Example calculations for walls

Example 4.5 Masonry cavity wall

Figure 4.4 shows a masonry cavity wall. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.35 W/m2K, and it is required to determine the necessary thickness of expanded polystyrene board (EPS) insulation, of thermal conductivity 0.040 W/mK. From Table 4.5, there are no AU reductions, and so:

Now use Tables 4.8 and 4.9 to find the base thickness and the allowable reductions. Note that the reduction for 150 mm of 600kg/m3 blockwork is 1.5 x 17 = 25.5mm. As this is a reduction, it should be rounded to 25mm.

From Table 4.8 Base thickness = 107 mm

From Table 4.9 Reduction for brickwork outer leaf = 5 mm cavity = 7 mm concrete blockwork = 25 mm

Fig. 4.4 Masonry cavity wall with internal insulation.

air space behind plasterboard = 5 mm plasterboard = 3 mm

Total reduction = 45 mm

Minimum insulation thickness = base thickness — total reduction = 107 — 45

The next available thickness of EPS board above the 62 mm minimum is likely to be 75 mm. This should have sufficient rigidity to be suitable for fixing, as shown in Fig. 4.4.

Example 4.6 Masonry wall with cavity fill

Figure 4.5 shows a masonry wall with the cavity completely filled with polyurethane foam insulation of thermal conductivity 0.040 W/mK. The brickwork and blockwork are tied with stainless steel vertical-twist ties. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.35 W/m2K, and it is required to determine the necessary thickness of insulation. From Table 4.5, there is a AU reduction of 0.02 W/m2K for the wall ties, and so:

Next, use Table 4.8 to find the base thickness of insulation. This requires an interpolation:

Fig. 4.5 Masonry wall with cavity fill.

At U = 0.33, thickness at 1 = 0.040 is 127 - (0.03/0.05) x (127 - 107)

= 115mm

Thus, from Table 4.8 Base thickness = 115mm Now use Table 4.9 to obtain allowable reductions:

From Table 4.9 Reduction for brickwork outer leaf = 5 mm concrete blockwork = 3 mm air space behind plasterboard = 5 mm plasterboard = 3 mm

Total reduction = 16 mm

Minimum insulation thickness = base thickness — total reduction = 115 — 16

This shows that the wall will have to be constructed with a 100 mm cavity. If this is not acceptable, an alternative solution would be to use concrete blocks with a lower density and/or a greater thickness. The calculation would have to be repeated to see if the reduction in the minimum insulation thickness is sufficient.

Example 4.7 Masonry cavity wall with partial cavity fill

Figure 4.6 shows a masonry wall with the cavity partially filled with insulation.

The brickwork and blockwork are tied with stainless steel vertical-twist ties. The

Fig. 4.6 Masonry with partial cavity fill.

maximum U-value allowed by the elemental method is, from Table 2.1, 0.35 W/ m2K, and it is required to determine the necessary thickness of insulation. From Table 4.5, there is a AU reduction of 0.02 W/m2K for the wall ties, and so:

Next, use Table 4.8 to find the base thickness of insulation. This requires an interpolation:

At U = 0.33, thickness at X = 0.025 is 79 — (0.03/0.05) x (79 — 67)

Thus, from Table 4.8 Base thickness = 72mm Now use Table 4.9 to obtain allowable reductions:

From Table 4.9

Reduction for brickwork outer leaf cavity concrete blockwork lightweight plaster

Total reduction

Minimum insulation thickness = base thickness — total reduction = 72 — 14

The minimum insulation thickness is much less than in the previous example. Nevertheless, if a 50 mm residual cavity is to be preserved, the overall cavity width will have to be of the order of 100 mm.

Example 4.8 Timber frame wall

Figure 4.7 shows a timber frame wall. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.35W/m2K. The 90mm timber frame has insulation that fully fills the space between the studs. It is required to determine the necessary thickness of a continuous insulation layer of EPS board that is to be fixed between the sheathing ply and the timber frame. From Table 4.5, there is a AU reduction of 0.01 for a timber frame where the insulation fully fills the space between the studs. However, the insulation layer which is being added is continuous, and so the AU reduction may be taken as zero.

Udesign = 0.35 — 0 = 0.35 W/m2K Now use Table 4.8 to find the base thickness:

From Table 4.8 Base thickness = 107 mm Now use Tables 4.9 and 4.10 to obtain allowable reductions:

Fig. 4.7 Timber frame wall.

From Table 4.9

From Table 4.10

Reduction for brickwork outer leaf = 5 mm cavity = 7 mm sheathing ply = 3 mm

2 sheets plasterboard = 6 mm

Reductions from Table 4.9 =21 mm Reduction for timber frame (73 x 90/100) = 66 mm

Total reduction = 87 mm

Minimum insulation thickness = base thickness — total reduction = 107 — 87

This thickness of insulation should provide a suitable solution.

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