Critical Radius of Insulation

Wrapping a cylinder in insulation does two opposite things: it adds conduction resistance, which cuts the loss, and it enlarges the outer surface, which raises the convective loss. Differentiate the total resistance and the two effects balance exactly at $r_{\mathrm{cr}} = k/h$. Below that radius, the first millimetres of lagging make the loss worse. For lagging at k = 0.05 W/(m·K) in still air at h = 10 W/(m²·K), $r_{\mathrm{cr}}$ = 5 mm, so any pipe larger than a 10 mm-diameter tube is already past the peak and insulation only helps.

The number matters far more in electrical work than in piping. Wire insulation has k ≈ 0.15 W/(m·K) and sits in near-still air at h ≈ 8, giving a critical radius near 19 mm: larger than most conductors, which means the plastic jacket on a small cable genuinely helps it run cooler while doubling as insulation. Deliberately exploiting this is standard practice for fine thermocouple leads and small transistors. The trap is applying the cylindrical result to a flat wall or a sphere: a plane wall has no critical thickness at all, and for a sphere the answer is 2k/h.