One-dimensional inviscid flow: Difference between revisions

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In Fig. \ref{fig:soundwave}, station 1 represents the flow state ahead of the sound wave and station 2 the flow state behind the sound wave. Set up the continuity equation for one-dimensional flows between 1 and 2. If we could change frame of reference and follow the sound wave, we would see fluid approaching the wave with the propagation speed of the wave, $a$, and behind the wave, the fluid would have a slightly modified speed, <math>a+da</math>. There would also be a slight in all other flow properties. Let's apply the one-dimensional continuity equation between station 1 and station 2.
In Fig. \ref{fig:soundwave}, station 1 represents the flow state ahead of the sound wave and station 2 the flow state behind the sound wave. Set up the continuity equation for one-dimensional flows between 1 and 2. If we could change frame of reference and follow the sound wave, we would see fluid approaching the wave with the propagation speed of the wave, <math>a</math>, and behind the wave, the fluid would have a slightly modified speed, <math>a+da</math>. There would also be a slight in all other flow properties. Let's apply the one-dimensional continuity equation between station 1 and station 2.


<math display="block"> \rho_1 u_1=\rho_2 u_2</math>
<math display="block"> \rho_1 u_1=\rho_2 u_2</math>
Line 192: Line 192:
</math>
</math>


\noindent If Eqn. \ref{eq:governing:energy:e} is evaluated in locations 1 and 2, we get\\
If Eqn. \ref{eq:governing:energy:e} is evaluated in locations 1 and 2, we get


\begin{equation}
<math display="block">
\begin{aligned}
\begin{aligned}
&a_1^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\\
&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
&a_2^2 = \frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2
\end{aligned}
\end{aligned}
\label{eq:governing:energy:f}
</math>
\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.\\
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 <math>a^*</math> will be constant.


\noindent Eqn. \ref{eq:governing:energy:f} inserted in \ref{eq:governing:mom:c} gives\\
Eqn. \ref{eq:governing:energy:f} inserted in \ref{eq:governing:mom:c} gives\\




\[\frac{1}{\gamma u_1}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_1^2\right)-\frac{1}{\gamma u_2}\left(\frac{\gamma+1}{2}{a^*}^2 - \frac{\gamma-1}{2}u_2^2\right)=u_2-u_1 \Rightarrow \]\\
<math display="block">
\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</math>


\[\left(\frac{\gamma+1}{2\gamma}\right){a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(\frac{\gamma+1}{2\gamma}\right)\left(u_2-u_1\right) \Rightarrow\]\\
<math display="block">
\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
</math>


\[{a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(u_2-u_1\right) \Rightarrow\]\\
<math display="block">
{a^*}^2\left(\frac{1}{u_1}-\frac{1}{u_2}\right)=\left(u_2-u_1\right) \Rightarrow
</math>


\[{a^*}^2\left(\frac{u_2}{u_1 u_2}-\frac{u_1}{u_1 u_2}\right)=\left(u_2-u_1\right) \Rightarrow\]\\
<math display="block">
{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
</math>


\[\frac{1}{u_1 u_2}{a^*}^2\left(u_2-u_1\right)=\left(u_2-u_1\right) \Rightarrow\]\\
<math display="block">
\frac{1}{u_1 u_2}{a^*}^2\left(u_2-u_1\right)=\left(u_2-u_1\right) \Rightarrow
</math>


\begin{equation}
<math display="block">
{a^*}^2=u_1 u_2
{a^*}^2=u_1 u_2
\label{eq:prandtl}
</math>
\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\\
Eqn. \ref{eq:prandtl} is sometimes referred to as the Prandtl relation. Divide the Prandtl relation by <math>{a^*}^2</math> on both sides gives


\[1=\frac{u_1}{a^*}\frac{u_2}{a^*}=M^*_1M^*_2\]\\
<math display="block">
1=\frac{u_1}{a^*}\frac{u_2}{a^*}=M^*_1M^*_2
</math>


or  
or  


\begin{equation}
<math display="block">
M^*_2=\frac{1}{M^*_1}
M^*_2=\frac{1}{M^*_1}
\label{eq:NormalMach}
</math>
\end{equation}\\


\noindent The relation between $M^*$ and $M$ is given by\\
The relation between <math>M^*</math> and <math>M</math> is given by


\begin{equation}
<math display="block">
{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}
{M^*}^2=\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}
\label{eq:MachStar}
</math>
\end{equation}\\


\noindent from which is can be seen that $M^*$ will follow the Mach number $M$ in the sense that\\
from which is can be seen that <math>M^*</math> will follow the Mach number <math>M</math> in the sense that


\begin{itemize}
* <math>M=1\Rightarrow M^*=1</math>
\item $M=1\Rightarrow M^*=1$
* <math>M<1\Rightarrow M^*<1</math>
\item $M<1\Rightarrow M^*<1$
* <math>M>1\Rightarrow M^*>1</math>
\item $M>1\Rightarrow M^*>1$
\end{itemize}


\noindent Eqn. \ref{eq:MachStar} inserted in Eqn. \ref{eq:NormalMach} gives\\
Eqn. \ref{eq:MachStar} inserted in Eqn. \ref{eq:NormalMach} gives




\[\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}=\frac{2+(\gamma-1)M_2^2}{(\gamma+1)M_2^2}\]\\
<math display="block">
\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}=\frac{2+(\gamma-1)M_2^2}{(\gamma+1)M_2^2}
</math>


\begin{equation}
<math display="block">
M^2_2=\frac{1+\left[(\gamma-1)/2\right]M^2_1}{\gamma M^2_1-(\gamma-1)/2}
M^2_2=\frac{1+\left[(\gamma-1)/2\right]M^2_1}{\gamma M^2_1-(\gamma-1)/2}
\label{eq:NormalMach:b}
</math>
\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.\\
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 <math>M_1=1.0</math> 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.
Analyzing the energy equation and the second law of thermodynamics shows that there is a direct relation between entropy increase and total pressure drop.


\[s_2-s_1=C_p\ln\dfrac{T_2}{T_1}-R\ln\dfrac{p_2}{p_1}\]
<math display="block">
s_2-s_1=C_p\ln\dfrac{T_2}{T_1}-R\ln\dfrac{p_2}{p_1}
</math>


\[s_2-s_1=C_p\ln\dfrac{T_2}{T_{o_2}}\dfrac{T_{o_1}}{T_1}\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_2}{p_{o_2}}\dfrac{p_{o_1}}{p_1}\dfrac{p_{o_2}}{p_{o_1}}\]
<math display="block">
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}}
</math>


\noindent using the isentropic relations we get
using the isentropic relations we get


\[s_2-s_1=C_p\ln\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_{o_2}}{p_{o_1}}\]
<math display="block">
s_2-s_1=C_p\ln\dfrac{T_{o_2}}{T_{o_1}}-R\ln\dfrac{p_{o_2}}{p_{o_1}}
</math>


\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
and since the process is adiabatic and thus <math>T_{o_2}=T_{o_1}</math> the change in entropy is directly related to the change in total pressure as


\[s_2-s_1=-R\ln\dfrac{p_{o_2}}{p_{o_1}}\]
<math display="block">
s_2-s_1=-R\ln\dfrac{p_{o_2}}{p_{o_1}}
</math>


or
or


\[\dfrac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}\]
<math display="block">
\dfrac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}
</math>


\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.
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.


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By rewriting the right-hand side of Eqn.\ref{eq:NormalMach:b}, it is easy to realize that the downstream Mach number <math>M_2</math> approaches a finite value for large values of the upstream Mach number, <math>M_1</math>.


<math display="block">
\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}
</math>


===Normal Shock Relations===


Rewriting the continuity equation (Eqn. \ref{eq:governing:cont})


 
<math display="block">
 
%\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
\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}
</math>
\end{equation}\\


\noindent Eqn. \ref{eq:MachStar} in Eqn. \ref{eq:Normal:density:a} gives\\
Eqn. \ref{eq:MachStar} in Eqn. \ref{eq:Normal:density:a} gives


\begin{equation}
<math display="block">
\frac{\rho_2}{\rho_1}=\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}
\frac{\rho_2}{\rho_1}=\frac{(\gamma+1)M_1^2}{2+(\gamma-1)M_1^2}
\label{eq:Normal:density:b}
</math>
\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})\\
To get a corresponding relation for the pressure ratio  over the shock, we go back to the momentum equation (Eqn. \ref{eq:governing:mom})


\[p_2-p_1=\rho_1 u^2_1 - \rho_2 u^2_2=\left\{\rho_1 u_1=\rho_2 u_1\right\}=\rho_1 u_1(u_1-u_2)=\rho_1 u^2_1\left(1-\frac{u_2}{u_1}\right)\]\\
<math display="block">
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)
</math>


\[\frac{p_2-p_1}{p_1}=\frac{\rho_1 u^2_1}{p_1}\left(1-\frac{u_2}{u_1}\right)=\left\{a_1=\sqrt{\frac{\gamma p_1}{\rho_1}}\right\}=\gamma\frac{u^2_1}{a^2_1}\left(1-\frac{u_2}{u_1}\right)=\gamma M^2_1\left(1-\frac{u_2}{u_1}\right)\]\\
<math display="block">
\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)
</math>


\[\frac{p_2}{p_1}-1=\gamma M^2_1\left(1-\frac{u_2}{u_1}\right)=\left\{\frac{u_2}{u_1}=\frac{\rho_1}{\rho_2}\right\}=\gamma M^2_1\left(1-\frac{2+(\gamma-1)M_1^2}{(\gamma+1)M_1^2}\right)\]\\
<math display="block">
\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)
</math>


\begin{equation}
<math display="block">
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M^2_1-1)
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M^2_1-1)
\label{eq:Normal:pressure}
</math>
\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.\\
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.


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\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 $p=\rho RT$\\
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 <math>p=\rho RT</math>


\begin{equation}
<math display="block">
\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)\left(\frac{\rho_1}{\rho_2}\right)
\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)\left(\frac{\rho_1}{\rho_2}\right)
\label{eq:Normal:temperature:a}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\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]
\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}
</math>
\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.\\
Figure~\ref{fig:normal:shock:relations} below shows how different flow properties change over a normal shock as a function of upstream Mach number.


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\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.
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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