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.

Brickwork Blockwork Cavity
Fig. 5.2 Brick and blockwork cavity wall.
Table 5.5 Thermal data for 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.
Table 5.6 Calculation of the upper resistance limit, cavity wall.

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

Table 5.7 Calculation of the lower resistance limit, cavity wall.

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:

Was this article helpful?

0 0

Post a comment