Two independent modeling questions
Incompressibility asks whether density variation is negligible; the continuum assumption asks whether molecular travel distances are small against the relevant length. Low Ma does not establish continuum behavior, and small Kn does not establish constant density.
The acoustic and stagnation relations below assume a calorically perfect ideal gas with constant γ (NASA Glenn gives γ = 1.4 for air at standard day conditions) and adiabatic, isentropic deceleration. They do not cover shocks, strong friction, heat input or rarefaction.
State variables and length scales
| Variable | SI unit | Meaning |
|---|---|---|
| γ | Dimensionless | Specific-heat ratio cₚ/cᵥ |
| R | J/(kg·K) | Specific gas constant |
| T, T₀ | K | Static and isentropic stagnation temperature |
| v, c | m/s | Local speed and local sound speed |
| ρ, ρ₀ | kg/m³ | Static and stagnation density |
| p, p₀ | Pa | Static and stagnation pressure |
| λ, L | m | Mean free path and relevant length |
Use kelvins in c = √(γRT) and a mass-specific R. Take λ at the relevant pressure and temperature; a channel width or gradient length may matter more than device length.
Deriving the stagnation-state comparison
For this ideal gas c² = γRT and Ma = v/c. Reversible adiabatic deceleration conserves energy: cₚT₀ = cₚT + v²/2. Inserting cₚ = γR/(γ−1) gives the first ratio below; p/ρᵞ = constant with the ideal-gas equation gives the other two.
ρ₀/ρ = (T₀/T)^(1/(γ−1))
p₀/p = (T₀/T)^(γ/(γ−1))
(p₀ − p)/(½ρv²) = (p₀/p − 1)/(γMa²/2)
The last line uses ½ρv² = γpMa²/2. It shows how far the isentropic pressure rise exceeds the Bernoulli value ½ρv², often the more practical error for a pitot reading or pressure coefficient. These ratios compare a local state with ideal deceleration, not inlet-to-outlet changes in a real duct.
Stagnation ratios at fixed temperature
Example 1. Ma = 0.3 and 0.8 at T = 300 K
Given: γ = 1.4, R = 287 J/(kg·K) and an assumed static temperature T = 300 K, so c = √120540 = 347.1887 m/s. The speeds are chosen to give Ma = 0.3 and 0.8; they are not atmospheric data.
| Ma / v | T₀/T | ρ₀/ρ | p₀/p |
|---|---|---|---|
| 0.3 / 104.1566 m/s | 1.018 | 1.04561 | 1.06443 |
| 0.8 / 277.7510 m/s | 1.128 | 1.35137 | 1.52434 |
(p₀ − p)/(½ρv²) = 0.06443/(0.7 × 0.09) = 1.0227 at Ma = 0.3
(p₀ − p)/(½ρv²) = 0.52434/(0.7 × 0.64) = 1.1704 at Ma = 0.8
At Ma = 0.3 the density ratio is about 4.56% above unity, yet the isentropic pressure rise exceeds ½ρv² by only about 2.3%. The answer to "is Ma = 0.3 incompressible?" therefore depends on whether density or pressure accuracy matters for your result.
Using standard-atmosphere data
Example 2. Ma = 0.8 at 11 000 m
Given: the standard atmosphere at 11 000 m from Table 1.1 of the Cornell MAE 5070 notes: T = 216.65 K, p = 22631.70 Pa, ρ = 0.363912 kg/m³, a = 295.07 m/s. The flight Mach number 0.8 is assumed.
Check: √(1.4 × 287 × 216.65) = 295.04 m/s; back-calculating R from the table row gives 287.05 J/(kg·K)
v = 0.8 × 295.07 = 236.06 m/s
T₀ = 216.65 × 1.128 = 244.38 K
p₀ = 22631.70 × 1.52434 = 34498 Pa; ρ₀ = 0.363912 × 1.35137 = 0.4918 kg/m³
½ρv² = 10139 Pa; isentropic p₀ − p = 11867 Pa
An incompressible pitot reduction would read 11867 Pa as dynamic pressure and overestimate v² by about 17%; the ratio 11867/10139 is the same 1.170 as Example 1 because it depends only on Ma and γ. The 0.01% gap between 295.04 and 295.07 m/s comes from rounding R to 287; it matters only when comparing sound speeds to five digits.
Check molecular scales after changing device size or pressure
Hold λ = 6.8×10⁻⁸ m fixed (NIST's 67.3 nm, rounded). At L = 0.001 m, Kn = 6.8×10⁻⁵; at L = 10⁻⁶ m, Kn = 0.068. Ma is unchanged at fixed speed, so size affects a separate assumption. At Kn = 0.068, assess wall slip and temperature jump before reusing a no-slip model.
| Regime | Kn range | Usual modeling consequence |
|---|---|---|
| Continuum | Kn < 0.01 | No-slip Navier–Stokes and Fourier conduction |
| Slip | 0.01–0.1 | Continuum equations with slip and temperature-jump walls |
| Transition | 0.1–10 | Kinetic methods such as DSMC |
| Free molecule | Kn > 10 | Molecule-surface collisions dominate |
The Utah/NASA survey states these ranges are approximate only, and the Mars Global Surveyor study uses 0.1–10 for transition and Kn > 10 for free molecular flow.
Example 3. A 10 µm channel as the pressure drops
Given: λ = 67.3 nm in air at 23 °C and 101.3 kPa (NIST). Channel width L = 10 µm and the pressures 10 kPa and 1 kPa are assumed. At fixed temperature, kinetic theory gives λ ∝ 1/p, so λ is scaled by 101.3/p.
101.3 kPa: Kn = 67.3 nm / 10 µm = 0.00673 (continuum)
10 kPa: λ = 0.6817 µm, Kn = 0.0682 (slip)
1 kPa: λ = 6.817 µm, Kn = 0.682 (transition)
Ma is unchanged by these pressure steps when T and v stay fixed, because c = √(γRT) does not depend on pressure. A low-speed vacuum line can therefore pass every compressibility check and still need a slip or kinetic model.
What to record when selecting a model
- State local temperature and the acoustic assumptions behind Ma.
- Choose an acceptable density or pressure error for an incompressible approximation.
- Explain which gas state and physical length λ and L represent.
- Do not apply isentropic ratios across a shock.
- Strong rarefaction or thermal nonequilibrium requires more than this ideal stagnation comparison.
Questions about Mach, Knudsen and model choice
Is Ma < 0.3 enough to treat a gas flow as incompressible?
Only if the error is acceptable: at Ma = 0.3 stagnation density is 4.56% above static density and the pressure rise is 2.3% above ½ρv².
Why does √(γRT) give 295.04 m/s at 11 000 m when the table lists 295.07?
The table row corresponds to R ≈ 287.05 J/(kg·K); using 287 lowers c by about 0.01%. Use the same R as your solver when comparing digits.
Does lowering the pressure change Ma or Kn?
At fixed temperature and speed, Ma stays the same because c depends on T, while λ grows as 1/p. In Example 3 a drop from 101.3 kPa to 1 kPa moves Kn from 0.00673 to 0.682.
Why do aerosol papers report a different Kn for the same particle?
The NIST slip-correction paper defines Kn = 2λ/Dp, using the particle radius rather than its diameter as the length. Check the definition before applying a regime boundary.
Can I use isentropic relations across a shock?
No. NASA Glenn notes that abrupt compression creates shocks, the flow becomes irreversible and the oblique or normal shock relations apply instead.
References
- NASA Glenn: Isentropic Flow Equations. Stagnation ratios; not valid across shocks.
- NASA Glenn: Specific Heats. γ = 1.4 for air at standard day conditions.
- Cornell University (MAE 5070): Introduction to Flight Dynamics. Table 1.1, standard atmosphere at 11 000 m.
- NIST: Slip Correction Measurements of Certified PSL Nanoparticles. λ = 67.3 nm at 23 °C and 101.3 kPa; Kn = 2λ/Dp.
- University of Utah for NASA: Survey of Literature on Convective Heat Transfer Coefficients and Recovery Factors for High Atmosphere Thermometry. Kn regime ranges, stated as approximate.
- NASA: Aerothermodynamics of the Mars Global Surveyor Spacecraft. Transition 0.1–10, free molecular Kn > 10.
- NASA (TM 107730): Reynolds Number Influences in Aeronautics. Knudsen-number and slip discussion, pp. 106–107.
The 300 K speeds, cruise Mach number, channel width and reduced pressures are our assumptions; printed results are recomputed by the site tests. Checked 2026-10-22.