Questions: Normal Shock Waves and Shock Analysis

Question 1

Across a normal shock wave, which of the following quantities is conserved (remains unchanged)?

• A. Static pressure — the shock is thin so static pressure must equalize
• B. Stagnation pressure — energy is conserved so total pressure must be constant
• C. Stagnation temperature — the process is adiabatic so total enthalpy is conserved
• D. Entropy — the flow through a shock is isentropic for a perfect gas

Answer: (C)

Stagnation temperature is conserved across a normal shock because the shock process is adiabatic — no heat is added or removed. Energy conservation (h₀ = h + V²/2 = constant) directly implies stagnation enthalpy and stagnation temperature are unchanged. However, stagnation pressure is NOT conserved: the irreversible viscous mixing inside the shock generates entropy, and entropy increase reduces stagnation pressure. Static pressure, density, and temperature all increase across the shock. A normal shock is decidedly not isentropic.

Question 2

A supersonic aircraft inlet must decelerate incoming flow from Mach 2.5 to subsonic before it enters the engine. Why do designers use a series of oblique shocks rather than a single normal shock?

• A. Oblique shocks are easier to create geometrically in an inlet duct
• B. A single normal shock at Mach 2.5 produces a greater entropy rise and stagnation pressure loss than multiple weaker shocks; more but weaker shocks means less total entropy production and higher pressure recovery
• C. Normal shocks always decelerate flow to exactly Mach 1, not to subsonic speeds, making them useless for inlet design
• D. Oblique shocks avoid the flow separation that always occurs with normal shocks

Answer: (B)

Entropy rise across a shock increases nonlinearly with upstream Mach number. A single strong normal shock at M=2.5 produces much more entropy (and stagnation pressure loss) than the sum of multiple weaker shocks that achieve the same total deceleration. By decelerating gradually through a sequence of oblique shocks, each at a lower Mach number, the designer keeps each individual shock weak, minimizing entropy production at each step. The cumulative stagnation pressure loss is substantially less, preserving more total pressure for the engine and improving thrust and efficiency.

Question 3

Stagnation temperature is conserved across a normal shock wave.

• T. True
• F. False

Answer: True

Stagnation temperature T₀ = T(1 + (γ−1)/2 · M²) is conserved because the flow through a shock is adiabatic — no heat is transferred. Energy conservation (h₀ = constant) directly implies constant stagnation temperature. This is distinct from stagnation pressure, which is NOT conserved because entropy increases. The distinction between stagnation temperature (conserved) and stagnation pressure (not conserved) is central to shock analysis.

Question 4

A stronger normal shock (higher upstream Mach number) produces less stagnation pressure loss than a weaker shock, because stronger shocks are more efficient at decelerating flow.

• T. True
• F. False

Answer: False

The opposite is true. Stronger shocks produce greater entropy rises, which directly correspond to greater stagnation pressure losses. At M₁ = 1 (vanishingly weak shock), stagnation pressure loss is zero and entropy rise is negligible. As M₁ increases, both entropy rise and stagnation pressure drop grow rapidly and nonlinearly. This is precisely why supersonic inlet design avoids strong normal shocks — a single shock at high Mach number wastes far more stagnation pressure than a series of weaker shocks achieving the same total deceleration.

Question 5

Why does stagnation pressure always drop across a normal shock, and why does this matter for jet engine performance?

Model answer: Stagnation pressure represents the maximum static pressure recoverable if the flow were decelerated isentropically to rest. Across a shock, the process is irreversible — violent viscous and thermal mixing generates entropy. The second law requires entropy to increase, and for an adiabatic process, entropy increase is thermodynamically equivalent to stagnation pressure decrease. Once stagnation pressure is lost, it cannot be recovered downstream. For a jet engine, the inlet must deliver air at the highest possible stagnation pressure to the compressor — higher inlet stagnation pressure means higher achievable compression ratio and greater thrust. Every unit of stagnation pressure lost in the inlet directly reduces engine performance.

This connection between irreversibility (entropy generation), stagnation pressure loss, and engine efficiency is the thermodynamic core of compressible flow analysis. It explains inlet design (minimize shock strength), nozzle design (avoid internal shocks), and why total pressure recovery is the key figure of merit for supersonic propulsion systems.