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.47.

Figure C.47: 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

$$df(t,x) = f(t,x)-f(t,x+dx) = f(t,x)-[f(t,x) + dx \cdot f'(x)]$$

$$= - dx \cdot f'(x)$$

$$= - dx \cdot[A(x)p(t,x)]'$$

$$= -dx \cdot[A' p + A'p],$$

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

$$\begin{eqnarray*} dM(t,x) &=& \int_x^{x+dx} \rho(t,\xi) A(\xi)\,d\xi \\ [5pt] &\approx& \int_x^{x+dx} \left[\rho(t,x)+\rho'(t,x)(\xi-x)][A(x)+A'(x)(\xi-x)\right]d\xi\\ [5pt] &\approx& \rho(t,x) A(x)\,dx + \int_x^{x+dx} \left[\rho'(t,x)A(x)+\rho(t,x)A'(x)\right](\xi-x)\,d\xi\\ [5pt] &=& \rho\,A\,dx + \left(\rho' A+\rho A' \right)\frac{(dx)^2}{2}\\ [5pt] &\approx& \rho\, A\,dx, \end{eqnarray*}$$

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.20 Applying Newton's second law $df = dM\cdot {\dot u}$ , we obtain

$$-dx \cdot (A'p + Ap') \eqsp \rho\, A\,dx\, {\dot u} \protect$$ (C.167)

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

$$p' + p \, \frac{A'}{A} \eqsp - \rho\,{\dot u}. \protect$$ (C.168)

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

$$p' + p \, 'A \eqsp - \rho\,{\dot u}. \protect$$ (C.169)

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.