Example calculations for roofs

Example 4.1 Pitched roof with insulation between the joists Figure 4.2a shows insulation laid between the ceiling joists of a pitched roof which is covered with 19 mm roof tiles. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.16W/m2K, and it is required to determine the necessary thickness of insulation. First, from Table 4.5, there is a AU correction of 0.01 W/m2K. The design (or 'look-up') U-value is therefore:

Joist Insulation

Joist Insulation

10mm plasterboard'

(a) Insulation between ceiling joists

10mm plasterboard'

(a) Insulation between ceiling joists

Ceiling Insulation Thickness

(c) Insulation between and over ceiling joists

(c) Insulation between and over ceiling joists

Insulation between joists Udesign = 0.16 — 0.01 = 0.15W/m2K

Next, use Table 4.6 to find the base thickness of insulation, and Table 4.7 to determine allowable reductions to the base thickness:

From Table 4.6 Base thickness = 557 mm

From Table 4.7 Reduction for 19 mm roof tiles = 1mm

Pitched roofspace = 6 mm

10 mm plasterboard = 2 mm

Total reduction = 9 mm

Minimum insulation thickness = base thickness — total reduction = 557 — 9

In this case, the ceiling joists create a thermal bridge that is not protected by the insulation layer, with the result that a very large thickness of insulation is required. As the joists are likely to be about 100 mm in depth, this thickness of insulation laid between them is not a practicable design solution.

Example 4.2 Pitched roof with insulation between the rafters Figure 4.2b shows a pitched roof with the insulation laid between the rafters. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.20W/m2K, and it is required to determine the necessary thickness of insulation. First, from Table 4.5, there is a AU correction of 0.01 W/m2K. The design (or 'look-up') U-value is therefore:

Insulation between rafters Udesign = 0.20 — 0.01 = 0.19W/m2K

Next, use Table 4.6 to find the base thickness of insulation. This requires an interpolation:

At U = 0.19, thickness at X = 0.030 is 557 — (0.04/0.05) x (557 — 269)

Thus, from Table 4.6 Base thickness = 327 mm

Table 4.7 is now used to determine allowable reductions to the base thickness:

From Table 4.7 Reduction for 19 mm roof tiles = 1mm

10 mm plasterboard = 2 mm Total reduction = 3 mm

Minimum insulation thickness = base thickness — total reduction = 327 — 3

As in the previous example, the bridging effect of the timber has produced a requirement for an insulation layer much thicker than depth of the rafters.

Unless the rafters themselves are also over 300 mm deep, there will be insufficient space between the plasterboard and the tiling for the insulation to be fitted.

Example 4.3 Pitched roof with insulation between and over ceiling joists Figure 4.2c shows a pitched roof with insulation between and over ceiling joists. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.16 W/m2K, and it is required to determine the necessary thickness of insulation. The insulation material is mineral wool quilt of thermal conductivity (Table 4.16) of 0.042 W/mK. First, from Table 4.5, the AU correction is zero. The design (or 'look-up') U-value is therefore:

Insulation between and over joists Udesign = 0.16 — 0 = 0.16W/m2K

Next, use Table 4.6 to find the base thickness of insulation. This requires a double interpolation:

AtU = 0.15, thickness at X = 0.042 is 277 + (0.002/0.005) x (307 — 277)

AtU = 0.20, thickness at X = 0.042 is 210 + (0.002/0.005) x (232 — 210)

At U = 0.16, thickness at X = 0.042 is 289 — (0.01/0.05) x (289 — 219)

Thus, from Table 4.6 Base thickness = 275 mm Table 4.7 is now used to determine allowable reductions to the base thickness:

From Table 4.7 Reduction for 19 mm roof tiles = 1mm

Pitched roofspace = 8 mm

10 mm plasterboard = 3 mm

Total reduction = 12 mm

Minimum insulation thickness = base thickness — total reduction = 275 — 12

In this case, the thermal bridge created by the ceiling joists is protected by part of the insulation. The resulting thickness of insulation is much less and more practicable than the first two examples, but is still considerable. If access to the roof space is required, then the problem of providing walkways over the insulation without compressing it or penetrating it with fixings will have to be solved.

Example 4.4 Flat roof

Figure 4.3 shows a concrete deck flat roof. The maximum U-value allowed by the elemental method is, from Table 2.1, 0.25W/m2K, and it is required to determine the necessary thickness of insulation. There are no AU corrections, and so:

Three layers of roofing felt Insulation, thermal conductivity 0.030W/mK

Three layers of roofing felt Insulation, thermal conductivity 0.030W/mK

150mm concrete, density 1100kg/m3

Fig. 4.3 Example 4.4, concrete deck roof.

150mm concrete, density 1100kg/m3

Fig. 4.3 Example 4.4, concrete deck roof.

Continuous layer of insulation Udesign = 0.25 — 0 = 0.25 W/m2K

Referring to Table 4.6 for a thermal conductivity of 0.030 W/mK, the base thickness of insulation is:

From Table 4.6 Base thickness = 116 mm

There are allowable reductions for the roofing felt and for the concrete deck itself. For 150 mm of concrete of density 1100 kg/m3, the reduction is 1.5 x 8 = 12 mm. Thus:

From Table 4.7 Reduction for 3 layers felt = 1mm

150 mm concrete deck = 12 mm Total reduction = 13 mm

Minimum insulation thickness = base thickness — total reduction = 116 — 13

A solution using this thickness of insulation is feasible but has a number of possible drawbacks. The insulation material itself is unlikely to have any significant compressive strength (otherwise it would not have a thermal conductivity as low as 0.030 W/mK), and so access to the roof deck could not be permitted. Also, because the insulation is immediately beneath the weatherproof membrane, solar radiation on the roof will generate very high surface temperatures, with the possibility of degradation of the membrane due to thermal movement or chemical action.

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