Critical Mach number — onset of local supersonic flow: Worked examples

3.1.24 · D3 · Physics › Compressible Flow & Aerodynamics › Critical Mach number — onset of local supersonic flow

Is page par critical Mach number ko har angle se drill kiya gaya hai jis tarah se woh problem set mein aa sakta hai: forward direction ( $M_{cr} \to C_{p, cr}$ ), backward direction ( $C_{p, 0} \to M_{cr}$ ), curve ke do limiting ends, degenerate inputs, ek real-world word problem, aur ek exam-style twist. Examples se pehle, hum ek scenario matrix banate hain taaki aap exactly dekh sakein ki har problem kaun sa "cell" fill karti hai — aur confirm karein ki koi case chhuta nahi.

Sab kuch do cheezein par tika hua hai jo aap parent note mein pehle hi dekh chuke hain. Hum unhe yahan dobara likhte hain taaki koi symbol use karne se pehle anchor na ho.

Definition Is page ke symbols, simple words mein

Inhe ek baar padhein — baad ki har line inhi ko reuse karti hai aur kuch nahi.

$M_{\infty}$ (bolo "M-infinity") — free-stream Mach number: poore aircraft ki flight speed ko local speed of sound se divide karo. Yeh batata hai aap kitni tez ud rahe hain.
$M_{cr}$ — $M_{\infty}$ ki critical value: woh ek khaas flight speed jis par wing ka sabse tez point pehli baar sound tak pahunchta hai. Isliye $M_{cr}$ , $M_{\infty}$ ki ek special reading hai, koi alag quantity nahi. Numerically $M_{cr} < 1$ hamesha.
$M_{l oc a l}$ — wing ki skin par ek single point par Mach number; yeh $M_{\infty}$ se zyada ho sakta hai kyunki curvature par hawa tez ho jaati hai.
$p$ — ek point par static pressure; $p_{\infty}$ — bahut dur upstream (undisturbed air) par static pressure; $p_{0}$ — stagnation pressure, woh pressure jo hawa pahunchti agar use smoothly rest par laaya jaaye (isentropic flow mein ek streamline ke along constant, per Isentropic Flow Relations); $p^{*}$ — us point par static pressure jahan $M_{l oc a l} = 1$ ho (woh "sonic" ya "starred" state).
$γ$ — ratio of specific heats (air ke liye $1.4$ , helium ke liye $5/3$ ).
$Λ$ (capital "lambda") — wing ki leading edge ka sweep angle, flow ke across straight se peeche ki taraf measure kiya gaya. $Λ = 0$ ek unswept wing hai; bada $Λ$ matlab wing aur peeche ki taraf angled hai. Yeh sirf Example 7 mein aata hai.

Definition Do curves, words mein

$C_{p, cr} (M_{cr})$ — gas-dynamics curve. "Surface pressure kitna neeche dip karna chahiye taaki wing ka sabse tez point just sound tak pahunche?" Yeh sirf $γ$ par depend karta hai. $C_{p, cr} = \frac{2}{γ M _{cr}^{2}} [(\frac{1 + \frac{γ - 1}{2} M _{cr}^{2}}{1 + \frac{γ - 1}{2}})^{\frac{γ}{γ - 1}} - 1]$
$C_{p, m i n} (M_{\infty}) = \frac{C _{p, 0}}{1 - M _{\infty}^{2}}$ — airfoil curve (ek Prandtl–Glauert estimate). "Speed $M_{\infty}$ par IS wing ka actual sabse kam pressure coefficient kya hai?" Yeh shape par depend karta hai ek single fixed number $C_{p, 0} < 0$ ke through.

Notation note: $C_{p, m i n}$ (airfoil ka lowest $C_{p}$ ek given speed par) aur $C_{p, cr}$ (gas-dynamics ki requirement) do alag curves hain. Yeh sirf crossing point par milte hain, aur woh crossing speed $M_{cr}$ hai. Hum neeche kahi bhi inhe alag abbreviate nahi karte.

Critical Mach number wahan hai jahan yeh do curves cross karti hain. Gas-dynamics requirement airfoil reality se milti hai.

Scenario matrix

Is topic ki har problem inhi cells mein se ek hoti hai. Neeche har example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Kya vary karta hai	Trap / lesson	Example
A Forward	Given $M_{cr}$ → find $C_{p, cr}$	Bas plug in karo — lekin saare digits rakho	Ex 1
B Backward	Given $C_{p, 0}$ → find $M_{cr}$	Do curves ko numerically intersect karna hoga	Ex 2
C Low-Mach limit	$M_{cr} \to 0$	$C_{p, cr} \to - \infty$ (infinite suction chahiye)	Ex 3
D High-Mach limit	$M_{cr} \to 1$	$C_{p, cr} \to 0$ (koi suction nahi chahiye)	Ex 3
E Degenerate airfoil	$C_{p, 0} = 0$ (flat plate at $0°$ )	Koi suction nahi ⇒ $M_{cr} = 1$	Ex 4
F Sign check	$C_{p, cr}$ hamesha $< 0$ ?	Haan $M_{cr} < 1$ ke liye; sign machinery verify karo	Ex 5
G Real-world word	Sea-level jet, altitude, speed	Compare karne se pehle speed → $M_{\infty}$ convert karo	Ex 6
H Design twist	Thinner / swept wing	$M_{cr}$ kis direction mein move karta hai?	Ex 7
I Non-air gas	$γ \neq = 1.4$ (e.g. helium $γ = 5/3$ )	"Universal" curve $γ$ ke saath shift karta hai	Ex 8

Neeche ki figure poora game board hai — do curves aur unka crossing. Har example uski ek move hai; Example 2 use detail mein walk karta hai.

Figure — Critical Mach number — onset of local supersonic flow

Intuition Figure kya dikhata hai (isko padhein chahe aap use dekh nahi sakte)

Horizontal axis flight speed $M_{\infty}$ hai ( $0.30$ se lekar lagbhag $1$ tak); vertical axis pressure coefficient $C_{p}$ hai, neeche negative (suction) values mein jaata hua. Do curves drawn hain:

Orange curve $C_{p, cr} (M_{\infty})$ hai, gas-dynamics ki requirement. Yeh left par bahut neeche se start hoti hai ( $M_{\infty} \to 0$ par $- \infty$ ki taraf plunges) aur smoothly rise karke right end par $0$ ko touch karti hai ( $M_{\infty} = 1$ ).
Teal curve airfoil ka khud ka minimum pressure $C_{p, 0} / 1 - M_{\infty}^{2}$ hai $C_{p, 0} = - 0.43$ ke liye (Example 2 ki wing). Yeh left par $- 0.43$ ke paas start hoti hai aur $M_{\infty}$ badhne par steadily sink karti hai.
Ek plum dot mark karta hai jahan orange aur teal cross karti hain, $M_{\infty} \approx 0.727$ par. Woh crossing speed hi $M_{cr}$ hai. Dot ke right mein teal (airfoil) line orange (requirement) line ke neeche dip karti hai — matlab wing kisi jagah already supersonic ho gayi hai. Left mein, airfoil requirement ke upar rehta hai — skin par abhi fully subsonic. Figure ke labels dono limits ko call out karte hain (left par $- \infty$ , $M_{\infty} = 1$ par $0$ ).

Worked examples

Cell A — forward direction

Worked example Example 1 — Given

$M_{cr}$ se $C_{p, cr}$ compute karo Given $M_{cr} = 0.75$ , $γ = 1.4$ . $C_{p, cr}$ find karo.

Forecast: Parent ke Example 1 mein, $M_{cr} = 0.70$ ne $C_{p, cr} \approx - 0.79$ diya. Humne $M_{cr}$ thoda badhaya. Kya aapko lagta hai $C_{p, cr}$ zyada negative hoga ya kam negative? (Padhne se pehle guess karo.)

Step 1. Inner ratio (numerator = free-stream side, denominator = sonic side): $\frac{1 + 0.2 ( 0.7 5 ^{2} )}{1 + 0.2} = \frac{1 + 0.2 ( 0.5625 )}{1.2} = \frac{1.1125}{1.2} = 0.927083$ Yeh step kyun? Free stream aur sonic point dono ek streamline par hain, isliye woh same stagnation pressure $p_{0}$ share karte hain (upar define kiya gaya). $p_{\infty} / p_{0}$ ko $p^{*} / p_{0}$ se likhne par $p_{0}$ cancel ho jaata hai aur exactly yahi ratio milta hai. Sonic point ( $M_{l oc a l} = 1$ ) denominator ko ek baar aur hamesha ke liye fix kar deta hai. Step 2. $\frac{γ}{γ - 1} = 3.5$ ki power tak raise karo (saare digits rakho): $0.92708 3^{3.5} = 0.76749$ Kyun? Isentropy us temperature/density-flavoured ratio ko pressure ratio $p / p_{\infty}$ mein badal deta hai. Step 3. Front factor aur 1 subtract karo: $C_{p, cr} = \frac{2}{1.4 ( 0.5625 )} (0.76749 - 1) = \frac{2}{0.7875} (- 0.23251) = - 0.59050$

Answer: $C_{p, cr} \approx - 0.590$ .

Verify: $M_{cr}$ ko $0.70$ ( $- 0.79$ ) se $0.75$ tak badhane par $C_{p, cr}$ kam negative ho gaya ( $- 0.59$ ). ✅ Yeh figure mein curve ke rise-toward-zero trend se match karta hai. Yeh parent ke doosre [!mistake] ko bhi debunk karta hai: higher $M_{cr}$ ⇒ aasaan suction requirement.

Cell B — backward direction (design question)

Worked example Example 2 — Airfoil ke

$C_{p, 0}$ se $M_{cr}$ find karo Ek airfoil ka incompressible minimum pressure coefficient $C_{p, 0} = - 0.43$ hai. $M_{cr}$ estimate karo ( $γ = 1.4$ ). Yeh wahi case hai jo figure mein drawn hai — padhte waqt wahan follow karo.

Forecast: Yeh suction peak parent ke $- 0.55$ se milder hai (jisne $M_{cr} \approx 0.70$ diya). Milder suction ko subsonic rakhna aasaan hai — toh kya aapko lagta hai $M_{cr}$ $0.70$ se zyada hoga ya kam?

Step 1. Dono curves ko $M_{\infty}$ ke function ke roop mein likho — figure mein yeh orange (gas) aur teal (airfoil) lines hain: $airfoil: C_{p, m i n} = \frac{- 0.43}{1 - M _{\infty}^{2}}, gas: C_{p, cr} (M_{\infty}) .$ Kyun? Inki crossing — figure mein plum dot — $M_{cr}$ ki definition hai. Do unknowns ek equation mein collapse ho jaate hain. Step 2. $M = 0.75$ probe karo: airfoil $= - 0.43/ 1 - 0.5625 = - 0.43/ 0.4375 = - 0.65017$ ; gas $C_{p, cr} (0.75) = - 0.59050$ . Figure mein teal line yahan orange ke neeche hai → suction already too strong → past sonic. Step 3. $M = 0.74$ probe karo: airfoil $= - 0.43/ 1 - 0.5476 = - 0.43/ 0.4524 = - 0.63930$ ; gas $C_{p, cr} (0.74) \approx - 0.60747$ . Abhi bhi thoda past. $M = 0.735$ probe karo: airfoil $= - 0.43/ 0.45978 \approx - 0.63417$ ; gas $C_{p, cr} (0.735) \approx - 0.61629$ . Do lines ab almost meet kar rahi hain — yahi plum dot hai.

Answer: $M_{cr} \approx 0.734$ — figure mein crossing se seedha padhein.

Verify: $0.734 > 0.70$ . ✅ Milder suction → higher $M_{cr}$ , exactly jaise forecast tha, aur "thinner wing ⇒ higher $M_{cr}$ " ke consistent.

Cells C & D — universal curve ki do limits

Worked example Example 3 —

$M_{cr} \to 0$ aur $M_{cr} \to 1$ par behaviour Calculator ke bina, $C_{p, cr}$ curve ke do ends describe karo.

Forecast: Ek end par aapko infinitely deep pressure well chahiye; doosre par essentially kuch nahi. Kaun sa kaun sa hai?

Step 1 (Cell D, $M_{cr} \to 1$ ). $M_{cr} = 1$ plug karo: inner ratio $= \frac{1 + 0.2}{1 + 0.2} = 1$ , toh $1^{3.5} - 1 = 0$ , giving $C_{p, cr} = 0$ . Figure mein yeh orange curve ka right end hai jo zero ko touch karta hai. Kyun? Agar free stream khud sonic hai, toh surface pressure ko $M_{l oc a l} = 1$ reach karne ke liye bilkul bhi dip nahi karna padta. Zero suction required. Step 2 (Cell C, $M_{cr} \to 0$ ). Jaise $M_{cr} \to 0$ , bracket $[\dots - 1]$ ek finite negative number ki taraf jaata hai, lekin front factor $\frac{2}{γ M _{cr}^{2}}$ $1/ M_{cr}^{2}$ ki tarah blow up karta hai. Isliye $C_{p, cr} \to - \infty$ (orange curve ka left end plunging). Kyun? Bahut slow speed par, hawa almost incompressible hoti hai; almost-still stream ko sound tak accelerate karne ke liye almost saara pressure remove karna padta hai — ek unbounded suction. Numerically $M_{cr} = 0.1$ par: inner $= \frac{1.002}{1.2} = 0.835$ ; $0.83 5^{3.5} = 0.531$ ; $C_{p, cr} = \frac{2}{1.4 ( 0.01 )} (- 0.469) = - 67.0$ .

Answer: $C_{p, cr} \to - \infty$ jaise $M_{cr} \to 0$ ; $C_{p, cr} \to 0^{-}$ jaise $M_{cr} \to 1$ . Poore band mein ek monotone rise.

Verify: $M = 0.1$ par value $- 67.0$ bahut bada aur negative hai (Cell C se match); $M = 1$ par exactly $0$ hai (Cell D). ✅ Dono endpoints reproduce ho gaye. Figure mein orange curve ke do ends dekho.

Cell E — degenerate airfoil (koi suction nahi)

Worked example Example 4 —

$C_{p, 0} = 0$ wali ek shape Zero angle of attack par ek vanishingly thin flat plate mein essentially koi pressure dip nahi hai: $C_{p, 0} = 0$ . $M_{cr}$ kya hai?

Forecast: Agar flow kahi bhi speed up nahi hoti, toh ek local point sound tak kab pahunchta hai? Algebra se pehle guess karo.

Step 1. Airfoil curve: $C_{p, m i n} = \frac{0}{1 - M _{\infty}^{2}} = 0$ sabhi $M_{\infty} < 1$ ke liye. Kyun? Prandtl–Glauert incompressible peak ko scale karta hai; zero ko scale karne par zero milta hai. Step 2. Gas curve ke barabar set karo: $0 = C_{p, cr} (M_{\infty})$ . Example 3 se, $C_{p, cr} = 0$ sirf $M_{\infty} = 1$ par hota hai. Kyun? Crossing answer hai, aur gas curve zero ko sirf unity par touch karta hai.

Answer: $M_{cr} = 1$ .

Verify: Koi suction nahi wali plate kisi bhi patch ko supersonic nahi bana sakti jab tak poori stream sonic na ho — isliye $M_{cr} = 1$ , theoretical ceiling. ✅ Yeh boundary case prove karta hai ki "less suction ⇒ higher $M_{cr}$ " apni limit tak pushed hai.

Cell F — sign machinery

Worked example Example 5 — Prove karo ki

$C_{p, cr} < 0$ har $M_{cr} \in (0, 1)$ ke liye Dikhao ki critical pressure coefficient hamesha negative hota hai $M_{\infty} = 1$ se neeche.

Forecast: Suction (free-stream se neeche pressure) matlab $C_{p} < 0$ . Kya aapko lagta hai gas curve $M = 1$ se pehle positive territory mein kabhi cross karega?

Step 1. Inner ratio $R = \frac{1 + \frac{γ - 1}{2} M _{cr}^{2}}{1 + \frac{γ - 1}{2}}$ dekho. $M_{cr} < 1$ ke liye numerator denominator se chota hai, isliye $R < 1$ . Kyun? Dono same additive $1$ share karte hain; sirf $M^{2}$ term alag hai, aur $M_{cr}^{2} < 1$ . Step 2. Phir $R^{γ / (γ - 1)} < 1$ , isliye bracket $[R^{3.5} - 1] < 0$ . Kyun? $(0, 1)$ mein kisi number ko positive power tak raise karne par woh $(0, 1)$ mein rahta hai. Step 3. Front factor $\frac{2}{γ M _{cr}^{2}} > 0$ . Positive $\times$ negative $=$ negative. Kyun? $γ > 0$ aur $M_{cr}^{2} > 0$ , isliye prefactor sign kabhi flip nahi karta.

Answer: $C_{p, cr} < 0$ strictly sabhi $0 < M_{cr} < 1$ ke liye, sirf $M_{cr} = 1$ par $0$ hit karta hai.

Verify: Middle point $M_{cr} = 0.5$ check karo: inner $= \frac{1.05}{1.2} = 0.875$ ; $0.87 5^{3.5} = 0.62666$ ; $C_{p, cr} = \frac{2}{1.4 ( 0.25 )} (- 0.37334) = - 2.1334 < 0$ . ✅ Negative, jaise prove kiya.

Cell G — real-world word problem

Worked example Example 6 — Altitude par cruise karta jet

Ek jet $V_{\infty} = 250 m/s$ par cruise karta hai jahan local speed of sound $a = 300 m/s$ hai. Uski wing ka $M_{cr} = 0.80$ hai. Kya wing par koi supersonic pocket present hai? Kitna margin hai?

Forecast: Flight Mach number ko $M_{cr}$ se compare karo. Guess karo ki $M_{\infty}$ $0.80$ se zyada hai ya nahi.

Step 1. Flight Mach number: $M_{\infty} = \frac{V _{\infty}}{a} = \frac{250}{300} = 0.8333.$ Kyun? Mach number local sound speed ke relative speed hai — $M_{cr}$ se compare karne se pehle hamesha convert karo. (Units cancel: m/s ÷ m/s = dimensionless. ✅) Step 2. Compare karo: $M_{\infty} = 0.833 > M_{cr} = 0.80$ . Kyun? $M_{cr}$ se upar, sabse tez surface point pehle hi $M_{l oc a l} = 1$ cross kar chuka hai → ek supersonic pocket exist karta hai, likely ek shock mein close hota hua, drag divergence drive karta hua. Step 3. Margin: $M_{\infty} - M_{cr} = 0.833 - 0.80 = 0.033$ .

Answer: Haan — wing critical se $0.033$ Mach upar hai; ek shock-terminated supersonic pocket present hai.

Verify: Agar pilot slow karke $V = 0.80 \times 300 = 240 m/s$ par aaye, toh $M_{\infty} = 0.80 = M_{cr}$ exactly — pocket abhi abhi gayab hoti hai. ✅ Consistent.

Cell H — design twist (sweep)

Worked example Example 7 — Wing ko

$30°$ peeche sweep karo Example 6 ki wing (straight $M_{cr} = 0.80$ ) ko $Λ = 30°$ se peeche sweep kiya gaya hai (sweep angle, glossary mein define kiya gaya). Aircraft ko ab kaun sa effective $M_{cr}$ milta hai, aur kya $M_{\infty} = 0.833$ par pocket abhi bhi banti hai?

Forecast: Sweep airfoil se speed chhupata hai. Kya poore aircraft ka $M_{cr}$ upar jaata hai ya neeche?

Step 1. Sirf leading edge ke normal velocity component se pressure dip drive hoti hai: $M_{eff} = M_{\infty} cos Λ$ . Kyun? Spanwise component wing ke along slide karta hai aur koi compressing nahi karta; simple-sweep theory sirf normal part rakhti hai. Step 2. Airfoil section tab bhi critical hoti hai jab uska normal Mach $0.80$ ho. Toh aircraft ka true critical Mach hai: $M_{cr}^{swept} = \frac{0.80}{c o s 30°} = \frac{0.80}{0.86603} = 0.92376.$ Kyun? $M_{\infty} cos Λ = 0.80$ ko $M_{\infty}$ ke liye rearrange karo. Step 3. Cruise $M_{\infty} = 0.833$ se compare karo: ab $0.833 < 0.924$ , toh koi pocket nahi.

Answer: Swept $M_{cr}$ $\approx 0.924$ tak rise karta hai; Ex 6 ka cruise ab comfortably subcritical hai — shock chala gaya.

Verify: Sweep ne $M_{cr}$ raise kiya ( $0.80 \to 0.924$ ), parent ke Example 3 forecast aur mnemonic "skinny swept wings sneak past sonic" se match karta hai. ✅ (Note: real sweep gains $1/ cos Λ$ se chote hote hain — fully swept ideal ek upper bound hai. Ek complementary transonic fix ke roop mein Area Rule (Transonic) bhi compare karo.)

Cell I — ek alag gas

Worked example Example 8 — Helium,

$γ = 5/3$ $M_{cr} = 0.70$ par $C_{p, cr}$ recompute karo lekin helium mein ( $γ = 5/3 \approx 1.6667$ ) air ki jagah. Kya "universal" curve move karta hai?

Forecast: Parent ne air mein $- 0.79$ paaya. Kya zyada $γ$ required suction ko deeper banayega ya shallower?

Step 1. Coefficient $\frac{γ - 1}{2} = \frac{0.6667}{2} = 0.33333$ compute karo aur inner ratio: $\frac{1 + 0.33333 ( 0.7 0 ^{2} )}{1 + 0.33333} = \frac{1 + 0.33333 ( 0.49 )}{1.33333} = \frac{1.16333}{1.33333} = 0.87250.$ Yeh step kyun? Air jaisi hi isentropic structure, lekin $\frac{γ - 1}{2}$ coefficient helium ke liye bada hai — yahi ek cheez hai jo tabdeel hoti hai jab hum gases swap karte hain. Step 2. Exponent $\frac{γ}{γ - 1} = \frac{1.66667}{0.66667} = 2.5$ . Raise karo: $0.8725 0^{2.5} = 0.71124.$ Kyun? Isentropy phir se ratio ko pressure ratio mein badalta hai — lekin helium ke exponent $2.5$ ke saath, air ke $3.5$ nahi. Step 3. Front factor $\frac{2}{γ M _{cr}^{2}} = \frac{2}{1.66667 ( 0.49 )} = \frac{2}{0.81667} = 2.44898$ , phir 1 subtract karo aur multiply karo: $C_{p, cr} = 2.44898 (0.71124 - 1) = 2.44898 (- 0.28876) = - 0.70717.$

Answer: Helium mein $C_{p, cr} \approx - 0.707$ , air ke $- 0.793$ ke versus, same $M_{cr} = 0.70$ par.

Verify: Curve $γ$ ke saath shift karta hai — yeh "universal" sirf ek fixed gas ke liye hai. Helium ( $- 0.707$ ) air ( $- 0.793$ ) se $M_{cr} = 0.70$ par kam negative hai. ✅ $γ$ par dependence (aur airfoil ke baare mein kuch nahi) exactly wahi hai jo parent ke [!formula] callout ne claim kiya tha.

Recall Kaun sa cell kaun sa tha?

Cell A forward — Ex1 answer ::: $C_{p, cr} \approx - 0.590$ at $M_{cr} = 0.75$ . Cell B backward — Ex2 answer ::: $M_{cr} \approx 0.734$ for $C_{p, 0} = - 0.43$ . Cell C limit — trend as $M_{cr} \to 0$ ::: $C_{p, cr} \to - \infty$ . Cell D limit — value at $M_{cr} = 1$ ::: $C_{p, cr} = 0$ (koi suction nahi chahiye). Cell E degenerate — flat plate $C_{p, 0} = 0$ ::: $M_{cr} = 1$ . Cell F sign check — sign of $C_{p, cr}$ for $M_{cr} \in (0, 1)$ , e.g. at $0.5$ ::: strictly negative; $C_{p, cr} (0.5) \approx - 2.133$ . Cell G word problem — kya $M_{\infty} = 0.833$ supercritical hai agar $M_{cr} = 0.80$ ? ::: Haan, $0.033$ Mach se. Cell H sweep $30°$ — new $M_{cr}$ ::: $\approx 0.924$ . Cell I helium — $C_{p, cr}$ at $M_{cr} = 0.70$ ::: $\approx - 0.707$ .

Connections

Universal curve Isentropic Flow Relations aur Pressure Coefficient Cp par tika hai.
Airfoil curve Prandtl–Glauert Compressibility Correction hai.
Crossing ke consequences: Drag Divergence Mach Number, Shock Waves & Normal Shocks.
Design fixes: Swept Wings & Transonic Design, Area Rule (Transonic).