Conduction Through a Pipe Wall

Q˙=2πkL ΔTln⁡(r2/r1)\dot{Q} = \frac{2 \pi k L \, \Delta T}{\ln(r_2 / r_1)}

Worked example: 10 m pipe, 50 → 100 mm lagging, k=0.05, 100 K → 453.2 W — press Try an example to run it live, then adjust anything.

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Conduction Through a Pipe Wall explained

r1r2kLQΔT

A pipe wall is not a flat slab: heat spreading outward keeps finding more area, so the resistance per unit thickness falls as you go, and the geometry hands you a logarithm instead of a simple L/kA. Ten metres of pipe lagged with 50 mm of k = 0.05 W/(m·K) insulation, from r₁ = 50 mm to r₂ = 100 mm with 100 K across it, loses 2π × 0.05 × 10 × 100 ÷ ln(2) = 453 W. Double the insulation thickness again, to r₂ = 150 mm, and the loss only falls to 279 W — the logarithm is a law of diminishing returns, and it is why insulation schedules stop where they do.

The trap is layering. Each layer needs its own ln(r₂/r₁) with its own k, and its own ΔT; you cannot average the conductivities across a jacketed system. In steam service the second trap is condensate: a 100 mm line at 180 °C bare loses roughly ten times what the lagged line does, and every watt of that loss is steam condensing somewhere it was not meant to, filling traps and hammering elbows. Run this page backwards on a measured surface temperature and you get the k your insulation is actually delivering, which after a decade of rain and mechanical damage is rarely the k in the catalogue.

Conduction Through a Pipe Wall formula

Q˙=2πkL ΔTln⁡(r2/r1)\dot{Q} = \frac{2 \pi k L \, \Delta T}{\ln(r_2 / r_1)}
Where
  • Q˙\dot{Q}= Heat flow rate (W)
  • kk= Thermal conductivity (W/(m·K))
  • LL= Pipe length (m)
  • ΔT\Delta T= Inner-to-outer ΔT (C°)
  • r1r_1= Inner radius (mm)
  • r2r_2= Outer radius (mm)