Visual walkthrough — Prandtl-Meyer expansion waves — isentropic, supersonic turning

3.1.16 · D2 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer expansion waves — isentropic, supersonic turni

Shuru karne se pehle, teen words jo hum constantly use karenge, seedhi simple bhasha mein:

Neeche sab kuch supersonic duniya $M>1$ mein hai, kyunki sirf wahan hi woh wafer-thin waves exist karti hain jo hum abhi draw karne wale hain.

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Step 1 — Ek wall bend = ek ripple

KYA. Figure dekho. Ek supersonic stream (moti kaali arrow) flat wall ke saath chal rahi hai. Ek jagah par wall thodi si flow se door mud jaati hai ek tiny angle par jise hum $d\theta$ kehte hain (Greek letter "theta", hamare naam ka turn angle; aage $d$ ka matlab hai "ek infinitesimally small amount of"). Ek akela, wafer-thin Mach wave (laal line) us kink se nikalta hai.

KYUN. Hum poore turn ko ek saath handle karne se mana karte hain. Bada turn mushkil hai; ek microscopic turn asaan hai kyunki ek thin wave ke paas se guzarte waqt almost kuch nahi badalta. Hum microscopic turn ko exactly handle karenge, phir unhe jod denge — millions baar (yahi integration Step 7 mein karega).

PICTURE. Laal Mach wave Mach angle $\mu$ (Greek "mu") par baithti hai, flow direction se naapi jaati hai. Mach-wave picture se, $\mu =\arcsin \left(\frac{1}{M}\right).$ Term by term padhte hain: $\arcsin$ poochta hai "kis angle ka yeh sine hai?"; $\frac{1}{M}$ ratio $a/V$ hai (signal speed over flow speed). Tez flow (bada $M$) chhota $\mu$ deta hai — ripple zyada flat hoti hai. Yeh dhyan rakhna; yahi wajah hai ki baad mein ek fan khulab jaata hai.

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Step 2 — Velocity ko ek triangle mein freeze karo

KYA. Laal wave ke paas se flow speed $V$ se $V+dV$ ho jaati hai aur direction $d\theta$ se swing karti hai. Hum har velocity ko do pieces mein todte hain: ek wave ke saath (ise $V_t$, tangential part, kaho) aur ek aarpaar (normal part). Figure mein "before" (kaala) aur "after" (laal) dono velocity arrows wave par resolve hote dikhte hain.

KYUN. Mach wave oblique shock ka sabse kamzor cousin hai. Iska ek hi rule hai: tangential velocity component unchanged rehta hai — sirf normal component thoda nudge hota hai. Agar ek component frozen hai, toh geometry plain right-triangle trigonometry mein collapse ho jaata hai, jise hum bina calculus ke solve kar sakte hain.

PICTURE. Figure mein $V_t$ (laal line par projection) before aur after identical hai — dashed guide lines usi length par aakar milti hain. Saara change us chhote normal wedge mein hai jo turn $d\theta$ produce karta hai.

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Step 3 — Kyun ek tiny wedge angle hota hi hai $d\theta$

KYA. Turn extract karne se pehle, ek cheez saaf kar lo: ek razor-thin wedge ke liye, angle aur uska tangent same number hote hain. Figure mein, ek small angle ke liye, teen cheezein overlay hoti hain jo identical lagti hain: seedha tangent side, curved arc, aur seedha chord.

KYUN. Angle $d\theta$ ko radians mein napo — unit circle par arc ki length. Us angle ka tangent tangent segment ki length hai. Jaise wedge shrink hota hai, arc, chord aur tangent sab ek hi length par collapse ho jaate hain, isliye $\tan (d\theta )\approx \sin (d\theta )\approx d\theta (\text{for tiny }d\theta ,\text{ in radians}).$

Term by term:

• $d\theta$ — unit circle par arc length, yani angle khud radians mein.
• $\tan (d\theta )$ — tangent segment ki length; bade angle ke liye yeh arc se door bulge karta hai, lekin tiny angle ke liye gap khatam ho jaata hai.

PICTURE. Figure mein laal tangent line ko dekho ki woh kaali arc ko kitna tightly hug karti hai jab angle $40^{\circ }$ se $5^{\circ }$ par aata hai. Yahi collapse hai jiske liye wedge angle ko as $d\theta$ likhne ki permission milti hai bina kisi correction term ke — valid sirf isliye kyunki Step 1 ne infinitesimal bend chuni.

Recall Hum

$\tan (d\theta )$ ko $d\theta$ se kyun replace kar sakte hain? Kyunki bend infinitesimal hai ::: ek tiny angle ke liye radians mein, $\tan (d\theta )\approx \sin (d\theta )\approx d\theta$.

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Step 4 — Triangle se $d\theta$ padho

KYA. "Tangential component unchanged" aur Step 3 ka small-angle fact use karke, before/after triangle speed mein kitna change hua aur direction mein kitna change hua ke beech ek clean link deta hai: $d\theta =\sqrt{M^2-1}\frac{dV}{V}.$

Term by term:

• $d\theta$ — tiny turn (jo hum accumulate karna chahte hain).
• $\frac{dV}{V}$ — speed ka fractional change: "kitne m/s" nahi balki "kitna percent." Yeh natural currency hai kyunki trig ratio sirf proportions ki parwah karta hai.
• $\sqrt{M^2-1}$ — conversion factor. Yeh $\cot \mu =1/\tan \mu$ ke barabar hai, laal wave ka "run over rise." Dhyan do ise $M>1$ chahiye, warna negative number ka square root aata hai aur koi real wave nahi hoti. Supersonic-only, algebra mein baked in.

KYUN YEH TOOLS. Humne $\tan \mu =\frac{1}{\sqrt{M^2-1}}$ use kiya (usi right triangle se jiska hypotenuse $M$ hai aur chhoti side $1$, kyunki $\sin \mu =1/M$). $\tan$ exactly "opposite over adjacent" hai us triangle par — woh ratio jo wave ki steepness measure karta hai, jo precisely speed change ko direction change mein convert karta hai.

PICTURE. Figure $\mu$-triangle draw karta hai: length $M$ ka hypotenuse, opposite side $1$, adjacent side $\sqrt{M^2-1}$ (laal). Woh laal side hi formula mein baithaa factor hai.

Recall

$\sqrt{M^2-1}$ kahan se aata hai? $\sin \mu =1/M$ se, adjacent side $\sqrt{M^2-1}$ hai ::: isliye $\cot \mu =\sqrt{M^2-1}$, exactly woh factor jo $dV/V$ ko multiply karta hai.

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Step 5 — "Speed change" ko "Mach change" se replace karo

KYA. Speed $V$ tabulate karna awkward hai; Mach number $M$ physicist ka ruler hai. Kyunki $V=Ma$, natural logarithm lo (likha jaata hai $\ln$; is page par "log" ka matlab hamesha $\ln$ hai, logarithm to base $e$) aur differentiate karo: $\ln V=\ln M+\ln a⟹\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}.$

KYUN. Natural log ek product ko sum mein badal deta hai, aur iska derivative friendly $d(\ln x)=dx/x$ hai — ek fractional change. Toh "ek product ka fractional change = fractional changes ka sum." Yeh dekhne ka sabse saaf tarika hai ki $V$ mein change ke do sources hain: flow tez ho sakta hai ($dM$), aur sound ki local speed badal sakti hai ($da$) kyunki gas thandi hoti hai.

Term by term:

• $\frac{dM}{M}$ — Mach number ka fractional change (jo cheez hum chahte hain).
• $\frac{da}{a}$ — local speed of sound ka fractional change. Yeh zero nahi hai: jaise gas expand hoti hai woh cool hoti hai, aur thandi gas mein sound slow chalti hai.

PICTURE. Figure mein bar chart ek $dV/V$ bar ko ek laal $dM/M$ piece aur ek kaale $da/a$ piece mein split karta dikhata hai. Hum $da/a$ piece ko bhool nahi sakte — yahi classic galati hai.

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Step 6 — Isentropic energy law se $da/a$ hatao

KYA — cancellation, line by line unpack ki. Ab hum teen chhote algebra ke pieces karte hain aur unhe jodenge. Figure wahi teen arrows dikhata hai jo ek clean result mein feed karte hain.

Piece A — $a$ ko $T$ se jodo. Perfect gas ke liye $a=\sqrt{\gamma RT}$, toh $a\propto \sqrt{T}$. $\ln$ lekar aur differentiate karke (same trick jaise Step 5): $\frac{da}{a}=\frac{1}{2}\frac{dT}{T}.$ (Half, kyunki $a$ $T$ ke square-root par depend karta hai.)

Piece B — $T$ ko $M$ se jodo. Isentropic energy relation, constant $T_0$ use karke, kehti hai $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$. Rearrange karo, $T=T_0/\left(1+\frac{\gamma -1}{2}{M}^2\right)$. Kyunki $T_0$ constant hai, $\ln$ lekar aur differentiate karo ise uda do aur bache: $\frac{dT}{T}=-\frac{(\gamma -1)MdM}{1+\frac{\gamma -1}{2}{M}^2}.$ (Minus sign physics hai: tez hona, yani $dM>0$, gas ko thanda karta hai, $dT<0$.)

Piece C — A aur B combine karo. B ko A mein substitute karo ($da/a=\frac{1}{2}dT/T$), toh factor $(\gamma -1)$ $\frac{\gamma -1}{2}$ ban jaata hai: $\frac{da}{a}=-\frac{\frac{\gamma -1}{2}MdM}{1+\frac{\gamma -1}{2}{M}^2}.$

Cancellation. Ab ise Step 5 ke $\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}$ mein feed karo. $\frac{dM}{M}=\frac{dM}{M}\cdot 1$ likho aur common denominator $D=1+\frac{\gamma -1}{2}{M}^2$ par combine karo:

=\frac{dM}{M}\cdot\frac{D-\frac{\gamma-1}{2}M^2}{D}.$$ Numerator dekho: $D-\tfrac{\gamma-1}{2}M^2 = \left(1+\tfrac{\gamma-1}{2}M^2\right)-\tfrac{\gamma-1}{2}M^2 = 1$. $\tfrac{\gamma-1}{2}M^2$ pieces **exactly cancel** ho jaate hain, sirf $1$ bachta hai: $$\boxed{\;\frac{dV}{V}=\frac{1}{1+\frac{\gamma-1}{2}M^2}\,\frac{dM}{M}.\;}$$ Bachne wale ka term by term: - $\gamma$ (Greek "gamma") — gas ka heat-capacity ratio; air ke liye $1.4$. - $R$ — har fractional relation se cancel ho jaata hai, toh final $\nu$ formula tak kabhi nahi pahunchta. - $1+\frac{\gamma-1}{2}M^2$ — "temperature-drop" denominator. Jaise $M$ badhta hai yeh bhi badhta hai, toh Mach number ka har extra bit **kam** extra velocity khareedta hai: flow ek ceiling speed ki taraf saturate ho jaata hai. **KYUN.** Hume ek equation chahiye thi jo $a$ (ya $T$) ko $M$ se jode. Ek adiabatic, isentropic stream (constant $T_0$) ke liye energy conservation exactly yahi deta hai, aur yeh poori derivation mein use hone wali *ekmaatra* extra physics hai. **PICTURE.** Figure A → B → C ko boxed cancellation mein walk karta hai, phir $1+\frac{\gamma-1}{2}M^2$ ko $M$ ke saath badhte dikhata hai; $dV/V$ aur $dM/M$ curves ke beech shrinking laal gap high $M$ par velocity flattenig out hona dikhata hai. --- ## Step 7 — Har ripple ko jodo: integrate karo **KYA.** Steps 4 aur 6 ko saath rakhke ek thin wave ki wajah se hone wala turn, poori tarah $M$ mein: $$d\theta=\frac{\sqrt{M^2-1}}{1+\frac{\gamma-1}{2}M^2}\,\frac{dM}{M}.$$ Ab sonic point $M=1$ se kisi bhi $M$ tak infinitely many aise micro-turns stack karo. "Infinitely many infinitesimals ko stack karna" exactly yahi hai jo integral $\int$ karta hai. Total ko **Prandtl–Meyer function** $\nu(M)$ define karo: $$\nu(M)=\int_{1}^{M}\frac{\sqrt{m^2-1}}{1+\frac{\gamma-1}{2}m^2}\,\frac{dm}{m}.$$ Integral carry out karne par (ek standard substitution) boxed result milta hai: > [!formula] Prandtl–Meyer function > $$\nu(M)=\sqrt{\frac{\gamma+1}{\gamma-1}}\;\arctan\!\sqrt{\frac{\gamma-1}{\gamma+1}\,(M^2-1)}\;-\;\arctan\!\sqrt{M^2-1},\qquad \nu(1)=0.$$ > Yahan $\nu$ **radians** mein aata hai (yeh $d\theta$ ka integral hai, ek angle). Degrees mein padhne ke liye > $180/\pi$ se multiply karo. ![[deepdives/dd-physics-3.1.16-d2-s07.png]] Term by term: - $\nu(M)$ — **running total turn**, radians mein, sonic ($M=1$) se $M$ tak accelerate karne ke liye zaroori. - Lower limit $1$ hi wajah hai ki $\nu(1)=0$: exactly sonic par tumne abhi kuch nahi moda. - Do $\arctan$ terms isliye aate hain kyunki integrand naturally do pieces mein split hota hai, har ek "$\dfrac{dx}{1+x^2}$" form ka — $\arctan$ ka fingerprint. Yeh fingerprint kyun? Kyunki agar $x=\tan\phi$ set karo, toh $dx=(1+x^2)\,d\phi$, isliye $\dfrac{dx}{1+x^2}=d\phi$ aur integral sirf angle $\phi=\arctan x$ hai (figure mein inset yeh substitution dikhata hai jo $1/(1+x^2)$ ke neeche area ko plain angle mein collapse karta hai). $\arctan$ exactly isliye aata hai kyunki hamare micro-turns Step 4 mein $\mu$ ke tangent se banaye gaye the. **KYUN.** $\nu$ ek *bookkeeping* quantity hai. Kyunki ise sonic se *measure* kiya jaata hai, kisi bhi real corner ki physics sirf $\nu$ ke **difference** ki parwah karti hai, toh angle $\theta$ ka corner bas add karta hai: > [!formula] Kaam ka rule > $$\nu(M_2)=\nu(M_1)+\theta.$$ > Phir saari properties constant-$T_0,p_0$ [[Isentropic flow relations]] se milti hain: > $$\frac{p_2}{p_1}=\left(\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\right)^{\!\frac{\gamma}{\gamma-1}},\qquad > \frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}.$$ **PICTURE.** Figure $\nu$-vs-$M$ curve hai: $(1,0)$ se shuru hoti hai, steeply badhti hai, phir horizontal ceiling ki taraf bend karti hai (Step 9). Laal bracket dikhata hai kaise $10^\circ$ ka $\theta$ tumhe curve par $M_1$ se $M_2$ tak slide karta hai. --- ## Step 8 — Ripples fan mein kyun phailti hain **KYA.** Step 1 se, $\mu=\arcsin(1/M)$. Expansion ke through $M$ **badhta** hai, toh $\mu$ **ghatta** hai. Pehli wave (at $M_1$) steep hai; aakhri wave (at $M_2$) shallow hai. Unke beech waves ka ek continuous spray hai — **fan**. ![[deepdives/dd-physics-3.1.16-d2-s08.png]] **KYUN.** Yeh parent note ka "diverging characteristics" claim hai, draw karke dikhaya gaya. Kyunki har wave infinitesimally weak hai aur woh *ek doosre se dur phailti hain* (compression ki tarah pile up karne ki jagah), koi wave kabhi doosri ko nahi pakadti. Kuch bhi discontinuity mein steep nahi hota, toh entropy rise $\Delta s\to 0$ rehti hai — process ==isentropic== hai. [[Oblique shock waves|Oblique shock]] se contrast karo, jahan converging characteristics ek lossy jump mein collide karti hain. **PICTURE.** Figure laal Mach lines ko steep se (leading edge, angle $\mu_1$) shallow tak (trailing edge, angle $\mu_2$) fan karta hai, flow arrows smoothly cross mein bend karte hain. --- ## Step 9 — Edge case: vacuum mein turning **KYA.** Corner ko aur aur zyada push karo. Jaise $M\to\infty$, dono square-root arguments blow up karte hain aur har $\arctan\to\tfrac{\pi}{2}$ (yeh $90^\circ$ hai). Formula ek hard ceiling par settle ho jaata hai. Radians mein yeh hai $$\nu_{\max}=\frac{\pi}{2}\!\left(\sqrt{\frac{\gamma+1}{\gamma-1}}-1\right).$$ Degrees mein convert karte hain ($180/\pi$ se multiply karo, toh leading $\pi/2$ $90^\circ$ ban jaata hai): $$\nu_{\max}\;\xrightarrow{\gamma=1.4}\;90^\circ\,(\sqrt6-1)\approx130.45^\circ.$$ ![[deepdives/dd-physics-3.1.16-d2-s09.png]] **KYUN.** Tum sonic stream ko $130.5^\circ$ se zyada nahi mod sakte. Usse aage gas ko $M=\infty$ chahiye hoga — infinite speed, zero pressure, zero temperature: actually vacuum. Flow wall se simply separate ho jaati hai; aur koi turning possible nahi. Yeh degenerate limit hai, aur yeh explain karta hai kyun over-expanded nozzles ki ek hard ceiling hoti hai (dekho [[Nozzle design and overexpansion]]). **PICTURE.** Figure Step 7 ki $\nu$-curve ko uski horizontal asymptote (laal dashed line) par $\nu_{\max}$ tak flatten hote dikhata hai, aakhri Mach wave flow ke almost flat lie karti hai. > [!mistake] "Wall ko bas zyada modte jao taaki flow hamesha turn karti rahe." > **Kyun sahi lagta hai:** Wall ka har extra degree ek $\theta$ degree add karta hai. > **Fix:** $\nu$ ki ek ceiling $\nu_{\max}$ hai. Jab $\nu(M_1)+\theta$ use exceed karne ki koshish kare, flow vacuum limit tak pahunch jaati hai aur detach ho jaati hai — aur koi isentropic turning exist nahi karti. --- ## Ek-picture summary Final figure poori chain ko ek canvas par stitch karta hai. Ek supersonic stream (kaali arrow) convex corner se milti hai, laal Mach waves ka **fan** throw off karti hai jo steep se (leading edge) shallow tak (trailing edge) jaati hain, aur neeche labelled dials record karte hain fan ke paas kya hota hai: $M\uparrow$, $\mu\downarrow$, $p\downarrow$, $T\downarrow$, jabki turning score exactly wall ke bend se badhta hai, $\nu(M_2)=\nu(M_1)+\theta$. Nau steps mein jo kuch tumne derive kiya — bend, frozen triangle, small-angle wedge, $\sqrt{M^2-1}$ factor, $M$ ka swap, cancellation, integral, fan, aur vacuum ceiling — sab ek single readable picture mein compress. ![[deepdives/dd-physics-3.1.16-d2-s10.png]] > [!recall]- Poori walkthrough ki Feynman retelling > Hum jaanna chahte the ki jab wall hawa se door bend ho toh tez hawa ki stream kitni speed up hoti hai. Hum seedha bada bend solve nahi kar sakte the, isliye wall ko *thoda sa* moda aur ek thin ripple dekha (Steps 1–2). > Kyunki bend tiny hai, chhote wedge ka angle *bas* $d\theta$ hai — uska tangent, arc aur chord sab > ek hi length hain (Step 3). Us ripple par velocity ka ek hissa wahan ka wahan raha, toh plain > triangle-trig ne bataya ki speed change ke har bit par direction kitna swing karti hai (Step 4). Speed > clumsy hai, isliye hum ne ise Mach number se replace kiya, yeh yaad rakhte hue ki hawa *thandi bhi hoti hai*, jo uski apni > sound ko slow karti hai (Steps 5–6); cooling term aur Mach term partly cancel karte hain, ek tidy denominator chodkar. > Phir hum ne millions of tiny bends ko integral se joda aur total ko $\nu$ naam diya — ek > "turning score" sound barrier se measure ki gayi, radians mein count ki gayi (Step 7). Waves fan out > karti hain kyunki tez hawa mein flatter ripples hoti hain (Step 8), aur poori cheez ki ek hard limit hai near > $130.5^\circ$, jahan hawa ko vacuum ban'na padega (Step 9). Use karne ke liye: apni wall ke > bend ko starting score mein add karo, new Mach number lookup karo, aur new pressure aur > temperature read off karo. > [!mnemonic] Chain chhe beats mein > **Bend → Wedge → Triangle → V ko M se Swap karo → Integrate karo → $\theta$ add karo.** --- ## Recall $\sqrt{M^2-1}$ factor flow ko supersonic kyun force karta hai? ::: $M<1$ ke liye yeh negative number ka root hai — koi real Mach wave exist nahi karti; sirf $M>1$ waves support karta hai. Step 5 mein hum $da/a$ term kyun rakhte hain? ::: Kyunki gas expand hone par cool hoti hai, toh sound ki local speed bhi badlti hai; ise ignore karna velocity ko pure Mach change ke roop mein double-count karta hai. Step 6 mein kya cancel hota hai jo tidy denominator chodta hai? ::: $dM/M$ ko common denominator $1+\frac{\gamma-1}{2}M^2$ par likhne par, $\frac{\gamma-1}{2}M^2$ terms cancel ho jaate hain, numerator $1$ chodke. Total (stagnation) temperature $T_0$ kya hai? ::: Woh temperature jo gas tab milti agar ise losslessly rest par laya jaaye; yeh ek isentropic expansion mein constant rehta hai. Hum tiny wedge angle ko exactly $d\theta$ kyun likh sakte hain? ::: Small-angle approximation $\tan(d\theta)\approx d\theta$, valid kyunki bend infinitesimal hai (angle radians mein). Integrate karte waqt $\arctan$ kyun aata hai? ::: Integrand $dx/(1+x^2)$ pieces mein split hota hai; $x=\tan\phi$ substitute karne par har ek $d\phi$ ban jaata hai, jiska integral $\arctan x$ hai. Air ke liye $\nu_{\max}$ kya hai aur physically iska kya matlab hai? ::: $\approx130.45^\circ$; sonic se sabse bada possible turn, sirf tab pohanchta hai jab $M\to\infty$ vacuum mein.