Question 3 (Part A, Question A3): Pitot-Static Tube in a Supersonic Stream: 22-Mec-B6 Advanced Fluid Mechanics, May 2013, Q3

Question 3 (Part A, Question A3): Pitot-Static Tube in a Supersonic Stream (20 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.

Quantity	Symbol	Value
Stagnation pressure at the nose	p₀₂	130 kPa
Static pressure near the tube	p₂	110 kPa
Static temperature near the tube	T₂	519 K
Case (a) gas	air	R = 287 J/(kg·K), γ = 1.4
Case (b) gas	helium	R = 2077 J/(kg·K), γ = 5/3

Find. The free-stream speed V₁ ahead of the probe for each gas.

Figure 3.1 — A pitot-static tube in supersonic flow. A detached bow shock stands ahead of the blunt nose; the nose port therefore reads the stagnation pressure behind the shock, and the static ports on the probe body read the local post-shock static conditions.

Check: the static pressure and temperature are taken as the local values behind the bow shock, as the phrase ‘in the proximity of the tube’ indicates and as the physical arrangement of a pitot-static probe requires. This reading is forced by the data: a normal shock always raises the ratio p₀₂/p₁ to at least 1.893 for air, whereas the measured ratio is 130/110 = 1.182, so if the 110 kPa were a free-stream value the flow could not be supersonic at all (it would be a subsonic M = 0.494). Treating both readings as post-shock removes the contradiction and yields a consistent supersonic solution.

Approach. Work backwards through the bow shock: use the isentropic relation between the nose reading and the local static pressure to get the post-shock Mach number, invert the normal-shock relation to recover the free-stream Mach number, use the shock temperature ratio to get the free-stream temperature, and finally multiply by the local speed of sound.

The two cases differ only in the gas constants, so the same four steps are carried out twice.

Part (a) — post-shock Mach number from the two pressure readings. Between the static ports and the nose the flow is decelerated isentropically, so the aid-sheet isentropic relation applies with the local Mach number: $$\begin{aligned} \frac{p_{02}}{p_2} &= \left(1 + \frac{\gamma-1}{2}M_2^{2}\right)^{\gamma/(\gamma-1)} \\ M_2 &= \sqrt{\frac{2}{\gamma-1}\left[\left(\frac{p_{02}}{p_2}\right)^{(\gamma-1)/\gamma} - 1\right]} \end{aligned}$$ With $p_{02}/p_2 = 130/110 = 1.1818$ and $\gamma = 1.4$ this gives $M_2 = 0.4944$. The value is subsonic, as it must be behind a normal shock, and it comfortably exceeds the limiting post-shock Mach number $\sqrt{(\gamma-1)/2\gamma} = 0.3780$ for air, so a genuine supersonic upstream state exists.
Part (a) — invert the normal-shock relation for the free-stream Mach number. The aid sheet gives the downstream Mach number in terms of the upstream one; solving that expression for $M_1$ yields $$\begin{aligned} M_2^{2} &= \frac{1 + \frac{\gamma-1}{2}M_1^{2}}{\gamma M_1^{2} - \frac{\gamma-1}{2}} \\ M_1^{2} &= \frac{1 + \frac{\gamma-1}{2}M_2^{2}}{\gamma M_2^{2} - \frac{\gamma-1}{2}} \end{aligned}$$ which is the same functional form, reflecting the fact that the normal-shock Mach relation is its own inverse. Substituting $M_2 = 0.4944$ gives $$M_1 = \boxed{2.716}$$ so the stream ahead of the probe is indeed supersonic.
Part (a) — free-stream static temperature. The shock heats the gas, and the temperature ratio across it is $$\frac{T_2}{T_1} = \frac{\left[2\gamma M_1^{2} - (\gamma-1)\right] \left[2 + (\gamma-1)M_1^{2}\right]}{(\gamma+1)^{2}M_1^{2}} = 2.3598$$ Since the ports read $T_2 = 519$ K, the free-stream temperature is $T_1 = 519 / 2.3598 = 219.94$ K. For completeness the static-pressure jump is $p_2/p_1 = 8.438$, giving a free-stream static pressure of only $13.036$ kPa — a thin, cold, fast stream, as one would expect at this Mach number.
Part (a) — convert to a speed. The free-stream speed of sound is $a_1 = \sqrt{\gamma R T_1}$, so $$a_1 = \sqrt{(1.4)(287)(219.94)} = 297.3\ \text{m/s}, \quad V_1 = M_1 a_1 = \boxed{807\ \text{m/s}}$$ in air.
Part (b) — post-shock Mach number from the two pressure readings. Between the static ports and the nose the flow is decelerated isentropically, so the aid-sheet isentropic relation applies with the local Mach number: $$\begin{aligned} \frac{p_{02}}{p_2} &= \left(1 + \frac{\gamma-1}{2}M_2^{2}\right)^{\gamma/(\gamma-1)} \\ M_2 &= \sqrt{\frac{2}{\gamma-1}\left[\left(\frac{p_{02}}{p_2}\right)^{(\gamma-1)/\gamma} - 1\right]} \end{aligned}$$ With $p_{02}/p_2 = 130/110 = 1.1818$ and $\gamma = 5/3$ this gives $M_2 = 0.4553$. The value is subsonic, as it must be behind a normal shock, and it comfortably exceeds the limiting post-shock Mach number $\sqrt{(\gamma-1)/2\gamma} = 0.4472$ for helium, so a genuine supersonic upstream state exists.
Part (b) — invert the normal-shock relation for the free-stream Mach number. The aid sheet gives the downstream Mach number in terms of the upstream one; solving that expression for $M_1$ yields $$\begin{aligned} M_2^{2} &= \frac{1 + \frac{\gamma-1}{2}M_1^{2}}{\gamma M_1^{2} - \frac{\gamma-1}{2}} \\ M_1^{2} &= \frac{1 + \frac{\gamma-1}{2}M_2^{2}}{\gamma M_2^{2} - \frac{\gamma-1}{2}} \end{aligned}$$ which is the same functional form, reflecting the fact that the normal-shock Mach relation is its own inverse. Substituting $M_2 = 0.4553$ gives $$M_1 = \boxed{9.365}$$ so the stream ahead of the probe is indeed supersonic.
Part (b) — free-stream static temperature. The shock heats the gas, and the temperature ratio across it is $$\frac{T_2}{T_1} = \frac{\left[2\gamma M_1^{2} - (\gamma-1)\right] \left[2 + (\gamma-1)M_1^{2}\right]}{(\gamma+1)^{2}M_1^{2}} = 28.2790$$ Since the ports read $T_2 = 519$ K, the free-stream temperature is $T_1 = 519 / 28.2790 = 18.35$ K. For completeness the static-pressure jump is $p_2/p_1 = 109.375$, giving a free-stream static pressure of only $1.006$ kPa — a thin, cold, fast stream, as one would expect at this Mach number.
Part (b) — convert to a speed. The free-stream speed of sound is $a_1 = \sqrt{\gamma R T_1}$, so $$a_1 = \sqrt{(5/3)(2077)(18.35)} = 252.1\ \text{m/s}, \quad V_1 = M_1 a_1 = \boxed{2,360\ \text{m/s}}$$ in helium.

Check: the helium answer sits very close to the limiting post-shock Mach number (M₂ = 0.4553 against a floor of 0.4472), which makes the inversion extremely stiff: a one per cent change in the measured pressure ratio moves M₁ by roughly ten per cent. The helium result should therefore be reported as an order-of-magnitude hypersonic value rather than a three-figure measurement. Air, by contrast, is far from the floor and the inversion is well conditioned.

Final Results.

Quantity	Symbol	(a) Air	(b) Helium
Local (post-shock) Mach number	M₂	0.4944	0.4553
Free-stream Mach number	M₁	2.716	9.365
Free-stream static temperature	T₁	219.9 K	18.35 K
Free-stream static pressure	p₁	13.04 kPa	1.006 kPa
Free-stream speed of sound	a₁	297.3 m/s	252.1 m/s
Free-stream speed	V₁	807 m/s	2,360 m/s

Concept & Theory

Topic. Supersonic pitot-static measurement: the detached bow shock and the Rayleigh pitot formula.

Key relations. Isentropic $p_0/p = (1 + \tfrac{\gamma-1}{2}M^{2})^{\gamma/(\gamma-1)}$; normal-shock Mach relation $M_2^{2} = \left(1 + \tfrac{\gamma-1}{2}M_1^{2}\right) / \left(\gamma M_1^{2} - \tfrac{\gamma-1}{2}\right)$; shock temperature ratio $T_2/T_1 = [2\gamma M_1^{2} - (\gamma-1)][2 + (\gamma-1)M_1^{2}] / [(\gamma+1)^{2}M_1^{2}]$; $a = \sqrt{\gamma R T}$.

Why this works. In supersonic flow a blunt probe cannot send pressure information upstream, so a curved detached shock forms ahead of it. Along the stagnation streamline that shock is locally normal, and the gas that reaches the nose port has therefore passed through a normal shock and then been compressed isentropically to rest. The nose reading is consequently the stagnation pressure behind the shock, which is lower than the true free-stream stagnation pressure by the shock total-pressure loss. Combining the isentropic relation downstream with the normal-shock relation across the wave gives the classical Rayleigh pitot formula, and because the shock Mach relation is an involution the inversion for the upstream Mach number takes exactly the same algebraic form as the forward relation.

Common pitfall. The dominant error is to apply the incompressible or the plain isentropic relation directly between the nose reading and the free stream, which ignores the shock entirely and gives a badly low Mach number. A second is to treat the static reading as a free-stream value; the static taps sit on the probe body, behind the shock, and using them as upstream data here would make the problem insoluble because the measured ratio lies below the M = 1 floor of 1.893 for air. A third is to forget that the post-shock Mach number cannot fall below √((γ−1)/2γ), and to accept an inverted result blindly when the data sit close to that floor — as they do for helium here.

Source. Anderson, Modern Compressible Flow, 3rd ed., §3.6 (the Rayleigh pitot tube) and §3.4; White, Fluid Mechanics, 8th ed., §9.5.

Question 3 of 6: Question 3 (Part A, Question A3): Pitot-Static Tube in a Supersonic Stream

Question 3 (Part A, Question A3): Pitot-Static Tube in a Supersonic Stream (20 marks)