ASCEE
/
lrftubes_doc
4
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity
lrftubes_doc/lrftubes.lyx

7350 lines
142 KiB
Plaintext
Raw Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
\save_transient_properties true
\origin unavailable
\textclass memoir
\begin_preamble
\input{tex/preamble.tex}
\end_preamble
\options a4paper
\use_default_options true
\maintain_unincluded_children false
\language american
\language_package babel
\inputencoding utf8
\fontencoding global
\font_roman "libertine" "FreeSerif"
\font_sans "default" "Courier New"
\font_typewriter "default" "default"
\font_math "libertine-ntxm" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\use_microtype false
\use_dash_ligatures false
\graphics default
\default_output_format pdf2
\output_sync 1
\output_sync_macro "\synctex=1"
\bibtex_command biber
\index_command default
\paperfontsize default
\spacing single
\use_hyperref true
\pdf_author "Dr.ir. J.A. de Jong - ASCEE"
\pdf_bookmarks true
\pdf_bookmarksnumbered false
\pdf_bookmarksopen false
\pdf_bookmarksopenlevel 1
\pdf_breaklinks false
\pdf_pdfborder true
\pdf_colorlinks true
\pdf_backref false
\pdf_pdfusetitle true
\papersize default
\use_geometry true
\use_package amsmath 1
\use_package amssymb 1
\use_package cancel 1
\use_package esint 1
\use_package mathdots 1
\use_package mathtools 1
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 0
\use_minted 0
\index Index
\shortcut idx
\color #008000
\end_index
\leftmargin 3cm
\topmargin 3cm
\rightmargin 2.5cm
\bottommargin 3.5cm
\headsep 1cm
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\end_header
\begin_body
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
thispagestyle{empty}
\end_layout
\end_inset
\end_layout
\begin_layout Title
LRFTubes documentation - v1.0
\end_layout
\begin_layout Author
Dr.ir.
J.A.
de Jong
\end_layout
\begin_layout Standard
\align center
\begin_inset Graphics
filename img_default/ascee_beeldmerk.pdf
width 65text%
\end_inset
\end_layout
\begin_layout Standard
\begin_inset VSpace vfill
\end_inset
\end_layout
\begin_layout Standard
\begin_inset VSpace bigskip
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Tabular
<lyxtabular version="3" rows="5" columns="2">
<features booktabs="true" tabularvalignment="middle" tabularwidth="100text%">
<column alignment="left" valignment="top" width="0pt">
<column alignment="right" valignment="top" width="9cm">
<row topspace="default" bottomspace="default" interlinespace="default">
<cell alignment="left" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
\end_layout
\end_inset
</cell>
<cell alignment="right" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
ASCEE
\begin_inset Newline newline
\end_inset
Vildersveenweg 19
\begin_inset Newline newline
\end_inset
7443 RZ Nijverdal
\begin_inset Newline newline
\end_inset
The Netherlands
\begin_inset Newline newline
\end_inset
\end_layout
\begin_layout Plain Layout
T:
\begin_inset space \hspace{}
\length -1.8cm
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
phantom{info@ascee.nl}
\end_layout
\end_inset
+31 6 18971622
\begin_inset Newline newline
\end_inset
E:
\begin_inset space \hspace{}
\length -1.8cm
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
phantom{+31 6 18971622}
\end_layout
\end_inset
info@ascee.nl
\end_layout
\end_inset
</cell>
</row>
<row topspace="default" bottomspace="default" interlinespace="default">
<cell alignment="left" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
Internal document ID:
\end_layout
\end_inset
</cell>
<cell alignment="right" valignment="top" topline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
628318
\end_layout
\end_inset
</cell>
</row>
<row topspace="default" bottomspace="default" interlinespace="default">
<cell alignment="left" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
Document status:
\end_layout
\end_inset
</cell>
<cell alignment="right" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
Draft
\end_layout
\end_inset
</cell>
</row>
<row topspace="default" bottomspace="default" interlinespace="default">
<cell alignment="left" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
Document revision:
\end_layout
\end_inset
</cell>
<cell alignment="right" valignment="top" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
1
\end_layout
\end_inset
</cell>
</row>
<row topspace="default" bottomspace="default" interlinespace="default">
<cell alignment="left" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
Revision history:
\end_layout
\end_inset
</cell>
<cell alignment="right" valignment="top" bottomline="true" usebox="none">
\begin_inset Text
\begin_layout Plain Layout
2018-02-21: rev.
1
\end_layout
\end_inset
</cell>
</row>
</lyxtabular>
\end_inset
\end_layout
\begin_layout Standard
Copyright (
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
textcopyright
\end_layout
\end_inset
)
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
the
\backslash
year
\backslash
\end_layout
\end_inset
ASCEE.
All rights reserved.
\end_layout
\begin_layout Standard
\begin_inset Newpage clearpage
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
pagestyle{fancy}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Newpage clearpage
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
%
\backslash
cleardoublepage
\end_layout
\begin_layout Plain Layout
\backslash
markboth{
\backslash
nomname}{
\backslash
nomname}% maybe with
\backslash
MakeUppercase
\end_layout
\begin_layout Plain Layout
\backslash
addcontentsline{toc}{chapter}{List of symbols}
\end_layout
\begin_layout Plain Layout
\backslash
setlength{
\backslash
nomitemsep}{-0.2pt}
\end_layout
\begin_layout Plain Layout
\backslash
thispagestyle{empty}
\end_layout
\begin_layout Plain Layout
% Roman (A)
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset Note Note
status open
\begin_layout Plain Layout
Unused:
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{e}_x$"
description "Unit vector in $x$-direction\\nomunit{-}"
literal "true"
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$H$"
description "Total enthalpy per unit mass \\nomunit{\\si{\\joule\\per\\kilogram}}"
literal "true"
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{I}$"
description "Identity tensor\\nomunit{-}"
literal "true"
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$h_\\nu$"
description "Viscothermal shape function for the velocity\\nomunit{-}"
literal "true"
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$h_\\kappa$"
description "Viscothermal shape function for the temperature\\nomunit{-}"
literal "true"
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$E$"
description "Total energy per unit mass \\nomunit{\\si{\\joule\\per\\kilogram}}"
literal "true"
\end_inset
\end_layout
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$c$"
description "Speed of sound\\nomunit{\\si{\\metre\\per\\second}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$c_p$"
description "Specific heat at constant pressure \\nomunit{\\si{\\joule\\per\\kilogram\\kelvin}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$c_s$"
description "Specific heat of the solid\\nomunit{\\si{\\joule\\per\\kilogram\\kelvin}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$c_v$"
description "Specific heat at constant density \\nomunit{\\si{\\joule\\per\\kilogram\\kelvin}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$D$"
description "Diameter\\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$f_\\kappa$"
description "Thermal Rott function \\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$f_\\nu$"
description "Viscous Rott function \\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$f$"
description "Frequency\\nomunit{\\si{\\hertz}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$i$"
description "Imaginary unit\\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$J_\\alpha$"
description "Bessel function of the first kind and order $\\alpha$\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$k$"
description "Wave number\\nomunit{\\si{\\radian\\per\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$L$"
description "Length\\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\ell$"
description "Characteristic length scale of a fluid space \\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$N$"
description "Number of\\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{n}$"
description "Normal vector pointing from the solid into the fluid\\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$p$"
description "Pressure, acoustic pressure \\nomunit{\\si{\\pascal}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$r_h$"
description "Hydraulic radius \\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{r}$"
description "Transverse position vector\\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$S$"
description "Cross-sectional area, surface area\\nomunit{\\si{\\square\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$t$"
description "Time \\nomunit{\\si{\\second}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$T$"
description "Temperature\\nomunit{\\si{\\kelvin}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{u}$"
description "Velocity vector\\nomunit{\\si{\\metre\\per\\second}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$u$"
description "Velocity in wave propagation direction\\nomunit{\\si{\\metre\\per\\second}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$U$"
description "Volume flow\\nomunit{\\si{\\cubic\\metre\\per\\second}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$V$"
description "Volume \\nomunit{\\si{\\cubic\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$\\mathbf{x}$"
description "Position vector \\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$z$"
description "Specific acoustic impedance\\nomunit{\\si{\\pascal\\second\\per\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "A"
symbol "$Z$"
description "Volume flow impedance\\nomunit{\\si{\\pascal\\second\\per\\cubic\\metre}}"
literal "true"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
% Greek (G)
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Note Note
status open
\begin_layout Plain Layout
Unused:
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\Delta$"
description "Difference\\nonomunit"
literal "true"
\end_inset
\end_layout
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\gamma$"
description "Ratio of specific heats\\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\Gamma$"
description "Viscothermal wave number for a prismatic duct \\nomunit{\\si{\\radian\\per\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\delta_{\\kappa}$"
description "Thermal penetration depth\\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\delta_{\\nu}$"
description "Viscous penetration depth\\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\epsilon_s$"
description "Ideal stack correction factor \\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\lambda$"
description "Wavelength \\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\pi$"
description "Ratio of the circumference to the diameter of a circle \\nomunit{-}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "G"
symbol "$\\Pi$"
description "Wetted perimeter (contact length between solid and fluid) \\nomunit{\\si{\\metre}}"
literal "true"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
% Miscellaneous symbols and operators (M)
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Note Note
status open
\begin_layout Plain Layout
Unused:
\end_layout
\begin_layout Plain Layout
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\otimes$"
description "Dyadic product\\nonomunit"
literal "true"
\end_inset
\end_layout
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\Re$"
description "Real part\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\Im$"
description "Imaginary part\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\nabla$"
description "Gradient \\nomunit{\\si{\\per\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\nabla^2$"
description "Laplacian\\nomunit{\\si{\\per\\square\\metre}}"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\sim$"
description "Same order of magnitude\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\left\\Vert \\bullet \\right\\Vert $"
description "Eucledian norm\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "d"
description "Infinitesimal\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\partial$"
description "Infinitesimal\\nonomunit"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "M"
symbol "$\\bullet$"
description "Placeholder for an operand\\nonomunit"
literal "true"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
% Subscripts (S)
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "wall"
description "At the wall"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "0"
description "Evaluated at the reference condition"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$f$"
description "Fluid"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$s$"
description "Solid"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$w$"
description "Wall"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$R$"
description "Right side"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$L$"
description "Left side"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$s$"
description "Solid"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$s$"
description "Squeeze"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$i$"
description "Inner"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$o$"
description "Outer"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "S"
symbol "$t$"
description "Tube"
literal "true"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
% Often used abbreviations (O)
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "O"
symbol "Sec(s)."
description "Section(s)"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "O"
symbol "Eq(s)."
description "Equation(s)"
literal "true"
\end_inset
\begin_inset CommandInset nomenclature
LatexCommand nomenclature
prefix "O"
symbol "LRF"
description "Low Reduced Frequency"
literal "true"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
printnomenclature[1.8cm]
\end_layout
\end_inset
\end_layout
\begin_layout Chapter
Overview of
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
lrftubes
\end_layout
\end_inset
\end_layout
\begin_layout Section
Introduction
\end_layout
\begin_layout Standard
Welcome to the documentation of
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
lrftubes
\end_layout
\end_inset
.
\begin_inset ERT
status collapsed
\begin_layout Plain Layout
\backslash
lrftubes
\backslash
\end_layout
\end_inset
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
\begin_inset CommandInset citation
LatexCommand cite
key "van_der_eerden_noise_2000"
literal "true"
\end_inset
,
\begin_inset CommandInset citation
LatexCommand cite
key "kampinga_viscothermal_2010"
literal "true"
\end_inset
and
\begin_inset CommandInset citation
LatexCommand cite
key "ward_deltaec_2017"
literal "true"
\end_inset
.
\end_layout
\begin_layout Standard
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
\begin_inset CommandInset href
LatexCommand href
name "README"
target "https://github.com/asceenl/lrftubes"
literal "false"
\end_inset
in the main repository.
\end_layout
\begin_layout Standard
This document is very brief on the theory and it is assumed that the reader
has some knowledge on the basics of acoustics in general and viscothermal
acoustics as well.
If you are not falling in this category, I would please refer you first
to the book of Swift
\begin_inset CommandInset citation
LatexCommand cite
key "swift_thermoacoustics:_2003"
literal "true"
\end_inset
.
A more detailed introduction to the notation used in this documentation
can be found in the PhD thesis of de Jong
\begin_inset CommandInset citation
LatexCommand cite
key "de_jong_numerical_2015"
literal "true"
\end_inset
.
\end_layout
\begin_layout Standard
Besides that, if you find the work interesting, but you are not sure how
to apply it, please contact ASCEE for more information.
\end_layout
\begin_layout Section
License and disclaimer
\end_layout
\begin_layout Standard
Redistribution and use in source and binary forms are permitted provided
that the above copyright notice and this paragraph are duplicated in all
such forms and that any documentation, advertising materials, and other
materials related to such distribution and use acknowledge that the software
was developed by the ASCEE.
The name of the ASCEE may not be used to endorse or promote products derived
from this software without specific prior written permission.
\begin_inset Newline newline
\end_inset
\end_layout
\begin_layout Standard
THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR IMPLIED WARRANTIE
S, INCLUDING, WITHOUT LIMITATION, THE IMPLIED WARRANTIES OF MERCHANTABILITY
AND FITNESS FOR A PARTICULAR PURPOSE.
\end_layout
\begin_layout Section
Features
\end_layout
\begin_layout Standard
Currently the
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
lrftubes
\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
Reference in New Issue View Git Blame Copy Permalink