One-dimensional inviscid flow: Difference between revisions

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\end{figure}
\end{figure}
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==Shock Wave==
<!--
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/shock-wave.pdf}
\caption{Stationary normal shock}
\label{fig:shock}
\end{center}
\end{figure}
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The starting point is to set up the governing equations for one-dimensional steady compressible flow over a control volume enclosing the normal shock (Fig. \ref{fig:shock:cv}).
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\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/shock-cv.pdf}
\caption{Normal shock control volume}
\label{fig:shock:cv}
\end{center}
\end{figure}
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continuity:
<math display="block">
\rho_1 u_1=\rho_2 u_2
</math>
momentum:
<math display="block">
\rho_1 u_1^2+p_1=\rho_2 u_2^2+p_2
</math>
energy:
<math display="block">
h_1 + \frac{1}{2}u_1^2=h_2 + \frac{1}{2}u_2^2
</math>
Divide the momentum equation by <math>\rho_1 u_1</math>
<math display="block">
\frac{1}{\rho_1 u_1}\left(\rho_1 u_1^2+p_1\right)=\frac{1}{\rho_1 u_1}\left(\rho_2 u_2^2+p_2\right)=\left\{\rho_1 u_1=\rho_2 u_2\right\}=\frac{1}{\rho_2 u_2}\left(\rho_2 u_2^2+p_2\right) \Rightarrow
</math>
<math display="block">
\frac{p_1}{\rho_1 u_1}-\frac{p_2}{\rho_2 u_2}=u_2-u_1
</math>
For a calorically perfect gas <math>a=\sqrt{\gamma p/\rho}</math>, which if implemented in Eqn. \ref{eq:governing:mom:b} gives
<math display="block">
\frac{a_1^2}{\gamma u_1}-\frac{a_2^2}{\gamma u_2}=u_2-u_1
</math>
The energy equation (Eqn. \ref{eq:governing:energy}) with <math>h=C_p T</math>
<math display="block">
C_p T_1 + \frac{1}{2}u_1^2=C_p T_2 + \frac{1}{2}u_2^2
</math>
Replacing <math>C_p</math> with <math>\gamma R/(\gamma-1)</math> gives
<math display="block">
\frac{\gamma RT_1}{\gamma-1} + \frac{1}{2}u_1^2=\frac{\gamma RT_2}{\gamma-1} + \frac{1}{2}u_2^2
</math>
With <math>a=\sqrt{\gamma RT}</math> this becomes
<math display="block">
\frac{a_1^2}{\gamma-1} + \frac{1}{2}u_1^2=\frac{a_2^2}{\gamma-1} + \frac{1}{2}u_2^2
</math>
Eqn. \ref{eq:governing:energy:d} can be set up between any two points in the flow. Specifically, we can use the relation to relate the flow velocity, <math>u</math>, and speed of sound, <math>a</math>, in any point to the corresponding flow properties at sonic conditions (<math>u=a=a^*</math>).
<math display="block">
\frac{a^2}{\gamma-1} + \frac{1}{2}u^2=\frac{\gamma+1}{2(\gamma-1)}{a^*}^2
</math>
\noindent If Eqn. \ref{eq:governing:energy:e} is evaluated in locations 1 and 2, we get\\
\begin{equation}
\begin{aligned}
&a_1^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\\
&a_2^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2
\end{aligned}
\label{eq:governing:energy:f}
\end{equation}\\
\noindent Since the change in flow conditions over the shock is adiabatic (no heat is added inside the shock), critical properties will be constant over the shock. Especially $a^*$ will be constant.\\
\noindent Eqn. \ref{eq:governing:energy:f} inserted in \ref{eq:governing:mom:c} gives\\
\[\frac{1}{\gamma u_1}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\right)-\frac{1}{\gamma u_2}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2\right)=u_2-u_1 \Rightarrow \]\\
\[\left(\frac{\gamma+1}{2\gamma}\right){a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(\frac{\gamma+1}{2\gamma}\right)\left(u_2-u_1\right) \Rightarrow\]\\
\[{a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(u_2-u_1\right) \Rightarrow\]\\
\[{a^*}^2\left(\frac{u_2}{u_1 u_2}-\frac{u_1}{u_1 u_2}\right)=\left(u_2-u_1\right) \Rightarrow\]\\
\[\frac{1}{u_1 u_2}{a^*}^2\left(u_2-u_1\right)=\left(u_2-u_1\right) \Rightarrow\]\\
\begin{equation}
{a^*}^2=u_1 u_2
\label{eq:prandtl}
\end{equation}\\
\noindent Eqn. \ref{eq:prandtl} is sometimes referred to as the Prandtl relation. Divide the Prandtl relation by ${a^*}^2$ on both sides gives\\
\[1=\frac{u_1}{a^*}\frac{u_2}{a^*}=M^*_1M^*_2\]\\
or
\begin{equation}
M^*_2=\frac{1}{M^*_1}
\label{eq:NormalMach}
\end{equation}\\
\noindent The relation between $M^*$ and $M$ is given by\\
\begin{equation}
{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}
\label{eq:MachStar}
\end{equation}\\
\noindent from which is can be seen that $M^*$ will follow the Mach number $M$ in the sense that\\
\begin{itemize}
\item $M=1\Rightarrow M^*=1$
\item $M<1\Rightarrow M^*<1$
\item $M>1\Rightarrow M^*>1$
\end{itemize}
\noindent Eqn. \ref{eq:MachStar} inserted in Eqn. \ref{eq:NormalMach} gives\\
\[\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}=\frac{2+(\gamma-1)M_2^2}{(\gamma+1)M_2^2}\]\\
\begin{equation}
M^2_2=\frac{1+\left[(\gamma-1)/2\right]M^2_1}{\gamma M^2_1-(\gamma-1)/2}
\label{eq:NormalMach:b}
\end{equation}\\
\noindent The Mach number relations above effectively show that if the Mach number upstream of the shock is greater than one, the downstream Mach number must be less than one and vice versa. We can also see that a sonic upstream flow $M_1=1.0$ gives sonic flow downstream of the shock. So, apparently the relation as such holds for both supersonic and subsonic upstream flow mathematically. The question is if it is also physically correct. For a supersonic upstream flow we will get a discontinuous compression and if the flow upstream of the control volume is subsonic we will instead get a discontinuous expansion inside the control volume but, again, is this physically correct? We will get the answer by analyzing the entropy change over the control volume.\\
\noindent Analyzing the energy equation and the second law of thermodynamics shows that there is a direct relation between entropy increase and total pressure drop.
\[s_2-s_1=C_p\ln\dfrac{T_2}{T_1}-R\ln\dfrac{p_2}{p_1}\]
\[s_2-s_1=C_p\ln\dfrac{T_2}{T_{o_2}}\dfrac{T_{o_1}}{T_1}\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_2}{p_{o_2}}\dfrac{p_{o_1}}{p_1}\dfrac{p_{o_2}}{p_{o_1}}\]
\noindent using the isentropic relations we get
\[s_2-s_1=C_p\ln\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_{o_2}}{p_{o_1}}\]
\noindent and since the process is adiabatic and thus $T_{o_2}=T_{o_1}$ the change in entropy is directly related to the change in total pressure as
\[s_2-s_1=-R\ln\dfrac{p_{o_2}}{p_{o_1}}\]
or
\[\dfrac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}\]
\noindent Figure~\ref{fig:shock:entropy} shows the entropy change over a normal shock. As can be seen in the figure, a subsonic upstream Mach number leads to a reduction of entropy, which once and for all rules out all such solutions as non-physical and thus the question about the upstream conditions can now be considered answered. This in turn implies that the Mach number downstream of a normal shock will always be subsonic, which can be seen in Fig~\ref{fig:shock:downstream:Mach} below.
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/shock-entropy.pdf}
\caption{Entropy change over a normal shock ($\Delta s$) as function of upstream Mach number ($M_1$)}
\label{fig:shock:entropy}
\end{center}
\end{figure}
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/shock-downstream-Mach.pdf}
\caption{Downstream Mach number ($M_2$) as function of upstream Mach number ($M_1$)}
\label{fig:shock:downstream:Mach}
\end{center}
\end{figure}
%\noindent The Mach number ahead of the shock must be greater than one and thus Eqn. \ref{eq:NormalMach} shows that the Mach number downstream of the shock must be less than one.\\
\noindent By rewriting the right-hand side of Eqn.\ref{eq:NormalMach:b}, it is easy to realize that the downstream Mach number $M_2$ approaches a finite value for large values of the upstream Mach number, $M_1$.
\[\left.M_2^2\right|_{M_1\rightarrow\infty}=\left.\dfrac{2/M_1^2+(\gamma-1)}{2\gamma-(\gamma-1)/M_1^2}\right|_{M_1\rightarrow\infty}=\dfrac{\gamma-1}{2\gamma}\]
\section{Normal Shock Relations}
\noindent Rewriting the continuity equation (Eqn. \ref{eq:governing:cont})\\
\begin{equation}
\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{u_1^2}{u_1 u_2}=\left\{{a^*}^2=u_1u_2\right\}=\frac{u_1^2}{{a^*}^2}={M^*_1}^2
\label{eq:Normal:density:a}
\end{equation}\\
\noindent Eqn. \ref{eq:MachStar} in Eqn. \ref{eq:Normal:density:a} gives\\
\begin{equation}
\frac{\rho_2}{\rho_1}=\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}
\label{eq:Normal:density:b}
\end{equation}\\
\noindent To get a corresponding relation for the pressure ratio  over the shock, we go back to the momentum equation (Eqn. \ref{eq:governing:mom})\\
\[p_2-p_1=\rho_1 u^2_1 - \rho_2 u^2_2=\left\{\rho_1 u_1=\rho_2 u_1\right\}=\rho_1 u_1(u_1-u_2)=\rho_1 u^2_1\left(1-\frac{u_2}{u_1}\right)\]\\
\[\frac{p_2-p_1}{p_1}=\frac{\rho_1 u^2_1}{p_1}\left(1-\frac{u_2}{u_1}\right)=\left\{a_1=\sqrt{\frac{\gamma p_1}{\rho_1}}\right\}=\gamma\frac{u^2_1}{a^2_1}\left(1-\frac{u_2}{u_1}\right)=\gamma M^2_1\left(1-\frac{u_2}{u_1}\right)\]\\
\[\frac{p_2}{p_1}-1=\gamma M^2_1\left(1-\frac{u_2}{u_1}\right)=\left\{\frac{u_2}{u_1}=\frac{\rho_1}{\rho_2}\right\}=\gamma M^2_1\left(1-\frac{2+(\gamma-1)M_1^2}{(\gamma+1)M_1^2}\right)\]\\
\begin{equation}
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M^2_1-1)
\label{eq:Normal:pressure}
\end{equation}\\
\noindent Figure~\ref{fig:shock:pressure:ratio} shows that the pressure must increase over the shock due to the fact that, based on the discussion above, the upstream Mach number must be greater than one and thus the shock is a discontinuous compression process.\\
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/shock-pressure-ratio.pdf}
\caption{Pressure ratio over a normal shock ($p_2/p_1$) as function of upstream Mach number ($M_1$)}
\label{fig:shock:pressure:ratio}
\end{center}
\end{figure}
\noindent The temperature ratio over the shock can be obtained using the already derived relations for pressure ratio and density ratio together with the equation of state $p=\rho RT$\\
\begin{equation}
\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)\left(\frac{\rho_1}{\rho_2}\right)
\label{eq:Normal:temperature:a}
\end{equation}\\
\begin{equation}
\frac{T_2}{T_1}=\left[1+\frac{2\gamma}{\gamma+1}(M^2_1-1)\right]\left[\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}\right]
\label{eq:Normal:temperature:b}
\end{equation}
\noindent Figure~\ref{fig:normal:shock:relations} below shows how different flow properties change over a normal shock as a function of upstream Mach number.\\
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter03/pdf/normal-shock-relations.pdf}
\caption{Changes of flow properties over a normal shock as a function of the upstream Mach number ($M_1$)}
\label{fig:normal:shock:relations}
\end{center}
\end{figure}
\noindent Now, one question remains. How come that we by analyzing the control volume using the upstream and downstream states get the normal shock relations. There is no way that the governing equations could have known about the fact that we assumed that there would be a shock inside of the control volume, or is it? The answer is that we have assumed that there will be a change in flow properties from upstream to downstream. We have further assumed that the flow is  adiabatic (we are using the adiabatic energy equation) so there is no heat exchange. We are, however, allowing for irreversibilities in the flow. The only way to accomplish a change in flow properties under those constraints is a formation of a normal shock (a discontinuity in flow properties - a sudden flow compression) between station 1 and station 2.
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