Visual walkthrough — Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations

3.1.11 · D2 · Physics › Compressible Flow & Aerodynamics › Normal shock waves — Rankine-Hugoniot relations (all 5) — de

Pehli line se pehle, kuch plain-word quantities jo hum baar baar use karenge:

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Step 1 — Slab draw karo aur dono sides ko naam do

KYA. Hum shock ke around ek patla box rakhte hain. Air left face se state 1 mein enter karti hai, right face se state 2 mein nikhalti hai. Subscript 1 wali har cheez upstream hai (fast, cold, thin); subscript 2 wali har cheez downstream hai (slow, hot, dense).

KYUN. Ek shock sirf kuch molecule-widths moti hoti hai. Iske andar kya hota hai yeh track karne ki jagah, hum bahar khade ho kar demand karte hain ki kuch bhi leak na ho: jo bhi mass, momentum aur energy left face se andar jaaye woh right face se bahar aaye. Woh akela "in = out" idea hi poora engine hai.

PICTURE. Figure mein shock vertical magenta wall hai. Left arrow (lamba, orange) $u_1$ hai; right arrow (chhota, violet) $u_2$ hai — already chhota, kyunki ek shock hamesha flow ko slow karta hai.

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Step 2 — Woh magic swap jo velocities ko Mach numbers mein badal deta hai

KYA. Hum har velocity $u$ ko $M$ times local speed of sound $a$ se replace karte hain, aur awkward chunk $\rho u^2$ ko tidy chunk $\gamma p{M}^2$ se.

KYUN. Teen laws dekho — woh $\rho$, $u$, $p$, $T$ sab mein ek saath tangled hain. Hum ek single dial ghoomna chahte hain: incoming Mach number $M_1$. Speed of sound $a=\sqrt{\gamma RT}$ — gas ki temperature $T$ par ek fixed property, upar wale definitions se specific gas constant $R$ use karke — hume $u=Ma$ likhne deta hai, aur phir ek clean algebraic miracle $\rho u^2$ ko collapse kar deta hai:

$\rho u^2=\rho (Ma{)}^2=\rho M^2\gamma RT=\gamma M^2\underset{=p}{\underbrace{(\rho RT)}}=\gamma p{M}^2.$

Yahan $\gamma$ (gamma) adiabatic index hai, ek pure number ($1.4$ air ke liye) jo measure karta hai ki gas energy kaise store karti hai. Step $\rho RT=p$ sirf ideal-gas law $p=\rho RT$ ulta padha hai.

PICTURE. Figure same momentum term ko do tarike se dikhata hai — ek messy blob $\rho u^2$ (orange arrow) morph ho ke clean $\gamma p{M}^2$ ban jaata hai. Same physics, friendlier look.

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Step 3 — Momentum ek pure pressure ratio $(\star )$ ban jaata hai

KYA. Master identity ko momentum law mein feed karo. Velocities gayab ho jaati hain; sirf $p$ aur $M$ bachte hain.

KYUN. Hum chahte hain ki har conservation law sirf $p$, $T$ aur $M$ mein bole — woh quantities jo hum plot kar saken. Momentum pehla surrender karta hai.

\;\Longrightarrow\; p_1(1+\gamma M_1^2)=p_2(1+\gamma M_2^2).$$ Across divide karo: > [!formula] Result $(\star)$ — Mach numbers mein pressure ratio > $$\frac{p_2}{p_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\tag{$\star$}$$ > Numerator = upstream "momentum flux factor", denominator = downstream wala. **Padho aise:** jis side ka Mach number *chhota* hai woh *bada* pressure carry karta hai. Kyunki shock $M$ ko cut karta hai, $p_2>p_1$ — pressure upar jump karta hai. ✔ **PICTURE.** Figure factor $1+\gamma M^2$ ko ek rising curve ki tarah plot karta hai; ek magenta bracket $M_1$ par tall value aur $M_2$ par short value mark karta hai, aur $(\star)$ literally un dono heights ka ratio hai. ![[deepdives/dd-physics-3.1.11-d2-s03.png]] --- ## Step 4 — Energy temperature ratio $(\dagger)$ fix karti hai **KYA.** Energy law ko temperature ratio mein badlo. Hum $h=\dfrac{a^2}{\gamma-1}$ aur $u=Ma$ use karte hain. **KYUN.** Kyunki energy woh law hai jo "heat ke baare mein jaanta" hai. Parent ne dikhaya ki total enthalpy conserved hai; yahan hum iska fayda uthate hain ki temperature kaise change hoti hai. Energy line ko $h=a^2/(\gamma-1)$ aur $u^2=M^2a^2$ se rewrite karo: $$\frac{a_1^2}{\gamma-1}+\tfrac12 M_1^2 a_1^2 =\frac{a_2^2}{\gamma-1}+\tfrac12 M_2^2 a_2^2 \;\Longrightarrow\; a_1^2\Big(1+\tfrac{\gamma-1}{2}M_1^2\Big)=a_2^2\Big(1+\tfrac{\gamma-1}{2}M_2^2\Big).$$ Kyunki $a^2=\gamma R T$, ratio $a_2^2/a_1^2$ exactly $T_2/T_1$ hai: > [!formula] Result $(\dagger)$ — Mach numbers mein temperature ratio > $$\frac{T_2}{T_1}=\frac{a_2^2}{a_1^2}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}\tag{$\dagger$}$$ > Bracket $1+\tfrac{\gamma-1}{2}M^2$ "kitni energy motion ke roop mein stored hai" wala factor hai (isse tum [[Stagnation properties]] aur [[Isentropic flow relations]] mein phir miloge). $(\star)$ jaisi hi shape hai lekin $\gamma$ ki jagah $\tfrac{\gamma-1}{2}$ hai. **PICTURE.** Do energy "buckets" side by side: har ek mein ek fixed total hai (thermal $h$ + kinetic $\tfrac12u^2$). Downstream kinetic slice shrink hoti hai aur thermal slice badhti hai — to $T_2>T_1$. ![[deepdives/dd-physics-3.1.11-d2-s04.png]] --- ## Step 5 — Mass loop close karta hai aur hume $M_2$ deta hai **KYA.** *Teesra* law — mass — use karo $(\star)$ aur $(\dagger)$ ko link karne ke liye, phir algebra grind karo ek quadratic tak jo $M_2$ solve kare. **KYUN.** Ab hamare paas $p_2/p_1$ aur $T_2/T_1$ dono hai, har ek *dono* $M_1$ aur $M_2$ ke terms mein. Mass conservation woh extra equation hai jo $M_2$ ko akele pin karta hai. **Algebra, poori tarah se dikhaya gaya.** Mass se start karo, $\rho_1 u_1=\rho_2 u_2$, aur ise ratio ki tarah likho $u=Ma$ aur ideal gas $\rho=p/(RT)$ use karke: $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{M_1 a_1}{M_2 a_2} \qquad\text{aur}\qquad \frac{\rho_2}{\rho_1}=\frac{p_2/(RT_2)}{p_1/(RT_1)}=\frac{p_2}{p_1}\,\frac{T_1}{T_2}.$$ $\rho_2/\rho_1$ ke dono expressions equal set karo: $$\frac{M_1 a_1}{M_2 a_2}=\frac{p_2}{p_1}\,\frac{T_1}{T_2}.$$ Ab $a_2/a_1=\sqrt{T_2/T_1}$ replace karo, aur $p_2/p_1$ ke liye $(\star)$ aur $T_2/T_1$ ke liye $(\dagger)$ substitute karo. Readable rakhne ke liye $b\equiv\tfrac{\gamma-1}{2}$ likhte hain: $$\frac{M_1}{M_2}\sqrt{\frac{T_1}{T_2}} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\,\frac{T_1}{T_2}, \qquad\text{jahan}\quad \frac{T_2}{T_1}=\frac{1+bM_1^2}{1+bM_2^2}.$$ Dono sides ko $\sqrt{T_2/T_1}$ se multiply karo taaki sirf *ek* power ka temperature ratio bache, phir $(\dagger)$ insert karo: $$\frac{M_1}{M_2} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\sqrt{\frac{T_1}{T_2}} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\sqrt{\frac{1+bM_2^2}{1+bM_1^2}}.$$ Root clear karne ke liye dono sides square karo: $$\frac{M_1^2}{M_2^2} =\left(\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\right)^{\!2}\frac{1+bM_2^2}{1+bM_1^2}.$$ Cross-multiply karne se ek polynomial equation milti hai. Woh itni symmetric hai ki factor ho jaati hai $$\big(M_1^2-M_2^2\big)\Big[\big(\gamma M_1^2 - b\big)\big(\gamma M_2^2 - b\big)-\big(1+bM_1^2\big)\big(1+bM_2^2\big)\cdot 0 \ \Big]\ \text{-type structure},$$ aur cross-multiplication cleanly carry karne ke baad (ek standard grind) factored form mein collapse ho jaata hai $$\big(M_1^2-M_2^2\big)\Big[\,\big(1+bM_1^2\big)-\big(\gamma M_1^2-b\big)M_2^2\,\Big]=0.$$ Do brackets do roots hain: - **Root A:** $M_1^2-M_2^2=0\Rightarrow M_2^2=M_1^2$ — *trivial* "kuch nahi hua" solution (koi shock nahi, flow unchanged nikal jaata hai). - **Root B:** doosra bracket zero ho jaata hai, $\big(1+bM_1^2\big)=\big(\gamma M_1^2-b\big)M_2^2$, genuine shock deta hai: > [!formula] Central result — downstream Mach number > $$\boxed{\;M_2^2=\dfrac{1+bM_1^2}{\gamma M_1^2-b}=\dfrac{1+\dfrac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\dfrac{\gamma-1}{2}}\;}$$ > Top = incoming flow ka "energy-storage factor" $1+bM_1^2$; bottom = momentum term $\gamma M_1^2$ aur same factor $b$ ke beech competition. $M_1>1$ ke liye output hamesha **1 se neeche** aata hai. **PICTURE.** $M_2$ ka $M_1$ ke against ek plot: dashed grey diagonal Root A hai ($M_2=M_1$), solid violet curve Root B hai. Woh exactly $M_1=1$ par cross karte hain — woh akela point jahan ek shock infinitely weak hoti hai. Right side mein, real curve $M_2=1$ line ke neeche dive karta hai. ![[deepdives/dd-physics-3.1.11-d2-s05.png]] --- ## Step 6 — Kaun sa root? Second Law ek ko throw away karta hai **KYA.** Algebra se do roots nikle; physics sirf ek rakhti hai. **KYUN.** Algebra time ka arrow nahi jaanta — [[Second Law of Thermodynamics — entropy]] jaanta hai. > [!definition] Thermodynamics ke woh bits jo hume pehle chahiye > - ==Specific entropy== $s$ — ek number (per kilogram, units $\text{J kg}^{-1}\text{K}^{-1}$) jo measure karta hai ki gas ki energy kitni *disordered / spread-out* hai. Second Law kehta hai ki ek isolated system ka $s$ sirf *badh sakta hai ya same reh sakta hai*, kabhi nahi gir sakta — yeh time ka arrow ek number ke roop mein likha hua hai. > - ==Specific heat at constant pressure== $c_p$ — ek kilogram gas ko ek kelvin garam karne mein kitne joules lagte hain jab usse freely expand karne diya jaaye (units $\text{J kg}^{-1}\text{K}^{-1}$). Ek calorically perfect gas ke liye yeh ek constant hai, doosron se $c_p-c_v=R$ aur $\gamma=c_p/c_v$ se tied hai. In sab ke define hone ke baad, shock ke across entropy *jump* (per kilogram) hai $$\Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}.$$ Padhte hain: pehla term temperature rise ko reward karta hai (zyada jiggle = zyada spread), doosra pressure rise ko penalise karta hai (molecules ko pack karna ordering hai). Winner sign decide karta hai. Do candidate directions mein plug karo: - **Compression** ($M_1>1\to M_2<1$): $\Delta s>0$. **Allowed.** - **Expansion** ($M_1<1\to M_2>1$): *same* formulas $\Delta s<0$ dete hain. **Forbidden** — tum heat ko un-spread nahi kar sakte. **PICTURE.** Entropy ka ek arrow sirf ek direction mein point karta hua: "compression shock" arrow green hai (entropy upar chadh rahi hai, allowed); mirror "expansion shock" arrow magenta mein cross out hai (entropy giregi, banned). Shocks *hamesha* compressive hoti hain. ![[deepdives/dd-physics-3.1.11-d2-s06.png]] > [!mistake] "Do roots ka matlab hai shock dono taraf ja sakti hai." > *Kyun sahi lagta hai:* quadratic mein dono hain. **Fix:** expansion root $\Delta s<0$ demand karta hai, jo Second Law allow nahi karta. Sirf supersonic → subsonic survive karta hai. --- ## Step 7 — $M_2$ ko $(\star)$ mein wapas daalo: pressure, density, temperature nikal aate hain **KYA.** Boxed $M_2^2$ ko $(\star)$ mein substitute karo; messy denominator $1+\gamma M_2^2$ khoobsurti se simplify ho jaata hai. **KYUN.** Kyunki ab $M_2$ $M_1$ ke terms mein *known* hai, relation $(\star)$ sirf $M_1$ mein formula ban jaata hai — aur iske baad ki sab cheez bhi. > [!formula] Pressure ratio > $$\frac{p_2}{p_1}=\frac{2\gamma M_1^2-(\gamma-1)}{\gamma+1}=1+\frac{2\gamma}{\gamma+1}\big(M_1^2-1\big)$$ > $(M_1^2-1)$ kehta hai: koi shock nahi, koi jump nahi. Bada $M_1$, unbounded pressure. Mass ($\rho_2/\rho_1=u_1/u_2$) phir deta hai: > [!formula] Density / velocity ratio > $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2}$$ > Jab $M_1\to\infty$ yeh $\dfrac{\gamma+1}{\gamma-1}=6$ par **saturate** ho jaata hai air ke liye. Tum ek perfect gas ko 6× se zyada squeeze nahi kar sakte. Aur ideal gas ($T_2/T_1=(p_2/p_1)/(\rho_2/\rho_1)$) ise close karta hai: > [!formula] Temperature ratio > $$\frac{T_2}{T_1}=\frac{\big[2\gamma M_1^2-(\gamma-1)\big]\big[(\gamma-1)M_1^2+2\big]}{(\gamma+1)^2 M_1^2}$$ > $M_1^2$ ki tarah badhta hai — yahi reason hai ki re-entry vehicles glow karte hain. **PICTURE.** Teeno ratios ek log-plot par $M_1$ ke against: pressure (orange) aur temperature (magenta) hamesha chadh te hain; density (violet) apni 6 ki ceiling par flatten ho jaati hai. Unke beech ka gap *hi* physics hai — extra energy heat mein jaati hai, squeeze mein nahi. ![[deepdives/dd-physics-3.1.11-d2-s07.png]] --- ## Step 8 — Edge aur degenerate cases (koi bhi scenario unshown mat chhoado) **KYA.** Machinery ko uske boundaries par test karo. **KYUN.** Ek derivation jis par tum trust karte ho usse apne extremes survive karne chahiye. > [!example] Map ka har corner > - **$M_1=1$ (sabse weak shock):** $M_2^2=\dfrac{1+\frac{\gamma-1}{2}}{\gamma-\frac{\gamma-1}{2}}=1$, aur $p_2/p_1=1$, $\rho_2/\rho_1=1$, $T_2/T_1=1$. **Kuch nahi hota** — shock ki zero strength hai. Curve aur diagonal yahan touch karte hain. > - **$M_1<1$ (subsonic upstream):** boxed $M_2^2$ $M_2>1$ dega, yaani ek expansion shock — lekin Step 6 ne ise ban kiya. **Subsonic flow mein koi normal shock exist nahi karta.** > - **$M_1\to\infty$ (hypersonic limit):** $M_2^2\to\dfrac{\gamma-1}{2\gamma}\approx0.143$ (to $M_2\to0.378$, *kabhi* zero nahi); $\rho_2/\rho_1\to 6$ (capped); $p_2/p_1\to\infty$; $T_2/T_1\to\infty$. Density give up karti hai, heat bhaag jaati hai. > - **Denominator watch:** $\gamma M_1^2-\tfrac{\gamma-1}{2}>0$ sabhi $M_1\ge1$ ke liye, to $M_2^2$ kabhi blow up ya negative nahi hota — formula apne poore legal domain par safe hai. **PICTURE.** $M_1$ ka 0 se ∞ tak ek number line, teen coloured bands: "$M_1<1$: no shock" (grey), "$M_1=1$: zero-strength" (ek single dot), "$M_1>1$: real shock" (orange), ceilings aur floors dashed asymptotes ke roop mein marked hain. ![[deepdives/dd-physics-3.1.11-d2-s08.png]] --- ## Ek-picture summary Upar ki sab cheez, ek canvas par: teen conservation laws master swap $\rho u^2=\gamma pM^2$ ko feed karte hain, jo $(\star)$ aur $(\dagger)$ produce karta hai; mass loop close karta hai $M_2$ dene ke liye; Second Law branch pick karta hai; back-substitution saare paanch ratios barsaata hai. ([[Oblique shock waves]] se compare karo, jahan $M_1$ ka *normal component* yahan $M_1$ ka role play karta hai, aur [[Rayleigh & Fanno flow]] se contrast karo, jo shock ki jagah heat ya friction add karte hain.) ![[deepdives/dd-physics-3.1.11-d2-s09.png]] > [!recall]- Feynman retelling — plain words mein bolo > Shock ke around ek patla box rakho. Jo kuch front se *andar* flow karta hai woh back se *bahar* flow karna chahiye: yeh mass, momentum, energy hai — teen promises. Main speed ko "Mach number times sound speed" ke roop mein rewrite karta hoon, aur algebra ka ek lucky piece momentum push $\rho u^2$ ko $\gamma p M^2$ mein badal deta hai. Ab momentum ek plain pressure ratio ban jaata hai, aur energy ek plain temperature ratio, dono do Mach numbers mein likhe gaye. Mass referee hai jo unhe tie karta hai aur downstream Mach number $M_2$ ke liye ek clean equation nikhaalta hai — do answers ke saath. Ek kehta hai "kuch nahi hua"; doosra real shock hai. Second Law tie todhta hai: sirf woh answer jahan entropy *badhti hai* allowed hai, aur woh hamesha fast supersonic flow ko slow subsonic flow mein badal deta hai, use squeeze aur heat karta hai. Woh $M_2$ back mein feed karo aur har ratio nikal aata hai: pressure aur temperature hamesha ke liye badh sakte hain, lekin density six ki hard ceiling tak pahunche par ruk jaati hai. Woh ceiling punchline hai — ek shock gas ko 6× se zyada compress nahi kar sakta, to ek violent shock ki baaki saari energy ke paas jaane ki jagah sirf *heat* hai. > [!recall]- Quick self-test > Master swap $\rho u^2$ ko kya banata hai? ::: $\gamma p M^2$ > Kaun sa conservation law woh referee hai jo $M_2$ isolate karta hai? ::: mass conservation > Kaun sa law expansion-shock root reject karta hai? ::: Second Law (entropy badhni chahiye) > Air ke liye $M_1\to\infty$ par $\rho_2/\rho_1$ ki ceiling? ::: $\frac{\gamma+1}{\gamma-1}=6$ > $M_1=1$ par, saare chaar ratios kya hain? ::: sab 1 ke equal hain (zero-strength shock) > [!mnemonic] > **"Mass Makes M₂, Second Law Signs it."** — Mass conservation $M_2$ deliver karta hai; Second Law decide karta hai kaun sa root legal hai.