Oblique shock waves — θ-β-M relation — Hinglish notes

3.1.13 · Physics › Compressible Flow & Aerodynamics

WHAT hai ek oblique shock, physically?

Key trick (WHY it works): upstream velocity ko shock ke normal aur tangential components mein decompose karo.

Normal component $u_{n 1} = V_{1} sin β$ → behave karta hai jaisa ek normal shock effective Mach number ke saath $M_{n 1} = M_{1} sin β$
Tangential component $w = V_{1} cos β$ → shock ke across unchanged rehta hai.

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

HOW derive karein θ-β-M relation scratch se

Hum sab kuch velocity triangle se build karte hain. Upstream flow wall ke along hai (shock ke saath angle $β$ ). Downstream yeh $θ$ se deflect hoti hai, toh yeh shock ke saath angle $(β - θ)$ banati hai.

Step 1 — Geometry se Normal components. $u_{n 1} = V_{1} sin β, u_{n 2} = V_{2} sin (β - θ)$ Yeh step kyun? Downstream flow shock ke saath angle $(β - θ)$ banati hai, toh uska normal component us angle ka use karta hai.

Step 2 — Tangential component conserved hai. $w = V_{1} cos β = V_{2} cos (β - θ)$ Yeh step kyun? Koi tangential force nahi (upar prove kiya), toh dono sides par same $w$ aata hai.

Step 3 — Ratio banao ( $V$ eliminate karo). Normal equations divide karo aur Step 2 use karo:

= \frac{\cos\beta}{\cos(\beta-\theta)}\cdot\frac{\sin(\beta-\theta)}{\sin\beta} = \frac{\tan(\beta-\theta)}{\tan\beta}$$ *Yeh step kyun?* Humne $V_2/V_1$ ko tangential identity $V_2 = V_1\cos\beta/\cos(\beta-\theta)$ se replace kiya. Ab sab kuch angles mein hai. **Step 4 — Normal-shock density (velocity) ratio use karo.** Ek normal shock ke across mass conservation + Rankine–Hugoniot dete hain $$\frac{u_{n2}}{u_{n1}} = \frac{\rho_1}{\rho_2} = \frac{(\gamma-1)M_{n1}^2 + 2}{(\gamma+1)M_{n1}^2}, \qquad M_{n1}=M_1\sin\beta$$ *Yeh step kyun?* Normal direction exactly normal-shock laws follow karta hai — decompose karne ka yahi toh poora point tha. **Step 5 — Steps 3 aur 4 equate karo:** $$\frac{\tan(\beta-\theta)}{\tan\beta} = \frac{(\gamma-1)M_1^2\sin^2\beta + 2}{(\gamma+1)M_1^2\sin^2\beta}$$ **Step 6 — $\tan\theta$ ke liye solve karo** ($\tan(\beta-\theta)$ expansion wali algebra se) standard explicit form paane ke liye: > [!formula] The θ-β-M relation > $$\boxed{\;\tan\theta = 2\cot\beta\,\dfrac{M_1^2\sin^2\beta - 1}{M_1^2(\gamma + \cos 2\beta) + 2}\;}$$ > - $\theta$: flow deflection, $\beta$: shock angle, $M_1$: upstream Mach, $\gamma$: ratio of specific heats (air ke liye $1.4$). > - $\beta = 90^\circ$ par **normal shock** ban jaata hai (tab $\tan\theta = 0 \Rightarrow \theta=0$, koi deflection nahi). > - Real shock ke liye $\sin\beta > 1/M_1$ zaroori hai (yani $M_{n1}>1$). $\beta=\mu=\arcsin(1/M_1)$ par, $\theta=0$ — yeh ek **Mach wave** hai. --- ## Curve padhna (isko use karne ka 80/20) Fixed $M_1$ ke liye, $\theta$ vs $\beta$ plot karne par ek **hump** milta hai: - Har $\theta < \theta_{max}$ ke liye **DO** solutions hote hain: - **Weak shock** (chhota $\beta$): downstream usually abhi bhi supersonic — *nature normally yahi choose karti hai*. - **Strong shock** (bada $\beta$): downstream subsonic. - $\theta_{max}$ = hump ki peak. Agar wedge $\theta > \theta_{max}$ demand kare, toh **koi attached oblique shock exist nahi karta** → shock **detach** ho jaata hai aur ek curved **bow shock** ban jaata hai jo body ke saamne khada rehta hai. > [!example] Worked Example 1 — wedge ke liye $\beta$ find karo > Air ($\gamma=1.4$), $M_1 = 3$, half-angle wedge $\theta = 20^\circ$. Weak-shock angle find karo. > > **Why:** wedge $\theta=20^\circ$ force karta hai; hum θ-β-M ko $\beta$ ke liye invert karte hain. > $\tan 20^\circ = 2\cot\beta\,\dfrac{9\sin^2\beta-1}{9(1.4+\cos2\beta)+2}$ mein plug karo aur numerically solve karo. > **Result:** $\beta \approx 37.8^\circ$ (weak). *Yeh root kyun?* Yeh dono mein se chhota hai; strong wala $\approx 82^\circ$ hai. > Phir $M_{n1}=3\sin37.8^\circ = 1.84$ → $p_2/p_1$, $M_{n2}$ ke liye normal-shock tables use karo, aur finally $M_2 = M_{n2}/\sin(\beta-\theta)$. > [!example] Worked Example 2 — Mach wave limit > $M_1 = 2$, $\theta \to 0$. **Why:** infinitesimally weak disturbance. > Numerator $M_1^2\sin^2\beta-1=0$ set karne par $\sin\beta = 1/M_1 = 0.5 \Rightarrow \beta=30^\circ$ milta hai. > *Interpretation:* yeh **Mach angle** $\mu=\arcsin(1/M)$ hai — sabse weakest oblique "shock" ka limiting case. > [!example] Worked Example 3 — downstream Mach number > Example 1 se: $M_{n1}=1.84 \Rightarrow M_{n2}\approx 0.608$ (normal-shock relation). > $$M_2 = \frac{M_{n2}}{\sin(\beta-\theta)} = \frac{0.608}{\sin(37.8^\circ-20^\circ)} = \frac{0.608}{\sin17.8^\circ}\approx 1.99.$$ > *$\sin(\beta-\theta)$ se divide kyun karte hain?* Kyunki $u_{n2}=V_2\sin(\beta-\theta)$, toh Mach version us geometry ko invert karta hai. Note karo $M_2>1$: yeh weak shock supersonic rehti hai. --- ## Common mistakes (Steel-manned) > [!mistake] "Normal-shock formula mein directly $M_1$ use karo." > **Kyun sahi lagta hai:** tum ek shock cross kar rahe ho, aur normal-shock formulas upstream Mach use karte hain. **Fix:** sirf **normal component** $M_{n1}=M_1\sin\beta$ shock ko normal shock ki tarah cross karta hai. Poora $M_1$ compression ko overstate karta hai. > [!mistake] "Shock ke baad downstream flow hamesha subsonic hoti hai." > **Kyun sahi lagta hai:** normal shocks ke liye true hai. **Fix:** **weak** oblique shock ke liye $M_2$ aksar abhi bhi **supersonic** rehta hai — sirf $M_{n2}<1$ guaranteed hai. > [!mistake] "Har deflection ke liye ek hi shock angle hota hai." > **Kyun sahi lagta hai:** geometry unique lagti hai. **Fix:** θ-β curve double-valued hai — **weak aur strong** roots hote hain. Context se specify karo (free flight → weak). > [!mistake] "Bada wedge → hamesha bada shock." > **Kyun sahi lagta hai:** zyada turning = zyada compression. **Fix:** sirf $\theta_{max}$ tak. Usse aage, shock **detach** ho jaata hai (bow shock) aur attached-shock formula ka koi real solution nahi hota. --- > [!recall]- Feynman: ek 12-saal ke bachche ko explain karo > Socho tum tez skating kar rahe ho aur ek tilted wall se takra rahe ho. Tumhari motion ka woh hissa jo wall ke *seedha andar* ja raha hai woh hard rok jaata hai (woh "shock" hai), lekin woh hissa jo wall ke *saath slide* kar raha hai woh same chalta rehta hai. Kyunki wall tilted hai, tum poori tarah nahi rukते — bas thoda sideways bend ho jaate ho aur slow ho jaate ho. Agar tum wall ko apne skates se handle ho sake se zyada steep karo, tum usse bilkul follow nahi kar sakte aur aage squished air ka ek bada curved cushion ban jaata hai. θ-β-M rule bas woh math hai jo kehta hai: "is speed aur is turn angle ke liye, shock exactly itni tilted honi chahiye." > [!mnemonic] Geometry yaad rakho > **"Normal Normally, Tangent Tags Along."** > **Normal** component **N**ormal-shock physics karta hai; **tangential** component conserved hoti hai (saath tags along, unchanged). Aur **"Weak is what nature seeks."** > [!recall] Active recall checkpoints > - Oblique shock ke across kya constant rehta hai? *(tangential velocity)* > - Normal-shock laws mein kaun sa Mach number jaata hai? *($M_{n1}=M_1\sin\beta$)* > - Jab $\theta>\theta_{max}$ ho toh kya hota hai? *(detached bow shock)* --- ### #flashcards/physics Oblique shock par flow ko hum kaun se do velocity components mein decompose karte hain? ::: Normal (shock ke perpendicular) aur tangential (shock ke along). Oblique shock ke across tangential velocity unchanged kyun rehti hai? ::: Shock face ke along koi tangential pressure force exist nahi karta, toh tangential momentum conserved rehta hai. Normal-shock physics kaun sa effective Mach number govern karta hai? ::: $M_{n1}=M_1\sin\beta$. θ-β-M relation state karo. ::: $\tan\theta = 2\cot\beta\,\dfrac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$. $\beta=90^\circ$ oblique shock ko kya reduce kar deta hai? ::: Ek normal shock (deflection $\theta=0$). Real oblique shock ke liye minimum $\beta$ kya hai? ::: Mach angle $\beta=\arcsin(1/M_1)$, jahan $M_{n1}=1$ aur $\theta=0$. Ek given $\theta<\theta_{max}$ ke liye kitne shock-angle solutions exist karte hain? ::: Do — ek weak (chhota $\beta$) aur ek strong (bada $\beta$) shock. External flow mein nature usually kaun sa solution choose karti hai? ::: Weak shock. Weak oblique shock ke liye downstream flow subsonic hoti hai kya? ::: Usually nahi — $M_2$ supersonic rehta hai, lekin $M_{n2}<1$ hamesha hota hai. Jab wedge angle $\theta_{max}$ se exceed ho toh kya hota hai? ::: Koi attached oblique shock nahi; body ke aage ek detached curved bow shock form hoti hai. $M_{n2}$ find karne ke baad $M_2$ kaise recover karte hain? ::: $M_2=M_{n2}/\sin(\beta-\theta)$. --- ### Connections - [[Normal Shock Waves]] — building block ($M_{n1}$ closure). - [[Mach Angle and Mach Waves]] — $\theta\to0$ limit. - [[Prandtl-Meyer Expansion]] — opposite case (flow door mudna, expansion fan). - [[Detached Bow Shocks]] — $\theta_{max}$ ke baad kya hota hai. - [[Rankine-Hugoniot Relations]] — Step 4 mein density/velocity ratio ka source. - [[Supersonic Wedge and Cone Flow]] — direct engineering application. ## 🖼️ Concept Map ```mermaid flowchart TD M1[Supersonic flow M1] -->|apni taraf muda| WEDGE[Wedge compression corner] WEDGE -->|flow ko theta se deflect kiya| OBS[Oblique shock at angle beta] OBS -->|velocity decompose karo| NORM[Normal component un1] OBS -->|velocity decompose karo| TANG[Tangential component w] NORM -->|behave karta hai jaisa| NS[Normal shock Mn1 = M1 sin beta] TANG -->|koi tangential force nahi| CONS[w conserved across shock] NS -->|geometry plus Rankine-Hugoniot| REL[theta-beta-M relation] CONS -->|tan ratio identity| REL REL -->|shock angle beta deta hai| DOWN[Downstream properties] DOWN -->|yield karta hai| PROPS[Pressure temperature M2] ```