Supersonic diffusers and supersonic wind tunnels

Why it matters

A supersonic wind tunnel spends most of its power making up the stagnation-pressure loss in its diffuser, and a supersonic aircraft intake is a diffuser whose losses come straight off the engine thrust. Knowing how a supersonic diffuser starts, why it needs a second throat and how much pressure it recovers lets you size tunnels and intakes and explain unstart.

Key ideas

The tunnel layout. Settling chamber (p₀, T₀) → C-D nozzle with first throat A_t1 → constant-area test section at the design Mach number M_T → supersonic diffuser with a second throat A_t2 → subsonic diffuser → exhaust (blowdown to atmosphere or vacuum) or return to the compressor (continuous tunnel).

The test-section area follows from isentropic expansion: A_T/A_t1 = (A/A*) at M_T. The nozzle contour is designed by the method of characteristics to give uniform, shock-free flow.

Why a diffuser is needed. The flow must be returned to low speed. Without any diffuser the flow would end in a normal shock at the test-section Mach number, losing a large fraction of p₀ (at M = 3 only 33% is recovered). The compressor must supply a pressure ratio of at least p₀₁/p₀₂ to drive the tunnel. A supersonic diffuser that slows the flow gradually, through a converging section to a second throat and then a diverging subsonic section, reduces this loss.

Ideal vs real. An isentropic supersonic diffuser (reverse nozzle, M → 1 at the throat, then subsonic deceleration) would recover all of p₀, but it cannot work in practice: it cannot be started, and slight disturbances make the shock unstable.

Starting problem. During start-up a normal shock travels from the nozzle through the test section. While the shock is in the test section at M_T, the mass flow downstream must pass the second throat with the reduced stagnation pressure p₀₂. Since ṁ ∝ p₀·A*, the second throat must satisfy A_t2/A_t1 ≥ p₀₁/p₀₂ (normal-shock value at M_T). If A_t2 is smaller, the shock cannot be "swallowed" and the tunnel will not start. Once the shock passes the second throat, the test section runs supersonic.

Running condition. After starting, the shock settles just downstream of the second throat, where the supersonic Mach number is set by A_t2/A_t1 acting as a local A/A*. This Mach number is lower than M_T, so the running loss is smaller than the starting loss. A variable-geometry second throat can be closed down after starting to reduce the shock Mach number further.

Oblique-shock diffusers and intakes. Compressing through one or more oblique shocks before a weak terminal normal shock recovers much more p₀ than a single normal shock. Intakes are rated by total-pressure recovery η = p₀,exit/p₀∞. Exceeding the allowable back pressure expels the terminal shock (unstart), cutting mass flow; boundary-layer bleed and variable ramps prevent this.

Real effects. Boundary-layer growth and shock–boundary-layer interaction in the diffuser make recovery lower than the inviscid estimates; designers use empirical factors from test data.

Formulas

A_T/A_t1 = (1/M_T)·[(2/(γ+1))(1 + (γ−1)/2·M_T²)]^((γ+1)/(2(γ−1))) — test-section to nozzle-throat area.

A_t2/A_t1 = p₀₁/p₀₂ (normal-shock ratio at M_T) — minimum second-throat area for starting.

p₀₂/p₀₁ = [ (γ+1)M²/(2 + (γ−1)M²) ]^(γ/(γ−1)) · [ (γ+1)/(2γM² − (γ−1)) ]^(1/(γ−1)) — normal-shock recovery.

η = p₀,out/p₀,in — total-pressure recovery of a diffuser or intake.

ṁ = 0.0404·p₀·A*/√T₀ — choked mass flow for air (SI units); continuity of p₀·A* at constant T₀.

Symbols: A_t1, A_t2 first and second throat areas (m²); A_T test-section area (m²); M_T test-section Mach number; p₀₁, p₀₂ stagnation pressures upstream and downstream of the normal shock (Pa); η recovery (dimensionless); T₀ stagnation temperature (K).

Worked examples

Example 1 (standard). A tunnel is to run at M_T = 2.5 with a first-throat area of 0.04 m². Find the test-section area and the minimum second-throat area for starting. Air, γ = 1.4.

1. A/A* at M = 2.5: (1/2.5) × [0.8333 × (1 + 0.2 × 6.25)]^3 = 0.4 × 1.875^3 = 2.637.
2. A_T = 2.637 × 0.04 = 0.1055 m².
3. Normal shock at M = 2.5: p₀₂/p₀₁ = 0.4990.
4. A_t2 ≥ A_t1/0.4990 = 0.04 × 2.004 = 0.0802 m².

Answer: A_T ≈ 0.105 m², A_t2,min ≈ 0.080 m² (about twice the first throat).

Example 2 (GATE level). A Mach 3 tunnel has a fixed second throat sized exactly for starting. Find the minimum compressor pressure ratio (a) without a supersonic diffuser and (b) with this diffuser running, assuming the shock stands at the second throat and the subsonic diffuser is loss-free.

1. Normal shock at M = 3: p₀₂/p₀₁ = 0.3283. (a) pressure ratio needed = 1/0.3283 = 3.05.
2. Starting throat: A_t2/A_t1 = 3.046.
3. Running: the flow from the test section to the second throat is isentropic with the original A* = A_t1, so at the second throat A/A* = 3.046, giving supersonic M = 2.65.
4. Normal shock at M = 2.65: p₀₂/p₀₁ = 0.4403. (b) pressure ratio = 1/0.4403 = 2.27.

Answer: (a) ≈ 3.05, (b) ≈ 2.27 — the second throat cuts the required pressure ratio by about a quarter.

Common mistakes

• Using A₁V₁ = A₂V₂ (incompressible continuity) in supersonic flow; use ρVA or p₀A* constant.
• Sizing the second throat smaller than the first; it must be larger because the shock lowers p₀.
• Thinking an area increase slows supersonic flow — in supersonic flow, a converging passage decelerates.
• Taking the running shock Mach number as M_T; once started, the shock sits at the second throat at a lower Mach number.
• Forgetting viscous losses: real recoveries are lower than inviscid values.

For GATE AE

Expect: test-section area from the area–Mach relation, second-throat sizing from the normal-shock p₀ ratio, compressor pressure ratio for starting and running, intake total-pressure recovery with oblique shocks plus a normal shock, and conceptual questions on starting, unstart and why an isentropic diffuser is impractical.

Quick check

1. Must A_t2 be larger or smaller than A_t1?
2. For M_T = 2, what is the minimum A_t2/A_t1?
3. Why does the running loss become smaller than the starting loss?
4. What is total-pressure recovery?
5. What is intake unstart?

Answers: 1. larger; 2. 1/0.7209 = 1.387; 3. the shock moves to the second throat where M is lower; 4. the ratio of exit to inlet stagnation pressure; 5. expulsion of the terminal shock ahead of the intake, causing a sudden drop in mass flow and recovery.