Worked examples — Shock wave angle, deflection angle

3.1.14 · D3 · Physics › Compressible Flow & Aerodynamics › Shock wave angle, deflection angle

Yeh page parent topic ki "practically use karo" companion hai. Hum yahan $\theta$–$\beta$–$M$ relation dobara derive nahi karenge — hum ise use karenge, har tarah ke input par. Shuru karne se pehle, ek reminder un do tools ka jo hum baar baar kaam mein lenge:

In saari formulas ki origin Normal Shock Waves hai jo perpendicular component par apply hoti hai, saath mein Rankine-Hugoniot Relations jo dono sides ko connect karti hain.

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The scenario matrix

Is topic ke har problem ka solution in cells mein se kisi ek mein aata hai. Neeche ke examples mein kaun si cell cover ho rahi hai clearly bataya gaya hai taaki tum dekh sako ki poora space filled hai.

#	Case class	Kya khaas baat hai	Example
A	Forward problem ($\beta$ given → $\theta$ nikalo)	Seedha plug in karo, koi root-finding nahi	Ex 1
B	Inverse, weak root ($\theta$ given → chhota $\beta$)	Usual external-aero case	Ex 2
C	Inverse, strong root ($\theta$ given → bada $\beta$)	Same $\theta$, subsonic exit	Ex 3
D	Degenerate: $\theta =0$	Do limits — Mach wave aur normal shock	Ex 4
E	Limiting: $\beta =90^{\circ }$	Exactly normal shock ban jaata hai	Ex 5
F	Over-turning: $\theta >{\theta }_{\max }$	Koi attached solution nahi → detached bow shock	Ex 6
G	Real-world word problem	Physical picture se angles nikalo	Ex 7
H	Exam twist: ${\theta }_{\max }$ khud nikalo	Jahan weak aur strong roots merge hote hain	Ex 8

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Example 1 — Cell A: forward problem ($\beta$ given)

Forecast: $\beta =40^{\circ }$ high Mach par ek kaafi shallow shock hai — ek achha turn expect karo, kahin $20$–$25^{\circ }$ range mein. Aage padhne se pehle guess karo.

1. Numerator piece $M_1^2{\sin }^2\beta -1$ compute karo. $M_1^2{\sin }^2\beta =9{\sin }^{2}40^{\circ }=9(0.6428{)}^2=3.719$, toh numerator hai $3.719-1=2.719$. Yeh step kyun? Yeh factor "shock strength" hai — yeh $M_{n1}^2-1$ hai, jo positive tabhi hoga jab normal Mach $1$ se zyada ho. Positive value confirm karta hai ki yeh genuine shock hai, Mach wave nahi.

2. Denominator $M_1^2(\gamma +\cos 2\beta )+2$ compute karo. $\cos 2\beta =\cos 80^{\circ }=0.1736$, toh $9(1.4+0.1736)+2=9(1.5736)+2=16.16$. Yeh step kyun? Yeh sirf compact formula ki algebra hai; koi physical interpretation nahi, lekin divide karne se pehle yeh banana zaroori hai.

3. $\tan \theta$ assemble karo. $\cot 40^{\circ }=1/\tan 40^{\circ }=1.1918$. $\tan \theta =2(1.1918)\frac{2.719}{16.16}=0.4011\Rightarrow \theta =\arctan (0.4011)=21.9^{\circ }.$ Yeh step kyun? Yeh relation seedha $\theta$ dene ke liye likhi gayi hai jab $M_1,\beta$ fixed hon — yahi "forward" direction ka poora point hai.

Verify: $\theta =21.9^{\circ }<\beta =40^{\circ }$ ✓ (turn hamesha chhota angle hota hai). Aur $M_{n1}=3\sin 40^{\circ }=1.93>1$ ✓ — ek real compression shock.

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Example 2 — Cell B: inverse problem, weak root

Forecast: $\beta$ Mach angle $\mu =\arcsin (1/2)=30^{\circ }$ aur $90^{\circ }$ ke beech hona chahiye. Weak root low end ke paas hoga. Guess ~$45^{\circ }$.

1. $\beta$ bracket karo. Mach angle $\mu =30^{\circ }$ floor hai; weak root pehla $\beta$ hai (jaise $\beta$ $30^{\circ }$ se upar jaata hai) jo $\tan 15^{\circ }=0.2679$ satisfy kare. Yeh step kyun? $\theta$–$\beta$ curve pehle utha phir girta hai, toh do $\beta$ same $\theta$ dete hain. Chhota wala weak branch hai — ${\theta }_{\max }$ ki location se neeche wala root lo.

2. Solve karo $0.2679=2\cot \beta \frac{4{\sin }^2\beta -1}{4(1.4+\cos 2\beta )+2}$. Numerically yeh $\beta =45.3^{\circ }$ (weak) deta hai. Yeh step kyun? $\beta$ ke liye koi closed form exist nahi karta, isliye hum root-find karte hain. Yahi branch real wedges par supersonic flight mein hoti hai.

3. Exit Mach number check karo. $M_{n1}=2\sin 45.3^{\circ }=1.421$. Normal-shock formula mein daalo: $M_{n2}^2=\frac{1.421^2+5}{7(1.421^2)-1}=\frac{7.019}{13.14}=0.534\Rightarrow M_{n2}=\mathrm{0.731.}$ Phir $M_2=\frac{0.731}{\sin (45.3^{\circ }-15^{\circ })}=\frac{0.731}{\sin 30.3^{\circ }}=\mathrm{1.45.}$ Yeh step kyun? $M_{n2}<1$ hamesha hota hai (koi bhi shock normal flow ko subsonic banata hai), lekin total $M_2=1.45>1$ — yahi weak branch ki pehchaan hai.

Verify: $M_{n1}=1.421>1$ ✓, $M_{n2}=0.731<1$ ✓, $M_2=1.45>1$ ✓ — teeno weak oblique shock ke saath consistent hain. Parent note se match karta hai.

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Example 3 — Cell C: inverse problem, strong root

Forecast: Strong root $90^{\circ }$ ke paas hoga (normal shock ke close). Guess ~$80^{\circ }$.

1. Usi equation ka doosra root lo $\tan 15^{\circ }=0.2679$. Curve ke peak se upar root-finding karne par $\beta =79.8^{\circ }$ (strong) milta hai. Yeh step kyun? Same $\theta$, same $M_1$, lekin shock ab nearly perpendicular hai — bahut zyada strong compression.

2. Upstream normal Mach. $M_{n1}=2\sin 79.8^{\circ }=1.969$. Yeh step kyun? Steeper shock ka matlab bada normal component, isliye stronger shock.

3. Exit Mach. $M_{n2}^2=\frac{1.969^2+5}{7(1.969^2)-1}=\frac{8.877}{26.14}=0.3396\Rightarrow M_{n2}=\mathrm{0.583.}$ $M_2=\frac{0.583}{\sin (79.8^{\circ }-15^{\circ })}=\frac{0.583}{\sin 64.8^{\circ }}=\mathrm{0.644.}$ Yeh step kyun? Yahan total exit Mach $1$ se neeche gir jaata hai — strong branch flow ko subsonic kar deta hai.

Verify: $M_2=0.644<1$ ✓ subsonic exit — strong branch ki defining feature. Same $\theta =15^{\circ }$ Ex 2 jaisa lekin bilkul alag flow state, exactly jaisa do-root picture predict karti hai.

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Example 4 — Cell D: degenerate input $\theta =0$

Forecast: Zero turning do extreme shock types par collapse ho jaani chahiye. Guess: Mach angle aur $90^{\circ }$.

1. Formula ka numerator zero set karo. $\tan \theta =0$ force karta hai ya toh $\cot \beta =0$ ya $M_1^2{\sin }^2\beta -1=0$. Yeh step kyun? Product tab zero hota hai jab koi factor zero ho — isse do limits saaf alag ho jaati hain.

2. Factor 1: $M_1^2{\sin }^2\beta =1$. $\sin \beta =1/M_1=1/2.5=0.4\Rightarrow \beta =\mu =23.6^{\circ }$ — Mach angle. Yeh step kyun? Yeh vanishingly weak "shock" hai — ek Mach wave jo kuch bhi nahi modta.

3. Factor 2: $\cot \beta =0\Rightarrow \beta =90^{\circ }$. Perpendicular shock — normal shock — jo zero turning bhi produce karta hai. Yeh step kyun? $\beta =90^{\circ }$ par poori velocity normal hai; redirect karne ke liye koi tangential component nahi, toh $\theta =0$.

Verify: Dono limits $\theta =0$ dete hain: $\mu =\arcsin (0.4)=23.6^{\circ }$ ✓ aur $\beta =90^{\circ }$ ✓. Yahi exactly parent note mein bataye gaye $\beta$ ke lower aur upper bounds hain.

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Example 5 — Cell E: limiting case $\beta =90^{\circ }$

Forecast: Shock perpendicular hone par $M_{n1}$ ko $M_1$ ke barabar hona chahiye, aur $M_2$ standard normal-shock answer (~$0.475$ for $M=3$) hona chahiye.

1. Deflection zero hai. Ex 4 ke Factor 2 se, $\beta =90^{\circ }\Rightarrow \theta =0$. Yeh step kyun? Confirm karta hai ki hum "normal shock" endpoint par hain — koi turning nahi.

2. Normal Mach = full Mach. $M_{n1}=3\sin 90^{\circ }=3=M_1$. Yeh step kyun? Jab shock flow ke seedha samne hoti hai, poori velocity normal hoti hai — projection kuch nahi karta.

3. Exit Mach. $M_{n2}^2=\frac{9+5}{7(9)-1}=\frac{14}{62}=0.2258\Rightarrow M_{n2}=\mathrm{0.4752.}$ $M_2=\frac{M_{n2}}{\sin (90^{\circ }-0^{\circ })}=\frac{0.4752}{1}=\mathrm{0.4752.}$ Yeh step kyun? $\beta -\theta =90^{\circ }$ hone par divisor $\sin (90^{\circ })=1$ ho jaata hai, toh $M_2=M_{n2}$ — oblique formula yahan normal-shock formula hi hai.

Verify: $M_2=0.475$ $M_1=3$ ke liye textbook normal-shock exit Mach hai ✓. Oblique framework mein normal shock $\beta =90^{\circ }$ special case ke roop mein contained hai.

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Example 6 — Cell F: over-turning ($\theta >{\theta }_{\max }$)

Forecast: Ex 8 se hum paayenge ki $M_1=2$ ke liye ${\theta }_{\max }\approx 23^{\circ }$ hai. Kyunki $25^{\circ }>23^{\circ }$, expect karo ki koi real $\beta$ nahi milega.

1. Maximum se compare karo. $M_1=2$ ke liye, ${\theta }_{\max }=22.97^{\circ }$ (Ex 8 mein compute kiya gaya). Demand $25^{\circ }$ isse exceed karta hai. Yeh step kyun? $\theta$–$\beta$ curve is Mach par kabhi $25^{\circ }$ reach nahi karta, toh equation $\tan 25^{\circ }=(\dots )$ ka $\beta$ mein koi real root nahi hai.

2. Physically interpret karo. Wedge flow ke liye itna blunt hai ki ek single straight attached shock se deal nahi ho sakta. Yeh step kyun? Flow is speed par ek step mein itna sharply turn nahi kar sakta, toh use koi aur arrangement dhundhi padegi.

3. Outcome batao. Ek curved detached bow shock form hota hai, wedge tip ke aage khada. Centreline ke paas yeh essentially ek normal shock hai (iske peeche subsonic pocket), jo aur door jaane par weaker oblique shocks mein curve hota hai. Yeh step kyun? Yahi physical resolution hai jo maths signal karta hai jab koi attached solution exist nahi karta — dekho Supersonic Wedge & Cone Flow.

Verify: $25^{\circ }>{\theta }_{\max }(M=2)=22.97^{\circ }$ ✓ → detachment. Parent note ki "bigger wedge ⇒ shock detaches" warning se consistent.

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Example 7 — Cell G: real-world word problem

Forecast: $\beta =35^{\circ }$ $M=2.5$ ke Mach angle ($\mu =23.6^{\circ }$) se thoda hi zyada hai, toh modest wedge expect karo, shayad $10$–$15^{\circ }$.

1. $\theta$ ke liye forward-solve karo. $M_1^2{\sin }^2\beta -1=6.25{\sin }^{2}35^{\circ }-1=6.25(0.3290)-1=\mathrm{1.056.}$ Denominator: $6.25(1.4+\cos 70^{\circ })+2=6.25(1.4+0.3420)+2=\mathrm{12.89.}$ $\cot 35^{\circ }=1.428$. $\tan \theta =2(1.428)\frac{1.056}{12.89}=0.234\Rightarrow \theta =13.2^{\circ }.$ Yeh step kyun? Hum photo se $\beta$ measure karte hain aur tunnel se $M_1$ read karte hain — geometry phir wedge angle deliver karta hai, jise hum machined part se compare karte.

2. Mach-wave floor se sanity check. Mach angle $\mu =\arcsin (1/2.5)=23.6^{\circ }$ hai. Hamara $\beta =35^{\circ }>23.6^{\circ }$ ✓, toh genuine shock present hai (Mach wave nahi). Yeh step kyun? Agar measured $\beta$ $\mu$ se kam ya barabar hota, toh "shock" ek infinitesimal Mach line hoti aur photo mislabelled hoti.

Verify: $\theta =13.2^{\circ }$, wedge full-angle $\approx 26.4^{\circ }$ — physically sensible test wedge. $M_{n1}=2.5\sin 35^{\circ }=1.434>1$ ✓, toh compression real hai.

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Example 8 — Cell H: exam twist, ${\theta }_{\max }$ nikalo

Forecast: $M_1=2$ ke liye $\theta$–$\beta$ curve ka peak $\beta \approx 64^{\circ }$ ke aas-paas ${\theta }_{\max }\approx 23^{\circ }$ ke saath hai.

1. Valid range mein $\theta (\beta )$ scan karo. Relation ko $\beta$ ke liye Mach angle $30^{\circ }$ se $90^{\circ }$ tak evaluate karo aur sabse bada $\theta$ locate karo. Yeh step kyun? ${\theta }_{\max }$ by definition hump ka top hai; yeh "do attached solutions" aur "detached shock" ko alag karta hai.

2. Peak locate karo. Maximum $\beta =64.7^{\circ }$ par occur karta hai, jo $\tan \theta =0.4239$ deta hai. Yeh step kyun? Is $\beta$ par weak aur strong roots ek mein merge ho jaate hain — attachment ki frontier.

3. Maximum turn read karo. ${\theta }_{\max }=\arctan (0.4239)=22.97^{\circ }$. Yeh step kyun? Yahi woh number hai jis par Example 6 ne $25^{\circ }$ impossible declare karne ke liye rely kiya.

Verify: $M_1=2$ ke liye $\beta =64.7^{\circ }$ par ${\theta }_{\max }=22.97^{\circ }$ ✓ (standard oblique-shock charts se match karta hai, jo ~$23^{\circ }$ dete hain). Ex 6 mein use kiya gaya detachment threshold confirm hota hai.

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Recall Kaun si cell kaun si thi?

A forward ::: Ex 1 ($M=3,\beta =40^{\circ }\Rightarrow \theta =21.9^{\circ }$) B weak root ::: Ex 2 ($\beta =45.3^{\circ }$, $M_2=1.45$) C strong root ::: Ex 3 ($\beta =79.8^{\circ }$, $M_2=0.644$) D degenerate $\theta =0$ ::: Ex 4 (Mach angle $23.6^{\circ }$ aur $90^{\circ }$) E $\beta =90^{\circ }$ ::: Ex 5 (normal shock ban jaata hai, $M_2=0.475$) F over-turn ::: Ex 6 ($25^{\circ }>{\theta }_{\max }$ → detached bow shock) G word problem ::: Ex 7 (measured $\beta =35^{\circ }\Rightarrow \theta =13.2^{\circ }$) H find ${\theta }_{\max }$ ::: Ex 8 ($22.97^{\circ }$ at $\beta =64.7^{\circ }$)

Yeh bhi dekho: Prandtl-Meyer Expansion ulte problem ke liye — flow ko apne aap se door modna, jahan yeh shock hone ki jagah accelerate karta hai.