Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁

1. The three conservation laws (first principles)

Take a control volume straddling the shock. State 1 = upstream (supersonic), state 2 = downstream (subsonic). Steady, 1-D, adiabatic, no area change, no friction work on the CV walls.

Why these three? Mass can't be created, momentum changes only via net force (here the pressure difference), and energy is conserved because no heat is added and the shock does no shaft work. Five unknowns ($P_2,T_2,\rho_2,u_2,M_2$), and these laws + equation of state close the system.

2. Deriving the downstream Mach number $M_2$ (the master result)

Everything is cleanest in terms of Mach number. Use $u = Ma$, $a^2=\gamma R T$, so $\rho u^2 = \rho \gamma R T M^2 = \gamma P M^2$.

Step A — rewrite momentum (2): Divide nothing, just substitute $\rho u^2=\gamma P M^2$: P_1(1+\gamma M_1^2) = P_2(1+\gamma M_2^2)\;\Rightarrow\; \frac{P_2}{P_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\tag{4} Why this step? Converts the momentum balance into a pure pressure ratio in Mach numbers.

Step B — rewrite energy (3): Stagnation temperature $T_0=T\left(1+\frac{\gamma-1}{2}M^2\right)$ is conserved, so: \frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\tag{5} Why this step? Energy conservation directly gives the temperature ratio once we know both Mach numbers.

Step C — rewrite mass (1): $\rho u = \frac{P}{RT}Ma = \frac{P M\sqrt{\gamma R T}}{RT}=\frac{PM\sqrt{\gamma}}{\sqrt{RT}}$. Setting state 1 = state 2: \frac{P_1 M_1}{\sqrt{T_1}}=\frac{P_2 M_2}{\sqrt{T_2}}\tag{6}

Step D — combine. Put (4) and (5) into (6). After algebra (square it to kill the roots) the $P$ and $T$ ratios collapse into a single equation relating $M_1$ and $M_2$, whose non-trivial root is:

Why this is beautiful: $M_2$ depends on $M_1$ only. The trivial root $M_2=M_1$ is the "no shock" solution; equation (7) is the genuine shock branch. For any $M_1>1$, equation (7) gives $M_2<1$ — supersonic always becomes subsonic.

3. The property ratios (all in terms of $M_1$)

Substitute (7) back into (4), (5), use mass for density. With $\gamma=1.4$ unless noted.

4. Stagnation pressure loss $P_{02}/P_{01}$ — the entropy fingerprint

$T_0$ is conserved (adiabatic), but $P_0$ is NOT — it drops because entropy rises. From the entropy change of an ideal gas $\Delta s = c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}$, and $P_0=P\left(1+\frac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)}$:

Since $\Delta s>0$ for all $M_1>1$, $P_{02}/P_{01}<1$ always. This loss is the engineer's enemy: in a supersonic inlet, a strong normal shock destroys stagnation pressure, hurting engine thrust. (Hence multiple weak oblique shocks are preferred.)

5. Worked examples

6. Common mistakes (Steel-man + fix)

Flashcards

Connections

• Isentropic Flow Relations — valid before/after, not across the shock.
• Speed of Sound and Mach Number — why $M>1$ is the trigger.
• Oblique Shock Waves — turning + weaker $P_0$ loss; built from normal-shock physics.
• Rankine–Hugoniot Relations — the $P$–$\rho$ form of these jumps.
• Entropy and Second Law — source of stagnation-pressure loss.
• Supersonic Inlets and Diffusers — engineering reason we minimise normal shocks.
• Stagnation Properties T0 and P0 — what survives vs what's lost.

Concept Map

Hinglish (regional understanding)