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Rankine–Hugoniot Conditions

Equations

Across a stationary normal shock (upstream 1 → downstream 2):

\[\text{mass:}\quad \rho_1 u_1 = \rho_2 u_2\]
\[\text{momentum:}\quad p_1 + \rho_1 u_1^2 = p_2 + \rho_2 u_2^2\]
\[\text{energy:}\quad h_1 + \tfrac12 u_1^2 = h_2 + \tfrac12 u_2^2, \qquad h = \frac{\gamma}{\gamma-1}\frac{p}{\rho}\]
\[\text{tangential velocity:}\quad v_{t1} = v_{t2}\]

Solved in terms of the upstream Mach number \(M_1 = u_1/c_1\), \(c_1 = \sqrt{\gamma p_1/\rho_1}\):

\[\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2 + 2} \qquad \frac{p_2}{p_1} = \frac{2\gamma M_1^2 - (\gamma-1)}{\gamma+1}\]
\[\frac{T_2}{T_1} = \frac{\left[2\gamma M_1^2 - (\gamma-1)\right]\left[(\gamma-1)M_1^2 + 2\right]}{(\gamma+1)^2 M_1^2} \qquad M_2^2 = \frac{M_1^2 + \frac{2}{\gamma-1}}{\frac{2\gamma}{\gamma-1}M_1^2 - 1}\]

Physical meaning

A shock is thin (∼ mean free path) but must still conserve mass, momentum and energy. These jump conditions are the conservation laws applied to a control volume straddling the front — the integral Reynolds transport theorem with the volume squeezed to zero thickness. Given the upstream state, the downstream state is fully determined.

Variables

\(\rho\) — density · \(u\) — normal velocity (shock frame) · \(p\) — pressure · \(h\) — specific enthalpy · \(T\) — temperature · \(M\) — Mach number · \(\gamma\) — specific-heat ratio.

Assumptions

Steady in the shock frame · ideal gas with constant \(\gamma\) · adiabatic (no external heat) · normal shock (oblique shocks: apply to the normal component, tangential velocity passes through unchanged).

Limits

Weak shock (\(M_1 \to 1\)): expanding to first order gives \(\Delta p = c_s^2\,\Delta\rho\) — an adiabatic sound wave.

Strong shock (\(M_1 \gg 1\)):

\[\frac{\rho_2}{\rho_1} \to \frac{\gamma+1}{\gamma-1}\ \ (\text{finite!}), \qquad \frac{p_2}{p_1} \approx \frac{2\gamma}{\gamma+1}M_1^2, \qquad \frac{T_2}{T_1} \approx \frac{2\gamma(\gamma-1)}{(\gamma+1)^2}M_1^2\]

Density saturates (4× for monatomic, 6× for diatomic); pressure and temperature grow without bound; downstream flow is subsonic.

Entropy selects the direction

\[\Delta S = C_v \ln\left[\frac{p_2}{p_1}\left(\frac{\rho_1}{\rho_2}\right)^\gamma\right] \geq 0\]

Only compressive solutions (\(\rho_2 > \rho_1\), \(p_2 > p_1\), \(T_2 > T_1\), \(u_2 < u_1\)) satisfy the second law — "expansion shocks" are forbidden. Dissipation inside the front converts bulk kinetic energy irreversibly into heat.

Applications

Supersonic intakes and nozzles · re-entry heating · blast waves (strong-shock limit) · astrophysical shocks · shock tubes.

Worked example

Air (\(\gamma = 1.4\)), \(M_1 = 2\): \(\rho_2/\rho_1 = \frac{2.4\times4}{0.4\times4+2} = 2.67\), \(p_2/p_1 = \frac{11.2-0.4}{2.4} = 4.5\), \(T_2/T_1 = 4.5/2.67 = 1.69\), and \(M_2 = 0.577\) — supersonic in, subsonic out.

Quiz

Q1 (computational). For a very strong shock in monatomic gas (\(\gamma = 5/3\)), what is the maximum density compression?

Answer

\((\gamma+1)/(\gamma-1) = \frac{8/3}{2/3} = 4\).

Q2 (conceptual). Why does the tangential velocity component pass through a shock unchanged?

Answer

The shock is thin and inviscid on either side: no shear stress acts along the front, so there is no force to change tangential momentum, while the tangential mass flux is continuous.

Q3 (multiple choice). Downstream of a normal shock, the flow is always:

  • (a) supersonic (b) sonic (c) subsonic (d) reversed
Answer

(c). \(M_2 < 1\) whenever \(M_1 > 1\) — shocks are nature's way of decelerating supersonic flow.