Speed of sound, Mach number and compressibility

Why it matters

Every result in compressible aerodynamics — shocks, nozzles, critical Mach number, wave drag — is written in terms of the Mach number, and the Mach number needs the local speed of sound. Knowing when density changes can be ignored (and when they cannot) decides whether you may use Bernoulli's equation or must switch to the compressible relations.

Key ideas

Speed of sound. A weak pressure disturbance (a sound wave) travels through a fluid at speed a. Applying mass and momentum conservation across a weak wave, and noting that a weak wave is so thin and fast that it is both adiabatic and reversible (isentropic), gives a² = (∂p/∂ρ)ₛ. For a calorically perfect gas, p/ρ^γ = constant along an isentrope, so a = √(γ·R·T).

• a depends only on temperature for a given gas — not on pressure or density separately.
• For air (γ = 1.4, R = 287 J/(kg·K)) at 288.15 K, a ≈ 340.3 m/s; at 216.65 K (the ISA tropopause) a ≈ 295.0 m/s.
• Light gases have high a: helium at 300 K has a ≈ 1019 m/s.
• Newton's original estimate used the isothermal value √(R·T) and came out about 15% low, because sound waves are isentropic, not isothermal.

Mach number. M = V/a, the ratio of the local flow speed to the local speed of sound. It is a local property: in a nozzle both V and a change along the duct. Physically, M² is proportional to the ratio of directed kinetic energy to random thermal energy of the molecules, which is why it governs compressibility.

Flow regimes (approximate boundaries):

• Incompressible: M < 0.3 — density changes below about 5%.
• Subsonic: 0.3 < M < 0.8 — compressible but no shocks; Prandtl–Glauert style corrections work.
• Transonic: 0.8 < M < 1.2 — mixed subsonic and supersonic regions, shocks on the body.
• Supersonic: 1.2 < M < 5 — wave drag, oblique shocks and expansion fans.
• Hypersonic: M > 5 — thin shock layers, high temperature effects (a gradual boundary, not a sharp one).

Compressibility. The compressibility of a fluid is τ = −(1/v)(∂v/∂p), the fractional volume change per unit pressure change. What matters in a flow is whether the pressure changes produced by the motion (of order ½ρV²) cause noticeable density changes. Combining the isentropic relations shows the fractional density change from stagnation is roughly Δρ/ρ ≈ M²/2 for small M. At M = 0.3 this is about 4.4%; at M = 0.5 it is about 11%. That is why M ≈ 0.3 is the usual limit for treating air as incompressible.

Mach waves. A small disturbance moving at M > 1 cannot send signals ahead of itself. The disturbances pile up on a cone of half-angle μ = sin⁻¹(1/M), the Mach angle. Outside the cone (the "zone of silence") the flow does not know the body is coming. At M < 1 the disturbances spread in all directions and the flow adjusts ahead of the body.

Formulas

a² = (∂p/∂ρ)ₛ — general definition, any fluid.

a = √(γ·R·T) — calorically perfect gas.

• a: speed of sound (m/s); γ = cp/cv (dimensionless, 1.4 for air); R: specific gas constant (J/(kg·K), 287 for air); T: static temperature (K, absolute).
• Equivalent form: a = √(γ·Rᵤ·T / Mₘ) with Rᵤ = 8314 J/(kmol·K) and molar mass Mₘ (kg/kmol).

M = V/a — V local flow speed (m/s), dimensionless result.

μ = sin⁻¹(1/M) — Mach angle, valid only for M ≥ 1.

ρ₀/ρ = (1 + (γ−1)/2 · M²)^(1/(γ−1)) — isentropic density ratio, used to judge compressibility; small-M limit Δρ/ρ ≈ M²/2.

Worked examples

Example 1 (standard). An aircraft flies at 250 m/s at sea level where T = 288.15 K. Find the speed of sound and the Mach number. Air: γ = 1.4, R = 287 J/(kg·K).

1. a = √(γ·R·T) = √(1.4 × 287 × 288.15) = √115 779 = 340.3 m/s.
2. M = V/a = 250 / 340.3 = 0.735.

Answer: a = 340.3 m/s, M = 0.735 (compressibility not negligible since M > 0.3).

Example 2 (GATE level). Up to what flight Mach number does the stagnation-point density differ from the free-stream density by no more than 5%? Then find the corresponding speed at 11 km altitude (T = 216.65 K). Air, γ = 1.4.

1. The condition is ρ/ρ₀ = 0.95, so ρ₀/ρ = 1/0.95 = 1.05263.
2. From ρ₀/ρ = (1 + 0.2·M²)^2.5: 1 + 0.2·M² = 1.05263^0.4 = 1.02073.
3. M² = 0.02073 / 0.2 = 0.1036, so M = 0.322.
4. a = √(1.4 × 287 × 216.65) = 295.0 m/s.
5. V = M·a = 0.322 × 295.0 = 95.0 m/s.

Answer: M ≈ 0.32, V ≈ 95 m/s — the familiar "M < 0.3" incompressible limit. The small-M estimate M²/2 = 0.05 gives M = 0.316, close to the exact value.

Common mistakes

• Using °C instead of K in a = √(γ·R·T). At 15 °C you must use 288.15 K.
• Mixing the universal and specific gas constants: either R = 287 J/(kg·K) alone, or Rᵤ/Mₘ — never 287 divided by molar mass again.
• Using the stagnation temperature instead of the static temperature for a. The Mach number uses the static (local) a.
• Assuming a changes with pressure at constant temperature. For an ideal gas it does not.
• Treating M = 0.3 as a sharp physical boundary; it is an engineering choice for about 5% density change.
• Applying μ = sin⁻¹(1/M) at M < 1 — there is no Mach cone in subsonic flow.

For GATE AE

Expect quick numericals: speed of sound at a given altitude temperature, Mach number from speed, the Mach angle from M (or M from a measured Mach cone angle), and statements about flow regimes. Practise converting °C to K, remembering ISA sea-level and tropopause temperatures, and using the isentropic density ratio to judge whether compressibility matters. Concept questions test why sound propagation is isentropic and why a depends on T only.

Quick check

1. What is the speed of sound in air at 250 K (γ = 1.4, R = 287 J/(kg·K))?
2. Does the speed of sound in an ideal gas change if pressure doubles at constant temperature?
3. What is the Mach angle at M = 2?
4. Roughly what percentage density change occurs at M = 0.3?
5. Why did Newton's isothermal estimate of the speed of sound come out low?

Answers: 1. 316.9 m/s; 2. No, it depends only on T; 3. 30°; 4. about 4.4%; 5. sound waves are isentropic, so γ (not 1) appears under the square root.