Civil Engineering Reference
In-Depth Information
In conclusion, for a porous material with porosity
φ
and flow resistivity
σ
,where
pores are identical parallel holes normal to the surface, the following expressions for the
effective mass, and the compressibility of air in the material, can be used:
ρ
o
1
jωρ
o
G
c
(s)
σφ
ρ
=
+
(4.125)
1
)F(B
2
ω)
]]
K
=
γP
o
/
[
γ
−
[
(γ
−
(4.126)
where
1
1
jB
2
ωρ
o
G
c
(Bs)
σφ
F(B
2
ω)
=
+
(4.127)
and
j
J
1
(s
√
−
1
−
J
1
(s
√
−
4
−
s
j)
2
s
√
−
j)
G
c
(s)
=−
J
o
(s
√
−
J
o
(s
√
−
(4.128)
j)
j
j)
In Equation (4.128),
s
is equal to
c
8
ωρ
o
σφ
1
/
2
s
=
(4.129)
(
8
η/(σφ))
1
/
2
/
¯
r
. Table 4.1 gives the values
and
c
depends on the hydraulic radius,
c
=
of c for various cross-sections.
4.8
Impedance of a layer with identical pores
perpendicular to the surface
4.8.1 Normal incidence
The layer of porous material represented in Figure 4.10 is placed on an impervious rigid
floor and is in contact with air in which a normal plane acoustic field is present. The
acoustic field in the air is plane up to a small distance
e
from the material and identical
waves propagate in each pore. The distance
e
is smaller than the distance between two
pores, and can be neglected for the more common acoustic materials. The porous material
of Figure 4.10 is represented on a macroscopic scale in Figure 4.11.
Two points
M
1
and
M
2
are selected at the surface of the material,
M
2
in free air and
M
1
in the porous material. Let
υ(M
1
)
be the mean velocity of the air in a pore close
to the surface and
υ(M
2
)
the velocity in free air in the plane field, the distance
e
being
neglected. The pressures
p(M
2
)
and
p(M
1
)
are the pressures in free air and in a pore,
respectively. The continuity of air flow and pressure at the surface of the porous material
implies the following two equations:
p(M
2
)
=
p(M
1
)
(4.130)
υ(M
2
)
=
υ(M
1
)φ
(4.131)
