Volume velocity

In plane wave propagation the continuity condition was that the volume velocity must be equal on $S_1$ and $S_2$ meaning that the mass of air flowing out of $S_1$ equals the flow of mass into $S_2$ at any given time. This implies that the velocity, which is assumed to be constant over the cross-section, is different on either side because of the difference of cross-sectional area. When we are treating the velocity field accurately in three dimensions it is clear that the velocity on the two surfaces should match and that the velocity into the $x$-$y$ plane wall on the larger cross-section is zero. For the case where $S_2>S_1$ we have

$$v_z^{(1)} = v_z^{(2)} \hspace{0.5cm} {\mathrm on} \hspace{0.5cm} S_1, \hspace{0.5cm} S_1<S_2$$

$$v_z^{(2)} = 0 \hspace{0.5cm} {\mathrm on} \hspace{0.5cm} S_2-S_1, \hspace{0.5cm} S_1<S_2$$ (B.10)

where $S_2-S_1$ is a shorthand for the $x$-$y$ plane wall resulting from the part of surface 2 which is not shared with surface 1. In terms of volume velocities this means that $U^{(1)}/S_1 = U^{(2)}/S_2$ on $S_1$ and $U^{(2)} = 0$ on $S_2-S_1$. Now we will use equation (2.27) (again ignoring the ${\mathrm exp}(i \omega t)$ time factor) and use the orthogonality of the modes. This time in order to include the fact that the volume velocity is zero on $S_2-S_1$ we must perform the integration over surface 2:

$$U_n^{(2)} = \int\limits_{S_2} \psi_n^{(2)} v_z^{(2)} dS = ... ..._n^{(2)} \sum\limits_{m=0}^{\infty} \psi_m^{(1)} U_m^{(1)} dS$$ (B.11)

which may be written as

$$U_n^{(2)} = \sum\limits_{m=0}^{\infty} F_{mn} U_m^{(1)}$$ (B.12)

where $F$ is given in equation (B.6). It should be noted that the integration in $F$ is this time over the $\psi_m$ on surface 1 and over $\psi_n$ on surface 2, hence $m$ and $n$ in the subscript to $F$ are swapped for equation (B.12). In matrix notation the result is

$${\mathbf U}^{(2)}= F^{\mathrm{T}} {\mathbf U}^{(1)}, \mbox{\hspace{1cm}} S_1 < S_2$$ (B.13)

proving equation (2.83). The swapping of indices in $F$ is denoted by the transpose operation represented by the superscript $\mathrm{T}$.

As with the pressure, when $S_1>S_2$ the volume velocity calculation can be performed simply be interchanging the labels 1 and 2 giving equation (2.84):

$${\mathbf U}^{(1)}= V^{\mathrm{T}} {\mathbf U}^{(2)}, \mbox{\hspace{1cm}} S_1 > S_2.$$ (B.14)

with $V$ given in equation (B.9).