Exercises — Shock wave angle, deflection angle

Wave angle ki sabse chhoti value Mach angle $\mu ={\sin }^{-1}(1/M_1)$ ho sakti hai jo upar define ki gayi hai — ek vanishingly weak shock sirf ek Mach wave hai (dekho Mach Angle and Mach Waves). $\mu ={\sin }^{-1}\left(\frac{1}{3}\right)=19.47^{\circ }$ Kyunki $15^{\circ }<19.47^{\circ }$, claimed $\beta$ impossible hai. Kisi bhi real shock ko $\mu \le \beta \le 90^{\circ }$ satisfy karna chahiye, isliye yahan $19.47^{\circ }\le \beta \le 90^{\circ }$.

Master relation mein $\theta =0$ set karne se numerator $M_1^2{\sin }^2\beta -1=0$, yaani $\sin \beta =1/M_1$, ya factor $\cot \beta =0$, yaani $\beta =90^{\circ }$ force hota hai.

• $\beta =\mu ={\sin }^{-1}(1/M_1)$ → ek Mach wave (infinitely weak, dekho Mach Angle and Mach Waves).
• $\beta =90^{\circ }$ → ek normal shock (dekho Normal Shock Waves). Dono endpoints flow ko kuch nahi mod'te; in dono ke beech sab kuch mod'ta hai.

Seedha master relation mein plug karo. Pehle pieces compute karo.

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

Sirf normal component shock strength ke liye matter karta hai: $M_{n1}=M_1\sin \beta =2.5\times \sin 40^{\circ }=2.5\times 0.6428=1.607$ Kyunki $M_{n1}=1.607>1$, perpendicular flow supersonic hai — ek real compression shock banti hai. (Agar $M_{n1}$ $\le 1$ aata toh humne Mach angle ke neeche ek $\beta$ choose kiya hota aur koi shock exist nahi karti.)

Humein $\beta$ chahiye jo satisfy kare $\tan 15^{\circ }=2\cot \beta \frac{4{\sin }^2\beta -1}{4(1.4+\cos 2\beta )+2},\tan 15^{\circ }=\mathrm{0.2679.}$ Mach angle $\mu ={\sin }^{-1}(1/2)=30^{\circ }$ aur $90^{\circ }$ ke beech search karo.

• $\beta =45^{\circ }$ try karo: $4{\sin }^{2}45^{\circ }-1=2-1=1$; $\cos 90^{\circ }=0$ isliye denominator $=4(1.4)+2=7.6$; $\cot 45^{\circ }=1$. RHS $=2(1)(1/7.6)=0.263$. Thoda $0.2679$ se neeche.
• $\beta =45.5^{\circ }$ try karo: RHS $\approx 0.270$ tak chadhta hai. Toh root ${\beta }_{\text{weak}}\approx 45.3^{\circ }$ par hai.
• Maximum ke baad continue karte hue, RHS wapas neeche aata hai aur $0.2679$ ko dobara cross karta hai ${\beta }_{\text{strong}}\approx 79.8^{\circ }$ ke paas — yahi strong shock hai (bada $\beta$, peeche flow subsonic ho jaata hai). Answer: ${\beta }_{\text{weak}}\approx 45.3^{\circ }$, ${\beta }_{\text{strong}}\approx 79.8^{\circ }$. External aerodynamics almost always weak branch select karta hai.

Step 1 — normal upstream Mach: $M_{n1}=2\sin 45.3^{\circ }=2\times 0.7108=1.4215$. Step 2 — page ke top ke paas wale box mein bataye formula se normal-shock $M_{n2}$: $M_{n2}^2=\frac{1+\frac{\gamma -1}{2}{M}_{n1}^2}{\gamma M_{n1}^2-\frac{\gamma -1}{2}}=\frac{1+0.2(2.0207)}{1.4(2.0207)-0.2}=\frac{1.4041}{2.6290}=0.5341$ toh $M_{n2}=0.7308$ (subsonic normal component — kisi bhi shock ke peeche expected). Step 3 — full downstream Mach recover karo $(\beta -\theta )=45.3^{\circ }-15^{\circ }=30.3^{\circ }$ se un-project karke: $M_2=\frac{M_{n2}}{\sin (\beta -\theta )}=\frac{0.7308}{\sin 30.3^{\circ }}=\frac{0.7308}{0.5045}=1.449$ $M_2\approx 1.45>1$: flow abhi bhi supersonic hai ⇒ yeh sach mein weak branch hai. ✓

$\theta (\beta )$ $\beta =\mu =30^{\circ }$ par 0 se rise karta hai, peak karta hai, phir $\beta =90^{\circ }$ par 0 tak fall karta hai (yeh exactly L3 figure mein blue curve hai). Kisi bhi smooth interior peak par derivative momentarily flat hoti hai, isliye maximum woh $\beta$ hai jahan derivative zero ho: $\frac{d\theta }{d\beta }=0.$ Yeh criterion kyun: $\theta (\beta )$ smooth hai aur rise phir fall karta hai, isliye iska single top wahan hai jahan yeh rise karna band kare aur fall shuru kare — wahan tangent horizontal hai, yaani slope zero. Yeh interior maximum ki calculus definition hai. Messy derivative haath se solve kiye bina locate kaise karein: bas $\beta$ values ki march par $\theta (\beta )$ evaluate karo aur dekho kahan turn around hota hai. Hum ek plausible peak ke paas start karte hain aur $1^{\circ }$ increments mein step karte hain; jab value rise phir fall ho, hum top ko bracket karke refine karte hain:

• $\beta =60^{\circ }\Rightarrow \theta =21.6^{\circ }$ (abhi bhi rising)
• $\beta =64^{\circ }\Rightarrow \theta =22.7^{\circ }$
• $\beta =65^{\circ }\Rightarrow \theta =22.77^{\circ }$ (essentially yahan flat — hum top ke paas hain)
• $\beta =66^{\circ }\Rightarrow \theta =22.71^{\circ }$ (ab falling — peak bracketed hai)
• $\beta =70^{\circ }\Rightarrow \theta =21.3^{\circ }$ Bracket $65^{\circ }$–$66^{\circ }$ ko fine grid par refine karne se turn-around $\beta \approx 64.7^{\circ }$ par milta hai jahan ${\theta }_{\max }\approx 22.97^{\circ }$ hai. Top ke paas values $0.01^{\circ }$ se kam change hoti hain, jo confirm karta hai ki march converge ho gayi hai. Answer: ${\theta }_{\max }\approx 23.0^{\circ }$ at $\beta \approx 64.7^{\circ }$. Matlab: half-angle $\approx 23^{\circ }$ tak ka ek wedge $M_1=2$ par ek attached oblique shock hold kar sakta hai. Isse zyaada turning maango aur koi attached solution exist nahi karega.

L4.1 se, ${\theta }_{\max }(M_1=2)\approx 23.0^{\circ }$. Required turn $25^{\circ }>23.0^{\circ }$ hai, isliye koi attached oblique shock flow ko itna nahi mod sakti. Isliye shock detach ho jaati hai: ek curved bow shock wedge nose ke aage khada rehta hai. Centreline ke paas bow shock essentially ek normal shock hai (wahan flow subsonic ho jaata hai), aur yeh sides par door kahin Mach wave tak curve ho jaata hai. Dekho Supersonic Wedge & Cone Flow ki kaise ek cone (jo 3-D mein turning relax karta hai) attached reh sakta hai jahan ek wedge nahi reh sakta.

${\theta }_{\max }$ $M_1$ ke saath badhta hai: faster flow attached rehte hue zyaada sharply turn kar sakti hai. Hum woh $M_1$ chahte hain jiska ceiling exactly $20^{\circ }$ ho. Hum $M_1$ mein march karte hain, aur har $M_1$ par hum L4.1 se wahi peak-finding march $\beta$ par ${\theta }_{\max }(M_1)$ nikalne ke liye run karte hain:

• $M_1=1.8\Rightarrow {\theta }_{\max }\approx 17.7^{\circ }$ (bahut kam)
• $M_1=1.9\Rightarrow {\theta }_{\max }\approx 20.0^{\circ }$ (essentially target)
• $M_1=2.0\Rightarrow {\theta }_{\max }\approx 23.0^{\circ }$ (zyaada) Crossing ${\theta }_{\max }=20^{\circ }$ $M_1\approx 1.90$ par land karti hai. Answer: shock $M_1\gtrsim 1.90$ ke liye attached hai aur $M_1<1.90$ ke liye detach ho jaati hai. Ek slower supersonic flow simply ek single oblique shock ke across $20^{\circ }$ nahi mod sakti.

Analytic: master relation mein poori right side $(M_1^2{\sin }^2\beta -1)$ ke proportional hai. Jab $\theta \to 0$, $\tan \theta \to 0$, isliye humein numerator $\to 0$ chahiye: $M_1^2{\sin }^2\beta -1=0\Rightarrow \sin \beta =\frac{1}{M_1}\Rightarrow \beta ={\sin }^{-1}\frac{1}{M_1}=\mu .$ Yeh exactly Mach angle hai — ek zero-turn "shock" ek infinitesimal wave hai. ($\tan \theta =0$ ka doosra root, yaani $\cot \beta =0\Rightarrow \beta =90^{\circ }$, strong-branch limit hai → ek full-strength normal shock jo flow ko kuch nahi mod'ti lekin use slam karke subsonic kar deti hai.) Numeric at $M_1=2$, $\theta =1^{\circ }$: Mach angle $\mu ={\sin }^{-1}(1/2)=30.00^{\circ }$ hai. Floor ke thoda upar wale small weak root ke liye master relation solve karo:

• $\beta =30.5^{\circ }$ try karo: 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 $=2(1.696)(0.0304/9.539)=0.0108$, deta hai $\theta =\arctan (0.0108)=0.62^{\circ }$ — abhi bhi $1^{\circ }$ se kam.
• $\beta =31.0^{\circ }$ try karo: 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 $=2(1.664)(0.0611/9.478)=0.02146$, deta hai $\theta =\arctan (0.02146)=1.23^{\circ }$ — thoda $1^{\circ }$ se upar. In brackets ke beech interpolate karne par, weak root $\beta \approx 30.9^{\circ }$ par milta hai, Mach angle $\mu =30^{\circ }$ se sirf $0.9^{\circ }$ upar. Conclusion: jaise wedge thin hota jaata hai ($\theta \to 0$), weak wave angle Mach angle par collapse ho jaata hai, exactly waise jaisa analytic argument predict karta hai. ✓

Neeche har line ek question aur uska answer hai, dash se alag — dash ke right side ko cover karo aur khud test karo.

• Attach floor? — $\mu ={\sin }^{-1}(1/M_1)$ sabse chhota possible $\beta$ hai.
• Detach kab? — required $\theta >{\theta }_{\max }(M_1)$ ⇒ curved bow shock banta hai.
• Ek θ ke liye do β? — weak (chhota $\beta$, $M_2>1$) aur strong (bada $\beta$, $M_2<1$).
• $M_2$ recover karo? — $M_2=M_{n2}/\sin (\beta -\theta )$, kabhi seedha $M_1$ nahi.