Prandtl-Meyer function ν(M): Worked examples — Hinglish deep dive

3.1.17 · D3 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer function ν(M)

Yeh page parent ν(M) note ki "sab kuch try karo" companion hai. Numbers grind karne se pehle, chalte hain har tarah ki situation ko naam dete hain jo ek Prandtl–Meyer problem mein aa sakti hai. Phir har worked example us matrix ke cell ke saath tagged hai jo woh cover karta hai, taaki end tak tumne sab dekh liya ho.

Neeche sab kuch air use karta hai, $γ = 1.4$ , jab tak aur bataya na jaye.

Hum teen formulas ka ek master toolbox rakhte hain:

Do words jo hum baar baar bolenge, seedhe define kiye hain:

Score $= ν (M)$ : turning ka running total jo ek flow ne "spent" kiya hai jab se woh barely supersonic thi ( $M = 1$ , score $= 0$ ). Apne aap se door turning score mein adds karta hai.
Fan edges: ek expansion Mach waves ka wedge hoti hai. Pehli (upstream) edge $μ_{1} = arcsin (1/ M_{1})$ par incoming flow se tilt hoti hai; aakhri (downstream) edge $μ_{2} = arcsin (1/ M_{2})$ par outgoing flow se tilt hoti hai. Ek Japanese hand fan ko khulte hue picture karo.

The scenario matrix

Yeh har distinct case class hai jo ek ν(M) problem tumhe de sakti hai. Baad ke har example mein ek ya zyada cells fill hote hain.

#	Cell (case class)	Isme kya special hai	Covered by
A	Forward expansion ( $M_{1}$ , $θ$ se $M_{2}$ nikalna)	Direct problem — bread and butter	Ex 1
B	Pressure/temperature after fan	Chain ν → isentropic ratios	Ex 2
C	Fan geometry (opening angle, edge tilts)	$μ_{1}, μ_{2}$ ko drawn wedge mein turn karta hai	Ex 3 (figure)
D	Degenerate: $θ = 0$	Koi turn nahi ⇒ Mach frozen; sanity anchor	Ex 4
E	Limiting: $M_{1} \to 1$	Fan edge $μ \to 9 0^{\circ}$ ; smallest score	Ex 4
F	Sign / direction trap (compression = subtract?)	Jab ν decrease hoti hai; jab yeh illegal hai	Ex 5
G	Vacuum limit ( $θ$ bahut bada, $M \to \infty$ )	$ν_{m a x}$ ceiling; void banta hai	Ex 6
H	Real-world word problem (jet exhaust plume)	Under-expanded nozzle lip par turning	Ex 7 (figure)
I	Exam twist: given $p_{2} / p_{1}$ , turn nikalna	Poori chain backwards chalana	Ex 8

Worked examples

Ex 1 — Forward expansion (Cell A)

Worked example Statement

$M_{1} = 2.0$ par air ek convex corner se guzarti hai jo wall ko flow se door $θ = 1 5^{\circ}$ turn karti hai. $M_{2}$ nikalo.

Forecast: Kya $M_{2}$ 2 se thoda upar hoga, ya bahut zyada? Padhne se pehle andaza lagao. (Hint: 15° woh ~26° score ka bada hissa hai jo $M = 2$ carry karta hai.)

Step 1. Incoming score compute karo $ν (M_{1}) = ν (2.0)$ . $ν (2) = 6 arctan \frac{0.4}{2.4} (3) - arctan 3 = 26.3 8^{\circ} .$ Yeh step kyun? Humein jaanna hai ki flow ne is corner se pehle kitna turning already spend kiya hai, kyunki ν ek per-corner quantity nahi, balki running total hai.

Step 2. Turn add karo: $ν (M_{2}) = ν (M_{1}) + θ = 26.3 8^{\circ} + 1 5^{\circ} = 41.3 8^{\circ}$ . Yeh step kyun? Expansion wall ko door turn karti hai, Mach waves spread karte hain — boxed rule $θ = ν (M_{2}) - ν (M_{1})$ (upar bataya) kehta hai score exactly $θ$ se upar jaata hai.

Step 3. $ν (M_{2}) = 41.3 8^{\circ}$ invert karo $\Rightarrow M_{2} \approx 2.60$ . Yeh step kyun? ν strictly increasing in $M$ hai, isliye ek score exactly ek Mach number ko point karta hai. Hum $M$ ke liye ν equation numerically solve karte hain.

Verify: $ν (2.60)$ directly recompute karo — yeh $41. 4^{\circ}$ return karna chahiye, aur $41.4 - 26.38 = 15. 0^{\circ}$ . ✓ Units: saare angles throughout degrees mein. Physically $M$ badha (expansion flow ko speed up karta hai), Cell A se match karta hai.

Ex 2 — Pressure after the fan (Cell B)

Worked example Statement

Ex 1 jaisa hi corner ( $M_{1} = 2 \to M_{2} = 2.6$ ). Static-pressure ratio $p_{2} / p_{1}$ nikalo.

Forecast: Expansion flow ko accelerate karta hai. Kya static pressure upar jaayegi ya neeche? Jawab commit karo.

Step 1. Kyunki fan isentropic hai, stagnation pressure $p_{0}$ (upar define kiya) unchanged hai. Toh likho $\frac{p _{2}}{p _{1}} = \frac{p _{2} / p _{0}}{p _{1} / p _{0}}$ . Yeh step kyun? Humein $p_{0}$ nahi pata, lekin woh cancel ho jaata hai. $p_{0}$ ki constancy exactly woh cheez hai jo ν ko usable banati hai — isliye hum smooth relations ko integrate kar sake.

Step 2. Pressure converter tool (2) use karke: $\frac{p _{1}}{p _{0}} = (1 + 0.2 \cdot 2^{2})^{- 3.5} = 0.1278, \frac{p _{2}}{p _{0}} = (1 + 0.2 \cdot 2. 6^{2})^{- 3.5} = 0.0501.$ Yeh step kyun? Yeh single formula har Mach number ko pressure fraction mein convert karta hai; higher $M$ matlab larger bracket, aur negative exponent ise smaller fraction banata hai.

Step 3. $\frac{p _{2}}{p _{1}} = \frac{0.0501}{0.1278} \approx 0.392$ . Yeh step kyun? Fractions ka ratio chain ν → M → p complete karta hai.

Verify: $0.392 < 1$ , toh pressure giri — expansion ke liye correct hai (flow work karta hai accelerate, cool, aur rare karne mein). Cell B confirmed.

Ex 3 — Fan geometry: yeh kitna wide khulta hai? (Cell C)

Worked example Statement

Ex 1 ke fan ke liye ( $M_{1} = 2 \to M_{2} = 2.6$ , turn $1 5^{\circ}$ ), pehle aur aakhri fan edges ka tilt nikalo aur woh total angular width nikalo jo fan lab frame mein sweep karta hai.

Forecast: Flow speed up hone par, kya har successive Mach wave flow direction ki taraf zyada lean karti hai ya door? Andaza lagao.

Step 1. Incoming flow se pehle edge ka tilt: $μ_{1} = arcsin (1/2) = 30.0 0^{\circ}$ . Yeh step kyun? Leading Mach wave upstream flow ke Mach angle par baithti hai — yahi Mach wave ki definition hai (Mach Waves and Mach Angle).

Step 2. Outgoing flow se aakhri edge ka tilt: $μ_{2} = arcsin (1/2.6) = 22.6 2^{\circ}$ . Yeh step kyun? Aakhri wave tak flow already $M_{2}$ reach kar chuki hai; wave us faster flow ke Mach angle par lean karti hai. Faster flow ⇒ smaller Mach angle ⇒ wave flow ke against flatter hoti hai.

Step 3. Dono edges ko original flow direction se measure karo. Pehla edge incoming flow se $μ_{1} = 3 0^{\circ}$ upar hai. Outgoing flow khud $θ = 1 5^{\circ}$ neeche rotate ho gayi hai, aur aakhri edge us rotated flow se $μ_{2} = 22.6 2^{\circ}$ upar hai. Toh aakhri edge original flow line se $μ_{2} - θ = 22.62 - 15 = 7.6 2^{\circ}$ upar baithti hai. Total fan sweep: $Δ = μ_{1} - (μ_{2} - θ) = 30 - 7.62 = 22.3 8^{\circ} .$ Yeh step kyun? Hum dono edges ko ek common reference (original flow) ke against line up karte hain taaki unka angular gap — fan wedge ki physical width — meaningful ho.

Verify: Fan $22.3 8^{\circ}$ span karta hai, $1 5^{\circ}$ turn se wider, jo sahi hai: fan mein dono Mach cones plus turning hona chahiye, sirf turning nahi. Cell C ✓.

Ex 4 — Degenerate & limiting inputs (Cells D and E)

Worked example Statement

(a) Ek "corner" wall ko $M_{1} = 2$ par $θ = 0^{\circ}$ bend karta hai. $M_{2}$ kya hai? (b) Ek flow just barely supersonic hai, $M_{1} = 1.0$ . Iska score $ν (1)$ kya hai, aur pehli Mach wave ka tilt $μ_{1}$ kya hai?

Forecast: (b) ke liye, guess karo ki jab flow exactly sonic ho toh Mach wave kahaan point karti hai — forward lean, sideways, ya straight across?

Step 1 (a). $ν (M_{2}) = ν (M_{1}) + 0 = ν (2)$ , toh $M_{2} = M_{1} = 2$ . Yeh step kyun? Zero turn score mein zero add karta hai; ν one-to-one hai, isliye Mach number move nahi kar sakta. Yeh sanity anchor hai: koi wall bend nahi, koi expansion nahi.

Step 2 (b). $ν (1) = 6 arctan (0) - arctan (0) = 0^{\circ}$ . Yeh step kyun? $M = 1$ woh reference state hai jo score $= 0$ define karta hai; dono arctan arguments mein $M^{2} - 1 = 0$ hai. Yeh domain ka bilkul edge hai ( $M \geq 1$ ).

Step 3 (b). $μ_{1} = arcsin (1/1) = arcsin (1) = 9 0^{\circ}$ . Yeh step kyun? Exactly $M = 1$ par Mach "wave" ek plane hai jo flow ke perpendicular khadi hai — yeh flow ko bilkul outrun nahi kar sakti, isliye stream mein flat pad jaati hai. Jaise $M$ badhta hai wave forward lean karti hai aur $μ$ shrink hota hai.

Verify: (a) $M_{2} = 2$ exactly — degenerate check pass. (b) $ν (1) = 0$ aur $μ (1) = 9 0^{\circ}$ ; yeh do boundary values hain jahan se har table start hoti hai. Cells D aur E ✓.

Ex 5 — The sign / direction trap (Cell F)

Worked example Statement

$M_{1} = 2.6$ par supersonic flow ek wall se milti hai jo flow ki taraf (concave) $1 0^{\circ}$ bend karti hai. Ek student likhta hai $ν (M_{2}) = ν (2.6) - 1 0^{\circ}$ aur $M_{2}$ read off karta hai. Yeh kab legal hai, aur yahan kya deta hai — aur trap kahaan hai?

Forecast: Flow mein into bend karne se woh speed up hogi ya slow down? Compute karne se pehle guess karo.

Step 1. $ν (2.6) = 41.3 8^{\circ}$ compute karo (Ex 1 se). Subtract karke: $ν (M_{2}) = 41.38 - 10 = 31.3 8^{\circ}$ . Yeh step kyun? Hum "compression = negative expansion" idea test kar rahe hain. Flow mein turning score lower karega, toh subtraction woh arithmetic hai jo yeh idea predict karta hai.

Step 2. $ν = 31.3 8^{\circ}$ invert karo $\Rightarrow M_{2} \approx 2.19$ . Toh flow slow down hogi. Yeh step kyun? Lower score ⇒ lower Mach number; ν monotonic. Compression par slowdown directionally correct hai.

Step 3 — the trap. Yeh sirf ek smooth, gradual concave wall ke liye valid hai (isentropic compression, fan ka mirror). Ek sharp concave corner ke liye Mach waves ek oblique shock mein pile up hoti hain — ek discontinuity jo entropy raise karti hai — aur ν apply nahi hota. Tumhe Oblique Shock Waves relations par switch karna hoga. Yeh step kyun? ν isentropic flow assume karke derive kiya gaya tha. Sharp compression woh assumption tod deta hai; wahan $ν - θ$ use karne se galat $M_{2}$ milega aur, worse, galat (bahut kam) entropy milegi.

Verify: Ek smooth wall ke liye $M_{2} \approx 2.19$ self-consistent hai: $ν (2.19) = 31. 4^{\circ}$ aur $41.38 - 31.4 = 1 0^{\circ}$ . ✓ Lekin flag: sharp corner ⇒ shock, yeh nahi. Cell F ✓.

Ex 6 — The vacuum ceiling (Cell G)

Worked example Statement

Ek flow $M_{1} = 3.0$ par enter karti hai. (a) Physically expansion se woh maximum kitna turn kar sakti hai? (b) Agar ek corner $9 5^{\circ}$ demand kare toh kya hota hai?

Forecast: Kya flow door turn karne ki koi hard maximum limit hai, ya woh forever turn kar sakti hai?

Step 1. $ν (3) = 6 arctan \frac{0.4}{2.4} (8) - arctan 8 = 49.7 6^{\circ}$ . Yeh step kyun? Humein already spend kiya gaya score jaanna hai; remaining budget ceiling minus yeh hai.

Step 2. Absolute ceiling hai $ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1) = 130.4 5^{\circ}$ , sirf $M \to \infty$ par reach hota hai. Yeh step kyun? $arctan$ $9 0^{\circ}$ par top out karta hai; closed form mein $M \to \infty$ feed karne se finite maximum score milta hai — flow is se zyada turning spend nahi kar sakti.

Step 3. Largest turn: $θ_{m a x} = ν_{m a x} - ν (3) = 130.45 - 49.76 = 80.6 9^{\circ}$ . Yeh step kyun? Is tak koi bhi turn $M$ ko infinity ki taraf aur $p \to 0$ drive karta hai.

Step 4 (b). Ek demanded $9 5^{\circ} > 80.6 9^{\circ}$ impossible hai: flow corner finish karne se pehle $M = \infty$ , $p = 0$ reach kar leti hai. Flow aur wall ke beech ek vacuum void ban jaata hai. Yeh step kyun? Physics score cap karta hai; wall flow ki gap fill karne ki ability ko "outrun" kar leti hai.

Verify: $80.6 9^{\circ}$ $ν_{m a x}$ se kam aur positive hai; $9 5^{\circ}$ ise exceed karta hai, toh impossibility flag fire hota hai. Cell G ✓.

Ex 7 — Real-world: nozzle lip par jet exhaust plume (Cell H)

Worked example Statement

Ek supersonic nozzle gas ko $M_{1} = 2.0$ par exit karta hai static pressure $p_{1}$ ke saath jo ambient se higher hai (ek under-expanded jet). Sharp lip par, jet boundary tab tak turn karti hai jab tak uska pressure ambient ke barabar na ho jaaye. Ambient pressure $p_{a} = 0.5 p_{1}$ hai. Plume lip par kitne angle se turn karta hai, aur kis Mach number tak?

Forecast: Under-expanded matlab gas outside ke liye "bahut moti" hai. Kya woh expand hokar outward bend karti hai, ya compress hokar lip par inward? Guess karo.

Step 1. Pressure demand ko Mach number mein convert karo. Kyunki $p_{0}$ fixed hai aur $p_{1} / p_{0} = 0.1278$ (Ex 2), target hai $p_{2} / p_{0} = (p_{a} / p_{1}) (p_{1} / p_{0}) = 0.5 \times 0.1278 = 0.0639$ . Yeh step kyun? Plume boundary ek free surface hai: yeh tab tak turning karti rehti hai jab tak uski static pressure ambient se match na kare. Pressure boundary condition hai, isliye hum ise $M_{2}$ mein translate karte hain.

Step 2. Solve karo $(1 + 0.2 M_{2}^{2})^{- 3.5} = 0.0639 \Rightarrow M_{2} \approx 2.43$ . Yeh step kyun? Isentropic pressure relation (tool 2) invert karo us Mach number ke liye jo ambient pressure produce karta hai.

Step 3. Turning angle: $θ = ν (M_{2}) - ν (M_{1}) = ν (2.43) - ν (2.0)$ . Numerically $ν (2.43) = 37.5 5^{\circ}$ , toh $θ = 37.55 - 26.38 = 11.1 7^{\circ}$ . Yeh step kyun? Plume expand hoti hai (pressure ambient ki taraf drop hoti hai), isliye boundary exactly ν gap se outward bend karti hai (boxed rule).

Verify: $p_{2} / p_{1} = 0.0639/0.1278 = 0.50 = p_{a} / p_{1}$ ✓ (ambient se match karta hai). Flow $2.0 \to 2.43$ accelerate hui aur $11.1 7^{\circ}$ outward bend hui — ek textbook under-expanded plume. Cell H ✓.

Ex 8 — Exam twist: chain ko backwards chalao (Cell I)

Worked example Statement

Ek fan static pressure ko $p_{2} / p_{1} = 0.30$ tak drop karta hai $M_{1} = 1.8$ se start karke. Turning angle $θ$ required nikalo.

Forecast: Tumhe pressure result diya gaya hai aur cause (turn) poochha gaya hai. Pehle kaun si quantity nikalte ho — $M_{2}$ ya $ν_{2}$ ?

Step 1. Tool (2) se $M_{1}$ ka pressure fraction lo: $p_{1} / p_{0} = (1 + 0.2 \cdot 1. 8^{2})^{- 3.5} = 0.1740$ . Yeh step kyun? Humein $p_{0}$ (isentropic fan ke across constant) se anchor karna hai pehle $M_{2}$ locate karne se.

Step 2. Target fraction: $p_{2} / p_{0} = 0.30 \times 0.1740 = 0.05219$ . $M_{2}$ get karne ke liye invert karo: $(1 + 0.2 M_{2}^{2})^{- 3.5} = 0.05219 \Rightarrow M_{2} \approx 2.575$ . Yeh step kyun? Pressure ratio, times $p_{1} / p_{0}$ , $p_{2} / p_{0}$ pin karta hai, aur isentropic relation ise $M_{2}$ mein convert karta hai. Yeh "backwards" leg hai jo exam test kar raha hai.

Step 3. Ab boxed rule use karo: $θ = ν (M_{2}) - ν (M_{1}) = ν (2.575) - ν (1.8)$ . Numerically $ν (1.8) = 20.7 3^{\circ}$ aur $ν (2.575) = 40.9 5^{\circ}$ , toh $θ = 40.9 5^{\circ} - 20.7 3^{\circ} = 20.2 2^{\circ} .$ Yeh step kyun? Dono Mach numbers known hone par, turn sirf score difference hai — boxed rule, pehle ki jagah last mein use kiya gaya.

Step 4. Answer state karo: fan ko woh pressure drop produce karne ke liye $θ \approx 20. 2^{\circ}$ wall turn chahiye tha. Yeh step kyun? Yeh woh quantity hai jo exam ne poochi; iske pehle sab kuch pressure ratio se angle tak jaane ka scaffolding tha.

Verify: Forward-check: $M_{1} = 1.8$ se $20.2 2^{\circ}$ turning $ν_{2} = 40.9 5^{\circ} \Rightarrow M_{2} = 2.575 \Rightarrow p_{2} / p_{0} = 0.0522$ deta hai, aur $0.0522/0.1740 = 0.30$ ✓. Poori chain apne aap par close hoti hai. Cell I ✓.

Recall

Recall Kya har cell cover hua?

Forecast test: naam batao kis example ne vacuum ceiling handle kiya. Which example? ::: Ex 6 (Cell G) — $θ_{m a x} = ν_{m a x} - ν (M_{1})$ . Frozen-Mach kaunse cell mein rehta hai ::: Cell D, Ex 4a — $θ = 0 \Rightarrow M_{2} = M_{1}$ . Backwards problem (pressure → turn) kaun sa example hai ::: Ex 8 (Cell I). $ν_{2} = ν_{1} - θ$ kab legal hai ::: sirf smooth isentropic concave wall ke liye; sharp corner shock banata hai (Ex 5, Cell F). ν(M) ka domain ::: sirf $M \geq 1$ (supersonic); sonic se neeche undefined hai.

Connections

Prandtl-Meyer function ν(M) — woh parent jise yeh page drill karta hai.
Isentropic Flow Relations — Ex 2, 7, 8 mein $p / p_{0}$ conversions.
Mach Waves and Mach Angle — Ex 3, 4 mein edge tilts $μ$ .
Oblique Shock Waves — correct tool jab Ex 5 ki compression sharp ho.
Expansion Fan / Centered Rarefaction — physical fan jo Ex 3 draw karta hai.
Nozzle Design (Supersonic) — Ex 7 ka plume.
Method of Characteristics — in fans ke downstream $ν \pm θ$ ko invariants ke roop mein use karta hai.