Estimate steady radial heat loss from a 100 ft pipe with 1 inch of insulation. The pipe outside diameter is 2.375 inches. Treat conductivity and the external surface coefficient as constant supplied values.
1. Calculator Inputs
Select the units below before entering the numbers. These are fixed benchmark inputs for checking the calculator.
| Calculator field | Value | Unit / setting |
|---|---|---|
| Pipe outside diameter | 2.375 | in |
| Radial insulation thickness | 1 | in |
| Insulated pipe length | 100 | ft |
| Pipe outer-wall temperature | 200 | °F |
| Ambient temperature | 70 | °F |
| Insulation conductivity, k | 0.025 | Btu/(h·ft·°F) |
| External surface coefficient, h | 2 | Btu/(h·ft²·°F) |
2. Step-by-step Math
Calculate the inner and outer insulation radii
r₁ = 2.375 / 24 ≈ 0.099 ft r₂ = r₁ + 1/12 ≈ 0.182 ft ΔT = 200 − 70 = 130°F
The insulation is 1 inch thick radially, so it increases the overall diameter by 2 inches to 4.375 inches.
Calculate thermal resistance per unit length
R′insulation = ln(r₂/r₁)/(2πk) = 3.889168009257 h·ft·°F/Btu R′outside = 1/(2πr₂h) = 0.436539272481 h·ft·°F/Btu R′total = 4.325707281738 h·ft·°F/Btu
Use the natural logarithm. These are resistance-per-length factors: dividing temperature difference by their sum gives heat rate per foot.
Calculate heat loss and outer-surface temperature
q′ = ΔT / R′total = 130 / 4.325707281738 q′ = 30.052888819 Btu/(h·ft) = 28.896431391 W/m Q = q′L = 30.052888819 × 100 = 3,005.288881863 Btu/h Tsurface = Tambient + q′R′outside Tsurface = 70 + 30.052888819 × 0.436539272481 Tsurface ≈ 83.1°F
The insulation outer surface lies between the ambient and pipe-wall temperatures.
Compare with a bare pipe at the same h
q′bare = 2πr₁hΔT = 161.661121966 Btu/(h·ft) Reduction = 100 × (1 − q′/q′bare) ≈ 81.4%
The comparison holds h and wall temperature fixed. A real bare pipe may have a different effective surface coefficient.
3. Compare Your Answers
Keep full precision during the calculation. The calculator may display fewer digits; compare answers rounded to three significant figures; retain unrounded intermediate values.
| Result | Expected answer | Your answer |
|---|---|---|
| Heat loss per length | 30.1 Btu/(h·ft) | __________ |
| Total heat loss | 3,010 Btu/h / 881 W | __________ |
| Outer-surface temperature | 83.1°F | __________ |
| Reduction relative to modelled bare pipe | 81.4% | __________ |
Second check
Set insulation thickness to zero. The cylindrical conduction resistance becomes zero, predicted heat loss becomes 162 Btu/(h·ft), and surface temperature equals the specified 200°F pipe-wall temperature.
4. Assumptions and Method
The model starts at the specified pipe outer-wall temperature, so it excludes internal convection and pipe-wall resistance. It assumes steady one-dimensional radial transfer, constant k and h, and no end or support losses. Radiation is not calculated separately; include it in an appropriate effective external coefficient if needed.
Equation definitions and implementation: Pipe Insulation Heat Loss Calculator. See also calculation methodology and other worked examples.