Insulation R-value calculator
Thermal resistances in series add. That is the whole basis of a build-up calculation.
W/(m K)
W/(m K)
W/(m K)
m2 K/W
K
U-value
0.3106W/(m2 K)
- Total R-value
- 3.2200 m2 K/W
- R-value, imperial units
- 18.28
- Layer 1 resistance
- 2.8571 m2 K/W
- Layer 2 resistance
- 0.1429 m2 K/W
- Layer 3 resistance
- 0.0500 m2 K/W
- Heat loss through the element
- 310.6 W
- Rough annual heat loss Taking 200 equivalent full heating days.
- 1491 kWh
Show the working
- Layer 1 thickness = 100
- Layer 1 conductivity = 0.035
- Layer 2 thickness = 100
- Layer 2 conductivity = 0.7
- Layer 3 thickness = 12.5
- Layer 3 conductivity = 0.25
- Surface resistances = 0.17
- Element area = 50
- Temperature difference = 20
- Layer 1 resistance = 100/1000 / 0.035 = 2.857 14
- Layer 2 resistance = 100/1000 / 0.7 = 0.142 857
- Layer 3 resistance = 12.5/1000 / 0.25 = 0.05
- Total R-value = 2.857 14 + 0.142 857 + 0.05 + 0.17 = 3.22
- U-value = 3.22 > 0 ? 1 / 3.22 : 0 = 0.310 559
- Heat loss through the element = 0.310 559 * 50 * 20 = 310.559
- R-value, imperial units = 3.22 * 5.678263398 = 18.284
- Rough annual heat loss = 310.559 * 24 * 200 / 1000 = 1490.68
Values are shown to six significant figures here; the arithmetic itself runs on exact fractions and is only rounded where a number is printed.
Scenarios & local history
Compare a saved scenario
Recent calculations
The formula
- R for a layer = thickness / conductivity
- total R = sum of the layers plus the surface resistances
- U = 1 / total R
This is a one-dimensional calculation. Thermal bridges through studs, joists and lintels can add 10 to 30% to the real heat loss.
Set a layer thickness to zero to leave it out.