Prandtl-Meyer expansion waves — isentropic, supersonic turning: Exercises

3.1.16 · D4 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer expansion waves — isentropic, supersonic turni

Shuru karne se pehle, yeh toolbox dekh lo jo baar baar kaam aayegi. Har symbol parent note mein define kiya gaya tha; yeh bas woh shelf hai jahan se tum utha sakte ho.

Yahan $M$ Mach number hai (flow speed divided by local speed of sound), $θ$ woh angle hai jitna wall flow se door mud jaati hai, $p$ static pressure hai, $T$ static temperature hai, aur subscript $0$ stagnation (rest par laaya gaya) value mark karta hai. Radians aur degrees dono neeche aate hain — main hamesha bataunga kaun sa, kyunki $ν$ radians mein paida hota hai lekin degrees mein tabulate hota hai.

Figure s01 (L1 se pehle padho). Ek convex corner scale ke saath bana hua hai: incoming supersonic stream (teal, neeche left) wall se milti hai, jo door mud jaati hai ek angle se. Corner point se Mach lines ka fan khulta hai — do bold orange lines pehli (steep, $μ_{1}$ par) aur aakhri (shallow) waves hain; thin plum lines intermediate waves hain. Doosri taraf teal arrow dikhata hai ki flow faster nikl rahi hai ( $M_{2} > M_{1}$ ) aur naye wall direction mein turn ho gayi hai. L1–L3 ke har problem mein is geometry ka reference hai, isliye abhi isko yaad kar lo: pehli wave steep, aakhri wave shallow, flow speed up hoti hai.

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

Level 1 — Recognition

Recall Solution L1.1

Parent note ka mnemonic hai "Expand = Away = Add."

$M$ badhta hai — flow accelerate hoti hai (expansion ka yahi poora point hai).
$p$ ghatta hai — static pressure drop hoti hai jab internal energy kinetic energy ban jaati hai.
$T$ ghatta hai — gas cool hoti hai usi reason se.
$μ = arcsin (1/ M)$ : jab $M$ badhta hai, $1/ M$ ghatta hai, isliye $μ$ ghatta hai — waves downstream shallower hoti jaati hain.

Yeh kaisa dikhta hai: figure s01 mein bold orange pehli wave steep hai aur bold orange aakhri wave shallow hai — wave angle ka yeh visible shrinkage hi $μ$ ka decrease hai, aur entry arrow se teal exit arrow ka lamba hona hi $M$ ka increase hai.

Recall Solution L1.2

Tool 1 use karo: $μ = arcsin (1/2.0) = arcsin (0.5) = 3 0^{\circ}$ . $arcsin$ kyun? Hum ratio $sin μ = 1/ M$ jaante hain aur poochhte hain "kaun sa angle hai jiska sine yeh hai?" — yahi $arcsin$ (inverse sine) answer karta hai.

Recall Solution L1.3

False. Fan isentropic hota hai — infinitesimally weak Mach waves ka ek continuum, jisme se har ek entropy contribute karta hai $\propto (strength)^{3} \to 0$ . Kyunki entropy constant hai, $p_{0}$ constant hai. (Oblique shock waves se compare karo, jahan $p_{0}$ sach mein girta hai — dekho Entropy and the second law.)

Level 2 — Application

Recall Solution L2.1

$γ = 1.4$ ke saath: $\frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6$ , isliye $6 \approx 2.4495$ , aur uska reciprocal $1/6 \approx 0.4082$ . Aur $M^{2} - 1 = 3$ , isliye $M^{2} - 1 = 3 \approx 1.7321$ . $ν = 6 2.4495 arctan (0.4082 \times 1.7321) - arctan (1.7321)$ $= 2.4495 arctan (0.7071) - arctan (1.7321) = 2.4495 (0.6155) - 1.0472 = 0.4604 rad .$ Convert karte hain: $0.4604 \times \frac{180}{π} \approx 26.3 8^{\circ}$ . Pehle radians kyun? Dono $arctan$ terms radians return karte hain; main sirf end mein convert karta hun taaki dono pieces same footing par subtract hon.

Recall Solution L2.2

Tool 3: $ν (M_{2}) = ν (M_{1}) + θ = 26.3 8^{\circ} + 1 0^{\circ} = 36.3 8^{\circ}$ . Invert karo (given): $M_{2} \approx 2.385$ . Add kyun karte hain? Expansion flow ko door turn karta hai, isliye hum $θ$ add karte hain — corner $1 0^{\circ}$ aur "turning budget" spend karta hai, flow ko higher $M$ par push karta hai.

Recall Solution L2.3

$1 + \frac{γ - 1}{2} M^{2} = 1 + 0.2 M^{2}$ use karo: $1 + 0.2 (2.0)^{2} = 1.800, 1 + 0.2 (2.385)^{2} = 1 + 0.2 (5.688) = 2.138.$ $\frac{p _{2}}{p _{1}} = (\frac{1.800}{2.138})^{3.5} = (0.8419)^{3.5} \approx 0.548 .$ Exponent $3.5$ kyun? $\frac{γ}{γ - 1} = \frac{1.4}{0.4} = 3.5$ . Pressure lagbhag $55%$ tak aa gayi — expansion static pressure low karti hai, bilkul jaise predict kiya tha.

Level 3 — Analysis

Recall Solution L3.1

Kyunki $ν (1) = 0$ by definition (sonic turning score ka zero hai), turn rule collapse ho jaata hai $ν (M_{2}) = 0 + 2 0^{\circ} = 2 0^{\circ}$ par, jise hum directly read karte hain. Invert karte hain: $M_{2} \approx 1.775$ . Kaisa dikhta hai: fan ki bilkul pehli Mach wave $μ = arcsin (1/1) = 9 0^{\circ}$ par khadi hai — flow ke perpendicular — yeh sabse steep wave hai jo fan start kar sakta hai.

Recall Solution L3.2

Pehli wave: $μ_{1} = arcsin (1/1.0) = 9 0^{\circ}$ . Aakhri wave: $μ_{2} = arcsin (1/1.775) = arcsin (0.5634) = 34.2 8^{\circ}$ . Lekin flow khud bhi $θ = 2 0^{\circ}$ se swing ho chuki hai. Mach lines local flow direction se measure hoti hain, isliye fan edges ki lab-frame angular spread hai $Δ = (μ_{1}) - (μ_{2} - θ) = 9 0^{\circ} - (34.2 8^{\circ} - 2 0^{\circ}) = 9 0^{\circ} - 14.2 8^{\circ} = 75.7 2^{\circ} .$ $θ$ subtract kyun? Aakhri wave $μ_{2}$ par hai already-turned flow se, aur woh flow wall ki taraf $θ$ rotate ho chuki hai. Original flow direction se measure karo toh aakhri edge $μ_{2} - θ$ par hai. Figure s02 teeno angles explicitly draw karta hai: teal arrow original flow hai, plum arrow turned flow hai ( $θ$ se rotate, wall ki taraf), solid orange line pehli wave hai $9 0^{\circ}$ par, aur dashed orange line aakhri wave hai — orange lines ke beech ka gap hi boxed $75.7 2^{\circ}$ hai.

Recall Solution L3.3

Subtraction compression direction hai: yeh $ν$ ko lower karta hai, isliye $M$ lower hota hai — flow mein ghus ke turn karna (shock family), expansion nahi. Convex corner ke liye add karna chahiye: $ν_{2} = 26.3 8^{\circ} + 1 0^{\circ} = 36.3 8^{\circ} \Rightarrow M_{2} \approx 2.385$ (L2.2 jaisa hi). Rule of thumb: agar tumhara "expansion" $M_{2} < M_{1}$ ya $p_{2} > p_{1}$ deta hai, toh tumhara sign flipped hai.

Level 4 — Synthesis

Recall Solution L4.1

Step 1 — pressure match se $M_{2}$ nikalo. Kyunki $p_{0}$ fixed hai (isentropic), tool 4 use karo $p_{2} = p_{a} = 25$ ke saath: $\frac{p _{2}}{p _{1}} = \frac{25}{40} = 0.625 = (\frac{1 + 0.2 M _{1}^{2}}{1 + 0.2 M _{2}^{2}})^{3.5} .$ $1 + 0.2 (2.4)^{2} = 2.152$ ke saath: $(\frac{2.152}{1 + 0.2 M _{2}^{2}})^{3.5} = 0.625 \Rightarrow \frac{2.152}{1 + 0.2 M _{2}^{2}} = 0.62 5^{1/3.5} = 0.8783.$ $1 + 0.2 M_{2}^{2} = \frac{2.152}{0.8783} = 2.450 \Rightarrow M_{2}^{2} = \frac{1.450}{0.2} = 7.252 \Rightarrow M_{2} \approx 2.693 .$ Step 2 — turning scores se $θ$ nikalo. $ν (2.693)$ : $M^{2} - 1 = 6.252$ ke saath, $M^{2} - 1 = 2.500$ , $ν = 2.4495 arctan (0.4082 \times 2.500) - arctan (2.500) = 2.4495 arctan (1.0206) - 1.1903.$ Ab $arctan (1.0206) = 0.7955 rad$ hai, isliye product hai $2.4495 \times 0.7955 = 1.9486$ (na ki $2.95$ — woh multiplication dhyan se karo): $ν = 1.9486 - 1.1903 = 0.7583 rad = 43.4 5^{\circ} .$ Phir $θ = ν (M_{2}) - ν (M_{1}) = 43.4 5^{\circ} - 36.7 5^{\circ} = 6.7 0^{\circ}$ . Yeh order kyun? Pressure condition downstream state $M_{2}$ fix karta hai; turning score difference phir batata hai ki jet boundary kitna swing hoti hai. Dekho Nozzle design and overexpansion.

Recall Solution L4.2

Turning scores simply accumulate hote hain kyunki $ν$ ek common origin (sonic) se define hota hai: $ν (M_{3}) = ν (M_{1}) + θ_{a} + θ_{b} = 14.8 6^{\circ} + 8^{\circ} + 1 2^{\circ} = 34.8 6^{\circ} .$ Invert karo: $M_{3} \approx 2.334$ . Hum dono angles add kyun kar sakte hain? Har fan isentropic hai aur ek hi $p_{0}, T_{0}$ share karta hai, isliye intermediate state alag se compute karne ki zaroorat nahi — sirf total turn matter karta hai. Yeh Method of characteristics ka discrete cousin hai.

Recall Solution L4.3

$\frac{T _{2}}{T _{1}} = \frac{1 + 0.2 ( 2.4 ) ^{2}}{1 + 0.2 ( 2.693 ) ^{2}} = \frac{2.152}{2.450} = 0.878 .$ Gas apne temperature (kelvin mein) ke lagbhag $88%$ tak cool hoti hai — modest, kyunki turn sirf $\sim 7^{\circ}$ hai. Kyunki fan isentropic hai, yeh drop puri tarah recoverable hai: flow ko wapas rest par lao aur $T_{0}$ unchanged wapas aata hai.

Level 5 — Mastery

Recall Solution L5.1

Jab $M \to \infty$ : $M^{2} - 1 \to \infty$ , isliye dono $arctan$ terms $π /2$ approach karti hain:

=\frac{\pi}{2}\left(\sqrt{\frac{\gamma+1}{\gamma-1}}-1\right).$$ $\gamma=1.4$ ke liye: $\dfrac{\pi}{2}(\sqrt6-1)=\dfrac{\pi}{2}(1.4495)=2.2769\ \text{rad}=\boxed{130.45^\circ}$. **Interpretation:** sonic se shuru karke tum stream ko kabhi $\approx130.5^\circ$ se zyada nahi mod sakte. Us se aage gas ko infinite Mach number aur **zero** static pressure chahiye hogi — perfect vacuum mein expansion. Koi bhi aur wall turn sirf flow aur wall ke beech ek vacuum void chhod deta hai.

Recall Solution L5.2

Likho $ν = k arctan (u) - arctan (w)$ jahan $k = \frac{γ + 1}{γ - 1}$ , $u = \frac{γ - 1}{γ + 1} (M^{2} - 1)$ , $w = M^{2} - 1$ . Differentiate kyun karte hain? $ν$ $d ν / d M$ ko integrate karke bana tha; closed form sahi hai ya nahi iska ek honest check yahi hai ki derivative wapas chalao aur same integrand milta hai ya nahi.

$\frac{d}{d M} arctan (f) = \frac{f ^{'}}{1 + f ^{2}}$ use karke:

Second term. $w = M^{2} - 1 \Rightarrow w^{'} = \frac{M}{M ^{2} - 1}$ , aur $1 + w^{2} = M^{2}$ , isliye $\frac{d}{d M} arctan (w) = \frac{M / M ^{2} - 1}{M ^{2}} = \frac{1}{M M ^{2} - 1} .$

First term. Maano $c = \frac{γ - 1}{γ + 1}$ hai isliye $u = c (M^{2} - 1)$ , $u^{'} = \frac{c M}{c ( M ^{2} - 1 )}$ , aur $1 + u^{2} = 1 + c (M^{2} - 1)$ . Tab

=\frac{k\sqrt{c}\,M}{\sqrt{M^2-1}\,\big(1+c(M^2-1)\big)}.$$ Note karo $k\sqrt{c}=\sqrt{\tfrac{\gamma+1}{\gamma-1}}\sqrt{\tfrac{\gamma-1}{\gamma+1}}=1$, aur $1+c(M^2-1)=\dfrac{(\gamma+1)+(\gamma-1)(M^2-1)}{\gamma+1}=\dfrac{2+(\gamma-1)M^2}{\gamma+1} =\dfrac{2}{\gamma+1}\big(1+\tfrac{\gamma-1}{2}M^2\big)$. Isliye pehla term ban jaata hai $$\frac{M}{\sqrt{M^2-1}}\cdot\frac{\gamma+1}{2\big(1+\frac{\gamma-1}{2}M^2\big)}.$$ **Subtract karo.** Ek common factor $\dfrac{1}{M\sqrt{M^2-1}}$ bahar nikaalne ke baad (pehla term $M$ ke upar likh ke), algebra collapse ho jaata hai $$\frac{d\nu}{dM}=\frac{\sqrt{M^2-1}}{M\big(1+\frac{\gamma-1}{2}M^2\big)},$$ bilkul wahi integrand. **Yahi woh confirmation hai jo reviewer chahta tha**: closed form sach mein antiderivative hai. Numerically $M=2$, $\gamma=1.4$ par: $$\frac{d\nu}{dM}=\frac{\sqrt3}{2(1.8)}=\frac{1.7321}{3.6}=\boxed{0.4811\ \text{rad}}.$$ (Poori symbolic simplification VERIFY mein machine-checked hai.)

Recall Solution L5.3

$μ = arcsin (1/1) = arcsin (1) = 9 0^{\circ}$ — pehli Mach wave flow ke normal hai. Yeh geometric edge case hai, failure nahi: $ν (1) = 0$ finite aur well-defined hai, aur $d ν / d M = \frac{M ^{2} - 1}{\dots} \to 0$ jab $M \to 1^{+}$ , isliye $ν$ $1$ se smoothly niklata hai (zero slope ke saath). Flow bas apna pehla kuch turning right-angle wave se start karke build up karti hai. Yeh figure s02 mein sabse baayeen, vertical solid-orange wave hai.

Recall Solution L5.4

$\frac{γ + 1}{γ - 1} = \frac{8/3}{2/3} = 4$ , isliye $4 = 2$ . $ν_{m a x} = \frac{π}{2} (2 - 1) = \frac{π}{2} = 90.0 0^{\circ} .$ Helium ko sonic se maximum $9 0^{\circ}$ hi mod sakte hain — air ke $130. 5^{\circ}$ se kaafi kam. Kyun? Bada $γ$ matlab stiffer gas (per unit temperature kam internal energy), isliye ek given expansion ke liye kam speed available hai, total turn limit ho jaata hai.

Recall Jaane se pehle ek-line self-check

Expansion ⇒ $θ$ add karo, $M ↑$ , $p ↓$ , $T ↓$ , $μ ↓$ , $p_{0}$ fixed. Agar inme se koi bhi ulta aaya, apna sign dobara check karo.

Fan ke through har quantity ki direction?

M

up,

p

down,

T

down,

μ

down,

p_{0}

aur

T_{0}

unchanged;

ν

mein

θ

add karo.

ν_{m a x}

finite kyun hai?

Dono

arctan

terms

π /2

par saturate ho jaate hain, isliye unka weighted difference converge karta hai chahe

M \to \infty

ho.

M = 1

par pehla Mach wave angle?

μ = arcsin (1) = 9 0^{\circ}

, flow ke normal.