Free Books

Momentum Conservation in Nonuniform Tubes

Newton's second law ``force equals mass times acceleration'' implies that the pressure gradient in a gas is proportional to the acceleration of a differential volume element in the gas. Let $ A(x)$ denote the area of the surface of constant phase at radial coordinate $ x$ in the tube. Then the total force acting on the surface due to pressure is $ f(t,x)=A(x)p(t,x)$, as shown in Fig.C.45.

Figure C.45: Differential volume element for the conical acoustic tube.
The net force $ df(t,x)$ to the right across the volume element between $ x$ and $ x+dx$ is then
$\displaystyle df(t,x) = f(t,x)-f(t,x+dx)$ $\displaystyle =$ $\displaystyle f(t,x)-[f(t,x) + dx \cdot f'(x)]$  
  $\displaystyle =$ $\displaystyle - dx \cdot f'(x)$  
  $\displaystyle =$ $\displaystyle - dx \cdot[A(x)p(t,x)]'$  
  $\displaystyle =$ $\displaystyle -dx \cdot[A' p + A'p],$  

where, when time and/or position arguments have been dropped, as in the last line above, they are all understood to be $ t$ and $ x$, respectively. To apply Newton's second law equating net force to mass times acceleration, we need the mass of the volume element
dM(t,x) &=& \int_x^{x+dx} \rho(t,\xi) A(\xi)\,d\xi \\ [5pt]
...\rho A' \right)\frac{(dx)^2}{2}\\ [5pt]
&\approx& \rho\, A\,dx,
where $ \rho(t,x)$ denotes air density. The center-of-mass acceleration of the volume element can be written as $ {\dot u}(t,x)$ where $ u(t,x)$ is particle velocity.C.16 Applying Newton's second law $ df = dM\cdot {\dot u}$, we obtain

$\displaystyle -dx \cdot (A'p + Ap') \eqsp \rho\, A\,dx\, {\dot u} \protect$ (C.146)

or, dividing through by $ -A\,dx$,

$\displaystyle p' + p \, \frac{A'}{A} \eqsp - \rho\,{\dot u}. \protect$ (C.147)

In terms of the logarithmic derivative of $ A$, this can be written

$\displaystyle p' + p \,$   ln$\displaystyle 'A \eqsp - \rho\,{\dot u}. \protect$ (C.148)

Note that $ p$ denotes small-signal acoustic pressure, while $ \rho$ denotes the full gas density (not just an acoustic perturbation in the density). We may therefore treat $ \rho$ as a constant.

Cylindrical Tubes

In the case of cylindrical tubes, the logarithmic derivative of the area variation, ln$ 'A(x) = A'/A$, is zero, and Eq.$ \,$(C.148) reduces to the usual momentum conservation equation $ p' = -\rho {\dot u}$ encountered when deriving the wave equation for plane waves [318,349,47]. The present case reduces to the cylindrical case when

$\displaystyle \frac{A'}{A} \;\ll\; \frac{p'}{p}

i.e., when the relative change in cross-sectional area is much less than the relative change in pressure along the tube. In other words, the tube area variation must be slower than the spatial variation of the wave itself. This assumption is also necessary for the ``one-parameter-wave'' approximation to hold in the first place. If we look at sinusoidal spatial waves, $ p=A_p e^{j k_p x}$ and $ A=A_A
e^{j k_A x}$, then $ A'/A = k_A$ and $ p'/p = k_p$, and the condition for cylindrical-wave behavior becomes $ k_A\ll k_p$, i.e., the spatial frequency of the wall variation must be much less than that of the wave. Another way to say this is that the wall must be approximately flat across a wavelength. This is true for smooth horns/bores at sufficiently high wave frequencies.
Next Section:
Wave Impedance in a Cone
Previous Section:
Conical Acoustic Tubes