Exercises — Shock wave angle, deflection angle

The smallest a wave angle can be is the Mach angle $\mu ={\sin }^{-1}(1/M_1)$ defined above — a vanishingly weak shock is just a Mach wave (see Mach Angle and Mach Waves). $\mu ={\sin }^{-1}\left(\frac{1}{3}\right)=19.47^{\circ }$ Since $15^{\circ }<19.47^{\circ }$, the claimed $\beta$ is impossible. Any real shock must satisfy $\mu \le \beta \le 90^{\circ }$, so here $19.47^{\circ }\le \beta \le 90^{\circ }$.

Setting $\theta =0$ in the master relation forces the numerator $M_1^2{\sin }^2\beta -1=0$, i.e. $\sin \beta =1/M_1$, or the factor $\cot \beta =0$, i.e. $\beta =90^{\circ }$.

• $\beta =\mu ={\sin }^{-1}(1/M_1)$ → a Mach wave (infinitely weak, see Mach Angle and Mach Waves).
• $\beta =90^{\circ }$ → a normal shock (see Normal Shock Waves). Both endpoints turn the flow by nothing; everything in between turns it.

Plug straight into the master relation. Compute pieces first.

• $\sin 40^{\circ }=0.6428\Rightarrow {\sin }^{2}40^{\circ }=0.4132$, so $M_1^2{\sin }^2\beta =6.25\times 0.4132=2.582$.
• Numerator: $M_1^2{\sin }^2\beta -1=2.582-1=1.582$.
• $\cos 2\beta =\cos 80^{\circ }=0.1736$, so $M_1^2(\gamma +\cos 2\beta )+2=6.25(1.4+0.1736)+2=6.25(1.5736)+2=11.835$.
• $\cot 40^{\circ }=1/\tan 40^{\circ }=1/0.8391=1.1918$. $\tan \theta =2(1.1918)\frac{1.582}{11.835}=2.3836\times 0.13367=0.3186$ $\theta =\arctan (0.3186)=17.67^{\circ }$ So the streamline bends by about $\theta \approx 17.7^{\circ }$.

Only the normal component matters for shock strength: $M_{n1}=M_1\sin \beta =2.5\times \sin 40^{\circ }=2.5\times 0.6428=1.607$ Since $M_{n1}=1.607>1$, the perpendicular flow is supersonic — a real compression shock forms. (If $M_{n1}$ came out $\le 1$ we would have picked a $\beta$ below the Mach angle and no shock would exist.)

We need $\beta$ solving $\tan 15^{\circ }=2\cot \beta \frac{4{\sin }^2\beta -1}{4(1.4+\cos 2\beta )+2},\tan 15^{\circ }=\mathrm{0.2679.}$ Search between the Mach angle $\mu ={\sin }^{-1}(1/2)=30^{\circ }$ and $90^{\circ }$.

• Try $\beta =45^{\circ }$: $4{\sin }^{2}45^{\circ }-1=2-1=1$; $\cos 90^{\circ }=0$ so denominator $=4(1.4)+2=7.6$; $\cot 45^{\circ }=1$. RHS $=2(1)(1/7.6)=0.263$. Slightly below $0.2679$.
• Try $\beta =45.5^{\circ }$: RHS climbs to $\approx 0.270$. So the root sits at ${\beta }_{\text{weak}}\approx 45.3^{\circ }$.
• Continuing past the maximum, the RHS comes back down and crosses $0.2679$ again near ${\beta }_{\text{strong}}\approx 79.8^{\circ }$ — that is the strong shock (larger $\beta$, flow behind goes subsonic). Answer: ${\beta }_{\text{weak}}\approx 45.3^{\circ }$, ${\beta }_{\text{strong}}\approx 79.8^{\circ }$. External aerodynamics almost always selects the weak branch.

Step 1 — normal upstream Mach: $M_{n1}=2\sin 45.3^{\circ }=2\times 0.7108=1.4215$. Step 2 — normal-shock $M_{n2}$ using the formula stated in the box near the top of the page: $M_{n2}^2=\frac{1+\frac{\gamma -1}{2}{M}_{n1}^2}{\gamma M_{n1}^2-\frac{\gamma -1}{2}}=\frac{1+0.2(2.0207)}{1.4(2.0207)-0.2}=\frac{1.4041}{2.6290}=0.5341$ so $M_{n2}=0.7308$ (subsonic normal component — expected behind any shock). Step 3 — recover the full downstream Mach by un-projecting through $(\beta -\theta )=45.3^{\circ }-15^{\circ }=30.3^{\circ }$: $M_2=\frac{M_{n2}}{\sin (\beta -\theta )}=\frac{0.7308}{\sin 30.3^{\circ }}=\frac{0.7308}{0.5045}=1.449$ $M_2\approx 1.45>1$: flow is still supersonic ⇒ this really is the weak branch. ✓

$\theta (\beta )$ rises from 0 at $\beta =\mu =30^{\circ }$, peaks, then falls to 0 at $\beta =90^{\circ }$ (this is exactly the blue curve in the L3 figure). At any smooth interior peak of a curve the slope is momentarily flat, so the maximum is the $\beta$ where the derivative vanishes: $\frac{d\theta }{d\beta }=0.$ Why this criterion: $\theta (\beta )$ is smooth and rises then falls, so its single top is where it stops rising and starts falling — the tangent there is horizontal, i.e. slope zero. That is the calculus definition of an interior maximum. How to locate it without solving the messy derivative by hand: just evaluate $\theta (\beta )$ on a march of $\beta$ values and watch where it turns around. We start near a plausible peak and step in $1^{\circ }$ increments; when the value rises then falls, we bracket the top and refine:

• $\beta =60^{\circ }\Rightarrow \theta =21.6^{\circ }$ (still rising)
• $\beta =64^{\circ }\Rightarrow \theta =22.7^{\circ }$
• $\beta =65^{\circ }\Rightarrow \theta =22.77^{\circ }$ (essentially flat here — we are near the top)
• $\beta =66^{\circ }\Rightarrow \theta =22.71^{\circ }$ (now falling — peak is bracketed)
• $\beta =70^{\circ }\Rightarrow \theta =21.3^{\circ }$ Refining the bracket $65^{\circ }$–$66^{\circ }$ on a fine grid gives the turn-around at $\beta \approx 64.7^{\circ }$ with ${\theta }_{\max }\approx 22.97^{\circ }$. The values change by less than $0.01^{\circ }$ near the top, confirming the march has converged. Answer: ${\theta }_{\max }\approx 23.0^{\circ }$ at $\beta \approx 64.7^{\circ }$. Meaning: a wedge of half-angle up to $\approx 23^{\circ }$ can hold an attached oblique shock at $M_1=2$. Ask for more turning and no attached solution exists.

From L4.1, ${\theta }_{\max }(M_1=2)\approx 23.0^{\circ }$. The required turn $25^{\circ }>23.0^{\circ }$, so no attached oblique shock can turn the flow that much. The shock therefore detaches: a curved bow shock stands off ahead of the wedge nose. Near the centreline the bow shock is essentially a normal shock (flow goes subsonic there), and it curves back to a weak Mach wave far off to the sides. See Supersonic Wedge & Cone Flow for how a cone (which relaxes the turning in 3-D) can stay attached where a wedge cannot.

${\theta }_{\max }$ grows with $M_1$: faster flow can turn more sharply while staying attached. We want the $M_1$ whose ceiling exactly equals $20^{\circ }$. We march in $M_1$, and at each $M_1$ we run the same peak-finding march over $\beta$ from L4.1 to get ${\theta }_{\max }(M_1)$:

• $M_1=1.8\Rightarrow {\theta }_{\max }\approx 17.7^{\circ }$ (too low)
• $M_1=1.9\Rightarrow {\theta }_{\max }\approx 20.0^{\circ }$ (essentially the target)
• $M_1=2.0\Rightarrow {\theta }_{\max }\approx 23.0^{\circ }$ (over) The crossing ${\theta }_{\max }=20^{\circ }$ lands at $M_1\approx 1.90$. Answer: the shock is attached for $M_1\gtrsim 1.90$ and detaches for $M_1<1.90$. A slower supersonic flow simply cannot bend $20^{\circ }$ across a single oblique shock.

Analytic: in the master relation the whole right side is proportional to $(M_1^2{\sin }^2\beta -1)$. As $\theta \to 0$, $\tan \theta \to 0$, so we need the numerator $\to 0$: $M_1^2{\sin }^2\beta -1=0\Rightarrow \sin \beta =\frac{1}{M_1}\Rightarrow \beta ={\sin }^{-1}\frac{1}{M_1}=\mu .$ That is exactly the Mach angle — a zero-turn "shock" is a infinitesimal wave. (The other root of $\tan \theta =0$, namely $\cot \beta =0\Rightarrow \beta =90^{\circ }$, is the strong-branch limit → a normal shock.) Numeric at $M_1=2$, $\theta =1^{\circ }$: the Mach angle is $\mu ={\sin }^{-1}(1/2)=30.00^{\circ }$. Solving the master relation for the small weak root just above the floor:

• Try $\beta =30.5^{\circ }$: numerator $4{\sin }^{2}30.5^{\circ }-1=4(0.5075{)}^2-1=0.0304$; denominator $4(1.4+\cos 61^{\circ })+2=4(1.4+0.4848)+2=9.539$; $\cot 30.5^{\circ }=1.696$. RHS $=2(1.696)(0.0304/9.539)=0.0108$, giving $\theta =\arctan (0.0108)=0.62^{\circ }$ — still short of $1^{\circ }$.
• Try $\beta =31.0^{\circ }$: numerator $4{\sin }^{2}31^{\circ }-1=4(0.5150{)}^2-1=0.0611$; denominator $4(1.4+\cos 62^{\circ })+2=4(1.4+0.4695)+2=9.478$; $\cot 31^{\circ }=1.664$. RHS $=2(1.664)(0.0611/9.478)=0.02146$, giving $\theta =\arctan (0.02146)=1.23^{\circ }$ — just over $1^{\circ }$. Interpolating between these brackets, the weak root sits at $\beta \approx 30.9^{\circ }$, only $0.9^{\circ }$ above the Mach angle $\mu =30^{\circ }$. Conclusion: as the wedge thins ($\theta \to 0$) the weak wave angle collapses onto the Mach angle, exactly as the analytic argument predicts. ✓

Each line below is a question and its answer, separated by a dash — cover the right of the dash and test yourself.

• Attach floor? — $\mu ={\sin }^{-1}(1/M_1)$ is the smallest possible $\beta$.
• Detach when? — required $\theta >{\theta }_{\max }(M_1)$ ⇒ curved bow shock forms.
• Two β for one θ? — weak (small $\beta$, $M_2>1$) and strong (large $\beta$, $M_2<1$).
• Recover $M_2$? — $M_2=M_{n2}/\sin (\beta -\theta )$, never $M_1$ directly.