Finite non-linear waves: Difference between revisions

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\section{Finite Nonlinear Waves}
Starting point: the governing flow equations on partial differential form


\noindent Starting point: the governing flow equations on partial differential form\\
Continuity equation:


\noindent Continuity equation:
<math display="block">
 
\begin{equation}
\frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+\rho\frac{\partial u}{\partial x}=0
\frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+\rho\frac{\partial u}{\partial x}=0
\label{eq:pde:cont}
</math>
\end{equation}\\


\noindent Momentum equation:
Momentum equation:


\begin{equation}
<math display="block">
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho}\frac{\partial p}{\partial x}=0
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho}\frac{\partial p}{\partial x}=0
\label{eq:pde:mom}
</math>
\end{equation}\\


\noindent Any thermodynamic property can be expressed as a function of two other thermodynamic properties. This means that we can get density as a function of pressure and entropy: $\rho=\rho(p,s)$ and therefore\\
Any thermodynamic property can be expressed as a function of two other thermodynamic properties. This means that we can get density as a function of pressure and entropy: <math>\rho=\rho(p,s)</math> and therefore


\begin{equation*}
<math display="block">
d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp+\left(\frac{\partial \rho}{\partial s}\right)_p ds
d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp+\left(\frac{\partial \rho}{\partial s}\right)_p ds
\end{equation*}\\
</math>


\noindent Assuming isentropic flow $ds=0$ gives\\
Assuming isentropic flow <math>ds=0</math> gives


\begin{equation*}
<math display="block">
d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp
d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp
\end{equation*}\\
</math>


\begin{equation}
<math display="block">
\begin{aligned}
\begin{aligned}
&\frac{\partial \rho}{\partial t}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial t}=\frac{1}{a^2}\frac{\partial p}{\partial t}\\
&\frac{\partial \rho}{\partial t}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial t}=\frac{1}{a^2}\frac{\partial p}{\partial t}\\
Line 44: Line 40:
&\frac{\partial \rho}{\partial x}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial x}=\frac{1}{a^2}\frac{\partial p}{\partial x}
&\frac{\partial \rho}{\partial x}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial x}=\frac{1}{a^2}\frac{\partial p}{\partial x}
\end{aligned}
\end{aligned}
\label{eq:rhotop}
</math>
\end{equation}\\


\noindent Now, insert \ref{eq:rhotop} in \ref{eq:pde:cont} gives\\
Now, insert \ref{eq:rhotop} in \ref{eq:pde:cont} gives


\begin{equation}
<math display="block">
\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}+\rho a^2\frac{\partial u}{\partial x}=0
\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}+\rho a^2\frac{\partial u}{\partial x}=0
\label{eq:pde:cont:b}
</math>
\end{equation}\\


\noindent Dividing \ref{eq:pde:cont:b} by $\rho a$ gives\\
Dividing \ref{eq:pde:cont:b} by <math>\rho a</math> gives


\begin{equation}
<math display="block">
\frac{1}{\rho a}\left(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}\right)+a\frac{\partial u}{\partial x}=0
\frac{1}{\rho a}\left(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}\right)+a\frac{\partial u}{\partial x}=0
\label{eq:pde:cont:c}
</math>
\end{equation}\\


\noindent A slightly modified form of the momentum equation is obtained by multiplying and dividing the last term by $a$\\
A slightly modified form of the momentum equation is obtained by multiplying and dividing the last term by <math>a</math>


\begin{equation}
<math display="block">
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0
\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0
\label{eq:pde:mom:c}
</math>
\end{equation}\\


\noindent If the continuity equation on the form \ref{eq:pde:cont:c} is added to the momentum equation on the form \ref{eq:pde:mom:c}, we get\\
If the continuity equation on the form \ref{eq:pde:cont:c} is added to the momentum equation on the form \ref{eq:pde:mom:c}, we get


\begin{equation}
<math display="block">
\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]=0
\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]=0
\label{eq:nonlin:a}
</math>
\end{equation}\\


\noindent If, instead, the continuity equation on the form \ref{eq:pde:cont:c} is subtracted from the momentum equation on the form \ref{eq:pde:mom:c}, we get\\
If, instead, the continuity equation on the form \ref{eq:pde:cont:c} is subtracted from the momentum equation on the form \ref{eq:pde:mom:c}, we get


\begin{equation}
<math display="block">
\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]=0
\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]=0
\label{eq:nonlin:b}
</math>
\end{equation}\\


\noindent Since $u=u(x,t)$, we have from the definition of a differential\\
Since <math>u=u(x,t)</math>, we have from the definition of a differential


\begin{equation}
<math display="block">
du=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dx=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}\frac{dx}{dt}dt
du=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dx=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}\frac{dx}{dt}dt
\label{eq:du}
</math>
\end{equation}\\


\noindent Now, let $dx/dt=u+a$\\
Now, let <math>dx/dt=u+a</math>


\begin{equation}
<math display="block">
du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]dt
du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]dt
\label{eq:du:b}
</math>
\end{equation}\\


\noindent which is the change of $u$ in the direction $dx/dt=u+a$\\
which is the change of <math>u</math> in the direction <math>dx/dt=u+a</math>


\noindent In the same way\\
In the same way


\begin{equation}
<math display="block">
dp=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}dx=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}\frac{dx}{dt}dt
dp=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}dx=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}\frac{dx}{dt}dt
\label{eq:dp}
</math>
\end{equation}\\


\noindent and thus, in the direction $dx/dt=u+a$\\
and thus, in the direction <math>dx/dt=u+a</math>


\begin{equation}
<math display="block">
dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]dt
dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]dt
\label{eq:dp:b}
</math>
\end{equation}\\


\noindent If we go back and examine Eqn. \ref{eq:nonlin:a}, we see that Eqns. \ref{eq:du:b} and \ref{eq:dp:b} appear in the equation and thus it can now be rewritten as follows\\
If we go back and examine Eqn. \ref{eq:nonlin:a}, we see that Eqns. \ref{eq:du:b} and \ref{eq:dp:b} appear in the equation and thus it can now be rewritten as follows


\begin{equation}
<math display="block">
\frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0
\frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0
\label{eq:nonlin:a:ode}
</math>
\end{equation}\\


\noindent Eqn. \ref{eq:nonlin:a:ode} applies along a $C^+$ characteristic, i.e., a line in the direction $dx/dt=u+a$ in $xt$-space and is called the compatibility equation along the $C^+$ characteristic. If we instead chose a $C^-$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space, we get\\
Eqn. \ref{eq:nonlin:a:ode} applies along a <math>C^+</math> characteristic, i.e., a line in the direction <math>dx/dt=u+a</math> in <math>xt</math>-space and is called the compatibility equation along the <math>C^+</math> characteristic. If we instead chose a <math>C^-</math> characteristic, i.e., a line in the direction <math>dx/dt=u-a</math> in <math>xt</math>-space, we get


\begin{equation}
<math display="block">
du=\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]dt
du=\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]dt
\label{eq:du:c}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
dp=\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]dt
dp=\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]dt
\label{eq:dp:c}
</math>
\end{equation}\\


\noindent which can be identified as subsets of Eqn. \ref{eq:nonlin:b} and thus\\
which can be identified as subsets of Eqn. \ref{eq:nonlin:b} and thus


\begin{equation*}
<math display="block">
\frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0
\frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0
\end{equation*}\\
</math>


\noindent In order to fulfill the relation above, either $du=dp=0$ or\\
In order to fulfil the relation above, either <math>du=dp=0</math> or


\begin{equation}
<math display="block">
du-\frac{dp}{\rho a}=0
du-\frac{dp}{\rho a}=0
\label{eq:nonlin:b:ode}
</math>
\end{equation}\\


\noindent Eqn. \ref{eq:nonlin:b:ode} applies along a $C^-$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space and is called the compatibility equation along the $C^-$ characteristic.\\
Eqn. \ref{eq:nonlin:b:ode} applies along a <math>C^-</math> characteristic, i.e., a line in the direction <math>dx/dt=u-a</math> in <math>xt</math>-space and is called the compatibility equation along the <math>C^-</math> characteristic.


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\noindent So, what we have done now is that we have have found paths through a point ($x_1$,$t_1$) along which the governing partial differential equations Eqns. \ref{eq:nonlin:a} and \ref{eq:nonlin:b} reduces to the ordinary differential equations \ref{eq:nonlin:a:ode} and \ref{eq:nonlin:b:ode}. The $C^+$ and $C^-$ characteristic lines are physically the paths of right- and left-running sound waves in the $xt$-plane.\\
So, what we have done now is that we have have found paths through a point (<math>x_1</math>, <math>t_1</math>) along which the governing partial differential equations Eqns. \ref{eq:nonlin:a} and \ref{eq:nonlin:b} reduces to the ordinary differential equations \ref{eq:nonlin:a:ode} and \ref{eq:nonlin:b:ode}. The <math>C^+</math> and <math>C^-</math> characteristic lines are physically the paths of right- and left-running sound waves in the <math>xt</math>-plane.


\subsection{Riemann Invariants}
==== Riemann Invariants ====


\noindent If the compatibility equations are integrated along respective characteristic line, i.e., integrate \ref{eq:nonlin:a:ode} along the $C^+$ characteristic and  \ref{eq:nonlin:b:ode} along the $C^-$ characteristic, we get the Riemann invariants $J^+$ and $J^-$.\\
If the compatibility equations are integrated along respective characteristic line, i.e., integrate \ref{eq:nonlin:a:ode} along the <math>C^+</math> characteristic and  \ref{eq:nonlin:b:ode} along the <math>C^-</math> characteristic, we get the Riemann invariants <math>J^+</math> and <math>J^-</math>.


\begin{equation}
<math display="block">
J^+=u+\int\frac{dp}{\rho a}=const
J^+=u+\int\frac{dp}{\rho a}=const
\label{eq:riemann:a}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
J^-=u-\int\frac{dp}{\rho a}=const
J^-=u-\int\frac{dp}{\rho a}=const
\label{eq:riemann:b}
</math>
\end{equation}\\


\noindent The Riemann invariants are constants along the associated characteristic line.\\
The Riemann invariants are constants along the associated characteristic line.


\noindent We have assumed isentropic flow and thus we may use the isentropic relations\\
We have assumed isentropic flow and thus we may use the isentropic relations


\begin{equation}
<math display="block">
p=C_1T^{\gamma/(\gamma-1)}=C_2a^{2\gamma/(\gamma-1)}
p=C_1T^{\gamma/(\gamma-1)}=C_2a^{2\gamma/(\gamma-1)}
\label{eq:isentropic:a}
</math>
\end{equation}\\


\noindent where $C_1$ and $C_2$ are constants. Differentiating Eqn. \ref{eq:isentropic:a} gives\\
where <math>C_1</math> and <math>C_2</math> are constants. Differentiating Eqn. \ref{eq:isentropic:a} gives


\begin{equation}
<math display="block">
dp=C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}da
dp=C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}da
\label{eq:isentropic:b}
</math>
\end{equation}\\


\noindent Now, if we further assume the gas to be calorically perfect\\
Now, if we further assume the gas to be calorically perfect


\begin{equation}
<math display="block">
a^2=\gamma RT=\frac{\gamma p}{\rho}\Rightarrow \rho=\frac{\gamma p}{a^2}
a^2=\gamma RT=\frac{\gamma p}{\rho}\Rightarrow \rho=\frac{\gamma p}{a^2}
\label{eq:isentropic:c}
</math>
\end{equation}\\


\noindent Eqn. \ref{eq:isentropic:a} in \ref{eq:isentropic:c} gives\\
Eqn. \ref{eq:isentropic:a} in \ref{eq:isentropic:c} gives


\begin{equation}
<math display="block">
\rho=C_2\gamma a^{[2\gamma/(\gamma-1)-2]}
\rho=C_2\gamma a^{[2\gamma/(\gamma-1)-2]}
\label{eq:isentropic:d}
</math>
\end{equation}\\


\noindent and thus\\
and thus


\begin{equation*}
<math display="block">
J^+=u+\int\frac{C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}}{C_2\gamma a^{[2\gamma/(\gamma-1)-2]}a}da=u+\left(\frac{2}{\gamma-1}\right)\int da
J^+=u+\int\frac{C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}}{C_2\gamma a^{[2\gamma/(\gamma-1)-2]}a}da=u+\left(\frac{2}{\gamma-1}\right)\int da
\end{equation*}\\
</math>


\begin{equation}
<math display="block">
J^+=u+\frac{2a}{\gamma-1}
J^+=u+\frac{2a}{\gamma-1}
\label{eq:riemann:a:b}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
J^-=u-\frac{2a}{\gamma-1}
J^-=u-\frac{2a}{\gamma-1}
\label{eq:riemann:b:b}
</math>
\end{equation}\\


\noindent Eqns. \ref{eq:riemann:a:b} and \ref{eq:riemann:b:b} are the Riemann invariants for a calorically perfect gas. The Riemann invariants are constants along $C^+$ and $C^-$ characteristics and if the situation shown in Fig. \ref{fig:characteristics} appears, that fact can be used to calculate the flow velocity and speed of sound in the location ($x_1$,$t_1$).\\
Eqns. \ref{eq:riemann:a:b} and \ref{eq:riemann:b:b} are the Riemann invariants for a calorically perfect gas. The Riemann invariants are constants along <math>C^+</math> and <math>C^-</math> characteristics and if the situation shown in Fig. \ref{fig:characteristics} appears, that fact can be used to calculate the flow velocity and speed of sound in the location (<math>x_1</math>, <math>t_1</math>).


\begin{equation}
<math display="block">
J^++J^-=u+\frac{2a}{\gamma-1}+u-\frac{2a}{\gamma-1}=2u\Rightarrow u=\frac{1}{2}(J^++J^-)
J^++J^-=u+\frac{2a}{\gamma-1}+u-\frac{2a}{\gamma-1}=2u\Rightarrow u=\frac{1}{2}(J^++J^-)
\label{eq:riemann:evaluation:a}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
J^+=u+\frac{2a}{\gamma-1}=\frac{1}{2}(J^++J^-)+\frac{2a}{\gamma-1}\Rightarrow a=\frac{\gamma-1}{4}(J^+-J^-)
J^+=u+\frac{2a}{\gamma-1}=\frac{1}{2}(J^++J^-)+\frac{2a}{\gamma-1}\Rightarrow a=\frac{\gamma-1}{4}(J^+-J^-)
\label{eq:riemann:evaluation:b}
</math>
\end{equation}\\
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