Prandtl-Meyer function ν(M): Visual walkthrough — Hinglish deep dive

3.1.17 · D2 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer function ν(M)

Kisi bhi calculus se pehle, humein teen plain-language pictures chahiye: supersonic flow arrow kya hai, Mach wave kya hoti hai, aur "turning away" ka matlab kya hai. Pehle woh banate hain.

Step 0 — Teen cheezein jo tumhe pehle se imagine karni chahiye

KYA HAI. Ek supersonic flow hawa ki ek dhara hai jo sound se tez chal rahi hai. Hum ise ek arrow ke roop mein draw karte hain jiska length $V$ hai (speed) aur jo us direction mein point karta hai jis direction mein hawa travel karti hai.

$V$ ::: flow speed (hawa kitne metres per second chal rahi hai).
$a$ ::: local speed of sound — ek tiny pressure ripple us hawa mein kitni tez travel karta hai.
$M = V / a$ ::: Mach number, "sound se kitne guna tez." $M > 1$ matlab supersonic.

YEH SYMBOLS KYUN. Aage ki poori kahaani arrow $V$ ke turning aur growing ke baare mein hai, aur ratio $M$ ke baare mein jo measure karta hai hum kitne supersonic hain. Jab tak $V$ , $a$ , aur $M$ named nahi hain, waves ki baat nahi ho sakti.

PICTURE. Ek single flow arrow, plus ek tiny circle jo dikhata hai ki sound ripple utne hi time mein kitni door spread hua — arrow circle se aage nikal jaata hai kyunki $M > 1$ .

Step 1 — Ek weak wave, aur velocity triangle

KYA HAI. Flow mein ek single Mach wave daalo. Hawa usse cross karti hai aur ek tiny angle $d θ$ se away bend karti hai. Hum claim karte hain: sirf woh velocity component change hota hai jo wave ke perpendicular hai; wave ke parallel component untouched rehta hai.

SIRF NORMAL COMPONENT KYUN. Ek Mach wave infinitely weak hoti hai — yeh sirf stacked sound ripples ki ek line hai. Sound ripples apne travel ki direction mein push karte hain (wavefront ke perpendicular). Toh wave flow ko sirf apni normal direction mein nudge kar sakti hai; sideways (tangential) motion seedha nikal jaati hai unchanged. Yeh ek fact poori derivation ka engine hai.

PICTURE. Incoming arrow $V$ ko wave line ke relative do pieces mein resolve karo: ek tangential piece $V_{t}$ (wave ke along) aur ek normal piece $V_{n}$ (wave ke across). Wave ke baad, $V_{t}$ same hai, lekin $V_{n}$ thoda sa grow kar gaya hai, toh resultant arrow dono longer bhi hai aur $d θ$ se rotate bhi hua hai.

Yahan ek crucial number nikalta hai. Normal Mach number hai $M_{n} = V_{n} / a = M sin μ = M \cdot \frac{1}{M} = 1$ . Flow ek Mach wave ko normal direction mein exactly sonic speed se cross karti hai — yahi literally ise Mach wave banata hai, shock nahi.

Step 2 — Triangle ko $d θ = \frac{M ^{2} - 1}{V} d V$ mein convert karo

KYA HAI. Us velocity triangle ki geometry padho taaki tiny bend $d θ$ ko tiny speed increase $d V$ se relate kar sako.

YEH EXACT FACTOR KYUN. Jab $V_{t}$ frozen hai aur $V_{n}$ grow karta hai $d V_{n}$ se, arrow ki tip wave ke perpendicular direction mein slide karti hai. Arrow tip par ek small right-triangle kehta hai: bend $d θ$ (arrow ka rotation) times length $V$ equals sideways slide $d V_{n}$ , geometry ke through project karke. Ise work out karne par (figure dekho) jo ratio appear karta hai woh $cot μ$ hai.

PICTURE. Arrow tip ka ek zoom: purana arrow, thoda-sa-longer-aur-rotated naya arrow, aur woh tiny right triangle jiske legs hain "flow ke along growth" aur "sideways swing $V d θ$ ."

Step 3 — $d V / V$ ko $d M / M$ se replace karo

KYA HAI. Hum answer Mach number $M$ mein chahte hain (tables yehi use karte hain), raw speed $V$ mein nahi. Toh $d V / V$ ko $d M / M$ mein convert karo.

YEH SIRF $d V / V = d M / M$ KYUN NAHI HAI. Kyunki $V = M a$ , aur sound speed $a$ drop karta hai jab flow accelerate karti hai (expanding gas cool hoti hai). Toh jab $M$ badhta hai, $V$ proportionally se kam badhta hai — "would-be" speed gain ka ek hissa falling $a$ kha jaata hai. Hume yeh account karna hoga.

PICTURE. Ek small acceleration ke liye do stacked bars: top bar $M$ ko ek full slice se grow karta dikhata hai; bottom bar $V$ ko ek smaller slice se grow karta dikhata hai, kyunki $a$ shrink hua. Shrink factor exactly neeche wala denominator hai.

$V = M a$ ke logs lo: $ln V = ln M + ln a$ , differentiate karo: $\frac{d V}{V} = \frac{d M}{M} + \frac{d a}{a}$ . Isentropic temperature law (dekho Isentropic Flow Relations) deta hai $a^{2} \propto T \propto (1 + \frac{γ - 1}{2} M^{2})^{- 1}$ , aur use differentiate karne se $d a / a$ milta hai. Result:

Step 4 — Ek clean differential mein combine karo

KYA HAI. Step 3 ko Step 2 ke boxed result mein substitute karo. Ab sab kuch sirf $M$ mein hai.

KYUN. Ab hamre paas $d θ$ purely $M$ ke through expressed hai — "Mach number mein tiny bump per flow kitna bend hota hai" ka ek self-contained rule. Yahi exactly woh cheez hai jise hum integrate kar sakte hain.

$d θ = call this f (M) \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M}$

Numerator $M^{2} - 1$ (Step 2 se) bending ko on karta hai; $M = 1$ par zero.
Denominator $1 + \frac{γ - 1}{2} M^{2}$ (Step 3 se) ise throttle karta hai jab $M$ badhta hai.
Unka ratio, $M$ se divide karke, "turning per unit Mach" hai.

PICTURE. Integrand $f (M) / M$ ko $M$ ke against plot karo: $0$ se start karta hai ( $M = 1$ par), ek hump tak rise karta hai, phir $0$ ki taraf decay karta hai jab $M \to \infty$ . Is curve ke neeche ka area total turning $ν$ hai — woh decaying tail yehi reason hai kyun total area finite hai.

Step 5 — Sonic se $M$ tak integrate karo: yahi $ν (M)$ hai

KYA HAI. Saare tiny $d θ$ 's ko $M = 1$ (jahan hum $θ = 0$ declare karte hain) se target $M$ tak add karo.

$M = 1$ SE START KYUN. Humein ek common zero-point chahiye taaki sabka "angle budget" ek hi jagah se measure ho. Sonic flow natural origin hai: usse neeche koi Mach wave nahi, koi fan nahi. $ν (1) = 0$ set karne se $ν$ ek universal ledger ban jaata hai.

$ν (M) = \int_{1}^{M} \frac{M ^{'2} - 1}{1 + \frac{γ - 1}{2} M ^{'2}} \frac{d M ^{'}}{M ^{'}} .$

Yeh standard integral carry out karne par (substitute karo taaki do arctangent pieces alag ho jaayein) closed form milta hai:

Step 6 — Edge case: $M \to \infty$ ek finite ceiling $ν_{m a x}$ deta hai

KYA HAI. $M$ ko infinity tak push karo. Dono arctangents apni maximum value $π /2$ par saturate ho jaate hain (kyunki $M^{2} - 1 \to \infty$ , aur $arctan (\infty) = π /2$ ).

YEH KYUN MATTER KARTA HAI. Tum expect kar sakte ho ki infinite speed ke liye infinite turning hogi. Lekin har arctangent top out kar jaata hai, toh total turning capped hai:

$ν_{m a x} = \frac{γ + 1}{γ - 1} \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1) \approx 130.4 5^{\circ} (γ = 1.4) .$

PICTURE. Poora $ν (M)$ curve: $M = 1$ se steeply rise karta hai, phir $130.4 5^{\circ}$ par horizontal asymptote ki taraf flatten hota hai. Asymptote ke aage ek shaded gap hai jis par "vacuum forms here" likha hai.

Step 7 — Ledger ko dono taraf padhna (fully annotated mini-example)

KYA HAI. $M_{1} = 2$ par $1 5^{\circ}$ expansion ke liye $ν$ ko two-way lookup ki tarah use karo (parent note se match karta hai).

KYUN. Yeh woh ek kaam dikhata hai jo tum $ν$ ke saath karte ho: $M \to ν$ convert karo, turn add/subtract karo, wapas convert karo.

$ν (2) = 26.3 8^{\circ}$ — $M = 2$ tak pahunchne mein pehle hi kharcha hua budget.
Expansion turn add karta hai: $ν (M_{2}) = 26.3 8^{\circ} + 1 5^{\circ} = 41.3 8^{\circ}$ .
Invert karo $ν = 41.3 8^{\circ} \Rightarrow M_{2} \approx 2.60$ (monotonic hai, toh unique hai).

PICTURE. $ν (M)$ curve ek horizontal "budget line" ke saath: $M = 2$ par upar read karo, $1 5^{\circ}$ upar step karo, $M_{2} \approx 2.6$ tak wapas neeche read karo.

Ek-picture summary

Upar sab kuch, compressed: ek supersonic arrow ek convex corner mein enter karta hai; corner ek fan of Mach waves spawn karta hai; arrow wave by wave bend aur lengthen karta hai; har tiny bend $d θ = f (M) d M / M$ hai; total bend, sonic se integrate karke, $ν (M_{2}) - ν (M_{1})$ hai. Dekho Expansion Fan / Centered Rarefaction aur Nozzle Design (Supersonic) jaahan yeh use hota hai.

Recall Poore walkthrough ki Feynman retelling

Socho tum sound se tez ek wall ke saath sprint kar rahe ho. Kyunki tum apni awaaz se aage nikal jaate ho, tumhari awaaz ek slanted line mein pile up ho jaati hai — ek Mach wave — angle $μ$ par tilted jahan $sin μ = 1/ M$ . Ab wall away bend karti hai. Tum ek aisi wave cross karte ho; woh tumhe sirf sideways shove kar sakti hai (khud ke perpendicular), tumhe uske along kabhi nahi kheeench sakti. Woh sideways shove tumhe thoda sa speed up karta hai aur tumhara direction thoda sa swing karta hai — woh "thoda sa" $d θ = M^{2} - 1 d V / V$ hai. Lekin tumhari speed tumhare Mach number se slower grow karti hai, kyunki hawa cool hoti hai aur sound slow ho jaata hai, toh hum $d M$ swap in karte hain aur ek throttle factor $1 + \frac{γ - 1}{2} M^{2}$ pick up karte hain. In saari tiny bends ka ek poore fan mein us moment se add up karna jab tum barely supersonic the, ek single running score deta hai, $ν (M)$ — do neat arctangents. Sabse best part: yeh $13 0^{\circ}$ par stall kar jaata hai. Infinite speed bhi tumhe sirf itna hi turn kar sakti hai; zyada maango toh flow mein vacuum tear ho jaata hai.

Active recall

sin μ = 1/ M

se shuru karke

cot μ = M^{2} - 1

prove karo.

cos μ = 1 - 1/ M^{2} = M^{2} - 1 / M

, toh

cot μ = cos μ / sin μ = M^{2} - 1

Integrand ka area finite kyun hai?

Tail

1/ M^{2}

ki tarah decay karta hai, toh

\int_{1}^{\infty}

converge karta hai — jo

ν_{m a x} \approx 130.4 5^{\circ}

deta hai.

arctan M^{2} - 1

geometrically kya equal karta hai?

9 0^{\circ} - μ

, flow ke relative Mach wave ka tilt.

Connections

Prandtl-Meyer function ν(M) — woh parent result jise yeh page visually derive karta hai.
Mach Waves and Mach Angle — Step 0 building blocks ( $sin μ = 1/ M$ ).
Isentropic Flow Relations — Step 3 mein use hone wala cooling law provide karta hai.
Expansion Fan / Centered Rarefaction — woh physical fan jo yeh waves build karte hain.
Oblique Shock Waves — compression counterpart (non-isentropic).
Method of Characteristics — $ν (M) \pm θ$ ko invariants ki tarah use karta hai.
Nozzle Design (Supersonic) — jaahan ledger apply hota hai.