How to use this calculator
Enter the inside and outside air temperatures and their convection coefficients, then the thickness and conductivity of each layer from the inside out. Set a convection coefficient to 0 when the temperature given is already the wall surface temperature, and a layer thickness to 0 when the layer is not there. Use an area of 1 m² for results per square meter.
Formulas
Worked example
20 cm brick, 5 cm insulation, 2 cm plaster; 20 °C inside, −10 °C outside (the default)
- R″ = 1/10 + 0.2/0.7 + 0.05/0.04 + 0.02/0.5 + 1/25 = 0.100 + 0.286 + 1.250 + 0.040 + 0.040 = 1.716 m²·K/W.
- q″ = (20 − (−10)) / 1.716 = 17.49 W/m²; U = 0.583 W/m²·K.
- The insulation, with 73% of the resistance, takes 21.9 °C of the 30 °C drop.
Steam pipe: 50 mm inside radius, 5 mm steel (k 45), 30 mm insulation (k 0.05); 200 °C steam (h 50), 20 °C air (h 10)
- Per meter: R = 1/(50·2π·0.05) + ln(55/50)/(2π·45) + ln(85/55)/(2π·0.05) + 1/(10·2π·0.085) = 0.0637 + 0.0003 + 1.3858 + 0.1872 = 1.637 K/W.
- q′ = 180/1.637 = 110 W per meter of pipe; the insulation does 85% of the work.
- Cylindrical layers use ln(r₂/r₁)/(2πkL) instead of L/(kA), because the area grows with radius.
Frequently asked questions
Why add resistances instead of conductivities?
The same heat flows through every layer, one after another, so each layer adds its own temperature drop, just like resistors in series carrying the same current.
What about a pipe with insulation?
Choose “Pipe or cylinder”. Cylindrical layers use R = ln(r₂/r₁)/(2πkL), and the calculator warns when thin insulation is below the critical radius k/h, where adding more would actually increase the heat loss.