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Consider a rectangular duct of half widths
and
terminated in an
infinite baffle, as shown in figure 3.4.
Figure 3.4:
Geometry of infinitely flanged rectangular duct
 |
Expressing (3.13) in rectangular coordinates for a rectangular duct
of half-widths
and
gives:
 |
(3.26) |
where
![\begin{displaymath}
h = [(x-x_0)^2 + (y-y_0)^2]^{\frac{1}{2}},
\end{displaymath}](img381.png) |
(3.27) |
and
 |
(3.28) |
with
 |
(3.29) |
 |
(3.30) |
 |
(3.31) |
Changing variables as in Swenson et al. [49] and Levine [51] to
,
,
, and
the integral becomes
 |
|
|
 |
|
|
|
|
(3.32) |
where
 |
(3.33) |
The quadruple integral can now be reduced to a double integral
by performing integration by
and
analytically.
The first step is to expand the cosines in equations (3.29)
and (3.30):
The second and third terms go to zero
since we are integrating over a symmetric interval in
.
Performing integration gives:
The integral for the impedance is then:
 |
(3.36) |
Changing variables to
and
means that the radiation
impedance is expressed in terms of the dimensionless variables
and
:
 |
(3.37) |
Note that if we put
into equation
(3.35) we obtain
 |
(3.38) |
The radiation impedance from equation (3.37) is then identical to
the radiation impedance of a rectangular piston in an
infinite baffle [47,48,49,50,51] (note that most authors
have used
and
as widths rather than half widths). Equation
(3.37) has a singularity at the origin if
and
which must be removed if the radiation impedance is to be calculated by
numerical integration. To do this the integral is first split into two parts:
where
 |
|
|
![$\displaystyle \left[{\mathrm sinc}((n_x+m_x)\pi)
+ {\mathrm sinc}((n_x-m_x)\pi)\right]
\left[{\mathrm sinc}((n_y+m_y)\pi)
+ {\mathrm sinc}((n_y-m_y)\pi)\right].$](img415.png) |
|
|
|
|
(3.40) |
The first part is non-singular and the singularity in the second half may be
removed by integration, giving:
Equation (3.41) may be evaluated by numerical integration to
provide the radiation impedance.
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Jonathan Kemp
2003-03-24