diff --git a/README.md b/README.md index 6976dae..9ca3ad3 100644 --- a/README.md +++ b/README.md @@ -1,3 +1,6 @@ # LRFTubes mathematical documentation [![Build Status](https://drone.ascee.nl/api/badges/ASCEE/lrftubes_doc/status.svg)](https://drone.ascee.nl/ASCEE/lrftubes_doc) + +This repository is the source code of the LRFTubes mathematical documentation. +Releases are automatically build and pushed at: diff --git a/lrftubes.lyx b/lrftubes.lyx index 6a8edbd..a29c534 100644 --- a/lrftubes.lyx +++ b/lrftubes.lyx @@ -5600,6 +5600,235 @@ literal "true" Exponential duct (horn) \end_layout +\begin_layout Standard +hor.py/class ExpHorn(Duct) +\end_layout + +\begin_layout Standard +An exponential horn can be modelled using ExpHorn. + The transfer matrix is derived from Webster's horn equation. + The solution for an exponential horn is +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +p\left(x\right)=C_{1}e^{E_{1}x}+C_{2}e^{E_{2}x},\qquad U\left(x\right)=i\frac{S\left(x\right)}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}x}+C_{2}E_{2}e^{E_{2}x}\right) +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +in which +\begin_inset Formula $S(x)$ +\end_inset + + is the local cross-sectional area, +\begin_inset Formula $\rho$ +\end_inset + + is the air density, +\begin_inset Formula $\omega$ +\end_inset + + is the angular frequency. + The exponent terms +\begin_inset Formula $E_{1}$ +\end_inset + + and +\begin_inset Formula $E_{2}$ +\end_inset + + are given by +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2} +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +in which +\begin_inset Formula $k$ +\end_inset + + is the wave number and +\begin_inset Formula $m$ +\end_inset + + is the flare rate, which is defined as +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right) +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +with +\begin_inset Formula $L$ +\end_inset + + being the horn length and +\begin_inset Formula $S_{R}$ +\end_inset + + and +\begin_inset Formula $S_{L}$ +\end_inset + + being the cross-sectional areas on the right-hand and left-hand side of + the horn respectively. + +\begin_inset Newline newline +\end_inset + + +\begin_inset Newline newline +\end_inset + +By setting the solutions on +\begin_inset Formula $x=0$ +\end_inset + + to +\begin_inset Formula $p_{L}$ +\end_inset + + and +\begin_inset Formula $U_{L}$ +\end_inset + + and those on +\begin_inset Formula $x=L$ +\end_inset + + to +\begin_inset Formula $p_{R}$ +\end_inset + + and +\begin_inset Formula $U_{R}$ +\end_inset + +, the following transfer matrix can be derived: +\end_layout + +\begin_layout Standard +\begin_inset Formula +\begin{equation} +T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc} +E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L} & i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)\\ +i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right) & -\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right) +\end{array}\right] +\end{equation} + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Note Note +status collapsed + +\begin_layout Plain Layout +\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$ +\end_inset + + +\begin_inset Newline newline +\end_inset + + +\begin_inset Newline newline +\end_inset + +Solve for constants +\begin_inset Newline newline +\end_inset + + +\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$ +\end_inset + + +\begin_inset Newline newline +\end_inset + + +\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$ +\end_inset + + +\begin_inset Newline newline +\end_inset + + +\begin_inset Newline newline +\end_inset + +Fill in in solution: +\begin_inset Newline newline +\end_inset + + +\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$ +\end_inset + + +\begin_inset Newline newline +\end_inset + + +\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$ +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + \begin_layout Standard \begin_inset Formula \begin{equation} @@ -5610,7 +5839,7 @@ S_{f}=\exp\left(\alpha x\right) \begin_inset Note Note -status collapsed +status open \begin_layout Plain Layout \begin_inset Formula $\frac{\mathrm{d}^{2}p}{\mathrm{d}x^{2}}+\alpha\frac{\mathrm{d}p}{\mathrm{d}x}+\Gamma^{2}p=0$ @@ -7922,239 +8151,6 @@ Z_{cpm}\left(f\right)=\frac{1}{i2\pi fC_{dyn}\left(f\right)} \end_inset -\end_layout - -\begin_layout Section -Exponential horn -\end_layout - -\begin_layout Standard -hon.py/class ExpHorn(Duct) -\end_layout - -\begin_layout Standard -An exponential horn can be modelled using ExpHorn. - The transfer matrix is derived from Webster's horn equation. - The solution for an exponential horn is -\end_layout - -\begin_layout Standard -\begin_inset Formula -\begin{equation} -p\left(x\right)=C_{1}e^{E_{1}x}+C_{2}e^{E_{2}x},\qquad U\left(x\right)=i\frac{S\left(x\right)}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}x}+C_{2}E_{2}e^{E_{2}x}\right) -\end{equation} - -\end_inset - - -\end_layout - -\begin_layout Standard -in which -\begin_inset Formula $S(x)$ -\end_inset - - is the local cross-sectional area, -\begin_inset Formula $\rho$ -\end_inset - - is the air density, -\begin_inset Formula $\omega$ -\end_inset - - is the angular frequency. - The exponent terms -\begin_inset Formula $E_{1}$ -\end_inset - - and -\begin_inset Formula $E_{2}$ -\end_inset - - are given by -\end_layout - -\begin_layout Standard -\begin_inset Formula -\begin{equation} -E_{1}=\frac{-m+\sqrt{-4k^{2}+m^{2}}}{2},\qquad E_{2}=-\frac{m+\sqrt{-4k^{2}+m^{2}}}{2} -\end{equation} - -\end_inset - - -\end_layout - -\begin_layout Standard -in which -\begin_inset Formula $k$ -\end_inset - - is the wave number and -\begin_inset Formula $m$ -\end_inset - - is the flare rate, which is defined as -\end_layout - -\begin_layout Standard -\begin_inset Formula -\begin{equation} -m=\frac{1}{L}\ln\left(\frac{S_{R}}{S_{L}}\right) -\end{equation} - -\end_inset - - -\end_layout - -\begin_layout Standard -with -\begin_inset Formula $L$ -\end_inset - - being the horn length and -\begin_inset Formula $S_{R}$ -\end_inset - - and -\begin_inset Formula $S_{L}$ -\end_inset - - being the cross-sectional areas on the right-hand and left-hand side of - the horn respectively. - -\begin_inset Newline newline -\end_inset - - -\begin_inset Newline newline -\end_inset - -By setting the solutions on -\begin_inset Formula $x=0$ -\end_inset - - to -\begin_inset Formula $p_{L}$ -\end_inset - - and -\begin_inset Formula $U_{L}$ -\end_inset - - and those on -\begin_inset Formula $x=L$ -\end_inset - - to -\begin_inset Formula $p_{R}$ -\end_inset - - and -\begin_inset Formula $U_{R}$ -\end_inset - -, the following transfer matrix can be derived: -\end_layout - -\begin_layout Standard -\begin_inset Formula -\begin{equation} -T_{ExpHorn}=\frac{1}{E_{2}-E_{1}}\left[\begin{array}{cc} -E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L} & i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)\\ -i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right) & -\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right) -\end{array}\right] -\end{equation} - -\end_inset - - -\end_layout - -\begin_layout Standard -\begin_inset Note Note -status open - -\begin_layout Plain Layout -\begin_inset Formula $p_{L}=p\left(0\right)=C_{1}+C_{2}$ -\end_inset - - -\end_layout - -\begin_layout Plain Layout -\begin_inset Formula $U_{L}=U\left(0\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}+C_{2}E_{2}\right)$ -\end_inset - - -\end_layout - -\begin_layout Plain Layout -\begin_inset Formula $p_{R}=p\left(L\right)=C_{1}e^{E_{1}L}+C_{2}e^{E_{2}L}$ -\end_inset - - -\end_layout - -\begin_layout Plain Layout -\begin_inset Formula $U_{R}=U\left(L\right)=i\frac{S_{L}}{\rho\omega}\left(C_{1}E_{1}e^{E_{1}L}+C_{2}E_{2}e^{E_{2}L}\right)$ -\end_inset - - -\begin_inset Newline newline -\end_inset - - -\begin_inset Newline newline -\end_inset - -Solve for constants -\begin_inset Newline newline -\end_inset - - -\begin_inset Formula $C_{1}=P_{L}-C_{2}\rightarrow U_{L}=i\frac{S_{L}}{\rho\omega}\left[E_{1}p_{L}+\left(E_{2}-E_{1}\right)C_{2}\right]$ -\end_inset - - -\begin_inset Newline newline -\end_inset - - -\begin_inset Formula $C_{1}=\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{2}P_{L}\right),\qquad C_{2}=-\frac{1}{E_{2}-E_{1}}\left(i\frac{\rho\omega}{S_{L}}U_{L}+E_{1}P_{L}\right)$ -\end_inset - - -\begin_inset Newline newline -\end_inset - - -\begin_inset Newline newline -\end_inset - -Fill in in solution: -\begin_inset Newline newline -\end_inset - - -\begin_inset Formula $p_{R}=\frac{1}{E_{2}E_{1}}\left[\left(E_{2}e^{E_{1}L}-E_{1}e^{E_{2}L}\right)P_{L}+i\frac{\rho\omega}{S_{L}}\left(e^{E_{1}L}-e^{E_{2}L}\right)U_{L}\right]$ -\end_inset - - -\begin_inset Newline newline -\end_inset - - -\begin_inset Formula $U_{R}=\frac{1}{E_{2}E_{1}}\left[i\frac{S_{R}}{\rho\omega}E_{1}E_{2}\left(e^{E_{1}L}-e^{E_{2}L}\right)P_{L}-\frac{S_{R}}{S_{L}}\left(E_{1}e^{E_{1}L}-E_{2}e^{E_{2}L}\right)U_{L}\right]$ -\end_inset - - -\end_layout - -\end_inset - - \end_layout \begin_layout Section @@ -8686,6 +8682,7 @@ The wavelength is much larger than the thermal penetration depth ( ). \end_layout +\begin_deeper \begin_layout Standard We can derive the following impedance boundary condition \begin_inset CommandInset citation @@ -8698,7 +8695,7 @@ literal "true" : \begin_inset Note Note -status collapsed +status open \begin_layout Plain Layout Delta EC User guide: @@ -8770,6 +8767,7 @@ Hence the impedance of a hard wall scales with . \end_layout +\end_deeper \begin_layout Standard \begin_inset Float figure wide false @@ -9727,7 +9725,7 @@ V_{\mathrm{in}}-V_{\mathrm{bemf}}=Z_{\mathrm{el}}I, \end_inset -where + where \begin_inset Formula $Z_{\mathrm{el}}$ \end_inset @@ -15015,8 +15013,8 @@ literal "true" \end_inset . - Assuming axial symmetrySuch that the acoustic pressure in for example tube - + Assuming axial symmetry, such that the acoustic pressure in for example + tube \begin_inset Formula $B$ \end_inset