Oblique shock waves — θ-β-M relation: Exercises

3.1.13 · D4 · Physics › Compressible Flow & Aerodynamics › Oblique shock waves — θ-β-M relation

Intuition Is page ko kaise use karein

Har problem ko solution kholne se pehle solve karo. Problems ek ladder climb karte hain: pehle tum sirf θ-β-M relation ke pieces ko pahchante ho, phir use apply karte ho, phir curves aur limits ko analyse karte ho, phir ek wedge ke across synthesise karte ho, aur finally shock detachment aur multi-shock reasoning ke saath mastery tak pahunchte ho (L5.1–L5.3).

Yahan use kiye gaye har symbol ki definition parent note mein hai. Quick reminder:

$M_{1}$ = upstream Mach number (flow speed ÷ speed of sound, shock se pehle).
$β$ = shock-wave angle (shock line ka tilt vs incoming flow).
$θ$ = flow deflection angle (streamline actually kitna bend karta hai).
$μ$ = Mach angle $= arcsin (1/ M_{1})$ — sabse kamzor possible wave (flow pe sawaar ek sound wave) ka tilt; yeh sabse chhota $β$ ho sakta hai kabhi bhi. Socho " $μ$ = ek whisper ka angle, $β$ = ek shove ka angle."
$M_{n 1} = M_{1} sin β$ = normal Mach component jo normal-shock physics karta hai.
$M_{n 2}$ = shock ke peeche normal Mach component; $M_{2}$ = full downstream Mach.
$V_{1}, V_{2}$ = actual flow speeds upstream/downstream; $u_{n 1}, u_{n 2}$ = unke pieces shock ke perpendicular; $w$ = shock ke saath wala piece (unchanged).
$p$ = pressure, $ρ$ = density, $a$ = local speed of sound (toh koi bhi Mach $= V / a$ ).
$γ = 1.4$ air ke liye (ratio of specific heats).

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

Master diagram ko abhi, line by line padho — har problem uspe lean karti hai. Har element ko colour aur shape/label dono se naam diya gaya hai, toh description grayscale mein bhi kaam karti hai:

Far left ke arrows, labelled " $M_{1}$ " (blue) incoming flow hain, sab horizontally point kar rahe hain.
Steeply rising solid line, labelled "shock" (red) shock hai; uske aur incoming flow ke beech ka angle $β$ hai.
Shallow solid line, labelled "wedge wall" (gray) wall hai; uske aur incoming flow ke beech ka angle $θ$ hai — deflected arrow labelled " $M_{2}$ " (green) is wall ke saath chalti hai.
Short arrow labelled " $M_{n 1}$ " (orange) incoming velocity ka woh piece hai jo seedha shock mein ja raha hai; uski length $M_{1} sin β = M_{n 1}$ hai. Woh chhota right triangle (horizontal side, normal side) hi poora reason hai ki $M_{n 1} = M_{1} sin β$ kyun hai: orange side us triangle mein angle $β$ ki "opposite" side hai.

Level 1 — Recognition

L1.1 — Teen angles ko naam do

Problem. Ek supersonic stream ek wedge se takraati hai. Master diagram mein teen angles dikh rahe hain: ek incoming flow aur shock ke beech, ek incoming flow aur deflected flow ke beech, aur ek deflected flow aur shock ke beech. Inhe θ-β-M language mein naam do, aur teesre ko baaki dono ke terms mein likho.

Recall Solution

Yeh padhte waqt upar master figure dekho.

Incoming flow (arrows labelled " $M_{1}$ ") ↔ shock (line labelled "shock"): yeh $β$ hai (shock angle).
Incoming flow ↔ deflected flow (arrow labelled " $M_{2}$ "): yeh $θ$ hai (deflection angle).
Deflected flow ↔ shock line: yeh $(β - θ)$ hai.

$(β - θ)$ kyun? Shock incoming flow se $β$ par hai. Flow phir shock ki taraf $θ$ se bend karti hai, toh gap close hota hai aur ab shock se chhota angle $β - θ$ banta hai. Yahan kuch naya nahi chahiye — yeh bas picture se seedha $β$ minus $θ$ padha jaata hai.

L1.2 — Shock ko kaun sa Mach number cross karta hai?

Problem. $M_{1} = 2.5$ aur $β = 4 0^{\circ}$ ke liye, woh Mach number compute karo jo actually normal-shock physics govern karta hai.

Recall Solution

Sirf normal component (master figure mein chhota arrow labelled " $M_{n 1}$ ") shock ko normal shock ki tarah cross karta hai: $M_{n 1} = M_{1} sin β = 2.5 sin 4 0^{\circ} = 2.5 (0.6428) = 1.607.$ Kyunki $M_{n 1} > 1$ hai, ek real shock form hoti hai.

L1.3 — Shock hai ya Mach wave?

Problem. $M_{1} = 2$ ke liye, kya $β = 2 5^{\circ}$ ek valid shock angle hai? Aur $β = 3 5^{\circ}$ ?

Recall Solution

Ek real shock ko chahiye $M_{n 1} = M_{1} sin β > 1$ , yaani $sin β > 1/ M_{1} = 0.5$ , yaani $β > 3 0^{\circ}$ .

$β = 2 5^{\circ}$ : $sin 2 5^{\circ} = 0.423 < 0.5$ → $M_{n 1} = 0.845 < 1$ . Shock nahi — angle bahut shallow hai, yeh Mach cone ke andar hai.
$β = 3 5^{\circ}$ : $sin 3 5^{\circ} = 0.574 > 0.5$ → $M_{n 1} = 1.147 > 1$ . Valid shock. Boundary $β = 3 0^{\circ}$ hi Mach angle $μ = arcsin (1/ M_{1})$ hai — upar symbol list se yaad karo ki $μ$ sabse kamzor possible wave (ek plain sound wave) ka tilt hai, sabse chhota tilt jo ek "shock" kabhi bhi le sakta hai. Dekho Mach Angle and Mach Waves.

Level 2 — Application

L2.1 — Known shock angle se deflection

Problem. Air, $M_{1} = 3$ , shock angle $β = 4 0^{\circ}$ . Is se produce hone wali deflection $θ$ nikalo.

Intuition θ-β-M relation ka har factor kahan se aata hai (plug in karne se pehle)

Relation $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2}$ koi magic formula nahi hai — har factor geometry ya physics ka ek piece hai jo tumne parent note mein already dekha hai. Ise teen chunks ke roop mein padho (aur master figure dekho):

Numerator mein $M_{1}^{2} sin^{2} β - 1 = M_{n 1}^{2} - 1$ . Yeh "shock strength" hai — normal Mach $1$ se kitna zyada hai. Agar $M_{n 1} = 1$ toh flow shock ke normal sirf sound barrier tak hi pahunchi, toh woh kuch nahi mudi: numerator zero ho jaata hai aur $θ = 0$ . Isliye Mach-wave limit numerator mein rehti hai.
$cot β$ prefactor. Turning isliye hoti hai kyunki normal speed girti hai jabki tangential slide fixed rehti hai. Velocity triangle par, dono components ke trade off ke saath ek given drop ke liye "bend" ki dikhayi dene wali amount cotangent of $β$ se scale hoti hai. Ek steep shock ( $β \to 9 0^{\circ}$ , $cot β \to 0$ ) mein almost koi tangential slide nahi hoti bend karne ke liye, toh woh flow ko barely turn karta hai — isliye $cot β$ sab kuch multiply karta hai aur $β = 9 0^{\circ}$ par deflection ko khatam kar deta hai.
Denominator $M_{1}^{2} (γ + cos 2 β) + 2$ . Yeh bas normal-shock density ratio substitute karne aur $tan (β - θ)$ expand karne ka algebraic residue hai (parent derivation ke Steps 4–6). Yeh real shock ke liye kabhi sign change nahi karta, toh yeh sirf $θ$ ki magnitude set karta hai, yeh nahi ki flow turn hogi ya nahi.

Toh turn ka sign aur existence numerator (shock strength) aur $cot β$ prefactor (geometry) se decide hoti hai; denominator bas scale karta hai.

Recall Solution

θ-β-M relation mein seedha plug karo (yahan inversion ki zaroorat nahi — $β$ diya hua hai): $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2} .$ Pieces:

$sin 4 0^{\circ} = 0.6428$ , toh $M_{1}^{2} sin^{2} β = 9 (0.4132) = 3.719$ ; numerator $= 3.719 - 1 = 2.719$ .
$cos 2 β = cos 8 0^{\circ} = 0.1736$ ; denominator $= 9 (1.4 + 0.1736) + 2 = 9 (1.5736) + 2 = 16.16$ .
$cot 4 0^{\circ} = 1/ tan 4 0^{\circ} = 1.1918$ . $tan θ = 2 (1.1918) \frac{2.719}{16.16} = 2.3836 (0.1683) = 0.4011 \Rightarrow θ = 21. 9^{\circ} .$

L2.2 — Downstream normal Mach aur pressure ratio

Problem. L2.1 ke liye ( $M_{1} = 3, β = 4 0^{\circ}$ ), $M_{n 1}$ nikalo, phir $M_{n 2}$ aur $p_{2} / p_{1}$ normal-shock relations use karke.

Intuition Pressure-ratio formula aisa kyun dikhta hai

Neeche figure mein chhota "shock box" dekho. Teen Rankine–Hugoniot jump laws mein se har ek kehta hai ki ek flux patli shock se thodi pehle aur thodi baad same hai:

Mass: $ρ_{1} u_{n 1} = ρ_{2} u_{n 2}$ (per second same amount of gas cross karta hai).
Momentum (normal): $p_{1} + ρ_{1} u_{n 1}^{2} = p_{2} + ρ_{2} u_{n 2}^{2}$ (pressure + "ram" conserved hai). Rearrange karo, pressure jump equal hai ram momentum ke loss ke, $p_{2} - p_{1} = ρ_{1} u_{n 1}^{2} - ρ_{2} u_{n 2}^{2}$ — yahi literally woh jagah hai jahan factor $(M_{n 1}^{2} - 1)$ paida hota hai.
Energy: total enthalpy conserved hai, jo temperature side pin karta hai. In teeno ko combine karo aur sab kuch $M_{n 1} = u_{n 1} / a_{1}$ ke saath likho toh compact result milta hai $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) .$ Physically padho: $M_{n 1} = 1$ par bracket $0$ hai toh $p_{2} / p_{1} = 1$ (koi shock nahi); normal Mach jitna zyada strong, pressure jump utna bada. Coefficient $2 γ / (γ + 1)$ bas yeh hai ki stiffness $γ$ wali gas us excess ram ko kitni efficiently pressure mein convert karti hai.

Recall Solution

Sirf normal component cross karta hai, toh $M_{n 1}$ har jagah use karo: $M_{n 1} = M_{1} sin β = 3 (0.6428) = 1.928.$ Normal-shock downstream Mach:

=\frac{0.4(3.718)+2}{2.8(3.718)-0.4}=\frac{3.487}{10.01}=0.3484,$$ toh $M_{n2}=0.590$. Pressure ratio (upar momentum jump se): $$\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M_{n1}^2-1)=1+\frac{2.8}{2.4}(3.718-1)=1+1.1667(2.718)=4.171.$$

L2.3 — True downstream Mach recover karo

Problem. L2.2 continue karo: $M_{2}$ (full downstream Mach) nikalo. Kya flow abhi bhi supersonic hai?

Recall Solution

Shock ke peeche normal component $u_{n 2} = V_{2} sin (β - θ)$ obey karta hai, toh local sound speed $a_{2}$ se divide karne par $M_{n 2} = M_{2} sin (β - θ)$ milta hai. Hum us geometry ko undo karte hain: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.590}{s i n ( 4 0 ^{\circ} - 21. 9 ^{\circ} )} = \frac{0.590}{s i n 18. 1 ^{\circ}} = \frac{0.590}{0.3107} = 1.899.$ $M_{2} = 1.90 > 1$ : abhi bhi supersonic — weak oblique shock ki hallmark. Bhale hi $M_{n 2} < 1$ hai, tangential slide itni speed carry karta hai ki total supersonic rahe.

Level 3 — Analysis

L3.1 — Kaun sa root kaun sa hai?

Problem. $M_{1} = 3$ , $θ = 2 0^{\circ}$ ke liye θ-β-M relation ke do roots hain, $β \approx 37. 8^{\circ}$ aur $β \approx 8 2^{\circ}$ . $M_{2}$ fully compute kiye bina argue karo kaun weak shock hai aur kaun strong shock, aur har ek ke liye $(M_{2} - 1)$ ka sign predict karo.

Recall Solution

Upar figure mein, teen humps directly curve par apne Mach number ke saath labelled hain (" $M_{1} = 2$ ", " $M_{1} = 3$ ", " $M_{1} = 5$ "); colour nahi, label use karo. " $M_{1} = 3$ " tagged curve follow karo. $θ = 2 0^{\circ}$ par horizontal dashed line ise do jagah kaatti hai — do arrows exactly roots $β = 37. 8^{\circ}$ (peak ke baayin) aur $β = 8 2^{\circ}$ (peak ke daayan) mark karte hain.

Chhota $β = 37. 8^{\circ}$ = peak ke baayin = weak shock. Chhota $β$ ⇒ chhota $M_{n 1}$ ⇒ milder compression ⇒ downstream usually supersonic ( $M_{2} > 1$ ).
Bada $β = 8 2^{\circ}$ = peak ke daayan = strong shock. Bada $β$ ⇒ $M_{n 1}$ $M_{1}$ ke karib ⇒ near-normal-shock ⇒ downstream subsonic ( $M_{2} < 1$ ). Nature normally free flight mein weak root select karta hai (downstream pressure artificially raise nahi hoti).

L3.2 — Do limiting shocks

Problem. Kisi bhi $M_{1}$ ke liye, hump $β$ ke do values par $θ = 0$ ko touch karta hai. Dono nikalo aur batao har ek physically kya hai.

Recall Solution

Usi figure par, notice karo har curve baayin taraf horizontal axis par start karti hai (apne Mach angle par) aur daayan taraf $β = 9 0^{\circ}$ par axis par wapas aati hai — woh do axis touchdowns exactly $θ = 0$ solutions hain. Algebraically, $tan θ = 0$ set karo; fraction zero (ya prefactor) tab hota hai jab:

Numerator zero: $M_{1}^{2} sin^{2} β - 1 = 0 \Rightarrow sin β = 1/ M_{1} \Rightarrow β = arcsin (1/ M_{1}) = μ$ . Yeh Mach wave hai — sabse kamzor possible disturbance, koi compression nahi (baayin touchdown).
$cot β = 0$ : $β = 9 0^{\circ}$ . Yeh normal shock hai — maximum compression, flow turn nahi hoti kyunki shock perpendicular hai (daayan touchdown). Toh hump baayin Mach wave aur daayan normal shock se bookended hai, dono opposite reasons se zero deflection dete hain.

L3.3 — Teen speeds par Mach angle

Problem. $M_{1} = 1.5, 2, 5$ ke liye Mach angle $μ = arcsin (1/ M_{1})$ compute karo. Kya trend dikh raha hai, aur fast vehicle ke liye physically kya matlab hai?

Recall Solution

$M_{1} = 1.5$ : $μ = arcsin (0.667) = 41. 8^{\circ}$ .
$M_{1} = 2$ : $μ = arcsin (0.5) = 30. 0^{\circ}$ .
$M_{1} = 5$ : $μ = arcsin (0.2) = 11. 5^{\circ}$ . Trend: jaise $M_{1}$ badhta hai, $μ$ shrink hota hai — Mach cone flight direction ke around tight squeeze ho jaata hai. Yeh θ-β figure par bhi dikh sakta hai: faster curve (labelled " $M_{1} = 5$ ") apna baayin touchdown aur baayin shuru karta hai (chhota $β$ ). Physically, tum jitna fast jaate ho, pressure disturbances utni hi thin sliver mein pile up hoti hain aage; sabse kamzor wave almost flow ke saath-saath hoti hai. Dekho Mach Angle and Mach Waves.

Level 4 — Synthesis

L4.1 — Full wedge solve

Problem. $M_{1} = 2.5$ par air ek symmetric wedge ke upar flow karti hai jiska half-angle $θ = 1 5^{\circ}$ hai. Weak shock angle $β$ nikalo, phir $M_{n 1}$ , $p_{2} / p_{1}$ , aur downstream Mach $M_{2}$ .

Intuition θ-β-M ko

$β$ ke liye actually kaise invert karo (root-finding, haath se ki gayi) Relation $β$ se $θ$ explicitly deta hai, lekin yahan hum $θ$ jaante hain aur $β$ chahiye — ek implicit equation. Koi clean closed form nahi hai, toh hum root dhundhte hain. "Residual" define karo: $f (β) = θ predicted by this β 2 cot β \frac{M _{1}^{2} sin ^{2} β - 1}{M _{1}^{2} ( γ + cos 2 β ) + 2} - tan θ_{wanted} .$ Hum chahte hain $f (β) = 0$ .

Pehle bracket karo. Kisi bhi real shock mein $μ \leq β \leq 9 0^{\circ}$ . Yahan $μ = arcsin (1/2.5) = 23. 6^{\circ}$ . $β = μ$ par, $f = - tan 1 5^{\circ} < 0$ (predicted $θ = 0$ ). $β = 9 0^{\circ}$ par, again predicted $θ = 0$ toh $f < 0$ . Beech mein $f$ $0$ se upar uthta hai (hump $1 5^{\circ}$ se zyada ho jaata hai) aur wapas neeche aata hai — toh $f = 0$ ke do roots hain, ek peak $θ_{ma x}$ ke har taraf.
Weak root chunno: woh hai jo chhote $β$ wala hai, yaani peak se pehle wala crossing. Use $μ = 23. 6^{\circ}$ aur peak location ke beech bracket karo ( $M_{1} = 2.5$ ke liye peak $β \approx 6 5^{\circ}$ ke karib hai, toh bracket $[2 4^{\circ}, 6 5^{\circ}]$ ).
Bisection. Midpoint par $f$ evaluate karo; woh aadha rakho jahan $f$ sign change kare; repeat karo. Kuch steps: $f (3 0^{\circ}) < 0$ , $f (4 0^{\circ}) > 0$ ⇒ root $[3 0^{\circ}, 4 0^{\circ}]$ mein; $f (3 7^{\circ}) < 0$ , $f (37. 5^{\circ}) > 0$ ⇒ root $36. 9^{\circ}$ ke karib. Yeh weak root $β \approx 36. 9^{\circ}$ par converge karta hai. (Bracket $[6 5^{\circ}, 9 0^{\circ}]$ lete toh strong root $8 2^{\circ}$ ke karib milta.)

Recall Solution

Step A — θ-β-M ko $β$ ke liye invert karo upar bracket-and-bisection recipe use karke, weak (chhote- $β$ wala) root: $β \approx 36. 9^{\circ}$ . Step B — normal component. $M_{n 1} = 2.5 sin 36. 9^{\circ} = 2.5 (0.6004) = 1.501$ . Step C — pressure ratio. $\frac{p _{2}}{p _{1}} = 1 + \frac{2.8}{2.4} (1.50 1^{2} - 1) = 1 + 1.1667 (1.253) = 2.462.$ Step D — downstream Mach. Pehle normal-shock relation se $M_{n 2}$ : $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{0.4 ( 2.253 ) + 2}{2.8 ( 2.253 ) - 0.4} = \frac{2.901}{5.908} = 0.4911,$ toh $M_{n 2} = 0.4911 = 0.7008$ . Phir full downstream Mach paane ke liye geometry undo karo: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.7008}{s i n ( 36. 9 ^{\circ} - 1 5 ^{\circ} )} = \frac{0.7008}{s i n 21. 9 ^{\circ}} = \frac{0.7008}{0.3730} = 1.879.$ Result: $β \approx 36. 9^{\circ}$ , $M_{n 1} \approx 1.50$ , $p_{2} / p_{1} \approx 2.46$ , $M_{n 2} \approx 0.701$ , $M_{2} \approx 1.88$ (abhi bhi supersonic). Yeh poora wedge-flow recipe end to end hai.

L4.2 — Compression vs expansion sign check

Problem. L4.1 mein flow apne andar turn hui aur pressure badha. Agar wall ki jagah door (ek convex corner) $1 5^{\circ}$ se turn karti, toh kya shock banta? Woh case kaun si physics govern karta hai?

Recall Solution

Ek convex (expansion) corner flow ko apne aap se door turn karta hai. Information escape kar sakti hai, toh koi shock nahi banta — flow Mach waves ke ek fan ke through smoothly accelerate karta hai (upar figure mein drawn hai: corner se fanout karte rays, har ek ek weak Mach line, gradually flow bend aur speed karta hai). Pressure aur temperature girte hain, Mach number badhta hai. Yeh Prandtl-Meyer Expansion hai, oblique shock ka mirror-image process: ek abrupt line ki jagah, turn poore fan mein smoothly share hoti hai. θ-β-M relation sirf concave/compression turns par apply hota hai jahan flow apne andar force hoti hai.

Level 5 — Mastery

L5.1 — Detachment threshold

Problem. $M_{1} = 2$ ke liye, maximum deflection $θ_{ma x} \approx 22.9 7^{\circ}$ hai. Is flow mein $2 5^{\circ}$ half-angle wala ek wedge rakha jaata hai. Kya hota hai, aur θ-β-M relation "fail" kyun karta hai jawab dene mein?

Recall Solution

Kyunki $2 5^{\circ} > θ_{ma x} = 22.9 7^{\circ}$ hai, wedge itna turning maang raha hai jo koi bhi attached oblique shock supply nahi kar sakta. θ-β-M equation ka koi real root nahi $β$ ke liye — algebraically required RHS $tan 2 5^{\circ}$ tak nahi pahunch sakti kisi bhi $β$ ke liye (poora hump $2 5^{\circ}$ se neeche peak karta hai). Physically shock detach ho jaata hai aur body ke saamne ek curved bow shock ke roop mein khada rehta hai: centreline ke karib yeh essentially ek normal shock hai (peeche subsonic), Mach wave tak curve karta hua far out. Formula ki "failure" hai mathematics ka sahi se batana ki assumed attached-shock geometry exist nahi kar sakti.

L5.2 — $θ_{ma x}$ speed-dependent hai

Problem. Qualitatively explain karo ki $θ_{ma x}$ kyun badhta hai jaise $M_{1}$ badhta hai (jaise $θ_{ma x} \approx 22.9 7^{\circ}$ $M_{1} = 2$ par lekin $\approx 34.0 7^{\circ}$ $M_{1} = 3$ par), aur iska wedge design ke liye kya matlab hai higher speeds par.

Recall Solution

Faster flow shock ke normal zyada momentum carry karti hai, toh geometry impossible hone se pehle use ek sharper turn se force kiya ja sakta hai. Isliye θ-β curve mein hump taller hota jaata hai jaise $M_{1}$ badhta hai: $θ_{ma x}$ badhta hai ( $M_{1} \to \infty$ hone par approximately $45. 6^{\circ}$ tak approach karta hai). Yeh seedha θ-β figure se padha ja sakta hai — " $M_{1} = 3$ " labelled curve " $M_{1} = 2$ " wale se zyada high peak karti hai. Design meaning: ek wedge jo low supersonic speed par shock detach kar deta woh same geometry higher Mach par clean attached shock ke saath run kar sakta hai — toh same geometry flight envelope mein differently behave karti hai.

L5.3 — Double-wedge / multi-shock reasoning

Problem. $M_{1} = 3$ par flow ek $2 0^{\circ}$ wedge se guzarti hai (parent ke Worked Example 1 se, jo $β = 37. 8^{\circ}$ , $M_{2} \approx 1.99$ deta hai). Downstream flow ko ek doosra $1 0^{\circ}$ compression turn milta hai. Setup karo (fully invert mat karo) ki tum doosra shock angle $β_{2}$ kaise dhundhoge, aur ek subtlety note karo.

Recall Solution

Setup. Doosre shock ke liye upstream Mach pehle shock ke baad local Mach hai, $M_{2} \approx 1.99$ , nahi $M_{1} = 3$ . θ-β-M dobara apply karo $M_{2}$ aur $θ_{2} = 1 0^{\circ}$ ke saath: $tan 1 0^{\circ} = 2 cot β_{2} \frac{1.9 9 ^{2} s i n ^{2} β _{2} - 1}{1.9 9 ^{2} ( 1.4 + c o s 2 β _{2} ) + 2},$ aur weak (chhote- $β_{2}$ wala) root lo, $μ_{2} = arcsin (1/1.99) = 30. 2^{\circ}$ aur doosre hump ke peak ke beech bracket karke. Numerically $β_{2} \approx 34. 3^{\circ}$ . Subtlety. $β_{2}$ pehle turn ke baad local flow direction ke relative measure kiya jaata hai, jo khud original freestream se already $2 0^{\circ}$ deflect hua hai. Lab frame mein shock draw karne ke liye woh $2 0^{\circ}$ wapas add karna padega. Yeh bhi check karo ki $M_{2} > 1$ abhi bhi hold karta hai aur $θ_{2} < θ_{ma x} (M_{2})$ toh doosra shock attached rahe (yahan $θ_{ma x} (1.99) \approx 22. 9^{\circ} > 1 0^{\circ}$ hai, toh rehta hai).

Recall Self-test summary (finish karne ke baad reveal karo)

$β$ ko kya bound karta hai? ::: $arcsin (1/ M_{1}) \leq β \leq 9 0^{\circ}$ . Normal-shock laws mein kaun sa Mach enter karta hai? ::: $M_{n 1} = M_{1} sin β$ . $M_{n 2}$ se $M_{2}$ kaise recover karo? ::: $M_{2} = M_{n 2} / sin (β - θ)$ . Agar $θ > θ_{ma x}$ toh kya? ::: Koi attached shock nahi — ek detached bow shock form hota hai. Weak vs strong root? ::: Chhota $β$ = weak (peeche aksar supersonic); bada $β$ = strong (peeche subsonic).