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On the transition between continuous and shocked solutions for resonant gas oscillations in the frustum of a closed cone

Published online by Cambridge University Press:  22 February 2022

D.E. Amundsen*
Affiliation:
School of Mathematics and Statistics, Carleton University, Ottawa, ON K1S 5B6, Canada
M.P. Mortell
Affiliation:
Department of Applied Mathematics, University College Cork, Cork, Ireland
B.R. Seymour
Affiliation:
Department of Mathematics, University of British Columbia, Vancouver, BC V6T 1Z2, Canada
*
Email address for correspondence: dave@math.carleton.ca

Abstract

Resonant gas oscillations in a closed straight tube contain shocks for frequencies near the linear resonant frequency. As the tube geometry changes from straight to a cone with slope $a,$ the shock strength decreases until, for large enough $a,$ the motion is continuous. The analytical result for small $a$ follows from a nonlinearization of the linear resonant response, while the result for large $a$ is a dominant single mode approximation. The connection between these two forms is analysed numerically. The shocked solutions change to multimode continuous solutions as $a$ increases to cross the curve $a_{s}\doteqdot 12.8M^{1/3}$, where $M\ll 1$ is the Mach number of the input. Then the amplitudes of the higher modes decrease as $a$ continues to increase until, having crossed $a_{m}\doteqdot 30M^{1/3}$, the single mode solution emerges.

Information

Type
JFM Papers
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
Copyright
© The Author(s), 2022. Published by Cambridge University Press
Figure 0

Figure 1. Polytropic gas in frustum of a closed cone and forced at one end.

Figure 1

Figure 2. Comparison of approximate solutions based on (3.15) (dashed lines) vs direct numerical simulation (solid) for $M=1.25\times 10^{-4}$ and (a) $a=0.05$, (b) $a=0.1$, (c) $a=0.15$, (d) $a=0.2$.

Figure 2

Figure 3. Comparison of approximate solutions based on (3.13) (dashed lines) vs direct numerical simulation (solid) for $M=1.25\times 10^{-4}$ and (a) $a=0.05$, (b) $a=0.1$, (c) $a=0.15$, (d) $a=0.2$.

Figure 3

Table 1. Relative 2-norm errors for approximations using (3.15), $\ell =0$ and (3.13), $\ell >0$, for $M=1.25\times 10^{-4}$.

Figure 4

Table 2. Values of $u(0.5,t)$ on left-hand ($u_l$) and right-hand ($u_r$) sides of upper continuous interval using approximation (3.13), with $M=1.25\times 10^{-4}$.

Figure 5

Figure 4. (a) Approximate solution $u(0.5,t)$ based on (3.21) for $i=30$ compared with linear solution. (b) Continuous and discontinous solutions $g(t)$ of (3.21) when $M=1.25\times 10^{-4}$, $a=0.1$ and $i=30,50,100$.

Figure 6

Figure 5. Comparison of approximate solutions based on (3.33) (dashed lines) vs direct numerical simulation (solid) for $M=1.25\times 10^{-4}$, at the edge of the resonant band $\varDelta =\varDelta ^{+}$ and for (a) $a=0$, (b) $a=0.1$, (c) $a=0.2$.

Figure 7

Figure 6. Deformation of resonant profiles $u(0.5,t)$ for $t\in [0,2]$ for increasing geometric parameter $a$, (a) $M=1.5625 \times 10^{-5}$, ($\nu =1 \times 10^{-5}$) and (b) $M=1.25 \times 10^{-4}$ ($\nu =2 \times 10^{-5}$).

Figure 8

Figure 7. Variation of qualitatively distinct resonant profiles in terms of forcing amplitude ($M$) and geometric variation ($a$).

Figure 9

Figure 8. Scaling of response $O(M^{\kappa })$ and relative modal amplitudes as $a$ varies for $M=1.25\times 10^{-4}$.

Figure 10

Table 3. Amplitudes for first harmonic $A_1$ and higher harmonics for $a=0,2$ and when $M=1.25\times 10^{-4}$, quantitatively showing how the higher modes attenuate in these distinct limiting cases.

Figure 11

Figure 9. Regions in $a$ vs $\varDelta$ where aNLGA (3.13), dominant first harmonic and exact linear approximations provide agreement to within $5\,\%$ for $M=1.25 \times 10^{-4}$.