Heat Conduction Plate Bench (3D)

The heat equation

Heat flows from hot to cold at a rate given by Fourier's law q = -k ∇T (W/m²). Conservation of energy then gives ρcp ∂T/∂t = ∇·(k ∇T), which for uniform material is ∂T/∂t = α ∇2T with thermal diffusivity α = k/(ρcp). Copper (α ≈ 117 mm²/s) equalises a hand-sized plate in minutes; wood (α ≈ 0.13 mm²/s) takes days. The characteristic time to diffuse across a length L is L²/α.

How the simulation works

The plate is cut into 80 x 60 cells; each cell exchanges heat with its four neighbours through the face conductivity 2k1k2/(k1+k2) (thermal resistances in series), which conserves energy exactly even at a material interface. Time stepping is explicit (forward Euler) and stable only when Δt ≤ ρcp / (2k (1/Δx² + 1/Δy²)); the page always sub-steps below this limit. A fixed-temperature edge conducts through half a cell, an insulated edge has zero flux, and surface losses to the air add -(2h/d)(T - Tamb)/ρcp for a plate of thickness d = 3 mm. Hot and cold spots are clamped to a fixed temperature, like a soldering iron or a cold finger.

Steady state and series resistance

With two fixed edges and insulated sides, the steady flux through two layers is q = (TL - TR) / (LA/kA + LB/kB): the layer with the lowest conductivity carries most of the temperature drop, which is why insulation works.

Honest simplifications

The plate is thin and the temperature is uniform through its thickness (small Biot number); material properties are constant (no temperature dependence); the surface loss is a single linear coefficient (no radiation or natural-convection law); contact resistance at the interface is zero. The 3D companion shows the same model on a bench rig with an infrared-camera view.

Model checks (run in the page on private grids): a two-layer steady wall against the series-resistance solution, the decay of the lowest sine mode against exp(-απ²t/L²), and conservation of total thermal energy on an insulated copper-glass plate.