One-dimensional inviscid flow

Acoustic Wave

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, $a+da$. 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.

$${\rho }_1{u}_1={\rho }_2{u}_2$$

$$\rho a=(\rho +d\rho )(a+da)$$

$$\cancel{\rho a}=\cancel{\rho a}+\rho da+ad\rho +\underset{\sim 0}{\underbrace{d\rho da}}\Rightarrow$$

$$a=-\rho \frac{da}{d\rho }$$

The one-dimensional momentum equation between station 1 and station 2 gives

$${\rho }_1{u}_1^2+p_1={\rho }_2{u}_2^2+p_2$$

$$\rho a^2+p=(\rho +d\rho )(a+da{)}^2+(p+dp)$$

$$\cancel{\rho a^2}+\cancel{p}=\cancel{\rho a^2}+2\rho ada+\underset{\sim 0}{\underbrace{\rho d{a}^2}}+d\rho a^2+\underset{\sim 0}{\underbrace{2d\rho ada}}+\underset{\sim 0}{\underbrace{d\rho d{a}^2}}+\cancel{p}+dp\Rightarrow$$

$$dp=-2\rho ada-d\rho a^2\Rightarrow$$

$$da=-\frac{dp+d\rho a^2}{2\rho a}=-\frac{d\rho }{2a\rho }(\frac{dp}{d\rho }+a^2)\Rightarrow$$

$$\frac{da}{d\rho }=-\frac{1}{2a\rho }(\frac{dp}{d\rho }+a^2)$$

Eqn. \ref{eq:speedofsound:b} in \ref{eq:speedofsound:a} gives

$$a=\frac{1}{2a}(\frac{dp}{d\rho }+a^2)\Rightarrow$$

$$a^2=\frac{dp}{d\rho }$$

Sound wave:

• gradients are small
• irreversible (dissipative effects are negligible)
• no heat addition

Thus, the change of flow properties as the sound wave passes can be assumed to be an isentropic process

$$a^2={\left(\frac{dp}{d\rho }\right)}_s$$

$$a=\sqrt{{\left(\frac{dp}{d\rho }\right)}_s}=\sqrt{\frac{1}{\rho {\tau }_s}}$$

where ${\tau }_s$ is the compressibility of the gas. Eqn. \ref{eq:speedofsound:d} is valid for all gases. It can be seen from the equation, that truly incompressible flow (${\tau }_s=0$) would imply infinite speed of sound.

Since the process is isentropic, we can use the isentropic relations if we also assume the gas to be calorically perfect\\

$$\frac{p_2}{p_1}={\left(\frac{{\rho }_2}{{\rho }_1}\right)}^{\gamma }\Rightarrow p=C{\rho }^{\gamma }$$

$$a^2={\left(\frac{dp}{d\rho }\right)}_s=\gamma C{\rho }^{\gamma -1}=\gamma \underset{=p}{\underbrace{\left[C{\rho }^{\gamma }\right]}}{\rho }^{-1}=\frac{\gamma p}{\rho }\Rightarrow$$

$$a=\sqrt{\frac{\gamma p}{\rho }}$$

or

$$a=\sqrt{\gamma RT}$$

From the relation above, it is obvious that the local speed of sound is related to the temperature of the flow, which in turn is a measure of the motion of elementary particles (atoms and/or molecules) of the fluid at a specific location. This stems from the fact that sound waves are propagated via interaction of these elementary particles. Since information in a flow is propagated via molecular interaction the relation between the speed at which this information is conveyed and the speed of the flow has important physical implications. Figure~\ref{fig:speed:of:sound} compares three sound wave patterns generated by a a beacon. In the left picture, the sound transmitter is stationary and thus the acoustic waves are centered around the transmitter. In the middle image, the transmitter is moving to the left at a speed less than the speed of sound and thus the transmitter will always be within all sound wave circles but it will be off-centered with a bias in the direction that the transmitter is moving. In the right image the transmitter is moving faster than the speed of sound and thus it will always be located outside of all acoustic waves. In a supersonic flow, no information can travel upstream and therefore there is no way for the flow to adjust to downstream obstacles. This is compensated for by the introduction of shocks in the flow. Over a shock flow properties changes discontinuity. An example is given in Figure~\ref{fig:supersonic:flow}.

Shock Wave

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}).

continuity:

$${\rho }_1{u}_1={\rho }_2{u}_2$$

momentum:

$${\rho }_1{u}_1^2+p_1={\rho }_2{u}_2^2+p_2$$

energy:

$$h_1+\frac{1}{2}{u}_1^2=h_2+\frac{1}{2}{u}_2^2$$

Divide the momentum equation by ${\rho }_1{u}_1$

$$\frac{1}{{\rho }_1{u}_1}({\rho }_1{u}_1^2+p_1)=\frac{1}{{\rho }_1{u}_1}({\rho }_2{u}_2^2+p_2)=\{{\rho }_1{u}_1={\rho }_2{u}_2\}=\frac{1}{{\rho }_2{u}_2}({\rho }_2{u}_2^2+p_2)\Rightarrow$$

$$\frac{p_1}{{\rho }_1{u}_1}-\frac{p_2}{{\rho }_2{u}_2}=u_2-u_1$$

For a calorically perfect gas $a=\sqrt{\gamma p/\rho }$, which if implemented in Eqn. \ref{eq:governing:mom:b} gives

$$\frac{a_1^2}{\gamma u_1}-\frac{a_2^2}{\gamma u_2}=u_2-u_1$$

The energy equation (Eqn. \ref{eq:governing:energy}) with $h=C_{p}T$

$$C_p{T}_1+\frac{1}{2}{u}_1^2=C_p{T}_2+\frac{1}{2}{u}_2^2$$

Replacing $C_p$ with $\gamma R/(\gamma -1)$ gives

$$\frac{\gamma R{T}_1}{\gamma -1}+\frac{1}{2}{u}_1^2=\frac{\gamma R{T}_2}{\gamma -1}+\frac{1}{2}{u}_2^2$$

With $a=\sqrt{\gamma RT}$ this becomes

$$\frac{a_1^2}{\gamma -1}+\frac{1}{2}{u}_1^2=\frac{a_2^2}{\gamma -1}+\frac{1}{2}{u}_2^2$$

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, $u$, and speed of sound, $a$, in any point to the corresponding flow properties at sonic conditions ($u=a=a^{*}$).

$$\frac{a^2}{\gamma -1}+\frac{1}{2}{u}^2=\frac{\gamma +1}{2(\gamma -1)}{a^{*}}^2$$

\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.