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Heat Flux Through Insulation (q = ΔT/R)

q′′=ΔTRq'' = \frac{\Delta T}{R}

Worked example: 22 K across RSI 3.5 → 6.286 W/m2 — press Try an example to run it live, then adjust anything.

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Heat Flux Through Insulation (q = ΔT/R) explained

ΔTq″R

Strip the area out of the heat-loss equation and what is left is flux: watts per square metre, which is the fairest way to compare two assemblies because it does not care how big the building is. RSI 3.5 with 22 K across it passes 22/3.5 = 6.3 W/m². An R-19 wall with 40 °F across it passes 40/19 = 2.1 BTU/(h·ft²), the same 6.6 W/m². Codes and mechanical insulation specifications are often written directly in these terms — a personnel-protection or condensation-control spec is really a flux limit in disguise.

The reason this page matters in the field is the heat flux meter. Tape one to the inside of a wall, log the surface-to-surface or air-to-air ΔT alongside it, and R = ΔT/q″ hands you the as-built resistance of the wall in front of you — including the framing, the settled insulation, the gap the electrician left, and the moisture. ISO 9869 covers the method and asks for at least 72 hours of averaging, because the wall stores heat and a short test measures the weather rather than the wall. Two traps: mount the meter away from studs unless the stud is what you are hunting, and use air-to-air ΔT only if you intend the surface films to be part of the R you report.

Heat Flux Through Insulation (q = ΔT/R) formula

q′′=ΔTRq'' = \frac{\Delta T}{R}
Where
  • q′′q''= Heat flux (W/m²)
  • ΔT\Delta T= Temperature difference (C°)
  • RR= R-value of the assembly (RSI (m²·K/W))

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