Oblique shock waves — θ-β-M relation: Worked examples

3.1.13 · D3 · Physics › Compressible Flow & Aerodynamics › Oblique shock waves — θ-β-M relation

Intuition Ye page kis liye hai

Parent note ne tumhe machinery di thi. Ye page usi ko har us case ke against drill karta hai jo appear ho sakta hai — θ-β "hump" ki har branch, har degenerate limit, ek word problem, aur ek exam twist. Agar tum neeche ke aathon examples kar sako, toh koi bhi oblique-shock question tumhe surprise nahi kar sakta.

Ye page self-contained hai. Jo kuch bhi hum use karte hain, woh yahan zaroorat padne par restate kiya gaya hai:

$θ$ = deflection angle (flow physically corner par kitna turn karta hai).
$β$ = shock-wave angle (shock line ka tilt incoming flow ke relative).
$M_{1}$ = upstream Mach number (flow speed ÷ upstream speed of sound).
$M_{n 1} = M_{1} sin β$ = upstream Mach number ka normal component — yahi ek cheez hai jo shock ko cross karti hai aur normal-shock physics "karti" hai.
$γ = 1.4$ = air ke liye ratio of specific heats (ek fixed gas property).
Inhe connect karne wala master rule, jo neeche plot kiya gaya hai, θ-β-M relation hai: $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2} .$
Do normal-shock facts jo hum reuse karenge (dono $M_{n 1}$ use karte hain, kabhi bhi full $M_{1}$ nahi): downstream normal Mach $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )}$ , pressure jump $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1)$ .

Scenario matrix

θ-β-M relation ko ek machine ki tarah socho jisme teen dials hain ( $θ$ , $β$ , $M_{1}$ ) aur ek knob ( $γ$ ). Har question kuch dials fix karta hai aur baaki ke liye poochta hai. Hump curve (θ utha, $θ_{ma x}$ par peak kiya, phir gira) woh map hai — neeche ka har example uske ek labelled spot par land karta hai.

Figure 1 — θ-β hump aur har case kahan hai.

Figure — Oblique shock waves — θ-β-M relation

Figure 1 kaise padhen. Do humps drawn hain. Pale-yellow curve $M_{1} = 3$ ke liye hai; chalk-blue curve $M_{1} = 2$ ke liye hai. Horizontal axis shock angle $β$ hai (degrees mein) aur vertical axis deflection $θ$ hai (degrees mein) jo flow undergo karta hai. Kisi bhi example se Figure 1 ka naam lekar refer karo:

Left foot (yellow curve axis ko $β = 19.4 7^{\circ}$ par chhorti hai, "Cell D" label hai) Mach-wave limit hai: $θ = 0$ , sabse weak possible shock.
Right foot (curve $β = 9 0^{\circ}$ par axis par vapas aati hai, "Cell E" label hai) normal shock hai: phir $θ = 0$ , lekin ab head-on.
Har curve ke top par filled dot $θ_{ma x}$ hai — peak ("peak = theta_max, Cell F if exceeded" label hai). Isse zyada turn maango aur curve ka koi bhi point tumhara nahi pakad paata, toh shock detach ho jaata hai.
Rising (left) branch par har weak root hota hai — "Cell A" $β = 37. 8^{\circ}, θ = 2 0^{\circ}$ par label hai.
Falling (right) branch par har strong root hota hai — "Cell B" usi $θ = 2 0^{\circ}$ ke liye $β = 82. 2^{\circ}$ par label hai, jo dikhata hai ki curve double-valued hai.
Cell C simply ek chosen $β$ par curve ki height padhta hai; Cells G aur H usi map ko alag Mach numbers par reuse karte hain. Figure 1 ko har example ke saath paas rakho.

Cell	Case class	Kya fix hai / kya degenerate hai	Example
A	Standard invert: weak $β$ dhundho	$M_{1}, θ$ diya, $θ < θ_{ma x}$	Ex 1
B	Same $θ$ , strong root	bada $β$ chuno	Ex 2
C	Forward: $β$ diya, $θ$ dhundho	koi inversion nahi chahiye	Ex 3
D	Degenerate low: $β = μ$ (Mach wave)	numerator $= 0$ , $θ = 0$	Ex 4
E	Degenerate high: $β = 9 0^{\circ}$	normal shock, $θ = 0$	Ex 5
F	Peak / over-limit: $θ > θ_{ma x}$	koi real $β$ nahi → detaches	Ex 6
G	Real-world word problem	full downstream chain $p_{2}, T_{2}, M_{2}$	Ex 7
H	Exam twist: wall se reflection	do shocks, doosre turn ka sign	Ex 8

Cell A — standard inversion (weak root)

Worked example Ex 1 ·

$M_{1} = 3$ , wedge half-angle $θ = 2 0^{\circ}$ , weak $β$ dhundho. Forecast: andaza lagao — kya $β$ Mach angle ( $arcsin \frac{1}{3} \approx 19. 5^{\circ}$ ) ke kareeb hoga ya $9 0^{\circ}$ ke? Aage padhne se pehle apna guess likho.

Step 1 — $β$ ke liye solve karne ki equation setup karo. $tan 2 0^{\circ} = 2 cot β \frac{9 s i n ^{2} β - 1}{9 ( 1.4 + c o s 2 β ) + 2} .$ Ye step kyun? Wedge $θ$ force karta hai; ek maatra unknown $β$ hai. Ye ek root-find hai, algebra solve nahi, kyunki $β$ ek saath teen trig functions ke andar baitha hai.

Step 2 — Left foot se upar scan karo. Left foot Mach angle $μ = arcsin (1/3) = 19.4 7^{\circ}$ hai jahan $θ = 0$ hai. Jaise $β$ $19.4 7^{\circ}$ se chadha, $θ$ ek peak tak badha phir $9 0^{\circ}$ par $0$ par wapas aa gaya. $θ = 2 0^{\circ}$ ki pehli crossing weak root hai. Ye step kyun? Weak matlab chota $β$ , Figure 1 ki rising (left) branch par. Free flight mein nature is kamzor root ko select karti hai.

Step 3 — Numerical root. Solve karne par: $β \approx 37.7 6^{\circ} .$

Verify: wapas plug karo: $9 sin^{2} 37.7 6^{\circ} = 9 (0.6124)^{2} = 3.375$ , toh numerator $= 2.375$ ; denominator $= 9 (1.4 + cos 75. 5^{\circ}) + 2 = 9 (1.4 + 0.2504) + 2 = 16.85$ ; $2 cot 37.7 6^{\circ} = 2 (1.292) = 2.585$ . Product $= 2.585 \times 2.375/16.85 = 0.364 = tan 20. 0^{\circ}$ . ✓

Cell B — same deflection, strong root

Worked example Ex 2 · Same

$M_{1} = 3, θ = 2 0^{\circ}$ — strong $β$ dhundho, phir prove karo ki ye subsonic hai. Forecast: ye $37. 8^{\circ}$ se bada aur $9 0^{\circ}$ se chota hona chahiye. Guess karo kahan.

Step 1 — Peak ke baad scan karte raho. Hump ke peak ke baad, $θ$ girta hai. Ye $2 0^{\circ}$ ko neeche jaate hue phir cross karta hai — woh doosri crossing strong root hai. Ye step kyun? θ-β curve double-valued hai: ek $θ$ , do $β$ . Same equation, doosra root, Figure 1 ki falling (right) branch par.

Step 2 — Root. $β \approx 82.1 5^{\circ} .$

Step 3 — Normal component. $M_{n 1} = 3 sin 82.1 5^{\circ} = 2.972$ — almost pura $M_{1}$ , toh ye shock almost head-on aur strongly compressive hai.

Step 4 — Actual downstream Mach $M_{2}$ compute karo (sirf "subsonic" assert mat karo). Pehle normal-shock relation se downstream normal component: $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{0.4 ( 8.833 ) + 2}{2.8 ( 8.833 ) - 0.4} = \frac{5.533}{24.33} = 0.2274,$ toh $M_{n 2} = 0.4769$ . Phir geometry restore karo — flow shock ko angle $(β - θ)$ par chhorti hai, toh $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.4769}{s i n ( 82.1 5 ^{\circ} - 2 0 ^{\circ} )} = \frac{0.4769}{s i n 62.1 5 ^{\circ}} = 0.5395.$ Ye step kyun? "Bada $β$ ⇒ subsonic" sirf ek heuristic hai; honest check $M_{2}$ compute karta hai aur paata hai $M_{2} = 0.54 < 1$ . Ab subsonic claim substantiated hai.

Verify: numerator $9 sin^{2} 82.1 5^{\circ} - 1 = 7.833$ ; denominator $9 (1.4 + cos 164. 3^{\circ}) + 2 = 5.934$ ; $2 cot 82.1 5^{\circ} = 0.2757$ . Product $= 0.364 = tan 2 0^{\circ}$ ✓. Aur $M_{2} = 0.54 < 1$ ✓ — genuinely subsonic.

Cell C — forward evaluation ( $β$ diya gaya)

Worked example Ex 3 ·

$M_{1} = 2.5$ , shock $β = 4 0^{\circ}$ par observe kiya gaya. $θ$ dhundho. Forecast: koi inversion nahi — bas plug karo. Kya $θ$ chota hoga (Mach wave ke kareeb) ya bada?

Step 1 — Pehle check karo ki shock exist bhi karta hai ya nahi. Chahiye $M_{n 1} = M_{1} sin β > 1$ : $2.5 sin 4 0^{\circ} = 1.607 > 1$ . ✓ Ye step kyun? Agar $M_{n 1} \leq 1$ toh "shock" actually ek Mach wave hai ya kuch nahi — formula $θ \leq 0$ wapas deta, jo shock kehna unphysical hai.

Step 2 — Direct substitution. $tan θ = 2 cot 4 0^{\circ} \frac{6.25 s i n ^{2} 4 0 ^{\circ} - 1}{6.25 ( 1.4 + c o s 8 0 ^{\circ} ) + 2} .$ Numerator $= 6.25 (0.4132) - 1 = 1.582$ ; denominator $= 6.25 (1.4 + 0.1736) + 2 = 11.83$ ; $2 cot 4 0^{\circ} = 2.384$ . Toh $tan θ = 2.384 \times 1.582/11.83 = 0.3189$ . Ye step kyun? $β$ aur $M_{1}$ dono known hain, toh θ-β-M relation ke right side par har symbol ab ek number hai — koi equation solve nahi karni, hum bas $M_{1} = 2.5$ hump par ek point par machine evaluate kar rahe hain. Ye Cell C hai: Figure 1 mein curve ki height padhna.

Step 3 — Result. $θ = arctan 0.3189 = 17.6 9^{\circ}$ . Arctan yahan kyun? Hum ratio $tan θ$ jaante hain; arctan jawab deta hai "kis angle ka ye tangent hai?" — ye tan ko undo karta hai, aur $0 < θ < 9 0^{\circ}$ hone se koi quadrant ambiguity nahi hai.

Verify: $tan 17.6 9^{\circ} = 0.3189$ ✓ aur $M_{n 1} = 1.607 > 1$ toh ye $β$ valid shock hai. ✓

Cell D — degenerate low limit (Mach wave)

Worked example Ex 4 ·

$M_{1} = 2$ , $θ \to 0$ upar se. $β$ kya hai? Forecast: shock infinitely weak ho jaata hai. Iska angle kahan settle hoga?

Step 1 — Numerator ko zero karo. $tan θ = 0$ chahiye $M_{1}^{2} sin^{2} β - 1 = 0$ ( $cot β$ factor yahan finite hai). Ye step kyun? $θ = 0$ ⇒ $tan θ = 0$ ⇒ fraction zero hai ⇒ iska numerator zero hai. Ye Figure 1 mein hump ka left foot hai.

Step 2 — Solve karo. $sin β = 1/ M_{1} = 0.5 \Rightarrow β = 3 0^{\circ}$ . Result: $β = 3 0^{\circ}$ , exactly Mach angle $μ = arcsin (1/ M)$ Mach Angle and Mach Waves se.

Step 3 — Meaning. Ek infinitesimal disturbance Mach wave ke roop mein $μ$ par travel karti hai; oblique shock ka sabse kamzor possible tilt ye angle hai. Iske neeche, $M_{n 1} < 1$ aur koi shock form nahi ho sakta.

Verify: $arcsin (0.5) = 3 0^{\circ}$ aur $β = 3 0^{\circ}$ par, $M_{n 1} = 2 sin 3 0^{\circ} = 1.000$ — exactly sonic, threshold. ✓

Cell E — degenerate high limit (normal shock)

Worked example Ex 5 ·

$M_{1} = 3$ , $β = 9 0^{\circ}$ set karo. Dikhao $θ = 0$ hai aur normal shock recover karo. Forecast: ek "vertical" shock — kya flow bilkul bhi turn karta hai?

Step 1 — Numerator ke cofactor ko evaluate karo. $β = 9 0^{\circ}$ par, $cot 9 0^{\circ} = 0$ , toh pura product $0$ hai chahe fraction kuch bhi ho ⇒ $tan θ = 0$ ⇒ $θ = 0$ . Ye step kyun? $cot β = cos β / sin β$ aur $cos 9 0^{\circ} = 0$ . Ye Figure 1 mein hump ka right foot hai.

Step 2 — Physics identify karo. $θ = 0$ ke saath flow deflect nahi hota: normal aur full velocity coincide karte hain, $M_{n 1} = M_{1} sin 9 0^{\circ} = M_{1}$ . Oblique shock hai ek normal shock. Ye step kyun? Ye sanity anchor hai: general relation mein special case hona chahiye, warna ye galat hai.

Step 3 — Downstream Mach (normal-shock relation). $M_{2}^{2} = \frac{( γ - 1 ) M _{1}^{2} + 2}{2 γ M _{1}^{2} - ( γ - 1 )} = \frac{0.4 ( 9 ) + 2}{2.8 ( 9 ) - 0.4} = \frac{5.6}{24.8} = 0.2258,$ toh $M_{2} = 0.4752$ — subsonic, jaisa normal shock demand karta hai.

Verify: $tan θ = 2 cot 9 0^{\circ} (\dots) = 0$ ✓ aur $M_{2} = 0.475$ standard Rankine–Hugoniot value se match karta hai $M_{1} = 3$ ke liye. ✓

Cell F — limit se upar (detachment) + $θ_{ma x}$ kaise dhundhen

Worked example Ex 6 ·

$M_{1} = 2$ , wedge half-angle $θ = 2 5^{\circ}$ . $β$ dhundho — aur pehle, $θ_{ma x}$ derive karo. Forecast: $M_{1} = 2$ ke liye peak $θ_{ma x} \approx 22.9 7^{\circ}$ hai. Hum flow se zyada turn maang rahe hain jo attached reh sake. Kya hota hai?

Step 1 — KYUN derivative? $d θ / d β = 0$ se peak locate karo. Hump ka top wahan hai jahan $θ$ badna band ho jaata hai aur girna shuru ho jaata hai — yaani jahan curve ki slope flat ho. Derivative woh tool hai jo "curve ki slope" measure karta hai. Toh peak $θ_{ma x}$ woh $β$ hai jahan $\frac{d θ}{d β} = 0.$ Ye tool kyun, koi doosra kyun nahi? Ek smooth function ka maximum exactly zero slope ka point hai; derivative ke alawa kuch bhi use usse pin nahi kar sakta. Isse zero karne par "top kahan hai?" ek solvable equation mein convert ho jaata hai.

Step 2 — Closed-form maximum condition. θ-β-M relation differentiate karke $d θ / d β = 0$ set karne par standard maximum-deflection condition milti hai ( $sin^{2} β$ mein ek quadratic): $sin^{2} β_{ma x} = \frac{1}{γ M _{1}^{2}} [\frac{γ + 1}{4} M_{1}^{2} - 1 + (γ + 1) (\frac{γ + 1}{16} M_{1}^{4} + \frac{γ - 1}{2} M_{1}^{2} + 1)] .$ Ye step kyun? Ye woh algebra hai jo $d θ / d β = 0$ solve karne par produce hoti hai — ek aisa formula jo tum evaluate kar sako, koi root-scanning ki zaroorat nahi.

Step 3 — $M_{1} = 2, γ = 1.4$ plug karo. Bracket $sin^{2} β_{ma x} = 0.8175$ deta hai, toh $β_{ma x} = 64.6 7^{\circ}$ . Ise θ-β-M relation mein wapas feed karne par peak height milti hai: $θ_{ma x} = 22.9 7^{\circ} .$ Ye step kyun? Ab hume peak bina guess kiye mili — aur $2 5^{\circ} > 22.9 7^{\circ}$ .

Step 4 — Conclude: koi real solution nahi ⇒ detachment. Kyunki demanded $2 5^{\circ}$ , $θ_{ma x} = 22.9 7^{\circ}$ se zyada hai, θ-β-M relation ka koi real $β$ nahi hai. Attached-shock assumption fail ho jaati hai aur shock detach ho jaata hai: ek curved bow shock wedge nose ke aage khada ho jaata hai — Figure 2 dekho.

Figure 2 — attached shock vs detached bow shock.

Figure 2 kaise padhen. Left panel: $θ < θ_{ma x}$ , ek straight oblique shock (yellow line) ek definite angle $β$ par wedge nose se chipka hua hai. Right panel: $θ > θ_{ma x}$ , shock attach nahi ho sakta — ek curved pink bow shock nose ke aage khada hai, axis par near-normal (subsonic pocket, shaded) aur flanks par door jaate jaate Mach wave tak kamzor hota jaata hai.

Verify: $θ_{ma x} (M_{1} = 2) = 22.9 7^{\circ}$ closed form se aur $2 5^{\circ} > 22.9 7^{\circ}$ ⇒ demanded > max attainable ⇒ detached. ✓

Cell G — real-world word problem (full downstream chain)

Worked example Ex 7 · Ek supersonic intake ramp

$M_{1} = 3$ , $p_{1} = 20 kPa$ , $T_{1} = 220 K$ par air ko $θ = 1 5^{\circ}$ turn karta hai. $β$ , $p_{2}$ , $T_{2}$ , aur $M_{2}$ dhundho (weak shock). Forecast: compression ⇒ $p_{2} > p_{1}$ , $T_{2} > T_{1}$ , aur $M_{2}$ drop karega lekin supersonic rehna chahiye (weak). $M_{2}$ guess karo: 2 ke upar ya neeche?

Definition Teen "downstream speed" symbols (use se pehle define kiye)

$V_{2}$ = downstream velocity ki magnitude — literally kitne metres per second air shock ke baad move karti hai, ek arrow ki tarah.
$a_{2}$ = downstream speed of sound — ek tiny pressure ripple post-shock gas mein kitni tez travel karta hai.
$M_{2} = V_{2} / a_{2}$ = downstream Mach number — donon ka ratio.

Geometry mein kahan aate hain ye (Figure 3 dekho): downstream velocity arrow $V_{2}$ shock ko angle $(β - θ)$ par chhorti hai. Iska component shock ke perpendicular $u_{n 2} = V_{2} sin (β - θ)$ hai. $a_{2}$ se divide karne par velocities Mach numbers ban jaati hain: $M_{n 2} = M_{2} sin (β - θ)$ , yaani $M_{2} = M_{n 2} / sin (β - θ)$ . Hume raw $V_{2}$ ya $a_{2}$ ki kabhi zaroorat nahi — ye cancel ho jaate hain — lekin inhe naam dena geometry ko honest banata hai.

Figure 3 — downstream velocity triangle.

Figure 3 kaise padhen. Blue arrow $V_{2}$ hai, shock (yellow line) ko angle $(β - θ)$ par chhorti hai. Shock par ek perpendicular giraao: pink arrow normal part $u_{n 2} = V_{2} sin (β - θ)$ hai, shock ke along white arrow tangential part hai (shock ke across unchanged). Har arrow ki length $a_{2}$ se divide karo aur wahi picture Mach numbers mein padha jaata hai.

Step 1 — Weak $β$ ke liye θ-β-M invert karo. $tan 1 5^{\circ} = 2 cot β \frac{9 sin ^{2} β - 1}{9 ( 1.4 + cos 2 β ) + 2}$ solve karne par $β \approx 32.2 4^{\circ}$ milta hai. Ye step kyun? Downstream properties sab normal component se key off hoti hain, jiske liye $β$ chahiye.

Step 2 — Normal Mach. $M_{n 1} = 3 sin 32.2 4^{\circ} = 1.600$ . Ye step kyun? Sirf $M_{n 1}$ shock ko ek normal shock ki tarah cross karta hai: full $M_{1}$ compression overstate karega.

Step 3 — Pressure ratio (Rankine–Hugoniot). $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) = 1 + \frac{2.8}{2.4} (2.56 - 1) = 2.820,$ toh $p_{2} = 2.820 \times 20 = 56.4 kPa$ . Ye step kyun? Pressure isse set hota hai ki normal momentum kitni hard roka jaata hai, toh Rankine–Hugoniot pressure jump $M_{n 1}$ use karta hai, $M_{1}$ nahi. Normal component jitni hard hit karti hai, pressure spike utna bada.

Step 4 — Temperature ratio. $\frac{T _{2}}{T _{1}} = \frac{[ 2 γ M _{n 1}^{2} - ( γ - 1 )] [( γ - 1 ) M _{n 1}^{2} + 2 ]}{( γ + 1 ) ^{2} M _{n 1}^{2}} = \frac{( 6.768 ) ( 3.024 )}{( 5.76 ) ( 2.56 )} = 1.388,$ toh $T_{2} = 1.388 \times 220 = 305.4 K$ . Ye step kyun? Roki gayi normal kinetic energy heat mein dump ho jaati hai, $T$ badh jaata hai — shock ke peeche ki squeeze hui air denser aur hotter dono hoti hai. Phir se sirf $M_{n 1}$ se driven.

Step 5 — Downstream Mach. Pehle shock ke baad normal component: $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{3.024}{6.768} = 0.4468 \Rightarrow M_{n 2} = 0.6684.$ Phir Figure 3 ke triangle ka use karke geometry restore karo: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.6684}{s i n ( 32.2 4 ^{\circ} - 1 5 ^{\circ} )} = \frac{0.6684}{s i n 17.2 4 ^{\circ}} = 2.256.$ $sin (β - θ)$ se kyun divide kiya? Kyunki $u_{n 2} = V_{2} sin (β - θ)$ ; $a_{2}$ se divide karne par $M_{n 2} = M_{2} sin (β - θ)$ milta hai, aur invert karne par $M_{2}$ isolate ho jaata hai.

Verify: $M_{2} = 2.256 > 1$ (weak supersonic rehta hai ✓), $p_{2} > p_{1}$ ✓, $T_{2} > T_{1}$ ✓. Saare ratios $M_{n 1} = 1.600$ use karte hain, self-consistent. Units: kPa×(dimensionless ratio)=kPa, K×(ratio)=K ✓.

Cell H — exam twist: wall se shock reflection

Worked example Ex 8 · Ek weak oblique shock (ek

$θ = 1 0^{\circ}$ ramp se, $M_{1} = 3$ ) ek flat wall jo original flow ke parallel hai se takraata hai. Woh deflection dhundho jo reflected shock produce karni chahiye, aur iska incident angle. Forecast: wall ko flow wapas apne aap ke parallel bhejna hai. Toh reflected shock ko turn undo karna hoga. Kitne se?

Step 1 — Incident shock. $M_{1} = 3, θ = 1 0^{\circ}$ ke liye, weak root $β_{1} \approx 27.3 8^{\circ}$ hai. Iske peeche $M_{n 1} = 3 sin 27.3 8^{\circ} = 1.380$ , $M_{n 2} = 0.7472$ deta hai aur $M_{2} = M_{n 2} / sin (β_{1} - 1 0^{\circ}) = 2.505$ . Ye step kyun? Hume reflection feed karne wala naya upstream Mach $M_{2}$ chahiye.

Step 2 — Reflected turn. Wall ek solid boundary hai: flow isse parallel nikal jaana chahiye. Pehle shock ne flow ko wall ki taraf $1 0^{\circ}$ turn kiya, toh reflected shock use $1 0^{\circ}$ wapas turn karni chahiye — same magnitude, turn ka opposite sign. Ye step kyun? Yehi poora trick hai: reflection ke liye deflection phir $θ = 1 0^{\circ}$ hai, lekin naye flow ke relative $M_{2} = 2.505$ par, $M_{1}$ nahi.

Step 3 — Reflected shock angle. θ-β-M ko $M = 2.505, θ = 1 0^{\circ}$ ke saath invert karo: weak root $β_{2} \approx 31.8 3^{\circ}$ hai (local post-first-shock flow direction se measure kiya gaya). Ye step kyun? Kum Mach ⇒ same turn ke liye steep shock — reflected shock incident shock se zyada inclined hai ( $31.8 3^{\circ} > 27.3 8^{\circ}$ ), ek classic exam catch.

Verify: incident $β_{1} = 27.3 8^{\circ}$ $M_{1} = 3$ par $θ = 1 0^{\circ}$ deta hai ✓; reflected $β_{2} = 31.8 3^{\circ}$ $M_{2} = 2.505$ par $θ = 1 0^{\circ}$ deta hai ✓; $31.8 3^{\circ} > 27.3 8^{\circ}$ confirm karta hai "lower $M$ ⇒ larger $β$ ." ✓

Recall Kaun sa cell kaun sa tha? (self-test)

Sirf "strong shock dhundho" diya — hump ka kaun sa foot/branch? ::: Right (falling) branch, bada $β$ (Ex 2 / Cell B). $θ_{ma x}$ generally kaise locate karte hain? ::: $d θ / d β = 0$ set karo; closed form $sin^{2} β_{ma x}$ deta hai, phir wahan θ-β-M evaluate karo (Ex 6 / Cell F). $θ$ demanded $θ_{ma x}$ se zyada — outcome aur note? ::: Detached bow shock, Detached Bow Shocks (Ex 6 / Cell F). Har downstream ratio mein kaun sa ek Mach number jaata hai? ::: $M_{n 1} = M_{1} sin β$ (Ex 7 / Cell G). Wall se reflection — doosra deflection kya set karta hai? ::: Wall flow ko phir parallel force karti hai, toh reflected turn incident $θ$ ke barabar hai lekin naye $M_{2}$ par (Ex 8 / Cell H).

Expansion turns ke liye (opposite sign of $θ$ , ek convex corner) Prandtl-Meyer Expansion dekho; 3-D nose analogue ke liye Supersonic Wedge and Cone Flow dekho.