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LRFTubes documentation - v1.0
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Dr.ir.
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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 
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.
 A more detailed introduction to the notation used in this documentation
 can be found in the PhD thesis of de Jong 
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.
\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
\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 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
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 $e^{+i\omega t}$
\end_inset

 convention is used.
 A common choice of state variables is such that their product has the unit
 of power.
 For the acoustic systems in this work the state variables are acoustic
 pressure 
\begin_inset Formula $p\left(\omega\right)$
\end_inset

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

.
 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 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 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 
\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 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 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=k\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 open

\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 Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout

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

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

:
\end_layout

\begin_layout Plain Layout
\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 Plain Layout
\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

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


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

\end_inset

where
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\begin{equation}
Z_{c,0}=\frac{kz_{0}}{\left(1-f_{\nu}\right)\pi r_{0}^{2}\Gamma_{0}}
\end{equation}

\end_inset


\end_layout

\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 eqation 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 Subsection
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
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
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.
\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}},
\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 Standard
\begin_inset Note Note
status open

\begin_layout Plain Layout
\begin_inset CommandInset bibtex
LatexCommand bibtex
bibfiles "lrftubes"
options "plain"

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
printbibliography
\end_layout

\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 CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$r$"
description "Radial position in cylindrical coordinates\\nomunit{\\si{\\m}}"
literal "true"

\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 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

 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 imcomplete.
\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


\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 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