Flow through converging-diverging nozzles

Why it matters

Rocket and jet nozzles, supersonic wind tunnels and steam turbines all rely on converging–diverging (C-D) nozzles. What the nozzle actually does — stay subsonic, carry a shock inside, or deliver a clean supersonic jet — depends on the back pressure, and getting this wrong costs thrust or wrecks a wind-tunnel test.

Key ideas

Set-up. A reservoir at p₀, T₀ feeds a C-D nozzle with throat area A_t and exit area A_e, discharging into a region at back pressure p_b. Flow is isentropic except across any shock. The area ratio A_e/A_t has two isentropic exit solutions: subsonic M_sub and supersonic M_sup.

Operating regimes as p_b is lowered from p₀:

1. p_b = p₀: no flow.
2. p₀ > p_b > p₁ (first critical): subsonic everywhere, a venturi. The throat is below M = 1; mass flow rises as p_b falls.
3. p_b = p₁: throat just sonic, subsonic diverging part (exit at M_sub). The nozzle is choked: mass flow is now fixed at its maximum, ṁ = 0.0404·p₀·A_t/√T₀ for air.
4. p₁ > p_b > p₂: supersonic flow after the throat ends in a normal shock inside the diverging part; the subsonic flow behind it decelerates to p_e = p_b. Lowering p_b moves the shock downstream.
5. p_b = p₂: the normal shock stands exactly at the exit plane.
6. p₂ > p_b > p₃: over-expanded — the flow inside is fully supersonic and isentropic with exit pressure p₃ < p_b; oblique shocks outside the exit raise the pressure to p_b.
7. p_b = p₃: design condition — exit pressure equals back pressure, shock-free jet.
8. p_b < p₃: under-expanded — exit pressure higher than p_b; expansion fans at the lip complete the expansion outside.

From regime 3 onward the flow upstream of the throat (and the mass flow) no longer changes with p_b: disturbances cannot travel upstream through the sonic throat.

Shock inside the nozzle. Across the shock T₀ is unchanged but p₀ drops, so the downstream sonic reference area grows: A₂* = A_t·(p₀₁/p₀₂). Behind the shock the flow is subsonic and follows the subsonic branch of A/A₂*.

Locating the shock (standard trick). Mass flow is the same at the throat and the exit: p₀₁·A_t = p₀ₑ·A₂* gives pₑ·Aₑ/(p₀₁·A_t) = (pₑ/p₀ₑ)·(Aₑ/A₂*), a function of Mₑ only. With pₑ = p_b known, solve for Mₑ, get p₀ₑ/p₀₁, then use the normal-shock table to find M₁ and the shock area.

Converging-only nozzle. It can reach M = 1 at most, at its exit. Below the critical back pressure p* = 0.528 p₀ (air) the exit stays at p* and the jet expands outside.

Formulas

ṁ = (p₀·A_t/√T₀)·√(γ/R)·(2/(γ+1))^((γ+1)/(2(γ−1))) — choked mass flow (kg/s); for air ṁ = 0.0404·p₀·A_t/√T₀ with p₀ in Pa, A in m², T₀ in K.

A/A* = (1/M)·[(2/(γ+1))(1 + (γ−1)/2·M²)]^((γ+1)/(2(γ−1))) — area–Mach relation.

p/p₀ = (1 + (γ−1)/2·M²)^(−γ/(γ−1)) — isentropic, using the local p₀.

A₂*/A₁* = p₀₁/p₀₂ — across a shock (constant T₀).

pₑ·Aₑ/(p₀·A_t) = (1/Mₑ)·(2/(γ+1))^((γ+1)/(2(γ−1)))·(1 + (γ−1)/2·Mₑ²)^(−1/2) — exit relation used to find Mₑ when a shock is inside.

Thrust of an ideal nozzle: F = ṁ·Vₑ + (pₑ − p_a)·Aₑ.

Symbols: p₀, T₀ reservoir stagnation pressure (Pa) and temperature (K); A_t throat area, Aₑ exit area (m²); p_b back pressure, pₑ exit pressure, p_a ambient (Pa); Mₑ exit Mach number; Vₑ exit velocity (m/s); F thrust (N); γ, R as usual.

Worked examples

Example 1 (standard). A C-D nozzle with Aₑ/A_t = 2.0 is fed with air at p₀ = 1.0 MPa, T₀ = 500 K, A_t = 0.01 m². Find the mass flow when choked and the three critical back pressures.

1. ṁ = 0.0404 × 1.0 × 10⁶ × 0.01 / √500 = 404/22.36 = 18.1 kg/s.
2. A/A* = 2.0 gives M_sub = 0.306 and M_sup = 2.197.
3. First critical (choked, subsonic exit): p₁ = p₀/(1 + 0.2 × 0.306²)^3.5 = 1000/1.067 = 937 kPa.
4. Design (fully supersonic): p₃ = 1000/(1 + 0.2 × 2.197²)^3.5 = 1000/10.646 = 93.9 kPa.
5. Shock at exit: p₂ = p₃ × (p₂/p₁)shock at M = 2.197: 1 + 1.1667 × (4.827 − 1) = 5.465, so p₂ = 93.9 × 5.465 = 513 kPa.

Answer: ṁ ≈ 18.1 kg/s; p₁ ≈ 937 kPa, p₂ ≈ 513 kPa, p₃ ≈ 93.9 kPa.

Example 2 (GATE level). The same nozzle discharges into p_b = 700 kPa. Locate the shock.

1. Since 937 > 700 > 513 kPa, a normal shock is inside the diverging section.
2. pₑAₑ/(p₀A_t) = 0.7 × 2.0 = 1.40.
3. Solve (1/Mₑ)·0.5787·(1 + 0.2Mₑ²)^(−0.5) = 1.40 (for air, (2/2.4)^3 = 0.5787): Mₑ = 0.407.
4. pₑ/p₀ₑ = (1 + 0.2 × 0.1654)^(−3.5) = 0.8923, so p₀ₑ = 700/0.8923 = 784.5 kPa, i.e. p₀₂/p₀₁ = 0.7845.
5. Normal-shock table: p₀₂/p₀₁ = 0.7845 at M₁ = 1.863.
6. A/A_t at M = 1.863 = 1.510, so the shock sits where the area is 1.51 × A_t (0.0151 m²).

Answer: shock at A ≈ 1.51 A_t, upstream Mach ≈ 1.86, exit Mach ≈ 0.41.

Common mistakes

• Swapping over- and under-expanded: over-expanded means exit pressure below ambient.
• Thinking a lower back pressure always raises mass flow — not once the throat is choked.
• Using the supersonic root of A/A* behind a shock; behind it the flow is subsonic and A* has grown.
• Keeping the upstream p₀ downstream of the shock.
• Forgetting that the normal shock can stand only in the diverging part, never in the converging part.
• Using p* = 0.528 p₀ as the exit pressure of a C-D nozzle; that is the throat pressure.

For GATE AE

Expect: choked mass flow, design exit Mach number and pressure from the area ratio, the three critical back pressures, identifying the regime for a given p_b, shock location problems, and nozzle thrust including the pressure term. Practise reading isentropic and normal-shock tables quickly and checking which branch applies.

Quick check

1. For air, what is p_t/p₀ at a choked throat?
2. Exit pressure is lower than ambient. Over- or under-expanded?
3. Does mass flow change when p_b drops from 600 kPa to 300 kPa in Example 1?
4. Where can a normal shock stand in a C-D nozzle?
5. Across a shock in the nozzle, does A* increase or decrease?

Answers: 1. 0.528; 2. over-expanded; 3. no, the nozzle is already choked; 4. only in the diverging section (or at the exit); 5. it increases, by the factor p₀₁/p₀₂.