Everything so far assumed a flat wall: the same area at the hot face and the cold face. Wrap a pipe and that stops being true — the outside of the lagging has far more surface than the inside, and the heat spreads out as it crosses. Integrate across that growing area and a logarithm falls out: $\dot{Q} = \dfrac{2\pi k L \, \Delta T}{\ln(r_2 / r_1)}$ — read aloud Q-dot equals two pi k L delta-T over the natural log of r-two over r-one.

Every letter, in words. $\dot{Q}$ is the heat flow in watts. $k$ is the lagging's thermal conductivity in W/(m·K). $L$ is the length of pipe in metres. $\Delta T$ is the temperature difference across THIS layer, inner face to outer face, in kelvin. And the subscript convention, stated once and never broken: $r_1$ is the inner radius of the layer, $r_2$ is its outer radius, both in metres. A nugget for free — they appear only as a ratio, so millimetres over millimetres works exactly as well, and diameters over diameters give the same logarithm too.

Then the result nobody believes the first time. On a small enough cylinder, adding insulation can INCREASE the heat loss. Lagging adds resistance, yes — but it also adds outer surface for the air to strip heat from, and on a thin pipe the second effect can win. The turning point is the critical radius: $r_{cr} = \dfrac{k}{h}$ — r-critical equals k over h, where $k$ is the insulation's conductivity in W/(m·K), $h$ is the outside film coefficient in W/(m²·K) (how briskly the surrounding air carries heat away), and $r_{cr}$ is a radius in metres.

Put numbers on it: ordinary lagging at 0.05 W/(m·K) in still air at 10 W/(m²·K) has a critical radius of 5 mm. Every steam main in the plant is already far outside that, which is why lagging works and why nobody thinks about this on pipework. Instrument leads, thermocouple wire and fine tubing are another matter entirely — and cable designers sometimes exploit it deliberately, jacketing a conductor to help it shed heat.