Layered Heat Conduction Calculator
Calculate steady conductive heat flow through layers in series. Inspect thermal resistance, heat flux, transferred energy and interface temperatures.
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Result
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Calculation steps
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How to calculate
Add each layer thickness divided by conductivity to obtain resistance per area. Divide the face-temperature difference by this resistance for heat flux; multiply by area and elapsed time for power and energy.
Rarea=Σ(d/k); flux=ΔT/Rarea; power=area×flux
Worked example
A 0.1 m layer with k=0.04 W/(m K), area 10 m² and a 20 K temperature difference conducts 80 W, or 0.08 kWh over one hour.
Assumptions and input details
One-dimensional steady conduction through 1–20 uniform planar layers in series with a common positive face area. Enter layer thickness 1e−12–1e6 m and positive conductivity up to 1e6 W/(m K). Entered face temperatures may be reversed; heat and energy then carry a negative sign from first toward last face. Temperatures cannot be below absolute zero. This omits convection, radiation, contact/film resistance, thermal bridges and transient storage. It is not a whole-building heating-load calculation.
References: OpenStax conduction
Unit definitions
Unit selection is independent of language. Each selector contains only units for its physical quantity. Calculations use defined base units internally.
Choose from 23 length definitions, including explicitly labelled regional measures. Area conversions square the length factor; geometric volumes cube it.
Regional measures are not interchangeable merely because their names look similar. This is a documented selection, not a claim to include every historical unit.
NIST measurement tables · Japanese traditional units · Mainland China and Hong Kong unit distinctions
Using this calculator
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Results use floating-point arithmetic and displayed rounding. Read the formula, assumptions and any stated data precision before interpreting the result.
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