Normal shock relations

Why it matters

Normal shocks sit in front of blunt bodies, in supersonic intakes and in over-expanded nozzles and wind tunnels. They are the simplest irreversible process in gas dynamics, and their stagnation-pressure loss is what makes supersonic intakes, diffusers and tunnels expensive. The same relations, applied to the normal component of velocity, give oblique shocks.

Key ideas

What a shock is. A shock is a very thin region (a few mean free paths, of order micrometres) across which the flow changes almost discontinuously. Viscosity and heat conduction act inside it, so it is irreversible, but it is treated as a discontinuity between two uniform states.

Control-volume analysis. For a stationary normal shock in steady, adiabatic flow with no work, apply between upstream state 1 and downstream state 2:

• mass: ρ₁·u₁ = ρ₂·u₂
• momentum: p₁ + ρ₁·u₁² = p₂ + ρ₂·u₂²
• energy: h₁ + u₁²/2 = h₂ + u₂²/2, so T₀₁ = T₀₂

With the perfect-gas equation these give every downstream ratio as a function of M₁ and γ only.

Results and physical meaning.

• M₁ > 1 always, and M₂ < 1. The flow becomes subsonic.
• Pressure, density and temperature rise; velocity falls. Speed of sound rises with T.
• T₀ is unchanged (adiabatic); p₀ falls (entropy rises).
• An "expansion shock" (M₁ < 1 → M₂ > 1) satisfies the conservation laws but would decrease entropy, so the second law forbids it.
• Weak shocks (M₁ just above 1) are nearly isentropic: the entropy rise is proportional to (M₁² − 1)³.
• Strong-shock limits (M₁ → ∞): ρ₂/ρ₁ → (γ+1)/(γ−1) = 6 for air and M₂ → √((γ−1)/(2γ)) = 0.378, while p₂/p₁ and T₂/T₁ grow without bound.

Prandtl relation. u₁·u₂ = a*², i.e. M₁*·M₂* = 1: if the upstream characteristic Mach number is above 1, the downstream one must be below 1.

Moving shocks. A shock moving into still gas is analysed by switching to a frame fixed to the shock; the relations above then apply to the relative velocities.

Formulas

M₂² = (1 + (γ−1)/2 · M₁²) / (γ·M₁² − (γ−1)/2)

p₂/p₁ = 1 + 2γ/(γ+1) · (M₁² − 1)

ρ₂/ρ₁ = u₁/u₂ = (γ+1)·M₁² / (2 + (γ−1)·M₁²)

T₂/T₁ = (p₂/p₁)·(ρ₁/ρ₂)

p₀₂/p₀₁ = [ (γ+1)M₁² / (2 + (γ−1)M₁²) ]^(γ/(γ−1)) · [ (γ+1) / (2γM₁² − (γ−1)) ]^(1/(γ−1))

s₂ − s₁ = −R·ln(p₀₂/p₀₁) (J/(kg·K))

u₁·u₂ = a*² (Prandtl relation)

Symbols: subscripts 1 and 2 = just upstream and just downstream of the shock; M Mach number; p pressure (Pa); ρ density (kg/m³); T temperature (K); u velocity normal to the shock (m/s); p₀ stagnation pressure (Pa); γ ratio of specific heats; R gas constant (J/(kg·K)). All valid for a calorically perfect gas with M₁ > 1.

Reference values for air (γ = 1.4): at M₁ = 2, M₂ = 0.5774, p₂/p₁ = 4.5, ρ₂/ρ₁ = 2.667, T₂/T₁ = 1.6875, p₀₂/p₀₁ = 0.7209.

Worked examples

Example 1 (standard). A normal shock stands in air at M₁ = 2.5, p₁ = 30 kPa, T₁ = 220 K. Find M₂, p₂, T₂ and u₂. γ = 1.4, R = 287 J/(kg·K).

1. M₂² = (1 + 0.2 × 6.25)/(1.4 × 6.25 − 0.2) = 2.25/8.55 = 0.2632, M₂ = 0.513.
2. p₂/p₁ = 1 + (2.8/2.4)(6.25 − 1) = 1 + 1.1667 × 5.25 = 7.125, so p₂ = 213.8 kPa.
3. ρ₂/ρ₁ = 2.4 × 6.25/(2 + 0.4 × 6.25) = 15/4.5 = 3.333.
4. T₂/T₁ = 7.125/3.333 = 2.1375, so T₂ = 470.3 K.
5. a₂ = √(1.4 × 287 × 470.3) = 434.7 m/s; u₂ = 0.513 × 434.7 = 223 m/s. Check: u₁ = 2.5 × √(1.4 × 287 × 220) = 743.3 m/s and u₁/u₂ = 3.333 gives u₂ = 223 m/s.

Answer: M₂ = 0.513, p₂ ≈ 214 kPa, T₂ ≈ 470 K, u₂ ≈ 223 m/s.

Example 2 (GATE level). For the same shock, find the stagnation-pressure ratio and entropy rise, and the static pressure downstream as a fraction of the upstream stagnation pressure.

1. ρ₂/ρ₁ = 3.333, so the first bracket is 3.333^3.5 = 67.6.
2. Second bracket: 2.4/(2.8 × 6.25 − 0.4) = 2.4/17.1 = 0.14035, raised to 2.5: 0.14035^2.5 = 0.007380.
3. p₀₂/p₀₁ = 67.6 × 0.007380 = 0.499.
4. s₂ − s₁ = −287 × ln 0.499 = 199.5 J/(kg·K).
5. Upstream p₀₁/p₁ = (1 + 0.2 × 6.25)^3.5 = 2.25^3.5 = 17.09. So p₂/p₀₁ = 7.125/17.09 = 0.417.

Answer: p₀₂/p₀₁ ≈ 0.499 (half the total pressure is lost), Δs ≈ 200 J/(kg·K), p₂/p₀₁ ≈ 0.417.

Common mistakes

• Assuming p₀ is conserved across the shock — only T₀ is.
• Using the downstream stagnation pressure with the upstream Mach number in isentropic relations.
• Forgetting that M₂ depends only on M₁: you cannot "choose" a downstream supersonic state.
• Inverting the density ratio (ρ₂/ρ₁ > 1, u₂/u₁ < 1).
• Applying the stationary-shock relations to a moving shock without changing to the shock-fixed frame.
• Thinking the density ratio can grow without limit; it tends to 6 for air.

For GATE AE

Expect direct use of the relations at a given M₁ (pressure, density, temperature ratios, M₂, p₀₂/p₀₁), combinations with isentropic flow before and after the shock (for example in nozzles or pitot probes), entropy-change questions, strong-shock limits, and statements about which quantities rise, fall or stay constant. Learn the M = 2 values; they appear often as a check.

Quick check

1. Find p₂/p₁ for M₁ = 3 in air.
2. Which quantity is unchanged across a stationary normal shock: p₀, T₀ or s?
3. What is the limiting density ratio across a very strong shock in air?
4. Find T₂/T₁ for M₁ = 3 in air.
5. Why is an expansion shock impossible?

Answers: 1. 10.33; 2. T₀; 3. 6; 4. 2.679; 5. it would reduce entropy in an adiabatic process, violating the second law.