From aff2c1256915d4f6a19fd5b96459e3bbec69b1d1 Mon Sep 17 00:00:00 2001
From: "J.A. de Jong - Redu-Sone B.V., ASCEE V.O.F" <j.a.dejong@ascee.nl>
Date: Mon, 6 Dec 2021 11:18:33 +0100
Subject: [PATCH] Added quadrupole, some minor changes

---
 img/quadrupole.pdf | Bin 0 -> 36247 bytes
 lrftubes.lyx       | 560 +++++++++++++++++++++++++++++++++++++++++++--
 2 files changed, 545 insertions(+), 15 deletions(-)
 create mode 100644 img/quadrupole.pdf

diff --git a/img/quadrupole.pdf b/img/quadrupole.pdf
new file mode 100644
index 0000000000000000000000000000000000000000..03768c134d70ee317e076b3bfcbcc8ce0992c2cc
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literal 0
HcmV?d00001

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