Questions: Normal Shock Wave Relations: Pressure, Temperature, and Density

Question 1

A normal shock passes through air at M₁ = 2.5. After the shock, which statement about stagnation quantities is correct?

• A. Both stagnation temperature and stagnation pressure are conserved because the shock does no work on the gas
• B. Stagnation temperature is conserved because the shock is adiabatic, but stagnation pressure decreases because entropy increases irreversibly inside the shock
• C. Stagnation pressure is conserved by energy conservation, but stagnation temperature drops because kinetic energy is lost in deceleration
• D. Both stagnation temperature and stagnation pressure decrease because the flow is decelerated and loses total energy

Answer: (B)

A normal shock is adiabatic — no heat crosses the shock boundary — so total enthalpy h₀ = c_p T₀ is conserved, and stagnation temperature T₀ is unchanged. However, entropy is generated irreversibly inside the shock (viscous dissipation in the thin shock zone), and stagnation pressure is directly linked to entropy via the second law: any entropy increase reduces stagnation pressure (p₀₂/p₀₁ < 1). Option A is the classic misconception: 'no work done' does not prevent entropy production. The shock generates entropy without doing work or transferring heat.

Question 2

As the upstream Mach number M₁ increases without limit, what happens to the density ratio ρ₂/ρ₁ across the normal shock?

• A. It increases without bound, proportionally to M₁², as the shock compresses the gas ever harder
• B. It approaches a finite limit of (γ+1)/(γ−1), approximately 6 for air (γ = 1.4)
• C. It approaches 1, as extremely strong shocks force both sides to the same density
• D. It grows linearly with M₁ because mass flux conservation requires proportional density increase

Answer: (B)

While pressure and temperature ratios grow without bound as M₁ → ∞, density has a finite upper limit (γ+1)/(γ−1) ≈ 6 for air. The physical reason: as shock strength increases, temperature rises so dramatically that further pressure increases are absorbed by temperature rather than density (via the ideal gas law p = ρRT). The gas becomes too hot to compress much further. This density limit is important for hypersonic reentry vehicles — the shock-compressed air cannot be packed arbitrarily densely, which constrains heat transfer models.

Question 3

Stagnation temperature is conserved across a normal shock, meaning if you brought the gas to rest isentropically before and after the shock, you would measure the same temperature in both cases.

• T. True
• F. False

Answer: True

True. A normal shock is adiabatic — no heat transfer occurs across the shock boundary. By the energy equation for adiabatic flow, total enthalpy h₀ = h + V²/2 is conserved. Since h₀ = c_p T₀ for a perfect gas, stagnation temperature T₀ is the same upstream and downstream. The shock converts kinetic energy to thermal energy (static temperature rises, velocity falls), but the total — what you'd measure if you decelerated the gas to rest — is unchanged. This is why stagnation temperature is also called 'total temperature': it includes both the static temperature and the kinetic energy equivalent.

Question 4

Stronger normal shocks (higher M₁) recover more stagnation pressure than weaker ones, which is why supersonic inlets use a single strong normal shock to efficiently decelerate incoming flow.

• T. True
• F. False

Answer: False

False — this is exactly backwards. Stronger shocks generate more entropy and therefore lose more stagnation pressure. At M₁ = 1.5, p₀₂/p₀₁ ≈ 0.93 (7% loss). At M₁ = 3.0, p₀₂/p₀₁ ≈ 0.33 (67% loss). This is precisely why high-performance supersonic inlets use a series of weaker oblique shocks to decelerate flow gradually, with each weak shock producing less entropy than one strong normal shock. Stagnation pressure recovery is a critical measure of inlet efficiency — losses here degrade engine thrust and fuel economy.

Question 5

Why do supersonic inlet designs use a series of oblique shocks rather than a single normal shock to decelerate flow from supersonic to subsonic speeds?

Model answer: Each shock wave generates entropy irreversibly, and stagnation pressure loss is proportional to entropy production. A single strong normal shock at high Mach number produces far more entropy — and loses far more stagnation pressure — than multiple weaker shocks accomplishing the same total deceleration in stages. By using oblique shocks to progressively reduce the Mach number in steps, each shock operates at a lower local Mach number and produces less entropy. The cumulative stagnation pressure recovered through a multi-shock system is substantially higher than any single-shock design. The final weak normal shock closes out the deceleration to subsonic. Higher stagnation pressure recovery means more thrust from the same fuel and air.

The mathematics confirms this: stagnation pressure ratio p₀₂/p₀₁ approaches 1 as shock strength approaches zero (isentropic limit). Splitting one strong shock into N weaker shocks, each operating at a lower Mach number, improves total p₀ recovery. In the limit of infinitely many infinitesimally weak shocks, the deceleration becomes isentropic — a theoretical maximum. Real inlets approximate this with 2–4 oblique shocks. The tradeoff is mechanical complexity against thermodynamic efficiency.