Example 5 - Masonry cavity wall with internal insulation
Determine the thickness of the insulation layer required to achieve a U-value of 0.45 W/m2K for the wall construction shown below.
diagram
Using Table E5, from column C, row 4 of the table, the base thickness of insulation layer is 51 mm.
The base thickness may be reduced by taking account of the other materials as follows from Table E6:
|
Brick outer leaf |
column C, row 2 |
= 3 mm |
|
Cavity |
column C, row 1 |
= 5 mm |
|
Plasterboard |
column C, row 6 |
= 2 mm |
and from Table E7:
|
Concrete block |
column C, row 1 |
|
|
adjusted for 150 mm |
= 17 mm |
|
|
block thickness (1.5 x 11) |
||
|
Total reduction |
= 27 mm |
|
The minimum thickness of the insulation layer required to achieve a U-value of 0 45 W/ m2K is therefore:
Base thickness less total reduction ie 51 — 27 = 24 mm.
Example 6 - Masonry cavity wall filled with insulation with plasterboard on dabs
Determine the thickness of the insulation layer required to achieve a U-value of 0.45 W/m2K for the wall construction shown below.
diagram
Using Table E5, from column E, row 4 of the table, the base thickness of the insulation layer is 71 mm.
The base thickness may be reduced by taking account of the other materials as follows:
|
from Table E6: |
||
|
Brick outer leaf |
column E, row 2 |
= 4 mm |
|
13 mm plasterboard |
column E, row 6 |
= 3 mm |
|
Airspace behind plasterboard |
column E, row 7 |
= 4 mm |
|
and from Table E7: |
||
|
Concrete block |
column E, row 6 |
= 5 mm |
|
Total reduction |
= 16 mm |
|
The minimum thickness of the insulation layer required to achieve a U-value of 0.45 W/ m2K is therefore:
Base thickness less total reduction ie 71 — 16 = 55 mm.
Example 7 - Masonry wall with partial cavity-fill
Determine the thickness of the insulation layer required to achieve a U-value of 0.45 W/m2K for the wall construction shown below.
diagram
Using Table E5, from column E, row 4 of the table, the base thickness of the insulation layer is 71 mm.
The base thickness may be reduced by taking account of the other materials as follows:
|
from Table E6: |
||
|
Brick outer leaf |
column E, row 2 |
= 4 mm |
|
Cavity |
column E, row 1 |
= 6 mm |
|
Plaster |
column E, row 3 |
= 1 mm |
|
and from Table E7: |
||
|
Concrete block |
column F, row 3 |
|
|
adjusted for 125 mm thickness (1.25 x 11) |
= 14 mm |
|
|
Total reduction |
= 25 mm |
|
The minimum thickness of the insulation layer required to achieve a U-value of 0.45 W/ m2K is therefore:
Base thickness less total reduction ie 71 — 25 = 46 mm.
Example 8 - Semi-exposed solid wall
Determine the thickness of the insulation layer required to achieve a U-value of 0.6 W/m2K for the wall construction shown below.
diagram
Using Table E5, from column B. row 5 of the table, the base thickness of the insulation layer is 30 mm.
The base thickness may be reduced by taking account of the other materials as follows:
|
from Table E6: |
||
|
Plasterboard |
column B, row 6 |
= 2 mm |
|
and from Table E7: |
||
|
Concrete block |
column B, row 7 |
|
|
adjusted for 215 mm thickness (2.15 x 8) |
= 17 mm |
|
|
Total reduction |
= 19 mm |
|
The minimum thickness of the insulation layer required to achieve a U-value of 0.6 W/m2K is therefore:
Base thickness less total reduction ie 30 — 19 = 11 mm.
Example 9 - Timber-frame wall
Determine the thickness of the insulation layer required to achieve a U-value of 0.45 W/m2K for the wall construction shown below.
Using Table E5, from column E, row 4 of the table, the base thickness of the insulation layer is 71 mm.
The base thickness may be reduced by taking account of the other materials as follows:
|
from Table E6: |
||
|
Brick outer leaf |
column E, row 2 |
= 4 mm |
|
Cavity |
column E, row 1 |
= 6 mm |
|
Sheathing ply |
column E, row 8 |
= 2 mm |
|
Plasterboard |
column E, row 6 |
= 3 mm |
|
and from Table E8: |
||
|
Timber frame adjusted for shallower member (0.9 x 74 mm) |
column E, row 1 |
= 67 mm |
|
Total reduction |
= 82 mm |
|
The reduction in base thickness is greater than the required thickness therefore no additional insulation is required.
The architect may wish to substitute a cheaper insulant with conductivity of 0.04 W/mK. In this case, the base thickness of insulation from Table A5 column F, row 4 would be 82 mm but the reductions due to the effects of the other materials would equal this and no additional insulation need be provided.
However, the designer may wish to increase the standard of insulation by reducing the U-value to 0.35 W/m2K. The base thickness of insulation from Table E5 column E, row 2 would be 94 mm. The total reduction in base thickness is as before, ie 82 mm and so the minimum thickness of insulation required in addition to the insulation between frame members to achieve a U-value of 0.35 W/m2K is therefore:
Base thickness less total reduction ie 94 — 82 = 12 mm.