Compressible Aerodynamics: Isentropic Flow, Shocks, Expansions, Nozzles and Duct Flows
1. Compressibility, the speed of sound and stagnation conditions
Small pressure disturbances travel at the speed of sound a = √(γRT) — 340.3 m/s at sea-level ISA and 295.1 m/s at 216.65 K — and the Mach number M = V/a measures how far the flow is from being able to "warn" the fluid ahead. For steady adiabatic flow the energy equation is c_pT + V²/2 = c_pT0, which gives the stagnation relations: T0/T = 1 + ((γ − 1)/2)M²; for isentropic deceleration p0/p = (T0/T)^(γ/(γ−1)) and ρ0/ρ = (T0/T)^(1/(γ−1)). At M = 1 these give the sonic (starred) ratios T/T0 = 2/(γ + 1) = 0.8333, p/p0 = 0.5283 and ρ*/ρ0 = 0.6339. The stagnation temperature is conserved in any adiabatic flow, shocks included; the stagnation pressure is conserved only if the flow is also reversible.
2. Isentropic flow in variable-area ducts — nozzles and diffusers
Combining continuity, momentum and the speed of sound gives the area-velocity relation dA/A = (M² − 1)dV/V. In subsonic flow (M < 1) a converging duct accelerates the flow; in supersonic flow (M > 1) a diverging duct does; and M = 1 can occur only at a throat, where dA = 0. So a flow is accelerated from rest to supersonic speed only by a converging-diverging (de Laval) nozzle with sonic flow at the throat. The isentropic area ratio is A/A* = (1/M)[(2/(γ + 1))(1 + ((γ − 1)/2)M²)]^((γ+1)/(2(γ−1))), which is 1 at M = 1, larger on either side, and 1.6875 at M = 2. Once the throat is sonic the mass flow is fixed by the reservoir: ṁ = (p0A/√T0)·√(γ/R)·(2/(γ + 1))^((γ+1)/(2(γ−1))) = 0.0404 p0A/√T0 for air in SI units — the nozzle is choked, and lowering the back pressure further cannot draw more flow.
- Back pressure slightly below p0: subsonic everywhere, a venturi; mass flow rises as p_b falls.
- First critical back pressure: the throat just reaches M = 1 and the divergent part decelerates the flow subsonically — the nozzle is now choked.
- Lower still: supersonic flow after the throat ends in a normal shock inside the divergent section, which moves towards the exit as p_b falls.
- Over-expanded: the flow is supersonic to the exit but p_e < p_b, so oblique shocks form outside to raise the pressure.
- Design condition: p_e = p_b, a clean supersonic jet.
- Under-expanded: p_e > p_b, and expansion fans outside the lip complete the expansion.
A diffuser does the reverse job, decelerating the flow and recovering pressure. A subsonic diffuser simply diverges gently enough to avoid separation. An ideal supersonic diffuser would be a converging-diverging passage that slows the flow isentropically to M = 1 at its throat and then subsonically, but it cannot be started: while the flow is being established a normal shock stands ahead of it, and swallowing that shock needs a throat larger than the isentropic one (or overspeeding, or variable geometry). Real supersonic diffusers and intakes therefore use a system of oblique shocks followed by a weak normal shock.
3. Normal shocks, oblique shocks and the Prandtl-Meyer expansion
A normal shock is a thin, irreversible jump from supersonic to subsonic flow. Mass, momentum and energy across it give, for upstream Mach number M1: M2² = (1 + ((γ − 1)/2)M1²)/(γM1² − (γ − 1)/2); p2/p1 = 1 + (2γ/(γ + 1))(M1² − 1); ρ2/ρ1 = V1/V2 = (γ + 1)M1²/(2 + (γ − 1)M1²); and T2/T1 = (p2/p1)/(ρ2/ρ1). The stagnation temperature is unchanged, the entropy rises and the stagnation pressure falls, s2 − s1 = −R ln(p02/p01). The Prandtl relation V1V2 = a*² is a compact check. At M1 = 2: M2 = 0.5774, p2/p1 = 4.5, ρ2/ρ1 = 2.667, T2/T1 = 1.6875 and p02/p01 = 0.7209.
When supersonic flow is turned into itself by a wedge or ramp of angle θ, an oblique shock forms at wave angle β. The tangential velocity is unchanged, and the normal component M1n = M1 sin β obeys the normal-shock relations, M2 = M2n/sin(β − θ). The θ-β-M relation tan θ = 2 cot β (M1² sin²β − 1)/(M1²(γ + cos 2β) + 2) has two solutions for each θ below a maximum θ_max: the weak shock, which is what normally occurs and usually leaves the flow supersonic, and the strong shock, which leaves it subsonic. Beyond θ_max no attached solution exists and a detached bow shock stands ahead of the body. As the deflection tends to zero the oblique shock weakens to a Mach wave at the Mach angle μ = sin⁻¹(1/M).
When supersonic flow turns away from itself round a convex corner it expands through a fan of Mach waves — a centred Prandtl-Meyer expansion. The process is isentropic, so p0 and T0 are constant, the Mach number rises and the pressure falls. The Prandtl-Meyer function ν(M) = √((γ + 1)/(γ − 1)) tan⁻¹√((γ − 1)(M² − 1)/(γ + 1)) − tan⁻¹√(M² − 1) is zero at M = 1, 26.38° at M = 2, and tends to 130.45° as M → ∞; a turn of θ takes the flow from ν(M1) to ν(M2) = ν(M1) + θ. Oblique shocks and expansion fans together give shock-expansion theory for supersonic aerofoils; for a thin flat plate at small α, linear theory gives c_l = 4α/√(M∞² − 1).
| Quantity | Normal or oblique shock | Prandtl-Meyer expansion |
|---|---|---|
| Mach number | Falls | Rises |
| Static pressure, density, temperature | Rise | Fall |
| Stagnation temperature | Constant | Constant |
| Stagnation pressure | Falls | Constant |
| Entropy | Rises | Constant |
4. Critical and drag-divergence Mach numbers
Flow over the upper surface of an aerofoil is faster than the free stream, so the local Mach number reaches 1 somewhere before the free stream does. The critical Mach number M_cr is the free-stream Mach number at which sonic flow first appears on the surface. It is found by intersecting two curves: the minimum pressure coefficient corrected for compressibility, which by the Prandtl-Glauert rule is C_p = C_p0/√(1 − M∞²), and the pressure coefficient at which the local flow is sonic, C_p,cr = (2/(γM∞²))[((2 + (γ − 1)M∞²)/(γ + 1))^(γ/(γ−1)) − 1]. Slightly above M_cr a pocket of supersonic flow terminated by a shock forms; as it strengthens the shock thickens the boundary layer and can separate it, and the drag coefficient rises steeply at the drag-divergence Mach number.
5. Fanno flow and Rayleigh flow
Two constant-area duct flows show what friction and heating do to a compressible stream. Fanno flow is adiabatic flow with wall friction: the stagnation temperature is constant, entropy rises along the duct, and friction always drives the Mach number towards 1 — a subsonic stream accelerates and a supersonic one decelerates. The maximum length that can be sustained before the flow chokes at the exit is given by 4f̄L/D = (1 − M²)/(γM²) + ((γ + 1)/(2γ)) ln[(γ + 1)M²/(2 + (γ − 1)M²)]. If the duct is longer than L, a subsonic flow reduces its mass flow and a supersonic flow forms a shock inside the duct.
Rayleigh flow is frictionless flow with heat addition or removal; p + ρV² is constant. Heating raises the stagnation temperature and drives the Mach number towards 1 from either side, just as friction does: subsonic flow accelerates and its static pressure falls, supersonic flow decelerates and its pressure rises. Cooling does the opposite. The maximum stagnation temperature occurs at M = 1, so there is a limit to the heat a given inlet state can accept — thermal choking. Curiously, in subsonic heating the static temperature peaks at M = 1/√γ ≈ 0.845 and then falls even though heat is still being added, the extra energy going into kinetic energy.
Key takeaways
- a = √(γRT); T0/T = 1 + 0.2M²; p0/p = (T0/T)^3.5; sonic ratios T/T0 = 0.8333, p/p0 = 0.5283. T0 is conserved in any adiabatic flow, p0 only in isentropic flow.
- dA/A = (M² − 1)dV/V: sonic only at a throat; A/A* = 1.6875 at M = 2; choked ṁ = 0.0404 p0A*/√T0; a converging nozzle chokes at p0/p_b = 1.893.
- Normal shock at M1 = 2: M2 = 0.577, p2/p1 = 4.5, ρ2/ρ1 = 2.667, p02/p01 = 0.721; T0 constant, entropy up.
- Oblique shock: use M1 sin β in normal-shock relations; weak solution normally occurs; beyond θ_max the shock detaches. Prandtl-Meyer expansion is isentropic, ν(2) = 26.38°, ν_max = 130.45°.
- M_cr from C_p0/√(1 − M²) meeting C_p,cr; thinness, supercritical sections and sweep (M cos Λ) raise M_cr and M_dd.
- Fanno (friction) and Rayleigh (heating) both drive M towards 1 and choke at M = 1; Fanno keeps T0 constant, Rayleigh changes it.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the speed of sound in air at 216.65 K, in m/s (to one decimal place)? Take γ = 1.4 and R = 287 J/(kg·K).
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Answer: 295
a = √(γRT) = √(1.4 × 287 × 216.65) = √87 052 = 295.04 ≈ 295 m/s. Using the sea-level temperature gives 340 m/s; the speed of sound depends only on temperature for a perfect gas, not on pressure.Air flows at Mach 2 with a static temperature of 250 K. What is its stagnation temperature, in K?
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Answer: 450
T0 = T(1 + ((γ − 1)/2)M²) = 250 × (1 + 0.2 × 4) = 250 × 1.8 = 450 K. Forgetting to square M gives 350 K.What is the isentropic area ratio A/A* for air (γ = 1.4) at Mach 2 (to three decimal places)?
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Answer: 1.688
A/A* = (1/M)[(2/(γ + 1))(1 + 0.2M²)]^3 = (1/2)[(1/1.2) × 1.8]^3 = 0.5 × 1.5³ = 0.5 × 3.375 = 1.6875. The exponent (γ + 1)/(2(γ − 1)) is exactly 3 for γ = 1.4, which is what makes this a clean number; using 3.5 (the pressure exponent) gives about 2.07.A nozzle is supplied with air at a stagnation pressure of 500 kPa and stagnation temperature of 300 K, and its throat of area 10 cm² is choked. Taking γ = 1.4 and R = 287 J/(kg·K), what is the mass flow rate, in kg/s (to two decimal places)?
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Answer: 1.17
ṁ = 0.04042 p0A/√T0 = 0.04042 × 5 × 10⁵ × 10⁻³/√300 = 20.21/17.32 = 1.167 kg/s. Check at the throat: T = 250 K, p* = 264.1 kPa, ρ* = 3.681 kg/m³, a* = 316.9 m/s, so ρaA* = 1.167 kg/s. Using stagnation density and speed instead of sonic values overestimates it.Supersonic flow in a duct is accelerated by:
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Answer: B — A diverging section
dA/A = (M² − 1)dV/V: with M > 1 the bracket is positive, so dV > 0 needs dA > 0 — a diverging section. Friction (Fanno) and heating (Rayleigh) both drive supersonic flow towards M = 1, that is, they decelerate it.A normal shock stands in air (γ = 1.4) with an upstream Mach number of 2. What is the static pressure ratio p2/p1 across it?
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Answer: 4.5
p2/p1 = 1 + (2γ/(γ + 1))(M1² − 1) = 1 + (2.8/2.4) × 3 = 1 + 3.5 = 4.5. The stagnation-pressure ratio across the same shock is 0.721, a loss; confusing the two gives answers below 1.For the same normal shock (M1 = 2, γ = 1.4), what is the downstream Mach number M2 (to three decimal places)?
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Answer: 0.577
M2² = (1 + 0.2M1²)/(1.4M1² − 0.2) = 1.8/5.4 = 0.3333, so M2 = 0.577. Check with ρ2/ρ1 = 2.667 and T2/T1 = 1.6875: M2 = M1(V2/V1)√(T1/T2) = 2 × 0.375 × 0.7698 = 0.577.Across a stationary normal shock in a perfect gas, which of the following hold?
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Answer: A — The stagnation temperature is unchanged; B — The stagnation pressure decreases; C — The entropy increases
The shock is adiabatic, so h0 and hence T0 are constant; it is irreversible, so entropy rises and p0 falls (s2 − s1 = −R ln(p02/p01)). Behind a normal shock the flow is always subsonic; only an oblique shock can leave it supersonic.For a supersonic flow deflected by a wedge whose half-angle exceeds the maximum deflection angle θ_max for the free-stream Mach number:
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Answer: B — A detached curved bow shock forms ahead of the wedge
The θ-β-M relation has no solution for θ > θ_max, so no straight attached shock can turn the flow; a detached bow shock stands off the nose, normal (and strong) on the centre-line and weakening outward. A compressive turn never produces an expansion fan.What is the Mach angle, in degrees, for a flow at Mach 2?
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Answer: 30
μ = sin⁻¹(1/M) = sin⁻¹(0.5) = 30°. Using tan⁻¹ or cos⁻¹ of 1/M gives 26.6° or 60°, and 26.4° is a different quantity altogether, the Prandtl-Meyer angle at Mach 2.A supersonic stream at Mach 2 (Prandtl-Meyer angle 26.38°) expands round a convex corner that turns it by 10°. What is the Prandtl-Meyer angle of the flow after the turn, in degrees (to two decimal places)?
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Answer: 36.38
For an expansion the turning angle adds to the Prandtl-Meyer function: ν2 = ν1 + θ = 26.38 + 10 = 36.38°, which corresponds to M2 ≈ 2.38. Subtracting (16.38°) is the rule for a compressive turn, which in fact produces a shock rather than a Prandtl-Meyer compression here.The minimum pressure coefficient on an aerofoil at low speed is −0.3. Using the Prandtl-Glauert rule, what is it at a free-stream Mach number of 0.6 (to three decimal places, with sign)?
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Answer: -0.375
C_p = C_p0/√(1 − M∞²) = −0.3/√(1 − 0.36) = −0.3/0.8 = −0.375. Compressibility amplifies the pressure coefficient, which is why the suction peak reaches sonic conditions at the critical Mach number; multiplying by 0.8 instead gives −0.24.A wing with 35° leading-edge sweep flies at a free-stream Mach number of 0.85. By simple sweep theory, the Mach number component that governs the onset of compressibility effects on its sections is closest to:
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Answer: B — 0.70
The sections respond to the component normal to the leading edge: M_n = M∞ cos Λ = 0.85 × cos 35° = 0.85 × 0.819 = 0.70. Using sin 35° gives 0.49, and dividing by cos Λ gives 1.04 — sweep reduces the effective Mach number, never increases it.Which of the following statements about constant-area duct flows are correct?
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Answer: A — In Fanno flow the stagnation temperature remains constant; B — Friction accelerates a subsonic flow towards Mach 1; D — In Rayleigh flow the maximum stagnation temperature occurs at Mach 1
Fanno flow is adiabatic, so T0 is constant, and friction drives M towards 1 from either side — a subsonic stream speeds up. Heating in Rayleigh flow also drives M towards 1, so it DECELERATES a supersonic stream; the third statement has the direction wrong. T0 peaks at M = 1, the thermal-choking limit.