Civil Engineering Reference
In-Depth Information
frequency ω. The radial surface velocity of the sphere may then be written as
ˆ
exp(
j
ω
,
which is also indicated in
Figure 3.3
.
uu
=⋅
p rt
(,)
G
r
uu
ω
=×
ˆ
e
jt
a
a
a
Figure 3.3
Sound radiation from a monopole source.
An outgoing wave is generated and the sound field must, due to symmetry, be equal
in all directions. Outside the sphere the sound pressure must satisfy the wave equation
using spherical coordinates. The solution must be of the same type as in Equation
(3.20)
so we may write
A
j(
ω−
tkr
)
(3.35)
prt
(,)
=⋅
e
,
r
where
r
is the distance to the centre of the sphere and
A
is unknown for the time being.
To determine the latter we shall again make use of the Euler equation
(3.3)
, which
connects the gradient of the pressure to the particle velocity
v
(
r,t
),
∂
prt
(,)
=−
j
ρω
⋅
vrt
(,)
=−
j
ρ
ck vrt
⋅
(,).
(3.36)
0
0
0
∂
r
We may then calculate the particle velocity, which at the surface of the sphere, i.e. when
the distance
r
is equal to the radius
a
, must be equal to the velocity
u
a
of the sphere. The
unknown quantity
A
is thereby determined, giving the pressure
2
ˆ
j
ρ
ckau
j(
ω
tkra
−−
(
))
00
a
prt
(,)
=
e
.
(3.37)
(1
+
j
ka r
)
The question is now how large the radiated sound power will be and, furthermore, what
are the controlling parameters? As we now have expressions both for the pressure and
the particle velocity we may calculate the intensity and by integrating the intensity over a
closed surface around the source we arrive at the total sound power. We shall perform
this exercise at the surface of the sphere where the pressure after some algebra may be
written as:
⎛
22
⎞
ka
ka
j
ω
t
ˆ
pat
(,)
=
ρ
cu
+
j
⋅
e
.
(3.38)
⎜
⎟
00
a
⎜
⎟
22
22
1
+
ka
1
+
ka
⎝
⎠
The sound pressure is then represented by two terms, the first term being in phase with
the velocity of the sphere and the other 90° out of phase. The latter term will be dominant
