U-Value Calculator (Wall, Roof & Floor Build-Ups)

How it works

R_layer = thickness (m) ÷ λ · R_total = Rsi + Σ R_layer + R_cavity + Rse · with framing: R_T = (R′ + R″) ÷ 2, where 1/R′ = f_a/R_a + f_b/R_b (parallel paths) and R″ uses λ″ = f_a λ + f_b λ_frame in the framed layer · ΔU_g = ΔU″ × (R_ins ÷ R_T)² · U = 1 ÷ R_T + ΔU_g · heat loss = U × area × ΔT · yearly kWh = U × area × degree-days × 24 ÷ 1000 · θ_si = θ_i − ΔT × Rsi ÷ R_total · extra insulation = (1 ÷ U_target − 1 ÷ U) × λ_ins

Heat flows through a wall the way current flows through resistors in series: each layer resists in proportion to its thickness and in inverse proportion to its conductivity λ, the still air films on either face add a little more (more on a floor, where warm air sits still, than on a ceiling), and the U-value is the reciprocal of the total — watts per square metre for each degree of temperature difference. Because conductivities span two orders of magnitude, a hundred millimetres of insulation resists heat as much as several metres of brick, which is why the build-up table shows masonry contributing almost nothing. Studs, joists and rafters cut through the insulation, so ISO 6946 averages two estimates — parallel heat paths through the insulation and through the timber, and a layer-by-layer average conductivity — and adds a correction if the insulation boards have gaps. The calculator then gives the heat loss for your area and temperatures, a rough yearly energy figure from heating degree-days, the temperature at each layer and the inside surface temperature against the room’s dew point, and how the result compares with UK Part L, Passivhaus and the Spanish CTE limits. Wall ties, lintels and junctions are linear thermal bridges outside a U-value calculation.

Worked example

A wall of 12.5 mm plasterboard, 100 mm mineral wool (λ 0.04), 100 mm aerated block (0.19), a 0.18 air cavity and 102 mm brick (0.77) has R = 0.13 + 0.06 + 2.50 + 0.53 + 0.18 + 0.13 + 0.04 = 3.57 m²K/W, a U-value of 0.28 W/m²K — meeting the 0.30 renovation value but not the 0.18 new-build target, which needs about (1/0.18 − 3.57) × 0.022 = 44 mm more PIR. At 20 °C inside and 0 °C outside it loses 5.6 W/m², or 168 W for 30 m², and about 0.28 × 30 × 2,500 × 24 ÷ 1000 = 504 kWh a year. If the mineral wool sat between timber studs taking 15% of the layer, the U-value would rise to about 0.33.

Sources: ISO 6946:2017 — Building components and building elements: thermal resistance and thermal transmittance · Approved Document L, volume 1: dwellings (2021 edition) · Código Técnico de la Edificación, DB-HE (2019)

Frequently asked questions