Civil Engineering Reference
In-Depth Information
p
1
X
e
_
p
2
Figure 4.8
A sample of porous material of thickness
e
and of unit cross-section, sub-
mitted to a constant differential pressure
p
2
−
p
1
. The mean molecular velocity in the
pores is
υ
.
The flow resistivity
σ
is given by
8
η
R
2
(nπR
2
)
σ
=
(4.78)
The quantity
nπR
2
is the porosity
φ
of the material, and Equation (4.78) may be
rewritten
8
η
R
2
φ
σ
=
(4.79)
Using Equation (4.79), Equation (4.16) can be rewritten
8
ωρ
o
σφ
1
/
2
s
=
(4.80)
Equations (4.18), (4.48) and (4.80) can be used to calculate
ρ
and
K
at a given angular
frequency
ω
in the porous material represented in Figure 4.8. The description of the
viscous force in the Newton equation (4.19) can be modified in the following way. The
quantity
ρ
o
jω/s
is given by
σφs(
−
j)
2
/
8
ρ
o
jω/s
=−
(4.81)
Substituting the right-hand side of Equation (4.81) for
ρ
o
jω/s
in Equation (4.19)
gives
∂p
∂x
=
−
jωρ
o
υ
+
σφυG
c
(s)
(4.82)
where
G
c
(s)
is given by
j
J
1
(s
√
−
1
−
J
1
(s
√
−
4
−
s
j)
2
s
√
−
j)
G
c
(s)
=−
J
o
(s
√
−
J
o
(s
√
−
(4.83)
j)
j
j)
The limit at low frequencies of
G
c
(s)
is 1 at
ω
=
0, and Eq. (4.82) becomes
∂p
∂x
=
−
σφυ
(4.84)
