Shock waves

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 ($u=a=a^{*}$).

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

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

$$\begin{array}{cc}& a_1^2=\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_1^2\\ & a_2^2=\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_2^2\end{array}$$

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.

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

$$\frac{1}{\gamma u_1}(\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_1^2)-\frac{1}{\gamma u_2}(\frac{\gamma +1}{2}{a^{*}}^2-\frac{\gamma -1}{2}{u}_2^2)=u_2-u_1\Rightarrow$$

$$\left(\frac{\gamma +1}{2\gamma }\right){a^{*}}^2(\frac{1}{u_1}-\frac{1}{u_2})=\left(\frac{\gamma +1}{2\gamma }\right)(u_2-u_1)\Rightarrow$$

$${a^{*}}^2(\frac{1}{u_1}-\frac{1}{u_2})=(u_2-u_1)\Rightarrow$$

$${a^{*}}^2(\frac{u_2}{u_1{u}_2}-\frac{u_1}{u_1{u}_2})=(u_2-u_1)\Rightarrow$$

$$\frac{1}{u_1{u}_2}{a^{*}}^2(u_2-u_1)=(u_2-u_1)\Rightarrow$$

$${a^{*}}^2=u_1{u}_2$$

Eqn. \ref{eq:prandtl} is sometimes referred to as the Prandtl relation. Divide the Prandtl relation by ${a^{*}}^2$ on both sides gives

$$1=\frac{u_1}{a^{*}}\frac{u_2}{a^{*}}=M_1^{*}{M}_2^{*}$$

or

$$M_2^{*}=\frac{1}{M_1^{*}}$$

The relation between $M^{*}$ and $M$ is given by

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

from which is can be seen that $M^{*}$ will follow the Mach number $M$ in the sense that

• $M=1\Rightarrow M^{*}=1$
• $M<1\Rightarrow M^{*}<1$
• $M>1\Rightarrow M^{*}>1$

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

$$\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}=\frac{2+(\gamma -1)M_2^2}{(\gamma +1)M_2^2}$$

$$M_2^2=\frac{1+[(\gamma -1)/2]M_1^2}{\gamma M_1^2-(\gamma -1)/2}$$

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.

Analyzing the energy equation and the second law of thermodynamics shows that there is a direct relation between entropy increase and total pressure drop.

$$s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{p_2}{p_1}$$

$$s_2-s_1=C_p\ln \frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}\frac{T_{o_2}}{T_{o_1}}-R\ln \frac{p_2}{p_{o_2}}\frac{p_{o_1}}{p_1}\frac{p_{o_2}}{p_{o_1}}$$

using the isentropic relations we get

$$s_2-s_1=C_p\ln \frac{T_{o_2}}{T_{o_1}}-R\ln \frac{p_{o_2}}{p_{o_1}}$$

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

$$s_2-s_1=-R\ln \frac{p_{o_2}}{p_{o_1}}$$

or

$$\frac{p_{o_2}}{p_{o_1}}=e^{-(s_2-s_1)/R}$$

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.

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, $M_1$.

$${M_2^2|}_{M_1\to \mathrm{\infty }}={\frac{2/M_1^2+(\gamma -1)}{2\gamma -(\gamma -1)/M_1^2}|}_{M_1\to \mathrm{\infty }}=\frac{\gamma -1}{2\gamma }$$