Room acoustics 123
Figure 4.8 Power attenuation coefficient m for atmospheric absorption at 20° Celsius. Calculated from ISO
9613–1.
0 10203040506070809
Relative humidity (%)00.000.010.020.030.040.050.06Power attenuation coefficientm(m-1)
8000 Hz4000 Hz2000 Hz1000This atmospheric or air absorption brings about a modification of the total
absorption area of a room by an added term 4mV, where V is the volume of the room.
Instead of Equation (4.35) we get
(^60)
0s
55.26
,
4
V
T
cAmV=⋅
+
(4.42)
where As represent the total absorption area in the room exclusive of the air absorption.
This added term may certainly also be included in other expressions for the reverberation
time by modifying the denominator in the Equations (4.36) and (4.38). (How should we
include the air absorption into Equation (4.40)?). Certainly, the air absorption will be
important in large rooms. However, at a relative humidity in the range 20–30 %, which is
not unusual at certain times of the year in some countries, one will find that the
reverberation time at frequencies above 6–8 kHz, even for moderate sized rooms, will be
considerably influenced by air absorption.
Example In a room of volume 100 m^3 one measures a reverberation time of 0.5
seconds in the one-third-octave band with centre frequency 8000 Hz. The relative
humidity is 20 %. Using Figure 4.8 we find that m is equal to 0.05 m-1 at the frequency
8000 Hz. (The figure applies to single frequencies but we shall use it to represent the
corresponding frequency band.) The air absorption alone then gives an absorption area of
20 m^2. Applying Equation (4.35) we find the total absorption area A of the room is
approximately 32.5 m^2. More than half of this absorption area is then due to air
absorption. Without this contribution, the reverberation time would be well over one
second.
Evidently, the air absorption may have important implications on the reverberation
time but also on sound pressure levels in rooms at sufficiently high frequencies. We