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Lossy propagation

So far we have discussed the behaviour of acoustic waves in tubes assuming that none of the acoustic energy is lost to heat. In reality there is a boundary layer immediately beside the tube walls in which viscous and thermal losses occur. It is possible to use a lossy boundary condition to give lossy versions of $\psi_n$ and $\alpha_n$ but the effect of losses will be noticeable in the $z$ direction only because we will be considering objects which are significantly longer than they are wide. The inclusion or exclusion of the effect of losses will therefore be represented entirely by the choice of $z$ direction wavenumber, $k_n$. Starting with a lossy boundary condition which allows a small acoustic particle velocity flow into the wall of the tube, Bruneau et al [44] have produced a complex $z$ direction wavenumber:

\begin{displaymath}
k_n =
\pm \sqrt{k^2 - \left(\frac{\gamma_n}{R}\right)^2
...
...R}\right)
[ \mbox{Im}(\epsilon_n)-i \mbox{Re}(\epsilon_n) ]}
\end{displaymath} (2.53)

where $\epsilon_n$ is the boundary specific admittance at the wall. Admittance is the reciprocal of the impedance, so gives a measure of the acoustic velocity into the wall for a given acoustic pressure. The implication is not that there is really a flow into the walls (which are rigid in reality) but that the loss of energy at the boundary layer is simulated by imagining that such a flow exists. The boundary specific admittance is given by [44]:
\begin{displaymath}
\epsilon_n =
\left( 1 - \gamma_n^2/(k^2 R^2) \right) \epsilon_v + \epsilon_t
\end{displaymath} (2.54)

with $\epsilon_v = (1+i) \mbox{ } 2.03 \times 10^{-5} f^{1/2}$ and $\epsilon_t = (1+i) \mbox{ } 0.95 \times 10^{-5} f^{1/2}$ under standard conditions. The full expressions for $\epsilon_v$ and $\epsilon_t$ in terms of the thermodynamic constants of air are given in [44].

The choice of signs is complicated by the fact that we are performing the square root operation on a complex number. To split $k_n$ into real and imaginary parts, it is helpful to first express (2.53) as follows

\begin{displaymath}
k_n = \pm \sqrt{A_n + I_n - iR_n}
\end{displaymath} (2.55)

where $A_n$ is the square of $k_n$ in the absence of losses:
\begin{displaymath}
A_n = k^2 - \left(\frac{\gamma_n}{R}\right)^2,
\end{displaymath} (2.56)

$R_n$ gives the imaginary part of the correction in $k_n^2$:
\begin{displaymath}
R_n = (2k/R)\mbox{Re}(\epsilon_n),
\end{displaymath} (2.57)

and $I_n$ is the real part of the correction in $k_n^2$:
\begin{displaymath}
I_n = (2k/R)\mbox{Im}(\epsilon_n).
\end{displaymath} (2.58)

Now we can express $k_n$ in terms of real and imaginary parts:
\begin{displaymath}
k_n = \chi_n + i\kappa_n.
\end{displaymath} (2.59)

Equating equations (2.55) and (2.59) we get
\begin{displaymath}
\chi_n^2 - \kappa_n^2 +2i\chi_n \kappa_n = A_n + I_n - i R_n
\end{displaymath} (2.60)

which can be solved by simultaneous equations for the real and imaginary parts giving [44]
\begin{displaymath}
\chi_n= \frac{1}{\sqrt{2}}
\sqrt{\left\{A_n + I_n + \sqrt{(A_n + I_n)^2 + R_n^2}\right\}}
\end{displaymath} (2.61)

and
\begin{displaymath}
\kappa_n=-\frac{1}{\sqrt{2}}
\sqrt{\left\{-(A_n + I_n) + \sqrt{(A_n + I_n)^2 + R_n^2}\right\}}.
\end{displaymath} (2.62)

Putting $n = 0$ this equation gives the imaginary part of the plane mode wavenumber as $-2.98 \times 10^{-5} f ^{1/2}/R$.


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: Solutions for a uniform Up: Solutions for a cylinder Previous: Loss-less propagation   Contents
Jonathan Kemp 2003-03-24