Civil Engineering Reference
In-Depth Information
Making use of
a
rJ
o
(r)
d
r
=
aJ
1
(a)
(4.14)
0
and substitution of Equation (4.11) into Equation (4.13) yields
1
−
J
1
(s
√
−
1
jωρ
o
∂p
∂x
3
2
s
√
−
j
j)
J
o
(s
√
−
j)
υ
3
=−
(4.15)
In this equation,
s
is equal to
ωρ
o
R
2
η
1
/
2
s
=
(4.16)
The effective density
ρ
of the air in the tube is defined by rewriting Equation (4.15)
in a more compact form as
∂p
∂x
3
=
−
jωρυ
3
(4.17)
with
ρ
o
1
−
J
1
(s
√
−
2
s
√
−
j)
ρ
=
J
o
(s
√
−
(4.18)
j
j)
On the other hand, the true density
ρ
o
can be used in the Newton equation (4.15)
J
1
(s
√
−
j)
J
o
(s
√
−
∂p
∂x
3
=
2
s
√
−
ρ
o
jωυ
3
−
jωρ
o
υ
3
+
J
1
(s
√
−
(4.19)
1
−
j
j)
2
s
√
−
j)
J
o
(s
√
−
j
j)
Slit
The geometry of the problem is represented in Figure 4.3. With the same simplifications
as in the case of cylindrical tubes having a circular cross-section, the Newton equation
(4.9) reduces to
η
∂
2
υ
3
∂x
1
ρ
o
∂υ
3
∂p
∂x
3
+
∂t
=−
(4.20)
where
υ
3
is not dependent on
x
2
. The velocity must vanish at the surface of the slit. The
solution of Equation (4.20), where the velocity vanishes at
x
1
=±
a
,is
1
cosh
(l
x
1
)
cosh
(l
a)
1
jωρ
o
∂p
∂x
3
υ
3
=−
−
(4.21)
