Civil Engineering Reference
In-Depth Information
Denoting the mean value of the particle velocity as
v
m
will give the following
equation of force
∂
−=
p
8
μ
j
ωρ
v
+
v
=
j
ωρ
v
+
r v
'
,
(5.43)
0m
m
0m
m
2
∂
x
a
where we have introduced the flow resistivity
r
'
in the tube, having the dimension
Pa⋅s/m
2
. For simplicity, we have also replaced the constant 4/3 in the mass term by 1.
The corresponding continuity equation, i.e. the equation stating the conservation of
mass, may in this one-dimensional case be written
ρ
∂
v
∂
ρ
1
∂
p
m
−
=
=
,
(5.44)
0
2
0
∂
x
∂
t
∂
t
c
when we take account of the relationship between the pressure and the density
fluctuations. We shall look for solutions to these equations expressing the sound pressure
and the particle velocity as
(
)
j
ω
tkx
−
'
ˆ
e
pp
=⋅
)
(5.45)
(
j
ω
tkx
−
'
ˆ
and
vv
=⋅
e
,
mm
where the wave number
k'
now is a complex quantity. Inserting these expressions into
kp
'
+−
( j '
r
ρω
)
v
=
0
0
m
(5.46)
2
00
and
ωρ
p
−
c
k v
'
=
0,
m
where a solution may only be found when the determinant is zero, i.e. when
ρω
ωρ
kr
'
j'
−
0
=
0.
2
00
−
ck
'
This gives us the following equation for the wave number
ω
r
'
k
'
=
1
−
j
(5.47)
c
ρω
0
0
and thereby the impedance
p
r
ρω
'
Z
'
==
ρ
c
1
−
j
.
(5.48)
00
v
m
0
We have then arrived at an expression giving the complex wave number and the
characteristic impedance in one tube in the matrix of tubes sketched above. Our task now
is to find the input impedance at the surface of the matrix. We have then to take account
of the contraction of the “stream lines” when entering the matrix from the outside
medium. This is accomplished by introducing the porosity σ of the matrix, the ratio of
the pore (or tube) volume to the total volume. For this simple case the porosity will be
