Civil Engineering Reference
In-Depth Information
The total energy density at a receiving point at a given distance
r
from the source is
assumed to be given by
www
=
+
s
,
(4.72)
tot
d
where
w
d
is the contribution from the direct sound, i.e. the non-scattered part, and
w
s
is
the contribution from the phonons arriving at the receiver position after one or more
collisions with the scattering objects. Without these scattering objects,
w
d
will be given
by Equation
(4.63)
but now we shall have to modify this expression by subtracting the
part being scattered or attenuated in other ways than by specular reflections from the
room boundaries. We shall start looking at the scattered sound.
4.8.1.1
Scattered sound energy
Starting from the probability density given in Equation
(4.70)
, Kuttruff (1981) calculated
the corresponding probability that a phonon after a time
t
should be at a distance
r
from
the source. In an infinite space, this probability density will be given by
3
−
3
q
r
ct
2
⎛
⎞
3
q
2
4
Prt
(,)
=
⋅
e
,
(4.73)
⎜
⎟
0
4
π
ct
⎝
⎠
0
assuming that
qc
0
t
>> 1,
w
hich implies that the travelled distance
c
0
t
must be much larger
than the mean free path
R
=1/
q
. It may also be mentioned that
P
is a solution of the so-
called diffusion equation used in fluid dynamics, which is
∂
∇=⋅
∂
1
Q
2
Q
,
(4.74)
Dt
when setting the diffusion constant
D
equal to
c
0
/(3
q
). The diffusion equation may e.g.
describe how the concentration
Q
of a fluid, such as a dye, when injected into another
fluid, changes with time. It should not be too difficult to envisage that this is a process
quite analogous to how sound particles or phonons diffuse into a space containing
scattering objects.
Jovicic assumes that the same probability
P
(
r,t
) applies to the phonons from the
image sources as all scattering objects are mirrored in the boundary surfaces (floor and
ceiling) as well. The predicted total probability applicable to the phonons sent out from
the original source and the image sources is then given by
−
3
q
2
⋅
r
3
q
4
ct
(4.75)
Prth
(,, )
=
e
,
0
4
π
cht
0
where
h
is the height of the room. Inside a small volume element, containing the
receiving position at a distance
r
from the source, we shall find phonons emitted from the
source (and the image sources) at different points in time, thereby having different
probability
P
(
r,t,h
) of arriving at the chosen volume element. The shortest time of arrival
will be
r
/
c
0
and the longest one will be infinity.
On their way, the phonons are losing their energy, partly by hitting the scattering
objects having absorption factor
α
s
, partly hitting the floor and ceiling having absorption
factors
α
f
and
α
c
, respectively. In addition, we have the excess attenuation due to air
