Thermal Resistance of a Plane Wall

R=LkAR = \frac{L}{k A}

Worked example: 100 mm concrete (k=1.4) over 12 m2 → 0.005952 K/W — press Try an example to run it live, then adjust anything.

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Thermal Resistance of a Plane Wall explained

AkLR

Joseph Fourier spent the years after 1807 being told by Lagrange and Laplace that his heat series were not rigorous; the Théorie analytique de la chaleur finally appeared in 1822 and gave physics both the conduction law and the Fourier series. Written as a resistance, his law becomes R = L/(kA) — thickness over conductivity times area — and once heat flow wears the clothes of Ohm's law you can stack, branch and total resistances exactly as an electrician would. A 100 mm concrete wall (k = 1.4 W/(m·K)) of 12 m² has R = 0.1/(1.4 × 12) = 0.00595 K/W, so 22 K across it drives 22/0.00595 ≈ 3700 W.

Two traps. First, this is the absolute resistance in K/W, not the building-trade R-value, which is per unit area (m²·K/W, or h·ft²·°F/BTU in the US) — multiply the RSI by area and invert to compare. Second, k is not a constant: mineral wool at −20 °C is not the mineral wool on the datasheet at 24 °C, and wet insulation can lose three-quarters of its resistance because water conducts 25 times better than the trapped air it displaced. Every insulated cold line that sweats is quietly converting itself into a bare pipe.

Thermal Resistance of a Plane Wall formula

R=LkAR = \frac{L}{k A}
Where
  • RR= Conduction resistance (K/W)
  • LL= Wall thickness (mm)
  • kk= Thermal conductivity (W/(m·K))
  • AA= Cross-sectional area (m²)

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