Heat Transfer · interactive worked example
An electrical current heats a copper rod. It loses that energy two ways at once — convection to the surrounding air, and radiation to its surroundings — and they don't add up the same way. Watch the rod heat up live, and see which mode actually controls the answer.
A real problem from the site author's own numerical-methods teaching notes: "Convection and Radiation Heat Transfer from a Heated Copper Rod."
Rod temperature T(t) — heating from ambient to steady state
—
Cu rod: D=0.015 m, L=1 m
ρ=8954 kg/m³, c_p=383.1 J/kg·K
R=125 Ω/m, I=2 A → I²R=500 W/m
T∞ = T_surr = 298.15 K (25°C)
Strictly, an exponential-type approach to equilibrium never exactly reaches it — so "time to steady state" always means "time to get within some tolerance of it." This lab uses 99.95% of the final temperature rise; the original paper used a looser visual read of its plot, which is why the two won't match to the second. The steady-state temperature itself has no such ambiguity — it's an exact root of the energy balance, and matches the paper to the figures it reported (786.2 K, 583.6 K, 308.8 K for the three cases below).
Convective loss is linear in (T−T∞); radiative loss is a difference of fourth powers, T⁴−T_surr⁴. At modest temperature rises above ambient, the radiative term is tiny compared to convection — copper's natural emissivity (ε≈0.04, a polished metal) barely radiates at all. That's why, at the paper's default settings, the rod heats up almost entirely by fighting convection, and the radiative term barely moves the answer.
Push ε up to 0.8 (an oxidized or painted surface) and radiative loss stops being negligible — it now competes with convection for carrying heat away, so the steady-state temperature drops substantially and the rod reaches equilibrium faster, since there's more total loss pulling it back down. Push h up to 1000 W/m²·K instead (forced convection — a fan blowing across the rod) and convection alone dwarfs both the heating and the radiative term: the rod barely warms above ambient at all, and reaches steady state in seconds rather than hours.
This is the same Stefan-Boltzmann physics as the Thermal Radiation lab and the same Newton's-law-of-cooling physics as the Convection lab — just forced to share one energy balance instead of being studied in isolation. Most real surfaces lose heat both ways simultaneously; which one actually matters for your design depends entirely on the temperature range and the surface finish, exactly as this lab shows.
EngineeringCandy · Nonlinear ODE solved live by RK4 · steady state found by root-bracketing