Question 2 (Part A, Question A2): Pitot-Static Airspeed at Altitude and the Compressibility Error: 22-Mec-B6 Advanced Fluid Mechanics, May 2014, Q2

Question 2 (Part A, Question A2): Pitot-Static Airspeed at Altitude and the Compressibility Error (14 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Free-stream static pressure \(P = 30\ \text{kPa}\); free-stream static temperature \(T = -55\,{}^{\circ}\text{C} = 218.15\ \text{K}\); measured stagnation pressure \(P_o = 51\ \text{kPa}\); air with \(R = 287\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}\) and \(\gamma = 1.4\).

Find. The true airspeed from the compressible (isentropic) relation, the speed an incompressible Bernoulli reading would give, and the resulting error.

Figure 2.1 — Pitot-static probe: the total port stagnates the flow isentropically to \(P_o\), the static ports read \(P\), and the instrument infers speed from the difference.

Approach. Check first that the pressure ratio is subsonic (no bow shock), then invert the isentropic stagnation-pressure relation for the Mach number, multiply by the local speed of sound, and compare against \(\sqrt{2\Delta P/\rho}\).

Confirm the flow is subsonic, so no shock stands ahead of the probe. The Rayleigh-pitot limit for sonic flight is \(P_o/P = (1+\tfrac{\gamma-1}{2})^{\gamma/(\gamma-1)} = 1.2^{3.5} = 1.893\). Here \[\frac{P_o}{P}=\frac{51}{30}=1.700 < 1.893,\] so the aircraft is subsonic, the stagnation process along the stagnation streamline is isentropic, and the plain isentropic relation applies. (Had the ratio exceeded 1.893 we would have needed the Rayleigh-pitot formula across a normal shock instead.)
Invert the isentropic stagnation relation for Mach number. The aid sheet gives \(P_o/P=\left(1+\tfrac{\gamma-1}{2}M^2\right)^{\gamma/(\gamma-1)}\), so \[\begin{aligned}M&=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_o}{P}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}\\&=\sqrt{\frac{2}{0.4}\left[1.700^{0.2857}-1\right]}=\sqrt{5(1.16371-1)}\end{aligned}\] \[\boxed{M = 0.9047}\] which is high subsonic — a realistic cruise Mach number for a transport aircraft at 11 km.
Compute the local speed of sound. With \(T = 218.15\ \text{K}\), \[a=\sqrt{\gamma R T}=\sqrt{(1.4)(287)(218.15)}=\sqrt{87\,653}=296.06\ \text{m/s}.\] The cold stratosphere is why a 268 m/s airspeed is already Mach 0.90; at sea level the same speed would be only Mach 0.79.
Assemble the true airspeed. \[\boxed{V = M\,a = (0.9047)(296.06)=267.9\ \text{m/s}\;(964\ \text{km/h})}\]
Repeat with the incompressible Bernoulli equation. Bernoulli in the form \(P_o = P+\tfrac12\rho V^2\) needs the free-stream density, \[\rho=\frac{P}{RT}=\frac{30\,000}{(287)(218.15)}=0.4792\ \text{kg/m}^3,\] whence \[V_{\text{Bern}}=\sqrt{\frac{2(P_o-P)}{\rho}}=\sqrt{\frac{2(21\,000)}{0.4792}}=296.1\ \text{m/s}.\]
Quantify the error. The incompressible reading overstates the speed by \[\Delta V = 296.1-267.9=28.2\ \text{m/s},\qquad \boxed{\frac{\Delta V}{V}=+10.5\ \text{per cent}}\] Bernoulli over-predicts because it ignores the density rise in the stagnation process: the real fluid packs itself denser at the probe mouth, so a given speed produces a larger pressure rise than the constant-density model expects, and reading that pressure rise with the constant-density formula returns too high a speed.

Quantity	Symbol	Value
Stagnation pressure ratio	\(P_o/P\)	1.700 (subsonic, no shock)
Flight Mach number	\(M\)	0.9047
Local speed of sound	\(a\)	296.06 m/s
True airspeed	\(V\)	267.9 m/s (964 km/h)
Free-stream density	\(\rho\)	0.4792 kg/m³
Bernoulli (incompressible) speed	\(V_{\text{Bern}}\)	296.1 m/s
Absolute error	\(\Delta V\)	+28.2 m/s
Relative error	\(\Delta V/V\)	+10.5 per cent

Topic. Compressible pitot-static airspeed measurement and the magnitude of the incompressible-Bernoulli error at high subsonic Mach number.

Key relations. Isentropic stagnation: \(P_o/P=(1+\tfrac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)}\) and \(T_o/T=1+\tfrac{\gamma-1}{2}M^2\); speed of sound \(a=\sqrt{\gamma RT}\); ideal gas \(\rho=P/RT\); incompressible Bernoulli \(V=\sqrt{2(P_o-P)/\rho}\); sonic pitot limit \(P_o/P=1.893\) for \(\gamma=1.4\).

Why this works. A pitot tube brings the flow to rest at its mouth. If that deceleration is adiabatic and reversible, entropy is constant along the stagnation streamline and the isentropic relation ties \(P_o/P\) uniquely to \(M\) — no velocity or density measurement is needed. Expanding that relation in \(M^2\) gives \(P_o - P = \tfrac12\rho V^2\left(1+\tfrac{M^2}{4}+\tfrac{2-\gamma}{24}M^4+\ldots\right)\), so Bernoulli is the leading term and the compressibility correction grows as \(M^2/4\). At \(M=0.9047\) that factor is about 1.22 in dynamic pressure, or roughly 10 per cent in speed — precisely the error found. Below about \(M=0.3\) the same expansion gives under 1 per cent, which is why the incompressible formula is safe for low-speed work.

Common pitfall. The commonest error is using standard sea-level density (1.225 kg/m³) instead of the local 0.479 kg/m³ in the Bernoulli check, which produces a "speed" of 185 m/s and an error of the wrong sign; the comparison must use the same free-stream state as the compressible calculation. The second is failing to test \(P_o/P\) against 1.893 first — above that value a bow shock stands ahead of the probe, the probe reads the post-shock stagnation pressure, and the isentropic formula silently returns a wrong Mach number. A third is forgetting that −55 °C must be converted to 218.15 K before it enters either \(a\) or \(\rho\).

Source. F. M. White, Fluid Mechanics, 8th ed., Ch. 9.3–9.4 (isentropic flow, stagnation properties, pitot measurement); J. D. Anderson, Modern Compressible Flow, 3rd ed., Ch. 3.4 and 3.6 (measurement of airspeed, Rayleigh pitot formula).

Question 2 of 7: Question 2 (Part A, Question A2): Pitot-Static Airspeed at Altitude and the Compressibility Error

Question 2 (Part A, Question A2): Pitot-Static Airspeed at Altitude and the Compressibility Error (14 marks)