; it decays to zero.
$T_1$
Chip temperature
The fixed temperature at the hot end.
$T_2$
Heatsink temperature
The fixed temperature at the cool end.
$L$
Bar length
10 cm here.
$k$
Thermal diffusivity
How fast heat spreads in the material: about 1.1 cm²/s for copper.
$B_n$
Mode amplitudes
How much of each sine mode the starting temperature contains.

The chip end is held at $T_1$ and the heatsink end at $T_2$ . Those are nonzero boundary conditions, so separation of variables can't be applied directly.
First find the equilibrium: with $u_t = 0$ the heat equation says $u_{xx} = 0$ , a straight line $u_E$ from $T_1$ to $T_2$ .
The leftover $v = u - u_E$ satisfies the heat equation with zero boundary values, so it's a sine series whose modes decay like $e^{-k(n\pi/L)^2t}$ , starting from $v(x, 0) = 25 - u_E(x)$ .
The slowest mode sets the warm-up time: it shrinks by a factor of $e$ every $L^2/(k\pi^2)$ , about 9 seconds for this 10 cm copper bar.

Learn the math

The Method of Separation of VariablesPDE · Lesson 6, Separation of Variables Solving the Heat EquationPDE · Lesson 7, Separation of Variables Nonzero Boundary TemperaturesPDE · Lesson 9, Separation of Variables