Exercises — Critical Mach number — onset of local supersonic flow

(a) False. Poora aircraft abhi bhi subsonic hai; sirf curved surface ke upar ek local patch sonic reach kiya hai. Air curvature par accelerate karti hai (thumb-on-hose picture), isliye $M_{local}>M_{\infty }$. Toh $M_{cr}<1$ hamesha hota hai — typically $0.7$–$0.85$. (b) True. Exactly $M_{\infty }=M_{cr}$ par, woh single lowest-pressure point (suction peak) abhi $M_{local}=1$ touch karta hai. Iske neeche: fully subsonic. Iske upar: ek badhta hua supersonic pocket. (c) False. $C_{p,cr}$ curve purely isentropic gas dynamics se aati hai — isme sirf $\gamma$ aur $M_{cr}$ hai. Air mein har airfoil ek hi same $C_{p,cr}$ curve share karta hai. Airfoils ke beech jo differ karta hai woh hai suction curve $C_{p,\min }$.

• Blue, $C_{p,cr}$: universal. Har $M_{\infty }$ par iska value woh pressure coefficient hai jise ek point ko sonic hone ke liye zaroor reach karna hoga. Yeh $M_{\infty }$ badhne par kam negative hoti hai (ek faster free stream ko Mach 1 locally hit karne ke liye ek chhote pressure dip ki zaroorat hai).
• Yellow, $C_{p,\min }$: airfoil-specific. Yeh airfoil ke incompressible suction value $C_{p,0}$ se shuru hoti hai aur $M_{\infty }$ badhne par $1/\sqrt{1-M_{\infty }^2}$ factor se zyada negative hoti jaati hai kyunki compressibility suction peak ko tighten karti hai. Crossing hi $M_{cr}$ hai.

Step 1 — inner ratio. $\frac{1+0.2(0.75^2)}{1.2}=\frac{1+0.2(0.5625)}{1.2}=\frac{1.1125}{1.2}=0.92708$. Kyun: yeh $p_{\infty }/p^{\ast }$ hai — free-stream pressure sonic pressure $p^{\ast }$ ke against measure kiya gaya, jo upar define kiya gaya (streamline par matched stagnation pressure). Step 2 — isentropic power. $0.92708^{3.5}=0.76505$. Step 3 — assemble karo. $C_{p,cr}=\frac{2}{1.4(0.5625)}(0.76505-1)=\frac{2}{0.7875}(-0.23495)=-0.5967$. Answer: $C_{p,cr}\approx -0.597$.

Step 1. $\frac{1+0.2(0.36)}{1.2}=\frac{1.072}{1.2}=0.89333$. Step 2. $0.89333^{3.5}=0.67243$. Step 3. $C_{p,cr}=\frac{2}{1.4(0.36)}(0.67243-1)=\frac{2}{0.504}(-0.32757)=-1.3000$. Answer: $C_{p,cr}\approx -1.300$. Indeed $-1.300<-0.597$: lower free-stream Mach par local sonic reach karne ke liye ek deeper pressure dip chahiye. Yeh blue curve ke downward-left rise ko confirm karta hai.

$C_{p,\min }=\frac{-0.43}{\sqrt{1-0.65^2}}=\frac{-0.43}{\sqrt{1-0.4225}}=\frac{-0.43}{\sqrt{0.5775}}=\frac{-0.43}{0.75993}=-0.5658$. Answer: $C_{p,\min }\approx -0.566$. Suction peak purely compressibility ki wajah se $-0.43$ se $-0.566$ tak gehra ho gaya. Dekho Prandtl–Glauert Compressibility Correction.

Humein chahiye $f(M)=\frac{-0.55}{\sqrt{1-M^2}}-C_{p,cr}(M)=0$.

• $M=0.70$: suction $=-0.55/\sqrt{0.51}=-0.7702$; required $C_{p,cr}(0.70)=-0.7790$. Phir $f=-0.7702-(-0.7790)=+0.0088>0$ → suction abhi itni deep nahi, abhi bhi subsonic.
• $M=0.71$: suction $=-0.55/\sqrt{0.4959}=-0.7810$; $C_{p,cr}(0.71)=-0.7376$. Phir $f=-0.7810-(-0.7376)=-0.0434<0$ → sonic se past. $f$ $0.70$ aur $0.71$ ke beech $+$ se $-$ flip karta hai, toh root (crossing) wahan hai, $0.70$ ke paas. Answer: $M_{cr}\approx 0.702$.

Prediction: thinner ⇒ gentler curvature ⇒ chhhota suction ⇒ iska yellow curve upar baithta hai (kam negative) ⇒ universal curve se milne se pehle ise aur right tak jaana padega ⇒ $M_{cr}$ rise karega. Numerics. $f(M)=-0.35/\sqrt{1-M^2}-C_{p,cr}(M)=0$ solve karo.

• $M=0.76$: suction $=-0.35/\sqrt{0.4224}=-0.5385$; $C_{p,cr}(0.76)=-0.5583$. Phir $f=-0.5385-(-0.5583)=+0.0198>0$ → abhi sonic nahi.
• $M=0.78$: suction $=-0.35/\sqrt{0.3916}=-0.5593$; $C_{p,cr}(0.78)=-0.4952$. Phir $f=-0.5593-(-0.4952)=-0.0641<0$ → sonic se past. $f$ $0.76$ aur $0.78$ ke beech sign flip karta hai, toh crossing wahan hai, $M_{cr}\approx 0.765$ ke paas. Answer: $M_{cr}\approx 0.77$ — A ke $0.702$ se zyada, jaise predict kiya. Exactly isiliye fast jets thin wings use karte hain.

$f(M)=-0.05/\sqrt{1-M^2}-C_{p,cr}(M)$ banao.

• $M=0.90$: suction $=-0.05/\sqrt{0.19}=-0.1147$; $C_{p,cr}(0.90)=-0.1876$. $f=+0.0729>0$.
• $M=0.95$: suction $=-0.05/\sqrt{0.0975}=-0.1601$; $C_{p,cr}(0.95)=-0.0863$. $f=-0.0738<0$. Sign flip hota hai, lekin sirf $M_{\infty }=1$ ke bahut paas: solve karne par $M_{cr}\approx 0.93$ milta hai. Edge-case lesson: itne thin section ke liye suction itni weak hai ki tiny suction peak ke Mach 1 reach karne se pehle free stream ko almost sonic push karna padta hai. Limit mein $C_{p,0}\to 0^{-}$ (zero incidence par flat plate — na thickness, na suction), yellow curve axis se chipki rehti hai, aur crossing $M_{cr}\to 1$ tak push ho jaata hai: aisa body kabhi local supersonic pocket develop nahi karta poora flow sonic hone se pehle. Toh $M_{cr}$ upar se $1$ tak bounded hai, jo sirf tab approach hota hai jab suction vanish hoti hai.

• $M=0.40$: suction $=-1.60/\sqrt{0.84}=-1.7457$; $C_{p,cr}(0.40)=-3.6607$. $f=+1.915>0$ → abhi sonic nahi.
• $M=0.55$: suction $=-1.60/\sqrt{0.6975}=-1.9158$; $C_{p,cr}(0.55)=-1.6606$. $f=-1.9158-(-1.6606)=-0.2552<0$ → sonic se past. Sign flip $0.40$ aur $0.55$ ke beech → $M_{cr}\approx 0.52$. Edge-case lesson: bahut gehra suction ($|C_{p,0}|$ bada) crossing ko low $M_{\infty }$ par kheench laata hai. Extreme mein, agar incompressible suction bahut low $M_{\infty }$ par $C_{p,cr}$ value se pehle hi exceed kar le, toh section fly karte hi almost critical hai — aisi shapes high-speed flight ke liye useless hain. Toh crossing ke useful $0.5$–$0.9$ band se bahar jaane ke do tarike hain: (i) suction itni weak ki crossing $M_{\infty }\to 1$ ki taraf race kare (L3.3), aur (ii) suction itni strong ki crossing low $M_{\infty }$ ki taraf collapse kare (yahaan).

Idea. Sirf woh velocity component jo leading edge ke normal ho, "airfoil dekhta hai." Sweep streamwise speed chhupa leta hai: effective normal Mach $M_{\infty }\cos \Lambda$ hai. Airfoil tab critical jaata hai jab yeh normal component straight-wing critical value ke barabar ho jaata hai: $M_{cr,\text{straight}}=M_{cr,\text{swept}}\cos \Lambda \Rightarrow M_{cr,\text{swept}}=\frac{M_{cr,\text{straight}}}{\cos \Lambda }.$ Compute karo. $\cos 30^{\circ }=0.86603$, toh $M_{cr,\text{swept}}=\frac{0.72}{0.86603}=0.8314$. Answer: $M_{cr}\approx 0.831$. Sweep ne roughly $+0.11$ ka critical Mach kharida — dekho Swept Wings & Transonic Design.

Sirf thinning: airfoil B deta hai $M_{cr}=0.77<0.80$ — falls short. A ko $25^{\circ }$ sweep karna: $\cos 25^{\circ }=0.90631$, toh $M_{cr}=0.702/0.90631=0.7746<0.80$ — yeh bhi falls short. Koi bhi single change $0.80$ nahi reach karta. Combine karo: airfoil B ko $25^{\circ }$ sweep karo: $M_{cr}=0.77/0.90631=0.8496$. Answer: koi bhi single change akele kaafi nahi; thin + swept combine karna $0.80$ comfortably clear karta hai ($\approx 0.85$). Real transonic wings exactly yahi karte hain — plus fuselage par Area Rule (Transonic).

Pehle bracket mein $M_{cr}=1$ set karo. Inner ratio ban jaata hai $\frac{1+0.2(1)}{1.2}=\frac{1.2}{1.2}=1$. $3.5$ tak raise karo: abhi bhi $1$. Toh bracket hai $1-1=0$. Prefactor. $\frac{2}{\gamma M_{cr}^2}=\frac{2}{1.4}$ finite rehta hai. Finite $\times$ zero $=0$. ${\lim }_{M_{cr}\to 1}{C}_{p,cr}=0.$ Neeche se approach ($M_{cr}<1$): inner ratio $<1$ hai, iska $3.5$ power $<1$ hai, toh bracket negative hai → $C_{p,cr}\to 0^{-}$. Physical meaning: jab free stream khud almost sonic ho, surface point ko Mach 1 reach karne ke liye barely accelerate karna padta hai — toh almost koi bhi pressure dip ki zaroorat nahi, $C_{p,cr}\to 0$. Blue curve $M_{\infty }=1$ par horizontal axis se milti hai, jo figure mein dikhta hai.

Crossing par, dono curves equal hain: $\frac{C_{p,0}}{\sqrt{1-M_{cr}^2}}=C_{p,cr}(M_{cr})$, toh $C_{p,0}=C_{p,cr}(M_{cr})\sqrt{1-M_{cr}^2}.$ $C_{p,cr}(0.68)$ compute karo. Inner ratio $=\frac{1+0.2(0.4624)}{1.2}=\frac{1.09248}{1.2}=0.9104$. Power: $0.9104^{3.5}=0.71611$. Phir $C_{p,cr}=\frac{2}{1.4(0.4624)}(0.71611-1)=\frac{2}{0.64736}(-0.28389)=-0.8771$. $\sqrt{1-0.68^2}$ se multiply karo. $\sqrt{1-0.4624}=\sqrt{0.5376}=0.73321$. Toh $C_{p,0}=-0.8771\times 0.73321=-0.6431$. Answer: $C_{p,0}\approx -0.643$. Airfoil ka incompressible suction peak approximately $-0.643$ se zyada deep nahi design karna chahiye taaki $M_{cr}=0.68$ hold kare.

Sweep rule rearrange karo. $M_{cr,\text{swept}}=\frac{M_{cr,\text{straight}}}{\cos \Lambda }\Rightarrow \cos \Lambda =\frac{M_{cr,\text{straight}}}{M_{cr,\text{swept}}}=\frac{0.702}{0.80}=0.8775$. Invert karo. $\Lambda =\arccos (0.8775)=28.71^{\circ }$. Answer: $\Lambda \approx 28.7^{\circ }$. $29^{\circ }$ ke paas ka sweep airfoil A ka critical Mach $0.702$ se $0.80$ tak lift karta hai — real transonic transports aur unhe jis Drag Divergence Mach Number margin ki zaroorat hai, uske consistent.

• L1: $M_{cr}<1$; blue curve universal, yellow curve airfoil-specific; answer = intersection.
• L2: $C_{p,cr}$ mein plug karo; $0.2$ factor se pehle $M$ ko square karo; $p^{\ast }$ = sonic-point pressure.
• L3: $f=$ (suction − requirement) ki sign flip se root bracket karo — intermediate-value guarantee; crossing $1$ ki taraf escape kar sakti hai (too thin) ya $0$ ki taraf (too cupped).
• L4: thinning $M_{cr}$ lift karta hai; sweep $1/\cos \Lambda$ se multiply karta hai; transonic wings ke liye combine karo.
• L5: limits ($C_{p,cr}\to 0^{-}$ at $M_{\infty }\to 1$), inverse design ($C_{p,0}=C_{p,cr}\sqrt{1-M_{cr}^2}$), sweep sizing.