3.1.11 · HinglishCompressible Flow & Aerodynamics
Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations
3.1.11· Physics › Compressible Flow & Aerodynamics
Ek normal shock ek razor-thin (kuch mean-free-paths ki) discontinuity hoti hai jo supersonic flow ke perpendicular khadi rehti hai. Iske across flow abruptly supersonic se subsonic ho jaati hai, pressure aur temperature shoot up karte hain, aur flow total pressure lose karti hai (entropy badhti hai). Paanch Rankine–Hugoniot relations hume ratios , , , , aur dete hain — purely upstream Mach number ke functions ke roop mein (calorically perfect gas ke liye).
1. The setup — WHAT we are doing
Teen master conservation laws (per unit area):
\underbrace{p_1 + \rho_1 u_1^2 = p_2 + \rho_2 u_2^2}_{\text{momentum}}\qquad \underbrace{h_1 + \tfrac12 u_1^2 = h_2 + \tfrac12 u_2^2}_{\text{energy}}$$ jahan $h=c_p T = \dfrac{\gamma}{\gamma-1}\dfrac{p}{\rho}$. **Yeh teen kyun?** Mass appear/vanish nahi ho sakta, momentum sirf slab faces pe pressure forces se change hoti hai, aur koi heat/work nahi hone se **total enthalpy** conserved rehti hai. Baaki sab kuch teen lines pe algebra hai. ![[3.1.11-Normal-shock-waves-—-Rankine-Hugoniot-relations-(all-5)-—-derivations.png]] --- ## 2. Master substitution: $u = M\sqrt{\gamma R T}$ aur $\rho u^2 = \gamma p M^2$ > [!intuition] > Woh trick jo sab kuch sirf $M$ ke functions mein collapse kar deti hai: speed of sound $a=\sqrt{\gamma R T}$, to $u=Ma$, aur > $$\rho u^2 = \rho (Ma)^2 = \rho M^2 \gamma R T = \gamma M^2 (\rho R T) = \gamma p M^2.$$ > To $\rho u^2 = \gamma p M^2$. Yeh ek identity poori derivation ka engine hai. --- ## 3. Relation 1 — Velocity / density ratio (Prandtl-style result) Hum pehle density (= velocity) ratio derive karte hain kyunki baaki sab isi pe depend karte hain. **Step A.** Momentum ko $\rho u^2=\gamma p M^2$ use karke rewrite karo: $$p_1(1+\gamma M_1^2) = p_2(1+\gamma M_2^2)\;\Rightarrow\; \frac{p_2}{p_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}.\tag{$\star$}$$ *Yeh step kyun?* Yeh momentum equation ko Mach numbers mein ek clean pressure ratio mein badal deta hai. **Step B.** Total enthalpy ke saath energy. Kyunki $h=\frac{a^2}{\gamma-1}$ aur $u=Ma$: $$\frac{a_1^2}{\gamma-1}+\frac12 M_1^2 a_1^2=\frac{a_2^2}{\gamma-1}+\frac12 M_2^2 a_2^2 \;\Rightarrow\; \frac{a_2^2}{a_1^2}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}=\frac{T_2}{T_1}.\tag{$\dagger$}$$ *Yeh step kyun?* Total enthalpy constant ⇒ $a^2(1+\frac{\gamma-1}{2}M^2)$ dono sides pe same hai. **Step C.** Mass + ideal gas: $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{M_1 a_1}{M_2 a_2}.$$ Aur ideal gas aur ($\star$) se: $\dfrac{\rho_2}{\rho_1}=\dfrac{p_2}{p_1}\dfrac{T_1}{T_2}=\dfrac{p_2/p_1}{a_2^2/a_1^2}$. Dono density expressions ko equal set karo aur ($\star$),($\dagger$) use karo. Simplification ke baad (algebra niche) tumhe famous **Mach-number relation** milti hai jo phir har ratio deti hai. --- ## 4. Relation 2 — Downstream Mach number $M_2$ ($\star$) aur mass/energy results ko combine karo. Sabse clean route: $\frac{a_2^2}{a_1^2}$ ko ($\dagger$) se substitute karo aur $\frac{u_1}{u_2}=\frac{\rho_2}{\rho_1}$ ko mass conservation mein likha hua use karo: $$\frac{\rho_2}{\rho_1}=\frac{M_1 a_1}{M_2 a_2}=\frac{M_1}{M_2}\sqrt{\frac{a_1^2}{a_2^2}}.$$ $\dfrac{p_2/p_1}{a_2^2/a_1^2}$ ke saath equate karo aur grind karo (ek quadratic jo factor hoti hai, trivial root $M_2=M_1$ woh no-shock case hai) toh surviving root milta hai: > [!formula] Downstream Mach number > $$\boxed{\,M_2^2=\dfrac{1+\dfrac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\dfrac{\gamma-1}{2}}\,}$$ > $M_1>1$ ke liye yeh $M_2<1$ deta hai (subsonic). $M_1\to\infty$ ke liye, $M_2^2\to\frac{\gamma-1}{2\gamma}$ (jaise $\approx0.378$ for $\gamma=1.4$). > [!intuition] > Hamesha do algebraic solutions hote hain: $M_2=M_1$ (no shock, flow bas continue karta hai) aur genuine shock. Second Law (entropy badhni chahiye) shock branch select karta hai aur reverse ko forbid karta hai (expansion shock $M_1<1\to M_2>1$). **Shocks hamesha compressive hote hain.** --- ## 5. Relation 3 — Pressure ratio $M_2^2$ expression ko ($\star$) mein daalo. Algebra ke baad denominator mein $1+\gamma M_2^2$ beautifully simplify ho jaata hai: > [!formula] Static pressure ratio > $$\boxed{\,\dfrac{p_2}{p_1}=1+\dfrac{2\gamma}{\gamma+1}\left(M_1^2-1\right)\,}=\dfrac{2\gamma M_1^2-(\gamma-1)}{\gamma+1}$$ > *Sanity check:* $M_1=1$ pe, $p_2/p_1=1$ (infinitely weak shock). $M_1\to\infty$ ke liye yeh unbounded badhta hai — strong shocks bahut bada pressure pile up karte hain. --- ## 6. Relation 4 — Density (velocity) ratio Mass se $\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}$, upar ke results substitute karo: > [!formula] Density / velocity ratio > $$\boxed{\,\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}=\dfrac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2}\,}$$ > [!intuition] **Density saturate hoti hai.** $M_1\to\infty$ pe, $\rho_2/\rho_1\to\frac{\gamma+1}{\gamma-1}$ ($=6$ for $\gamma=1.4$). Shock chahe kitni bhi strong ho, perfect gas ko 6× se zyada compress nahi kar sakte. Extra energy **temperature** mein jaati hai, density mein nahi. --- ## 7. Relation 5 — Temperature ratio Ideal gas: $\dfrac{T_2}{T_1}=\dfrac{p_2/p_1}{\rho_2/\rho_1}$. Relations 3 aur 4 multiply karo: > [!formula] Static temperature ratio > $$\boxed{\,\dfrac{T_2}{T_1}=\dfrac{\big[2\gamma M_1^2-(\gamma-1)\big]\big[(\gamma-1)M_1^2+2\big]}{(\gamma+1)^2 M_1^2}\,}$$ > Density ke unlike, temperature bade $M_1$ ke liye $M_1^2$ ki tarah badhti hai — isliye re-entry vehicles jalte hain. --- ## 8. Bonus — Total (stagnation) pressure ratio & entropy Total enthalpy conserved hai ⇒ $T_{01}=T_{02}$, to $T_{0}$ **unchanged** rehta hai. Lekin entropy badhti hai, to $p_0$ **drop** karta hai: $$\frac{p_{02}}{p_{01}}=\left(\frac{\rho_2}{\rho_1}\right)^{\gamma/(\gamma-1)}\!\!\left(\frac{T_1}{T_2}\right)^{1/(\gamma-1)}\!\!\cdot(\text{from }s),\qquad \Delta s = c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}>0.$$ $$\boxed{\,\dfrac{p_{02}}{p_{01}}=\exp\!\left(-\dfrac{\Delta s}{R}\right)\le 1\,}$$ $p_0$-loss woh price hai (drag, inefficiency) jo shock cross karne ka milta hai. --- ## 9. Worked examples > [!example] (a) Air, $\gamma=1.4$, $M_1=2$ > - $M_2^2=\dfrac{1+0.2(4)}{1.4(4)-0.2}=\dfrac{1.8}{5.4}=0.3333\Rightarrow M_2=0.577$. *Kyun:* $M_1^2=4$ plug kiya. > - $p_2/p_1=1+\frac{2.8}{2.4}(4-1)=1+1.1667(3)=4.50$. *Kyun:* Relation 3. > - $\rho_2/\rho_1=\frac{2.4(4)}{0.4(4)+2}=\frac{9.6}{3.6}=2.667$. *Kyun:* Relation 4. > - $T_2/T_1=4.50/2.667=1.687$. *Kyun:* ideal gas $T=p/\rho$ ratio. > [!example] (b) Strong-shock limit, $M_1=10$, $\gamma=1.4$ > - $\rho_2/\rho_1=\frac{2.4(100)}{0.4(100)+2}=\frac{240}{42}=5.71$ (cap 6 ke paas). *Kyun:* density saturate hoti hai. > - $T_2/T_1=\frac{[2.8(100)-0.4][0.4(100)+2]}{2.4^2(100)}=\frac{279.6\times42}{576}=20.4$. *Kyun:* energy heat mein dump ho jaati hai. > - Notice karo ki $T$ blow up karta hai lekin $\rho$ example (a) se barely aage badhta hai — yeh steel-man insight quantitatively sabit hai. --- ## 10. Common mistakes (Steel-man + fix) > [!mistake] "Total pressure conserved hai kyunki flow adiabatic hai." > *Kyun sahi lagta hai:* adiabatic ⇒ koi heat nahi ⇒ "energy conserved" ⇒ lagta hai $p_0$ stay karna chahiye. **Fix:** adiabatic total **enthalpy/temperature** ($T_0$) conserve karta hai, total **pressure** nahi. Shock *irreversible* hai (internal viscosity/conduction), entropy badhti hai, to $p_0$ **girta hai**. Adiabatic ≠ isentropic. > [!mistake] "Ek shock subsonic flow ko supersonic tak speed up kar sakta hai." > *Kyun sahi lagta hai:* math do roots deta hai, ek reverse jaisa lagta hai. **Fix:** woh root entropy *lower* kar deta — 2nd Law se forbidden. Shocks sirf supersonic→subsonic jaate hain (compression). Expansion shocks exist nahi karte. > [!mistake] $\rho_2/\rho_1=p_2/p_1$ use karna ("pressure aur density dono equally saath badhte hain"). > *Kyun sahi lagta hai:* dono shock ke across increase hote hain. **Fix:** yeh isothermal nahi hai! $T$ bhi jump karta hai, to $\rho_2/\rho_1=(p_2/p_1)(T_1/T_2)$. Density $\frac{\gamma+1}{\gamma-1}$ pe saturate hoti hai jabki pressure unbounded hai. --- ## 11. Recall > [!recall]- Active recall — answers cover karo > - Shock relations konse teen conservation laws se aate hain? ⇒ mass, momentum, energy. > - Shock ke across kya constant rehta hai, aur kya drop karta hai? ⇒ $T_0$ (total temp) constant; $p_0$ drops. > - $\gamma=1.4$ ke liye $M_1\to\infty$ pe $\rho_2/\rho_1$ ki limit? ⇒ 6. > - $M_1>1$ ke liye $M_2<1$ guaranteed kyun hai? ⇒ entropy badhni chahiye; subsonic root hi physical hai. > [!recall]- Feynman — ek 12-saal ke bacche ko explain karo > Highway pe cars ko imagine karo jo achanak ek traffic jam mein ghus jaati hain. Tez jaane ki wajah se unhe koi warning nahi milti (koi honk untak nahi pahunchta kyunki woh awaaz se tez hain), to woh ek patli packed wall mein jam jaati hain. Us wall ke baad cars slowly chalti hain, ek doosre ke paas pack hoti hain (high density), aur bumping se garam ho jaati hain (high temperature). Woh packed wall hi **shock** hai. Cars ko sirf itna hi squish kar sakte ho (density limit), lekin bumping ki heat aur badhti rahti hai jitna tez woh aayi hain. Aur kuch "orderliness" crash mein hamesha ke liye kho jaati hai — yahi lost total pressure hai. > [!mnemonic] > **"My Mom Eats Pancakes Daily, Temperature Sky-high"** → > **M**ass, **M**omentum, **E**nergy dete hain → **P**ressure, **D**ensity, **T**emperature ratios. > Aur yaad rakho **"$T_0$ stays, $p_0$ pays."** --- ## 12. Connections - [[Speed of sound and Mach number]] - [[Isentropic flow relations]] (contrast: reversible, $p_0$ conserved) - [[Second Law of Thermodynamics — entropy]] (shock branch select karta hai) - [[Oblique shock waves]] (normal shock = oblique with deflection 0) - [[Rayleigh & Fanno flow]] (heat/friction-driven Mach changes) - [[Stagnation properties]] #flashcards/physics Rankine–Hugoniot relations ke peeche kaunse assumptions hain? ::: Steady, 1-D, adiabatic, no body forces/external work, calorically perfect gas; viscosity/conduction sirf thin shock ke andar. Woh identity jo sab kuch M ke functions mein reduce karti hai ::: $\rho u^2=\gamma p M^2$ (kyunki $u=M\sqrt{\gamma RT}$). Downstream Mach number formula ::: $M_2^2=\dfrac{1+\frac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\frac{\gamma-1}{2}}$. Normal shock ke across static pressure ratio ::: $\dfrac{p_2}{p_1}=1+\dfrac{2\gamma}{\gamma+1}(M_1^2-1)$. Density/velocity ratio ::: $\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}=\dfrac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2}$. Temperature ratio ::: $\dfrac{T_2}{T_1}=\dfrac{[2\gamma M_1^2-(\gamma-1)][(\gamma-1)M_1^2+2]}{(\gamma+1)^2M_1^2}$. γ=1.4 ke liye maximum density ratio ::: 6 ($\frac{\gamma+1}{\gamma-1}$ ki limit as $M_1\to\infty$). Shock ke across kya conserved hota hai vs kya lost hota hai? ::: Total temperature $T_0$ conserved (adiabatic); total pressure $p_0$ decrease hota hai (entropy badhti hai). Expansion shocks kyun nahi hote? ::: Woh entropy lower kar denge, 2nd Law violate hoga; shocks compressive hi hone chahiye. M1=2, γ=1.4 ke liye: M2, p2/p1, ρ2/ρ1, T2/T1 ::: 0.577, 4.50, 2.667, 1.687. ## 🖼️ Concept Map ```mermaid flowchart TD A[Normal Shock: supersonic to subsonic] B[Assumptions: steady 1D adiabatic no-work perfect gas] C[Mass: rho1 u1 = rho2 u2] D[Momentum: p + rho u^2] E[Energy: h + half u^2] F[Identity rho u^2 = gamma p M^2] G[Pressure ratio p2/p1] H[Density/velocity ratio] I[Temperature ratio T2/T1] J[Downstream Mach M2] K[Total pressure loss p02/p01] L[Entropy rises] B -->|justify| C B -->|justify| D B -->|justify| E C --> F D -->|combined with| F F -->|transforms| G E -->|total enthalpy| I G --> H D -->|yields| H E -->|gives| J G -->|with H| I J -->|function of M1| G K -->|implies| L G -->|feeds| K ```