## Example calculations

The calculation method is most conveniently explained by means of examples. Example 5.1 Cavity wall

Figure 5.2 shows a cavity wall consisting of external brickwork, cavity, lightweight blockwork, mineral wool insulation within a timber sub-frame, and internal plasterboard. The blockwork and the mineral wool are the main providers of thermal insulation in this construction, and both suffer from thermal bridging. The blockwork is bridged by the mortar joints, and the mineral wool is bridged by the timber frame. In each case, the proportion of the area bridged is:

Blockwork 93% of area mortar joints 7% of area Mineral wool 88% of area timber battens 12% of area.

Table 5.5 gives the thermal data for the wall.

Fig. 5.2 Brick and blockwork cavity wall.
 Thermal Thermal Material Thickness conductivity resistance mm W/mK m2K/W External surface — — 0.040 Outer brickwork 102 0.77 — Cavity, unvented — — 0.180 AAC blocks 100 0.11 0.909 Mortar 100 0.88 0.114 Mineral wool insulation 89 0.038 2.342 Timber battens 89 0.13 0.685 Plasterboard 12.5 0.25 0.050 Internal surface — — 0.130

The upper resistance limit, Rupper

Each possible heat flow path through the wall is considered separately, and in this case it can be seen that there are four such paths, as shown in Fig. 5.3. The resistance of each path is calculated on the basis that the materials are in series, and then the four paths are combined on the basis that they are in parallel. The first part of the calculation is illustrated in Table 5.6. The four paths are then combined in parallel to find Rupper:

Fig. 5.3 Brick and blockwork cavity wall - upper resistance limit.
 Thermal resistance, m2K/W Path 1 Path 2 Path 3 Path 4 External surface resistance 0.040 0.040 0.040 0.040 Resistance of brickwork 0.132 0.132 0.132 0.132 Resistance of cavity 0.180 0.180 0.180 0.180 Resistance of AAC blocks 0.909 — 0.909 — Resistance of mortar — 0.114 — 0.114 Resistance of mineral wool 2.342 2.342 — — Resistance of timber — — 0.685 0.685 Resistance of plasterboard 0.050 0.050 0.050 0.050 Internal surface resistance 0.130 0.130 0.130 0.130 Total thermal resistance of path 3.783 2.988 2.126 1.331 Fractional area of path 93% x 88% 7% x 88% 93% x 12% 7% x 12% = 0.818 = 0.062 = 0.112 = 0.008

Rupper = 3.382m2K/W The lower resistance limit, Riower

Each thermal bridge in the construction element is first converted to a single combined resistance, as shown in Fig. 5.4. Using these combined resistances, the construction can then be considered as a single heat flow path with all components in series. Thus in the present example, the AAC blocks and the mortar form one thermal bridge, and their combined resistance, Rbm, is found from:

Fig. 5.4 Brick and blockwork cavity wall - lower resistance limit.

1 Fblocks , Fmortar 0.93 0.07

1.637

Rbm Rblocks Rmortar 0.909 0.114 Rbm = 0.611 m2K/W

The mineral wool insulation in its timber frame form another thermal bridge, and their combined resistance, Rit, is found from:

1 _ Finsulation Ftimber _ 0.88 0.12 _ 0 5509 Rit Rinsulation Rtimber 2.342 0.685

The combined resistances may now be used with the other resistances in the chain to find the lower resistance limit, as shown in Table 5.7.

Note that Rupper is an overestimate of the true resistance, whereas Rlower is an

 Thermal resistances, m2 Thermal bridges K/W Components Combined External surface resistance Resistance of brickwork Resistance of cavity Resistance of AAC blocks (93%) Resistance of mortar (7%) Resistance of mineral wool (88%) Resistance of timber (12%) Resistance of plasterboard Internal surface resistance 0.909J 0.114 J 2.3421 0.685 J 0.611 1.815 0.050 0.130 Total thermal resistance, Rlower 2.958

underestimate. The average of these is very close to the true value. Hence, the total resistance of the wall is found from

Rt = 2 (Rupper + Rlower) = 2 (3-382 + 2.958) = 3.170 m2K/W

and the U-value is

Corrections to the U-value for air gaps and mechanical fixings If there are small air gaps or mechanical fixings (such as wall ties) penetrating the insulation layer, it may be necessary to add a correction, AUg , to the U-value. The correction is required if AUg is 3% or more of the uncorrected U-value, but may be ignored if it is less than 3%. The correction is calculated from: