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Analysis

Consider a rectangular duct of half widths $a$ and $b$ terminated in an infinite baffle, as shown in figure 3.4.

Figure 3.4: Geometry of infinitely flanged rectangular duct
\begin{figure}\begin{center}
\epsfig{figure=chapter3/rect.eps, width=.5\linewidth}\end{center}\end{figure}

Expressing (3.13) in rectangular coordinates for a rectangular duct of half-widths $a$ and $b$ gives:

\begin{displaymath}
Z_{nm} =
\frac{i\omega\rho}{2\pi S^2}
\int\limits_{-a}^{...
..._{-b}^{b} dy_0
\psi_m(x_0,y_0) \psi_n(x,y) \frac{e^{-ikh}}{h}
\end{displaymath} (3.26)

where
\begin{displaymath}
h = [(x-x_0)^2 + (y-y_0)^2]^{\frac{1}{2}},
\end{displaymath} (3.27)

and
\begin{displaymath}
\psi_n = \phi_{n_x}\sigma_{n_y},
\end{displaymath} (3.28)

with
\begin{displaymath}
\phi_{n_x} = N_{n_x}\cos \left(\frac{n_x \pi x}{a} \right),
\end{displaymath} (3.29)


\begin{displaymath}
\sigma_{n_y} = N_{n_y}\cos \left(\frac{n_y \pi y}{b} \right),
\end{displaymath} (3.30)


\begin{displaymath}
N_{n_x} = \left\{ \begin{array}
{r@{\quad:\quad}l}
1 & n_x = 0, \\
\sqrt{2} & n_x > 0. \\
\end{array} \right.
\end{displaymath} (3.31)

Changing variables as in Swenson et al. [49] and Levine [51] to $\xi = x - x_0$, $\eta = x + x_0$, $\zeta = y - y_0$, and $\mu = y + y_0$ the integral becomes

$\displaystyle {Z_{nm} =}$
    $\displaystyle \frac{i\omega\rho}{2\pi S^2}
\int\limits_0^{2a} d\xi
\int\limits_...
...hi_{m_{x0}}
\int\limits_{-2b+\zeta}^{2b-\zeta}d\mu
\sigma_{n_y} \sigma_{m_{y0}}$  
      (3.32)

where
\begin{displaymath}
h = \sqrt{\xi^2 + \zeta^2}.
\end{displaymath} (3.33)

The quadruple integral can now be reduced to a double integral by performing integration by $\eta$ and $\mu$ analytically. The first step is to expand the cosines in equations (3.29) and (3.30):
$\displaystyle {\phi_{n_x} \phi_{m_{x0}}= N_n N_m \times}$
    $\displaystyle \scriptstyle
\Big[
\cos(\frac{n_x\pi\xi}{2a})\cos(\frac{m_x\pi\xi...
...in(\frac{m_x\pi\xi}{2a})
\cos(\frac{n_x\pi\eta}{2a})\sin(\frac{m_x\pi\eta}{2a})$  
    $\displaystyle \scriptstyle
- \sin(\frac{n_x\pi\xi}{2a})\cos(\frac{m_x\pi\xi}{2a...
...c{m_x\pi\xi}{2a})
\sin(\frac{n_x\pi\eta}{2a})\sin(\frac{m_x\pi\eta}{2a})
\Big].$  
      (3.34)

The second and third terms go to zero since we are integrating over a symmetric interval in $\eta$. Performing integration gives:
$\displaystyle {
G(n_x,m_x,\xi,a) = \int\limits_{-2a+\xi} ^{2a-\xi} d\eta
\phi_{n_x} \phi_{m_{x0}}
}$
    $\displaystyle = N_{n_x} N_{m_x} (2a - \xi)
\Bigg[
{\mathrm sinc}\left((n_x + m_...
...t(1-\frac{\xi}{2a}\right)\right)
\cos\left(\frac{(n_x-m_x) \pi \xi}{2a}\right)+$  
    $\displaystyle \mbox{\hspace{3cm}}
{\mathrm sinc}\left((n_x - m_x)\pi\left(1-\frac{\xi}{2a}\right)\right)
\cos\left(\frac{(n_x+m_x) \pi \xi}{2a}\right) \Bigg].$ (3.35)

The integral for the impedance is then:
\begin{displaymath}
Z_{nm} = \frac{i\omega\rho}{2\pi S^2}
\int\limits_0^{2a} d\...
...eta
\frac{e^{-ikh}}{h}
G(n_x,m_x,\xi,a)G(n_y,m_y,\zeta,b).
\end{displaymath} (3.36)

Changing variables to $u = k\xi$ and $v = k\zeta$ means that the radiation impedance is expressed in terms of the dimensionless variables $ka$ and $kb$:

\begin{displaymath}
Z_{nm} = \frac{i \rho c}{2\pi S}
\int\limits_0^{2ka} du
...
...1}{2}\right)
G\left(n_y,m_y,\frac{v}{2kb},\frac{1}{2}\right).
\end{displaymath} (3.37)

Note that if we put $n_x = m_x = n_y = m_y = 0$ into equation (3.35) we obtain
\begin{displaymath}
G\left(n_x,m_x,\frac{u}{2ka},\frac{1}{2}\right)
G\left(n_y...
...
= 4\left(1-\frac{u}{2ka}\right)\left(1-\frac{v}{2kb}\right).
\end{displaymath} (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 $a$ and $b$ as widths rather than half widths). Equation (3.37) has a singularity at the origin if $n_x = m_x$ and $n_y = m_y$ 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:
$\displaystyle {Z_{nm} =}$
    $\displaystyle \frac{i\rho c}{2\pi S}
\int\limits_0^{2ka} du
\int\limits_0^{2kb} dv
\frac{(1 - \frac{u}{2ka}) (1 - \frac{v}{2kb})}{\sqrt{u^2+v^2}}$  
    $\displaystyle \mbox{\hspace{4cm}}
\left[e^{-i\sqrt{u^2+v^2}}
\frac{G(n_x,m_x,\f...
...frac{G(n_y,m_y,\frac{v}{2kb},\frac{1}{2})}{(1 - \frac{v}{2kb})}
- f(n,m)\right]$  
    $\displaystyle +\frac{i\rho c}{2\pi S}
\int\limits_0^{2ka} du
\int\limits_0^{2kb} dv
\frac{(1 - \frac{u}{2ka}) (1 - \frac{v}{2kb})}{\sqrt{u^2+v^2}}
f(n,m)$ (3.39)

where
$\displaystyle {f(n,m) = N_{n_x}N_{m_x}N_{n_y}N_{m_y} \times}$
    $\displaystyle \left[{\mathrm sinc}((n_x+m_x)\pi)
+ {\mathrm sinc}((n_x-m_x)\pi)\right]
\left[{\mathrm sinc}((n_y+m_y)\pi)
+ {\mathrm sinc}((n_y-m_y)\pi)\right].$  
      (3.40)

The first part is non-singular and the singularity in the second half may be removed by integration, giving:
$\displaystyle {Z_{nm} =}$
    $\displaystyle \frac{i\rho c}{2\pi S}
\int\limits_0^{2ka} du
\int\limits_0^{2kb} dv
\frac{(1 - \frac{u}{2ka}) (1 - \frac{v}{2kb})}{\sqrt{u^2+v^2}} \times$  
    $\displaystyle \mbox{\hspace{4cm}}
\left[e^{-i\sqrt{u^2+v^2}}
\frac{G(n_x,m_x,\f...
...frac{G(n_y,m_y,\frac{v}{2kb},\frac{1}{2})}{(1 - \frac{v}{2kb})}
- f(n,m)\right]$  
    $\displaystyle + \frac{i\rho c}{2\pi S}\int\limits_0^{2ka} du
\left(1 - \frac{u}{2ka}\right) \times$  
    $\displaystyle \mbox{\hspace{2cm}}
\left[
\ln{\left(2kb + \sqrt{u^2+(2kb)^2}\right)} + \frac{u}{2kb}
- \frac{1}{2kb}\sqrt{u^2 + (2kb)^2}
\right] f(n,m)$  
    $\displaystyle + \frac{i\rho c}{2\pi S}
\left[-ka\ln{(2ka)} + \frac{3}{2}ka \right]
f(n,m).$ (3.41)

Equation (3.41) may be evaluated by numerical integration to provide the radiation impedance.


This thesis has moved to Jonathan Kemp Thesis at http://www.kempacoustics.com/thesis
Please change your bookmark/reference to reflect this change as this site may be discontinued
next up previous contents
Next: Results Up: Multimodal radiation impedance of Previous: Multimodal radiation impedance of   Contents
Jonathan Kemp 2003-03-24