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

3.1.17 · D4 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer function ν(M)

Poore page mein, air ke saath $γ = 1.4$ . Teen formulas jo tum baar baar use karoge:

Yahan $M$ = Mach number (flow speed ÷ local sound speed), $θ$ = wall ka turning angle, aur $p_{0}$ = reservoir (stagnation) pressure, jo kisi bhi isentropic fan mein constant rehta hai.

Intuition

$ν (M)$ ya pressure relation ko invert kaise karein Dono $ν (M)$ aur $p / p_{0}$ monotonic hain $M$ mein $M \geq 1$ ke liye ( $ν$ badhta hai, $p / p_{0}$ girta hai), toh har output exactly ek $M$ se correspond karta hai — inverse exist karta hai aur unique hai. Us $M$ ko haath se dhundne ke liye tumhare paas do honest routes hain:

Table + interpolation. Apni target value ko do table rows ke beech bracket karo aur linearly interpolate karo. Example: $ν = 36.3 8^{\circ}$ invert karne ke liye, note karo $ν (2.0) = 26.38$ aur $ν (2.5) = 39.12$ ; linear estimate $M \approx 2.0 + 0.5 \cdot \frac{36.38 - 26.38}{39.12 - 26.38} = 2.0 + 0.5 (0.785) = 2.39$ . (Exact root $2.385$ hai — interpolation close hai kyunki $ν$ yahan almost straight hai.)
Equation solve karo. Closed form ko target ke barabar set karo aur numerically solve karo (Newton's method / ek solver), kyunki koi elementary algebraic inverse nahi hai. Pressure relation ke liye inverse algebraically possible hai: $p / p_{0} = (1 + \frac{γ - 1}{2} M^{2})^{- γ / (γ - 1)}$ se dono sides ko $- \frac{γ - 1}{γ}$ power par raise karo bracket ko free karne ke liye, phir $M$ ke liye solve karo.

Ek short reference table (γ = 1.4) taaki tum inversions haath se check kar sako:

$M$	$ν (M)$ (deg)
1.0	0.00
1.5	11.91
2.0	26.38
2.5	39.12
3.0	49.76
3.5	58.53
4.0	65.78
$\infty$	130.45

Level 1 — Recognition

Exercise 1.1

Ek ek sentence mein batao, (a) kaunsa reference Mach number $ν = 0$ deta hai, aur (b) kya $ν$ badhta hai ya ghatta hai jab ek supersonic flow accelerate karta hai.

Recall Solution

(a) $ν = 0$ at $M = 1$ (sonic). Function define hi is liye kiya gaya hai ki "turning budget" ki counting sonic state se shuru ho, toh jo flow exactly sonic hai usne zero turning spend ki hai. (b) $ν$ $M$ ke saath badhta hai. Derivation se differential relation dekho, $d θ = \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M}$ . $M > 1$ ke liye numerator $M^{2} - 1$ real aur positive hai aur baaki sab positive hai, toh $d ν > 0$ jab bhi $d M > 0$ . Accelerate karo ⇒ $ν$ badhta hai.

Exercise 1.2

Ek flow $M = 2.0$ par ek tiny expansion corner se gujarta hai aur $M = 2.1$ tak speed up hota hai. Kisi bhi formula ke bina batao ki uska static pressure upar gaya ya neeche, aur $ν$ upar gaya ya neeche.

Recall Solution

Expansion = flow accelerates = $ν$ upar, static pressure neeche. Mnemonic yaad karo E.A.V.: Expansion → $ν$ Adds → Velocity (Mach) upar. Pressure hamesha expansion mein girta hai kyunki gas rarefy hoti hai jab woh speed up hone ke liye kaam karti hai.

Level 2 — Application

Exercise 2.1

$M_{1} = 2.0$ par air ek convex (expansion) corner se $θ = 1 0^{\circ}$ par guzarti hai. $M_{2}$ dhundho.

Recall Solution

Step 1 — starting budget lo. $ν (M_{1}) = ν (2.0) = 26.3 8^{\circ}$ . Kyun: ye woh angle hai jo $M = 2$ tak pohonchne mein already "spend" ho chuki hai. Step 2 — turn add karo. Expansion add karta hai ( $θ > 0$ ): $ν (M_{2}) = 26.3 8^{\circ} + 1 0^{\circ} = 36.3 8^{\circ}$ . Kyun: boxed rule $θ = ν (M_{2}) - ν (M_{1})$ . Step 3 — invert karo. Hume woh $M$ chahiye jiska $ν = 36.3 8^{\circ}$ ho. Kyun inversion zaruri hai: $ν (M)$ ka koi elementary algebraic inverse nahi hai, lekin ye monotonic hai (Ex. 1.1b), toh exactly ek $M$ kaam karta hai. Table se bracket karo: $ν (2.0) = 26.38$ , $ν (2.5) = 39.12$ , toh $M$ $(2.0, 2.5)$ mein hai. Closed form ko numerically solve karna (Newton se interpolated guess $2.39$ se) deta hai $M_{2} \approx 2.385$ . Answer: $M_{2} \approx 2.39$ .

Exercise 2.2

$M_{1} = 1.5$ par ek flow ko $M_{2} = 3.0$ tak expand karna hai. Iske liye corner angle $θ$ kya chahiye?

Recall Solution

Ye forward direction hai — koi inversion nahi chahiye, bas dono budgets evaluate karo. $θ = ν (3.0) - ν (1.5) = 49.7 6^{\circ} - 11.9 1^{\circ} = 37.8 5^{\circ} .$ Answer: $θ \approx 37.8 5^{\circ}$ . Kyun: turning angle literally do budgets ka difference hai.

Exercise 2.3

Exercise 2.1 ( $M_{1} = 2.0 \to M_{2} = 2.385$ ) ki flow ke liye, static-pressure ratio $p_{2} / p_{1}$ dhundho.

Recall Solution

Step 1. Fan isentropic hai ⇒ $p_{0}$ constant ⇒ $\frac{p _{2}}{p _{1}} = \frac{p _{2} / p _{0}}{p _{1} / p _{0}}$ . Step 2. $\frac{p _{1}}{p _{0}} = (1 + 0.2 \cdot 2^{2})^{- 3.5} = (1.8)^{- 3.5} = 0.1278$ . Step 3. $\frac{p _{2}}{p _{0}} = (1 + 0.2 \cdot 2.38 5^{2})^{- 3.5} = (2.1378)^{- 3.5} = 0.07031$ . Step 4. $\frac{p _{2}}{p _{1}} = 0.07031/0.1278 \approx 0.550.$ Answer: $p_{2} / p_{1} \approx 0.55$ — pressure gira, jaisa expansion demand karta hai.

Level 3 — Analysis

Exercise 3.1

$M_{1} = 2.0$ par air ek $1 5^{\circ}$ corner ke around expand karti hai (toh $M_{2} \approx 2.60$ , $ν_{2} = 41.3 8^{\circ}$ ). Expansion ek single line nahi balki Mach waves ka ek fan hai. Dhundho ki fan ka har edge incoming flow direction se kitne angle par hai, aur fan space mein kitni total angular width sweep karta hai. Figure dekho.

Neeche di gayi figure is geometry ko sketch karti hai: upar ki horizontal kali line incoming wall/flow hai ( $M_{1} = 2.0$ ), neeche ki kali line woh wall hai jo $θ = 1 5^{\circ}$ neeche bend hoti hai, aur corner origin par hai. Straight fan rays corner se nikalti hain. Sabse upar wali ray leading Mach wave hai jo incoming flow se $3 0^{\circ}$ upar hai; red ray trailing Mach wave hai jo incoming flow se $7.6 2^{\circ}$ upar hai — figure ka key object. Beech ki faint rays intermediate Mach waves hain. Do curved short arrows incoming aur outgoing flow directions ko mark karti hain.

Recall Solution

Fan mein ek leading Mach wave hoti hai (set by $M_{1}$ ) aur ek trailing Mach wave (set by $M_{2}$ ), dono apni khud ki local flow direction se $μ = arcsin (1/ M)$ use karke measure ki jaati hain (valid kyunki dono $M \geq 1$ ). Leading edge: $μ_{1} = arcsin (1/2.0) = 30. 0^{\circ}$ — incoming flow se measure kiya. Trailing edge: $μ_{2} = arcsin (1/2.6) = 22.6 2^{\circ}$ — lekin ye outgoing flow se measure kiya gaya hai, jo khud incoming direction se $θ = 1 5^{\circ}$ door turn ho chuki hai. $ϕ_{2}$ denote karo trailing wave ka angle incoming flow direction se measure karke (taaki dono fan edges ek common reference share karein). Kyunki outgoing flow incoming flow se $θ$ neeche tilt ho gayi hai, hum woh tilt subtract karte hain: $ϕ_{2} = μ_{2} - θ = 22.6 2^{\circ} - 1 5^{\circ} = 7.6 2^{\circ} .$ Fan ki total angular width (leading edge minus trailing edge, dono ab incoming flow se referenced): $Δ = μ_{1} - ϕ_{2} = 30. 0^{\circ} - 7.6 2^{\circ} = 22.3 8^{\circ} .$ Answers: leading wave $30. 0^{\circ}$ par, trailing wave $ϕ_{2} = 7.6 2^{\circ}$ incoming flow line se upar, fan $\approx 22.3 8^{\circ}$ span karta hai.

Exercise 3.2

Differential relation se dikhao ki $M = 1$ par expansion "infinitely eager" hoti hai: ek vanishing turn $d θ$ se $M$ mein ek finite jump hota hai — yani $d M / d θ \to \infty$ jab $M \to 1^{+}$ .

Recall Solution

$d θ = \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M}$ ko rearrange karke: $\frac{d M}{d θ} = \frac{M ( 1 + \frac{γ - 1}{2} M ^{2} )}{M ^{2} - 1} .$ Jab $M \to 1^{+}$ , numerator $\to 1 \cdot (1 + \frac{γ - 1}{2})$ = finite $= 1.2$ (for $γ = 1.4$ ), jabki denominator $M^{2} - 1 \to 0^{+}$ . Toh $d M / d θ \to + \infty$ . Matlab: bilkul sonic ke paas, choti si corner bhi acceleration ka ek burst unleash kar deti hai — fan steeply khul jaata hai. Yehi reason hai ki $ν (M)$ ka shape $M = 1$ ke paas vertical-tangent (square-root) jaisa hai.

Level 4 — Synthesis

Exercise 4.1

Ek symmetric supersonic nozzle exit pressure $p_{1}$ par $M_{1} = 2.0$ deliver karta hai. Exit ek back-pressure region mein flow karta hai jahan $p_{b} = 0.30 p_{1}$ hai. Jet edge expand hoti hai (Prandtl–Meyer) jab tak uska static pressure $p_{b}$ se match nahi ho jaata. Outer streamline ka exit Mach number $M_{2}$ aur woh angle $θ$ dhundho jis se jet boundary turn hoti hai. (Nozzle Design (Supersonic) aur Expansion Fan / Centered Rarefaction se connects karta hai.)

Recall Solution

Step 1 — $p_{2} / p_{0}$ pin karo. Isentropic hai, toh $p_{0}$ constant. Pehle $p_{1} / p_{0} = (1.8)^{- 3.5} = 0.1278$ . Target static pressure $p_{2} = 0.30 p_{1}$ , toh $p_{2} / p_{0} = 0.30 \times 0.1278 = 0.03834$ . Step 2 — $M_{2}$ ke liye pressure relation invert karo. Kyun ye inversion algebraic hai: dono sides ko $- \frac{γ - 1}{γ} = - \frac{1}{3.5}$ par raise karo bracket se exponent strip karne ke liye: $1 + 0.2 M_{2}^{2} = (0.03834)^{- 1/3.5} = 2.5399.$ $M_{2}^{2} = (2.5399 - 1) /0.2 = 7.6995 \Rightarrow M_{2} = 2.775.$ Step 3 — $ν$ se turn angle. $ν (2.0) = 26.3 8^{\circ}$ , $ν (2.775) = 44.8 7^{\circ}$ . $θ = ν (M_{2}) - ν (M_{1}) = 44.8 7^{\circ} - 26.3 8^{\circ} = 18.4 9^{\circ} .$ Answers: $M_{2} \approx 2.78$ , jet boundary $θ \approx 18. 5^{\circ}$ turn karti hai. Kyun ye kaam karta hai: pressure-match condition isentropic relations ke through $M_{2}$ fix karti hai; $ν$ phir batata hai ki boundary ko wahan pohonchne ke liye kitna bend karna pada.

Exercise 4.2

$M_{1} = 2.5$ par ek flow ek $1 2^{\circ}$ expansion corner se gujarti hai, phir turant ek doosre $1 2^{\circ}$ expansion corner se. Final Mach number $M_{3}$ dhundho, aur confirm karo ki ye ek single $2 4^{\circ}$ turn ke result ke barabar hai.

Recall Solution

$ν$ additive hai, toh do turns bas add karte rehte hain: $ν (M_{1}) = ν (2.5) = 39.1 2^{\circ} .$ Dono corners ke baad: $ν (M_{3}) = 39.1 2^{\circ} + 1 2^{\circ} + 1 2^{\circ} = 63.1 2^{\circ} .$ Invert karo (monotonic $ν$ , toh unique root; bracket $ν (3.5) = 58.53$ , $ν (4.0) = 65.78$ se): $M_{3} \approx 3.757.$ Single-turn check: $ν (2.5) + 2 4^{\circ} = 39.1 2^{\circ} + 2 4^{\circ} = 63.1 2^{\circ}$ — identical, toh $M_{3} \approx 3.76$ dono taraf se. Lesson: kyunki turning angle $ν$ -differences se map hota hai, sequential expansions exactly ek bade expansion ki tarah add hote hain. Path matter nahi karta, sirf total angle matter karta hai.

Level 5 — Mastery

Exercise 5.1

Ek flow $M_{1} = 3.0$ par enter karti hai. Wall ek 95° expansion turn demand karti hai. Kya ye physically possible hai? Agar nahi, toh actually kya hota hai, aur flow le sakti hai sabse bada turn kaunsa hai?

Recall Solution

Step 1 — ceiling. Maximum $ν$ (jab $M \to \infty$ ) hai $ν_{m a x} = 130.4 5^{\circ}$ for $γ = 1.4$ . Step 2 — remaining budget. $M_{1} = 3.0$ se shuru karke, $ν (3.0) = 49.7 6^{\circ}$ , toh flow ab bhi at most turn kar sakti hai $θ_{m a x} = ν_{m a x} - ν (M_{1}) = 130.4 5^{\circ} - 49.7 6^{\circ} = 80.6 9^{\circ} .$ Step 3 — verdict. Wall $9 5^{\circ} > 80.6 9^{\circ}$ maangti hai, toh demanded turn budget exceed kar raha hai — ye physically impossible hai ki flow wall se poori tarah chipki rahe. Gas jitna ho sake utna expand hoti hai, poore $80.6 9^{\circ}$ turn karke, jis point par $M \to \infty$ aur static pressure $p \to 0$ . Ye aur aage nahi turn kar sakta kyunki speed mein convert karne ke liye pressure khatam ho gayi. Consequently streamline wall se separate ho jaati hai: last expanded streamline aur remaining wall surface ke beech wedge mein ek vacuum (void) khul jaata hai, aur flow simply us void ke past seedha coast karti hai. Answer: possible nahi; maximum turn $θ_{m a x} \approx 80.6 9^{\circ}$ hai, jiske aage streamline separate ho jaati hai aur wall ke saath vacuum form ho jaata hai.

Exercise 5.2

Prove karo ki $ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1)$ closed form mein $M \to \infty$ le jaake, phir $γ = 1.4$ ke liye degrees mein evaluate karo.

Recall Solution

Jab $M \to \infty$ , dono square-root arguments blow up karte hain: $\frac{γ - 1}{γ + 1} (M^{2} - 1) \to \infty$ aur $M^{2} - 1 \to \infty$ . Kyunki $arctan (x) \to \frac{π}{2}$ jab $x \to \infty$ , har arctan $\frac{π}{2}$ hit karta hai: $ν_{m a x} = \frac{γ + 1}{γ - 1} \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1) .$ $γ = 1.4$ ke liye: $2.4/0.4 = 6 = 2.4495$ , toh $ν_{m a x} = \frac{π}{2} (1.4495) = 2.2770$ rad $= 130.4 5^{\circ}$ . ✓

Exercise 5.3

$M_{1} = 2.5$ par ek supersonic flow ek concave wall se milti hai jo use khud mein $θ = 1 0^{\circ}$ se turn karti hai (ek compression). (a) Agar wall smooth hai (gradual concave curve, isentropic compression), $θ < 0$ branch use karke downstream Mach number $M_{2}$ dhundho. (b) Explain karo ki physically kya change hota hai agar wall ek single sharp $1 0^{\circ}$ concave corner ho, aur kyun $ν$ wahan ab use nahi ho sakti. (Oblique Shock Waves dekho.)

Recall Solution

(a) Smooth isentropic compression. Compression budget se subtract karta hai, toh boxed rule $θ = ν (M_{2}) - ν (M_{1})$ mein $θ = - 1 0^{\circ}$ ke saath: $ν (M_{2}) = ν (M_{1}) + θ = ν (2.5) - 1 0^{\circ} = 39.1 2^{\circ} - 1 0^{\circ} = 29.1 2^{\circ} .$ Invert karo (monotonic $ν$ , unique root; bracket $ν (2.0) = 26.38$ , $ν (2.5) = 39.12$ se): $M_{2} \approx 2.125$ . Flow slow hoti hai aur uska pressure badhta hai, bilkul expansion ka mirror image. Ye legitimate hai kyunki ek gradual concave wall compression Mach waves ko spread out rakhti hai — woh kabhi coalesce nahi karte, toh flow isentropic rehti hai aur $ν$ still apply hota hai. (b) Sharp corner. Ek single sharp concave corner par compression Mach waves ek oblique shock mein coalesce ho jaati hain (parent note, "why a fan not a shock"). Shock ke across, entropy badhti hai, toh reservoir pressure $p_{0}$ girta hai — flow ab isentropic nahi rahi. Kyunki $ν (M)$ isentropic assumption ke under derive kiya gaya tha, ye silently us entropy jump ko ignore karta hai aur wrong downstream state deta hai. Sharp compression ke liye tumhe oblique-shock relations use karne padhenge, $ν$ nahi. Answers: (a) $M_{2} \approx 2.13$ (isentropic, valid); (b) ek shock form hota hai, entropy badhti hai, $ν$ invalid hai — oblique-shock relations use karo.

Connections

Expansion Fan / Centered Rarefaction — woh physical structure jo fan-geometry problems describe karte hain.
Mach Waves and Mach Angle — Level 3 mein use kiya gaya $μ = arcsin (1/ M)$ supply karta hai.
Isentropic Flow Relations — yahan har pressure problem ke liye $p / p_{0}$ tool.
Oblique Shock Waves — irreversible compression counterpart (Exercise 5.3).
Nozzle Design (Supersonic) — jahan Level 4 jet-boundary matching rehta hai.
Method of Characteristics — $ν \pm θ$ invariants se poore flow fields build karta hai.