Building Services Engineering 5th Edition Handbook

Heat loss calculations 75
U =
33.33
[21 − (−2)] W/m2 K
= 1.45 W/m 2 K
This is an elderly wall, which has a higher thermal transmittance than for modern standards.
Consideration can be given as to how much additional thermal insulation is possible. The thermal
resistance of the existing structure, without the surface film resistances, can be found from:
Q = 1 W
R m 2 K × A m2 (t 2 − t 3 ) K
33.33 = 1 W
R m 2 K × 1m2 × (17 − 0) K
(17 − 0)
R =
33.33 m2 K/ W
= 0.51 m 2 K/ W
When the thermal transmittance is known from design calculations or in situ measurements,
the thickness of additional thermal insulation that is needed to reduce heat loss can be calculated.
This may be desirable in order to align the building with current regulations and improve
its energy-using efficiency. Outdated building designs will be less attractive to potential users
than new or recently refurbished, low-energy consumption residential, commercial and industrial
alternative sites.
The wall U value that was considered here could be lowered from 1.45 to, say, 0.4 W/m 2 Kby
the addition of thermal insulation. If the insulation can be injected into the wall cavity no further
constructional measures are needed. Where there is no cavity, or if rainwater penetration could
result, then an additional internal or exterior layer of material is required. Thermal insulation may
not be structurally rigid and it often does not provide a hard-wearing or weatherproof surface.
Adding layers to either side of a wall necessitates architectural changes, particularly to fenestration
and doorways. If polyurethane board and an internal surface finish of 10 mm plasterboard can be
fitted to the interior surfaces, the necessary thickness of insulation can be calculated as follows.
From Table 3.1, the thermal conductivities are:
plasterboard: λ = 0.16 W/mK
polyurethane board: λ = 0.025 W/mK
new thermal resistance of whole structure R n = 1 U n
m 2 K
W
= 1
0.4
m 2 K
W
= 2.5 m 2 K/W
resistance of plasterboard = 0.01
0.16
resistance of existing wall = 1
1.45
mK
W
= 0.0625 m 2 K/W
m 2 K
W
= 0.69 m 2 K/W