Prandtl-Meyer expansion waves — isentropic, supersonic turning: Worked examples

3.1.16 · D3 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer expansion waves — isentropic, supersonic turni

Koi bhi symbol aane se pehle, un quantities ka ek reminder jo hum juggle karenge, simple words mein:

Definition The characters

$M$ = Mach number = (flow speed) ÷ (local speed of sound). $M = 1$ bilkul sound ki speed hai; $M > 1$ supersonic hai. Yeh ek pure number hai, koi units nahi.
$θ$ = deflection angle = wall (aur flow) kitne degrees bend hoti hai. Hum ise yahan degrees mein measure karte hain, lekin $ν$ formula radians produce karta hai — hum convert karte hain.
$ν (M)$ = Prandtl–Meyer function = "sonic ( $M = 1$ ) se current $M$ tak accelerate karne mein aapne abhi tak kitne degrees turning spend kar diye hain." Ise turning ke liye ek fuel gauge samjho. $ν (1) = 0$ .
$p$ = static pressure = woh pressure jo ek sensor gas ke saath drift karte hue read karta. Iske values $p_{1}, p_{2}$ woh hain jo fan ke across change hote hain aur jo hum neeche compute karte hain.
$p_{0}$ = stagnation (total) pressure = woh pressure jo gas tab reach karti agar use smoothly rest par laya jaaye. Isentropic fan mein $p_{0}$ constant rehta hai; sirf static $p$ (aur $p_{0}$ se uska ratio) change hota hai. Dekho Isentropic flow relations.

Throughout, air ek perfect gas hai jisme $γ = 1.4$ hai (ratio of specific heats — woh number jo batata hai ki gas energy kaise store karti hai). $γ = 1.4$ ke saath useful constants hain $\frac{γ + 1}{γ - 1} = 6$ aur $\frac{γ - 1}{2} = 0.2$ .

The scenario matrix

Is topic ka har problem in cells mein se ek hai. Neeche ke examples us cell ke saath tagged hain jo woh cover karte hain.

Cell	Kya special hai	Covered by
A. Standard turn	Ordinary $M_{1} > 1$ , moderate $θ$ , find $M_{2}$ aur $p$ ratio	Ex 1
B. Start at sonic	$M_{1} = 1$ toh $ν_{1} = 0$ — degenerate lower limit	Ex 2
C. Vacuum / max turn	$M \to \infty$ , $θ \to ν_{m a x}$ limiting behaviour	Ex 3
D. Wrong sign trap	Koi $θ$ subtract karta hai (compression confusion)	Ex 4
E. Inverse question	Given $M_{1}, M_{2}$ , find the required $θ$	Ex 5
F. Real-world word problem	Airfoil / nozzle lip, physical pressure numbers	Ex 6
G. Over-turn / exam twist	$θ$ itna bada ki $ν_{2} > ν_{m a x}$ → impossible	Ex 7
H. Mach-angle tracking	$μ$ ko fan ke across follow karo (first vs last wave)	Ex 8

Saara kaam do formulas karte hain. Inhe saamne rakho:

Cell A — the standard turn

Worked example Example 1 — Air at

$M_{1} = 2.0$ turns through $θ = 1 0^{\circ}$ Forecast: aage padhne se pehle, guess karo: kya $M$ upar jaayega ya neeche? Kya $p$ badhega ya ghategaa? Kya doosri Mach wave pehli se steeper hogi ya shallower?

Step 1 — $ν (M_{1})$ find karo. $M_{1} = 2$ plug karo: $M^{2} - 1 = 3$ , toh $ν (2) = 6 arctan 0.5 - arctan 3 = 0.4604 rad = 26.3 8^{\circ}$ . Yeh step kyun? $ν (M_{1})$ woh turning hai jo $M = 2$ tak pahunchne mein already "spend" ho chuki hai. Corner ki turn add karne se pehle hume yeh jaanna zaroori hai.

Step 2 — Deflection add karo. Expansion ⇒ add (mnemonic yaad karo "Expand = Away = Add"): $ν (M_{2}) = 26.3 8^{\circ} + 1 0^{\circ} = 36.3 8^{\circ}$ . Yeh step kyun? $ν$ sonic se total turn measure karta hai; corner exactly $θ$ aur contribute karta hai.

Step 3 — $M_{2}$ paane ke liye invert karo. $ν$ monotonically $M$ ke saath badhta hai, isliye exactly ek $M$ hai jo $36.3 8^{\circ}$ deta hai. Numerically solve karne par: $M_{2} \approx 2.385$ . Yeh step kyun? Nayi Mach number chahiye taaki baaki saari properties read kar sakein.

Step 4 — Static-pressure ratio. $0.2 M_{1}^{2} = 0.8$ aur $0.2 M_{2}^{2} = 0.2 (2.385)^{2} = 1.138$ ke saath: $\frac{p _{2}}{p _{1}} = (\frac{1.8}{2.138})^{3.5} \approx 0.55.$ Yeh step kyun? Kyunki fan isentropic hai, stagnation pressure $p_{0}$ fixed rehta hai (jaisa upar define kiya); sirf static pressures $p_{1}, p_{2}$ — woh pressures jo gas ke saath travel karti hain — girti hain. Common $p_{0}$ se unka ratio wahi hai jo isentropic relation deta hai, aur un dono ratios ko divide karne par upar waala formula milta hai.

Step 5 — Forecast ka wave question answer karo. Mach angle $μ = arcsin (1/ M)$ hai. Pehli wave: $arcsin (1/2.0) = 30. 0^{\circ}$ ; aakhri wave: $arcsin (1/2.385) = 24. 8^{\circ}$ . Kyunki $M$ badha, $μ$ gira, toh doosri (aakhri) wave shallower hai pehli se. Yeh step kyun? Forecast ka loop close karna — fan downstream mein khulta hai, bilkul jaisa Ex 8 detail mein study karta hai.

Verify: $M_{2} > M_{1}$ ✓ (flow speed up hua), $p_{2} / p_{1} < 1$ ✓ (static pressure gira), $μ_{2} < μ_{1}$ ✓ (aakhri wave shallower) — bilkul wahi jo "expansion" karna chahiye. Magnitude ka sanity check: $M = 2$ ke paas $1 0^{\circ}$ turn typically $M$ ko kuch tenths upar dhakelta hai; $2.0 \to 2.39$ reasonable hai.

Cell B — exactly sonic par start karna

Worked example Example 2 —

$M_{1} = 1$ par flow $θ = 2 0^{\circ}$ turn karti hai Forecast: $ν (M_{1})$ kya hoga jab flow sound ki speed par hi start karti hai?

Step 1 — $ν (1)$ evaluate karo. $M = 1$ par, $M^{2} - 1 = 0$ , toh dono $arctan$ terms $arctan 0 = 0$ hain. Isliye $ν (1) = 0$ . Yeh step kyun? $M = 1$ hi $ν$ scale ka built-in zero hai — degenerate lower edge. Ise confirm karna matlab hai ki hamaara fuel gauge start mein empty read karta hai.

Step 2 — Turn add karo. $ν (M_{2}) = 0 + 2 0^{\circ} = 2 0^{\circ}$ . Yeh step kyun? Kuch spend nahi hua, toh poora $2 0^{\circ}$ flow ko accelerate karne ke liye available hai.

Step 3 — Invert karo. $ν (M_{2}) = 2 0^{\circ}$ solve karo: $M_{2} \approx 1.775$ . Yeh step kyun? $ν = 2 0^{\circ}$ se seedha $M$ read karna tabhi legal hai jab $ν_{1} = 0$ ho — sabse simple case.

Verify: $M_{2} \approx 1.78 > 1$ ✓ — sonic stream $2 0^{\circ}$ turn karne ke baad ab comfortably supersonic hai. Ex 3 ki logic se cross-check karo: yahi poori $ν$ curve ka entry point hai.

Cell C — vacuum limit (maximum turn)

Worked example Example 3 — Ek sonic stream kitna turn kar sakti hai gas "khatam" hone se pehle?

Forecast: degrees mein maximum turn guess karo. Right angle se zyada ya kam? $9 0^{\circ}$ se zyada?

Step 1 — $ν$ mein $M \to \infty$ lo. Jab $M \to \infty$ , $M^{2} - 1 \to \infty$ , dono $arctan$ arguments blow up karte hain, aur $arctan (\infty) = \frac{π}{2}$ . Toh $ν_{m a x} = 6 \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} (6 - 1) .$ Yeh step kyun? Infinite Mach number physical ceiling hai — gas ne apni saari internal energy kinetic energy mein convert kar li, toh pressure $\to 0$ (vacuum). Aur turning possible nahi hai.

Step 2 — Numbers. $\frac{π}{2} (6 - 1) = 9 0^{\circ} (6 - 1) \approx 130.4 5^{\circ}$ . Yeh step kyun? Yeh ek number har problem ko cap karta hai: $M = 1$ se start hone wala koi bhi expansion kabhi bhi $\approx 130. 5^{\circ}$ turn se zyada nahi kar sakta.

Verify: $6 \approx 2.449$ , toh $2.449 - 1 = 1.449$ , times $9 0^{\circ} \approx 130. 4^{\circ}$ ✓. Yeh $9 0^{\circ}$ se bada hai — supersonic gas vacuum mein expand hote hue right angle se bhi aage turn kar sakti hai.

Cell D — wrong-sign trap

Worked example Example 4 — Koi Ex 1 ke turn ke liye

$ν (M_{2}) = ν (M_{1}) - θ$ likhta hai. Kya galat hota hai? Forecast: agar hum add ki jagah subtract karein, toh predicted $M_{2}$ bahut bada aayega ya bahut chhota?

Step 1 — Galat rule apply karo. $ν (M_{2}) = 26.3 8^{\circ} - 1 0^{\circ} = 16.3 8^{\circ}$ . Yeh step kyun? Hum jaanbujhkar galti follow kar rahe hain taaki iska consequence dekh sakein, sirf batane ke liye nahi.

Step 2 — Invert karo. $ν = 16.3 8^{\circ}$ solve karne par $M_{2} \approx 1.686$ milta hai. Yeh step kyun? Yeh contradiction expose karta hai: "expanded" flow $M_{1} = 2$ se slow nikla.

Step 3 — Absurdity detect karo. Ek expansion (convex corner, flow bahar bend hoti hai) flow ko accelerate karna chahiye. $M_{2} < M_{1}$ matlab humne accidentally ek compression model kar li, jo supersonic flow ke liye ek shock hai — isentropic nahi, fan nahi. Yeh step kyun? Sign check built-in error alarm hai: agar stated expansion ke liye $M_{2} < M_{1}$ aaye, toh sign flip ho gayi hai.

Verify: $1.686 < 2.0$ ✓ (galat answer really flow slow karta hai). Correct rule (Ex 1) ne $2.385 > 2.0$ diya. Rule: Expand = Away = Add.

Cell E — inverse question

Cell F — real-world word problem

Worked example Example 6 — Wind-tunnel nozzle lip

Air ek nozzle se $M_{1} = 2.4$ aur static pressure $p_{1} = 40 kPa$ par nikalti hai, phir ek lip ke paas se gujarti hai jo wall ko flow se $1 5^{\circ}$ door bend karti hai. $M_{2}$ aur downstream pressure $p_{2}$ find karo. Forecast: kya $p_{2}$ , $40 kPa$ se upar hoga ya neeche? Atmospheric ( $101 kPa$ ) se upar ya neeche?

Step 1 — $ν (M_{1})$ . $M_{1}^{2} - 1 = 4.76$ , toh $ν (2.4) = 0.6265 rad = 35.9 0^{\circ}$ . Yeh step kyun? Har baar pehla move same hota hai: starting fuel gauge read karo.

Step 2 — Lip turn add karo. $ν (M_{2}) = 35.9 0^{\circ} + 1 5^{\circ} = 50.9 0^{\circ}$ . Yeh step kyun? Convex lip ek expansion hai, toh add karo.

Step 3 — Invert karo. $ν = 50.9 0^{\circ} \Rightarrow M_{2} \approx 3.061$ . Yeh step kyun? Koi bhi pressure calculation karne se pehle $M_{2}$ chahiye.

Step 4 — Static pressure. $0.2 M_{1}^{2} = 1.152$ , $0.2 M_{2}^{2} = 0.2 (3.061)^{2} = 1.874$ : $p_{2} = 40 (\frac{1 + 1.152}{1 + 1.874})^{3.5} = 40 (0.7487)^{3.5} \approx 14.5 kPa .$ Yeh step kyun? Isentropic ratio Mach change ko ek real static pressure mein convert karta hai, usi kPa units mein jo humein di gayi thi (stagnation pressure $p_{0}$ fan ke across unchanged hai).

Verify: Units: kPa × (dimensionless) $^{3.5}$ = kPa ✓. $14.5 < 40$ ✓ (expansion pressure drop karti hai). Yeh gas ab atmospheric se kaafi neeche hai — ek classic overexpanded exit, jo Nozzle design and overexpansion ke liye relevant hai.

Cell G — impossible over-turn (exam twist)

Worked example Example 7 —

$M_{1} = 4.0$ par Air se $θ = 12 0^{\circ}$ turn karne ko kaha jaata hai. Possible hai? Forecast: hum jaante hain ceiling $ν_{m a x} \approx 130. 5^{\circ}$ hai. Kya $M_{1} = 4$ already uske paas hai?

Step 1 — $ν (M_{1})$ . $M_{1}^{2} - 1 = 15$ , toh $ν (4) = 1.1441 rad = 65.5 5^{\circ}$ . Yeh step kyun? $M = 4$ par $130. 5^{\circ}$ ka bahut saara budget already spend ho chuka hai — check karo kitna bacha hai.

Step 2 — Required $ν (M_{2})$ . $65.5 5^{\circ} + 12 0^{\circ} = 185.5 5^{\circ}$ . Yeh step kyun? Seedha add karo — lekin ab ceiling se compare karo.

Step 3 — $ν_{m a x}$ se compare karo. $185.5 5^{\circ} > 130.4 5^{\circ}$ . Impossible. Yeh step kyun? $ν$ kabhi bhi $ν_{m a x}$ exceed nahi kar sakta; isse zyada maangna matlab hai flow ko $M > \infty$ aur negative pressure chahiye. Physically gas turn complete karne se pehle separate / void ho jaati hai.

Verify: $M = 4$ se remaining budget $130.4 5^{\circ} - 65.5 5^{\circ} = 64.9 0^{\circ}$ hai. Maanga gaya $12 0^{\circ}$ isse exceed karta hai ✓ — correct exam answer hai "nahi, flow itna far turn nahi kar sakti; $M = 4$ se maximum further turn $\approx 64. 9^{\circ}$ hai."

Cell H — fan ke across Mach angle track karna

Worked example Example 8 — Ex 1 ke turn (

$M_{1} = 2.0 \to M_{2} \approx 2.385$ ) ke liye, pehli aur aakhri wave ka Mach angle find karo, aur confirm karo ki fan forward khulta hai. Forecast: kaunsi wave steeper hai — pehli ( $M_{1}$ par) ya aakhri ( $M_{2}$ par)?

Step 1 — Mach angle recall karo. $μ = arcsin (1/ M)$ : woh angle jo har Mach wave local flow ke saath banati hai (parent note mein sound-circle picture se built in). Yeh step kyun? Fan in dono waves se bounded hai; unke angles iska geometry define karte hain.

Step 2 — Pehli wave. $μ_{1} = arcsin (1/2.0) = arcsin (0.5) = 30. 0^{\circ}$ . Yeh step kyun? Yahi fan ki leading, steepest edge hai.

Step 3 — Aakhri wave. $μ_{2} = arcsin (1/2.385) = arcsin (0.4193) = 24.7 9^{\circ}$ . Yeh step kyun? Kyunki $M$ badha, $1/ M$ gira, toh $μ$ gira — trailing wave shallower hai.

Verify: $24.7 9^{\circ} < 30. 0^{\circ}$ ✓ — waves downstream mein shallower hoti jaati hain, toh fan flow direction mein khulta hai. Yeh mistake-buster " $μ$ fan ke through decrease karta hai" se match karta hai.

Neeche ki figure exactly yahi draw karti hai. Upstream wall horizontal hai; corner wall ko $θ = 1 0^{\circ}$ neeche turn karta hai. Red line pehli Mach wave hai incoming flow se $μ_{1} = 3 0^{\circ}$ upar; green line aakhri wave hai, already-turned downstream flow se $μ_{2} = 24. 8^{\circ}$ par draw ki gayi — toh yeh red line se neeche hai, aur beech mein dashed blue lines fan ki intermediate waves hain. Red-to-green spread ka khulna notice karo: woh khulna hi physical expansion fan hai, aur green ka red se shallower hona wahi $μ$ decrease hai jo humne abhi compute ki. Corner par yellow arc $1 0^{\circ}$ deflection mark karta hai.

Figure — Prandtl-Meyer expansion waves — isentropic, supersonic turning

Recall Matrix par quick self-test

Cell B answer — $ν (1)$ kitna hota hai? ::: $0$ , kyunki $M^{2} - 1 = 0$ dono arctangents ko zero banata hai. Cell C answer — $γ = 1.4$ ke liye $ν_{m a x}$ ? ::: $\approx 130.4 5^{\circ} = 9 0^{\circ} (6 - 1)$ . Cell D detector — flipped sign kaise pakdoge? ::: Agar "expanded" flow mein $M_{2} < M_{1}$ aaye, toh tumne add ki jagah subtract kiya. Cell G rule — turn impossible kab hota hai? ::: Jab $ν (M_{1}) + θ > ν_{m a x}$ ; flow ko $M = \infty$ , $p = 0$ chahiye hoga. Cell H trend — fan ke through $μ$ kya karta hai? ::: Decrease karta hai, kyunki $M$ badhta hai aur $μ = arcsin (1/ M)$ .

See also: Mach waves and Mach cone, Isentropic flow relations, Oblique shock waves, Method of characteristics, Entropy and the second law.