Building Services Engineering 5th Edition Handbook

Condensation in buildings 249
Also:
Q f = 1
R so
(t 4 − t 5 )
and,
t 5 = t 4 − Q f R so
Calculating the outdoor air temperature t 5 is a check on the accuracy of the calculations and
method. It should agree with the original value used in finding Q f to within ±1%.
To find t 3 use:
Q f = 2λ (t
l 2 − t 3 )
Hence:
( ) l
t 3 = t 2 −
2λ Q f
EXAMPLE 10.7
Calculate the temperature gradient through a medium-weight concrete block wall 100 mm
thick. The internal and external air temperatures are 20 ◦ C d.b. and −1 ◦ C d.b.
From Table 3.1 thermal conductivity λ = 0.51 W/mK, from Table 3.2 R si = 0.12 m 2 K/W and
from Table 3.3 R so = 0.06 m 2 K/W. Then the thermal transmittance is given by:
1
U =
R si + (l/λ) + R so
1
=
0.12 + (0.1/0.51) + 0.06 W/m2 K
= 2.66 W/m 2 K
For a wall area of 1 m 2 :
Q 2 = 2.66 ×[22 − (−1)] W
= 61.16 W
Using the numbered locations in Fig. 10.2:
t 1 = 22 ◦ C, t 5 =−1 ◦ C
t 2 = 22 − 61.16 × 0.12 = 14.66 ◦ C
t 4 = 14.66 − 0.1
0.51 × 61.16 = 2.67◦ C