Building Services Engineering 5th Edition Handbook

250 Condensation in buildings
and,
t 5 = 2.67 − 61.16 × 0.06 =−1 ◦ C
which agrees with the input data. Also,
( )
0.1
t 3 = 14.66 −
2 × 0.51 × 61.16 = 8.66 ◦ C
which should be the temperature midway between t 2 and t 4 that is (14.66 ± 2.67)/2 = 8.67 ◦ C,
which it is.
EXAMPLE 10.8
Calculate the temperature gradient through a cavity wall consisting of 13 mm lightweight
plaster, 100 mm lightweight concrete block, 40 mm mineral fibre slab, 10 mm air space
and 105 mm brickwork. Internal and external air temperatures are 22 ◦ C and 0 ◦ C.
From Table 3.1, the thermal conductivities are as follows: plaster, λ 1 = 0.16 W/mK; concrete,
λ 2 = 0.19 W/mK; mineral fibre, λ 3 = 0.035 W/mK; brickwork, λ 4 = 0.84 W/mK. From Tables 3.2,
3.3 and 3.4, R si = 0.12 m 2 K/W, R so = 0.06 m 2 K/W and R a = 0.18 m 2 K/W. Then
U =
=
(
R si + l 1
+ l 2
+ l 3
+ R a + l ) −1
4
+ R so
λ 1 λ 2 λ 3 λ 4
(
0.12 + 0.013
0.16 + 0.10
0.19 + 0.04
)
0.105 −1
+ 0.18 +
0.035 0.84 + 0.06 W/m 2 K
= 0.45 W/m 2 K
For a wall area of 1 m 2 :
Q f = 0.45 × (0.22 − 0) = 9.9 W
Using the notation from Fig. 10.3, temperatures are calculated as follows:
t 2 = 22 − (0.12 × 9.9) = 20.81 ◦ C
( )
0.013
t 3 = 20.81 −
0.16 × 9.9 = 20.01 ◦ C
( 0.01
t 4 = 20.01 −
0.19 × 9.9 )
= 14.8 ◦ C
( )
0.04
t 5 = 14.8 −
0.035 × 9.9 = 3.49 ◦ C
t 6 = 3.49 − (0.18 × 9.9) = 1.71 ◦ C
( )
0.105
t 7 = 1.71 −
0.84 × 9.9 = 0.47 ◦ C
t 8 = 0.47 − (0.06 × 9.9) = 0.12 ◦ C