Annex 6.B Worked examples of U-value calculations using the combined method
6.B.0 Introduction
6.B.1 The procedure
6.B.2 Timber framed wall example
6.B.3 Cavity wall with lightweight masonry leaf and insulated dry-lining example

annex 6.B

Worked examples of U-value calculations using the Combined Method
[Appendix B]

6.B.0 Introduction

For building elements which contain repeating thermal bridges, such as timber ceiling ties or joists between insulation in a roof or floor, timber studs in a wall, or mortar joints in lightweight blockwork, the effect of thermal bridges should be taken into account when calculating the U-value. The calculation method, known as the Combined Method, is set out in BS EN ISO 6946 and the following examples illustrate the use of the method for typical wall, roof and floor designs.

In cases where the ceiling ties, studs or joists in roof, wall or floor constructions project beyond the surface of the insulation the depths of these components should be taken to be the same as the thickness of insulation for the purposes of the U-value calculation (as specified in BS EN ISO 6946).

Conductivity values for common building materials can be obtained from the CIBSE Guide Section A3 or from BS EN ISO 12524. For specific insulation products, however, data should be obtained from manufacturers. Table 6.A.18 (Annex A) gives typical thermal conductivities for some common construction materials.

The procedure in this Annex does not address elements containing metal connecting paths. For built-up sheet metal walls and roofs, BRE IP 5/98 may be used. For curtain walling, the reader is directed to the CAB publication (Guide for assessment of the thermal performance of aluminium curtain wall framing" (September 1996). For ground floors and basements the reader is directed to Annex C.

6.B.1 The procedure

The U-value is calculated by applying the following steps:

a. Calculate the upper resistance limit (R_upper) by combining in parallel the total resistances of all possible heat-flow paths (i.e. sections) through the plane building element.

b. Calculate the lower resistance limit (R_lower) by combining in parallel the resistances of the heat flow paths of each layer separately and then summing the resistances of all layers of the plane building element.

c. Calculate the U-value of the element from U = 1 / R_T,

where

d. Where appropriate, add a correction for air gaps and mechanical fasteners (including wall ties) as described in BS EN ISO 6946 Appendix D.

6.B.2 Timber framed wall example

In this example there is a single bridged layer in the wall, involving insulation bridged by timber studs. The construction consists of outer leaf brickwork, a clear ventilated cavity, 19 mm plywood, 38 x 140 mm timber framing with 140 mm of mineral wool quilt insulation between the timber studs and 2 sheets of plasterboard each 12.5 mm thick.

(Total thickness: 336 mm; U-value: 0.29 W/m²K)

The thicknesses of each layer, together with the thermal conductivities of the materials in each layer, are shown below. The internal and external surface resistances are those appropriate for wall constructions. Layer 4 is thermally bridged and two thermal conductivities are given for this layer, one for the unbridged part and one for the bridging part of the layer. For each homogeneous layer and for each section through a bridged layer, the thermal resistance is calculated by dividing the thickness (in metres) by the thermal conductivity.

Calculation of thermal resistance (timber frame)

Both the upper and the lower limits of thermal resistance are calculated by combining the alternative resistances of the bridged layer in proportion to their respective areas, as illustrated below. The method of combining differs in the two cases.

Upper resistance limit

When calculating the upper limit of thermal resistance, the building element is considered to consist of two thermal paths (or sections). The upper limit of resistance is calculated from:

Conceptual illustration of how to calculate the upper limit of thermal resistance

where F₁ and F₂ are the fractional areas of the two sections (paths) and R₁ and R₂ are the total resistances of the two sections. The method of calculating the upper resistance limit is illustrated conceptually below:

Resistance through the section containing insulation

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.090

Resistance of plywood

= 0.146

Resistance of mineral wool (90.5%)

= 3.333

Resistance of plasterboard

= 0.100

Internal surface resistance

= 0.130

Total (R₁)

= 3.971 m²K/W

Fractional area F₁ = 0.905 (90.5%)

Resistance through the section containing timber stud

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.090

Resistance of plywood

= 0.146

Resistance of timber studs (9.5%)

= 1.077

Resistance of plasterboard

= 0.100

Internal surface resistance

= 0.130

Total (R₂)

= 1.715 m²K/W

Fractional area F₂ = 0.095 (9.5%)

The upper limit of resistance is then:

Lower resistance limit

When calculating the lower limit of thermal resistance, the resistance of a bridged layer is determined by combining in parallel the resistances of the unbridged part and the bridged part of the layer. The resistances of all the layers in the element are then added together to give the lower limit of resistance.

The resistance of the bridged layer is calculated using:

Conceptual illustration of how to calculate the lower limit of thermal resistance

The method of calculating the lower limit of resistance is illustrated conceptually below.

The lower limit of resistance is then obtained by adding up the resistances of all the layers:

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.090

Resistance of plywood

= 0.146

Resistance of bridged layer =

= 2.780

Resistance of plasterboard = 0.100

Internal surface resistance = 0.130

Total (R_lower) = 3.418 m²K/W

Total resistance of wall (not allowing for air gaps around the insulation)

The total resistance of the wall is the average of the upper and lower resistance limits:

Correction for air gaps

If there are small air gaps penetrating the insulating layer a correction should be applied to the U-value to account for this. The correction for air gaps is ∆U_g where

∆U_g = ∆U'' x (R_I / R_T)²and where R_I is the thermal resistance of the layer containing gaps, R_T is the total resistance of the element and ∆U'' is a factor which depends upon the way in which the insulation is installed. In this example R_I is 2.780 m²K/W, R_T is 3.474 m²K/W and ∆U'' is 0.01 (i.e. correction level 1). The value of ∆U_g is then:

∆U_g = 0.01 x (2.780 / 3.474)² = 0.006 W/m²K

U-value of the wall

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / R_T + ∆U_g (if ∆U_g is not less than 3% of 1 / R_T)

U = 1 / R_T (if ∆U_g is less than 3% of 1 / R_T)

In this case∆U_g = 0.006 W/m²K and 1 / R_T = 0.288 W/m²K. Since ∆U_g is less than 3% of (1 / R_T),

U = 1 / R_T = 1 / 3.474 = 0.29 W/m²K.

Notes:

1. In the above calculation it is assumed that the dwangs do not penetrate the whole of the insulation. If the dwangs do penetrate the whole of the insulation thickness they should be included as part of the timber percentage used in the calculation.

2. In this example correction level 1 is appropriate. This is because air gaps are likely to exist, in some cases, between the insulation and the timber framing.

3. The additional timbers at the junctions of plane elements, for example wall/wall, wall/floor, and wall ceiling junctions, and the additional timbers surrounding openings are taken account of in the treatment of such details and so are not taken into account in the calculation of the U-value of the wall.

4. BS EN ISO 6946 states that if the insulation is installed in such a way that no air circulation is possible on the warm side of the insulation then ∆U'' is set to 0.01 W/m²K. If, on the other hand, air circulation is possible on the warm side then it should be set to 0.04 W/m²K. The possible correction levels and correction factors are summarised as follows:

Correction for air gaps

6.B.3 Cavity wall with lightweight masonry leaf and insulated dry-lining example

In this example there are two bridged layers - insulation bridged by timber and lightweight blockwork bridged by mortar. The construction consists of outer leaf brickwork, a clear cavity, 125 mm AAC blockwork, 38 x 89 mm timber studs (400 mm centre-to-centre spacing) with insulation between the studs and one sheet of 12.5 mm plasterboard.

The thicknesses of each layer, together with the thermal conductivities of the materials, are shown below, with appropriate internal and external surface resistances, these being, for a wall, 0.13 m²K/W and 0.04 m²K/W. Layers 3 and 4 are both thermally bridged and two thermal conductivities are given for each layer to reflect the bridged part and the bridging part in each case. For each homogeneous layer and for each section through a bridged layer the thermal resistance is calculated by dividing the thickness (expressed in metres) by the thermal conductivity.

Calculation of thermal resistance (cavity wall)

Both the upper and lower limits of thermal resistance are calculated by combining the alternative resistances of the bridged layer in proportion to their respective areas, as illustrated below. The method of combining differs in the two cases.

Upper resistance limit

When calculating the upper limit of thermal resistance, the building element is considered to consist of a number of thermal paths (or sections). In this example there are four sections (or paths) through which heat can pass. The upper limit of resistance, R_upper, is given by

Conceptual illustration of how to calculate the upper limit of thermal resistance

where F₁, F₂, F₃ and F₄ are the fractional areas of sections 1, 2, 3 and 4 respectively and R₁, R₂, R₃ and R₄ are the corresponding total thermal resistances of the sections. A conceptual illustration of the method of calculating the upper limit of resistance is shown in the figure below:

Resistance through section containing AAC blocks and mineral wool

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.180

Resistance of AAC blocks (93.4%)

= 1.136

Resistance of mineral wool (90.5%)

= 2.342

Resistance of plasterboard

= 0.050

Internal surface resistance

= 0.130

Total thermal resistance (R₁)

= 4.010 m²K/W

Fractional area F₁ = 0.845 (93.4% x 90.5%)

Resistance through the section containing mortar and mineral wool

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.180

Resistance of mortar (6.6%)

= 0.142

Resistance of mineral wool (90.5%)

= 2.342

Resistance of plasterboard

= 0.050

Internal surface resistance

= 0.130

Total thermal resistance (R₂)

= 3.016 m²K/W

Resistance through section containing AAC blocks and timber

Fractional area F₂ = 0.060 (6.6% x 90.5%)

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.180

Resistance of AAC blocks (93.4%)

= 1.136

Resistance of timber (9.5%)

= 0.685

Resistance of plasterboard

= 0.050

Internal surface resistance

= 0.130

Total thermal resistance (R₃)

= 2.353 m²K/W

Fractional area F₃ = 0.089 (93.4% x 9.5%)

Resistance through section containing mortar and timber

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.180

Resistance of mortar (6.6%)

= 0.142

Resistance of timber (9.5%)

= 0.685

Resistance of plasterboard

= 0.050

Internal surface resistance

= 0.130

Total thermal resistance (R₄)

= 1.359 m²K/W

Fractional area F₄ = 0.006 (6.6% x 9.5%)

Combining these resistances we obtain:

Conceptual illustration of how to calculate the lower limit of thermal resistance

When calculating the lower limit of thermal resistance, the resistance of a bridged layer is determined by combining in parallel the resistances of the unbridged part and the bridged part of the layer. The resistances of all the layers in the element are then added together to give the lower limit of resistance. A conceptual illustration of the method of calculating the lower limit of resistance is shown below:

The resistances of the layers are added together to give the lower limit of resistance. The resistance of the bridged layer consisting of AAC blocks and mortar is calculated using:

and the resistance of the bridged layer consisting of insulation and timber is calculated using:

The lower limit of resistance is then obtained by adding together the resistances of all the layers:

External surface resistance

= 0.040

Resistance of bricks

= 0.132

Resistance of air cavity

= 0.180

Resistance of first bridged layer

= 0.777

Resistance of second bridged layer

= 1.904

Resistance of plasterboard

= 0.050

Internal surface resistance

= 0.130

Total (R_lower)

= 3.213 m²K/W

Total resistance of wall

The total resistance of the wall is the average of the upper and lower resistance limits:

Correction for air gaps between the timber studs

Since the insulation is entirely between studs (i.e. there is no continuous layer of insulation) a correction should be applied to the U-value in order to account for air gaps. The overall U-value of the wall should include a term ΔU_g, where

ΔU_g = ΔU'' x (R_I / R_T)²

and where ΔU'' = 0.01 (referred to in BS EN ISO 6946 as correction level 1), R_I is the thermal resistance of the layer containing the gaps and R_T is the total resistance of the element. ΔUg is therefore:

ΔU_g = 0.01 x (1.904 / 3.439)² = 0.003 W/m²K

The effect of air gaps or mechanical fixings should be included in the U-value unless they lead to an adjustment in the U-value of less than 3%.

U = 1 / R_T + ΔU_g                 (if ΔU_g is not less than 3% of 1 / R_T)

U = 1 / R_T                           (if ΔU_g is less than 3% of 1 / R_T)

In this case ΔU_g = 0.003 W/m²K and 1 / R_T = 0.291 W/m²K.
Since ΔU_g is less than 3% of (1 / R_T),

U = 1 / 3.439 = 0.29 W/m²K.

Notes:

1. Since the cavity wall ties do not penetrate any insulation no correction need be applied to the U-value to take account of them.

2. In the above calculation it is assumed that the dwangs do not penetrate the whole of the insulation. If the dwangs do penetrate the whole of the insulation thickness they should be included as part of the timber percentage used in the calculation.