Calculating thermal resistance and U value

The thermal conductivity of a material and its thermal resistance are related by the equation:

Thermal resistance (R_c) = Thickness of material/Thermal conductivity.

For example consider a 0.22 m thick brick wall with a thermal conductivity of 0.6 Wm^-1oC^{- 1}.
Thermal resistance = 0.22/0.6 = 0.367 m^{2 o}CW^-1.

The U value of a barrier is defined as:

U value = 1/(Thermal resistance) = Thermal conductivity/Thickness of material

So for the brick wall the U value would be 1/0.367 = 2.72 Wm^{-2 o}C^-1. If the wall had a lower thermal conductivity its thermal resistance would have been greater.

Material	Description	U value (Wm-2 oC-1)
Wall (outside)	22cm solid brick	2.2
	26 cm brick-block cavity - unfilled	1.0
	26 cm brick-block cavity - insulated	0.6
Walls (internal)	plaster, 10cm lightweight block, plaster	1.2
	plasterboard, 10cm studding, plasterboard	1.8
Floor (Ground)	solid concrete	0.8
	suspended - timber	0.7
Floor (other than ground)	Plasterboard/ 8 inch joist space/ T&g boards - heat flow up	1.7
Roof	pitched with felt, 100mm insulation	0.3
	flat, 50mm insulation	0.7
Window	wooden/uvpc frame, single glazed	5.0
(standard after April 2002)	wooden/uvpc frame, double glazed - 20mm gap, Low-E	1.7
	metal frame, single glazed	5.8
Door	external solid timber	2.4