Civil Engineering Reference
In-Depth Information
∂
p
ρ
∂
v
x
= −
+ ⋅
rv
.
(3.24)
0
x
∂
x
∂
t
Combining this equation with the corresponding one-dimensional versions of Equations
impedance
Z
s
can be written as
ω
r
Γ=
j
1j
−
a d
c
ρω
0
0
(3.25)
p
r
Z
==
ρ
c
1j
−
.
s
0
0
v
ρω
x
0
If we compare with the corresponding expressions for a lossless plane wave (for
Z
s
see
Equation
(3.17)
), we have now got an additional complex root expression. A medium
having this property is moreover the simplest model for a porous material, a
Rayleigh
model
(named after physicist Lord Rayleigh). This model will, together with other
models for porous material, be treated in Chapter 5. At this point, however, we shall only
give an example of the attenuation brought about by such a resistive component. We
shall assume a high frequency and/or a low flow resistivity such that the imaginary part
in the root expression is << 1. The propagation coefficient will then be
⎛
⎞
ω
r
ω
r
r
ω
Γ=
j
1j
−
≈
j
1j
−
=
+
j
.
(3.26)
⎜
⎟
c
ρω
c
2
ρω
2
ρ
c
c
⎝
⎠
0
0
0
0
0
0
0
The attenuation coefficient α is the real part of the expression, and the attenuation Δ
L
in
decibels per metre will be given by
r
(
)
Δ=
L
8.69
dB/m .
(3.27)
2
ρ
c
00
Having a flow resistivity of 1000 Pa⋅s/m
2
will then give us an attenuation ≈ 10 dB/m. We
may add that the normal quality of mineral wool used in buildings has a flow resistivity
10 times higher but we shall be reminded of the assumption introduced above.
(You may try the analogous calculation assuming a high flow resistivity and/or low
frequencies. Hint:
(
)
for
x
>> 1.)
1j
−⋅ ≈ −
x
1j
x
/2
3.3
SOUND INTENSITY AND SOUND POWER
A sound wave involves transport of energy and the energy flow per unit time through a
given surface is called
sound power.
If this surface encloses a given sound source
completely we will determine the total power, which is a characteristic quantity of the
source.
