Building Services Engineering 5th Edition Handbook

328 Room acoustics
Table 14.2 Solution to Example 14.1.
Surface
Absorption data at frequency
125 Hz 250 Hz 500 Hz 1 kHz 2 kHz 4 kHz
Floor α 0.02 0.02 0.02 0.04 0.05 0.05
Ceiling α 0.15 0.4 0.75 0.85 0.8 0.85
Walls α 0.05 0.04 0.04 0.03 0.03 0.02
Floor (S × α) 0.24 0.24 0.24 0.48 0.6 0.6
Ceiling (S × α) 1.8 4.8 9.0 10.2 9.6 10.2
Walls (S × α) 1.75 1.4 1.4 1.05 1.05 0.7
ᾱ 0.064 0.109 0.18 0.199 0.191 0.195
Room constant R m 2 4.03 7.21 12.95 14.66 13.93 14.29
Reverberation T s 1.28 0.75 0.45 0.41 0.43 0.42
The surface absorption coefficients are selected from Table 14.1. It can be seen that there will
be a different room constant and reverberation time for each frequency. The solution is presented
in Table 14.2.
Room volume V = 4 × 3 × 2.5 m 2
= 30 m 3
floor area = 12 m 2
ceiling area = 35 m 2
wall area = 35 m 2
Room surface area A = (2 × 4 × 3) + (4 + 4 + 3 + 3) × 2.5 m 2
= 59 m 2
For 125 Hz, the mean absorption coefficient is,
12 × 0.02 + 12 × 0.15 + 35 × 0.05
α =
12 + 12 + 35
= 0.064
room constant R = S × α
1 − α m2
59 × 0.064
=
1 − 0.064 m2
= 4.03 m 2
reverberation time T = 0.161 × V
S × α
0.161 × 30
=
59 × 0.064 s
= 1.28 s