Question bank — Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations

A normal shock always slows a supersonic flow to subsonic. ::: True. For any $M_1>1$ the surviving root of the shock equations gives $M_2<1$; the only other root ($M_2=M_1$) is the no-shock case. A shock can accelerate a subsonic flow ($M_1<1$) up to supersonic. ::: False. That "expansion shock" root would lower entropy, which the Second Law forbids. Real shocks are always compressive. Across an adiabatic shock the total pressure $p_0$ is conserved. ::: False. Adiabatic conserves total enthalpy (so $T_0$ is constant), not total pressure. The shock is irreversible, entropy rises, and $p_0$ therefore drops. Static temperature $T$ and stagnation temperature $T_0$ both jump across the shock. ::: Half true: static $T$ jumps up, but $T_0$ stays exactly the same because total enthalpy is conserved ($h_1+\frac{1}{2}{u}_1^2=h_2+\frac{1}{2}{u}_2^2$). Entropy increases across every real normal shock. ::: True. The irreversible viscous/conductive dissipation inside the shock guarantees $\Delta s>0$; this is precisely the law that picks the compression root. As $M_1\to \infty$ the density ratio ${\rho }_2/{\rho }_1$ also grows without bound. ::: False. It saturates at $\frac{\gamma +1}{\gamma -1}$ (= 6 for air). A perfect gas cannot be compressed past that ratio no matter how strong the shock. As $M_1\to \infty$ the pressure ratio $p_2/p_1$ grows without bound. ::: True. $p_2/p_1=1+\frac{2\gamma }{\gamma +1}(M_1^2-1)$ has no ceiling — pressure and temperature keep climbing even after density has capped out. A shock of Mach number exactly $M_1=1$ produces a finite pressure jump. ::: False. At $M_1=1$ every ratio collapses to 1 — it is an infinitely weak shock, i.e. no jump at all. A "sonic shock" carries zero strength. The flow crosses the shock at constant total temperature but not constant total pressure. ::: True. This is the defining signature of a shock: $T_0$ preserved (adiabatic, no heat/work), $p_0$ lost (irreversible). Adiabatic $\ne$ isentropic. Because the shock conserves mass, momentum and energy, it must be reversible. ::: False. Those three conservation laws hold across any control volume, reversible or not. Reversibility is a separate condition (constant $s$), which the shock violates.

"The gas is compressed, so ${\rho }_2/{\rho }_1=p_2/p_1$." ::: Wrong — that assumes constant temperature. Since $T$ also jumps, ${\rho }_2/{\rho }_1=(p_2/p_1)(T_1/T_2)$; density saturates while pressure does not. "The flow is adiabatic, so it must be isentropic across the shock." ::: Wrong — adiabatic means no heat added, but internal viscous dissipation still generates entropy. Adiabatic + reversible would give isentropic; the shock is adiabatic + irreversible. "Downstream Mach number: $M_2=1/M_1$." ::: Wrong — there is no simple reciprocal rule. The correct relation is $M_2^2=\frac{1+\frac{\gamma -1}{2}{M}_1^2}{\gamma M_1^2-\frac{\gamma -1}{2}}$, which only roughly mirrors 1 near $M_1=1$. "Since $T_0$ is unchanged, the flow could be brought to rest with no loss." ::: Wrong — $T_0$ is unchanged but $p_0$ has fallen. The stagnation state is degraded; you would recover the same temperature but a lower pressure, meaning lost usable work. "$\rho u^2$ has units of pressure only by coincidence." ::: Wrong — $\rho u^2$ is a genuine momentum flux and shares pressure's units by physics, which is exactly why $p+\rho u^2$ (the momentum equation) is meaningful. Using $\rho u^2=\gamma p{M}^2$ is the derivation's engine. "A weak shock ($M_1$ just above 1) still loses a lot of total pressure." ::: Wrong — the $p_0$ loss scales like $(M_1^2-1{)}^3$ for weak shocks, so it is third-order small. Weak shocks are nearly isentropic; big losses need strong shocks. "Because velocity drops across the shock, kinetic energy is destroyed." ::: Wrong — kinetic energy is not destroyed but converted into enthalpy (thermal energy); total enthalpy $h+\frac{1}{2}{u}^2$ is conserved. The disorder of that conversion is what raises entropy.

Why must a supersonic flow use a shock rather than a smooth pressure rise to meet an obstacle? ::: Pressure signals travel at the speed of sound; upstream molecules moving faster than sound cannot "hear" the obstacle, so information can't propagate ahead and the adjustment must happen abruptly. Why does the shock select the compression root and not the expansion root? ::: Both roots satisfy mass, momentum and energy, but only compression gives $\Delta s>0$. The Second Law is the tie-breaker that the three conservation laws alone cannot supply. Why does density saturate but temperature keep rising as $M_1\to \infty$? ::: A perfect gas has a fixed limit on how tightly molecules can pack under these relations, so extra incoming kinetic energy has nowhere to go but into random thermal motion — i.e. temperature, not further compaction. Why is $T_0$ constant across the shock even though everything else changes? ::: Because the shock adds no heat and does no work, so total enthalpy $h_0=c_p{T}_0$ is conserved; $T_0$ measures that conserved total enthalpy directly. Why does re-entry heating come from the shock, not from surface friction alone? ::: The bow shock converts the vehicle's enormous relative kinetic energy into $T_2/T_1\sim M_1^2$ heating; at hypersonic $M_1$ this thermal jump dwarfs boundary-layer friction. Why does the substitution $\rho u^2=\gamma p{M}^2$ make the algebra collapse to functions of $M$ alone? ::: It rewrites the only velocity-and-density term in the momentum equation purely in terms of $p$ and $M$, so once $T$ ratios are handled via $a=\sqrt{\gamma RT}$, everything reduces to Mach number. Why is the total-pressure loss called the "price" of a shock? ::: Lost $p_0$ is lost ability to do useful work (thrust, recovery); an engine inlet with a strong shock wastes stagnation pressure, showing up as drag and reduced efficiency.

What happens to all five ratios in the limit $M_1\to 1^{+}$? ::: Every ratio ($p_2/p_1$, ${\rho }_2/{\rho }_1$, $T_2/T_1$, $p_{02}/p_{01}$) $\to 1$ and $M_2\to 1$. The shock vanishes into an infinitely weak Mach wave. What is the limiting downstream Mach number $M_2$ as $M_1\to \infty$? ::: $M_2^2\to \frac{\gamma -1}{2\gamma }$ ($\approx 0.378$, so $M_2\approx 0.615$ for air). Even an infinitely strong shock leaves a finite subsonic exit Mach number. Can $M_1=1$ exactly produce a shock? ::: No — $M_1=1$ is the degenerate boundary where shock strength is zero. A genuine shock needs $M_1>1$ strictly. Is a normal shock possible in a purely subsonic flow anywhere? ::: No. A stationary normal shock requires the upstream flow to be supersonic; with $M_1<1$ the compression root would violate the Second Law. In the strong-shock limit, does $p_{02}/p_{01}$ approach zero or a finite floor? ::: It approaches zero — entropy generation grows without bound, so $p_{02}/p_{01}=\exp (-\Delta s/R)\to 0$. Total-pressure recovery collapses for very strong shocks. What distinguishes a normal shock's downstream state from an oblique shock's on the same $M_1$? ::: An oblique shock processes only the flow's normal component, so it is effectively a weaker "normal shock" at $M_1\sin \beta$; the tangential velocity passes through unchanged, giving a milder jump and possibly still-supersonic downstream flow. If you insisted $\gamma \to 1$, what happens to the density-saturation limit? ::: $\frac{\gamma +1}{\gamma -1}\to \infty$, meaning a gas with many internal energy modes (low $\gamma$) can be compressed arbitrarily — the "6× cap" is specific to $\gamma =1.4$.