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Cylindrical insulation heat loss through a resistance network

A pipe’s heat loss follows a series resistance path from the fluid to ambient. Cylindrical layers use logarithmic radius ratios, not the flat-wall thickness/area expression. State whether the result is per metre or for the full pipe length.

Worked numbers are illustrative. Use your own operating conditions and documented material/process data.

Required inputs

  1. Inside fluid and ambient temperatures, pipe inner/outer radii and length.
  2. Each layer’s conductivity at a relevant mean temperature and its outer radius.
  3. Internal/external heat-transfer coefficients; explicitly decide how radiation is treated.

Governing equations

Rfilm = 1/(h 2πrL)

Rlayer = ln(ro/ri)/(2πkL)

Q = (Ti − Ta)/ΣR; Tinterface = Ti − QΣRupstream

Variables and units

  1. r, L: m; k: W/(m·K); h: W/(m²·K); R: K/W; Q: W.
  2. Use kelvin for radiation absolute temperatures; °C differences equal kelvin differences for conduction.

Assumptions

  1. Steady constant-property radial flow; no end loss or thermal bridges.
  2. Illustrative case omits radiation and pipe axial conduction; coefficients are user-entered assumptions.

Calculation workflow

  1. Convert every radius and thickness to metres.
  2. Assemble films and each cylindrical layer in series on the same length basis.
  3. Calculate heat rate, then interface temperatures by subtracting each resistance drop.

Worked example

  1. Ti = 90°C, Ta = 25°C, L = 1 m; ri = 0.050 m, pipe ro = 0.056 m, insulation ro = 0.106 m.
  2. Entered kpipe = 50, kins = 0.045 W/(m·K), hi = 2000 and ho = 15 W/(m²·K).
  3. Rin = 0.001592, Rpipe = 0.000361, Rins = 2.256773, Rout = 0.100097 K/W.
  4. Q = 65/2.358822 = 27.556124 W per metre; insulation surface ≈ 27.758°C.

Limitations

  1. Condensation, weather, temperature-dependent properties, radiation iteration and personnel safety require appropriate additional inputs/checks.
  2. ISO 12241 gives relevant industrial-insulation calculation scope; this educational network is not a claim of compliance with its full procedure.

Common mistakes

  1. Entering diameter as radius.
  2. Combining per-metre resistance with a full-length heat rate.
  3. Using a room-temperature k across a large temperature range without verification.

Practical interpretation

The low-conductivity insulation dominates this example. Do not assume every added cylindrical layer reduces heat loss: the simplified cylinder has a critical outer radius k/h when increasing area competes with conduction resistance. In this example k/h = 3 mm, well below the existing outer radius.

Related formulas

The governing equations, units and calculation steps above provide the formula basis for this guide. The related workspace also shows its calculation trace.

Conduction and convection resistance equations →

Related workspace

Open Thermal Engineering →

Sources and reference basis

Equations and worked examples are independently explained here. References reviewed 2026-10-17; no copyrighted standard tables are reproduced.