diff --git a/img/quadrupole.pdf b/img/quadrupole.pdf new file mode 100644 index 0000000..03768c1 Binary files /dev/null and b/img/quadrupole.pdf differ diff --git a/lrftubes.lyx b/lrftubes.lyx index 363853f..d3db5f1 100644 --- a/lrftubes.lyx +++ b/lrftubes.lyx @@ -2444,7 +2444,7 @@ literal "false" \begin_layout Standard \begin_inset Formula \begin{equation} -k_{mix}=\sum_{α=0}^{N-1}\frac{x_{α}k_{α}}{\sum_{β=0}^{N-1}\Phi_{αβ}x_{β}}\label{eq:kappamix} +k_{\mathrm{mix}}=\sum_{α=0}^{N-1}\frac{x_{α}k_{α}}{\sum_{β=0}^{N-1}\Phi_{αβ}x_{β}}\label{eq:kappamix} \end{equation} \end_inset @@ -8699,7 +8699,7 @@ where \begin_layout Standard After some algebraic manipulations we find: \begin_inset Note Note -status collapsed +status open \begin_layout Plain Layout \begin_inset Formula $z_{m}u=\left(p_{l}-p_{r}\right)S+B\ell I$ @@ -8721,6 +8721,30 @@ where \end_inset +\end_layout + +\begin_layout Plain Layout +Units of +\begin_inset Formula $\left[B\ell\right]=\frac{N}{A}=\frac{\mathrm{kg}\mathrm{m}s}{\mathrm{s}^{2}C}$ +\end_inset + +, knowing that +\begin_inset Formula $V=\frac{J}{C}$ +\end_inset + +, we can write this as: +\begin_inset Formula $\frac{\mathrm{kg}\mathrm{m}s}{\mathrm{s}^{2}C}=\frac{V\mathrm{kg}\mathrm{m}s}{\mathrm{s}^{2}J}=\frac{Vs}{\mathrm{m}}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +And +\begin_inset Formula $\left[\frac{B\ell^{2}}{Z_{\mathrm{el}}}\right]=\left[\frac{Vs}{\mathrm{m}}\frac{N}{A}\frac{A}{V}\right]=\left[\frac{s}{\mathrm{m}}\frac{N}{A}\right]$ +\end_inset + + \end_layout \begin_layout Plain Layout @@ -8728,7 +8752,14 @@ Results in: \end_layout \begin_layout Plain Layout -\begin_inset Formula $\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)u=\left(p_{l}-p_{r}\right)S+\frac{B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}}$ +\begin_inset Formula $z_{m}u=\left(p_{l}-p_{r}\right)S+B\ell\frac{V_{\mathrm{in}}-V_{\mathrm{bemf}}}{Z_{\mathrm{el}}}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $\frac{B\ell^{2}u}{Z_{\mathrm{el}}}+z_{m}u=\left(p_{l}-p_{r}\right)S+\frac{B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}}$ \end_inset @@ -9939,7 +9970,7 @@ For square holes: \end_layout \begin_layout Standard -where +where \begin_inset Formula \begin{equation} \xi^{2}=\frac{\pi D^{2}}{4P^{2}} @@ -10979,7 +11010,402 @@ Z_{\mathrm{tr}}=\frac{p_{\mathrm{DRP}}}{U_{\mathrm{coupler,entrance}}} \end_layout \begin_layout Chapter -Kampinga's SLNS model in our notation +Standard acoustic solutions +\end_layout + +\begin_layout Section +Spherically symmetric breathing ball (monopole) +\end_layout + +\begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Plain Layout +From Rienstra and Hirschberg: +\begin_inset Formula +\begin{equation} +\hat{p}(r)=-z_{0}c_{0}k\frac{\hat{v}}{i\omega}\frac{k^{2}a_{0}^{2}}{1+ika_{0}}\frac{\exp\left(-i\left(kr-a_{0}\right)\right)}{kr} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +To our definitions and a bit of rewriting: +\begin_inset Formula +\[ +\hat{p}(r)=\frac{i\rho_{0}c_{0}ka^{2}}{1+ika}\frac{\exp\left(-i\left(kr-a\right)\right)}{r}\hat{v} +\] + +\end_inset + + +\end_layout + +\end_inset + +Radiation from a compact monopole with radius +\begin_inset Formula $a$ +\end_inset + + and +\begin_inset Quotes eld +\end_inset + +breathing +\begin_inset Quotes erd +\end_inset + + velocity amplitude +\begin_inset Formula $\hat{v}$ +\end_inset + +: +\begin_inset Formula +\begin{equation} +\hat{p}(r)=\frac{iz_{0}ka^{2}}{1+ika}\frac{\exp\left(-i\left(kr-a\right)\right)}{r}\hat{v}. +\end{equation} + +\end_inset + +Small source limit ( +\begin_inset Formula $ka\ll1$ +\end_inset + +): +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +\hat{p}(r)\approx iz_{0}\frac{ka^{2}}{r}\left[\exp\left(-i\left(kr-a\right)\right)\right]\hat{v}. +\end{equation} + +\end_inset + +In terms of the transfer impedance ( +\begin_inset Formula $\hat{U}=4\pi a^{2}\hat{v}$ +\end_inset + +): +\begin_inset Note Note +status collapsed + +\begin_layout Plain Layout +\begin_inset Formula +\[ +\hat{p}(r)=\frac{i\rho_{0}c_{0}ka^{2}}{1+ika}\frac{\exp\left(-i\left(kr-a\right)\right)}{r}\frac{\hat{U}}{4\pi a^{2}} +\] + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula +\[ +\hat{p}(r)=\frac{iz_{0}k}{4\pi\left(1+ika\right)r}\left[\exp\left(-i\left(kr-a\right)\right)\right]\hat{U} +\] + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +which is also: +\begin_inset Formula +\[ +\hat{p}(r)\approx\frac{iz_{0}}{2\lambda r}\left[\exp\left(-i\left(kr-a\right)\right)\right]\hat{U} +\] + +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula +\begin{equation} +\hat{p}(r)=\underbrace{\frac{iz_{0}k}{4\pi\left(1+ika\right)r}\left[\exp\left(-i\left(kr-a\right)\right)\right]}_{Z_{\mathrm{tr}}(r)}\hat{U}, +\end{equation} + +\end_inset + +For easy estimations, in the small source ( +\begin_inset Formula $ka\ll1$ +\end_inset + +) and far field limit ( +\begin_inset Formula $kr\gg1$ +\end_inset + +): +\begin_inset Formula +\begin{equation} +\hat{p}(r)\approx\frac{iz_{0}}{2\lambda r}\hat{U}\left[\exp\left(-ikr\right)\right]. +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Section +Dipoles +\end_layout + +\begin_layout Subsection +Translating sphere, exact solution +\end_layout + +\begin_layout Standard +\begin_inset Formula $\theta$ +\end_inset + +: pole angle. + Then the velocity follows: +\begin_inset Formula +\begin{equation} +\hat{v}(\theta)=\hat{v}_{0}\cos\left(\theta\right). +\end{equation} + +\end_inset + +After performing analysis, we find for the pressure: +\begin_inset Formula +\begin{equation} +\hat{p}(r,\theta)=\frac{-i\omega\rho_{0}\hat{v}_{0}a^{3}\cos\theta}{2\left(1+ika\right)-\left(ka\right)^{2}}\frac{\partial}{\partial r}\left\{ \frac{\exp\left(-ik\left(r-a\right)\right)}{r}\right\} . +\end{equation} + +\end_inset + +In the small source limit ( +\begin_inset Formula $ka\ll1$ +\end_inset + +): +\begin_inset Note Note +status open + +\begin_layout Plain Layout +\begin_inset Formula $\hat{p}(r,\theta)=-\hat{v}_{0}\frac{z_{0}k^{2}a^{3}\cos\theta}{2r}\left(1+\frac{1}{ikr}\right)e^{-ik\left(r-a\right)}$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula +\begin{equation} +\hat{p}(r,\theta)\approx-\frac{z_{0}k^{2}a^{3}\cos\theta}{2r}\left(\frac{1+ikr}{ikr}\right)\left[\exp\left(-ik\left(r-a\right)\right)\right]\hat{v}_{0}.\label{eq:dipole_transl_sphere} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +Small source limit, far field ( +\begin_inset Formula $ka\ll1$ +\end_inset + +, +\begin_inset Formula $kr\gg1$ +\end_inset + +): +\begin_inset Formula +\begin{equation} +\hat{p}(r,\theta)\approx-\hat{v}_{0}\frac{z_{0}k^{2}a^{3}\cos\theta}{2r}e^{-ikr}. +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Subsection +Perfect dipole from two compact monopoles +\end_layout + +\begin_layout Standard +Distance between sources: +\begin_inset Formula $d\ll\lambda$ +\end_inset + +. + Volume flow from a single pole: +\begin_inset Formula $\hat{U}$ +\end_inset + +. + From the other source +\begin_inset Formula $-\hat{U}$ +\end_inset + +. + The angle +\begin_inset Formula $\theta$ +\end_inset + + is 0 at positions where the positive source is the closest to the listening + point. + Distance between the sources is +\begin_inset Formula $d$ +\end_inset + +. + Then the sound pressure is +\begin_inset Formula +\begin{equation} +\hat{p}(r,\theta)\approx-k^{2}z_{0}\frac{\exp\left(-ikr\right)\cos\theta}{4\pi r}\left(\frac{1+ikr}{ikr}\right)\hat{U}d +\end{equation} + +\end_inset + +Comparing this equation to Eq. +\begin_inset space ~ +\end_inset + + +\begin_inset CommandInset ref +LatexCommand ref +reference "eq:dipole_transl_sphere" + +\end_inset + +, we find that for the same acoustic pressure of a perfect dipole vs. + a translating sphere: +\begin_inset Formula +\begin{equation} +2\pi a^{2}\hat{v}_{0}a=\hat{U}d. +\end{equation} + +\end_inset + +So if we set the volume flow of a translating sphere equal to the frontal + area of +\begin_inset Formula $\pi a^{2}$ +\end_inset + +, the effective dipole distance is +\begin_inset Formula $2a$ +\end_inset + +, which corresponds to the diameter of the sphere! +\begin_inset Note Note +status open + +\begin_layout Plain Layout +\begin_inset Formula $\frac{a^{3}}{2}\hat{v}_{0}=\frac{1}{4\pi}\hat{U}d$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +Hence: if we set +\begin_inset Formula $\hat{U}_{\mathrm{tr\,sphere}}=\pi a^{2}\hat{v}$ +\end_inset + +: the effective distance +\begin_inset Formula $d$ +\end_inset + + of a translating sphere is: +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $2\pi a^{2}\hat{v}_{0}a=\hat{U}d$ +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Section +Compact quadrupole +\end_layout + +\begin_layout Standard +\begin_inset Float figure +wide false +sideways false +status open + +\begin_layout Plain Layout +\noindent +\align center +\begin_inset Graphics + filename img/quadrupole.pdf + width 60text% + +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Caption Standard + +\begin_layout Plain Layout +Schematic of the quadrupole. +\end_layout + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Standard +A compact square-shaped quadrupole with distances of +\begin_inset Formula $d$ +\end_inset + + between each pole, distance +\begin_inset Formula $kd\ll1$ +\end_inset + +. + Volume flow from a single pole: +\begin_inset Formula $\hat{U}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +\hat{p}(x,y)=-ik^{3}z_{0}\hat{U}d^{2}\frac{xy\exp\left(-ikr\right)}{4\pi r^{3}}\left(1+\frac{3}{ikr}-\frac{3}{\left(kr\right)^{2}}\right). +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Chapter +3D (FEM) Models \end_layout \begin_layout Standard @@ -11224,7 +11650,7 @@ From \end_layout \begin_layout Section -Model +SLNS model \end_layout \begin_layout Standard @@ -11256,7 +11682,98 @@ The velocity is: \end_layout \begin_layout Standard -With boundary conditions: +Comsol writes for the effective density: +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +\left(-\frac{1}{\rho_{c}}\nabla p\right)=i\omega\boldsymbol{u}, +\end{equation} + +\end_inset + +such that +\begin_inset Note Note +status open + +\begin_layout Plain Layout +\begin_inset Formula $\frac{1}{\rho_{c}}=\frac{1-h_{\nu}}{\rho_{0}},$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula +\begin{equation} +\rho_{c}=\frac{\rho_{0}}{1-h_{\nu}}, +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +And: +\begin_inset Formula +\begin{equation} +\nabla\cdot\left(-\frac{1}{\rho_{c}}\nabla p_{t}\right)-\frac{\omega^{2}}{c^{2}\rho_{c}}p=Q_{m}, +\end{equation} + +\end_inset + +Filling in: +\begin_inset Note Note +status collapsed + +\begin_layout Plain Layout +\begin_inset Formula $\nabla\cdot\left(-\frac{1}{\rho_{c}}\nabla p_{t}\right)-\frac{\omega^{2}}{c^{2}\rho_{c}}p=Q_{m}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $\nabla\cdot\left(-\frac{\left(1-h_{\nu}\right)}{\rho_{m}}\nabla p\right)-\frac{k^{2}}{\rho_{m}}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p=0$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +Makes: +\begin_inset Formula $c^{2}\rho_{c}=\frac{c_{m}^{2}\rho_{m}}{1+\left(\gamma-1\right)h_{\kappa}}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $c^{2}=\frac{c_{m}^{2}\left(1-h_{\nu}\right)}{1+\left(\gamma-1\right)h_{\kappa}}$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula +\begin{equation} +c^{2}=\frac{c_{m}^{2}\left(1-h_{\nu}\right)}{1+\left(\gamma-1\right)h_{\kappa}} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +With boundary conditions at isothermal no-slip wall: \end_layout \begin_layout Standard @@ -11269,6 +11786,21 @@ h_{\kappa} & =1\qquad\mathrm{at\,the\,wall} \end_inset +\end_layout + +\begin_layout Standard +Symmetry / inlet outlet: +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +h_{\nu}=h_{\kappa}=0 +\end{equation} + +\end_inset + + \end_layout \begin_layout Standard @@ -11285,7 +11817,7 @@ For pressure / velocity b.c.'s \begin_layout Standard \begin_inset Note Note -status open +status collapsed \begin_layout Plain Layout Combine with pressure acoustics: @@ -11672,6 +12204,10 @@ We hebben altijd op een rand: \end_layout +\begin_layout Standard +We can write this as a weak contribution: +\end_layout + \begin_layout Standard Weak contribution in pressure acoustics interface: \end_layout @@ -11684,13 +12220,7 @@ acpr.delta/acpr.rho_c \end_layout \begin_layout Standard -\begin_inset Formula -\begin{equation} -\end{equation} - -\end_inset - - +Or we could write this with a custom density and speed of sound <— TODO! \end_layout \begin_layout Standard