U values

Building construction information often mentions U-values. What are they and how can they be useful?

Lower is better

The U-value is a measure of the heat loss through a construction. The lower the value, the lower the heat loss.

The actual heat loss is found by multiplying the U-value by the area and the temperature difference. For example, a door of U = 2.0 W/(m2K) and area 2.0 m2 separating a house of 20°C from a freezing outside of 0°C will be losing 2.0 W/(m2K) * 2.0 m2 * (20°C – 0°C) = 2.0 * 2.0 * 20 = 80 W. That’s a bit like the heat from an old incandescent light bulb being lost continuously through this door.

The U-value is calculated from consideration of all the materials used in the construction

How to calculate the U-value from a construction?

It is easiest to do this by considering the thermal resistances of all the materials in the construction. The thermal resistance, or R-value, is the inverse of the U-value, namely U = 1/R.

For a layered construction we can simply sum the resistances. We also include the thermal resistance of the inner and outer surfaces. Here is an example. Here is how to calculate the thermal resistance of a cavity wall.

Each element has a thermal resistance. The outer surface, inner surface and cavities have resistances that have been measured and can be simply copied over. The solid materials have resistances that are the thickness of that layer (in metres) divided by the thermal conductivity of that layer (in Watts per metre per Kelvin).

Calculating the thermal resistances in an example cavity wall

The total thermal resistance of this example is 1.622 m2K/W . The U value is 1 / 1.622 = 0.62 W/(m2K). This wall example is simplified from a property built in the 1990s. Today regulations in the UK require the walls to achieve a U-value of 0.3 or lower.

Here are some useful numbers for typical resistances

DescriptionThermal Resistance (m2K/W)
Outside surface (any exposed)0.04
Inside surface (ceiling)0.10
Inside surface (wall)0.13
Plaster dabs cavity 0.17
Batten cavity0.18
Cavity (unventilated, wall)0.18
Cavity next to foil-faced insulation (roof)0.34
Cavity next to foil-faced insulation (wall)0.44
A sample of typical thermal resistances for surfaces and cavities (for more see [1] or manufacturer technical specfications)

and here are some useful conductivities of common materials

DescriptionThermal Conductivity (W/(mK))
Dense concrete blocks (structural)~1
Lightweight concrete blocks~0.4
Wood (typical softwood)0.13
Aerated blocks (best thermal performance)0.11
Glass wool0.04
Mineral wool0.035
PIR panel0.022
Phenolic panel (best performance)0.018
Vacuum insulated panel0.007
A sample of typical thermal conductivities for building materials (see manufacturer technical specfications for measured values)

What happens with a more complex construction?

If we have thermal bridging (that is a material in parallel with our main construction that conducts heat more easily) then we can include that effect by calculating the two thermal conduction paths separately and using what we call an area-weighted average of them to calculate the overall U-value. This is used for structural elements such as steel beams in a building or wooden battens behind plasterboard.

Here is an example with a steel beam in the block wall

Calculating the thermal resistances in an example cavity wall with a steel beam alongside the blocks

The steel beam is of a small area compared to the rest of the block wall (10%) so the effect on the U-value is small. The resistance changes from 1.622 to 1.531 and so the U-value increases from 0.62 to 0.65.

For readers interested in more detail and for understanding how to assess the U-values of other more complicated constructions, please see the BRE report [1].

Example 1. Loft Insulation

Most properties have at least one layer of loft insulation. What is the effect of a top-up layer?

Calculating the thermal resistances of basic loft insulation with glass wool between joists

This example has only a single layer of glass wool between the wooden joists (the joists are assumed to be 35 mm wide and are spaced apart by 600 mm so that the area fraction for the timber is 35/600 or 5.8%). The upper surface resistance is not for an exposed surface because there is a ventilated roof structure above it. The thermal resistance here is 2.648 giving a U-value of 0.38.

Adding a generous top-up layer of 150mm thick across the top of the joists helps deal with the thermal bridging too.

Calculating the thermal resistances in a loft with top-up insulation

The effect on the thermal resistance is a change from 2.648 to 6.398 – so the U-value reduces from 0.38 to 0.16. That’s why topping up the loft insulation is such a good idea.

What would be the effect of no insulation at all?

Calculating the thermal resistances in a loft with no insulation

A loft with no insulation has a thermal resistance of 0.293 corresponding to a U-value of 3.4! Because the U-value is so high, any areas of the loft that are bare with no insulation will be a source of major heat loss. Pay close attention to ceilings above dormer windows: accessing these areas from the main loft may be tricky and so they could be missing insulation.

Example 2. Under Rafter Insulation

Insulation between the rafters can be improved by using a multi-foil or insulated plasterboard or both.

Here is a basic approach relying on PIR between the rafters.

Calculating the thermal resistances of a simplified approximation of PIR panels between the rafters

The U-value of using a 50mm foil-faced PIR panel is around 0.30. This case could be analysed with the rafters cold-bridging the PIR panels and adjacent cavities and doing so increases the U-value to 0.31 (assuming 35mm wide timber at 600mm centre-spacing then 0.34 + 2.273 + 0.34 will be replaced by 2.838).

What is the effect of adding a multi-foil layer and insulated plasterboard?

Calculating the thermal resistances when adding multi-foil and insulated plasterboard

The resistances of the multi-foil and adjacent cavities comes from one multi-foil manufacturer’s datasheet. The resistance of the roof has increased from 3.336 to 6.118, giving a low U-value of 0.16.

Example 3. External Wall Insulation

What would happen if we add 50mm of polystyrene on to the outside of the example wall construction above?

The effect of 50mm Expanded Polystyrene on the example wall

In this example the effect of 50mm XPS changes the resistance from 1.622 to 3.197 and so the U-value halves from 0.62 to 0.31.

Most external wall insulation projects are applied to properties with solid walls. In this example we would need a thicker layer of insulation to achieve a U-value below 0.30.

Example 4. Internal Wall Insulation

What would happen if we replace the existing plasterboard with insulated plasterboard on the inside of the example wall construction above?

The effect of insulated plasterboard on the example wall

In this example the effect of the new insulated plasterboard is to change the resistance from 1.622 to 2.424 and so the U-value reduces from 0.62 to 0.41. A higher performing plasterboard (such as a foil-backed phenolic insulation) would help as would increasing its thickness. The foil-backing would push the cavity resistance up from 0.17 to around 0.4 and the phenolic core would push up the resistance of the insulation from 0.82 to 1.5 (giving a U-value of around 0.3).

Example 5. Cavity Wall Insulation

What would happen if we retrofit cavity wall insulation into the example wall construction above?

Thermal Resistance of cavity wall insulation
The effect of retrofit cavity wall insulation on the example wall

In this example the effect of fully-filling the cavity changes the resistance from 1.622 to 3.442 and so the U-value reduces from 0.62 to 0.29.

A new-build construction with a 100 mm cavity filled with foil-faced rigid PIR panels (conductivity 0.022 W/(m K)), or better still foil-faced phenolic panels (best conductivity 0.018 W/(m K)) could easily achieve 0.30. For PIR, take off 2.00 from 3.442 above and add 0.100/0.022 or 4.55 giving 3.442 – 2.00 + 4.55 = 5.99 (U-value of 0.17). For phenolic panels, take off 2.00 from 3.442 above and add 0.100/0.018 or 5.56 giving 3.442 – 2.00 + 5.56 = 7.00 (U-value of 0.14).

What U-values mean when it comes to your fuel bill

The heat loss in the property comes through the heat loss through the external surfaces plus the air changes due to ventilation or draughts. Heat loss through each external surface is its U-value multiplied by its area multiplied by the average temperature difference between inside and outside.

A rule of thumb in the UK for a typical dwelling house is that heat loss through walls is 35%, heat loss through the roof is 25%, 10% is lost through the floor and 30% is for heat loss through windows, doors and draughts (air changes). So if we improved the roof U-value from 0.3 to 0.2 then roughly-speaking that is worth 8%. And if we halved the U-value of the walls then that is worth approximately 17%. So if you spend £1000 on gas per year then these changes may be worth £80 and £170 per year.

A better approximation comes from estimating the heat loss through your property. If the inside is hotter than outside by an average of 10°C over a year (probably not far off for central UK) then a wall of U-value 0.62 extending over 100 m2 will lose 0.62 * 100 * 10 = 620 W. There are 8760 hours in a year so the energy lost over a year is 5430 kWh. If your boiler is 85% efficient then you need to buy 6400 kWh of gas (to balance the heat loss through the walls) and at 6 p/kWh that is £384. So halving the U-value is worth £192 per year. As gas prices rise, so will the saving from insulation.

Click here to go to the useful knowledge article about domestic insulation.

[1] Anderson, B., “Conventions for U-value calculations”, 2006 edition, BRE Scotland, BR 446:2006. (*An updated 2019 version is also available online).