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Added quadrupole, some minor changes

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Anne de Jong 2021-12-06 11:18:33 +01:00
parent a25e2e2685
commit aff2c12569
2 changed files with 545 additions and 15 deletions
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@ -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