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LRFTubes documentation - v1.1
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Dr.ir.
 J.A.
 de Jong
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Ir.
 C.
 Jansen
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Overview of 
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lrftubes
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Introduction
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Welcome to the documentation of 
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lrftubes
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.
 
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lrftubes
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 is a numerical code to solve one-dimensional acoustic duct systems using
 the transfer matrix method.
 Segments can be connected to generate simple one-dimensional acoustic systems
 to model acoustic propagation problems in ducts in the frequency domain.
 Viscothermal dissipation mechanisms are taken into account such that the
 damping effects can be modeled accurately, below the cut-on frequency of
 the duct.
 For more information regarding the models and the theory behind the models,
 the reader is referred to the work of 
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, 
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 and 
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.
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This documentation serves as a reference for the implemented models.
 For examples on how to use the code, please take a look at the example
 models as worked out in the IPython Notebooks.
 For installation instructions, please refer the the 
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 in the main repository.
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This document is very brief on the theory and it is assumed that the reader
 has some knowledge on the basics of acoustics in general and viscothermal
 acoustics as well.
 If you are not falling in this category, I would please refer you first
 to the book of Swift 
\begin_inset CommandInset citation
LatexCommand cite
key "swift_thermoacoustics:_2003"
literal "true"

\end_inset

.
 A more detailed introduction to the notation used in this documentation
 can be found in the PhD thesis of de Jong 
\begin_inset CommandInset citation
LatexCommand cite
key "de_jong_numerical_2015"
literal "true"

\end_inset

.
\end_layout

\begin_layout Standard
Besides that, if you find the work interesting, but you are not sure how
 to apply it, please contact ASCEE for more information.
\end_layout

\begin_layout Section
License and disclaimer
\end_layout

\begin_layout Standard
Redistribution and use in source and binary forms are permitted provided
 that the above copyright notice and this paragraph are duplicated in all
 such forms and that any documentation, advertising materials, and other
 materials related to such distribution and use acknowledge that the software
 was developed by the ASCEE.
 The name of the ASCEE may not be used to endorse or promote products derived
 from this software without specific prior written permission.
\begin_inset Newline newline
\end_inset


\end_layout

\begin_layout Standard
THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR IMPLIED WARRANTIE
S, INCLUDING, WITHOUT LIMITATION, THE IMPLIED WARRANTIES OF MERCHANTABILITY
 AND FITNESS FOR A PARTICULAR PURPOSE.
\end_layout

\begin_layout Section
Features
\end_layout

\begin_layout Standard
Currently the 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\backslash

\end_layout

\end_inset

 code provides acoustic models for the following physical entities:
\end_layout

\begin_layout Itemize
Prismatic ducts with circular cross section,
\end_layout

\begin_layout Itemize
Prismatic ducts with triangular cross section,
\end_layout

\begin_layout Itemize
Prismatic ducts with parallel plate cross section,
\end_layout

\begin_layout Itemize
Prismatic ducts with square cross section,
\end_layout

\begin_layout Itemize
Acoustic compliance volumes
\end_layout

\begin_layout Itemize
Discontinuity correction
\end_layout

\begin_layout Itemize
End correction for a baffled piston
\end_layout

\begin_layout Itemize
Lumped series impedance
\end_layout

\begin_layout Standard
These segments can be connected to form one-dimensional acoustic systems
 to model wave propagation below the cut-on frequency of higher order modes.
 For a circular cross section, the cut-on frequency is 
\begin_inset CommandInset citation
LatexCommand cite
key "van_der_eerden_noise_2000"
literal "true"

\end_inset

:
\begin_inset Formula 
\begin{equation}
f_{c}\approx\frac{c_{0}}{3.4r},
\end{equation}

\end_inset

where 
\begin_inset Formula $r$
\end_inset

 is the tube radius and 
\begin_inset Formula $c_{o}$
\end_inset

 is the speed of sound.
 Above the cut-on frequency, besides evanescent waves, there are also propagatin
g waves with a non-constant pressure distribution along the cross section
 of the duct.
\end_layout

\begin_layout Subsection
Limitations and future features
\end_layout

\begin_layout Standard
The current version of has some limitations that will be resolved in a future
 release.
 These are:
\end_layout

\begin_layout Subsubsection
Ducts with (turbulent) flow
\end_layout

\begin_layout Standard
For thermoacoustic and HVAC (Heating, ventilation and Air Conditioning)
 duct modeling it is imperative that mean flows can be taken into account.
 An acoustic wave superimposed on a mean flow results in asymmetric wave
 propagation.
 More specifically, the phase velocity is higher in the direction of the
 mean flow, and slower in the opposite direction.
 In a future release, we will provide models for ducts including a mean
 flow.
 
\end_layout

\begin_layout Subsubsection
Porous acoustic absorbers
\end_layout

\begin_layout Standard
To model absorption of sound, a one-dimensional porous material model should
 be implemented.
 This work has been postponed to a later stage.
\end_layout

\begin_layout Standard
Prismatic and spherical ducts filled with porous material are defined in
 dbmduct.py.
 These use the Delaney-Bazley-Miki model.
\end_layout

\begin_layout Section
Overview of this documentation
\end_layout

\begin_layout Standard
The next chapter of this documentation will describe the basic framework
 of the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 code: the transfer matrix method.
 After that, in Chapter 
\begin_inset CommandInset ref
LatexCommand ref
reference "chap:Provided-acoustic-models"

\end_inset

, an overview of the provided acoustic models is given, with which acoustic
 networks can be built.
 For each of the segments, the resulting transfer matrix model is derived.
\end_layout

\begin_layout Chapter
Material properties
\end_layout

\begin_layout Section
Air
\end_layout

\begin_layout Standard
Nonlinearity parameter:
\end_layout

\begin_layout Section
Exhaust gas
\end_layout

\begin_layout Subsection
Composition
\end_layout

\begin_layout Standard
Definitions:
\end_layout

\begin_layout Itemize
\begin_inset Formula $\omega_{i}$
\end_inset

 mass fraction of species 
\begin_inset Formula $i$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $x_{i}$
\end_inset

 molar / volume fraction of species 
\begin_inset Formula $i$
\end_inset

 (assuming ideal gas behavior)
\end_layout

\begin_layout Itemize
\begin_inset Formula $\overline{M}$
\end_inset

 average molar mass of (exhaust gas) mixture
\end_layout

\begin_layout Itemize
\begin_inset Formula $M_{i}$
\end_inset

 molar mass of species 
\begin_inset Formula $i$
\end_inset


\end_layout

\begin_layout Standard
The following equations hold in a mixture:
\begin_inset Formula 
\begin{align}
\sum_{i}\omega_{i} & =1\\
\sum_{i}x_{i} & =1\\
\overline{M} & =\sum\nolimits _{i}x_{i}M_{i}\label{eq:molar_mass_comp}
\end{align}

\end_inset

We can convert mass fractions to mole fractions with the following rule:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
The total mass is (
\begin_inset Formula $N$
\end_inset

) is the total number of moles
\begin_inset Formula 
\[
m=x_{i}M_{i}N
\]

\end_inset


\end_layout

\begin_layout Plain Layout
The total number of moles is:
\begin_inset Formula 
\[
N=\frac{m}{\overline{M}}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
The average molar mass is:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\overline{M}=\frac{m}{N}=\sum_{i}x_{i}M_{i}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
The mass fraction to mole fraction is:
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\omega_{i}=x_{i}\frac{M_{i}}{\overline{M}}\qquad\Longleftrightarrow\qquad x_{i}=\omega_{i}\frac{\overline{M}}{M_{i}}\label{eq:massfr_to_molarfr_viceversa}
\end{equation}

\end_inset

Henceforth, what is often used, is to compute the average molar mass given
 only the mass fractions:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\omega_{i}m=N_{i}M_{i}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\omega_{i}}{M_{i}}=\frac{N_{i}}{m}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\sum_{i}\frac{\omega_{i}}{M_{i}}=\frac{N}{m}=\frac{1}{\overline{M}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\overline{M}=\frac{1}{\sum\nolimits _{i}\frac{\omega_{i}}{M_{i}}}\label{eq:molar_mass_vs_massfr}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Mixing of mixtures
\end_layout

\begin_layout Standard
Suppose we mix two mixtures of substances, mixture 1, and mixture 2.
 We want to know the final concentrations / mass fraction in the mixed mixture.
 Mix 1 comprises mass fractions 
\begin_inset Formula $\omega_{1,i}$
\end_inset

, and mix 2 comprises mass fractions 
\begin_inset Formula $\omega_{2,j}$
\end_inset

.
 We assume that 
\begin_inset Formula $i$
\end_inset

 and 
\begin_inset Formula $j$
\end_inset

 can interfere.
 For example, mixing air with Dutch natural gas, both contain nitrogen.
 The first step is to determine the mass flow of the two, called 
\begin_inset Formula $m_{1}$
\end_inset

 and 
\begin_inset Formula $m_{2}$
\end_inset

.
 Then, assuming mass conservation under chemically inert conditions:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
m_{1}\omega_{1,i}+m_{2}\omega_{2,i}=m\omega_{i}.
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection*
Mixing air with natural gas
\end_layout

\begin_layout Standard
The air factor 
\begin_inset Formula $\lambda$
\end_inset

 (not to be confused with wavelength in an acoustic context), is defined
 as the ratio of air to the stoichiometric ratio.
 The stoichiometric ratio can be determined by calculating the required
 moles of oxygen such that all carbon atoms can become CO
\begin_inset Formula $_{2}$
\end_inset

 , and 
\emph on
half of
\emph default
 all hydrogen atoms can become H
\begin_inset Formula $_{2}$
\end_inset

.
\end_layout

\begin_layout Subsection
Ideal gas mixtures
\end_layout

\begin_layout Standard
For an ideal gas, the components of a gas mixture can be represented by
 their 
\begin_inset Quotes eld
\end_inset

partial pressure
\begin_inset Quotes erd
\end_inset

, which is the total pressure times the volume fraction of the component
 in the mixture.
 For an ideal gas, the volume fraction equals to mole fraction.
 Hence:
\begin_inset Formula 
\begin{equation}
\frac{V_{i}}{V}\overset{\mathrm{ideal\,gas}}{=}x_{i}=\frac{p_{i}}{R_{u}T}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The mass fraction can be computed from the mole fraction.
\end_layout

\begin_layout Subsection
Transport properties
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="5">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top" width="0pt">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Substance
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $M$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $T_{c}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $G$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $C_{r}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Carbon dioxide
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{44.01e-3}{kg
\backslash
per
\backslash
mole}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{304}{
\backslash
K}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
44.6
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.766
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Oxygen
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{32.00e-3}{kg
\backslash
per
\backslash
mole}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{154}{
\backslash
K}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
32.8
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.712
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Nitrogen
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{28.02e-3}{kg
\backslash
per
\backslash
mole}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{126}{
\backslash
K}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
24.6
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0.881
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Water vapor
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{18.02e-3}{kg
\backslash
per
\backslash
mole}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
SI{647}{
\backslash
K}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
52.2
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
1.018
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Critical values and constants of common diatomic gases
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "tab:crit_values_diatom_gas"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Dynamic viscosity of pure gases
\end_layout

\begin_layout Standard
Here we assume the dynamic viscosity of a pure substance can be modeled
 using Sutherland's equation:
\begin_inset Formula 
\begin{equation}
\mu=\mu_{c}\left(\frac{T_{0}+C}{T+C}\right)\left(\frac{T}{T_{0}}\right)^{3/2},
\end{equation}

\end_inset

where the subscript 
\begin_inset Formula $c$
\end_inset

 denotes the value at its 
\begin_inset Quotes eld
\end_inset

critical point
\begin_inset Quotes erd
\end_inset

.
 In convenient form we solve:
\begin_inset Formula 
\begin{equation}
\mu=\mu_{c}\mu_{r},
\end{equation}

\end_inset

where 
\begin_inset Formula $\mu_{c}$
\end_inset

 is the critical viscosity and 
\begin_inset Formula $\mu_{r}$
\end_inset

 is the 
\begin_inset Quotes eld
\end_inset

reduced viscosity defined as 
\begin_inset Formula $\mu/\mu_{c}$
\end_inset

.
 For 
\begin_inset Formula $\mu_{c}$
\end_inset

 we have the reduced form of Sutherland's equation:
\begin_inset Formula 
\begin{equation}
\mu_{c}=\frac{1+C_{r}}{T_{r}+C_{r}}T_{r}^{3/2}
\end{equation}

\end_inset

 The value for 
\begin_inset Formula $\mu_{c}$
\end_inset

 can be calculated as:
\begin_inset Formula 
\begin{equation}
\mu_{c}=\num{3.5e-6}G
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Values for 
\begin_inset Formula $T_{r}$
\end_inset

, 
\begin_inset Formula $C_{r}$
\end_inset

 and 
\begin_inset Formula $G$
\end_inset

 are listed in Table 
\begin_inset CommandInset ref
LatexCommand ref
reference "tab:crit_values_diatom_gas"

\end_inset

 
\begin_inset CommandInset citation
LatexCommand cite
key "licht_variation_1944"
literal "false"

\end_inset

.
\end_layout

\begin_layout Subsubsection
Dynamic viscosity of a gas mixture
\end_layout

\begin_layout Standard
The dynamic viscosity of a gas mixture can be derived from the dynamic viscositi
es of pure gases as 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 27"
key "bird_transport_2007"
literal "false"

\end_inset

:
\begin_inset Formula 
\begin{equation}
\mu_{\mathrm{mix}}=\sum_{α=0}^{N-1}\frac{x_{α}\mu_{α}}{\sum_{β=0}^{N-1}\Phi_{αβ}x_{β}},\label{eq:mumix}
\end{equation}

\end_inset

where 
\begin_inset Formula $\mu_{α}$
\end_inset

 is the dynamic viscosity of pure chemical species 
\begin_inset Formula $α$
\end_inset

 and 
\begin_inset Formula $x_{α}$
\end_inset

denotes its mole fraction in the mixture.
 
\begin_inset Formula $\Phi_{αβ}$
\end_inset

 is defined as:
\begin_inset Formula 
\begin{equation}
\Phi_{αβ}=\frac{1}{\sqrt{8}}\left(1+\frac{M_{α}}{M_{β}}\right)^{-1/2}\left[1+\left(\frac{\mu_{α}}{\mu_{β}}\right)^{1/2}\left(\frac{M_{β}}{M_{α}}\right)^{1/4}\right]^{2},\label{eq:Phi_mn}
\end{equation}

\end_inset

where 
\begin_inset Formula $M_{α}$
\end_inset

 is the molar mass of species 
\begin_inset Formula $α$
\end_inset

.
 The denominator of Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:mumix"

\end_inset

 can efficiently be solved by noting that 
\begin_inset Formula $d_{α}=\sum_{β=0}^{N-1}\Phi_{αβ}x_{β}$
\end_inset

 is a matrix-vector product, which can be written as 
\begin_inset Formula $\boldsymbol{d}=\boldsymbol{\Phi}\cdot\boldsymbol{x}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Thermal conductivity of a gas mixture
\end_layout

\begin_layout Standard
The thermal conductivity of a gas mixture can be derived from the thermal
 conductivities of pure gases as 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 276"
key "bird_transport_2007"
literal "false"

\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
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


\end_layout

\begin_layout Standard
where 
\begin_inset Formula $k_{α}$
\end_inset

 is the thermal conductivity of pure chemical species 
\begin_inset Formula $α$
\end_inset

 and 
\begin_inset Formula $x_{α}$
\end_inset

denotes its mole fraction in the mixture and 
\begin_inset Formula $\Phi_{αβ}$
\end_inset

is identical to that appearing in the viscosity equation, see 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Phi_mn"
plural "false"
caps "false"
noprefix "false"

\end_inset

.
\end_layout

\begin_layout Subsection
Combustion
\end_layout

\begin_layout Standard
To compute the gas constant, first the mixture components of the exhaust
 gas need to be computed.
 We assume that the oxidizer is air with 79% vol of nitrogen (molecules)
 and 21% oxygen molecules.
 The tiny part of argon and other components is ignored.
 Then, the gross formula for combustion is:
\begin_inset Formula 
\begin{equation}
\underbrace{x_{f,C}C+x_{f,O}O+x_{H,f}H+x_{f,N}N}_{\mathrm{fuel}}+\underbrace{y_{\mathrm{ox}}\left(0.79N_{2}+0.21O_{2}\right)}_{\mathrm{oxidizer}}\rightarrow\underbrace{y_{g,\mathrm{water}}H_{2}O+y_{g,CO_{2}}CO_{2}+y_{g,N_{2}}N_{2}}_{\mathrm{exhaust\,gas}}.\label{eq:combustion}
\end{equation}

\end_inset

Above reaction formula can be read as: 
\begin_inset Quotes eld
\end_inset

take 
\begin_inset Formula $x_{f,C}$
\end_inset

 moles of carbon in the fuel, add 
\begin_inset Formula $y_{\mathrm{ox}}$
\end_inset

 moles of air, and it should result in 
\begin_inset Formula $y_{g,CO_{2}}$
\end_inset

 moles of 
\begin_inset Formula $CO_{2}$
\end_inset


\begin_inset Quotes erd
\end_inset

 And so on for the other elements.
 The mole fractions in the fuel composition can be derived from its mass
 fractions, upon utilizing Eqs.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:massfr_to_molarfr_viceversa"

\end_inset

 and 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:molar_mass_vs_massfr"

\end_inset

.
 From Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:combustion"

\end_inset

, the following system of equations can be created:
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
x_{f,C}\\
x_{f,O}\\
x_{f,H}\\
x_{f,N}
\end{array}\right\} +\left[\begin{array}{cccc}
0 & 0 & -1 & 0\\
2\times0.21 & -1 & -2 & 0\\
0 & -2 & 0 & 0\\
2\times0.79 & 0 & 0 & -2
\end{array}\right]\left\{ \begin{array}{c}
y_{\mathrm{ox}}\\
y_{g,\mathrm{water}}\\
y_{g,CO_{2}}\\
y_{g,N_{2}}
\end{array}\right\} =\left\{ \begin{array}{c}
0\\
0\\
0\\
0
\end{array}\right\} 
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Solving this results in:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $x_{f,O}+2\times0.21y_{\mathrm{ox}}-y_{g,\mathrm{water}}-2y_{g,CO_{2}}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $2\times0.21y_{\mathrm{ox}}=\frac{1}{2}x_{f,H}+2x_{f,C}+x_{f,O}$
\end_inset


\end_layout

\begin_layout Plain Layout
–
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $x_{f,N}+2\times0.79y_{\mathrm{ox}}-2y_{g,N_{2}}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $y_{g,N_{2}}=0.79y_{\mathrm{ox}}+\frac{1}{2}x_{f,N}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
y_{g,CO_{2}} & =x_{f,C}\\
y_{g,\mathrm{water}} & =\frac{1}{2}x_{f,H}\\
y_{\mathrm{ox}}= & \frac{\frac{1}{2}x_{f,H}+2x_{f,C}-x_{f,O}}{2\times0.21}\\
y_{g,N_{2}}= & 0.79y_{\mathrm{ox}}+\frac{1}{2}x_{f,N}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
Note that the mole fractions are 
\emph on
unnormalized
\emph default
 (that is why we use symbol 
\begin_inset Formula $y$
\end_inset

, not 
\begin_inset Formula $x$
\end_inset

): they denote the number of moles required to burn 1 mole of fuel.
 To compute the mole fractions in the exhaust gas, 
\begin_inset Formula 
\begin{equation}
x_{g,\mathrm{water}}=\frac{y_{1}}{y_{1}+y_{2}+y_{3}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Table
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "tab:fuel_components"

\end_inset

 gives an overview of the composition of typical combustion fuels.
 Once the molar fractions of the exhaust gas are known, the average molar
 mass can be computed using Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:molar_mass_comp"

\end_inset

.
 Then, the specific gas constant can be computed according to:
\begin_inset Formula 
\begin{equation}
R_{s}=\frac{R_{u}}{\overline{M}},
\end{equation}

\end_inset

where 
\begin_inset Formula $R_{u}$
\end_inset

 is the universal gas constant.
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="3">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top" width="0pt">
<column alignment="right" valignment="top" width="0pt">
<column alignment="right" valignment="top" width="0pt">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Mass fraction
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Wood
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
https://www.engineeringtoolbox.com/co2-emission-fuels-d_1085.html
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Dutch Natural gas
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Carbon
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
50 %
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Oxygen
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
42 %
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0 %
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Hydrogen
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
6 %
\end_layout

\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Nitrogen
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
0 %
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Mixture mass composition of fuels
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "tab:fuel_components"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Specific heat ratio
\end_layout

\begin_layout Standard
The specific heat is build-up according to mass percentages of the flue
 gas.
 Carbon dioxide has a 
\begin_inset Formula $c_{p}$
\end_inset

 of 840 J/kg/K, water vapor of 1930:
\begin_inset Formula 
\begin{equation}
\overline{c}_{p}=\sum\nolimits _{i}\omega_{i}c_{p,i}.
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Sound absorbing solid materials
\end_layout

\begin_layout Standard
High porosity soft materials can be modeled adequately with the Delaney-Bazley-M
iki model.
 The model has a single input, namely the static flow resistivity.
 Table
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="2">
<features booktabs="true" tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Name
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Basotect TG
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
A.k.a.Flamex Basic (akoestiekwinkel.nl)
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Description
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Melamine resin foam (fire retardant)
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Density [
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{
\backslash
kg
\backslash
per
\backslash
cubic
\backslash
m}
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
https://www.forman.co.nz/media/emizen_banner/b/a/basf_basotect_datasheet.pdf
\end_layout

\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Flow resistivity [
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{
\backslash
pascal
\backslash
s
\backslash
per
\backslash
meter}
\end_layout

\end_inset

]
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
num{8.5e3}
\end_layout

\end_inset

, source: 
\begin_inset CommandInset citation
LatexCommand cite
key "kino_investigation_2009"
literal "false"

\end_inset

, Table 2 average value.
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Resistivity values are given for room temperature
\end_layout

\end_inset


\end_layout

\end_inset

Conversion
\end_layout

\begin_layout Chapter
The transfer matrix method
\end_layout

\begin_layout Section
Introduction
\end_layout

\begin_layout Standard
Each part of an acoustic system in 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 is modeled using a so-called transfer matrix.
 A transfer matrix maps the state quantities on one side of the segment
 (node) to the other side of the segment (node).
\end_layout

\begin_layout Standard
For one-dimensional wave propagation, analytical solutions for the velocity,
 temperature and density field in the transverse direction can be found.
 The state variables in frequency domain satisfy a system of first order
 ordinary differential equations.
 Once the solution is known on one end of a segment, the solution on the
 other end can be deduced.
 The transfer matrix couples the state variables 
\begin_inset Formula $\boldsymbol{\phi}$
\end_inset

 on one end of a segment to the other end, in frequency domain:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\boldsymbol{\phi}_{R}(\omega)=\boldsymbol{T}(\omega)\boldsymbol{\phi}_{L}(\omega)+\mathbf{s}(\omega),
\end{equation}

\end_inset

where 
\begin_inset Formula $L$
\end_inset

 and 
\begin_inset Formula $R$
\end_inset

 denote the left and right side, respectively, 
\begin_inset Formula $\boldsymbol{T}$
\end_inset

 denotes the transfer matrix and 
\begin_inset Formula $\boldsymbol{s}$
\end_inset

 is a source term.
 In the code and in this documentation 
\begin_inset Formula $e^{{\color{red}+}i\omega t}$
\end_inset

 convention is used.
 A common choice of state variables is such that their product has the unit
 of power.
 For all systems in this code, the state variables satisfy this property.
 For example in an acoustic segment, the power is the product of acoustic
 pressure 
\begin_inset Formula $p\left(\omega\right)$
\end_inset

 and volume flow 
\begin_inset Formula $U\left(\omega\right)$
\end_inset

.
 For complex phasors and, the acoustic power flow can then be computed as:
\begin_inset Formula 
\begin{equation}
E=\frac{1}{2}\Re\left[pU^{*}\right],
\end{equation}

\end_inset

where 
\begin_inset Formula $\Re[\bullet]$
\end_inset

 denotes the real part of 
\begin_inset Formula $\bullet$
\end_inset

, and * denotes the complex conjugation.
\end_layout

\begin_layout Section
Example transfer matrix of an acoustic duct
\end_layout

\begin_layout Standard
This section will provide the derivation of the transfer matrix of a simple
 acoustic duct.
 Starting with the isentropic acoustic continuity and momentum equation
 :
\begin_inset Formula 
\begin{align}
\frac{1}{c_{0}^{2}}\frac{\partial\hat{p}}{\partial\hat{t}}+\rho_{0}\nabla\cdot\hat{\boldsymbol{u}} & =0,\\
\rho_{0}\frac{\partial\hat{\boldsymbol{u}}}{\partial t}+\nabla\hat{p} & =0.
\end{align}

\end_inset

The next step is to transform these equations to frequency domain and assuming
 only wave propagation in the 
\begin_inset Formula $x-$
\end_inset

direction, integrating over the cross section we find:
\begin_inset Formula 
\begin{align}
\frac{i\omega}{c_{0}^{2}}p+\frac{\rho_{0}}{S_{f}}\frac{\mathrm{d}U}{\mathrm{d}x} & =0,\label{eq:contU}\\
\rho_{0}i\omega U+S_{f}\frac{\mathrm{d}p}{\mathrm{d}x} & =0,\label{eq:momU}
\end{align}

\end_inset

where 
\begin_inset Formula $U$
\end_inset

 denotes the acoustic volume flow in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{
\backslash
cubic
\backslash
metre
\backslash
per
\backslash
second}
\end_layout

\end_inset

.
 Eqs.
\begin_inset space ~
\end_inset

(
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:contU"

\end_inset

-
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:momU"

\end_inset

) is a coupled set of ordinary differential equations, which can be solved
 for the acoustic pressure to find
\begin_inset Formula 
\begin{equation}
p(x)=A\exp\left(-ikx\right)+B\exp\left(ikx\right),\label{eq:HH_sol_prismaticinviscid}
\end{equation}

\end_inset

where 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

 are constants, to be determined from the boundary conditions.
 Setting 
\begin_inset Formula $p=p_{L}$
\end_inset

, and 
\begin_inset Formula $U=U_{L}$
\end_inset

 at 
\begin_inset Formula $x=0$
\end_inset

, we can solve for the acoustic pressure, upon using Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:momU"

\end_inset

 as:
\begin_inset Formula 
\begin{equation}
p(x)=p_{L}\cos\left(kx\right)-iZ_{0}\sin\left(kx\right)U_{L},
\end{equation}

\end_inset

and for the acoustic volume flow we find:
\begin_inset Formula 
\begin{equation}
U(x)=U_{L}\cos\left(kx\right)-\frac{i}{Z_{0}}\sin\left(kx\right)p_{L}.
\end{equation}

\end_inset

Now, we have all ingredients to derive the transfer matrix of an acoustic
 duct.
 Setting 
\begin_inset Formula $p(x=L)=p_{R}$
\end_inset

, and 
\begin_inset Formula $U(x=L)=U_{R}$
\end_inset

, we find the following two-port coupling between the pressure and the velocity
 from the left side of the duct to the right side of the duct:
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} =\left[\begin{array}{cc}
\cos\left(kL\right) & -iZ_{0}\sin\left(kL\right)\\
-iZ_{0}^{-1}\sin\left(kL\right) & \cos\left(kL\right)
\end{array}\right]\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} .\label{eq:transfer_inviscid}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Setting up the system of equations
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\backslash

\end_layout

\end_inset

 has been set up to solve systems of acoustic segments such as this prismatic
 duct.
 The advantage of the transfer matrix method is the ease with which mixed
 (impedance/pressure/velocity) boundary conditions can be implemented.
\end_layout

\begin_layout Standard
In this section, the assembly of the global system of equations is explained.
 The state variables of each segment are stacked in a column vector 
\series bold

\begin_inset Formula $\boldsymbol{\phi}_{\mbox{sys}}$
\end_inset


\series default
, which has the size of 
\begin_inset Formula $4N_{\mbox{segs}}$
\end_inset

, where 
\begin_inset Formula $N_{\mbox{segs}}$
\end_inset

 denotes the number of segments in the system.
 The coupling equations between the nodes of each segment, are the transfer
 matrices.
 Since the transfer matrices are 
\begin_inset Formula $2\times2$
\end_inset

, this fills only half of the required amount of equations.
 The other half is filled with boundary conditions.
 Each segments transfer matrix can be regarded as the element matrix, which
 all have a form like:
\begin_inset Formula 
\begin{equation}
\boldsymbol{\phi}_{R}=\boldsymbol{T}\cdot\boldsymbol{\phi}_{L}+\boldsymbol{s},
\end{equation}

\end_inset

where 
\begin_inset Formula $\boldsymbol{\phi}_{L},\boldsymbol{\phi}_{R}$
\end_inset

 are the state vectors on the left and right sides of the segment, respectively,
 
\begin_inset Formula $\boldsymbol{T}$
\end_inset

 is the transfer matrix, and 
\begin_inset Formula $\boldsymbol{s}$
\end_inset

 is a source term.
\end_layout

\begin_layout Standard
There are two kind of boundary conditions, called external and internal
 boundary conditions.
 External boundary conditions apply where a prescribed condition is given,
 such as a prescribed pressure, voltage, volume flow, current or acoustic/electr
ic impedance.
 Internal boundary conditions are used to couple different segments at a
 connection point, which is recognized by a shared node number.
 At a connection point, the effort variable is shared, which means that
 the pressure at the node is equal for each connected segment sharing the
 node.
 The flow variable is conserved, so the sum of the volume flow out of all
 segments connected at the node is 0.
\end_layout

\begin_layout Subsection*
Example: two ducts
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/tfm_expl.pdf
	width 80text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Example of two simple duct segments connected together.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:coupling_example"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
This procedure of creating a system matrix is explained by an example where
 only two ducts are coupled.
 A schematic of the situation is depicted in Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:coupling_example"

\end_inset

.
 For the example situation, at the left node of segment (1), an impedance
 boundary 
\begin_inset Formula $Z_{L}$
\end_inset

 is prescribed.
 The right node of segment (1) is connected to the left node of segment
 (2), and at the right side of segment (2), a volume flow boundary condition
 is prescribed of 
\begin_inset Formula $U_{R}$
\end_inset

.
 The corresponding system of equations for this case is
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\left[\begin{array}{cccc}
\mathbf{T}_{1} & -\mathbf{I} & \mathbf{0} & \mathbf{0}\\
\mathbf{0} & \mathbf{0} & \mathbf{T}_{2} & -\mathbf{I}\\
\mathbf{0} & \left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right] & \left[\begin{array}{cc}
-1 & 0\\
0 & -1
\end{array}\right] & \mathbf{0}\\
\left[\begin{array}{cc}
1 & Z_{L}\\
0 & 0
\end{array}\right] & \mathbf{0} & \mathbf{0} & \left[\begin{array}{cc}
0 & 0\\
0 & 1
\end{array}\right]
\end{array}\right]\left\{ \begin{array}{c}
p_{1L}\\
U_{1L}\\
p_{1R}\\
U_{1R}\\
p_{2L}\\
U_{2L}\\
p_{2R}\\
U_{2R}
\end{array}\right\} =\left\{ \begin{array}{c}
0\\
0\\
0\\
0\\
0\\
0\\
0\\
U_{R}
\end{array}\right\} ,
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
In this system matrix, 
\begin_inset Formula $\mathbf{0}$
\end_inset

 denotes a 
\begin_inset Formula $2\times2$
\end_inset

 sub matrix of zeros and 
\begin_inset Formula $\mathbf{I}$
\end_inset

 denotes a 
\begin_inset Formula $2\times2$
\end_inset

 identity sub matrix.
 
\begin_inset Formula $\mathbf{T}_{i}$
\end_inset

 is the transfer matrix of the 
\begin_inset Formula $i$
\end_inset

-th segment.
 The solution can be obtained by Gaussian elimination, for which in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 the 
\family typewriter
numpy.linalg.solve()
\family default
 solver is used.
 Once the solution on the nodes is known, the solution in each segment can
 be computed as a post processing step.
 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 provides some post processing routines to aid in visualization of the acoustic
 field inside a non-lumped segment, such as an acoustic duct.
\end_layout

\begin_layout Section
Input impedance, output impedance
\end_layout

\begin_layout Standard
The acoustic input impedance 
\begin_inset Formula $Z_{\mathrm{in}}\equiv p_{L}/U_{L}$
\end_inset

 on the left side of a segment is defined as the impedance a connecting
 segment 
\begin_inset Quotes eld
\end_inset

feels
\begin_inset Quotes erd
\end_inset

 for a certain boundary condition on the right side.
 
\begin_inset Foot
status open

\begin_layout Plain Layout
Note that the definitions of open and closed below are relating to electrical
 circuits, not open or closed in the acoustical sense.
 I.e.
 an open impedance corresponds to a hard acoustic wall (which is acoustically
 closed).
\end_layout

\end_inset

 There are two special load cases for the segment, either on the right side,
 the circuit is open, resulting in 
\begin_inset Formula $U_{R}=0$
\end_inset

, or the circuit is shorted, which results in 
\begin_inset Formula $p_{R}=0$
\end_inset

.
 For the open circuit, the input impedance can be computed from the transfer
 matrix as:
\begin_inset Note Note
status open

\begin_layout Plain Layout
Open case: (
\begin_inset Formula $U_{R}=0$
\end_inset

_
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{R}=T_{11}p_{L}+T_{12}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=0=T_{21}p_{L}+T_{22}U_{L}\Rightarrow\frac{p_{L}}{U_{L}}=-\frac{T_{22}}{T_{21}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Shorted case (
\begin_inset Formula $p_{R}=0$
\end_inset

 ):
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $0=T_{11}p_{L}+T_{12}U_{L}\Rightarrow\frac{p_{L}}{U_{L}}=-\frac{T_{12}}{T_{11}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
Z_{\mathrm{in},\mathrm{open}} & =-\frac{T_{22}}{T_{21}}\\
Z_{\mathrm{in},\mathrm{short}} & =-\frac{T_{12}}{T_{11}}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
For a passive component (and passive load on the right side), the real part
 of the input impedance should be positive:
\begin_inset Formula 
\begin{equation}
\Re\left[Z_{\mathrm{in}}\right]\geq0.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The acoustic output impedance 
\begin_inset Formula $Z_{\mathrm{out}}\equiv p_{R}/U_{R}$
\end_inset

 on the right side of a segment is defined as the impedance a connecting
 segment 
\begin_inset Quotes eld
\end_inset

feels
\begin_inset Quotes erd
\end_inset

 for a certain boundary condition on the left side.
 
\begin_inset Formula 
\begin{align}
Z_{\mathrm{out},\mathrm{open}} & =\frac{T_{11}}{T_{21}}\\
Z_{\mathrm{out},\mathrm{short}} & =\frac{T_{\mathrm{12}}}{T_{22}}
\end{align}

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
Open case left side, means 
\begin_inset Formula $U_{L}=0$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{R}=T_{11}p_{L}+T_{12}U_{L}$
\end_inset

 –> 
\begin_inset Formula $p_{R}=T_{11}p_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=T_{21}p_{L}+T_{22}U_{L}$
\end_inset

 –> 
\begin_inset Formula $U_{R}=T_{21}p_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
==================================
\end_layout

\begin_layout Plain Layout
Shorted case, means 
\begin_inset Formula $p_{L}=0$
\end_inset

,
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{R}=T_{11}p_{L}+T_{12}U_{L}$
\end_inset

 –> 
\begin_inset Formula $p_{R}=T_{12}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=T_{21}p_{L}+T_{22}U_{L}$
\end_inset

 –> 
\begin_inset Formula $U_{R}=T_{22}U_{L}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
For passive segments, the real part of the output impedance should be 
\emph on
negative:
\emph default

\begin_inset Formula 
\begin{equation}
\Re\left[Z_{\mathrm{out}}\right]\leq0.
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
Segment properties and arguments
\begin_inset CommandInset label
LatexCommand label
name "chap:Segment-properties"

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
Dit is meer een gebruikershandleiding van de technische achtergrond.
 Misschien kunnen we hier een apart deel voor maken in de documentatie.
\end_layout

\end_inset


\end_layout

\begin_layout Section
Introduction
\end_layout

\begin_layout Standard
Each segment has various properties and user-set parameters.
 This chapter provides a basic reference to these items.
 Which properties and parameters are available differs for each segment.
\end_layout

\begin_layout Section
Parameters
\end_layout

\begin_layout Subsection
Radius, diameter, area
\end_layout

\begin_layout Standard
Most segments require a measure for the cross-sectional area.
 For ease of use, this can be entered as either the radius
\begin_inset space ~
\end_inset


\begin_inset Formula $r$
\end_inset

, diameter
\begin_inset space ~
\end_inset


\begin_inset Formula $D$
\end_inset

 or cross-sectional area
\begin_inset space ~
\end_inset


\begin_inset Formula $S$
\end_inset

.
 The letter is followed by either a capital
\begin_inset space ~
\end_inset


\begin_inset Formula $L$
\end_inset

 for the left node or 
\begin_inset Formula $R$
\end_inset

 for the right, e.g.
\begin_inset space ~
\end_inset


\begin_inset Formula $rL=0.2$
\end_inset

.
 To avoid interpretation errors, they should be entered as keyword arguments,
 e.g.
\begin_inset space ~
\end_inset


\begin_inset Formula $\mathrm{ConeDuct}(L=1,rL=0.2,rR=0.3)$
\end_inset

 instead of 
\begin_inset Formula $\mathrm{ConeDuct}(1,0.2,0.3)$
\end_inset

.
\end_layout

\begin_layout Section
Properties
\end_layout

\begin_layout Standard
Once a segment has been constructed, its arguments become properties and
 can be adjusted later on.
 For example, say 'duct' is an instance of the class PrsDuct and we want
 to change its radius.
 Then 
\begin_inset Formula $\mathrm{duct}.r=2$
\end_inset

 will set it to 
\begin_inset Formula $2$
\end_inset

 m.
 Retrieving any other measure of cross-sectional area is also possible:
 
\begin_inset Formula $x=\mathrm{duct}.D$
\end_inset

 will retrieve the diameter and save it to 
\begin_inset Formula $x$
\end_inset

.
\end_layout

\begin_layout Standard
On top of that, segments contain calculated properties.
 These are listed in Table
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "tab:calc_seg_properties"
plural "false"
caps "false"
noprefix "false"

\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Float table
wide false
sideways false
status collapsed

\begin_layout Plain Layout
\noindent
\align center
\begin_inset Tabular
<lyxtabular version="3" rows="4" columns="2">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Property
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Explanation
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $T(\omega)$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Transfer matrix
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $Vf$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
Volume of segment filled with fluid [
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{m}
\end_layout

\end_inset

]
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $Z(\omega)$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
(series) impedance [
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{Pa*s/m^3}
\end_layout

\end_inset

]
\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Calculated properties of segments
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
\noindent
\align center
\begin_inset CommandInset label
LatexCommand label
name "tab:calc_seg_properties"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Chapter
Provided acoustic models
\begin_inset CommandInset label
LatexCommand label
name "chap:Provided-acoustic-models"

\end_inset


\end_layout

\begin_layout Section
Introduction
\end_layout

\begin_layout Standard
This chapter provides a concise overview of the provided acoustic models
 implemented in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\end_layout

\end_inset

.
\end_layout

\begin_layout Section
Prismatic duct
\begin_inset CommandInset label
LatexCommand label
name "subsec:Prismatic-duct"

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/prsduct.pdf
	width 80text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Geometry of the prismatic duct
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:prsduct"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
A prismatic duct is used to model one-dimensional acoustic wave propagation.
 The prismatic duct is implemented in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 in the 
\family typewriter
PrsDuct
\family default
 class.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:prsduct"

\end_inset

 shows this segment schematically.
 In the thermal boundary layer, heat and momentum diffuse to the wall.
 The thermal boundary layer can be a small layer w.r.t.
 to the transverse characteristic length scale of the tube, or can fully
 occupy the tube.
 In the latter case, the solution converges to the classic laminar Poisseuille
 flow solution.
 The basic assumptions behind this model are
\end_layout

\begin_layout Itemize
Prismatic cross sectional area.
\end_layout

\begin_layout Itemize
\begin_inset Formula $L\gg r_{h}$
\end_inset

, (tube is long compared to its transverse length scale).
\end_layout

\begin_layout Itemize
Radius is much smaller than the wave length.
\end_layout

\begin_layout Itemize
Wave length is much larger than viscous penetration depth.
\end_layout

\begin_layout Itemize
End effects and entrance effects are negligible.
\end_layout

\begin_layout Standard
For a formal derivation of the model for prismatic cylindrical tubes, the
 reader is referred to the work of Tijdeman 
\begin_inset CommandInset citation
LatexCommand cite
key "tijdeman_propagation_1975"
literal "true"

\end_inset

 and Nijhof 
\begin_inset CommandInset citation
LatexCommand cite
key "nijhof_viscothermal_2010"
literal "true"

\end_inset

.
 For a somewhat more pragmatic derivation, we would like to refer to the
 work of Swift 
\begin_inset CommandInset citation
LatexCommand cite
key "swift_thermoacoustics:_2003,swift_thermoacoustic_1988"
literal "true"

\end_inset

 and Rott 
\begin_inset CommandInset citation
LatexCommand cite
key "rott_damped_1969"
literal "true"

\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\frac{\mathrm{d}p}{\mathrm{d}x} & =\frac{\omega\rho_{0}}{i\left(1-f_{\nu}\right)S_{f}}U,\label{eq:momentum_LRF}\\
\frac{\mathrm{d}U}{\mathrm{d}x} & =\frac{k}{iZ_{0}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p,\label{eq:continuity_LRF}
\end{align}

\end_inset

where 
\begin_inset Formula $S_{f}$
\end_inset

 is the cross-sectional area filled with fluid, 
\begin_inset Formula $k$
\end_inset

 is the inviscid wave number, and 
\begin_inset Formula $Z_{0}$
\end_inset

 the inviscid characteristic impedance of a tube (
\begin_inset Formula $Z_{0}=z_{0}/S_{f}$
\end_inset

).
 
\begin_inset Formula $f_{\nu}$
\end_inset

 and 
\begin_inset Formula $f_{\kappa}$
\end_inset

 are the viscous and thermal Rott functions, respectively 
\begin_inset CommandInset citation
LatexCommand cite
key "rott_damped_1969"
literal "true"

\end_inset

.
 They model the viscous and thermal effects with the wall.
 For circular tubes, the 
\begin_inset Formula $f$
\end_inset

's are defined as 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 88"
key "swift_thermoacoustics:_2003"
literal "true"

\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
f_{j,\mathrm{circ}}=\frac{J_{1}\left[\left(i-1\right)\frac{2r_{h}}{\delta_{j}}\right]}{\left(i-1\right)\frac{r_{h}}{\delta}J_{0}\left[\left(i-1\right)\frac{2r_{h}}{\delta_{j}}\right]},\label{eq:f_cylindrical}
\end{equation}

\end_inset


\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$j$"
description "Index, subscript placeholder\\nomunit{-}"
literal "true"

\end_inset

where 
\begin_inset Formula $\delta_{j}=\delta_{\nu}$
\end_inset

 for 
\begin_inset Formula $f_{\nu,\mathrm{circ}}$
\end_inset

 and 
\begin_inset Formula $\delta_{j}=\delta_{\kappa}$
\end_inset

 for 
\begin_inset Formula $f_{\kappa,\mathrm{circ}}$
\end_inset

.
 
\begin_inset Formula $J_{\alpha}$
\end_inset

 denotes the cylindrical Bessel function of the first kind and order 
\begin_inset Formula $\alpha$
\end_inset

.
 
\begin_inset Formula $r_{h}$
\end_inset

 is the hydraulic radius, defined as the ratio of the cross sectional area
 to the 
\begin_inset Quotes eld
\end_inset

wetted perimeter
\begin_inset Quotes erd
\end_inset

:
\begin_inset Formula 
\begin{equation}
r_{h}=S_{f}/\Pi.
\end{equation}

\end_inset

Note that for a circular tube with diameter 
\begin_inset Formula $D$
\end_inset

, 
\begin_inset Formula $r_{h}=\nicefrac{D}{4}$
\end_inset

.
 The parameter 
\begin_inset Formula $\epsilon_{s}$
\end_inset

 in Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:continuity_LRF"

\end_inset

 is the ideal solid correction factor, which corrects for solids that have
 a finite heat capacity.
 This parameter is dependent on the thermal properties and the geometry
 of the solid.
 An example of 
\begin_inset Formula $\epsilon_{s}$
\end_inset

 is derived in Section 
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Thermal-relaxation-effect"

\end_inset

.
 For the case of an thermally ideal solid, 
\begin_inset Formula $\epsilon_{s}$
\end_inset

 can be set to 0.
\end_layout

\begin_layout Subsection
Other cross-sectional geometries
\end_layout

\begin_layout Subsubsection
Rectangular duct
\end_layout

\begin_layout Standard
Analytical functions exist for prismatic geometries, such as parallel plates,
 rectangular holes, and even triangular holes.
 For parallel plates with sides 
\begin_inset Formula $2y_{0}\times2z_{0}$
\end_inset

, the Rott function reads:
\begin_inset Formula 
\begin{equation}
f=1-\frac{64}{\pi^{4}}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{\left(2m-1\right)^{2}}\frac{1}{\left(2n-1\right)^{2}C_{mn}},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
C_{mn}=1-\frac{i\pi^{2}\delta^{2}}{8y_{0}^{2}z_{0}^{2}}\left(\left(2m-1\right)^{2}z_{0}^{2}+\left(2n-1\right)^{2}y_{0}^{2}\right).
\end{equation}

\end_inset

The hydraulic radius is related to 
\begin_inset Formula $y_{0}$
\end_inset

 and 
\begin_inset Formula $z_{0}$
\end_inset

 as:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $r_{h}=\frac{S}{\Pi}=\frac{4y_{0}z_{0}}{4y_{0}+4z_{0}}=$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
r_{h}=\frac{y_{0}z_{0}}{y_{0}+z_{0}}
\end{equation}

\end_inset

Defining the aspect ratio as 
\begin_inset Formula $\AR=z_{0}/y_{0}$
\end_inset

, a useful equation is to derive 
\begin_inset Formula $y_{0}$
\end_inset

 and 
\begin_inset Formula $z_{0}$
\end_inset

 from 
\begin_inset Formula $r_{h}$
\end_inset

 and 
\begin_inset Formula $\AR$
\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $r_{h}=\frac{y_{0}A}{\left(1+A\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $y_{0}=r_{h}\frac{\left(1+A\right)}{A}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z_{0}=r_{h}\left(1+A\right)$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
y_{0} & =r_{h}\frac{\left(1+\AR\right)}{\AR}\\
z_{0} & =r_{h}\left(1+\AR\right)
\end{align}

\end_inset


\end_layout

\begin_layout Subsubsection
Annular ring
\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $i\omega\rho_{0}u=-\frac{\mathrm{d}p}{\mathrm{d}x}+\mu_{0}\nabla_{\perp}^{2}u$
\end_inset


\end_layout

\begin_layout Plain Layout
Fill in:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
u=\frac{i}{\omega\rho_{0}}\left(1-h_{\nu}\right)\frac{\mathrm{d}p}{\mathrm{d}x}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
Note that 
\begin_inset Formula $h_{\nu}|_{\mathrm{wall}}\equiv1$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $0=h_{\nu}+\frac{i\delta_{\nu}^{2}}{2}\nabla_{\perp}^{2}h_{\nu}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $h_{\nu}+\frac{i\mu_{0}}{\omega\rho_{0}}\nabla_{\perp}^{2}h_{\nu}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
-
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{i\mu_{0}}{\omega\rho_{0}}\nabla_{\perp}^{2}h_{\nu}+h_{\nu}=0$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
The differential equation that is required to be solved
\begin_inset Formula 
\begin{equation}
\frac{i\mu_{0}}{\omega\rho_{0}}\nabla_{\perp}^{2}h_{\nu}+h_{\nu}=0,\qquad h_{\nu|\mathrm{wall}}=0
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
For an annular duct the Rott function reads:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
h_{\nu}=\frac{\left(J_{0}\left(\frac{r_{0}\left(1-i\right)}{\delta_{\nu}}\right)-J_{0}\left(\frac{r_{1}\left(1-i\right)}{\delta_{\nu}}\right)\right)Y_{0}\left(\frac{r\left(1-i\right)}{\delta_{\nu}}\right)+\left(Y_{0}\left(\frac{r_{1}\left(1-i\right)}{\delta_{\nu}}\right)-Y_{0}\left(\frac{r_{0}\left(1-i\right)}{\delta_{\nu}}\right)\right)J_{0}\left(\frac{r\left(1-i\right)}{\delta_{\nu}}\right)}{J_{0}\left(\frac{r_{0}\left(1-i\right)}{\delta_{\nu}}\right)Y_{0}\left(\frac{r_{1}\left(1-i\right)}{\delta_{\nu}}\right)-J_{0}\left(\frac{r_{1}\left(1-i\right)}{\delta_{\nu}}\right)Y_{0}\left(\frac{r_{0}\left(1-i\right)}{\delta_{\nu}}\right)}
\]

\end_inset


\end_layout

\begin_layout Standard
Where
\begin_inset Formula 
\begin{align*}
\alpha_{0} & =\frac{r_{0}\left(1-i\right)}{\delta_{i}}\\
\alpha_{1} & =\frac{r_{1}\left(1-i\right)}{\delta_{i}}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
And:
\begin_inset Formula 
\begin{align}
C_{1} & =\frac{Y_{0}\left(\alpha_{1}\right)-Y_{0}\left(\alpha_{0}\right)}{J_{0}\left(\alpha_{0}\right)Y_{0}\left(\alpha_{1}\right)-J_{0}\left(\alpha_{1}\right)Y_{0}\left(\alpha_{0}\right)}\\
C_{2} & =\frac{J_{0}\left(\alpha_{0}\right)-J_{0}\left(\alpha_{1}\right)}{J_{0}\left(\alpha_{0}\right)Y_{0}\left(\alpha_{1}\right)-J_{0}\left(\alpha_{1}\right)Y_{0}\left(\alpha_{0}\right)}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
f_{i} & =\delta_{i}\left(1+i\right)\left[\frac{\left\{ H_{0}^{(1)}\left(\alpha_{0}\right)-H_{0}^{(1)}\left(\alpha_{1}\right)\right\} \left[r_{0}H_{-1}^{(2)}\left(\alpha_{0}\right)-r_{1}H_{-1}^{(2)}\left(\alpha_{1}\right)\right]}{\left(r_{1}^{2}-r_{0}^{2}\right)\left[H_{0}^{(1)}\left(\alpha_{0}\right)H_{0}^{(2)}\left(\alpha_{1}\right)-H_{0}^{(1)}\left(\alpha_{1}\right)H_{0}^{(2)}\left(\alpha_{0}\right)\right]}+\right.\\
 & \qquad\qquad\qquad\left.\frac{\left\{ H_{0}^{(2)}\left(\alpha_{0}\right)-H_{0}^{(2)}\left(\alpha_{1}\right)\right\} \left[r_{1}H_{-1}^{(1)}\left(\alpha_{1}\right)-r_{0}H_{-1}^{(1)}\left(\alpha_{0}\right)\right]}{\left(r_{1}^{2}-r_{0}^{2}\right)\left[H_{0}^{(1)}\left(\alpha_{0}\right)H_{0}^{(2)}\left(\alpha_{1}\right)-H_{0}^{(1)}\left(\alpha_{1}\right)H_{0}^{(2)}\left(\alpha_{0}\right)\right]}\right]
\end{align}

\end_inset


\end_layout

\begin_layout Subsection
Transfer matrix
\end_layout

\begin_layout Standard
Upon solving for Eqs.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:momentum_LRF"

\end_inset

-
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:continuity_LRF"

\end_inset

, a transfer matrix can be derived which couples the pressure and volume
 flow on the left side to the right side as:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
\frac{\mathrm{d}p}{\mathrm{d}x} & =\frac{\omega\rho_{0}}{i\left(1-f_{\nu}\right)S_{f}}U,\\
\frac{\mathrm{d}U}{\mathrm{d}x} & =\frac{k}{iZ_{0}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p,
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
We know the solution for 
\begin_inset Formula $p$
\end_inset

 is
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=A\exp\left(-i\Gamma x\right)+B\exp\left(i\Gamma x\right)$
\end_inset

 where 
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)}{1-f_{\nu}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Then
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U=\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\Gamma i\left(-A\exp\left(-i\Gamma x\right)+B\exp\left(i\Gamma x\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U=-\frac{\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\Gamma\left(B\exp\left(i\Gamma x\right)-A\exp\left(-i\Gamma x\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Now: 
\begin_inset Formula $p(x=0)=p_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
And: 
\begin_inset Formula $U(x=0)=U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
Then:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{L}=\frac{\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\Gamma\left(A-B\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{L}=A+B\Rightarrow B=p_{L}-A$
\end_inset


\end_layout

\begin_layout Plain Layout
Hence:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{L}=\frac{\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\Gamma\left(2A-p_{L}\right)$
\end_inset

 or 
\begin_inset Formula $A=\frac{1}{2}p_{L}+\frac{1}{2}\frac{\omega\rho_{0}}{\left(1-f_{\nu}\right)S_{f}\Gamma}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
And: 
\begin_inset Formula $B=p_{L}-A=\frac{1}{2}p_{L}-\frac{1}{2}\frac{\omega\rho_{0}}{\left(1-f_{\nu}\right)S_{f}\Gamma}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
So, finally for 
\begin_inset Formula $p$
\end_inset

 we find:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=\left(\frac{1}{2}p_{L}+\frac{1}{2}\frac{\omega\rho_{0}}{\left(1-f_{\nu}\right)S_{f}\Gamma}U_{L}\right)\exp\left(-i\Gamma x\right)+\left(\frac{1}{2}p_{L}-\frac{1}{2}\frac{\omega\rho_{0}}{\left(1-f_{\nu}\right)S_{f}\Gamma}U_{L}\right)\exp\left(i\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=\left(\frac{1}{2}p_{L}+\frac{1}{2}Z_{c}U_{L}\right)\exp\left(-i\Gamma x\right)+\left(\frac{1}{2}p_{L}-\frac{1}{2}Z_{c}U_{L}\right)\exp\left(i\Gamma x\right)$
\end_inset

 where 
\begin_inset Formula $Z_{c}=\frac{kZ_{0}}{\left(1-f_{\nu}\right)\Gamma}$
\end_inset


\end_layout

\begin_layout Plain Layout
Or, working to transfer matrices
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=\frac{1}{2}p_{L}\exp\left(-i\Gamma x\right)+\frac{1}{2}Z_{c}U_{L}\exp\left(-i\Gamma x\right)+\frac{1}{2}p_{L}\exp\left(i\Gamma x\right)-Z_{c}U_{L}\exp\left(i\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=p_{L}\cos\left(\Gamma x\right)+\frac{1}{2}Z_{c}U_{L}\exp\left(-i\Gamma x\right)-Z_{c}U_{L}\exp\left(i\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Using the rule: 
\begin_inset Formula $\sin\left(x\right)=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p=p_{L}\cos\left(\Gamma x\right)-iZ_{c}U_{L}\sin\left(\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Using 
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U=\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{0}}\frac{\mathrm{d}p}{\mathrm{d}x}=\frac{i}{Z_{c}}\left[-p_{L}\sin\left(\Gamma x\right)-iZ_{c}U_{L}\cos\left(\Gamma x\right)\right]=\left[-\frac{i}{Z_{c}}p_{L}\sin\left(\Gamma x\right)+U_{L}\cos\left(\Gamma x\right)\right]$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} =\left[\begin{array}{cc}
\cos\left(\Gamma L\right) & -iZ_{c}\sin\left(\Gamma L\right)\\
-iZ_{c}^{-1}\sin\left(\Gamma L\right) & \cos\left(\Gamma L\right)
\end{array}\right]\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} ,\label{eq:transfer_matrix_prismatic_duct}
\end{equation}

\end_inset

where 
\begin_inset Formula $Z_{c}$
\end_inset

 is the characteristic impedance of the duct, i.e.
 the impedance 
\begin_inset Formula $p/U$
\end_inset

 of a plane (although damped) propagating wave:
\begin_inset Formula 
\begin{equation}
Z_{c}=\frac{kZ_{0}}{\left(1-f_{\nu}\right)\Gamma}.\label{eq:Z_c_prismduct}
\end{equation}

\end_inset

The parameter 
\begin_inset Formula $\Gamma$
\end_inset

 in Eqs.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:transfer_matrix_prismatic_duct"

\end_inset

 and 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Z_c_prismduct"

\end_inset

 is the viscothermal wave number, i.e.
 the wave number corrected for viscothermal losses:
\begin_inset Formula 
\begin{equation}
\Gamma=\frac{\omega}{c_{0}}\sqrt{\frac{1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\epsilon_{s}}}{1-f_{\nu}}}.\label{eq:Gamma}
\end{equation}

\end_inset

Due to the numerical implementation of the Bessel functions in many libraries,
 the 
\begin_inset Formula $f_{j}$
\end_inset

 function for cylindrical ducts (Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:f_cylindrical"

\end_inset

) cannot be computed for high 
\begin_inset Formula $r_{h}/\delta$
\end_inset

 by computing this ratio 
\begin_inset Formula $J_{1}/J_{0}$
\end_inset

.
 The numerical result starts to break down at 
\begin_inset Formula $r_{h}/\delta\sim100$
\end_inset

.
 To resolve this problem, the 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 code applies a smooth transition from the Bessel function ratio to the
 boundary layer limit solution for 
\begin_inset Formula $f$
\end_inset

:
\begin_inset Formula 
\begin{equation}
f_{j,\mathrm{bl}}=\frac{\left(1-i\right)\delta_{j}}{2r_{h}}
\end{equation}

\end_inset

in the range of 
\begin_inset Formula $100<r_{h}/\delta\leq200$
\end_inset

.
\end_layout

\begin_layout Standard
Note that in the limit of 
\begin_inset Formula $r_{h}\to\infty$
\end_inset

, or 
\begin_inset Formula $\kappa$
\end_inset

 and 
\begin_inset Formula $\mu$
\end_inset

 
\begin_inset Formula $\to0$
\end_inset

, 
\begin_inset Formula $\Re\left[\Gamma\right]\to k$
\end_inset

 and 
\begin_inset Formula $\Re\left[Z_{c}\right]\to Z_{0}$
\end_inset

 whereas 
\begin_inset Formula $\Im\left[\Gamma\right]$
\end_inset

 and 
\begin_inset Formula $\Im\left[Z_{c}\right]$
\end_inset

 
\begin_inset Formula $\to0$
\end_inset

.
 Hence in these limits the lossless wave equation is resolved from the result.
 This is not true in the limit of 
\begin_inset Formula $\omega\to\infty$
\end_inset

, as in that limit it can be computed that 
\begin_inset Formula $\Re\left[\Gamma\right]\to k$
\end_inset

, while the imaginary part 
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\Gamma=k\sqrt{\frac{1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\epsilon_{s}}}{1-f_{\nu}}}$
\end_inset

filling in 
\begin_inset Formula $f_{\mathrm{bl}}$
\end_inset

 
\begin_inset Formula $f_{\nu}=\frac{\left(1-i\right)\delta_{\nu}}{2r_{h}}$
\end_inset

 and 
\begin_inset Formula $f_{\kappa}=\frac{\left(1-i\right)\delta_{\nu}}{2\sqrt{\Pr}r_{h}}$
\end_inset

, 
\begin_inset Formula $\epsilon_{s}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{1+\left(\gamma-1\right)\frac{\left(1-i\right)\delta_{\nu}}{2\sqrt{\Pr}r_{h}}}{1-\frac{\left(1-i\right)\delta_{\nu}}{2r_{h}}}$
\end_inset

 
\end_layout

\begin_layout Plain Layout
Using 
\begin_inset Formula $\alpha=\frac{1}{\sqrt{\Pr}}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{r_{h}+\frac{1}{2}\alpha\left(\gamma-1\right)\left(1-i\right)\delta_{\nu}}{r_{h}-\frac{1}{2}\left(1-i\right)\delta_{\nu}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Multiply numerator and denominator with 
\begin_inset Formula $r_{h}+\frac{1}{2}\left(-i-1\right)\delta_{\nu}$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{\left(r_{h}+\frac{1}{2}\left(-i-1\right)\delta_{\nu}\right)\left(r_{h}+\frac{1}{2}\left(-i-1\right)\delta_{\nu}\right)}{\left[r_{h}-\frac{1}{2}\left(1-i\right)\delta_{\nu}\right]\left(r_{h}+\frac{1}{2}\left(-i-1\right)\delta_{\nu}\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{r_{h}^{2}+\frac{1}{2}r_{h}\delta_{\nu}\left[\alpha\left(\gamma-1\right)-1-i\left(1+\alpha\left(\gamma-1\right)\right)\right]+-\frac{1}{2}\alpha\delta_{\nu}^{2}\left(\gamma-1\right)}{r_{h}^{2}-r_{h}\delta_{\nu}+\frac{1}{2}\delta_{\nu}^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Leaving terms of 
\begin_inset Formula $\mathcal{O}\left(\delta_{\nu}^{0}\right)$
\end_inset

 in the denominator and 
\begin_inset Formula $\mathcal{O}\left(\delta_{\nu}^{1}\right)$
\end_inset

 in the numerator:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\frac{r_{h}^{2}+\frac{1}{2}r_{h}\delta_{\nu}\left[\alpha\left(\gamma-1\right)-1-i\left(1+\alpha\left(\gamma-1\right)\right)\right]}{r_{h}^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Removing from the real part the small stuff:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\left(1-i\frac{1}{2}\frac{\delta_{\nu}}{r_{h}}\left(1+\alpha\left(\gamma-1\right)\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma^{2}=k^{2}\left(1-ix\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
where 
\begin_inset Formula $x=\frac{\delta_{\nu}}{2r_{h}}\left[\left(1+\left(\gamma-1\right)\sqrt{\Pr^{-1}}\right)\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
Taking the square root:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Gamma=\sqrt{k^{2}\left(1-ix\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
Take the imaginary part: 
\begin_inset Formula $\Im\left[\sqrt{a}\right]=\sqrt{|a|\frac{\Im\left[a\right]}{|a|}}=\sqrt{|a|}\frac{\Im\left[a\right]}{2|a|}$
\end_inset


\end_layout

\begin_layout Plain Layout
Now we assume: 
\begin_inset Formula $\Im\left[a\right]/|a|\ll1$
\end_inset

, such that:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Im\left[\sqrt{a}\right]\approx\frac{1}{2}\sqrt{|a|}\frac{\Im\left[a\right]}{|a|}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\Im\left[\Gamma\right]\approx k\frac{1}{2}\frac{-k^{2}x}{k^{2}}=-\frac{1}{2}kx=-k\frac{\delta_{\nu}}{4r_{h}}\left[1+\frac{\left(\gamma-1\right)}{\sqrt{\Pr}}\right]$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
-\Im\left[\Gamma\right]\to\sqrt{\omega}\frac{\sqrt{\frac{1}{8}\frac{\mu}{\rho_{0}}}}{c_{0}r_{h}}\left[1+\frac{\left(\gamma-1\right)}{\sqrt{\Pr}}\right].\label{eq:hf_limit_im_gamma}
\end{equation}

\end_inset

In other words the imaginary part of the wave number keeps growing, although
 with a smaller rate than real part of the wave number.
 So the higher the frequency, the smaller the viscothermal damping per wavelengt
h, but the higher the viscothermal damping per meter of duct.
\end_layout

\begin_layout Standard
Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:im_gamma"

\end_inset

 shows the imaginary part of the wave number as a function of the frequency.
 As visible, the magnitude of the viscothermal damping grows monotonically
 with frequency.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status collapsed

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/im_Gamma.pdf
	width 80text%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Logarithmic plot of the negative of imaginary part of the viscothermal wave
 number 
\begin_inset Formula $\left(-\Im\left[\Gamma\right]\right)$
\end_inset

, for a tube with a diameter of 1 mm.
 In blue, the full 
\begin_inset Formula $f_{\nu}$
\end_inset

 and 
\begin_inset Formula $f_{\kappa}$
\end_inset

 of Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Gamma"

\end_inset

 and 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:f_cylindrical"

\end_inset

 is used.
 The orange curve corresponds to Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:hf_limit_im_gamma"

\end_inset

.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:im_gamma"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Section

\series bold
Duct with varying cross-sectional area
\end_layout

\begin_layout Standard
For ducts with variation in the cross-sectional area, an approximately valid
 ordinary differential equation can be derived, which is a viscothermal
 correction to Webster's horn equation 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 181"
key "rienstra_introduction_2015"
literal "true"

\end_inset

:
\begin_inset Formula 
\begin{equation}
\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\frac{1}{S_{f}}\frac{\mathrm{d}S_{f}}{\mathrm{d}x}\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Exponential duct (horn)
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
S_{f}=\exp\left(\alpha x\right)
\end{equation}

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\alpha\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling in: 
\begin_inset Formula $p=a\exp\left(\beta x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\beta^{2}+\alpha\beta+\Gamma^{2}=0
\]

\end_inset


\end_layout

\begin_layout Plain Layout
Solving for 
\begin_inset Formula $\beta$
\end_inset

:
\begin_inset Formula 
\[
\beta=\frac{1}{2}\left(-\alpha\pm\sqrt{\alpha^{2}-4\Gamma^{2}}\right)
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsection
Conical ducts
\end_layout

\begin_layout Standard
For conical ducts, i.e.
 ducts with quadratic variation in the cross-sectional area (linear variation
 in the diameter, or cross-sectional length scale), 
\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\begin{eqnarray*}
\frac{dp_{1}}{dx} & = & \frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)S_{f}}U_{1},\\
\frac{dU_{1}}{dx} & = & \frac{\omega S_{f}}{i\gamma p_{m}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p_{1},\\
 &  & +\tfrac{f_{\kappa}-f_{\nu}}{\left(1-\Pr\right)\left(1+\epsilon_{s}\right)}\frac{1}{T_{m}}\frac{dT_{m}}{dx}U_{1},
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Neglect dTmdx part, assume Sf not consant:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dp_{1}}{dx}=\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)S_{f}}U_{1}$
\end_inset

 so 
\begin_inset Formula $U_{1}=\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{m}}\frac{dp_{1}}{dx}$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $\frac{d^{2}p_{1}}{dx^{2}}=\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)}\left(\frac{1}{S_{f}}\frac{dU_{1}}{dx}-\frac{U_{1}}{S_{f}^{2}}\frac{dS_{f}}{dx}\right)$
\end_inset

———< fill in one below
\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $\frac{dU_{1}}{dx}=\frac{\omega S_{f}}{i\gamma p_{m}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p_{1}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
————-
\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $\frac{d^{2}p_{1}}{dx^{2}}=\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)}\left(\frac{1}{S_{f}}\left(\frac{\omega S_{f}}{i\gamma p_{m}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p_{1}\right)-\frac{1}{S_{f}^{2}}\frac{dS_{f}}{dx}\left(\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{m}}\frac{dp_{1}}{dx}\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $\frac{d^{2}p_{1}}{dx^{2}}=\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)}\frac{\omega}{i\gamma p_{m}}\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)p_{1}-\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)}\frac{1}{S_{f}}\frac{i\left(1-f_{\nu}\right)}{\omega\rho_{m}}\frac{dS_{f}}{dx}\frac{dp_{1}}{dx}$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $\frac{d^{2}p_{1}}{dx^{2}}+\frac{1}{S_{f}}\frac{dS_{f}}{dx}\frac{dp_{1}}{dx}+\frac{\omega^{2}}{c_{m}^{2}}\frac{\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)}{\left(1-f_{\nu}\right)}p_{1}=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Makes:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\frac{d^{2}p_{1}}{dx^{2}}+\frac{1}{S_{f}}\frac{dS_{f}}{dx}\frac{dp_{1}}{dx}+\Gamma^{2}p_{1}=0
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\Gamma^{2}=\frac{\omega^{2}}{c_{m}^{2}}\frac{\left(1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\varepsilon_{s}}\right)}{\left(1-f_{\nu}\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $r=r_{0}+\alpha x$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $S=\pi\left(r_{0}+\alpha x\right)^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dS_{f}}{dx}=2\alpha\pi\left(r_{0}+\alpha x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{1}{S_{f}}\frac{dS_{f}}{dx}=\frac{2\alpha}{\left(r_{0}+\alpha x\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
For this horn,
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\frac{d^{2}p_{1}}{dx^{2}}+\frac{2\alpha}{\left(r_{0}+\alpha x\right)}\frac{dp_{1}}{dx}+\Gamma^{2}p_{1}=0
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
And we find volume flow from
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dp_{1}}{dx}=\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)S_{f}}U_{1}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{i\left(1-f_{\nu}\right)\pi\left(r_{0}+\alpha x\right)^{2}}{\omega\rho_{m}}\frac{dp_{1}}{dx}=U_{1}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
such that for a conical tube the radius 
\begin_inset Formula $r(x)$
\end_inset

 varies as:
\begin_inset Formula 
\begin{equation}
r(x)=r_{0}+\eta x,
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
\eta=\frac{x}{L}\left(r_{1}-r_{0}\right)
\end{equation}

\end_inset

Filling in for 
\begin_inset Formula $S_{f}=\pi\left(r_{0}+\eta x\right)^{2}$
\end_inset

 yields
\begin_inset Formula 
\begin{equation}
\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\frac{2\eta}{r_{0}+\eta x}\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0,
\end{equation}

\end_inset

for which the solution is:
\begin_inset Formula 
\begin{equation}
p=\frac{C_{1}\exp\left(-i\Gamma x\right)+C_{1}\exp\left(-i\Gamma x\right)}{r_{0}+\eta x}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
If we assume 
\begin_inset Formula $S_{f}=\pi\left(r_{0}+\eta x\right)^{2}$
\end_inset

, where 
\begin_inset Formula $\eta$
\end_inset

 is the radius variation factor, this can be written as
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{equation}
\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\frac{2\eta}{\left(r_{0}+\eta x\right)}\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0.
\end{equation}

\end_inset

Now assume that
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{equation}
\Gamma(x)\approx\Gamma(x=0)\equiv\Gamma_{0},
\end{equation}

\end_inset

or, the variation in the viscothermal wave number is negligible.
 We can find the solution to this differential equation to be
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
Solution:
\end_layout

\begin_layout Plain Layout

\lang english
Try: 
\begin_inset Formula $p_{1}=Ae^{kx}\frac{1}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{d}{dx}p_{1}=Ae^{kx}\left(\frac{k}{r_{0}+\alpha x}-\frac{\alpha}{\left(r_{0}+\alpha x\right)^{2}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{d^{2}}{dx^{2}}p_{1}=Ae^{kx}\left(\frac{k^{2}}{r_{0}+\alpha x}-\frac{\alpha k}{\left(r_{0}+\alpha x\right)^{2}}\right)+Ae^{kx}\left(-\frac{\alpha k}{\left(r_{0}+\alpha x\right)^{2}}+\frac{2\alpha^{2}}{\left(r_{0}+\alpha x\right)^{3}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
——————-Substitution in
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{d^{2}p_{1}}{dx^{2}}+\frac{2\alpha}{\left(r_{0}+\alpha x\right)}\frac{dp_{1}}{dx}+\Gamma^{2}p_{1}=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{k^{2}}{r_{0}+\alpha x}-\frac{2\alpha k}{\left(r_{0}+\alpha x\right)^{2}}+\frac{2\alpha^{2}}{\left(r_{0}+\alpha x\right)^{3}}+\frac{2\alpha}{\left(r_{0}+\alpha x\right)}\left(\frac{k}{r_{0}+\alpha x}-\frac{\alpha}{\left(r_{0}+\alpha x\right)^{2}}\right)+\Gamma^{2}\frac{1}{r_{0}+\alpha x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{k^{2}}{r_{0}+\alpha x}-\frac{2\alpha k}{\left(r_{0}+\alpha x\right)^{2}}+\frac{2\alpha^{2}}{\left(r_{0}+\alpha x\right)^{3}}+\frac{2\alpha k}{\left(r_{0}+\alpha x\right)^{2}}-\frac{2\alpha^{2}}{\left(r_{0}+\alpha x\right)^{3}}+\Gamma^{2}\frac{1}{r_{0}+\alpha x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{k^{2}}{r_{0}+\alpha x}+\Gamma^{2}\frac{1}{r_{0}+\alpha x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $k^{2}=-\Gamma^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Resulting in:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=p^{+}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+p^{-}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
p_{1}=C_{1}\frac{e^{-i\Gamma_{0}x}}{r_{0}+\eta x}+C_{2}\frac{e^{i\Gamma_{0}x}}{r_{0}+\eta x},
\end{equation}

\end_inset

where 
\begin_inset Formula $C_{1}$
\end_inset

 and 
\begin_inset Formula $C_{2}$
\end_inset

 are constants to be determined from the boundary conditions.
 Upon filling in the boundary conditions, we can derive a transfer matrix
 for a conical tube:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
Derivation transfer matrix:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=\frac{i\left(1-f_{\nu}\right)S_{f}}{\omega\rho_{m}}\frac{dp_{1}}{dx}=\frac{i\left(1-f_{\nu}\right)}{kZ_{0}}\frac{dp_{1}}{dx}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
And:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=C_{1}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+C_{2}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dp_{1}}{dx}=C_{1}\left(-i\Gamma\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}-\alpha\frac{e^{-i\Gamma x}}{\left(r_{0}+\alpha x\right)^{2}}\right)+C_{2}\left(i\Gamma\frac{e^{i\Gamma x}}{r_{0}+\alpha x}-\frac{\alpha e^{i\Gamma x}}{\left(r_{0}+\alpha x\right)^{2}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
So:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\left(-C_{1}\left(i\Gamma\left(r_{0}+\alpha x\right)e^{-i\Gamma x}+\alpha e^{-i\Gamma x}\right)+C_{2}\left(\left(r_{0}+\alpha x\right)i\Gamma e^{i\Gamma x}-\alpha e^{i\Gamma x}\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
————-
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\left(-C_{1}\left(i\Gamma\left(r_{0}+\alpha x\right)e^{-i\Gamma x}+\alpha e^{-i\Gamma x}\right)+C_{2}\left(\left(r_{0}+\alpha x\right)i\Gamma e^{i\Gamma x}-\alpha e^{i\Gamma x}\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
———————-
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=C_{1}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+C_{2}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{L}=\frac{1}{r_{0}}\left(C_{1}+C_{2}\right)\Rightarrow C_{2}=r_{0}p_{L}-C_{1}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{L}=U_{1}(0)=\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\left(r_{0}p_{L}\left(r_{0}i\Gamma-\alpha\right)-2C_{1}r_{0}i\Gamma\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $r_{0}p_{L}\left(r_{0}i\Gamma-\alpha\right)-\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)\pi}U_{L}=2C_{1}r_{0}i\Gamma$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
So:
\begin_inset Formula $C_{1}=\frac{r_{0}p_{L}\left(r_{0}i\Gamma-\alpha\right)-\frac{\omega\rho_{m}}{i\left(1-f_{\nu}\right)\pi}U_{L}}{2i\Gamma r_{0}}=\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}+\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
And:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $C_{2}=r_{0}p_{L}-C_{1}=r_{0}p_{L}-\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}-\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Makes finally:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=C_{1}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+C_{2}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=\left(\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}+\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\right)\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+\left(r_{0}p_{L}-\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}-\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\right)\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\frac{e^{-i\Gamma x}}{r_{0}+\alpha x}+r_{0}p_{L}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}-\frac{p_{L}\left(r_{0}i\Gamma-\alpha\right)}{2i\Gamma}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}-\frac{\omega\rho_{m}}{2\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\frac{e^{i\Gamma x}}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=p_{L}\left[\frac{r_{0}\cos\left(\Gamma x\right)}{r_{0}+\alpha x}+\frac{\alpha}{\Gamma}\frac{\sin\left(\Gamma x\right)}{r_{0}+\alpha x}\right]-\frac{i\omega\rho_{m}}{\Gamma\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\frac{\sin\left(\Gamma x\right)}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Check with prismatic:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Formula $p_{1}=p_{L}\cos\left(\Gamma x\right)-\frac{\omega\rho_{m}i}{\left(1-f_{\nu}\right)S_{f}\Gamma}U_{L}\sin\left(\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Check!
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Check one: p(0):
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}\left(0\right)=p_{L}$
\end_inset

 check
\end_layout

\begin_layout Plain Layout

\lang english
Now: U1:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dp_{1}}{dx}=-\Gamma p_{L}\frac{r_{0}\sin\left(\Gamma x\right)}{r_{0}+\alpha x}-p_{L}\alpha\frac{r_{0}\cos\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}+p_{L}\alpha\frac{\cos\left(\Gamma x\right)}{r_{0}+\alpha x}-p_{L}\frac{\alpha^{2}}{\Gamma}\frac{\sin\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}-\frac{i\omega\rho_{m}}{\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\frac{\cos\left(\Gamma x\right)}{r_{0}+\alpha x}+\frac{i\omega\rho_{m}}{\Gamma\pi r_{0}\left(1-f_{\nu}\right)}\alpha U_{L}\frac{\sin\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\frac{dp_{1}}{dx}=-\Gamma p_{L}\frac{r_{0}\sin\left(\Gamma x\right)}{r_{0}+\alpha x}-p_{L}\frac{\alpha^{2}}{\Gamma}\frac{\sin\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}-p_{L}\alpha\frac{r_{0}\cos\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}+p_{L}\alpha\frac{\cos\left(\Gamma x\right)}{r_{0}+\alpha x}-\frac{i\omega\rho_{m}}{\pi r_{0}\left(1-f_{\nu}\right)}U_{L}\frac{\cos\left(\Gamma x\right)}{r_{0}+\alpha x}+\frac{i\omega\rho_{m}}{\Gamma\pi r_{0}\left(1-f_{\nu}\right)}\alpha U_{L}\frac{\sin\left(\Gamma x\right)}{\left(r_{0}+\alpha x\right)^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=\frac{i\left(1-f_{\nu}\right)\pi\left(r_{0}+\alpha x\right)^{2}}{\omega\rho_{m}}\frac{dp_{1}}{dx}=\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\left(-p_{L}\left(\Gamma\left(r_{0}^{2}+\alpha r_{0}x\right)+\frac{\alpha^{2}}{\Gamma}\right)\sin\left(\Gamma x\right)+p_{L}\alpha^{2}x\cos\left(\Gamma x\right)+U_{L}\left(-\left(r_{0}+\alpha x\right)\frac{i\omega\rho_{m}}{\pi r_{0}\left(1-f_{\nu}\right)}\cos\left(\Gamma x\right)+\frac{i\omega\rho_{m}}{\Gamma\pi r_{0}\left(1-f_{\nu}\right)}\alpha\sin\left(\Gamma x\right)\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=-p_{L}\left(\Gamma\left(r_{0}^{2}+\alpha r_{0}x\right)+\frac{\alpha^{2}}{\Gamma}\right)\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\sin\left(\Gamma x\right)+p_{L}\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\alpha^{2}x\cos\left(\Gamma x\right)+U_{L}\left(\frac{\left(r_{0}+\alpha x\right)}{r_{0}}\cos\left(\Gamma x\right)-\frac{\alpha}{\Gamma r_{0}}\sin\left(\Gamma x\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
\begin_inset Note Note
status open

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none
\lang english
Introducing:
\begin_inset Formula $\delta=\frac{i\omega\rho_{m}}{\left(1-f_{\nu}\right)S_{f}\Gamma}\Rightarrow\delta_{0}=\frac{i\omega\rho_{m}}{\left(1-f_{\nu}\right)\pi r_{0}^{2}\Gamma}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Check for 
\begin_inset Formula $U_{1}(0)$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}(0)=U_{L}$
\end_inset

 check!!
\end_layout

\begin_layout Plain Layout

\lang english
————————– Simpler form of 
\begin_inset Formula $U_{1}$
\end_inset

?
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=p_{L}\left(\Gamma\left(r_{0}^{2}+\alpha r_{0}x\right)+\frac{\alpha^{2}}{\Gamma}\right)\frac{\left(1-f_{\nu}\right)\pi}{i\omega\rho_{m}}\sin\left(\Gamma x\right)+p_{L}\frac{i\left(1-f_{\nu}\right)\pi}{\omega\rho_{m}}\alpha^{2}x\cos\left(\Gamma x\right)+U_{L}\left(\frac{\left(r_{0}+\alpha x\right)}{r_{0}}\cos\left(\Gamma x\right)-\frac{\alpha}{\Gamma r_{0}}\sin\left(\Gamma x\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Prismatic tube check:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1pris}=p_{L}\frac{\Gamma\left(1-f_{\nu}\right)\pi r_{0}^{2}}{i\omega\rho_{m}}\sin\left(\Gamma x\right)+U_{L}\cos\left(\Gamma x\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
with:
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\uuline off
\uwave off
\noun off
\color none

\begin_inset Formula $\frac{\left(1-f_{\nu}\right)\Gamma S_{f}}{i\omega\rho_{m}}p_{L}\sin\left(\Gamma x\right)+U_{L}\cos\left(\Gamma x\right)$
\end_inset

 << from previous derivation!
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=p_{L}\left[\left(1+\frac{\alpha x}{r_{0}}+\frac{\alpha^{2}}{\Gamma^{2}r_{0}^{2}}\right)\frac{1}{\delta_{0}}\sin\left(\Gamma x\right)-\frac{\alpha^{2}x\cos\left(\Gamma x\right)}{r_{0}^{2}\Gamma\delta_{0}}\right]+U_{L}\left[\left(1+\frac{\alpha x}{r_{0}}\right)\cos\left(\Gamma x\right)-\frac{\alpha}{\Gamma r_{0}}\sin\left(\Gamma x\right)\right]$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=p_{L}\left[\frac{r_{0}\cos\left(\Gamma x\right)}{r_{0}+\alpha x}+\frac{\alpha}{\Gamma}\frac{\sin\left(\Gamma x\right)}{r_{0}+\alpha x}\right]-\delta_{0}U_{L}\frac{r_{0}\sin\left(\Gamma x\right)}{r_{0}+\alpha x}$
\end_inset


\end_layout

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $U_{1}=p_{L}\left[\left(1+\frac{\alpha x}{r_{0}}+\frac{\alpha^{2}}{\Gamma^{2}r_{0}^{2}}\right)\frac{1}{\delta_{0}}\sin\left(\Gamma x\right)-\frac{\alpha^{2}x\cos\left(\Gamma x\right)}{r_{0}^{2}\Gamma\delta_{0}}\right]+U_{L}\left[\left(1+\frac{\alpha x}{r_{0}}\right)\cos\left(\Gamma x\right)-\frac{\alpha}{\Gamma r_{0}}\sin\left(\Gamma x\right)\right]$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{1}=p_{L}\left[\frac{r_{0}\cos\left(\Gamma x\right)}{r_{0}+\alpha x}+\frac{\alpha}{\Gamma}\frac{\sin\left(\Gamma x\right)}{r_{0}+\alpha x}\right]-\delta_{0}U_{L}\frac{r_{0}\sin\left(\Gamma x\right)}{r_{0}+\alpha x}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\left\{ \begin{array}{c}
p_{1}\\
U_{1}
\end{array}\right\} _{R}=\left[\begin{array}{cc}
\left[\frac{r_{0}\cos\left(\Gamma L\right)}{r_{0}+\alpha L}+\frac{\alpha}{\Gamma}\frac{\sin\left(\Gamma L\right)}{r_{0}+\alpha L}\right] & \left[-\delta_{0}\frac{r_{0}\sin\left(\Gamma L\right)}{r_{0}+\alpha L}\right]\\
\left[\left(1+\frac{\alpha L}{r_{0}}+\frac{\alpha^{2}}{\Gamma^{2}r_{0}^{2}}\right)\frac{1}{\delta_{0}}\sin\left(\Gamma L\right)-\frac{\alpha^{2}L\cos\left(\Gamma L\right)}{r_{0}^{2}\Gamma\delta_{0}}\right] & \left[\left(1+\frac{\alpha L}{r_{0}}\right)\cos\left(\Gamma L\right)-\frac{\alpha}{\Gamma r_{0}}\sin\left(\Gamma L\right)\right]
\end{array}\right]\left\{ \begin{array}{c}
p_{1}\\
U_{1}
\end{array}\right\} _{L}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\delta_{0}=i\frac{\omega\rho_{m}}{\left(1-f_{\nu}\right)\pi r_{0}^{2}\Gamma}=iZ_{c,0}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
According to sympy, for pR:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{\Gamma p_{L}r_{0}\cos\left(\Gamma L\right)+\eta p_{L}\sin\left(\Gamma L\right)}{\Gamma r_{1}}-iZ_{c,0}\frac{r_{0}\sin\left(\Gamma L\right)}{r_{1}}U_{L}$
\end_inset


\end_layout

\end_inset

, klopt!
\end_layout

\begin_layout Plain Layout
According to sympy, for UR:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
U_{R} & =-i\frac{\Gamma\left(L\eta+r_{0}\right)\left(\Gamma p_{L}r_{0}\sin{\left(\Gamma L\right)}+\left(i\Gamma U_{L}Z_{c0}r_{0}-\eta p_{L}\right)\cos{\left(\Gamma L\right)}\right)+\eta\left(\Gamma p_{L}r_{0}\cos{\left(\Gamma L\right)}-\left(i\Gamma U_{L}Z_{c0}r_{0}-\eta p_{L}\right)\sin{\left(\Gamma L\right)}\right)}{\Gamma^{2}Z_{c0}r_{0}^{2}}\\
\end{align*}

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\mathbf{T}_{\mbox{cone}}=\left[\begin{array}{cc}
\frac{\Gamma r_{0}\cos\left(\Gamma L\right)+\eta\sin\left(\Gamma L\right)}{\Gamma r_{1}} & -iZ_{c,0}\frac{kr_{0}\sin\left(\Gamma L\right)}{\Gamma r_{1}}\\
\frac{iL\eta^{2}\cos\left(\Gamma L\right)}{\Gamma Z_{c0}r_{0}^{2}}-\frac{i}{Z_{c0}}\left(\frac{r_{1}}{r_{0}}+\frac{\eta^{2}}{\Gamma^{2}r_{0}^{2}}\right)\sin\left(\Gamma L\right)\,\,\,\, & \frac{r_{1}}{r_{0}}\cos\left(\Gamma L\right)-\frac{\eta\sin\left(\Gamma L\right)}{\Gamma r_{0}}
\end{array}\right],
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
Z_{c,0}=\frac{\omega\rho_{0}}{\left(1-f_{\nu}\right)S_{f,0}\Gamma_{0}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
and 
\begin_inset Formula $\Gamma$
\end_inset

 is defined in eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Gamma"
plural "false"
caps "false"
noprefix "false"

\end_inset

 and can be approximated as 
\begin_inset Formula $\Gamma\approx k$
\end_inset

 under some undefined and guessed circumstances AANVULLEN, in which 
\begin_inset Formula $k=\frac{\omega}{c_{0}}$
\end_inset

 .
\end_layout

\begin_layout Section
Prismatic lined circular duct
\end_layout

\begin_layout Standard
The Fourier transformed wave equation in axisymmetric cylindrical coordinates
 can be written as:
\begin_inset Formula 
\begin{equation}
\frac{\partial^{2}p}{\partial r^{2}}+\frac{1}{r}\frac{\partial p}{\partial r}+\frac{\partial^{2}p}{\partial x^{2}}+k^{2}p=0,
\end{equation}

\end_inset

Using separation of variables:
\begin_inset Formula 
\begin{equation}
p=\rho(r)\xi(x),
\end{equation}

\end_inset

this can be written as:
\begin_inset Formula 
\begin{equation}
\frac{\rho^{''}}{\rho}+\frac{1}{r}\frac{\rho'}{\rho}+\frac{\xi^{''}}{\xi}+k^{2}=0
\end{equation}

\end_inset

Solutions:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\frac{1}{r}\frac{\rho'}{\rho}+\frac{\rho^{''}}{\rho}=-k^{2}-\frac{\xi^{''}}{\xi}=-\epsilon^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout
Try:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\xi=A\exp\left(\alpha x\right)$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $-k^{2}-\alpha^{2}=-\epsilon^{2}$
\end_inset

 Or: 
\begin_inset Formula $\alpha^{2}=\epsilon^{2}-k^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout
And
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{1}{r}\frac{\rho'}{\rho}+\frac{\rho^{''}}{\rho}=-\epsilon^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout
Means:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $r\frac{\rho'}{\rho}+r^{2}\frac{\rho^{''}}{\rho}+r^{2}\epsilon^{2}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Which has solution:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\rho=J_{0}\left(\epsilon r\right)$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
\xi & =\exp\left(-i\alpha x\right),\\
\rho & =J_{0}\left(\epsilon r\right),
\end{align}

\end_inset

such that the solution for the pressure is:
\begin_inset Formula 
\begin{equation}
p=J_{0}\left(\epsilon r\right)\exp\left(\alpha x\right)
\end{equation}

\end_inset

under the condition:
\begin_inset Formula 
\begin{equation}
\alpha^{2}=k^{2}-\epsilon^{2}.
\end{equation}

\end_inset

At 
\begin_inset Formula $r=R$
\end_inset

 we have the boundary condition that 
\begin_inset Formula $Z_{0}\zeta_{R}u=p$
\end_inset

.
 After filling in and using the rule 
\begin_inset Formula $J_{0}'(x)=J_{-1}(x)$
\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\frac{i\zeta_{R}}{k}\frac{\partial p}{\partial x}|_{r=R}=p|r=R$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{i\zeta_{R}}{k}\epsilon J'_{0}\left(\epsilon r\right)=J_{0}\left(\epsilon r\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Or:
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\epsilon R\frac{J_{-1}\left(\epsilon R\right)}{J_{0}\left(\epsilon R\right)}=-i\upsilon,
\end{equation}

\end_inset

where 
\begin_inset Formula $\upsilon=\frac{kR}{\zeta_{R}}$
\end_inset

.
 This is the characteristic equation for 
\begin_inset Formula $\epsilon R$
\end_inset

.
 Solutions for 
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\Im\left[\epsilon R\right]<3$
\end_inset


\end_layout

\begin_layout Plain Layout
And 
\begin_inset Formula $\Re\left[2\right]<2$
\end_inset


\end_layout

\begin_layout Plain Layout
Using 
\begin_inset Formula $\upsilon=\frac{kR}{\zeta_{R}}$
\end_inset

.
\end_layout

\begin_layout Plain Layout
Solution:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\epsilon R\right)^{2}\approx\frac{96+36i\upsilon\pm\sqrt{9216+2304i\upsilon-912\upsilon^{2}}}{12+i\upsilon}$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling in for 
\begin_inset Formula $U$
\end_inset

:
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\epsilon\approx+\frac{1}{R}\sqrt{\frac{96+36i\upsilon\pm\sqrt{9216+2304i\upsilon-912\upsilon^{2}}}{12+i\upsilon}}
\end{equation}

\end_inset

where 
\begin_inset Formula $0\leq\Re[\epsilon R]\leq2$
\end_inset

 and 
\begin_inset Formula $0\leq\Im\left[\epsilon R\right]\leq3$
\end_inset

 should be satisfied in order to guarantee precision, see Mechel, p.
 630.
\end_layout

\begin_layout Section
Prismatic duct with flow
\end_layout

\begin_layout Itemize
Assuming fully developed plug flow in a duct the linearized governing equations
 in frequency domain read:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
i\omega\rho+\rho_{0}\frac{\mathrm{d}u}{\mathrm{d}x}+u_{0}\frac{\mathrm{d}\rho}{\mathrm{d}x} & =0\\
i\rho_{0}\omega u+\rho_{0}u_{0}\frac{\mathrm{d}u}{\mathrm{d}x}+\frac{\mathrm{d}p}{\mathrm{d}x} & =0\\
p & =c_{0}^{2}\rho
\end{align}

\end_inset


\end_layout

\begin_layout Itemize
With subscript 0 are the mean flow variables.
 Eliminating 
\begin_inset Formula $\rho$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\frac{1}{c_{0}^{2}}\left(i\omega p+u_{0}\frac{\mathrm{d}p}{\mathrm{d}x}\right)+\rho_{0}\frac{\mathrm{d}u}{\mathrm{d}x} & =0\\
\rho_{0}\left(i\omega u+u_{0}\frac{\mathrm{d}u}{\mathrm{d}x}\right)+\frac{\mathrm{d}p}{\mathrm{d}x} & =0
\end{align}

\end_inset


\end_layout

\begin_layout Itemize
Taking spatial derivative of momentum and subtracting the convective derivative
 of the continuity equation from it yields the convective wave equation:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $i\omega\rho_{0}\frac{\mathrm{d}u}{\mathrm{d}x}+u_{0}\rho_{0}\frac{\mathrm{d}^{2}u}{\mathrm{d}^{2}x}+\frac{\mathrm{d}^{2}p}{\mathrm{d}^{2}x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Take the convected time derivative of the continuity:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{2}\frac{1}{c_{0}^{2}}p+\rho_{0}\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)\frac{\mathrm{d}u}{\mathrm{d}x}$
\end_inset


\end_layout

\begin_layout Plain Layout
–
\end_layout

\begin_layout Plain Layout
Subtract:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{2}\frac{1}{c_{0}^{2}}p-\frac{\mathrm{d}^{2}p}{\mathrm{d}^{2}x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Try
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{2}\frac{1}{c_{0}^{2}}p-\frac{\mathrm{d}^{2}p}{\mathrm{d}^{2}x}=0
\end{equation}

\end_inset

For constant 
\begin_inset Formula $u_{0}$
\end_inset

, we try solutions of the form:
\begin_inset Formula 
\begin{equation}
p=A\exp\left(\alpha x\right),
\end{equation}

\end_inset

which yields the characteristic equation for 
\begin_inset Formula $\alpha$
\end_inset

:
\begin_inset Formula 
\begin{equation}
\underbrace{\left(M^{2}-1\right)}_{a}\alpha^{2}+\underbrace{2Mki}_{b}\alpha\underbrace{-k^{2}}_{c}=0,
\end{equation}

\end_inset

where 
\begin_inset Formula $M$
\end_inset

 denotes the Mach number 
\begin_inset Formula $u_{0}/c_{0}$
\end_inset

.
 The solutions for 
\begin_inset Formula $\alpha$
\end_inset

 are:
\begin_inset Formula 
\begin{equation}
\alpha=i\frac{Mk\pm k}{1-M^{2}}=\pm ik\frac{1}{1\mp M}
\end{equation}

\end_inset

Written out:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)^{2}\frac{1}{c_{0}^{2}}A\exp\left(\alpha x\right)-\frac{\mathrm{d}^{2}A\exp\left(\alpha x\right)}{\mathrm{d}^{2}x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(i\omega+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\right)\left(i\omega\frac{1}{c_{0}^{2}}A\exp\left(\alpha x\right)+\frac{u_{0}}{c_{0}^{2}}\alpha A\exp\left(\alpha x\right)\right)-\frac{\mathrm{d}^{2}A\exp\left(\alpha x\right)}{\mathrm{d}^{2}x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(i\omega i\omega\frac{1}{c_{0}^{2}}\exp\left(\alpha x\right)+u_{0}\alpha i\omega\frac{1}{c_{0}^{2}}\exp\left(\alpha x\right)\right)+\left(i\omega\frac{u_{0}}{c_{0}^{2}}\alpha\exp\left(\alpha x\right)+u_{0}\frac{u_{0}}{c_{0}^{2}}\alpha^{2}\exp\left(\alpha x\right)\right)-\frac{\mathrm{d}^{2}\exp\left(\alpha x\right)}{\mathrm{d}^{2}x}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(M^{2}-1\right)\alpha^{2}+2Mk\alpha i-k^{2}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Regular form:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\underbrace{\left(M^{2}-1\right)}_{a}\alpha^{2}+\underbrace{2Mki}_{b}\alpha\underbrace{-k^{2}}_{c}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Solutions are:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\alpha=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-2Mki\pm\sqrt{\left(2Mki\right)^{2}+4\left(M^{2}-1\right)k^{2}}}{2\left(M^{2}-1\right)}=\frac{\pm ik-Mki}{\left(M^{2}-1\right)}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
p=A\exp\left(-\frac{ik}{1+M}x\right)+B\exp\left(\frac{ik}{1-M}x\right),
\end{align}

\end_inset

and the volume flow:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{\mathrm{d}u}{\mathrm{d}x}=-\frac{1}{\rho_{0}c_{0}^{2}}\left(i\omega\left(A\exp\left(-\frac{ik}{1+M}x\right)+B\exp\left(\frac{ik}{1-M}x\right)\right)+u_{0}\frac{\mathrm{d}}{\mathrm{d}x}\left(A\exp\left(-\frac{ik}{1+M}x\right)+B\exp\left(\frac{ik}{1-M}x\right)\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\mathrm{d}u}{\mathrm{d}x}=-\frac{1}{\rho_{0}c_{0}^{2}}i\omega\left(A\exp\left(-\frac{ik}{1+M}x\right)+B\exp\left(\frac{ik}{1-M}x\right)\right)+-\frac{1}{\rho_{0}c_{0}^{2}}u_{0}\left(-\frac{ik}{1+M}A\exp\left(-\frac{ik}{1+M}x\right)+B\frac{ik}{1-M}\exp\left(\frac{ik}{1-M}x\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $u=\frac{A}{z_{0}}\exp\left(-\frac{ik}{1+M}x\right)-\frac{B}{z_{0}}\exp\left(\frac{ik}{1-M}x\right)+C$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Section
Cremers impedance
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\frac{kR}{\zeta}=2.9803824+1.2796025i
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Or:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\zeta=y_{cr}\pi\frac{kR}{y_{cr}\pi}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\zeta=kR\left(0.28-0.12i\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Attenuation reached when the liner impedance equals Cremer's impedance is
 around 15 dB per unit of radius maximum.
 It decreases with increasing frequency, when 
\begin_inset Formula $fR\approx100$
\end_inset

.
\end_layout

\begin_layout Subsection
Locally reacting lining with back-volume
\end_layout

\begin_layout Standard
Impedance of concentric liner, outer radius is 
\begin_inset Formula $R_{o}$
\end_inset

, inner radius is 
\begin_inset Formula $R_{i}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\zeta_{\mathrm{back}}=i\frac{H_{0}^{(1)}\left(kR_{i}\right)-\frac{H_{1}^{(1)}\left(kR_{o}\right)}{H_{1}^{(2)}\left(kR_{o}\right)}H_{0}^{(2)}\left(kR_{i}\right)}{H_{1}^{(1)}\left(kR_{i}\right)-\frac{H_{1}^{(1)}\left(kR_{o}\right)}{H_{1}^{(2)}\left(kR_{o}\right)}H_{1}^{(2)}\left(kR_{i}\right)}
\end{equation}

\end_inset

Such that the total impedance is
\begin_inset Formula 
\begin{equation}
\zeta=\zeta_{\mathrm{back}}+\zeta_{\mathrm{MPP}}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Cavity silencer
\end_layout

\begin_layout Standard
-
\end_layout

\begin_layout Section
Compliance volume
\begin_inset CommandInset label
LatexCommand label
name "subsec:Compliance-volume"

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/volume.pdf
	width 30text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic of the compliance volume segment.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:compliance"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:compliance"

\end_inset

 gives a schematic of the compliance volume.
 A compliance volume is implemented in the 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 code in the 
\family typewriter
Volume
\family default
 class.
 A compliance volume is a volume (tank) which is small compared to the wavelengt
h.
 Hence, we can assume that the acoustic pressure is constant throughout
 the volume 
\begin_inset Formula $V$
\end_inset

.
 As thermal relaxation still occurs, the model for this segment takes into
 account thermal relaxation due to temperature oscillations.
 The basic assumptions behind the model are:
\end_layout

\begin_layout Itemize
The characteristic length scale of volume is small compared to the wavelength.
\end_layout

\begin_layout Itemize
The characteristic length scale of volume is large compared to thermal penetrati
on depth.
\end_layout

\begin_layout Standard
The lower the frequency, the more the second assumption is violated, while
 the higher the frequency, the more the first assumption is violated.
 In practice, violating the first assumption has a larger impact.
 For a compliance, the following governing equations can be derived 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 156"
key "ward_deltaec_2017"
literal "true"

\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Derivation of the capacitance:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=U_{L}-i\frac{k}{z_{0}}\left(V-\frac{i}{2}\frac{\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)p,$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{U_{R}}{i\omega}=\frac{U_{L}}{i\omega}-\frac{1}{z_{0}c_{0}}\left(V-\frac{i}{2}\frac{\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)p$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{U_{R}}{i\omega}=\frac{U_{L}}{i\omega}-\frac{1}{z_{0}c_{0}}\left(V-\frac{i}{2}\frac{\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)p$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\xi_{L}-\xi_{R}=\frac{1}{z_{0}c_{0}}\left(V-\frac{i}{2}\frac{\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)p$
\end_inset


\end_layout

\begin_layout Plain Layout
Using 
\begin_inset Formula $C=\frac{\xi_{L}-\xi_{R}}{p}$
\end_inset

 in that case: 
\begin_inset Formula $C=\frac{1}{z_{0}c_{0}}\left(V-\frac{i}{2}\frac{\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Such that:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=U_{L}-i\omega C_{c}p$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
p_{L} & =p=p_{R},\\
U_{R} & =U_{L}-i\omega C_{c}p,
\end{align}

\end_inset

in which 
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$C_c$"
description "Acoustic capacitance of a compliance volume\\nomunit{\\si{\\cubic\\metre\\per\\pascal}}"
literal "true"

\end_inset


\begin_inset Formula $C_{c}$
\end_inset

 is the acoustic 
\begin_inset Quotes eld
\end_inset

capacitance
\begin_inset Quotes erd
\end_inset

:
\begin_inset Formula 
\begin{equation}
C_{c}=\frac{1}{z_{0}c_{0}}\left(V+\frac{1}{2}\frac{\left(1-i\right)\left(\gamma-1\right)}{1+\epsilon_{s,0}}S\delta_{\kappa}\right)
\end{equation}

\end_inset

where 
\begin_inset Formula $V$
\end_inset

 is the volume, 
\begin_inset Formula $S$
\end_inset

 the surface area of the volume in contact with a wall, and
\begin_inset Formula 
\begin{equation}
\epsilon_{s,0}=\sqrt{\frac{\kappa\rho_{0}c_{p}}{\kappa_{s}\rho_{s}c_{s}}}.
\end{equation}

\end_inset

It should be noticed that in practice, a compliance volume often functions
 as the end of an acoustic system.
 In that case, either 
\begin_inset Formula $U_{L}$
\end_inset

 or 
\begin_inset Formula $U_{R}$
\end_inset

 is 0.
\end_layout

\begin_layout Section
Membrane
\end_layout

\begin_layout Standard
A membrane is a mechanical 
\end_layout

\begin_layout Section
Circular plate membrane
\end_layout

\begin_layout Standard
series_impedance/class CircPlateMembrane(SeriesImpedance)
\end_layout

\begin_layout Standard
A thin circular plate can be modeled using CircPlateMembrane.
 It behaves like an acoustic compliance.
 A typical use is the attenuation of acoustic pressure by combining it with
 an enclosed volume.
\end_layout

\begin_layout Standard
Two boundary condition cases can be applied: fixed/clamped edges and simply
 supported edges.
 The general equation for the static displacement of the plate is given
 by 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 487"
key "young_roarks_2002"
literal "false"

\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
y\left(r\right)=y_{c}+\frac{M_{c}r^{2}}{2D\left(1+\nu\right)}+LT_{y}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $y_{c}$
\end_inset

 and 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none

\begin_inset Formula $M_{c}$
\end_inset

 are
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
 the displacement and moment at the center of the plate, 
\begin_inset Formula $LT_{y}$
\end_inset

 is the load term in the y-direction, 
\begin_inset Formula $\nu$
\end_inset

 is the Poisson's ratio of the plate material and 
\begin_inset Formula $D$
\end_inset

 is the flexural stiffness of the plate, which is given by the equation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
D=\frac{Et^{3}}{12\left(1-\nu^{2}\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $E$
\end_inset

 is the Young's modulus of the plate material and 
\begin_inset Formula $t$
\end_inset

 is the plate thickness.
 Substituting 
\begin_inset Formula $D$
\end_inset

, 
\begin_inset Formula $y_{c}$
\end_inset

, 
\begin_inset Formula $M_{c}$
\end_inset

 and 
\begin_inset Formula $LT_{y}$
\end_inset

 for this specific load case (uniform load/pressure) and boundary conditions
 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 458 & p. 488"
key "young_roarks_2002"
literal "false"

\end_inset

 and simplifying yields the following equations for the static plate deflection:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
In these equations the distributed load 
\begin_inset Formula $q$
\end_inset

 is replaced by 
\begin_inset Formula $-p$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
y_{ss}\left(r\right)=\frac{3p\left(1-\nu^{2}\right)}{16Et^{3}\left(1+\nu\right)}\left(a^{2}\left[a^{2}\left\{ 5+\nu\right\} -2r^{2}\left\{ 3+\nu\right\} \right]+r^{4}\left[1+\nu\right]\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
y_{fix}\left(r\right)=\frac{3p\left(1-\nu^{2}\right)}{16Et^{3}}\left(a^{4}-2a^{2}r^{2}+r^{4}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
In which 
\begin_inset Formula $a$
\end_inset

 is the radius of the plate and 
\begin_inset Formula $r$
\end_inset

 is the radial coordinate.
 The static acoustic compliance of the plate is given by the equation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
c_{stat}\left(r\right)=\frac{y\left(r\right)}{p}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The static acoustic volume compliance for both cases can be calculated by
 integrating over the surface of the plate:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
C_{stat}=2\pi\int_{0}^{a}c_{stat}\left(r\right)rdr
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Performing this integration for both boundary condition cases yields:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
C_{stat,ss}=\frac{\pi a^{6}}{16Et^{3}}\left(7-6\nu-\nu^{2}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
C_{stat,fix}=\frac{\pi a^{6}}{16Et^{3}}\left(1-\nu^{2}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The dynamic acoustic volume compliance of the plate is given by the equation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
C_{dyn}\left(f\right)=\frac{C_{stat}}{1-\left(\frac{f}{f_{r}}\right)^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $f$
\end_inset

 is the frequency in Hz and 
\begin_inset Formula $f_{r}$
\end_inset

 is the resonance frequency of the plate in Hz.
 The resonance frequency for the simply supported plate is given by the
 equation 
\begin_inset CommandInset citation
LatexCommand citeyear
key "calcdevice"
literal "false"

\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
This is an approximation from an online calculator.
 A more exact equation like the one for the fxed case should be found.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
f_{r,ss}=\frac{0.8}{a^{2}}\sqrt{\frac{D}{\rho t}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $\rho$
\end_inset

 is the density of the plate material.
 The resonance frequency for the fixed plate is given by the equation 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 430"
key "leniowska_plate_resonance_1999"
literal "false"

\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
f_{r,fix}=\frac{\gamma_{1}^{2}}{a^{2}}\sqrt{\frac{D}{\rho t}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $\gamma_{1}$
\end_inset

 is the first solution to the following equation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
J_{0}\left(\gamma_{m}\right)I_{1}\left(\gamma_{m}\right)+J_{1}\left(\gamma_{m}\right)I_{0}\left(\gamma_{m}\right)=0\label{eq:gamma}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $J_{n}\left(\gamma_{m}\right)$
\end_inset

 and 
\begin_inset Formula $I_{n}\left(\gamma_{m}\right)$
\end_inset

 are the Bessel function of the first kind and modified Bessel functions
 of order 
\begin_inset Formula $n$
\end_inset

.
 Solving equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:gamma"
plural "false"
caps "false"
noprefix "false"

\end_inset

 yields 
\begin_inset Formula $\gamma_{1}=3.196$
\end_inset

.
 The impedance is given by the equation:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
Z_{cpm}\left(f\right)=\frac{1}{i2\pi fC_{dyn}\left(f\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Exponential horn
\end_layout

\begin_layout Standard
hon.py/class ExpHorn(Duct)
\end_layout

\begin_layout Standard
An exponential horn can be modelled using ExpHorn.
 The transfer matrix is derived from Webster's horn equation.
 The solution for an exponential horn is
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p\left(x\right)=C_{1}e^{E_{1}x}+C_{2}e^{E_{2}x},\qquad U\left(x\right)=i\frac{S\left(x\right)}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}x}+C_{2}E_{2}e^{E_{2}x}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $S(x)$
\end_inset

 is the local cross-sectional area, 
\begin_inset Formula $\rho$
\end_inset

 is the air density, 
\begin_inset Formula $\omega$
\end_inset

 is the angular frequency.
 The exponent terms 
\begin_inset Formula $E_{1}$
\end_inset

 and 
\begin_inset Formula $E_{2}$
\end_inset

 are given by
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $k$
\end_inset

 is the wave number and 
\begin_inset Formula $m$
\end_inset

 is the flare rate, which is defined as
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
with 
\begin_inset Formula $L$
\end_inset

 being the horn length and 
\begin_inset Formula $S_{R}$
\end_inset

 and 
\begin_inset Formula $S_{L}$
\end_inset

 being the cross-sectional areas on the right-hand and left-hand side of
 the horn respectively.
 
\begin_inset Newline newline
\end_inset


\begin_inset Newline newline
\end_inset

By setting the solutions on 
\begin_inset Formula $x=0$
\end_inset

 to 
\begin_inset Formula $p_{L}$
\end_inset

 and 
\begin_inset Formula $U_{L}$
\end_inset

 and those on 
\begin_inset Formula $x=L$
\end_inset

 to 
\begin_inset Formula $p_{R}$
\end_inset

 and 
\begin_inset Formula $U_{R}$
\end_inset

, the following transfer matrix can be derived:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc}
E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L} & i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)\\
i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right) & -\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)
\end{array}\right]
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$
\end_inset


\begin_inset Newline newline
\end_inset


\begin_inset Newline newline
\end_inset

Solve for constants
\begin_inset Newline newline
\end_inset


\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$
\end_inset


\begin_inset Newline newline
\end_inset


\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$
\end_inset


\begin_inset Newline newline
\end_inset


\begin_inset Newline newline
\end_inset

Fill in in solution:
\begin_inset Newline newline
\end_inset


\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$
\end_inset


\begin_inset Newline newline
\end_inset


\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Section
Holes in plate
\end_layout

\begin_layout Standard
series_impedance.py/class CircHoleNeck(SeriesImpedance)
\end_layout

\begin_layout Standard
A plate with several holes can be modelled using CircHoleNeck.
 It behaves like an acoustic mass with losses and can represent the neck
 of a Helmholtz resonator.
 Typical uses are to connect volumes to eachother or volumes to ducts, to
 form Helmholtz resonators.
\end_layout

\begin_layout Standard
Limitations are that hole-hole interaction is neglected and that the resistance
 term is an approximation for holes with diameter >> length.
\end_layout

\begin_layout Standard
Impedance is given by the equation:
\end_layout

\begin_layout Standard
\noindent
\align center
\begin_inset Formula 
\begin{equation}
Z_{holes}=\frac{1}{N_{h}}\left(R_{v}+i\omega M_{A}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $N_{h}$
\end_inset

 is the number of holes, 
\begin_inset Formula $R_{v}$
\end_inset

 the acoustic resistance as described in equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Rv_hole"
plural "false"
caps "false"
noprefix "false"

\end_inset

, 
\begin_inset Formula $\omega$
\end_inset

 the angular frequency and 
\begin_inset Formula $m_{a}$
\end_inset

 the acoustic mass as described in equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:acoustic_mass"
plural "false"
caps "false"
noprefix "false"

\end_inset

, except without Karal's discontinuity factor.
\end_layout

\begin_layout Section
End corrections and discontinuities
\begin_inset CommandInset label
LatexCommand label
name "subsec:End-corrections-and"

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/discontinuity.pdf
	width 60text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic of a waveguide discontinuity.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:karal"

\end_inset


\end_layout

\end_inset

For discontinuities in the cross section of a waveguide, and the case of
 inviscid adiabatic wave propagation, an exact expression is available for
 the added acoustic mass 
\begin_inset CommandInset citation
LatexCommand cite
key "karal_analogous_1953"
literal "true"

\end_inset

.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:karal"

\end_inset

 gives a schematic of the situation.
 The model is implemented in the 
\family typewriter
Discontinuity
\family default
 class in the 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 code.
  The assumptions behind the model are:
\end_layout

\begin_layout Itemize
Both tubes on either side of the discontinuity are cylindrical.
 The tubes are co-axially connected.
\end_layout

\begin_layout Itemize
The wavelength is larger than transverse characteristic length scale (no
 propagating modes expect for the plane waves).
\end_layout

\begin_layout Itemize
Other discontinuities are far away from the current one.
\end_layout

\begin_layout Itemize
Inviscid and adiabatic wave propagation (Helmholtz equation).
\end_layout

\begin_layout Standard
The ratio of tube radii 
\begin_inset Formula $a_{L}/a_{R}$
\end_inset

 is denoted by 
\begin_inset Formula $\alpha$
\end_inset


\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\alpha$"
description "Ratio of tube radii\\nomunit{-}"
literal "true"

\end_inset

.
 It turns out that a surface area discontinuity only generates an acoustic
 pressure discontinuity.
 The volume flow is preserved.
 Hence:
\begin_inset Formula 
\begin{align}
U_{R} & =U_{L}\\
p_{R} & =p_{L}-i\omega M_{A}U_{L}
\end{align}

\end_inset

where 
\begin_inset Formula $M_{A}$
\end_inset

 is the so-called added acoustic mass in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
si{
\backslash
kg
\backslash
per
\backslash
metre
\backslash
tothe{4}}
\end_layout

\end_inset

, which equals
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$M_A$"
description "Acoustic mass\\nomunit{\\si{\\kg\\per\\metre\\tothe{4}}}"
literal "true"

\end_inset


\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$a$"
description "Tube radius\\nomunit{\\si{\\metre}}"
literal "true"

\end_inset


\begin_inset Formula 
\begin{equation}
M_{A}=\chi(\alpha,k)\frac{8\rho_{0}}{3\pi^{2}a_{L}},\label{eq:acoustic_mass}
\end{equation}

\end_inset

where 
\begin_inset Formula $\chi$
\end_inset

 
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\chi$"
description "Karal's discontinuity factor\\nomunit{-}"
literal "true"

\end_inset

is Karal's discontinuity factor, which is in general a function of the tube
 radii and the wave number.
\end_layout

\begin_layout Standard
For 
\begin_inset Formula $\lambda\gg a_{R}$
\end_inset

, the dependency of 
\begin_inset Formula $\chi$
\end_inset

 on the wave number 
\begin_inset Formula $k$
\end_inset

 can be neglected, which lowers the computational burden significantly,
 as 
\begin_inset Formula $\chi$
\end_inset

 has to be computed only once.
 For the case 
\begin_inset Formula $\alpha\to0$
\end_inset

 (by letting 
\begin_inset Formula $a_{R}\to\infty$
\end_inset

), 
\begin_inset Formula $\chi\to1$
\end_inset

.
 In case of 
\begin_inset Formula $\alpha\to1$
\end_inset

, the acoustic mass gradually reduces to zero as 
\begin_inset Formula $\chi\to0$
\end_inset

.
 When 
\begin_inset Formula $\alpha=1$
\end_inset

, there is no continuity left, such that 
\begin_inset Formula $M_{A}=0$
\end_inset

.
\end_layout

\begin_layout Standard
The derivation of the coefficient 
\begin_inset Formula $\chi$
\end_inset

 is documented in Appendix 
\begin_inset CommandInset ref
LatexCommand ref
reference "chap:Derivation-of-Karal's"

\end_inset

, except of the following information.
 To solve the curve of 
\begin_inset Formula $\chi$
\end_inset

, a system of infinite equations has to be solved for an infinite number
 of unknowns.
 In the 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\end_layout

\end_inset

 code, as a standard this system is truncated up to 
\begin_inset Formula $N=$
\end_inset

100 equations and 100 unknowns.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:chi_vs_alpha"

\end_inset

 shows the effect of truncating this infinite system of equations.
 As visible for the case of 100 equations, the curves start to deviate from
 each other for lower values of 
\begin_inset Formula $\alpha$
\end_inset

.
 Assuming that convergence is obtained as 
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$N$"
description "Number\\nomunit{-}"
literal "true"

\end_inset

 
\begin_inset Formula $N\to\infty$
\end_inset

, the curve of 
\begin_inset Formula $N=100$
\end_inset

 has acceptable accuracy for 
\begin_inset Formula $\alpha>0.07$
\end_inset

.
 To limit possible faulty results, the 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubess
\end_layout

\end_inset

 code gives a warning when the tube ratio is chosen such that an invalid
 
\begin_inset Formula $\chi$
\end_inset

 is computed.
 When an 
\begin_inset Formula $\alpha<0.07$
\end_inset

 is desired, the user should choose a higher value of 
\begin_inset Formula $N$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/chi_vs_alpha.pdf
	width 90text%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
\begin_inset Formula $\chi$
\end_inset

 vs 
\begin_inset Formula $\alpha$
\end_inset

 for different truncations 
\begin_inset Formula $\left(N\right)$
\end_inset

 of the infinite system of equations.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:chi_vs_alpha"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Section
Hard wall
\end_layout

\begin_layout Standard
A hard wall is the wall perpendicular to the wave propagation direction.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:hardwall"

\end_inset

 shows the schematic configuration for this segment.
 Due to thermal relaxation a hard wall consumes acoustic energy is consumed.
 The hard wall segment models this thermal relaxation loss.
 The assumptions behind the model are:
\end_layout

\begin_layout Itemize
Normal incident waves.
\end_layout

\begin_layout Itemize
Uniform normal velocity.
\end_layout

\begin_layout Itemize
The wavelength is much larger than the thermal penetration depth (
\begin_inset Formula $\lambda\gg\delta_{\kappa}$
\end_inset

).
\end_layout

\begin_layout Standard
We can derive the following impedance boundary condition 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 157"
key "ward_deltaec_2017"
literal "true"

\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Delta EC User guide:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
U_{R}=U_{L}-\frac{\omega p}{\rho_{0}c_{0}^{2}}\frac{\gamma-1}{1+\epsilon_{s}}S\frac{\delta_{\kappa}}{2}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
Or:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
U_{L}=\frac{k}{z_{0}}\frac{\gamma-1}{1+\epsilon_{s}}S\frac{\delta_{\kappa}}{2}p
\]

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
U=k\delta_{\kappa}\frac{S}{z_{0}}\frac{\left(\gamma-1\right)\left(1+i\right)}{2\left(1+\epsilon_{s}\right)}p.
\end{equation}

\end_inset

Hence the impedance of a hard wall scales with 
\begin_inset Formula $Z\sim Z_{0}\frac{\lambda}{\delta_{\kappa}}$
\end_inset

.
 For 1 kHz, this results in 
\begin_inset Formula $\sim4100Z_{0}$
\end_inset

, which is practically already close to 
\begin_inset Formula $\infty$
\end_inset

.
 Except for really high frequencies this segment can often be replaced with
 a boundary condition of 
\begin_inset Formula $U=0$
\end_inset

.
 An important point to make here is that this boundary condition is inconsistent
 with the LRF solution for 1D wave propagation in ducts, as the velocity
 profile in a duct is not uniform.
 This is especially true for the case of small ducts where 
\begin_inset Formula $r_{h}\sim\delta$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/hardwall.pdf
	width 50text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic of a hard acoustic wall where the thermal boundary layer dissipates
 a bit of the acoustic energy (
\begin_inset Formula $Z\neq\infty$
\end_inset

).
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:hardwall"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Section
Spherical wave propagation models
\end_layout

\begin_layout Standard
For spherical waves, the Helmholtz equation reads
\begin_inset Formula 
\begin{equation}
\left(\frac{\mathrm{d}^{2}}{\mathrm{d}r^{2}}+\frac{2}{r}\frac{\mathrm{d}}{\mathrm{d}r}+\Gamma^{2}\right)p=0.\label{eq:hh_spher}
\end{equation}

\end_inset

The solution of Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:hh_spher"

\end_inset

 reads:
\begin_inset Formula 
\begin{equation}
p=\frac{C_{1}\exp\left(-i\Gamma r\right)+C_{2}\exp\left(-i\Gamma r\right)}{r}.
\end{equation}

\end_inset

The acoustic volume flow can be computed as
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $u_{r}=\frac{i}{\omega\rho_{0}}\frac{\mathrm{d}p}{\mathrm{d}r}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $u_{r}=\frac{i}{kz_{0}}\frac{\mathrm{d}p}{\mathrm{d}r}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{r}=4\pi r^{2}\alpha\frac{i}{kz_{0}}\frac{\mathrm{d}p}{\mathrm{d}r}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
U=i\frac{\alpha4\pi r^{2}}{\Gamma z_{c}}\frac{\mathrm{d}p}{\mathrm{d}r},
\end{equation}

\end_inset

where 
\begin_inset Formula $\alpha=1$
\end_inset

 for a full sphere and 
\begin_inset Formula $\alpha=\frac{1}{2}$
\end_inset

 for a hemisphere.
 We can derive the following transfer matrix for 
\begin_inset Formula $p$
\end_inset

 and 
\begin_inset Formula $U$
\end_inset

:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{C_{1}\exp\left(-i\Gamma r_{L}\right)+C_{2}\exp\left(-i\Gamma r_{L}\right)}{\Gamma r_{L}}=p_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{C_{1}\exp\left(-i\Gamma r_{R}\right)+C_{2}\exp\left(-i\Gamma r_{R}\right)}{\Gamma r_{R}}=p_{R}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(U_{L}\frac{e^{-i\Gamma\left(r_{L}+r_{R}\right)}}{8\pi\Gamma\alpha r_{L}r_{R}}\left(i\Gamma z_{c}e^{2i\Gamma r_{L}}-i\Gamma z_{c}e^{2i\Gamma r_{R}}\right)+p_{L}\left(\frac{r_{L}e^{i\Gamma\left(r_{L}+r_{R}\right)}}{2r_{R}}+\frac{i}{2\Gamma r_{R}}\left(e^{i\Gamma\left(r_{L}-r_{R}\right)}-e^{i\Gamma\left(r_{R}+r_{L}\right)}\right)\right)\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{R}=\frac{iU_{L}z_{c}}{4\pi\alpha r_{L}r_{R}}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)+p_{L}\left[\frac{r_{L}}{r_{R}}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)-\frac{1}{\Gamma r_{R}}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
and:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{R}=U_{L}\left(\frac{r_{R}}{r_{L}}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)+\frac{1}{\Gamma r_{L}}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)\right)+\frac{4i\pi\alpha}{z_{c}}p_{L}\left[\left(r_{L}r_{R}+\frac{1}{\Gamma^{2}}\right)\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)+\frac{\left(r_{R}-r_{L}\right)}{\Gamma}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)\right]$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{R}=\left[\begin{array}{cc}
M_{11} & M_{12}\\
M_{21} & M_{22}
\end{array}\right]\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{L},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{align}
M_{11} & =\frac{r_{L}}{r_{R}}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)-\frac{1}{\Gamma r_{R}}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right),\\
M_{12} & =\frac{iz_{c}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)}{4\pi\alpha r_{L}r_{R}},\\
M_{21} & =\frac{4\pi i\alpha}{z_{c}}\left[\left(r_{L}r_{R}+\frac{1}{\Gamma^{2}}\right)\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right)+\frac{r_{R}-r_{L}}{\Gamma}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)\right]\\
M_{22} & =\frac{r_{R}}{r_{L}}\cos\left(\Gamma\left(r_{L}-r_{R}\right)\right)+\frac{1}{\Gamma r_{L}}\sin\left(\Gamma\left(r_{L}-r_{R}\right)\right),
\end{align}

\end_inset


\end_layout

\begin_layout Section
Boundary conditions
\end_layout

\begin_layout Subsection
Radiation impedance of a baffled piston
\end_layout

\begin_layout Itemize
\begin_inset Formula $a$
\end_inset

: radius of the exit [m]
\end_layout

\begin_layout Itemize
\begin_inset Formula $S$
\end_inset

: 
\begin_inset Formula $\pi a^{2}$
\end_inset

 cross sectional area [m
\begin_inset Formula $^{2}$
\end_inset

]
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p=Z_{\mathrm{rad}}U,
\end{equation}

\end_inset


\begin_inset Formula 
\begin{equation}
Z_{\mathrm{rad}}=\frac{z_{0}}{S}\left[1-\frac{2J_{1}\left(2ka\right)}{2ka}+i\frac{2H_{1}(2ka)}{2ka}\right]
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
In the low frequency range, a power series expansion of 
\begin_inset Formula $H_{1}$
\end_inset

 yields [Aarts]:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
H_{1}(x)=\frac{2}{\pi}\left[\frac{x^{2}}{3}-\frac{x^{4}}{45}+\frac{x^{6}}{1575}-\dots\right]
\end{equation}

\end_inset

Filling this in, we obtain the following low-frequency approximation to
 
\begin_inset Formula $Z_{\mathrm{rad}}$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
Z_{\mathrm{rad}}=\frac{z_{0}}{S}\left[i\frac{8ka}{3\pi}+\frac{1}{2}\left(ka\right)^{2}+\mathcal{O}\left(\left(ka\right)^{3}\right)\right]\label{eq:Zrad-baffled-piston}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Incident plane wave on small port in infinite baffle
\end_layout

\begin_layout Standard
Situation: an acoustic system, which is connected to the outside world though
 a port, ending in an infinite wall
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:bc_planewave_port"
plural "false"
caps "false"
noprefix "false"

\end_inset

.
 There is an incident plane wave with specified amplitude and frequency.
 It would be beneficial for computing time to replace the outside world
 by a boundary condition on the port.
 Here it is approached as a scattering problem.
 More information is described in 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 132-134"
key "zwikker_sound_1949"
literal "false"

\end_inset

.
 The pressure field can be written as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{t}=p_{i}+p_{s}\label{eq:scattering-problem}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $p_{t}$
\end_inset

 is the total pressure field, 
\begin_inset Formula $p_{i}$
\end_inset

 the incident pressure field (the field as if there were only an infinite
 wall) and 
\begin_inset Formula $p_{s}$
\end_inset

 the scattered pressure field.
 The combination of the incident and scattered field combined result in
 the total pressure field.
 All depend on both position and time (or frequency).
 If only the infinite wall is taken into account and the port and system
 behind it are ignored, the amplitude of the incident plane wave is:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{i}(x,\omega)=\begin{cases}
P_{i}\cdot\cos(kx) & x<=0\\
0 & x>0
\end{cases}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $P_{i}$
\end_inset

 is the amplitude of the incident plane wave at the wall (resulting in sound
 pressure 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none

\begin_inset Formula $P_{i}$
\end_inset


\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
 on the surface of a reflecting wall), 
\begin_inset Formula $k$
\end_inset

 is the wave number and 
\begin_inset Formula $x$
\end_inset

 the position into the wall.
 There is no scattered pressure field, so this is the total pressure field
 right away.
 When the port and system behind it are added, the total pressure field
 is no longer equal to the incident pressure field: a correction must be
 added, which is captured in 
\begin_inset Formula $p_{s}$
\end_inset

.
 The correction is due to the air slug within the port moving.
 At 
\begin_inset Formula $x<0$
\end_inset

, this has the same effect as a baffled piston.
 On the condition that the wavelength is much larger than the port size,
 the scattered field near the boundary (but still outside of the port) is
 given by:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{s}(x=0^{-})=-Z_{\mathrm{rad}}U
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $Z_{\mathrm{rad}}$
\end_inset

 is the radiation impedance of a baffled piston and 
\begin_inset Formula $U$
\end_inset

 is the acoustic volume flow rate.
 Note the minus sign, which stems from the direction in which 
\begin_inset Formula $U$
\end_inset

 is defined.
 The same convention is taken as in COMSOL: velocity 
\begin_inset Formula $v$
\end_inset

 is positive when inwards, so inwards 
\begin_inset Formula $U$
\end_inset

 is positive.
 Filling in equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:scattering-problem"
plural "false"
caps "false"
noprefix "false"

\end_inset

, just outside of the port at 
\begin_inset Formula $x=0^{-}$
\end_inset

, yields:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{t}(x=0^{-})=P_{i}-Z_{rad}U
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
It is questionable whether the port acoustically ends at the boundary, so
 this might be an approximation.
 In COMSOL, the pressure is continuous, to it is fine to apply it at 
\begin_inset Formula $x=0$
\end_inset

 instead of 
\begin_inset Formula $x=0^{-}$
\end_inset

.
 
\begin_inset Formula $U$
\end_inset

 can be found by integrating the inner product of velocity and the normal
 vector over the boundary, while adding a minus sign because the normal
 vector points outwards.
 In COMSOL it is more convenient to use 
\emph on
specific
\emph default
 impedances and 
\emph on
velocities
\emph default
.
 Then the equation is slightly modified to:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{t}(x=0)=P_{i}-z_{\mathrm{rad}}v\label{eq:bc-planewave-port-pressure}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $z_{\mathrm{rad}}$
\end_inset

 is the specific radiation impedance of a baffled piston and 
\begin_inset Formula $v$
\end_inset

 the acoustic velocity (inwards).
 This equation can be applied as a 
\emph on
pressure 
\emph default
boundary condition in COMSOL.
 The required 
\begin_inset Formula $v$
\end_inset

 can be 
\begin_inset Quotes eld
\end_inset

measured
\begin_inset Quotes erd
\end_inset

 by averaging the normal component of the velocity and adding a minus sign
 to make it inwards.
 Alternatively, the equation can be solved for 
\begin_inset Formula $v$
\end_inset

 to obtain a 
\emph on
velocity 
\emph default
boundary condition:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
v=\frac{P_{i}-p_{t}(x=0)}{z_{\mathrm{rad}}}\label{eq:bc-planewave-port-velocity}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $p_{t}(x=0)$
\end_inset

 can be 
\begin_inset Quotes eld
\end_inset

measured
\begin_inset Quotes erd
\end_inset

 by averaging it over the port's boundary.
 The LRFTubes implementation of this 
\emph on
mixed
\emph default
 boundary condition is for a left wall:
\begin_inset Formula 
\begin{equation}
p_{L}+Z_{\mathrm{rad}}U_{L}=P_{i},
\end{equation}

\end_inset

and the same on a right wall:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{R}-Z_{\mathrm{rad}}U_{R}=P_{i}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
TO DO: redraw image and list what approximations are used
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/bc_planewave_port.jpg
	lyxscale 10
	width 50text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic view of incident wave (green) on an infinite wall (blue) containing
 a port with a system connected to it.
 The location of the boundary condition is shown in red.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:bc_planewave_port"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Chapter
Thermoacoustic segments
\end_layout

\begin_layout Standard
For relatively small temperature gradients, Swift's thermoacoustic equations
 
\begin_inset CommandInset citation
LatexCommand cite
after "p. 91"
key "swift_thermoacoustics:_2003"
literal "false"

\end_inset

:
\begin_inset Formula 
\begin{align}
\frac{\mathrm{d}p}{\mathrm{d}x} & =-\frac{-i\omega p_{m}}{R_{s}T_{m}S_{\mathrm{gas}}\left(1-f_{\nu}\right)}U,\\
\frac{\mathrm{d}U}{\mathrm{d}x} & =\frac{-i\omega S_{\mathrm{gas}}}{\gamma p_{m}}\left[1+\left(\gamma-1\right)f_{\kappa}\right]p+\frac{f_{\kappa}-f_{\nu}}{\left(1-f_{\nu}\right)\left(1-\Pr\right)}\frac{1}{T_{m}}\frac{\mathrm{d}T_{m}}{\mathrm{d}x}U,
\end{align}

\end_inset

can be integrated.
 Assuming 
\begin_inset Formula $\frac{\mathrm{d}T_{m}}{\mathrm{d}x}L\ll T_{m}$
\end_inset

.
 Then we find for the solution::
\begin_inset Formula 
\begin{equation}
p(x)=C_{1}\exp\left(\Gamma_{1}x\right)+C_{2}\exp\left(\Gamma_{2}x\right),
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
\Gamma_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\label{eq:Gammasol}
\end{equation}

\end_inset

wheren 
\begin_inset Formula $1$
\end_inset

 denotes to the 
\begin_inset Formula $+$
\end_inset

 and 2 to the 
\begin_inset Formula $-$
\end_inset

 sign.
 In Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Gammasol"

\end_inset

, 
\begin_inset Formula $a,$
\end_inset


\begin_inset Formula $b,$
\end_inset

 and 
\begin_inset Formula $c$
\end_inset

 are defined as:
\begin_inset Formula 
\[
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{\left(\Gamma_{1}f_{\nu}p_{L}-\Gamma_{1}p_{L}+iU_{L}\omega\rho_{m}\right)e^{\Gamma_{2}L}+\left(-\Gamma_{2}f_{\nu}p_{L}+\Gamma_{2}p_{L}-iU_{L}\omega\rho_{m}\right)e^{\Gamma_{1}L}}{\Gamma_{1}f_{\nu}-\Gamma_{1}-\Gamma_{2}f_{\nu}+\Gamma_{2}}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\begin{array}{c}
p_{R}\\
U_{R}
\end{array}=\left[\begin{array}{cc}
\frac{\left(\Gamma_{1}f_{\nu}p_{L}-\Gamma_{1}p_{L}\right)e^{\Gamma_{2}L}+\left(-\Gamma_{2}f_{\nu}p_{L}+\Gamma_{2}p_{L}\right)e^{\Gamma_{1}L}}{\left(\Gamma_{2}-\Gamma_{1}\right)\left(1-f_{\nu}\right)} & i\frac{\left(iU_{L}\omega\rho_{m}\right)e^{\Gamma_{2}L}+\left(-iU_{L}\omega\rho_{m}\right)e^{\Gamma_{1}L}}{\left(\Gamma_{2}-\Gamma_{1}\right)\left(1-f_{\nu}\right)}\\
\\
\end{array}\right]\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} 
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
Speaker segment
\end_layout

\begin_layout Section
As an active element, with voltage control
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/spk.pdf
	width 100text%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Electrical and mechanical model of the speaker
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
The speaker generates electromotive force
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
F_{\mathrm{emf}} & =B\ell I,
\end{align}

\end_inset

where 
\begin_inset Formula $B\ell$
\end_inset

 is the 
\begin_inset Quotes eld
\end_inset

motor constant
\begin_inset Quotes erd
\end_inset

, or force factor, in units 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
si{
\backslash
newton
\backslash
per
\backslash
ampere}
\end_layout

\end_inset

, or 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
si{
\backslash
volt
\backslash
second
\backslash
per
\backslash
meter}
\end_layout

\end_inset

.
 The back-emf 
\begin_inset Quotes eld
\end_inset

force
\begin_inset Quotes erd
\end_inset

:
\begin_inset Formula 
\begin{equation}
V_{\mathrm{bemf}}=B\ell u
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The 
\begin_inset Quotes eld
\end_inset

circuit equation
\begin_inset Quotes erd
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
V_{\mathrm{in}}-V_{\mathrm{bemf}}=Z_{\mathrm{el}}I,
\end{equation}

\end_inset

where 
\begin_inset Formula $Z_{\mathrm{el}}$
\end_inset

 is the equivalent impedance of the electrical circuit in 
\begin_inset Formula $\Omega$
\end_inset

.
 The mechanical impedance comprises a stiffness part, a damping part and
 a mass part.
 The equation of motion is:
\begin_inset Formula 
\begin{equation}
z_{m}u=F_{\mathrm{emf}}+p_{l}S-p_{r}S,
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
where 
\begin_inset Formula $u$
\end_inset

 denotes the velocity phasor of the membrane.
 The mechanical impedance 
\begin_inset Formula $z_{m}$
\end_inset

 is defined as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
z_{m}=i\omega m_{m}+r_{m}+\frac{k_{m}}{i\omega},
\end{equation}

\end_inset

where 
\begin_inset Formula $m_{m}$
\end_inset

 is the moving mass, 
\begin_inset Formula $r_{m}$
\end_inset

 the damping force and 
\begin_inset Formula $k_{m}$
\end_inset

 the spring constant.
 
\begin_inset Formula $z_{m}$
\end_inset

 can equivalently be written as:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=i\omega m+R_{m}+\frac{K_{m}}{i\omega}$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=m\left(i\omega+\frac{R_{m}}{m}+\frac{\omega_{r}^{2}}{i\omega}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=m\left(i\omega+\frac{R_{m}}{m}+\frac{\omega_{r}^{2}}{i\omega}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
using: 
\begin_inset Formula $\omega_{r}^{2}=\frac{K_{m}}{m}\Rightarrow m=\frac{K_{m}}{\omega_{r}^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=\frac{m}{i\omega}\left(-\omega^{2}+i\omega\frac{R_{m}}{m}+\omega_{r}^{2}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=\frac{m}{i\omega}\left(-\omega^{2}+i\omega\frac{R_{m}\omega_{r}^{2}}{K_{m}}+\omega_{r}^{2}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Now, writing 
\begin_inset Formula $R_{m}$
\end_inset

 as: 
\begin_inset Formula $R_{m}=2\zeta\sqrt{K_{m}m}$
\end_inset

: 
\begin_inset Formula $\zeta=\frac{1}{2}\frac{r_{m}}{\sqrt{k_{m}m_{m}}}\Rightarrow\zeta=\frac{1}{2}\frac{r_{m}}{\sqrt{k_{m}m_{m}}}=\frac{1}{2}\frac{r_{m}}{\omega_{r}m_{m}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $z_{m}=\frac{m}{i\omega}\left(\omega_{r}^{2}-\omega^{2}+2i\omega\zeta\omega_{r}\right)$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
z_{m}=m\left(i\omega+2\zeta\omega_{r}+\frac{\omega_{r}^{2}}{i\omega}\right),
\end{equation}

\end_inset

where 
\begin_inset Formula 
\begin{equation}
\omega_{r}^{2}=\frac{k_{m}}{m_{m}}\qquad;\qquad\zeta=\frac{r_{m}}{2\sqrt{k_{m}m_{m}}}=\frac{r_{m}}{2\omega_{r}m_{m}}=\frac{\omega_{r}r_{m}}{2k_{m}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
After some algebraic manipulations we find:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $z_{m}u=\left(p_{l}-p_{r}\right)S+B\ell I$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{1}{Z_{\mathrm{el}}}\left(V_{\mathrm{in}}-V_{\mathrm{bemf}}\right)=I$
\end_inset


\end_layout

\begin_layout Plain Layout
where 
\begin_inset Formula $V_{\mathrm{bemf}}=B\ell u$
\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
Results in:
\end_layout

\begin_layout Plain Layout
\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


\end_layout

\begin_layout Plain Layout
To acoustic variables
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\frac{1}{S}\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)U=\left(p_{l}-p_{r}\right)S+\frac{B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
To transfer matrix notation:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $p_{r}=p_{l}-\frac{1}{S^{2}}\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)U+\frac{B\ell}{Z_{\mathrm{el}}S}V_{\mathrm{in}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
\frac{1}{S_{l}}\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)U_{l} & =p_{l}S_{l}-p_{r}S_{r}+\frac{B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}},\label{eq:U_vs_V}\\
U_{r}-U_{l} & =0,
\end{align}

\end_inset

which is in transfer matrix notation:
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{r}\\
U_{r}
\end{array}\right\} =\boldsymbol{T}\left\{ \begin{array}{c}
p_{l}\\
U_{l}
\end{array}\right\} +\boldsymbol{s},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
\boldsymbol{T}=\left[\begin{array}{cc}
1 & -\frac{1}{S^{2}}\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)\\
0 & 1
\end{array}\right]\qquad;\qquad\boldsymbol{s}=\left\{ \begin{array}{c}
\frac{B\ell}{Z_{\mathrm{el}}S}V_{\mathrm{in}}\\
0
\end{array}\right\} 
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Computing the voltage input for given velocity
\end_layout

\begin_layout Standard
Suppose we know the membrane velocity, and we want to know the corresponding
 driving voltage.
 For that we can rearrange Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:U_vs_V"

\end_inset

 a bit:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\frac{1}{S_{l}}\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}\right)U_{l}=p_{l}S_{l}-p_{r}S_{r}+\frac{B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling in 
\begin_inset Formula $S_{l}$
\end_inset

 is 
\begin_inset Formula $S_{r}$
\end_inset

 = 
\begin_inset Formula $S_{d}$
\end_inset

 and 
\begin_inset Formula $\frac{p_{r}-p_{l}}{U}=Z_{\mathrm{ac}}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(z_{m}+\frac{\left(B\ell\right)^{2}}{Z_{\mathrm{el}}}+Z_{\mathrm{ac}}S\right)U=\frac{S_{d}B\ell}{Z_{\mathrm{el}}}V_{\mathrm{in}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Or:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\frac{B\ell}{S_{d}}+\frac{Z_{\mathrm{el}}\left(Z_{\mathrm{ac}}+z_{m}/S_{d}\right)}{B\ell}\right)U=V_{\mathrm{in}}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
V_{\mathrm{in}}=\left(\frac{B\ell}{S_{d}}+\frac{Z_{\mathrm{el}}\left(Z_{\mathrm{ac}}+z_{m}/S_{d}\right)}{B\ell}\right)U,
\end{equation}

\end_inset

or equivalently in terms of the mechanical velocity:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{B\ell^{2}+Z_{\mathrm{el}}\left(Z_{\mathrm{ac}}S_{d}+z_{m}\right)}{B\ell}u=V_{\mathrm{in}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
V_{\mathrm{in}}=\frac{B\ell^{2}+Z_{\mathrm{el}}\left(Z_{\mathrm{ac}}S_{d}+z_{m}\right)}{B\ell}u
\end{equation}

\end_inset

For a COMSOL implementation, in terms of the computed acoustic pressure
 and derivatives thereof (to create a linear system of equations):
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $V_{\mathrm{in}}=\frac{B\ell^{2}u+Z_{\mathrm{el}}\left(p+z_{m}u\right)}{B\ell}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $V_{\mathrm{in}}=\left(B\ell+\frac{Z_{\mathrm{el}}z_{m}}{B\ell}\right)u+\frac{Z_{\mathrm{el}}}{B\ell}p$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
V_{\mathrm{in}}=\left(B\ell+\frac{Z_{\mathrm{el}}z_{m}}{B\ell}\right)u+\frac{Z_{\mathrm{el}}}{B\ell}F_{\mathrm{spk}},
\end{equation}

\end_inset

where 
\begin_inset Formula $F_{\mathrm{spk}}$
\end_inset

 is the net force the speaker exerts 
\emph on
on the fluid
\emph default
.
\end_layout

\begin_layout Section
As antireciprocal segment
\end_layout

\begin_layout Standard
As antireciprocal segment, a voltage controlled speaker has electrical connectio
ns on the left side, and acoustical connections on the right side:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{R}=\boldsymbol{T}_{\mathrm{spk}}\left\{ \begin{array}{c}
V\\
I
\end{array}\right\} _{L}.
\end{equation}

\end_inset

A model us used for the back cavity pressure build-up which can be added
 as an extra impedance, placed in series with the effective acoustic impedance
 of the front side, hence the force balance reads:
\begin_inset Formula 
\begin{equation}
F_{\mathrm{emf}}=Z_{\mathrm{back}}U+Z_{\mathrm{front}}U
\end{equation}

\end_inset

The transfer matrix reads:
\begin_inset Formula 
\begin{equation}
\boldsymbol{T}_{\mathrm{spk}}=\left[\begin{array}{cc}
-\frac{S^{2}Z_{\mathrm{back}}+z_{m}}{SB\ell} & \frac{\left(B\ell\right)^{2}+Z_{\mathrm{el}}\left(z_{m}+S^{2}Z_{\mathrm{back}}\right)}{B\ell S}\\
\frac{S}{B\ell} & -\frac{SZ_{\mathrm{el}}}{B\ell}
\end{array}\right]
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
Determinant:
\begin_inset Formula 
\[
\frac{Z_{\mathrm{el}}\left(S^{2}Z_{\mathrm{back}}+z_{m}\right)}{B\ell^{2}}-\left(1+\frac{Z_{\mathrm{el}}\left(S^{2}Z_{\mathrm{back}}+z_{m}\right)}{B\ell^{2}}\right)=-1
\]

\end_inset


\end_layout

\begin_layout Plain Layout
For a closed back-cavity volume, the back-cavity is:
\end_layout

\begin_layout Plain Layout
Then again:
\end_layout

\begin_layout Plain Layout
Compute determinant:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\mathrm{det}=-S$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Chapter
(Micro)-perforated plate design
\end_layout

\begin_layout Standard
Given 
\begin_inset Formula $\beta$
\end_inset

, 
\begin_inset Formula $\zeta$
\end_inset

 and 
\begin_inset Formula $\omega_{r}$
\end_inset

, a proper acoustic mass has to be chosen.
 Given the resonator equations
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
C & =\frac{V}{\rho_{0}z_{0}},\\
m_{\mathrm{neck}} & =\frac{1}{\omega_{r}^{2}C}\\
\zeta & =\frac{1}{2}\omega_{r}CR_{v}.
\end{align*}

\end_inset

and 
\begin_inset Formula 
\[
Z_{h}=\frac{1}{\omega_{r}C}\left(\frac{\omega_{r}}{i\omega}+2\zeta+\frac{i\omega}{\omega_{r}}\right)
\]

\end_inset


\end_layout

\end_inset

, the viscous resistance and required acoustic mass can be determined.
 This results in requirements for the (effective) acoustic mass and resistance
 of the perforate.
 For arbitrary hole sizes, the definition of the acoustic impedance of a
 perforate is:
\begin_inset Formula 
\begin{equation}
z=\frac{\Delta p}{\overline{u}}.\label{eq:perforate_impedance_definition}
\end{equation}

\end_inset

where 
\begin_inset Formula $\overline{u}$
\end_inset

 denotes the acoustic volume flow per unit of area through the perforate
 (uncorrected yet for porosity), such that the area-averaged velocity 
\emph on
in a hole
\emph default
 is 
\begin_inset Formula $u_{h}=\overline{u}/\phi$
\end_inset

, where 
\begin_inset Formula $\phi$
\end_inset

 denotes the porosity.
 In Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:perforate_impedance_definition"

\end_inset

, it is assumed that the acoustic wavelength is typically much larger than
 the length scale(s) of the perforate.
 The model for the impedance of a perforate, in the linear range is 
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
The COMSOL language, partially translated:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z=-\rho_{0}c_{0}\frac{2i\sin\left(\frac{k_{c}t_{p}}{2}\right)}{\sqrt{\left(\gamma-\left(\gamma-1\right)\Psi_{h}\right)\Psi_{v}}}-\rho_{0}c_{0}\frac{i\omega}{c_{0}C_{D}\phi}\frac{2\delta}{\Psi_{v}}f_{\mathrm{int}},$
\end_inset


\end_layout

\begin_layout Plain Layout
Using the fact that: 
\begin_inset Formula $\Psi_{v}\equiv f_{\nu}-1$
\end_inset

 and equivalently: 
\begin_inset Formula $\Psi_{h}\equiv f_{\kappa}-1$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z=\rho_{0}c_{0}\frac{2i\sin\left(\Gamma\frac{t_{w}}{2}\right)}{\sqrt{\left(\gamma-\left(\gamma-1\right)\left(f_{\kappa}-1\right)\right)\left(f_{\nu}-1\right)}}+\rho_{0}c_{0}\frac{i\omega}{c_{0}C_{D}\phi}\frac{2\delta}{1-f_{\nu}}f_{\mathrm{int}},$
\end_inset


\end_layout

\begin_layout Plain Layout
where 
\begin_inset Formula $k_{c}$
\end_inset

 is our 
\begin_inset Formula $\Gamma$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z=\rho_{0}c_{0}\frac{2i\sin\left(\Gamma\frac{t_{w}}{2}\right)}{\sqrt{\left(1+\left(\gamma-1\right)f_{\kappa}\right)\left(1-f_{\nu}\right)}}+\frac{i\omega}{c_{0}C_{D}\phi}\frac{2\delta}{1-f_{\nu}}f_{\mathrm{int}},$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\delta$
\end_inset

 is the end correction length for one side: 
\begin_inset Formula $\delta=4D/(3\pi)$
\end_inset

.
 For small plate thicknesses:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z=\rho_{0}c_{0}\frac{2i\sin\left(\Gamma\frac{t_{w}}{2}\right)}{\sqrt{\left(1+\left(\gamma-1\right)f_{\kappa}\right)\left(1-f_{\nu}\right)}}+\rho_{0}c_{0}\frac{i\omega}{c_{0}C_{D}\phi}\frac{2\delta}{1-f_{\nu}}f_{\mathrm{int}},$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
Z_{c}=\frac{kZ_{0}}{\left(1-f_{\nu}\right)\Gamma}.
\]

\end_inset

viscothermal wave number, i.e.
 the wave number corrected for viscothermal losses:
\begin_inset Formula 
\[
\Gamma=\frac{\omega}{c_{0}}\sqrt{\frac{1+\tfrac{\left(\gamma-1\right)f_{\kappa}}{1+\epsilon_{s}}}{1-f_{\nu}}}.
\]

\end_inset


\end_layout

\begin_layout Plain Layout
For small plate thicknesses:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $z=i\omega\rho_{0}\frac{t_{w}+\frac{2\delta f_{\mathrm{int}}}{C_{D}}}{\left(1-f_{\nu}\right)},$
\end_inset


\end_layout

\end_inset

:
\begin_inset Formula 
\begin{equation}
z=\frac{i\omega\rho_{0}}{\phi}\left[\frac{t_{w}}{\left(1-f_{\nu}\right)}+2\delta f_{\mathrm{int}}\right]+\alpha\frac{\rho_{0}\omega\delta_{\nu}}{\phi},
\end{equation}

\end_inset

where 
\begin_inset Formula $f_{\mathrm{int}}$
\end_inset

 is the hole-hole interaction function which 
\begin_inset Formula $\to1$
\end_inset

 for 
\begin_inset Formula $\phi\to0$
\end_inset

, and 
\begin_inset Formula $\delta$
\end_inset

 is the single-sided hole (therefore, the factor 2 in front) end correction
 due to the added mass effect, for the situation of negligible hole-hole
 interaction.
 [Paper: Tayong, 2013].
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
f_{\mathrm{int}}(\phi) & =1-1.4092\sqrt{\phi}+0.33818\sqrt{\phi}^{3}+0.06793\sqrt{\phi}^{5}.\\
 & -0.02287\sqrt{\phi}^{6}+0.063015\sqrt{\phi}^{7}-0.01614\sqrt{\phi}^{8}
\end{align}

\end_inset

For square holes:
\end_layout

\begin_layout Standard
where
\begin_inset Formula 
\begin{equation}
\xi^{2}=\frac{\pi D^{2}}{4P^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\frac{D}{P}=\sqrt{\frac{4\phi}{\pi}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
For circular large holes with diameter 
\begin_inset Formula $D$
\end_inset

, the end correction for both sides is
\begin_inset Formula 
\begin{equation}
2\delta=\frac{8}{3\pi}D\approx0.85D.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Here we use a more advanced model, which includes the shear wave number.
 For unrounded edges and a perforate thickness of 
\begin_inset Formula $t_{p}$
\end_inset

, the added mass end correction can be computed as:
\begin_inset Note Note
status open

\begin_layout Plain Layout
Equation according to Temiz for added mass effect:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $2\delta=\frac{\delta_{\mathrm{temiz}}}{2}D$
\end_inset


\end_layout

\begin_layout Plain Layout
Where:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\delta_{\mathrm{temiz}}=0.97\exp\left(-0.2S_{h}\right)+1.54-0.003\frac{D}{t_{p}}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $S_{h}=\frac{D}{2}\sqrt{\frac{\rho_{0}\omega}{\mu_{0}}}$
\end_inset


\end_layout

\begin_layout Plain Layout
——–
\end_layout

\begin_layout Plain Layout
Ours:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $2\delta=\frac{8}{3\pi}D$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
2\delta=\frac{1}{2}\left[0.97\exp\left(-0.14\frac{D}{\delta_{\nu}}\right)+1.54-0.003\frac{D}{t_{p}}\right]D
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The factor 
\begin_inset Formula $\alpha$
\end_inset

 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\alpha=5.08\left(\frac{D}{\sqrt{2}\delta_{\nu}}\right)^{-1.45}+1.70-0.002\frac{D}{t_{p}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Tuning the hole diameter for large holes and the negligible hole-hole interactio
n
\end_layout

\begin_layout Standard
The coarse impedance of a Helmholtz resonator repeated here:
\begin_inset Formula 
\begin{equation}
Z(\omega)=\underbrace{i\omega m_{A}+R_{v}}_{Z_{h}}+\frac{\rho_{0}c_{0}^{2}}{i\omega V},
\end{equation}

\end_inset

The resistive and reacting part 
\begin_inset Formula $i\omega m_{A}+R_{v}$
\end_inset

 is due to the resonator holes,
\begin_inset Formula 
\begin{equation}
Z_{h}=i\omega m_{A}+R_{v}\approx\frac{1}{S}\left[\frac{i\omega\rho_{0}}{\phi}\left[\frac{t_{w}}{\left(1-f_{\nu}\right)}+2\delta f_{\mathrm{int}}\right]+\frac{\alpha\rho_{0}\omega\delta_{\nu}}{\phi}\right].\label{eq:Zhole}
\end{equation}

\end_inset

In the large hole limit, or high shear wave number:
\begin_inset Formula 
\[
\Re\left[i\omega m_{A}+R_{v}\right]\approx\frac{\rho_{0}\delta_{\nu}\omega}{\phi S}\left[\alpha+\frac{2t_{w}}{\left(D-4\delta_{\nu}\right)}\right]\underbrace{\propto}_{\mathrm{approx}.}\sqrt{\omega}.
\]

\end_inset


\end_layout

\begin_layout Standard
In the large hole limit, without hole-hole interaction and 
\begin_inset Formula $\delta_{\nu}\to0$
\end_inset

, we the resonance frequency of the system is:
\begin_inset Formula 
\begin{equation}
\omega_{r,\mathrm{lh}}^{2}=\frac{\phi Sc_{0}^{2}}{V\left(1.54D+t_{w}\right)}\label{eq:omgr_largeholes}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\frac{c_{0}^{2}\rho_{0}}{V\omega_{\mathrm{r,lh}}^{2}}\left[\frac{\omega_{\mathrm{r,lh}}^{2}}{i\omega}+\omega\left\{ \frac{\alpha\delta_{\nu}}{2\delta f_{\mathrm{int}}+t_{w}}+i\frac{Dt_{w}+2\delta f_{\mathrm{int}}\left(D-2\delta_{\nu}\left(1-i\right)\right)}{\left(D-2\delta_{\nu}\left(1-i\right)\right)\left(2\delta f_{\mathrm{int}}+t_{w}\right)}\right\} \right]$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{c_{0}^{2}\rho_{0}}{V\omega_{\mathrm{r,lh}}^{2}}\left[\frac{\omega_{\mathrm{r,lh}}^{2}}{i\omega}+\frac{\omega\alpha\delta_{\nu}}{2\delta f_{\mathrm{int}}+t_{w}}+i\omega\left(1+t_{w}\frac{2\delta_{\nu}\left(1-i\right)}{\left(D-2\delta_{\nu}\left(1-i\right)\right)\left(2\delta f_{\mathrm{int}}+t_{w}\right)}\right)\right]$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
Z_{\mathrm{large\,holes},\mathrm{res}}(\omega)=\frac{c_{0}^{2}\rho_{0}}{V\omega_{r,\mathrm{lh}}^{2}}\left[\frac{\omega_{r,\mathrm{lh}}^{2}}{i\omega}+\frac{i\omega t_{w}}{\left\{ 1+2\frac{\delta_{\nu}\left(i-1\right)}{D}\right\} \left(2\delta f_{\mathrm{int}}+t_{w}\right)}+\frac{i\omega\left[2\delta f_{\mathrm{int}}-i\delta_{\nu}\alpha\right]}{2\delta f_{\mathrm{int}}+t_{w}}\right]\label{eq:Zlargeholes_forres}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
COMSOL boundary condition to useful
\end_layout

\begin_layout Standard
When using COMSOL to compute Helmholtz resonances, the added mass effect
 is included just by solving the Helmholtz equation.
 Therefore, to model the holes, only the final wall thickness part of the
 added mass (and hole-hole interaction), and the resistive part of the impedance
 should be added to the simulation.
 If we look at Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Zhole"

\end_inset

, it means only the following part:
\begin_inset Formula 
\begin{equation}
z_{\mathrm{bc,\,COMSOL}}=i\omega\rho_{0}\frac{t_{w}}{1-f_{\nu}}+\alpha\rho_{0}\omega\delta_{\nu}.
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Porosity estimator constraint
\end_layout

\begin_layout Standard
An estimation for the porosity is a good requirement, as a too large porosity
 leads to too much hole-hole interaction and shift away from proper Helmholtz
 resonators.
 First of all, we set the surface area at the inner duct, which is available
 for holes as
\begin_inset Formula 
\begin{equation}
S=\Pi L_{h},
\end{equation}

\end_inset

and we fix 
\begin_inset Formula $L_{h}$
\end_inset

 to
\begin_inset Formula 
\begin{equation}
L_{h}=\lambda_{r}/20=\frac{2\pi c_{0}}{20\omega_{r,\mathrm{lh}}}=\frac{\pi c_{0}}{10\omega_{r,\mathrm{lh}}}.
\end{equation}

\end_inset

Rewriting Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:omgr_largeholes"

\end_inset

 to 
\begin_inset Formula $\phi$
\end_inset

 yields
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\phi\approx\frac{V\left(1.54D+t_{w}\right)\omega_{r,\mathrm{lh}}^{2}}{Sc_{0}^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout
Fill in for 
\begin_inset Formula $S=\Pi L_{h}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\phi\approx\frac{V\left(1.54D+t_{w}\right)\omega_{r,\mathrm{lh}}^{2}}{\Pi L_{h}c_{0}^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout
And for 
\begin_inset Formula $L_{h}$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $L_{h}=\frac{\pi c_{0}}{10\omega_{r,\mathrm{lh}}}.$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\phi\approx\frac{10}{\pi}\frac{V\left(1.54D+t_{w}\right)\omega_{r,\mathrm{lh}}^{3}}{\Pi c_{0}^{3}}$
\end_inset


\end_layout

\end_inset

:
\begin_inset Formula 
\begin{equation}
\phi_{\mathrm{estimation}}\approx\frac{10}{\pi}\frac{V\left(1.54D+t_{w}\right)\omega_{r,\mathrm{lh}}^{3}}{\Pi c_{0}^{3}}\leq0.1
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
See what this constraint does...*
\end_layout

\begin_layout Section
Large hole (boundary layer) limit
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\phi=\frac{S_{\mathrm{hole}}}{S_{\mathrm{tot}}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta_{\nu}\ll D$
\end_inset

.
 Given 
\begin_inset Formula $\zeta$
\end_inset

 and 
\begin_inset Formula $\omega_{r}$
\end_inset

.
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Now, the following substitutions are made:
\begin_inset Formula 
\begin{align*}
C & =\frac{V}{\rho_{0}z_{0}},\\
m_{\mathrm{neck}} & =\frac{1}{\omega_{r}^{2}C}\\
\zeta & =\frac{1}{2}\omega_{r}CR_{v}.
\end{align*}

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $ $
\end_inset


\end_layout

\end_inset

such that we can write:
\begin_inset Formula 
\[
Z_{h}=\frac{1}{\omega_{r}C}\left(\frac{\omega_{r}}{i\omega}+2\zeta+\frac{i\omega}{\omega_{r}}\right)
\]

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
Z_{h}=\frac{1}{\omega_{r}C}\left(\frac{\omega_{r}}{i\omega}+2\zeta+\frac{i\omega}{\omega_{r}}\right)=\frac{1}{i\omega C}+\frac{2\zeta\omega_{r}}{C}
\]

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{2\zeta}{\omega_{r}C}=R_{v}.$
\end_inset

 Or: 
\begin_inset Formula 
\[
\zeta=\frac{1}{2}\omega_{r}R_{v}C
\]

\end_inset

But: 
\begin_inset Formula $\frac{1}{Cm_{A}}=\omega_{r}^{2}$
\end_inset

Such that:
\begin_inset Formula $\frac{1}{C}=\omega_{r}^{2}m_{A}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\zeta=\frac{1}{2}\frac{R_{v}}{\omega_{r}m_{A}}
\]

\end_inset


\end_layout

\end_inset

 Note that:
\begin_inset Formula 
\begin{equation}
\zeta=\frac{1}{2}\frac{R}{m_{A}\omega_{r}}\approx\frac{1}{2}\frac{\Re\left[z\right]}{\Im\left[z\right]}
\end{equation}

\end_inset

Procedure:
\end_layout

\begin_layout Standard
In the boundary layer limit: 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
f_{\nu}=\frac{\left(1-i\right)\delta_{\nu}}{2r_{h}},
\end{equation}

\end_inset

such that:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
z_{\mathrm{perforate}}=\frac{i\omega\rho_{0}}{\phi}\frac{t_{w}+2\delta f_{\mathrm{int}}}{\left(1-\frac{\delta_{\nu}}{2r_{h}}+\frac{i\delta_{\nu}}{2r_{h}}\right)}
\end{equation}

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $z=\frac{i\omega\rho_{0}}{\phi}\frac{t_{w}+\frac{2\delta f_{\mathrm{int}}}{C_{D}}\left(1-\frac{\delta_{\nu}}{2r_{h}}+\frac{i\delta_{\nu}}{2r_{h}}\right)}{\left(1-\frac{\delta_{\nu}}{2r_{h}}+\frac{i\delta_{\nu}}{2r_{h}}\right)}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Typical resistance: fill in 
\begin_inset Formula $\omega=\omega_{r}$
\end_inset

.
 Filling in:
\begin_inset Formula 
\begin{equation}
\zeta\approx\frac{\delta_{\nu}}{D}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The real part of the perforate impedance is the resistive part.
 In a 3D simulation, this impedance can be added to a surface of the hole,
 to model the hole 
\emph on
resistance
\emph default
 in an otherwise inviscid simulation.
 The real part is:
\begin_inset Formula 
\begin{equation}
\frac{}{}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Lots of holes
\end_layout

\begin_layout Standard
Hereby, once we know the hole diameter, the required acoustic mass can be
 tuned using the porosity:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
m_{A}\approx\frac{\Im\left[z(\omega=\omega_{r}\right]}{\omega S_{\mathrm{t}}}\approx\frac{1}{S_{\mathrm{tot}}\phi}\left(\frac{\rho_{0}8Df_{\mathrm{int}}(\phi)}{3\pi}+\rho_{0}t_{w}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
So that the porosity can be computed as:
\begin_inset Formula 
\begin{equation}
\phi\approx F(\phi)=\frac{D\rho_{0}\left(D-2\delta_{\nu}\right)\left(8Df_{\mathrm{int}}+3\pi t_{w}\right)}{3\pi S_{\mathrm{tot}}m_{A}\left(D^{2}-4D\delta_{\nu}+8\delta_{\nu}^{2}\right)}\approx\frac{\rho_{0}\left(8Df_{\mathrm{int}}(\phi)+3\pi t_{w}\right)}{3\pi S_{\mathrm{tot}}m_{A}}.
\end{equation}

\end_inset

Note that this is a trancendental equation in 
\begin_inset Formula $\phi$
\end_inset

, which can easily be solved by iterating 
\begin_inset Formula $\phi$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\phi_{1} & =F(1)\\
\phi_{2} & =F(\phi_{1})\\
\phi_{3} & =F(\phi_{2})\\
\vdots & =\vdots
\end{align}

\end_inset


\end_layout

\begin_layout Subsection
Some holes
\end_layout

\begin_layout Standard
For only 
\begin_inset Quotes eld
\end_inset

some holes
\begin_inset Quotes erd
\end_inset

, far away from each other, we fill in for 
\begin_inset Formula $\phi=\frac{1}{4}N_{\mathrm{hole}}\pi D^{2}/S_{\mathrm{tot}}$
\end_inset

:
\begin_inset Formula 
\begin{equation}
m_{A}\approx\frac{\rho_{0}}{3\pi N_{\mathrm{hole}}D}\left(\frac{32}{\pi}+\frac{12t_{w}}{D}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
So the number of holes can be chosen as:
\begin_inset Formula 
\begin{equation}
N_{\mathrm{holes}}\approx\frac{4\rho_{0}\left(8Df_{\mathrm{int}}+3\pi t_{w}\right)}{3\pi^{2}D^{2}m_{A}}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Small hole limit
\end_layout

\begin_layout Standard
In the small hole limit,
\begin_inset space ~
\end_inset


\begin_inset Formula 
\begin{equation}
f_{\nu}\approx1-\frac{iD^{2}}{16\delta_{\nu}^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Such that:
\begin_inset Formula 
\begin{equation}
\zeta=\frac{1}{2}\frac{R}{m_{A}\omega_{r}}\approx\frac{1}{2}\frac{\Re\left[z(\omega=\omega_{r}\right]}{\Im\left[z(\omega=\omega_{r}\right]}\approx\frac{3\pi\delta_{\nu}^{2}t_{w}}{D^{3}f_{\mathrm{int}}}
\end{equation}

\end_inset

Such that:
\begin_inset Formula 
\begin{equation}
D=\sqrt[3]{\frac{6\pi\delta_{\nu}^{2}t_{w}}{6\zeta}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
And:
\begin_inset Formula 
\begin{equation}
m_{A}=\rho_{0}\frac{8Df_{\mathrm{int}}}{3\pi S_{\mathrm{tot}}\phi}
\end{equation}

\end_inset

Such that:
\begin_inset Formula 
\begin{equation}
\phi\approx\rho_{0}\frac{8Df_{\mathrm{int}}}{3\pi S_{\mathrm{tot}}m_{A}}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Geometry of hole patterns
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/hexagonal_pattern.pdf
	width 50text%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Geometry details of a hexagonal hole pattern
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:hexagonal_pitch"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
For a square hole pattern, with hole-hole pitch 
\begin_inset Formula $P$
\end_inset

, the overall surface of a unit cell 
\begin_inset Formula $S_{\mathrm{unit}}=P^{2}$
\end_inset

.
 For a certain porosity, the pitch can then be computed as:
\begin_inset Formula 
\begin{equation}
P=\sqrt{\frac{\pi}{4\phi}}D.
\end{equation}

\end_inset

For a hexagonal hole pattern (Fig.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "fig:hexagonal_pitch"

\end_inset

) with hole-hole pitch 
\begin_inset Formula $P$
\end_inset

, the overall surface of a unit cell 
\begin_inset Formula $S_{\mathrm{unit}}=\frac{\sqrt{3}}{2}P^{2}$
\end_inset

.
 Henceforth, the pitch can be computed from the porosity and the hole diameter
 as:
\begin_inset Formula 
\begin{equation}
P=\sqrt{\frac{\sqrt{3}\pi}{6\phi}}D.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The most important design parameters of a perforate are the porosity and
 the hole diameter.
\end_layout

\begin_layout Section
Addition of acoustic hole resistance in an otherwise inviscid simulation
\end_layout

\begin_layout Standard
We assume that in a 3D FEM simulation, the imaginary acoustic impedance
 of a single hole
\begin_inset Formula 
\begin{equation}
Z_{\mathrm{hole}}=i\omega\rho_{0}\frac{4}{\pi D^{2}}\left[\frac{t_{w}}{\left(1-f_{\nu}\right)}+\frac{8Df_{\mathrm{int}}}{3\pi C_{D}}\right],
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\Re[z_{\mathrm{hole}}]=\frac{2D\delta_{\nu}\omega\rho_{0}t_{w}}{\left(4\delta_{\nu}^{2}+\left(D-2\delta_{\nu}\right)^{2}\right)},\label{eq:Rv_hole}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Over-all transmission matrix
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
 &  &  &  & \left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{1} & = & \boldsymbol{T}_{1}\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} _{1}\\
 &  & \left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{2} &  & =\boldsymbol{T}_{2}\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{1}\\
\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{3} & =\boldsymbol{T}_{3} & \left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{2}\\
\end{align}

\end_inset

, hence 
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} _{3}=\underbrace{\boldsymbol{T}_{3}\cdot\boldsymbol{T}_{2}\cdot\boldsymbol{T}_{1}}_{\boldsymbol{T}}\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} _{1}
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
Miscellaneous models for acoustic components
\end_layout

\begin_layout Section
Acoustic impedance of small orifices
\end_layout

\begin_layout Subsection
Rectangular orifice
\end_layout

\begin_layout Subsection
Slit orifice
\end_layout

\begin_layout Standard
====================
\end_layout

\begin_layout Standard
Lookup model
\end_layout

\begin_layout Section
COMSOL model
\end_layout

\begin_layout Standard
\align left
LRFTubes allows importing transfer matrix data from externally computed
 sources (i.e.
 finite element model results).
 We focus on the use of COMSOL Multiphysics here.
 The output data from COMSOL should be created using the 
\begin_inset Quotes eld
\end_inset

Port Sweep
\begin_inset Quotes erd
\end_inset

 functionality.
 Implementation is only for 2 ports, as this is the only case for which
 COMSOL is able to export data.
 In COMSOL, the transfer matrix is defined as:
\end_layout

\begin_layout Standard
\align center
\begin_inset Graphics
	filename img/comsol_transfermatrix.png

\end_inset


\end_layout

\begin_layout Standard
\align left
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{i}\\
Q_{i}
\end{array}\right\} =\left[\begin{array}{cc}
T_{11} & T_{12}\\
T_{21} & T_{22}
\end{array}\right]\left\{ \begin{array}{c}
p_{o}\\
Q_{o}
\end{array}\right\} ,\label{eq:transfer_matrix_COMSOL}
\end{equation}

\end_inset

hence the transfer matrix definition of 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\end_layout

\end_inset

 is the 
\emph on
inverse
\emph default
 of the definition of COMSOL Multiphysics:
\begin_inset Formula 
\begin{equation}
\boldsymbol{T}_{\mathrm{\lrftubes}}=\boldsymbol{T}_{\mathrm{COMSOL}}^{-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
To properly use the Lookup model, in COMSOL port 1 should be corresponding
 to the LEFT side of a segment, and port 2 should be corresponding to the
 RIGHT side of a segment.
 Then, the data should be exported to a 
\emph on
txt
\emph default
 file with the columns in the following order: frequency, T11, T12, T21,
 T22.
 A file of this format, as exported by COMSOL can be passed to the constructor
 of 
\family typewriter
\emph on
LookupModel
\family default
.
\end_layout

\begin_layout Subsection
SPICE model
\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/two_port_probing.pdf
	width 90text%

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Two-port model, probing the transfer matrix by computing the simulation
 output.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:2-port-probing"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
A SPICE model can be created from a COMSOL model, by performing a circuit
 analysis of the system in two cases, one is the situation providing a voltage
 source on one side, and measuring the current going in, and the current
 going out on the other side, while the element is short-circuited.
 The other is similar, only in this case the segment is 
\emph on
open
\emph default
 on the other side.
 Fig.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "fig:2-port-probing"

\end_inset

 shows the schematic of the two cases that need to be computed.
 If we assume:
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{R}=\left[\begin{array}{cc}
A & B\\
C & D
\end{array}\right]\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{L},
\end{equation}

\end_inset

for the components of the transfer matrix, we can set the following equations:
\begin_inset Formula 
\begin{align}
U_{R}^{(1)} & =C+DU_{L}^{(1)},\\
0 & =A+BU_{L}^{(1)},\\
0 & =C+DU_{L}^{(2)},\\
p_{R}^{(2)} & =A+BU_{L}^{(2)},
\end{align}

\end_inset

which gives four equations, for the four unknown transfer matrix coefficients.
 We can directly perform this computation using the method 
\family typewriter
LookupModel.from_pU
\family default
 in 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
lrftubes
\end_layout

\end_inset

.
\end_layout

\begin_layout Chapter
Measuring the transmission matrix using the four microphone method
\end_layout

\begin_layout Standard
Based on Brüel Kjaer - Transmission loss in impedance tube.pdf in /home/anne/next
cloud/wip_redusone/2021-Steegmuller/measurement_setup
\end_layout

\begin_layout Standard
Modifications: volume flow U instead of velocity v; impedance Z instead
 of characteristic impedance z; transfer functions Hir instead of cross
 correlations (?).
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
TO DO:
\end_layout

\begin_layout Plain Layout
draw own image image
\end_layout

\begin_layout Plain Layout
fix citation
\end_layout

\begin_layout Plain Layout
Transfer matrix according to our own definition instead of the definition
 of Bruel & Kjaer = definition of COMSOL
\end_layout

\begin_layout Plain Layout
Consistently use Q or U for volume flow? Also in text above about COMSOL.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
The transfer matrix of a device can be measured using a four microphone
 setup as shown in figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:meas_transmatrix_4mic"
plural "false"
caps "false"
noprefix "false"

\end_inset

.
 The microphones record acoustic pressure and plane waves are assumed.
 In the following equations, time dependency 
\begin_inset Formula $\exp(+j*\omega*t)$
\end_inset

 is not shown.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/Bruel_Kjaer_fig1.png
	lyxscale 50
	width 80text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Experimental setup to measure the transfer matrix, using the four microphone
 method
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:meas_transmatrix_4mic"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
The transfer matrix coefficients are calculated based on sound pressure
 
\begin_inset Formula $p$
\end_inset

 and volume velocity 
\begin_inset Formula $U$
\end_inset

, as related by equation 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:transfer_matrix_COMSOL"
plural "false"
caps "false"
noprefix "false"

\end_inset

.
 Note that this definition is different than the definition used in LRFtubes
 and therefore 
\begin_inset Formula $T$
\end_inset

 should be inverted for further use.
 Subscrips 
\begin_inset Formula $i$
\end_inset

 and 
\begin_inset Formula $d$
\end_inset

 refer to 
\begin_inset Formula $x=0$
\end_inset

 and 
\begin_inset Formula $x=d$
\end_inset

 respectively.
 There are two equations and four unknowns, so two sets of measurements
 are required.
 The second set, indicated by superscript 
\begin_inset Formula $*$
\end_inset

, must be performed with a different acoustic termination.
 Together this results in four equations for four unknowns.
\end_layout

\begin_layout Standard
\align left
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{i}\\
Q_{i}
\end{array}\begin{array}{c}
p_{i}^{*}\\
Q_{i}^{*}
\end{array}\right\} =\left[\begin{array}{cc}
T_{11} & T_{12}\\
T_{21} & T_{22}
\end{array}\right]\left\{ \begin{array}{c}
p_{o}\\
Q_{o}
\end{array}\begin{array}{c}
p_{o}^{*}\\
Q_{o}^{*}
\end{array}\right\} ,\label{eq:transfer_matrix-double}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Solving for 
\begin_inset Formula $T$
\end_inset

 yields:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\left[\begin{array}{cc}
T_{11} & T_{12}\\
T_{21} & T_{22}
\end{array}\right]=\frac{1}{p_{d}Q_{d}^{*}-p_{d}^{*}Q_{d}}\left[\begin{array}{cc}
p_{i}Q_{d}^{*}-p_{i}^{*}Q_{d} & -p_{i}p_{d}^{*}+p_{i}^{*}p_{d}\\
Q_{i}Q_{d}^{*}-Q_{i}^{*}Q_{d} & -p_{d}^{*}Q_{i}+p_{d}Q_{i}^{*}
\end{array}\right]
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $p$
\end_inset

 and 
\begin_inset Formula $Q$
\end_inset

 at 
\begin_inset Formula $x=0$
\end_inset

 and 
\begin_inset Formula $x=d$
\end_inset

 can be calculated from travelling 
\begin_inset Formula $A$
\end_inset

, 
\begin_inset Formula $B$
\end_inset

, 
\begin_inset Formula $C$
\end_inset

 and 
\begin_inset Formula $D$
\end_inset

.
 The calculation of their second measurement counterparts 
\begin_inset Formula $*$
\end_inset

 goes analogously and uses 
\begin_inset Formula $A^{*}$
\end_inset

, 
\begin_inset Formula $B^{*}$
\end_inset

, 
\begin_inset Formula $C^{*}$
\end_inset

 and 
\begin_inset Formula $D^{*}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{i}=A+B
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
Q_{i}=\frac{A-B}{Z_{0}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{d}=C\cdot e^{-jkd}+D\cdot e^{jkd}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
Q_{d}=\frac{C\cdot e^{-jkd}-D\cdot e^{jkd}}{Z_{0}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
in which 
\begin_inset Formula $Z_{0}=\frac{z_{0}}{S}$
\end_inset

 is the impedance of an infinite duct, with 
\begin_inset Formula $z_{0}$
\end_inset

 the characteristic impedance and 
\begin_inset Formula $S$
\end_inset

 the cross-sectional area, 
\begin_inset Formula $j=\sqrt{-1}$
\end_inset

, 
\begin_inset Formula $k$
\end_inset

 the wavenumber.
 Travelling waves 
\begin_inset Formula $A$
\end_inset

, 
\begin_inset Formula $B$
\end_inset

, 
\begin_inset Formula $C$
\end_inset

 and 
\begin_inset Formula $D$
\end_inset

 can be calculated from transfer functions 
\begin_inset Formula $H_{ir}$
\end_inset

 from reference signal 
\begin_inset Formula $r$
\end_inset

, as sent to the loudspeaker, to the recorded signal of microphone 
\begin_inset Formula $i$
\end_inset

.
 The calculation of their second measurement counterparts 
\begin_inset Formula $*$
\end_inset

 goes analogously and uses 
\begin_inset Formula $H_{ir}^{*}$
\end_inset

.
\begin_inset Formula 
\begin{equation}
A=\frac{j\left(H_{1r}\cdot e^{jkx_{2}}-H_{2r}\cdot e^{jkx_{1}}\right)}{2\sin\left(k\left(x_{1}-x_{2}\right)\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
B=\frac{j\left(H_{2r}\cdot e^{-jkx_{1}}-H_{1r}\cdot e^{-jkx_{2}}\right)}{2\sin\left(k\left(x_{1}-x_{2}\right)\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
C=\frac{j\left(H_{3r}\cdot e^{jkx_{4}}-H_{4r}\cdot e^{jkx_{3}}\right)}{2\sin\left(k\left(x_{3}-x_{4}\right)\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
D=\frac{j\left(H_{4r}\cdot e^{-jkx_{3}}-H_{3r}\cdot e^{-jkx_{4}}\right)}{2\sin\left(k\left(x_{3}-x_{4}\right)\right)}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\sqrt{G_{rr}}$
\end_inset

 has been removed from the equations because Caspers thinks that 
\begin_inset Formula $H_{ir}$
\end_inset

 refers to the cross spectrum instead of the transfer function.
 If the transfer function is used, then 
\begin_inset Formula $\sqrt{G_{rr}}$
\end_inset

 shall be left out.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Note: if no reference signal has been recorded, the reference signal can
 be set to the signal captured by microphone 1.
 The equations have no way to figure out whether the loudspeaker really
 was driven by such a signal.
 Then a requirement is that all microphones are recorded simultaneously
 and with synchronized ADC clocks.
\end_layout

\begin_layout Chapter
IEC Coupler impedances
\end_layout

\begin_layout Standard
The Comsol model with which this data is gathered exports the input impedance
 correctly, but the transfer impedance is actually the 
\emph on
negative 
\emph default
of the actual transfer impedance.
 This is due to Comsol, which was only interested in the magnitude of the
 impedance values, and due to us (sloppy work).
 The input impedance is defined as:
\begin_inset Formula 
\begin{equation}
Z_{\mathrm{in}}=\frac{p_{\mathrm{coupler,entrance}}}{U_{\mathrm{coupler,entrance}}}
\end{equation}

\end_inset

and the transfer impedance as:
\begin_inset Formula 
\begin{equation}
Z_{\mathrm{tr}}=\frac{p_{\mathrm{DRP}}}{U_{\mathrm{coupler,entrance}}}
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
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
Optimized reactive silencers
\end_layout

\begin_layout Section
Parallel Helmholtz resonator transfer function and transmission loss
\end_layout

\begin_layout Standard
Equations for a side branch Helmholtz resonator:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
p_{R} & =p_{L},\\
U_{R} & =U_{L}-p_{L}/Z_{h},
\end{align}

\end_inset

where 
\begin_inset Formula $Z_{h}$
\end_inset

 is the side branch impedance of the Helmholtz resonator, defined as
\begin_inset Formula 
\begin{equation}
Z_{h}=\left(\frac{\rho_{0}z_{0}}{i\omega V}+R_{v}+i\omega m_{\mathrm{neck}}\right),
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
m_{\mathrm{neck}}=\frac{\rho_{0}\ell_{\mathrm{eff},\mathrm{neck}}}{S_{\mathrm{neck}}},\label{eq:acoustic_mass_neck}
\end{equation}

\end_inset

and for relatively large holes, air at STP, the resistance term can be estimated
 as [SOURCE HERE!]:
\begin_inset Formula 
\begin{equation}
R_{v}\approx7.2\times10^{-3}z_{0}/S_{h},
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Now, the following substitutions are made:
\begin_inset Formula 
\begin{align}
C & =\frac{V}{\rho_{0}z_{0}},\\
m_{\mathrm{neck}} & =\frac{1}{\omega_{r}^{2}C}\\
\zeta & =\frac{1}{2}\omega_{r}CR_{v}.
\end{align}

\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{2\zeta}{\omega_{r}C}=R_{v}.$
\end_inset


\end_layout

\end_inset

such that we can write:
\begin_inset Formula 
\begin{equation}
Z_{h}=\frac{1}{\omega_{r}C}\left(\frac{\omega_{r}}{i\omega}+2\zeta+\frac{i\omega}{\omega_{r}}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The quality factor of the resonator is the ratio of the resonance frequency
 to its bandwidth measure.
 If we take 
\begin_inset Formula 
\begin{equation}
Q\overset{\mathrm{def}}{=}\frac{f_{r}}{\Delta f},
\end{equation}

\end_inset

where 
\begin_inset Formula $\Delta f$
\end_inset

 is the full width at half the maximum value, i.e.
 the frequency distance between two points lying at 
\begin_inset Formula $-3$
\end_inset


\begin_inset space ~
\end_inset

dB w.r.t.
 the maximum value.
 The damping ratio 
\begin_inset Formula $\zeta$
\end_inset

 is related to 
\begin_inset Formula $Q$
\end_inset

 as:
\begin_inset Formula 
\begin{equation}
\zeta=\frac{1}{2Q}=\frac{1}{2}\frac{\Delta f}{f_{r}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Assembling the transfer matrix
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{R}=\left[\begin{array}{cc}
T_{11} & T_{12}\\
T_{21} & T_{22}
\end{array}\right]\left\{ \begin{array}{c}
p\\
U
\end{array}\right\} _{L},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{align}
T_{11} & =1\\
T_{12} & =0\\
T_{21} & =-Z_{h}^{-1}\\
T_{22} & =1
\end{align}

\end_inset


\end_layout

\begin_layout Subsection
Transmission loss
\end_layout

\begin_layout Standard
The transmission coefficient can be computed as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\tau=\frac{C}{A}=\frac{Z_{0}\left(T_{21}p_{L}+T_{22}U_{L}\right)}{\frac{1}{2}\left(p_{L}+Z_{0}U_{L}\right)},
\end{equation}

\end_inset

using
\begin_inset Formula 
\begin{equation}
T_{11}p_{L}+T_{12}U_{L}=p_{R}=Z_{0}U_{R}=Z_{0}\left(T_{21}p_{L}+T_{22}U_{L}\right),
\end{equation}

\end_inset

we get
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $Z_{0}\left(T_{21}p_{L}+T_{22}U_{L}\right)=T_{11}p_{L}+T_{12}U_{L}$
\end_inset


\end_layout

\begin_layout Plain Layout
–
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $U_{L}=\frac{\left(T_{11}-Z_{0}T_{21}\right)}{\left(Z_{0}T_{22}-T_{12}\right)}p_{L}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
U_{L}=\frac{\left(T_{11}-Z_{0}T_{21}\right)}{\left(Z_{0}T_{22}-T_{12}\right)}p_{L},
\end{equation}

\end_inset

filling in:
\begin_inset Formula 
\begin{equation}
\tau=\frac{2}{Z_{0}}\frac{T_{11}T_{22}-T_{12}T_{21}}{T_{11}-T_{12}/Z_{0}-T_{21}Z_{0}+T_{22}},
\end{equation}

\end_inset

assuming that the determinant of the transfer matrix be unity 
\begin_inset Formula $(T_{11}T_{22}-T_{12}T_{21}\equiv1$
\end_inset

) [THIS IS TRUE, BUT WHERE DOES THIS ASSUMPTION COME FROM??], this can be
 further simplified:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\tau=\frac{2}{T_{11}-T_{12}/Z_{0}-T_{21}Z_{0}+T_{22}},
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
For a Helmholtz resonator, this results in:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula 
\[
\tau=\frac{2}{T_{11}-T_{12}/Z_{0}-T_{21}Z_{0}+T_{22}},
\]

\end_inset


\end_layout

\begin_layout Plain Layout
Filling in: 
\begin_inset Formula $T_{11}=1$
\end_inset

, 
\begin_inset Formula $T_{12}=0$
\end_inset

, 
\begin_inset Formula $T_{21}=-1/Z_{h}$
\end_inset

 
\begin_inset Formula $T_{22}=1$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\tau=\frac{2Z_{h}}{2Z_{h}+Z_{0}},
\]

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\tau(\omega)=\frac{2Z_{h}(\omega)}{Z_{0}+2Z_{h}(\omega)},
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Filling in the Helmholtz resonator equation:
\begin_inset Formula 
\begin{equation}
\tau(\omega)=\frac{2\left(1+2\frac{\omega}{\omega_{r}}\zeta-\left(\frac{\omega}{\omega_{r}}\right)^{2}\right)}{2\left(1+2\frac{\omega}{\omega_{r}}\zeta-\left(\frac{\omega}{\omega_{r}}\right)^{2}\right)+i\frac{\omega}{\omega_{r}}\left(\frac{Cz_{0}\omega_{r}}{S}\right)}\label{eq:tau_hhres}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\left(\frac{Cz_{0}\omega_{r}}{S}\right)=\left(\frac{V\omega_{r}}{c_{0}S}\right)$
\end_inset


\end_layout

\end_inset

The peak height, filling in for 
\begin_inset Formula $\omega/\omega_{r}=1$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\tau=\frac{4\zeta}{4\zeta+\beta},
\end{equation}

\end_inset

where 
\begin_inset Formula $\beta$
\end_inset

 is defined as the resonator strength:
\begin_inset Formula 
\begin{equation}
\beta=\frac{V\omega_{r}}{Sc_{0}}
\end{equation}

\end_inset

In terms of transmission loss:
\begin_inset Formula 
\begin{equation}
\mathrm{TL}_{\omega=\omega_{r}}=20\log\left(\frac{\beta+4\zeta}{4\zeta}\right)
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
In case of weak damping (
\begin_inset Formula $\zeta\ll1$
\end_inset

), Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:tau_hhres"

\end_inset

 can be reduced to:
\begin_inset Formula 
\begin{equation}
\tau(\omega)=\frac{1-\left(\frac{\omega}{\omega_{r}}\right)^{2}}{1-\left(\frac{\omega}{\omega_{r}}\right)^{2}+\frac{1}{2}i\frac{\omega}{\omega_{r}}\beta}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The width of the peak over which a certain transmission loss is higher than
 a value of 
\begin_inset Formula $\mathrm{TL_{\mathrm{min}}}$
\end_inset


\begin_inset space ~
\end_inset

dB, can be computed as:
\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $|\tau(\omega_{r}+\Delta\omega)|=|\frac{1-\left(\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\right)^{2}}{1-\left(\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\right)^{2}+\frac{1}{2}i\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\beta}|=10^{\frac{\mathrm{TL}_{\mathrm{min}}}{20}}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $|\tau(\omega_{r}+\Delta\omega)|=|\frac{1-\left(\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\right)^{2}}{1-\left(\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\right)^{2}+\frac{1}{2}i\frac{\omega_{r}+\Delta\omega}{\omega_{r}}\beta}|=10^{\frac{\mathrm{TL}_{\mathrm{min}}}{20}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\beta=\frac{\Delta\omega}{\omega_{r}}4\sqrt{10^{^{\frac{\mathrm{TL_{\mathrm{min}}}}{10}}}-1}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
The peak half width is the distance over which the transmission loss has
 dropped 3
\begin_inset space ~
\end_inset

dB w.r.t.
 the transmission loss at the resonance frequency.
 This is an important design parameter.
 We can compute it by setting:
\begin_inset Formula 
\begin{equation}
|\frac{\tau|_{\omega_{r}+\Delta\omega}}{\tau|_{\omega_{r}}}|=\sqrt{2},
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
So given the -3
\begin_inset space ~
\end_inset

dB point, and the maximum required transmission loss, we can compute 
\begin_inset Formula $\zeta$
\end_inset

 and 
\begin_inset Formula $\beta$
\end_inset

:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Eq 1:
\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
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\begin_inset Formula $\frac{\alpha_{-3\mathrm{dB}}-1}{\zeta}=\sqrt{2}\Rightarrow\zeta=\frac{\alpha_{-3\mathrm{dB}}-1}{\sqrt{2}}$
\end_inset


\begin_inset Newline newline
\end_inset

Eq 2:
\end_layout

\begin_layout Plain Layout

\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $\mathrm{TL}_{\mathrm{max}}=20\log\left(\frac{\beta+4\zeta}{4\zeta}\right)\Rightarrow\frac{\beta+4\zeta}{4\zeta}=10^{\frac{\mathrm{TL}_{\mathrm{max}}}{20}}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\beta=4\zeta\left(10^{\frac{\mathrm{TL}_{\mathrm{max}}}{20}}-1\right)$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $\zeta=\frac{\alpha_{-3\mathrm{dB}}-1}{\sqrt{2}}$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $\beta=4\zeta\left(10^{\frac{\mathrm{TL}_{\mathrm{max}}}{20}}-1\right)$
\end_inset


\end_layout

\begin_layout Standard
Required volume in terms of resonator strength:
\begin_inset Formula 
\begin{equation}
V=\frac{Sc_{0}\beta}{\omega_{r}}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Insertion loss
\end_layout

\begin_layout Standard
For computation of the insertion loss, we require two more parameters:
\end_layout

\begin_layout Itemize
The load impedance at the downstream end of the silencer
\end_layout

\begin_layout Itemize
The output impedance of the source (
\begin_inset Formula $Z_{\mathrm{rad}}$
\end_inset

)
\end_layout

\begin_layout Standard
Suppose the source strength is defined by 
\begin_inset Formula $\mathcal{S}$
\end_inset

.
 Situation without silencer:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
U_{L} & =\mathcal{S}/\left(Z_{s}+Z_{l}\right),\\
U_{R} & =U_{L},\\
p_{R} & =Z_{\mathrm{rad}}U_{R},
\end{align}

\end_inset

where 
\begin_inset Formula $Z_{s}$
\end_inset

 denotes the source output impedance, and 
\begin_inset Formula $Z_{l}$
\end_inset

 denotes the load impedance as felt by the source.
\end_layout

\begin_layout Standard
For the reference case, the load impedance equals the radiation impedance,
 and the radiated acoustic power is:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $P_{\mathrm{ref}}=\frac{1}{2}\Re\left[p_{R}U_{R}^{*}\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $P_{\mathrm{ref}}=\frac{1}{2}\Re\left[Z_{\mathrm{rad}}\left(\mathcal{S}/Z_{s}\right)\left(\mathcal{S}/Z_{s}\right)^{*}\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $P_{\mathrm{ref}}=\frac{1}{2}|\mathcal{S}/Z_{s}|^{2}\Re\left[Z_{\mathrm{rad}}\right]$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
P_{\mathrm{ref}}=\frac{1}{2}\frac{|\mathcal{S}|^{2}}{|Z_{\mathrm{rad}}+Z_{s}|^{2}}\Re\left[Z_{\mathrm{rad}}\right]
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Now, situation including silencer, with in general, transfer matrix 
\begin_inset Formula $\boldsymbol{T}$
\end_inset

.
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $P_{\mathrm{ref}}=\frac{1}{2}\Re\left[p_{R}U_{R}^{*}\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $P_{\mathrm{ref}}=\frac{1}{2}\Re\left[Z_{\mathrm{rad}}U_{R}U_{R}^{*}\right]$
\end_inset


\end_layout

\begin_layout Plain Layout
Using:
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
P_{\mathrm{with\,silencer}}=\frac{1}{4}|\mathcal{S}|^{2}\frac{\Re\left[Z_{\mathrm{rad}}\right]}{|T_{22}Z_{\mathrm{rad}}-T_{12}+Z_{s}\left(T_{11}-T_{21}Z_{\mathrm{rad}}\right)|^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
From that, computing the power ratio, that 
\begin_inset Formula $\det\boldsymbol{T}\equiv1$
\end_inset

 for a reciprocal system:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
R_{P}=\frac{P_{\mathrm{with\,silencer}}}{P_{\mathrm{ref}}}=\frac{|Z_{\mathrm{rad}}+Z_{s}|^{2}}{|T_{22}Z_{\mathrm{rad}}-T_{12}+Z_{s}\left(T_{11}-T_{21}Z_{\mathrm{rad}}\right)|^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsection
Insertion loss for a Helmholtz side branch resonator
\end_layout

\begin_layout Standard
Filling in for a simple Helmholtz side branch resonator:
\begin_inset Formula 
\begin{equation}
R_{P,\mathrm{Helmholtz}}=\frac{|Z_{\mathrm{rad}}+Z_{s}|^{2}}{|Z_{\mathrm{rad}}+Z_{s}\left(1+\frac{Z_{\mathrm{rad}}}{Z_{h}}\right)|^{2}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Comparing this to the transmission loss curve:
\begin_inset Formula 
\begin{equation}
|\tau|_{\mathrm{Helmholtz}}^{2}=\frac{4|Z_{h}|^{2}}{|2Z_{h}+Z_{0}|^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
High output impedance limit 
\begin_inset Formula $(Z_{s}\gg Z_{\mathrm{rad}})$
\end_inset

, volume flow source
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
R_{P,\mathrm{Helmholtz}}=\frac{|Z_{h}|^{2}}{|Z_{h}+Z_{\mathrm{rad}}|^{2}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Low output impedance limit 
\begin_inset Formula $(Z_{s}\ll Z_{\mathrm{rad}})$
\end_inset

, pressure source
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
R_{P,\mathrm{Helmholtz}}=\frac{|Z_{h}|^{2}}{|Z_{h}+Z_{s}|^{2}}
\end{equation}

\end_inset


\end_layout

\begin_layout Subsubsection
Special case: barrier in an infinite space 
\begin_inset Formula $(Z_{s}=Z_{\mathrm{rad}})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
R_{P,\mathrm{Helmholtz}}=\frac{|Z_{h}|^{2}}{|Z_{h}+\frac{1}{2}Z_{\mathrm{rad}}|^{2}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Comparing limits to power transmission ratio
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
|\tau|^{2}=\frac{|Z_{h}|^{2}}{|Z_{h}+\frac{1}{2}Z_{0}|^{2}},
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
So the transmission loss is the reduction in transmitted sound power for
 the situation where the source output impedance equals the radiation impedance
 on the other side of the silencer.
\end_layout

\begin_layout Subsection
Multiple Helmholtz resonators at a single inlet
\end_layout

\begin_layout Standard
In case multiple resonators are connected to the same inlet, the parallel
 impedance can be computed by computing the equivalent parallel impedance:
\begin_inset Formula 
\begin{equation}
\frac{1}{Z_{h,\mathrm{tot}}}=\frac{1}{Z_{h,1}}+\frac{1}{Z_{h,2}}+\dots
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Transmission of the duct
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\left\{ \begin{array}{c}
p_{R}\\
U_{R}
\end{array}\right\} =\left[\begin{array}{cc}
\cos\left(kL\right) & -iZ_{0}\sin\left(kL\right)\\
-iZ_{0}^{-1}\sin\left(kL\right) & \cos\left(kL\right)
\end{array}\right]\left\{ \begin{array}{c}
p_{L}\\
U_{L}
\end{array}\right\} 
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
3D (FEM) Models
\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Apply equation of state:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{align*}
i\omega\rho+\rho_{0}\nabla\cdot\boldsymbol{u} & =0\\
i\omega\rho_{0}\boldsymbol{u} & =-\nabla p+\mu_{0}\nabla^{2}\boldsymbol{u}+\left(\frac{1}{3}\mu+\zeta\right)\nabla\left(\nabla\cdot\boldsymbol{u}\right)\\
i\omega\rho_{0}c_{p}T & =i\omega p+\kappa\nabla^{2}T\\
\frac{\rho}{\rho_{0}} & =\frac{p}{p_{0}}-\frac{T}{T_{0}}
\end{align*}

\end_inset


\end_layout

\begin_layout Plain Layout
Solving for 
\begin_inset Formula $i\omega\rho_{0}c_{p}T=i\omega p+\kappa\nabla^{2}T$
\end_inset


\end_layout

\begin_layout Plain Layout
: 
\begin_inset Formula $T=\frac{1}{\rho_{0}c_{p}}\left(1-h_{\kappa}\right)p$
\end_inset


\end_layout

\begin_layout Plain Layout
Where
\begin_inset Formula $\frac{i\delta_{\kappa}^{2}}{2}\nabla^{2}h_{\kappa}+h_{\kappa}=0$
\end_inset

 and
\end_layout

\begin_layout Plain Layout
Same for velocity, negliging 
\begin_inset Quotes eld
\end_inset

bulk
\begin_inset Quotes erd
\end_inset

 viscosity terms:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $i\omega\rho_{0}\boldsymbol{u}=-\nabla p+\mu_{0}\nabla^{2}\boldsymbol{u}$
\end_inset


\end_layout

\begin_layout Plain Layout
More or less solution:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\boldsymbol{u}=\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p$
\end_inset


\end_layout

\begin_layout Plain Layout
Where
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{2i}{\delta_{\nu}^{2}}\nabla^{2}h_{\nu}+h_{\nu}=0$
\end_inset

 and 
\begin_inset Formula $h_{\nu}|_{\mathrm{wall}}=1$
\end_inset

 for a no-slip b.c.
 and 0 for a slip b.c.
\end_layout

\begin_layout Plain Layout
Filling in the expression for eq of state,
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\rho=\frac{1}{c_{0}^{2}}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p$
\end_inset


\end_layout

\begin_layout Plain Layout
Substituting that one, for 
\begin_inset Formula $\rho$
\end_inset

 in continuity eq:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $i\omega\frac{1}{c_{0}^{2}}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p+\rho_{0}\nabla\cdot\boldsymbol{u}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\rho_{0}\nabla\cdot\boldsymbol{u}+i\frac{k}{c_{0}}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Fill in for momentum:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\rho_{0}\nabla\cdot\left(\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p\right)+i\frac{k}{c_{0}}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\nabla\cdot\left(\left(1-h_{\nu}\right)\nabla p\right)+k^{2}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Multiplying with weight factor, applying greens theorem:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}p_{w}k\left(1+\left(\gamma-1\right)h_{\kappa}\right)p-iz_{0}\nabla\cdot\boldsymbol{u}p_{w}\mathrm{d}V=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}p_{w}k\left(1+\left(\gamma-1\right)h_{\kappa}\right)p+iz_{0}\nabla p_{w}\cdot\boldsymbol{u}\mathrm{d}V=iz_{0}\oint_{S}p_{w}\boldsymbol{u}\cdot\boldsymbol{n}\mathrm{d}S$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}p_{w}k\left(1+\left(\gamma-1\right)h_{\kappa}\right)p+iz_{0}\nabla p_{w}\cdot\boldsymbol{u}\mathrm{d}V=iz_{0}\oint_{S}p_{w}\boldsymbol{u}\cdot\boldsymbol{n}\mathrm{d}S$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling in 
\begin_inset Formula $\boldsymbol{u}$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{i}{\rho_{0}\omega}\nabla p\left(1-\psi_{v}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}p_{w}k^{2}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p-\left(1-\psi_{v}\right)\nabla p_{w}\cdot\nabla p\mathrm{d}V=ikz_{0}\oint_{S}p_{w}\boldsymbol{u}\cdot\boldsymbol{n}\mathrm{d}S$
\end_inset


\end_layout

\begin_layout Plain Layout
Axially symmetric: 
\begin_inset Formula $\int_{z}\int_{r=0}^{a}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{z}\int_{r=0}^{a}\left(p_{w}k^{2}\left(1+\left(\gamma-1\right)\psi_{T}\right)p-\left(1-\psi_{v}\right)\nabla p_{w}\cdot\nabla p\right)2\pi r\mathrm{d}r\mathrm{d}z=ikz_{0}\oint_{S}p_{w}\boldsymbol{u}\cdot\boldsymbol{n}\mathrm{d}S$
\end_inset


\end_layout

\begin_layout Plain Layout
—– Which
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $i\omega\frac{1}{c_{0}^{2}}p\left(1+\left(\gamma-1\right)h_{\kappa}\right)+\rho_{0}\nabla\cdot\left(\frac{i}{\rho_{0}\omega}\nabla p\left(1-\psi_{v}\right)\right)=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $k^{2}p\left(1+\left(\gamma-1\right)h_{\kappa}\right)+\rho_{0}\nabla\cdot\left(\nabla p\left(1-\psi_{v}\right)\right)=0$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
From
\end_layout

\end_inset


\end_layout

\begin_layout Section
SLNS model
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{align}
\nabla^{2}h_{v}+\frac{2}{i\delta_{\nu}^{2}}h_{v} & =0,\\
\nabla^{2}h_{\kappa}+\frac{2}{i\delta_{\kappa}^{2}}h_{\kappa} & =0,\\
\frac{1}{k}\nabla\cdot\left(\left(1-h_{\nu}\right)\nabla p\right)+k\left(1+\left(\gamma-1\right)h_{\kappa}\right)p & =0\label{eq:slns}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
The velocity is:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\boldsymbol{u}=\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
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
\begin_inset Formula 
\begin{align}
h_{\nu} & =1\qquad\mathrm{at\,the\,wall}\\
h_{\kappa} & =1\qquad\mathrm{at\,the\,wall}
\end{align}

\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
For pressure / velocity b.c.'s
\begin_inset Formula 
\begin{equation}
\boldsymbol{u}=\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
Combine with pressure acoustics:
\end_layout

\begin_layout Plain Layout
Weak form:
\end_layout

\begin_layout Plain Layout
(-acpr.gradpx*acpr.gradtestpx-acpr.gradpy*acpr.gradtestpy-acpr.gradpz*acpr.gradtestpz-
acpr.p_t*test(pac)*acpr.ik^2)*acpr.delta/acpr.rho_c
\end_layout

\begin_layout Plain Layout
(-acpr.gradpx*acpr.gradtestpx-acpr.gradpy*acpr.gradtestpy-acpr.gradpz*acpr.gradtestpz-
acpr.p_t*test(pac)*acpr.ik^2)*acpr.delta/acpr.rho_c
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}\left[-\nabla p_{t}\cdot\nabla p-p_{t}p\left(ik\right)\right]\frac{\delta}{\rho_{0}c_{0}}\mathrm{dV}$
\end_inset


\end_layout

\begin_layout Plain Layout
Weak form of SLNS:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}p_{t}\left[\nabla\cdot\left(\left(1-h_{\nu}\right)\nabla p\right)+k^{2}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p\right]\frac{\delta}{\rho_{0}c_{0}}\mathrm{d}V$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}\left[-\nabla p_{t}\cdot\left(\left(1-h_{\nu}\right)\nabla p\right)+p_{t}k^{2}\left(1+\left(\gamma-1\right)h_{\kappa}\right)p\right]\frac{\delta}{\rho_{0}c_{0}}\mathrm{d}V$
\end_inset

+Boundary term.
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}\left[\underbrace{-\nabla p_{t}\cdot\nabla p-p_{t}\left(ik\right)^{2}p}_{\mathrm{already\,there}}+\nabla p_{t}\cdot\left(h_{\nu}\nabla p\right)-p_{t}\left(ik\right)^{2}p\left(\left(\gamma-1\right)h_{\kappa}\right)\right]\frac{\delta}{\rho_{0}c_{0}}\mathrm{d}V$
\end_inset


\end_layout

\begin_layout Plain Layout
Makes the weak contribution equal to:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\int_{V}\left[\nabla p_{t}\cdot\left(h_{\nu}\nabla p\right)+p_{t}\left(ik\right)^{2}p\left(\left(1-\gamma\right)h_{\kappa}\right)\right]\frac{\delta}{\rho_{0}c_{0}}\mathrm{d}V$
\end_inset


\end_layout

\begin_layout Plain Layout
Written out:
\end_layout

\begin_layout Plain Layout
(hnu*(test(px)*px+test(py)*py+pz*test(pz))+test(p)*p*acpr.ik^2*(1-gamma)*hkappa)*
acpr.delta/acpr.rho_c
\end_layout

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
DEPRECATED, we doen het met de pressure acoustics interface en een enkele
 weak contribution!
\end_layout

\begin_layout Section
Comsol implementation - General Form PDE
\end_layout

\begin_layout Plain Layout
Model in Comsol:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{equation}
e_{a}\frac{\partial^{2}p}{\partial t^{2}}+d_{a}\frac{\partial p}{\partial t}+\nabla\cdot\boldsymbol{\Gamma}=f
\end{equation}

\end_inset


\end_layout

\begin_layout Plain Layout
Comparing with Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:slns"

\end_inset

 results in:
\begin_inset Formula 
\begin{align}
\boldsymbol{\Gamma} & =\frac{1}{k}\left(1-h_{\nu}\right)\nabla p\\
f & =-k\left(1+\left(\gamma-1\right)h_{\kappa}\right)p
\end{align}

\end_inset


\end_layout

\begin_layout Section
Comsol implementation - prescribed velocity
\end_layout

\begin_layout Plain Layout
Flux / source term form in Comsol:
\begin_inset Formula 
\begin{equation}
-\boldsymbol{n}\cdot\boldsymbol{\Gamma}=g-qp
\end{equation}

\end_inset

From the mathematics, we find:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $k\boldsymbol{\Gamma}=\left(1-h_{\nu}\right)\nabla p$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\boldsymbol{u}=\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p$
\end_inset


\end_layout

\begin_layout Plain Layout
– Combine:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\rho_{0}\omega}{i}\boldsymbol{u}=\left(1-h_{\nu}\right)\nabla p$
\end_inset


\end_layout

\begin_layout Plain Layout
–
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\boldsymbol{\Gamma}=-iz_{0}\boldsymbol{u}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
-\boldsymbol{n}\cdot\boldsymbol{\Gamma}=iz_{0}\boldsymbol{u}\cdot\boldsymbol{n}\label{eq:Gam_vs_un}
\end{equation}

\end_inset

Such that:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\boldsymbol{u}\cdot\boldsymbol{n}=\frac{i}{\rho_{0}\omega}\left(1-h_{\nu}\right)\nabla p\cdot\boldsymbol{n}$
\end_inset


\end_layout

\begin_layout Plain Layout
Note that:
\begin_inset Formula 
\[
k\boldsymbol{\Gamma}=\left(1-h_{\nu}\right)\nabla p
\]

\end_inset


\end_layout

\begin_layout Plain Layout
Fill in:
\begin_inset Formula $iz_{0}\boldsymbol{u}\cdot\boldsymbol{n}=-\boldsymbol{\Gamma}\cdot\boldsymbol{n}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{align}
q & =0\\
g & =iz_{0}\boldsymbol{u}\cdot\boldsymbol{n}
\end{align}

\end_inset


\end_layout

\begin_layout Plain Layout
Moreover, at such a boundary, we need to set 
\begin_inset Formula $h_{\nu}$
\end_inset

 and 
\begin_inset Formula $h_{\kappa}$
\end_inset

 to 0.
\end_layout

\begin_layout Section
Normal impedance b.c.
\end_layout

\begin_layout Plain Layout
We set 
\begin_inset Formula 
\begin{equation}
z\boldsymbol{u}\cdot\boldsymbol{n}=p
\end{equation}

\end_inset


\end_layout

\begin_layout Plain Layout
Upon using Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Gam_vs_un"

\end_inset

, we find:
\end_layout

\begin_layout Plain Layout
Yields:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\frac{i}{z_{0}}\boldsymbol{n}\cdot\boldsymbol{\Gamma}=\boldsymbol{u}\cdot\boldsymbol{n}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
-\boldsymbol{n}\cdot\boldsymbol{\Gamma}=-i\frac{z_{0}}{z}p
\end{equation}

\end_inset


\end_layout

\begin_layout Plain Layout
Such that:
\begin_inset Formula 
\begin{align}
q & =i\frac{z_{0}}{z}\\
g & =0
\end{align}

\end_inset


\end_layout

\begin_layout Section
Interior impedance jump
\end_layout

\begin_layout Plain Layout
Equation:
\begin_inset Formula 
\begin{equation}
p_{\mathrm{up}}-p_{\mathrm{down}}=z\boldsymbol{u}\cdot\boldsymbol{n}_{\mathrm{up}}
\end{equation}

\end_inset


\end_layout

\begin_layout Plain Layout
It should be implemented as a 
\begin_inset Quotes eld
\end_inset

weak contribution
\begin_inset Quotes erd
\end_inset

.
 For that we refer the the weak form equation:
\begin_inset Note Note
status open

\begin_layout Plain Layout
Reverse engineering comsols weak contribution of such a split:
\end_layout

\begin_layout Plain Layout
-acpr.delta*acpr.iomega*(down(acpr.p_t)-up(acpr.p_t))*(down(test(acp))-up(test(acp))
)/acpr.Zi
\end_layout

\begin_layout Plain Layout
waar: delta = 1/omega^2
\end_layout

\begin_layout Plain Layout
Leest:
\end_layout

\begin_layout Plain Layout
-i/omega*(down(p)-up(p))*(down(test(p))-up(test(p))) /z
\end_layout

\begin_layout Plain Layout
We hebben altijd op een rand:
\begin_inset Formula 
\[
\]

\end_inset


\end_layout

\end_inset


\end_layout

\end_inset


\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

\begin_layout Standard

\family typewriter
(hnu*(test(px)*px+test(py)*py+pz*test(pz))+test(p)*p*acpr.ik^2*(1-gamma)*hkappa)*
acpr.delta/acpr.rho_c
\end_layout

\begin_layout Standard
Or we could write this with a custom density and speed of sound <— TODO!
\end_layout

\begin_layout Standard
2D Axisymmetric:
\end_layout

\begin_layout Standard

\family typewriter
(hnu*(test(pr)*pr+pz*test(pz))+test(p)*p*acpr.ik^2*(1-gamma)*hkappa)*acpr.delta/ac
pr.rho_c
\end_layout

\begin_layout Standard
\begin_inset CommandInset bibtex
LatexCommand bibtex
btprint "btPrintCited"
bibfiles "lrftubes"
options "plain"

\end_inset


\end_layout

\begin_layout Chapter
\start_of_appendix
Thermal relaxation in thick tubes
\end_layout

\begin_layout Section
\begin_inset CommandInset label
LatexCommand label
name "subsec:Thermal-relaxation-effect"

\end_inset

Thermal relaxation effect in thick tubes
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/prsduct_thermal_relax.pdf
	width 80text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic situation of a tube surrounded by a thick solid.
 Note that the transverse acoustic temperature is drawn to be not zero at
 the wall.
 This happens in case of thermal interaction with a solid with finite thermal
 effusivity.
\end_layout

\end_inset


\begin_inset CommandInset label
LatexCommand label
name "fig:prsduct-heatwave-solid"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
In this section, a formulation for 
\begin_inset Formula $\epsilon_{s}$
\end_inset

 is given for tubes where the temperature wave in the solid is present.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:prsduct-heatwave-solid"

\end_inset

 shows a schematic overview of the situation.
 As shown in the figure, the temperature wave accompanied with an acoustic
 wave results in heat conduction to/from the wall of the tube.
 To solve this interaction mathematically, the heat equation in the solid
 has to be solved.
 For constant thermal conductivity, density and heat capacity the heat equation
 of the solid is
\begin_inset Formula 
\begin{equation}
\rho_{s}c_{s}\frac{\partial\tilde{T}_{s}}{\partial t}=\kappa_{s}\nabla^{2}\tilde{T}_{s},
\end{equation}

\end_inset

where 
\begin_inset Formula $\rho_{s},c_{s},\tilde{T}_{s}$
\end_inset

 and 
\begin_inset Formula $\kappa_{s}$
\end_inset

 are the density, specific heat, temperature and thermal conductivity of
 the solid, respectively.
 In frequency domain and using cylindrical coordinates, assuming axial symmetry,
 this can be written as
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$r$"
description "Radial position in cylindrical coordinates\\nomunit{\\si{\\m}}"
literal "true"

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\left(r^{2}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{\partial^{2}}{\partial x^{2}}\right)+r\frac{\partial}{\partial r}+\frac{2}{i\delta_{s}^{2}}r^{2}\right)T_{s}=0,
\end{equation}

\end_inset

where 
\begin_inset Formula $\delta_{s}$
\end_inset

 is
\begin_inset Formula 
\begin{equation}
\delta_{s}=\sqrt{\frac{2\kappa_{s}}{\rho_{s}c_{s}\omega}}.
\end{equation}

\end_inset

Now, since 
\begin_inset Formula $\partial T_{s}/\partial x\sim\frac{\delta_{s}}{\lambda}\frac{\partial T_{s}}{\partial r}$
\end_inset

, the second order derivative of the temperature in the axial direction
 can be neglected.
 In that case, the differential equation to solve for is
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\rho_{s}c_{s}i\omega T_{s}=\kappa_{s}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}\right)T_{s}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $-\kappa_{s}\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}\right)T_{s}+\rho_{s}c_{s}i\omega T_{s}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}\right)T_{s}+2\frac{\rho_{s}c_{s}\omega}{2\kappa_{s}i}T_{s}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\delta_{s}^{2}=\frac{2\kappa_{s}}{\rho_{s}c_{s}\omega}$
\end_inset

<<< subst
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}\right)T_{s}+\frac{2}{i\delta_{s}^{2}}T_{s}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Multiply with 
\begin_inset Formula $r^{2}$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(r^{2}\frac{\partial^{2}}{\partial r^{2}}+r\frac{\partial}{\partial r}+\frac{2}{i\delta_{s}^{2}}r^{2}\right)T_{s}=0$
\end_inset


\end_layout

\begin_layout Plain Layout
Say: 
\begin_inset Formula $\xi^{2}=\frac{2}{i\delta_{s}^{2}}r^{2}\Rightarrow$
\end_inset


\end_layout

\begin_layout Plain Layout
Then:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\partial^{2}}{\partial r^{2}}=$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\left(r^{2}\frac{\partial^{2}}{\partial r^{2}}+r\frac{\partial}{\partial r}+\frac{2}{i\delta_{s}^{2}}r^{2}\right)T_{s}=0,
\end{equation}

\end_inset

which is a Bessel differential equation of the zero'th order in 
\begin_inset Formula $T_{s}$
\end_inset

.
 The solutions is sought in terms of traveling cylindrical waves:
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset Formula $\sqrt{\frac{2}{i}}=\sqrt{-2i}=\pm\left(i-1\right)$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
T_{s}=C_{1}H_{0}^{(1)}\left(\left(i-1\right)\frac{r}{\delta_{s}}\right)+C_{2}H_{0}^{(2)}\left(\left(i-1\right)\frac{r}{\delta_{s}}\right),
\end{equation}

\end_inset

where 
\begin_inset Formula $C_{1}$
\end_inset

 and 
\begin_inset Formula $C_{2}$
\end_inset

 constants to be determined from the boundary conditions, and 
\begin_inset Formula $H_{\alpha}^{(i)}$
\end_inset

 is the cylindrical Hankel function of the 
\begin_inset Formula $(i)^{\mathrm{th}}$
\end_inset

 kind and order 
\begin_inset Formula $\alpha$
\end_inset

.
 If we require 
\begin_inset Formula $T_{s}\to0$
\end_inset

 as 
\begin_inset Formula $r\to\infty$
\end_inset

, the constant 
\begin_inset Formula $C_{2}$
\end_inset

 is required to be 
\begin_inset Formula $0$
\end_inset

.
 From the acoustic energy equation, a similar differential equation can
 be found for the acoustic temperature in the fluid:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $\rho_{0}c_{p}i\omega T=i\omega\alpha_{P}T_{0}p+\kappa\nabla^{2}T$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\nabla^{2}-2\frac{\omega\rho_{0}c_{p}}{2\kappa}i\right)T=-\frac{1}{\kappa}i\omega\alpha_{P}T_{0}p$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\left(\nabla^{2}+\frac{2}{i\delta_{\kappa}^{2}}\right)T=\frac{2}{i\delta_{s}^{2}}\frac{\alpha_{P}T_{0}}{\rho_{0}c_{p}}p$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\[
\left(r^{2}\frac{\partial^{2}}{\partial r^{2}}+r\frac{\partial}{\partial r}+\frac{2}{i\delta_{s}^{2}}r^{2}\right)T=\frac{2}{i\delta_{s}^{2}}\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p,
\]

\end_inset

for which the (partial) solution is 
\begin_inset Formula 
\begin{equation}
T=\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\left(1-C_{3}J_{0}\left(\left(i-1\right)\frac{r}{\delta_{\kappa}}\right)\right).\label{eq:temp_partial_sol}
\end{equation}

\end_inset

To attain at Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:temp_partial_sol"

\end_inset

, use has been made of the fact that the temperature should be finite at
 
\begin_inset Formula $r=0$
\end_inset

.
 
\begin_inset Formula $C_{3}$
\end_inset

 is a constant that is to be determined from the boundary conditions at
 the solid-fluid interface.
 These boundary conditions are:
\begin_inset Formula 
\begin{align}
T_{s}|_{r=a} & =T|_{r=a},\\
-\kappa_{s}\frac{\partial T_{s}}{\partial r}|_{r=a} & =-\kappa\frac{\partial T}{\partial r}|_{r=a},
\end{align}

\end_inset

i.e.
 continuity of the temperature and the heat flux at the interface.
 This yields two equations for two unknowns (
\begin_inset Formula $C_{1}$
\end_inset

 and 
\begin_inset Formula $C_{3}$
\end_inset

, 
\begin_inset Formula $C_{2}$
\end_inset

 is already argued to be 
\begin_inset Formula $0$
\end_inset

).
 Solving for the acoustic temperature yields:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
\begin_inset Formula $T|_{r=a}=T_{s}|_{r=a}$
\end_inset


\end_layout

\begin_layout Plain Layout
–
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $C_{1}H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)=\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\left(1-C_{3}J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)\right)\Rightarrow C_{1}=\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\frac{\left(1-C_{3}J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)\right)}{H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}$
\end_inset

 (1)
\end_layout

\begin_layout Plain Layout
Derivative b.c.
\end_layout

\begin_layout Plain Layout
– 
\begin_inset Formula $-\frac{\partial T}{\partial r}|_{r=a}=-\frac{\kappa_{s}}{\kappa}\frac{\partial T_{s}}{\partial r}|_{r=a}$
\end_inset


\end_layout

\begin_layout Plain Layout
where 
\begin_inset Formula $-\frac{\partial T}{\partial r}|_{r=a}=\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\frac{\left(i-1\right)}{\delta_{\kappa}}C_{3}J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
using 
\begin_inset Formula $\frac{\partial H_{0}^{(1)}(z)}{\partial z}=-H_{1}^{(1)}(z)$
\end_inset

 ==> 
\begin_inset Formula $-\frac{\kappa}{\kappa_{s}}\frac{\partial T_{s}}{\partial r}|_{r=a}=\frac{\kappa}{\kappa_{s}}C_{1}\frac{\left(i-1\right)}{\delta_{s}}H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Such that: 
\begin_inset Formula $\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\frac{\left(i-1\right)}{\delta_{\kappa}}C_{3}J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)=\frac{\kappa_{s}}{\kappa}C_{1}\frac{\left(i-1\right)}{\delta_{s}}H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling in
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\frac{\left(i-1\right)}{\delta_{\kappa}}C_{3}J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)=\frac{\kappa_{s}}{\kappa}\left(\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}p\frac{\left(1-C_{3}J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)\right)}{H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}\right)\frac{\left(i-1\right)}{\delta_{s}}H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Solving for 
\begin_inset Formula $C_{3}$
\end_inset

 gives:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $C_{3}=\frac{1}{\left[\frac{J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}{\frac{\kappa_{s}}{\kappa}\frac{\delta_{\kappa}}{\delta_{s}}H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}+J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)\right]}$
\end_inset


\end_layout

\begin_layout Plain Layout
or:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $C_{3}=\frac{1}{\left[\left(1+\epsilon_{s}\right)J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)\right]}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\epsilon_{s}=\frac{\kappa\delta_{s}}{\delta_{\kappa}\kappa_{s}}\frac{H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)}{H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{\kappa\delta_{s}}{\delta_{\kappa}\kappa_{s}}=\sqrt{\frac{\kappa^{2}\delta_{s}^{2}}{\kappa_{s}^{2}\delta_{\kappa}^{2}}}=\sqrt{\frac{\kappa\rho_{0}c_{p}}{\kappa\rho_{s}c_{s}}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\[
T=\frac{\alpha_{p}T_{0}}{\rho_{0}c_{p}}\left(1-\frac{1}{\left(1+\epsilon_{s}\right)}\frac{J_{0}\left(\left(i-1\right)\frac{r}{\delta_{\kappa}}\right)}{J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)}\right)p,
\]

\end_inset

where
\begin_inset Formula 
\begin{equation}
\epsilon_{s}=\frac{e_{f}}{e_{s}}\frac{J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}{J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)},
\end{equation}

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout
-
\end_layout

\begin_layout Plain Layout
-Asymptotic form of the Hankel function for large argument, and
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $-\pi<\arg(z)<2\pi$
\end_inset

:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $H_{\alpha}^{(1)}(z)\sim\sqrt{\frac{2}{\pi z}}e^{i\left(z-\pi\frac{1+2\alpha}{4}\right)}$
\end_inset


\end_layout

\begin_layout Plain Layout
And for
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $J_{\alpha}(z)\sim\sqrt{\frac{2}{\pi z}}\cos\left(z-\pi\frac{1+2\alpha}{4}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout
Filling this in into
\end_layout

\begin_layout Plain Layout
\begin_inset Formula $\frac{e_{f}}{e_{s}}\cdot-ii=\frac{e_{f}}{e_{s}}$
\end_inset


\end_layout

\end_inset

where 
\begin_inset Formula $e_{f}$
\end_inset

 is the thermal effusivity
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$e$"
description "Thermal effusivity\\nomunit{\\si{\\joule\\per\\square\\metre\\kelvin\\second\\tothe{ \\frac{1}{2}  } }}"
literal "true"

\end_inset


\end_layout

\end_inset

 of the fluid, and 
\begin_inset Formula $e_{s}$
\end_inset

 the thermal effusivity of the solid, such that the ratio is
\begin_inset Formula 
\begin{equation}
\frac{e_{f}}{e_{s}}=\sqrt{\frac{\kappa\rho_{0}c_{p}}{\kappa_{s}\rho_{s}c_{s}}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Note that for large 
\begin_inset Formula $a/\delta_{\kappa}$
\end_inset

: 
\begin_inset Formula 
\begin{equation}
\frac{J_{1}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)}{J_{0}\left(\left(i-1\right)\frac{a}{\delta_{\kappa}}\right)}\to i,
\end{equation}

\end_inset

and for large 
\begin_inset Formula $a/\delta_{s}$
\end_inset


\begin_inset Formula 
\begin{equation}
\frac{H_{0}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}{H_{1}^{(1)}\left(\left(i-1\right)\frac{a}{\delta_{s}}\right)}\to-i,
\end{equation}

\end_inset

such that for both numbers large
\begin_inset Formula 
\begin{equation}
\epsilon_{s}\to\frac{e_{f}}{e_{s}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Chapter
Derivation of Karal's discontinuity factor
\begin_inset CommandInset label
LatexCommand label
name "chap:Derivation-of-Karal's"

\end_inset


\end_layout

\begin_layout Standard

\series bold
Note: this documentation is incomplete.
\end_layout

\begin_layout Standard
\begin_inset Float figure
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename img/discontinuity_appendix.pdf
	width 60text%

\end_inset


\begin_inset Caption Standard

\begin_layout Plain Layout
Schematic of a discontinuity at the interface between two tubes with different
 radius.
 Domain B is the smaller tube and domain C is the larger tube.
 The radius of the tube in domain B is 
\begin_inset Formula $b$
\end_inset

, and the radius of the tube in domain C is 
\begin_inset Formula $c$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Plain Layout
\align center
\begin_inset CommandInset label
LatexCommand label
name "fig:karal-1"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
This appendix describes the derivation of Karal's discontinuity factor.
 The following assumptions underlie the model:
\end_layout

\begin_layout Itemize
\begin_inset Formula $z=0$
\end_inset

 : position of the discontinuity
\end_layout

\begin_layout Itemize
Assume 
\begin_inset Formula $f\ll f_{c}$
\end_inset

, such that far away from the discontinuity, only propagating modes exist.
\end_layout

\begin_layout Itemize
Assume axial symmetry, so dependence of 
\begin_inset Formula $\theta$
\end_inset

 is dropped
\end_layout

\begin_layout Standard
In cylindrical coordinates, the solution of the Helmholtz equation can be
 written in terms of cylindrical harmonics 
\begin_inset CommandInset citation
LatexCommand cite
key "blackstock_fundamentals_2000"
literal "true"

\end_inset

.
 Assuming axial symmetrySuch that the acoustic pressure in for example tube
 
\begin_inset Formula $B$
\end_inset

 can be written as:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
p_{B}=\left\{ \begin{array}{c}
J_{m}\left(k_{r}r\right)\\
N_{m}\left(k_{r}r\right)
\end{array}\right\} \left\{ \begin{array}{c}
e^{im\phi}\\
e^{-im\phi}
\end{array}\right\} \left\{ \begin{array}{c}
e^{\beta z}\\
e^{-\beta z}
\end{array}\right\} 
\end{equation}

\end_inset

where 
\begin_inset Formula $J_{m}$
\end_inset

 is the cylindrical Bessel function of order
\begin_inset Formula 
\begin{equation}
k_{r}^{2}-\beta^{2}=k^{2}.
\end{equation}

\end_inset

Using the boundary condition that
\begin_inset Formula 
\begin{equation}
\frac{\partial p_{B}}{\partial r}|_{r=b}=0,
\end{equation}

\end_inset

and assuming axial symmetry (only the 
\begin_inset Formula $m=0$
\end_inset

 modes) it follows that
\begin_inset Formula 
\begin{equation}
\frac{\partial J_{0}}{\partial r}\left(k_{r}b\right)|_{r=b}=0.
\end{equation}

\end_inset

Assuming that 
\begin_inset Formula $\alpha_{k}$
\end_inset

 is the 
\begin_inset Formula $k^{\mathrm{th}}$
\end_inset

 zero of 
\begin_inset Formula $J_{0}^{'}(x)$
\end_inset

, we can write for 
\begin_inset Formula $k_{r,k}$
\end_inset

:
\begin_inset Formula 
\begin{equation}
k_{r,k}=\frac{\alpha_{k}}{b}.
\end{equation}

\end_inset

Hence we find the following reduced expression for the pressure in tube
 
\begin_inset Formula $B$
\end_inset

: 
\begin_inset Formula 
\begin{equation}
p_{B}=B_{0}^{0}\exp\left(ikz\right)+B_{0}^{1}\exp\left(-ikz\right)+\sum_{n=1}^{\infty}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)\left\{ \begin{array}{c}
e^{\beta_{n}z}\\
e^{-\beta_{n}z}
\end{array}\right\} ,
\end{equation}

\end_inset

where accordingly, 
\begin_inset Formula 
\begin{equation}
\beta_{k}^{2}=\left(\frac{\alpha_{k}}{b}\right)^{2}-k^{2}\label{eq:beta_k}
\end{equation}

\end_inset

For 
\begin_inset Formula $k^{2}<\left(\alpha_{k}/b\right)^{2}$
\end_inset

, 
\begin_inset Formula $\beta_{k}^{2}>0$
\end_inset

, the modes are evanescent.
 And since we only allow finite solutions for 
\begin_inset Formula $z\leq0$
\end_inset

, the final results for 
\begin_inset Formula $p_{B}$
\end_inset

 is
\begin_inset Formula 
\begin{equation}
p_{B}=B_{0}^{0}\exp\left(ikz\right)+B_{0}^{1}\exp\left(-ikz\right)+\sum_{n=1}^{\infty}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)e^{\beta_{n}z},
\end{equation}

\end_inset

where 
\begin_inset Formula $\beta_{n}$
\end_inset

 is defined as the positive root of the r.h.s.
 of Eq.
\begin_inset space ~
\end_inset


\begin_inset CommandInset ref
LatexCommand ref
reference "eq:beta_k"

\end_inset

.
 We simplify this relation to:
\begin_inset Formula 
\begin{equation}
p_{B}(z)=p_{B}^{0}(z)+\sum_{n=1}^{\infty}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)e^{\beta_{n}z}.
\end{equation}

\end_inset

For the velocity we find
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u=\frac{i}{\omega\rho_{0}}\frac{\partial p_{B}}{\partial z}=u_{B}^{0}(z)+\sum_{n=1}^{\infty}\frac{i\beta_{n}}{\omega\rho_{0}}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)e^{\beta_{n}z}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
u_{B}(z)=u_{B}^{0}(z)+\sum_{n=1}^{\infty}Y_{B,n}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)e^{\beta_{n}z},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
Y_{B,n}=\frac{i\beta_{n}}{\omega\rho_{0}}.
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Similarly, for the positive 
\begin_inset Formula $z$
\end_inset

 we find
\begin_inset Formula 
\begin{equation}
p_{C}(z)=P_{C}^{0}(z)+\sum_{m=1}^{\infty}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)e^{-\gamma_{m}z},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
\gamma_{m}=\sqrt{\left(\frac{\alpha_{m}}{c}\right)^{2}-k^{2}}.
\end{equation}

\end_inset

and
\begin_inset Formula 
\begin{equation}
u_{C}(z)=u_{C}^{0}(z)+\sum_{m=1}^{\infty}Y_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)e^{-\gamma_{m}z},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
Y_{C,m}=-\frac{i\gamma_{m}}{\omega\rho_{0}}
\end{equation}

\end_inset


\end_layout

\begin_layout Section
Boundary conditions
\end_layout

\begin_layout Standard
At the interface (
\begin_inset Formula $z=0$
\end_inset

), the following boundary conditions are valid:
\begin_inset Formula 
\begin{align}
u_{B}|_{z=0} & =u_{C}|_{z=0} & 0\leq r\leq b\label{eq:derivative1bc}\\
u_{C}|_{z=0} & =0 & b\leq r\leq c\label{eq:derivative2bc}\\
p_{B} & =p_{C} & 0\leq r\leq b\label{eq:continuitybc}
\end{align}

\end_inset

Taking Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:derivative1bc"

\end_inset

, multiply by 
\begin_inset Formula $r$
\end_inset

 and integrating from 
\begin_inset Formula $0$
\end_inset

 to 
\begin_inset Formula $c$
\end_inset

 , taking into account Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:derivative2bc"

\end_inset

 yields:
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{B}(z)=u_{B}^{0}(z)+\sum_{n=1}^{\infty}\zeta_{B,n}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)e^{\beta_{n}z}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Integrating from 0 to 
\begin_inset Formula $b$
\end_inset

 for 
\begin_inset Formula $u_{B}$
\end_inset

 and integrating from 0 to 
\begin_inset Formula $c$
\end_inset

 for 
\begin_inset Formula $u_{C}$
\end_inset

 cancels out the Bessel functions, as the primitive of 
\begin_inset Formula $J_{0}(x)x$
\end_inset

 is 
\begin_inset Formula $J_{1}(x)x$
\end_inset

, for which due to the no-slip b.c.
 the resulting integral is zero, and at 
\begin_inset Formula $r=0$
\end_inset

, the integral is zero as well.
 Hence we obtain only the propagating mode contribution to the volume flow.
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
b^{2}u_{B}^{0}=c^{2}u_{C}^{0}
\end{equation}

\end_inset

We require one more equation at the interface, which is found from the continuit
y boundary conditions as well.
 Multiplying Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:derivative1bc"

\end_inset

 with 
\begin_inset Formula $J_{0}(\alpha_{q}\frac{r}{c})r$
\end_inset

 and integrating setting 
\begin_inset Formula $q=m$
\end_inset

 and dividing by 
\begin_inset Formula $bc$
\end_inset

 yields:
\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{B}=u_{B}^{0}+\sum_{n=1}^{\infty}\zeta_{B,n}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{C}=u_{C}^{0}+\sum_{m=1}^{\infty}\zeta_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
–
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{b}u_{B}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
– Work out stuff, first line:
\end_layout

\begin_layout Plain Layout

\lang english
- Using the rule:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\int J_{0}(C_{1}x)J_{0}(C_{2}x)x\mathrm{d}x=x\frac{C_{1}J_{1}(C_{1}x)J_{0}(C_{2}x)-C_{2}J_{0}\left(C_{1}x\right)J_{1}(C_{2}x)}{C_{1}^{2}-C_{2}^{2}}
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
–> 
\begin_inset Formula $C_{1}=\frac{\alpha_{q}}{c}$
\end_inset

; 
\begin_inset Formula $C_{2}=\frac{\alpha_{n}}{b}$
\end_inset


\begin_inset Formula $x=b$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{b}u_{B}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=u_{B}^{0}J_{1}(\alpha_{q}\frac{b}{c})\frac{bc}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}b\frac{\frac{\alpha_{q}}{c}J_{1}(\frac{\alpha_{q}}{c}b)J_{0}(\frac{\alpha_{n}}{b}b)-\frac{\alpha_{n}}{b}J_{0}\left(\frac{\alpha_{q}}{c}b\right)J_{1}(\frac{\alpha_{n}}{b}b)}{\left(\frac{\alpha_{q}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}=$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Using: 
\begin_inset Formula $J_{1}\left(\alpha_{i}\right)=0$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{b}u_{B}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=u_{B}^{0}J_{1}(\alpha_{q}\frac{b}{c})\frac{bc}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{b}{\left(\frac{\alpha_{q}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}\frac{\alpha_{q}}{c}J_{1}(\frac{\alpha_{q}}{c}b)J_{0}(\frac{\alpha_{n}}{b}b)=$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Using: 
\begin_inset Formula $\rho=\frac{b}{c}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{b}u_{B}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=u_{B}^{0}J_{1}(\alpha_{q}\frac{b}{c})\frac{bc}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{\alpha_{q}\rho}{\left(\frac{\alpha_{q}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\frac{\alpha_{q}}{c}b)J_{0}(\frac{\alpha_{n}}{b}b)=$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Setting: 
\begin_inset Formula $q=m$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{b}u_{B}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{bc}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{\alpha_{m}\rho}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
———————————————————————
\end_layout

\begin_layout Plain Layout

\lang english
And the rhs:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=\int\limits _{0}^{c}\left[u_{C}^{0}J_{0}(\alpha_{q}\frac{r}{c})r+\sum_{m=1}^{\infty}Y_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)J_{0}(\alpha_{q}\frac{r}{c})r\right]\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=\int\limits _{0}^{c}\left[\sum_{m=1}^{\infty}Y_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)J_{0}(\alpha_{q}\frac{r}{c})r\right]\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Setting: 
\begin_inset Formula $q=m$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=\int\limits _{0}^{c}\left[\sum_{m=1}^{\infty}Y_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)J_{0}(\alpha_{m}\frac{r}{c})r\right]\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Using the rule:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\int J_{0}(C_{1}x)^{2}x\mathrm{d}x=\frac{1}{2}x^{2}\left(J_{0}(C_{1}x)^{2}+J_{1}(C_{1}x)^{2}\right)
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $C_{1}=\alpha_{m}\frac{r}{c}$
\end_inset

, 
\begin_inset Formula $x=c$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=Y_{C,m}C_{m}\frac{1}{2}c^{2}\left(J_{0}(\alpha_{m}\frac{c}{c})^{2}+J_{1}(\alpha_{m}\frac{c}{c})^{2}\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int\limits _{0}^{c}u_{C}J_{0}(\alpha_{q}\frac{r}{c})r\mathrm{d}r=Y_{C,m}C_{m}\frac{1}{2}c^{2}J_{0}(\alpha_{m})^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
— OR:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{bc}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{\alpha_{m}\rho}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})=Y_{C,m}C_{m}\frac{1}{2}c^{2}J_{0}(\alpha_{m})^{2}
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
– Divide by bc:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{1}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{\alpha_{m}\rho}{\left[\rho\alpha_{m}^{2}-\rho^{-1}\alpha_{n}^{2}\right]}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})=Y_{C,m}C_{m}\frac{1}{2}\rho^{-1}J_{0}(\alpha_{m})^{2}
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
- Deel teller en noemer in breuk door 
\begin_inset Formula $\rho$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{1}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}B_{n}\frac{\alpha_{m}}{\alpha_{m}^{2}-\frac{\alpha_{n}^{2}}{\rho^{2}}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})=Y_{C,m}C_{m}\frac{1}{2}\rho^{-1}J_{0}(\alpha_{m})^{2}
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{1}{\alpha_{q}}+\sum_{n=1}^{\infty}Y_{B,n}T_{mn}B_{n}=Y_{C,m}\frac{1}{2}\rho^{-1}J_{0}(\alpha_{m})^{2}C_{m},
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{equation}
T_{mn}=\frac{\alpha_{m}}{\alpha_{m}^{2}-\frac{\alpha_{n}^{2}}{\rho^{2}}}J_{0}\left(\alpha_{n}\right)J_{1}\left(\alpha_{m}\rho\right).
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
Setting 
\begin_inset Formula $p_{B}=p_{C}$
\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{B}(z=0)r\mathrm{d}r=\int_{0}^{b}\left[p_{B}^{0}+\sum_{n=1}^{\infty}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)\right]r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{B}(z=0)r\mathrm{d}r=\frac{b^{2}}{2}p_{B}^{0}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
———————————————–
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)r\mathrm{d}r=\int_{0}^{b}\left[p_{C}^{0}+\sum_{m=1}^{\infty}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)\right]r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)r\mathrm{d}r=\frac{b^{2}}{2}p_{C}^{0}+\sum_{m=1}^{\infty}\frac{bc}{\alpha_{m}}C_{m}J_{1}\left(\alpha_{m}\rho\right)$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Such that
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\frac{b^{2}}{2}p_{B}^{0}=\frac{b^{2}}{2}p_{C}^{0}+\sum_{m=1}^{\infty}\frac{bc}{\alpha_{m}}C_{m}J_{1}\left(\alpha_{m}\rho\right)
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Divide by 
\begin_inset Formula $\frac{b^{2}}{2}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
p_{B}^{0}=p_{C}^{0}+2\sum_{m=1}^{\infty}\frac{J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}C_{m}
\]

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
p_{B}^{0}=p_{C}^{0}+2\sum_{m=1}^{\infty}\frac{J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}C_{m}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{B}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\int_{0}^{b}\left[p_{B}^{0}J_{0}\left(\alpha_{p}\frac{r}{b}\right)r+\sum_{n=1}^{\infty}B_{n}J_{0}\left(\alpha_{n}\frac{r}{b}\right)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\right]\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{B}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\sum_{n=1}^{\infty}B_{n}\int_{0}^{b}J_{0}\left(\alpha_{n}\frac{r}{b}\right)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Setting 
\begin_inset Formula $p=n$
\end_inset

 en
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\int J_{0}(C_{1}x)^{2}x\mathrm{d}x=\frac{1}{2}x^{2}\left(J_{0}(C_{1}x)^{2}+J_{1}(C_{1}x)^{2}\right)
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $C_{1}=\frac{\alpha_{n}}{b}$
\end_inset

 en 
\begin_inset Formula $x=b$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{B}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=B_{n}\frac{1}{2}b^{2}J_{0}(\alpha_{n})^{2}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
– Zelfde voor integraal voor 
\begin_inset Formula $p_{C}$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\int_{0}^{b}\left[P_{C}^{0}+\sum_{m=1}^{\infty}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)\right]J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\sum_{m=1}^{\infty}C_{m}\int_{0}^{b}J_{0}\left(\alpha_{m}\frac{r}{c}\right)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Gebruik de regel:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
\int J_{0}(C_{1}x)J_{0}(C_{2}x)x\mathrm{d}x=x\frac{C_{1}J_{1}(C_{1}x)J_{0}(C_{2}x)-C_{2}J_{0}\left(C_{1}x\right)J_{1}(C_{2}x)}{C_{1}^{2}-C_{2}^{2}}
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Waarbij: 
\begin_inset Formula $C_{1}=\frac{\alpha_{m}}{c}$
\end_inset

 , 
\begin_inset Formula $C_{2}=\frac{\alpha_{p}}{b}$
\end_inset

 ; 
\begin_inset Formula $x=b$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\sum_{m=1}^{\infty}C_{m}b\frac{\frac{\alpha_{m}}{c}J_{1}(\frac{\alpha_{m}}{c}b)J_{0}(\frac{\alpha_{p}}{b}b)-\frac{\alpha_{p}}{b}J_{0}\left(\frac{\alpha_{m}}{c}x\right)J_{1}(\frac{\alpha_{p}}{b}b)}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{p}}{b}\right)^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\sum_{m=1}^{\infty}C_{m}\frac{\rho\alpha_{m}}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{p}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{p})$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Zet 
\begin_inset Formula $p=n$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\int_{0}^{b}p_{C}(z=0)J_{0}\left(\alpha_{p}\frac{r}{b}\right)r\mathrm{d}r=\sum_{m=1}^{\infty}C_{m}\frac{\rho\alpha_{m}}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Zodat:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
B_{n}\frac{1}{2}b^{2}J_{0}(\alpha_{n})^{2}=\sum_{m=1}^{\infty}C_{m}\frac{\rho\alpha_{m}}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})
\]

\end_inset


\end_layout

\end_inset


\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
B_{n}\frac{1}{2}b^{2}J_{0}(\alpha_{n})^{2}=\sum_{m=1}^{\infty}C_{m}\frac{\rho\alpha_{m}}{\left(\frac{\alpha_{m}}{c}\right)^{2}-\left(\frac{\alpha_{n}}{b}\right)^{2}}J_{1}(\alpha_{m}\rho)J_{0}(\alpha_{n})
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Deel linker en rechterzijde door 
\begin_inset Formula $\frac{1}{2}b^{2}$
\end_inset

:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
B_{n}J_{0}(\alpha_{n})^{2}=2\sum_{m=1}^{\infty}\rho^{-1}C_{m}\frac{\alpha_{m}}{\alpha_{m}^{2}-\frac{\alpha_{n}^{2}}{\rho^{2}}}J_{0}(\alpha_{n})J_{1}(\alpha_{m}\rho)
\]

\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Oftewel:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula 
\[
B_{n}J_{0}(\alpha_{n})^{2}=\frac{2}{\rho}\sum_{m=1}^{\infty}T_{mn}C_{m}
\]

\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
B_{n}J_{0}(\alpha_{n})^{2}=\frac{2}{\rho}\sum_{m=1}^{\infty}T_{mn}C_{m}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $B_{n}=\frac{2}{\rho J_{0}(\alpha_{n})^{2}}\sum_{q=1}^{\infty}T_{qn}C_{q}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{1}{\alpha_{m}}+\sum_{n=1}^{\infty}Y_{B,n}T_{mn}\frac{2}{\rho J_{0}(\alpha_{n})^{2}}\sum_{q=1}^{\infty}T_{qn}C_{q}=Y_{C,m}\frac{1}{2\rho}J_{0}(\alpha_{m})^{2}C_{m}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{2Y_{B,n}}{J_{0}(\alpha_{n})^{2}}T_{mn}\sum_{q=1}^{\infty}T_{qn}C_{q}-\frac{1}{2}Y_{C,m}J_{0}(\alpha_{m})^{2}C_{m}=-u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{\rho}{\alpha_{m}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
—————Setting ——- 
\begin_inset Formula $C_{m}=ikbu_{B}^{0}z_{0}D_{m}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{2Y_{B,n}}{J_{0}(\alpha_{n})^{2}}ikbu_{B}^{0}z_{0}T_{mn}\sum_{q=1}^{\infty}T_{qn}D_{q}-\frac{1}{2}Y_{C,m}ikbD_{m}u_{B}^{0}z_{0}J_{0}(\alpha_{m})^{2}D_{m}=-u_{B}^{0}J_{1}(\alpha_{m}\rho)\frac{\rho}{\alpha_{q}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Using: 
\begin_inset Formula $z_{0}Y_{B,n}=\frac{i\beta_{n}}{k}$
\end_inset

 and 
\begin_inset Formula $z_{0}Y_{C,m}=-\frac{i\gamma_{m}}{k}$
\end_inset

 and ,
\begin_inset Formula $\gamma_{m}=\sqrt{\left(\frac{\alpha_{m}}{c}\right)^{2}-k^{2}}$
\end_inset

 and 
\begin_inset Formula $\beta_{n}=\sqrt{\left(\frac{\alpha_{n}}{b}\right)^{2}-k^{2}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{2}{J_{0}(\alpha_{n})^{2}}\sqrt{\left(\frac{\alpha_{n}}{bk}\right)^{2}-1}kbT_{mn}\sum_{q=1}^{\infty}T_{qn}D_{q}+\sqrt{\left(\frac{\alpha_{m}}{kc}\right)^{2}-1}\frac{1}{2}kbD_{m}J_{0}(\alpha_{m})^{2}D_{m}=+J_{1}(\alpha_{m}\rho)\frac{\rho}{\alpha_{m}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
When 
\begin_inset Formula $kc\sim kb\ll1$
\end_inset

, this can be rewritten to:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{2\alpha_{n}}{J_{0}(\alpha_{n})^{2}}T_{mn}\sum_{q=1}^{\infty}T_{qn}D_{q}+\frac{\alpha_{m}\rho}{2}D_{m}J_{0}(\alpha_{m})^{2}D_{m}=J_{1}(\alpha_{m}\rho)\frac{\rho}{\alpha_{m}}$
\end_inset


\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\sum_{n=1}^{\infty}\frac{2\alpha_{n}}{J_{0}(\alpha_{n})^{2}}T_{mn}\sum_{q=1}^{\infty}T_{qn}D_{q}+\frac{1}{2}\rho\alpha_{m}J_{0}(\alpha_{m})^{2}D_{m}=J_{1}(\alpha_{m}\rho)\frac{\rho}{\alpha_{m}},\label{eq:D_meq}
\end{equation}

\end_inset

where 
\begin_inset Formula 
\begin{equation}
D_{m}=\frac{C_{m}}{ikbu_{B}^{0}z_{0}}
\end{equation}

\end_inset

Eq.
 
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:D_meq"

\end_inset

 is a set of infinite equations in terms of an infinite number of unknowns
 for 
\begin_inset Formula $D_{m}$
\end_inset

.
 In matrix algebra for a finite set, this can be written as
\begin_inset Formula 
\begin{equation}
(\boldsymbol{M}_{1}\cdot\boldsymbol{M}_{2}+\boldsymbol{K})\cdot\boldsymbol{D}=\boldsymbol{R}
\end{equation}

\end_inset

where
\begin_inset Formula 
\begin{align}
M_{1,ij} & =\frac{2\alpha_{j}}{J_{0}(\alpha_{j})^{2}}T_{ij}\\
M_{2,ij} & =T_{ji}\\
K_{ij} & =\frac{1}{2}\rho\alpha_{j}J_{0}(\alpha_{j})^{2} & ;\quad i=j\\
K_{ij} & =0 & ;\quad i\neq j\\
R_{i} & =J_{1}(\alpha_{i}\rho)\frac{\rho}{\alpha_{q}}
\end{align}

\end_inset


\end_layout

\begin_layout Standard
Finally, the added acoustic mass,
\begin_inset Formula 
\begin{equation}
p_{C}^{0}=p_{B}^{0}-i\omega M_{A}U_{B},
\end{equation}

\end_inset

can be computed as 
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{B}^{0}=p_{C}^{0}+\sum_{m=1}^{\infty}\frac{2J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}C_{m}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{B}^{0}=p_{C}^{0}+ikbu_{B}^{0}z_{0}\sum_{m=1}^{\infty}\frac{2J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}D_{m}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Filling in:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{C}^{0}=p_{B}^{0}-i\omega M_{A}U_{B}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
Then:
\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $p_{B}^{0}=p_{C}^{0}+i\omega M_{A}U_{B}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
or: 
\begin_inset Formula $i\omega M_{A}U_{B}=ikbu_{B}^{0}z_{0}\sum_{m=1}^{\infty}\frac{2J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}D_{m}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
And since:
\begin_inset Formula $M_{A}=\chi(\alpha)\frac{8\rho_{0}}{3\pi^{2}a_{L}}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $\chi(\alpha)=\frac{3\pi}{4}\sum_{m=1}^{\infty}\frac{J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}D_{m}$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{equation}
\rho_{0}\sum_{m=1}^{\infty}\frac{2}{\pi b}\frac{J_{1}\left(\alpha_{m}\rho\right)}{\rho\alpha_{m}}D_{m}
\end{equation}

\end_inset


\end_layout

\begin_layout Standard
For a given velocity 
\begin_inset Formula $u_{C,0}$
\end_inset

 the velocity profile at 
\begin_inset Formula $z=0$
\end_inset

 is
\begin_inset Note Note
status collapsed

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{C}(z)=u_{C}^{0}(z)+\sum_{m=1}^{\infty}Y_{C,m}C_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)e^{-\gamma_{m}z}$
\end_inset


\end_layout

\begin_layout Plain Layout

\lang english
\begin_inset Formula $u_{C}=u_{C}^{0}+u_{B}^{0}\sum_{m=1}^{\infty}\gamma_{m}bD_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)$
\end_inset


\end_layout

\end_inset

 
\begin_inset Formula 
\begin{equation}
u_{C}=u_{C}^{0}+bu_{B}^{0}\sum_{m=1}^{\infty}\gamma_{m}D_{m}J_{0}\left(\alpha_{m}\frac{r}{c}\right)
\end{equation}

\end_inset


\end_layout

\end_body
\end_document