Detached bow shock: Exercises — Physics deep dive

Throughout: $γ = 1.4$ for air, and every symbol keeps the meaning it had in the parent note —

$M_{1}$ = upstream (incoming) Mach number = flow speed ÷ speed of sound,
$θ$ = the turning angle the body demands of the flow,
$θ_{ma x} (M_{1})$ = the biggest turn an attached oblique shock can give at that $M_{1}$ ,
$β$ = the shock-wave angle (angle between the shock and the incoming flow),
$M_{n 1} = M_{1} sin β$ = the part of the Mach number pointing straight into the shock.

Level 1 — Recognition

L1.1

State the one-line rule that decides whether a shock stays attached to a sharp body or detaches into a bow shock.

Recall Solution

Compare the demanded turn to the ceiling: $θ > θ_{ma x} (M_{1}) \Rightarrow detached (bow shock), θ \leq θ_{ma x} (M_{1}) \Rightarrow attached.$ The number $θ_{ma x}$ is the tallest peak of the $θ$ -vs- $β$ curve at that $M_{1}$ . Ask for more turning than that peak and the θ–β–M equation simply has no real $β$ — the shock cannot stay glued on. See L1.2 for the picture.

L1.2

On the centreline (the stagnation streamline that runs straight into the nose), what is $β$ , and is the flow just behind that point subsonic or supersonic?

Recall Solution

On the centreline the shock stands perpendicular to the incoming flow, so $β = 9 0^{\circ}$ and $M_{n 1} = M_{1} sin 9 0^{\circ} = M_{1}$ . A shock with $β = 9 0^{\circ}$ is a normal shock, the strongest kind — it always drops the flow to subsonic ( $M_{2} < 1$ ). That subsonic pocket is exactly what lets the air peel gently around a blunt nose.

Level 2 — Application

L2.1

Air at $M_{1} = 2.0$ meets a wedge of half-angle $θ = 3 0^{\circ}$ . Given $θ_{ma x} (2.0) \approx 22.9 7^{\circ}$ , does the shock attach or detach?

Recall Solution

Compare: $3 0^{\circ} > 22.9 7^{\circ}$ . The demand exceeds the ceiling, so the shock detaches — a bow shock forms. Mechanically, the θ–β–M relation $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2}$ has no real $β$ that produces $θ = 3 0^{\circ}$ at $M_{1} = 2$ .

L2.2

Keep the same wedge ( $θ = 3 0^{\circ}$ ) but speed up to $M_{1} = 4.0$ , where $θ_{ma x} (4.0) \approx 38. 8^{\circ}$ . Attach or detach?

Recall Solution

Now $3 0^{\circ} < 38. 8^{\circ}$ : the demand is below the ceiling, so the shock stays attached and oblique. Faster flow raises $θ_{ma x}$ , giving the flow more "turning capacity." The moral: detachment depends on both $θ$ and $M_{1}$ — speeding up can re-attach a shock that was detached.

L2.3

Centreline of a bow shock at $M_{1} = 3$ . Using the normal-shock relation $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}},$ find $M_{2}$ behind the shock.

Recall Solution

On the centreline $β = 9 0^{\circ}$ , so $M_{n 1} = M_{1} = 3$ . Then $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{1 + 1.8}{12.6 - 0.2} = \frac{2.8}{12.4} = 0.2258,$ so $M_{2} = 0.2258 \approx 0.475$ . Subsonic, exactly as a normal shock demands.

Level 3 — Analysis

L3.1

Use the sphere stand-off correlation $\frac{Δ}{R _{n}} \approx 0.143 exp (\frac{3.24}{M _{1}^{2}})$ to compute $Δ/ R_{n}$ at $M_{1} = 2$ and $M_{1} = 4$ , and explain the trend physically.

Recall Solution

At $M_{1} = 2$ : exponent $= 3.24/4 = 0.81$ , so $Δ/ R_{n} = 0.143 e^{0.81} = 0.143 (2.248) \approx 0.322$ . At $M_{1} = 4$ : exponent $= 3.24/16 = 0.2025$ , so $Δ/ R_{n} = 0.143 e^{0.2025} = 0.143 (1.224) \approx 0.175$ . The gap shrinks as $M_{1}$ rises. Why: higher $M_{1}$ → stronger compression → denser, thinner shock layer → the shock sits closer to the nose. (The parent's quoted $\approx 0.41$ at $M_{1} = 2$ uses a slightly different rounding; the trend — smaller $Δ$ at higher Mach — is the point.)

L3.2

As $M_{1} \to \infty$ , what limiting value does $Δ/ R_{n}$ approach, and what does that tell you about hypersonic shock layers?

Recall Solution

As $M_{1} \to \infty$ , the exponent $3.24/ M_{1}^{2} \to 0$ , so $exp (\cdot) \to 1$ and $\frac{Δ}{R _{n}} \to 0.143.$ The stand-off never reaches zero — it settles at a small finite floor. Physically, at hypersonic speeds the shock hugs the nose in a thin, dense shock layer, but a nonzero gap always remains because the subsonic pocket at the stagnation point must exist to divert the flow.

Level 4 — Synthesis

L4.1

At $M_{1} = 2.0$ a body demands $θ = 3 0^{\circ}$ (detaches). You need it attached. From L2 you know $θ_{ma x} (4.0) \approx 38. 8^{\circ}$ . Between $M_{1} = 2$ and $M_{1} = 4$ , $θ_{ma x} (M_{1})$ is monotonically increasing. Estimate the minimum Mach number at which $θ_{ma x}$ first reaches $3 0^{\circ}$ , given the data points $θ_{ma x} (2.5) \approx 29.8 0^{\circ}$ and $θ_{ma x} (2.6) \approx 31.0 3^{\circ}$ .

Recall Solution

We want the $M_{1}$ where $θ_{ma x} = 3 0^{\circ}$ . It lies between the two given points. Linear interpolation on $θ_{ma x}$ vs $M_{1}$ : $M_{1}^{⋆} \approx 2.5 + (2.6 - 2.5) \cdot \frac{30 - 29.80}{31.03 - 29.80} = 2.5 + 0.1 \cdot \frac{0.20}{1.23} \approx 2.5 + 0.0163 \approx 2.516.$ So at roughly $M_{1} \approx 2.52$ the shock re-attaches. Below that it stays detached; above it, attached. This ties together the detachment criterion with the fact that $θ_{ma x}$ rises with $M_{1}$ .

L4.2

Take $M_{1} = 3$ and follow the curved bow shock outward from the centreline. At the centreline $β = 9 0^{\circ}$ (normal shock, $M_{2} \approx 0.475$ ). Far out, $β$ approaches the Mach angle $μ = arcsin (1/ M_{1})$ . Compute $μ$ , and describe what happens to $M_{2}$ between $β = 9 0^{\circ}$ and $β = μ$ .

Recall Solution

Mach angle: $μ = arcsin (1/3) = arcsin (0.3333) \approx 19.4 7^{\circ}$ . Sweeping from $β = 9 0^{\circ}$ down to $β = μ \approx 19.4 7^{\circ}$ , $M_{n 1} = M_{1} sin β$ falls from $3$ toward $M_{1} sin μ = 3 \cdot (1/3) = 1$ . The shock weakens continuously:

near the centreline $M_{2} < 1$ (subsonic),
at the sonic line $M_{2} = 1$ ,
beyond it $M_{2} > 1$ (supersonic again),
in the limit $β \to μ$ , $M_{n 1} \to 1$ and the "shock" fades into a zero-strength Mach wave. One curved shock thus contains an entire family of oblique strengths, from strongest (normal) to vanishing.

Level 5 — Mastery

L5.1

Inverse problem. Behind the centreline normal shock of a bow shock you measure $M_{2} = 0.50$ . Recover the upstream $M_{1}$ (i.e. $M_{n 1}$ , since $β = 9 0^{\circ}$ ). Use $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}} .$

Recall Solution

Set $M_{2}^{2} = 0.25$ and solve for $x = M_{n 1}^{2}$ . With $γ = 1.4$ : $0.25 = \frac{1 + 0.2 x}{1.4 x - 0.2} .$ Cross-multiply: $0.25 (1.4 x - 0.2) = 1 + 0.2 x \Rightarrow 0.35 x - 0.05 = 1 + 0.2 x$ . $0.15 x = 1.05 \Rightarrow x = 7.0 \Rightarrow M_{n 1} = 7 \approx 2.646.$ Since $β = 9 0^{\circ}$ , $M_{1} = M_{n 1} \approx 2.65$ . So a centreline $M_{2} = 0.50$ implies an incoming Mach of about $2.65$ .

L5.2

Stagnation check. For the flow of L5.1 ( $M_{1} \approx 2.646$ , $M_{2} \approx 0.50$ ), find the stagnation-pressure ratio $p_{02} / p_{01}$ across the centreline normal shock. Use $\frac{p _{02}}{p _{01}} = [\frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2}]^{\frac{γ}{γ - 1}} [\frac{γ + 1}{2 γ M _{n 1}^{2} - ( γ - 1 )}]^{\frac{1}{γ - 1}} .$

Recall Solution

Use $M_{n 1}^{2} = 7$ , $γ = 1.4$ , so $γ / (γ - 1) = 3.5$ and $1/ (γ - 1) = 2.5$ . First bracket: $\frac{2.4 \cdot 7}{0.4 \cdot 7 + 2} = \frac{16.8}{4.8} = 3.5$ . Second bracket: $\frac{2.4}{2.8 \cdot 7 - 0.4} = \frac{2.4}{19.6 - 0.4} = \frac{2.4}{19.2} = 0.125$ . $\frac{p _{02}}{p _{01}} = 3. 5^{3.5} \cdot 0.12 5^{2.5} .$ $3. 5^{3.5} \approx 80.21$ and $0.12 5^{2.5} \approx 5.524 \times 1 0^{- 3}$ , so $\frac{p _{02}}{p _{01}} \approx 80.21 \times 0.005524 \approx 0.443.$ About 44% of the stagnation pressure survives — the rest is lost to the strong shock. Bow shocks are expensive: the stronger the centreline shock, the larger the total-pressure penalty.

Flashcards

Attachment condition (one line)?

Shock attaches iff

θ \leq θ_{ma x} (M_{1})

; otherwise it detaches into a bow shock.

Centreline shock angle and flow regime?

β = 9 0^{\circ}

(normal shock), flow behind is subsonic.

Stand-off floor as

M_{1} \to \infty

(sphere correlation)?

Δ/ R_{n} \to 0.143

What is conserved across a shock —

T_{0}

p_{0}

T_{0}

(stagnation temperature);

p_{0}

drops because a shock is irreversible.

Mach angle at

M_{1} = 3

μ = arcsin (1/3) \approx 19.4 7^{\circ}