Normal shock relations

Remarks

The proof starts from the Rankine–Hugoniot conditions $\rho_1 u_1 = \rho_2 u_2$, $p_1 + \rho_1 u_1^2 = p_2 + \rho_2 u_2^2$ and $h_1 + u_1^2/2 = h_2 + u_2^2/2$, with $h = \gamma p / ((\gamma - 1)\rho)$. Eliminating $p_2$ and $\rho_2$ gives a quadratic equation for $u_2$. One root is $u_2 = u_1$, the trivial solution without a jump, and the other root gives relation 1. Relations 2 to 4 follow, and the condition that the gas is compressed across the shock is exactly $M_1 > 1$.

Example: for air, $\gamma = 1.4$, at $M_1 = 2$ the relations give $\rho_2/\rho_1 = 8/3$, $p_2/p_1 = 4.5$, $T_2/T_1 = 1.6875$ and $M_2 = 1/\sqrt{3} \approx 0.577$.

As $M_1 \to \infty$ the density ratio tends to $(\gamma + 1)/(\gamma - 1)$, which is $6$ for air, while the pressure ratio grows without bound. The algebraic formulas also give a solution for $M_1 < 1$, but then $\rho_2 < \rho_1$ and the entropy would decrease, so that jump is not a shock wave.

Going deeper

For a student

The relations are the Rankine–Hugoniot conditions solved explicitly for a polytropic gas, with the trivial solution removed. They are tabulated in engineering as normal shock tables. An oblique shock obeys the same relations with $M_1$ replaced by the normal component $M_1 \sin\beta$, where $\beta$ is the shock angle. In a moving frame they give the speed of a shock running into gas at rest, as in the shock tube problem.