Thermal engineering · Heat transfer · conduction

Cylindrical conduction calculator — pipe heat loss

Calculate steady conduction through one homogeneous cylindrical layer from its inner diameter, thickness, conductivity and the temperature difference between its surfaces.

Created v1.0

By Thibaut Grzelak, Mechanical Analysis Engineer

Heat transfer · conduction
01Calculation inputs
TₕT𝒸q′D₂D₁ek
R′ = ln(r₂/r₁)/(2πk)Calculate steady conduction through one homogeneous cylindrical layer from its inner diameter, thickness, conductivity and the temperature difference between its surfaces.
W/(m·K)
02

Results

q′
Heat-transfer rate Q̇
—

Calculated steady heat-transfer rate through the wall.

Resistance per length R′
—K·m/W

Cylindrical conductive resistance normalized by pipe length.

Thermal resistance R
—K/W

Total conductive thermal resistance over the entered length.

Outer diameter D₂
—

Calculated outer diameter from D₂ = D₁ + 2e.

Steady-state, one-dimensional radial conduction. One homogeneous cylindrical layer with constant thermal conductivity k.

01Formulas and symbols

Formulas used

R′ln(r₂ / r₁) / (2πk)Resistance per length R′
q′ΔT / R′Heat loss per length q′
Q̇q′ · LHeat-transfer rate Q̇
RR′ / LThermal resistance R
D₂D₁ + 2eOuter diameter D₂
02Assumptions and limits

Scope of validity

  • Steady-state, one-dimensional radial conduction.
  • One homogeneous cylindrical layer with constant thermal conductivity k.
  • D₁ is the diameter of the conducting layer’s inner surface; for insulation, use the pipe outside diameter beneath the insulation.
  • Tₕ and T𝒸 are layer surface temperatures, not fluid and ambient-air temperatures.
  • Convection, radiation, contact resistance, supports and axial conduction are not included.
  • This is a conduction-only physical estimate, not a regulatory or code-compliance check.
03Validation example

Reference numerical case

  1. Reference case: D₁ = 0.10 m, e = 0.05 m, k = 0.04 W/(m·K), L = 10 m and ΔT = 20 K.
  2. D₂ = 0.20 m and R′ = ln(0.20/0.10)/(2π×0.04) = 2.757945 K·m/W.
  3. q′ = 20/2.757945 = 7.251776 W/m.
  4. Q̇ = q′L = 72.517762 W and R = 0.2757945 K/W.
04References

Technical references

  1. Incropera et al. — Fundamentals of Heat and Mass Transfer: steady radial conduction through cylinders.
  2. NIST Special Publication 811 — SI units and temperature intervals.