Example calculations for floors

Example 4.9 Solid floor in contact with the ground

Figure 4.11 shows a solid floor. The maximum U-value allowed by the elemental method, from Tables 2.1 and 3.1, is 0.25 W/m2K, and it is desired to find the necessary thickness of insulation of thermal conductivity 0.025 W/mK. Table 4.11 must be used, and this requires the perimeter to area ratio to be calculated:

Floor perimeter = 6 + 2 + 4 + 4+10 + 6 = 32m Floor area = (6 x 6) + (4 x 4) = 52 m2

P 32

Perimeter to area ratio, P/A: — = — = 0.615 = 0.6

A 52

From Table 4.11, the necessary minimum thickness of insulation is 68 mm.

Fig. 4.11 Solid floor.

Fig. 4.11 Solid floor.

Example 4.10 Suspended timber ground floor

If the floor shown in Fig. 4.11 was a suspended timber floor of the same perimeter shape and dimensions, the perimeter to area ratio would still be 0.6, but it would be necessary to use Table 4.12. To achieve the same U-value of 0.25W/m2K, and using insulation of thermal conductivity 0.025 W/mK, the necessary minimum thickness of insulation placed between the joists is 98 mm.

Example 4.11 Upper floor, timber construction

Figure 4.12 shows an upper floor in timber construction. The joists are 45 mm wide and are at 400 mm centres. The maximum U-value allowed by the elemental method is, from Tables 2.1 and 3.1, 0.25 W/m2K, and it is desired to find the necessary thickness of insulation of thermal conductivity 0.030W/mK. The proportion by area of structural timber in the floor is (45/400) x 100 = 11.25%. As this is less than 12%, Table 4.14 will very slightly overestimate the insulation thickness and therefore can still be used. If the proportion of structural timber had been greater than 12%, Table 4.14 would be inappropriate and a suitable calculation method [17, 18] would have to be used instead. Using Tables 4.14 and 4.15:

From Table 4.14 Base thickness = 163 mm

From Table 4.15 Reduction for 10 mm plasterboard = 2 mm

19mm floorboards = 4mm

Total reductions = 6 mm

Minimum insulation thickness = base thickness — total reduction = 163 — 6

Fig. 4.12 Upper floor, timber construction.

Floor joist

Fig. 4.12 Upper floor, timber construction.

Floor joists are typically not less than 175 mm deep, and so this thickness of insulation can be accommodated.

Example 4.12 Upper floor, concrete

Figure 4.13 shows an upper floor in concrete construction. The required U-value is 0.25 W/m2K, and insulation of thermal conductivity 0.030 W/mK is to be used.

.m^wm^M^ss^^rn lnsulation

Fig. 4.13 Upper floor - concrete.

From Table 4.14 Base thickness = 112mm From Table 4.15 Reduction for 50 mm screed = 4 mm

Total reduction = 4 mm

Minimum insulation thickness = base thickness — total reduction = 112 — 4

The stability and compressive strength of this thickness of insulation immediately beneath the floor finishes must be considered.

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