Mach number M = V - a — subsonic ( - 1), transonic (~1), supersonic ( - 1), hypersonic ( - 5): Worked examples

3.1.4 · D3 · Physics › Compressible Flow & Aerodynamics › Mach number M = V - a — subsonic ( - 1), transonic (~1), sup

Yeh page Mach number ke liye "koi surprise nahi" wali drill hai. Hum har tarah ke question ke through march karte hain jo yeh topic throw kar sakta hai: har flow regime, degenerate cases ( $M \to 0$ , $M = 1$ , $M \to \infty$ ), Mach-cone geometry, ek real-world word problem, aur ek exam-style twist. Yahan use kiya gaya har symbol parent note mein build kiya gaya tha; jahan koi naya idea ghus aata hai, hum usse pehle scratch se build karte hain.

Scenario matrix

Har row ek alag case class hai. Jo worked examples follow karte hain woh us cell se tagged hain jo woh cover karte hain.

#	Case class	Kya tricky hai	Covered by
A	Subsonic, $M$ find karo $V, T$ se	operations ka order ( $a$ pehle)	Ex 1
B	Supersonic, $M$ find karo	regime boundary recognize karna	Ex 2
C	Same $V$ , alag $T$ → alag $M$	$M$ gas state track karta hai	Ex 3
D	Inverse problem: $M, T$ diya, $V$ find karo	formula rearrange karna	Ex 4
E	Limiting case $M \to 0$ (incompressible)	density change kab ignorable hai	Ex 5
F	Degenerate case $M = 1$ (sonic)	Mach angle $9 0^{\circ}$ tak collapse hota hai	Ex 6
G	Mach-cone geometry ( $sin μ = 1/ M$ ) + limit $M \to \infty$	angle → 0, needle cone	Ex 7
H	Real-world word problem: sonic-boom delay	geometry + kinematics mix	Ex 8
I	Exam twist: local vs free-stream Mach	ek number ≠ poora field	Ex 9

Ex 1 — Subsonic: $M$ find karo (Cell A)

Worked example Ek light aircraft

$V = 110 m/s$ pe fly karta hai jahan $T = 280 K$ hai. Air: $γ = 1.4$ , $R = 287 J/(kg\cdotK)$ . $M$ find karo aur regime ka naam batao.

Forecast: Guess karo — kya $M$ $0.3$ ke paas hai ya $0.9$ ke? (Everyday ke chhote planes slow hote hain.)

Step 1 — Pehle local sound speed compute karo. $a = γ R T = 1.4 \cdot 287 \cdot 280 = 112, 504 \approx 335.4 m/s$ Yeh step kyun? $M$ ek ratio hai; jab tak messenger speed $a$ nahi jaante, tum ise form nahi kar sakte. Aur $a$ sirf temperature pe depend karta hai (parent note), isliye bas $T$ chahiye.

Step 2 — Ratio form karo. $M = \frac{V}{a} = \frac{110}{335.4} \approx 0.328$ Yeh step kyun? $M$ answer karta hai "main local sound speed ke kitne multiples pe move kar raha hoon?"

Conclusion: $M \approx 0.33$ → subsonic, aur $M \approx 0.3$ line se thoda upar jahan compressibility matter karna shuru hoti hai (dekho Compressibility & Bernoulli's Limits).

Verify: Units: $(J/(kg\cdotK)) \cdot K = J/kg = m^{2} / s^{2} = m/s$ ✓. Aur $M < 1$ "slow plane" se match karta hai ✓.

Ex 2 — Supersonic: $M$ find karo (Cell B)

Ex 3 — Same $V$ , alag $T$ → alag $M$ (Cell C)

Worked example Ek cargo jet true airspeed

$V = 240 m/s$ pe fly karta hai. Iska Mach number compare karo (a) ground ke paas, $T = 300 K$ , aur (b) cruise pe, $T = 220 K$ .

Forecast: Speedometer dono times same read karta hai. Kya Mach number bhi same hoga? Computing se pehle guess karo.

Step 1 — Ground sound speed. $a_{gnd} = 1.4 \cdot 287 \cdot 300 = 120, 540 \approx 347.2 m/s$ Yeh step kyun? Garm air → faster molecular collisions → faster sound (parent note ka $T$ scaling).

Step 2 — Cruise sound speed. $a_{cr} = 1.4 \cdot 287 \cdot 220 = 88, 396 \approx 297.3 m/s$

Step 3 — Do Mach numbers. $M_{gnd} = \frac{240}{347.2} \approx 0.691, M_{cr} = \frac{240}{297.3} \approx 0.807$ Yeh step kyun? Same $V$ , lekin $a$ chhota ho gaya, toh ratio badh gaya.

Conclusion: Airspeed identical hai, phir bhi plane $M \approx 0.69$ (comfortably subsonic) se $M \approx 0.81$ (transonic ka edge) tak climb karta hai — sirf isliye ki air thandi ho gayi. Mach gas state ke baare mein hai, sirf tumhari speed ke nahi.

Verify: $a_{gnd} > a_{cr}$ kyunki $300 > 220$ , isliye $M_{gnd} < M_{cr}$ ✓ (bada denominator → chhota ratio). Dekho Speed of Sound in Gases.

Ex 4 — Inverse problem: $M$ aur $T$ diya, $V$ find karo (Cell D)

Ex 5 — Limiting case $M \to 0$ : flow "incompressible" kab hai? (Cell E)

Worked example Air

$V = 90 m/s$ pe $T = 288 K$ pe move karti hai. Fractional density change estimate karo aur decide karo ki Bernoulli ka incompressible formula safe hai ya nahi.

Forecast: Famous $M \approx 0.3$ threshold se neeche ya upar?

Step 1 — Sound speed aur Mach number. $a = 1.4 \cdot 287 \cdot 288 = 115, 718 \approx 340.2 m/s, M = \frac{90}{340.2} \approx 0.265$ Yeh step kyun? Density-change rule $M$ ke terms mein likhi hai, isliye pehle $M$ chahiye.

Step 2 — Density-change estimate apply karo. Parent note deta hai $\frac{d ρ}{ρ} \sim \frac{1}{2} M^{2} .$ Yeh tool kyun aur directly Bernoulli kyun nahi? Yeh ek number batata hai gas kitna squish hota hai; yeh yardstick hai ki constant- $ρ$ assumption jo incompressible Bernoulli ke behind hai woh acceptable hai ya nahi. $\frac{d ρ}{ρ} \sim \frac{1}{2} (0.265)^{2} \approx 0.0351 \Rightarrow \approx 3.5%$

Conclusion: Roughly $3.5%$ density change — $M = 0.3$ pe $\sim 5%$ line se neeche. Incompressible Bernoulli yahan fair approximation hai. Yeh matrix ka == $M \to 0$ == corner hai: compressible terms vanish ho jaate hain.

Verify: Threshold pe $M = 0.3$ : $\frac{1}{2} (0.3)^{2} = 0.045 = 4.5%$ ✓ (parent ke "~5%" se match karta hai). Humara $3.5% < 4.5%$ , toh $M = 0.3$ se neeche hona consistent hai ✓. Dekho Compressibility & Bernoulli's Limits.

Ex 6 — Degenerate case $M = 1$ : sonic wall (Cell F)

Worked example Exactly

$M = 1$ pe, Mach-cone half-angle $μ$ kya hai, aur wavefront picture kaisi dikhti hai?

Forecast: Jab body sound ki taraf speed up karti hai, kya cone sharper hota hai ya flatter? $M = 1$ pe angle predict karo.

Step 1 — Mach-angle relation yaad karo. Parent note se, $sin μ = \frac{1}{M} .$ Yeh tool kyun? $sin$ wavefront geometry (sides $a t$ aur $V t$ wala right triangle) ko ek angle mein turn karta hai — yeh cone ke liye "opposite over hypotenuse" ka ratio hai (figure mein triangle dekho).

Step 2 — $M = 1$ set karo. $sin μ = \frac{1}{1} = 1 \Rightarrow μ = arcsin (1) = 9 0^{\circ}$ Yeh step kyun? $arcsin$ poochta hai "kaunse angle ka yeh sine hai?" — $sin$ ka inverse — aur answer right angle hai.

Conclusion: $M = 1$ pe "cone" puri tarah $9 0^{\circ}$ tak khul jaata hai: wavefronts ek single flat wall of sound mein pile ho jaate hain motion ke perpendicular (figure ka left panel). Yeh transonic knife-edge hai.

Verify: $sin (9 0^{\circ}) = 1 = 1/ M$ with $M = 1$ ✓. $M$ thoda $1$ se upar, say $M = 1.05$ , $sin μ = 0.952$ , $μ \approx 72. 2^{\circ} < 9 0^{\circ}$ — wall abhi cone mein tilt hona shuru karti hai ✓.

Ex 7 — Mach cone geometry aur $M \to \infty$ limit (Cell G)

Worked example (a)

$M = 2$ , (b) $M = 5$ , (c) limit $M \to \infty$ ke liye Mach half-angle find karo.

Forecast: Kya speed badhne pe cone narrower hota hai ya wider? Extreme hypersonic speed pe shape guess karo.

Step 1 — $M = 2$ . $sin μ = \frac{1}{2} = 0.5 \Rightarrow μ = arcsin (0.5) = 3 0^{\circ}$ Yeh step kyun? Wavefront-envelope geometry ka seedha application (figure).

Step 2 — $M = 5$ (hypersonic ka start). $sin μ = \frac{1}{5} = 0.2 \Rightarrow μ = arcsin (0.2) \approx 11.5 4^{\circ}$ Yeh step kyun? Same tool, bada $M$ ⇒ chhota $sin μ$ ⇒ chhota angle.

Step 3 — Limit. Jab $M \to \infty$ , $sin μ = 1/ M \to 0$ , toh $μ \to 0^{\circ}$ . Yeh step kyun? Hum ek limit lete hain — "jab $M$ bina bound ke badhta hai toh $μ$ kya approach karta hai?" — hypersonic behaviour describe karne ke liye jo hum exactly kabhi reach nahi kar sakte lekin approach kar sakte hain.

Conclusion: Faster ⇒ narrower cone. Flat wall se $M = 1$ pe ( $9 0^{\circ}$ ) → $3 0^{\circ}$ at $M = 2$ → $\approx 11. 5^{\circ}$ at $M = 5$ → ek needle-thin cone jo flight path se chipka rehta hai jab $M \to \infty$ . Dekho Oblique Shocks & Mach Cone.

Verify: Angles monotonically decrease karte hain jab $M$ badhta hai: $9 0^{\circ} > 3 0^{\circ} > 11. 5^{\circ} > 0^{\circ}$ ✓. Aur $sin (3 0^{\circ}) = 0.5 = 1/2$ ✓.

Ex 8 — Real-world word problem: sonic-boom delay (Cell H)

Worked example Ek jet altitude

$h = 1500 m$ pe level fly karta hai aur $M = 1.8$ pe (local $a = 320 m/s$ ). Yeh ek observer ke directly overhead se time $t = 0$ pe guzarta hai. Observer sonic boom kitne time baad suntha hai?

Forecast: Kya boom plane ke overhead hone se pehle sunai deta hai, tab, ya baad mein? (Plane apni khud ki sound se aage nikal gaya...)

Step 1 — Mach angle find karo. $sin μ = \frac{1}{M} = \frac{1}{1.8} \approx 0.5556 \Rightarrow μ = arcsin (0.5556) \approx 33.7 5^{\circ}$ Yeh step kyun? Boom tab arrive karta hai jab trailing cone surface observer ke upar se sweep karti hai; cone ka half-angle uski geometry set karta hai (figure).

Step 2 — Flight speed. $V = M a = 1.8 \times 320 = 576 m/s$ Yeh step kyun? Hume chahiye ki plane kitni fast travel karta hai taaki hum horizontal distance ko time mein convert kar sakein.

Step 3 — Horizontal distance jo plane ne fly kiya jab cone ground point tak reach karta hai. Cone flight path se angle $μ$ pe peeche trail karta hai; jab observer (flight line se vertical distance $h$ neeche) cone se just touch hota hai, plane horizontal distance $d$ aage hai, jahan $tan μ = \frac{h}{d} \Rightarrow d = \frac{h}{t a n μ} .$ Yahan tangent kyun (sine kyun nahi)? Hamare paas vertical leg $h$ hai (angle ke opposite) aur horizontal leg $d$ chahiye (adjacent) — woh ratio "opposite over adjacent" exactly wahi hai jo $tan$ figure mein right triangle pe encode karta hai. $d = \frac{1500}{t a n ( 33.7 5 ^{\circ} )} = \frac{1500}{0.6682} \approx 2245 m$

Step 4 — Distance ko time mein convert karo. $t = \frac{d}{V} = \frac{2245}{576} \approx 3.90 s$ Yeh step kyun? Plane $d$ cover karta hai speed $V$ pe; time = distance / speed.

Conclusion: Observer boom roughly ==3.9 s baad mein== suntha hai jab jet overhead hota hai — boom hamesha lag karta hai, kyunki aircraft apne khud ke pressure signals se aage nikal gaya.

Verify: Geometry check — $tan μ = h / d = 1500/2245 \approx 0.668$ , aur $tan (33.7 5^{\circ}) \approx 0.668$ ✓. Units: $m / (m/s) = s$ ✓. "Overhead ke baad" supersonic-lag intuition se match karta hai ✓.

Ex 9 — Exam twist: local vs free-stream Mach (Cell I)

Worked example Ek wing free-stream

$M_{\infty} = 0.85$ pe fly karta hai air mein jahan $a = 300 m/s$ hai. Wing ke thickest part ke upar flow $V_{local} = 345 m/s$ tak accelerate karta hai (aur wahan local sound speed $a_{local} = 298 m/s$ tak drop ho gayi hai). Kya flow ka koi part supersonic hai?

Forecast: Poora plane subsonic hai ( $0.85 < 1$ ). Trick question — kya air phir bhi kahin supersonic ja sakti hai?

Step 1 — Free-stream Mach (given). $M_{\infty} = 0.85 < 1$ → wing ke aage subsonic. Yeh step kyun? "Ek number" establish karo jo sab quote karte hain — aur dikhao ki yeh poori story nahi hai.

Step 2 — Wing crest ke upar local Mach. $M_{local} = \frac{V _{local}}{a _{local}} = \frac{345}{298} \approx 1.158$ Yeh step kyun? Air curved upper surface ke upar speed up karti hai (Bernoulli), aur local sound speed fall kar sakti hai jab gas us fast region mein cool hoti hai — dono local ratio ko $1$ se upar push karte hain.

Conclusion: Even though $M_{\infty} = 0.85$ , local flow $M \approx 1.16$ reach karta hai → ek supersonic pocket ek chhote shock mein end hoti hai. Yeh exactly transonic ki difficulty hai: free-stream Mach ek number hai; field mein bahut saare hain. Isliye designers Prandtl–Glauert Correction use karte hain $M \approx 1$ ke paas.

Verify: $M_{local} = 345/298 \approx 1.158 > 1$ jabki $M_{\infty} = 0.85 < 1$ ✓ — overall subsonic, locally supersonic, exactly transonic paradox.

Active recall

Recall Pehle predict karo, phir reveal karo

$M$ find karte waqt pehle kya aata hai — $a$ compute karna ya divide karna? ::: Pehle $a = γ R T$ compute karo, phir $M = V / a$ . Same true airspeed, thandi air — $M$ badhta hai ya ghadta hai? ::: Badhta hai, kyunki $a = γ R T$ fall karta hai, denominator chhota hota hai. $M = 1$ pe Mach half-angle kya hai? ::: $9 0^{\circ}$ (sound ki flat wall; $sin μ = 1$ ). $M \to \infty$ pe Mach cone half-angle kya approach karta hai? ::: $0^{\circ}$ (needle-thin cone; $sin μ = 1/ M \to 0$ ). $M$ aur $a$ diye hain, $V$ kaise get karte hain? ::: $V = M a$ . Sonic boom observer tak plane ke overhead hone se pehle pahunchta hai ya baad mein? ::: Baad mein — aircraft apni khud ki sound se aage nikal gaya; trailing cone baad mein arrive karta hai. Kya subsonic free-stream mein supersonic local flow ho sakti hai? ::: Haan — air wing ke upar accelerate karti hai; local $M$ 1 se exceed kar sakta hai (transonic paradox).