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

A normal shock is a razor-thin (few mean-free-paths) discontinuity standing perpendicular to a supersonic flow. Across it the flow jumps abruptly from supersonic to subsonic, pressure and temperature shoot up, and the flow loses total pressure (entropy rises). The five Rankine–Hugoniot relations tell us the ratios $\rho_2/\rho_1$, $p_2/p_1$, $T_2/T_1$, $M_2$, and $p_{02}/p_{01}$ purely as functions of the upstream Mach number $M_1$ (for a calorically perfect gas).

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1. The setup — WHAT we are doing

The three 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}}$$ with $h=c_p T = \dfrac{\gamma}{\gamma-1}\dfrac{p}{\rho}$. **Why these three?** Mass can't appear/vanish, momentum changes only by the pressure forces on the slab faces, and with no heat/work the **total enthalpy** is conserved. Everything below is algebra on these three lines. ![[3.1.11-Normal-shock-waves-—-Rankine-Hugoniot-relations-(all-5)-—-derivations.png]] --- ## 2. Master substitution: $u = M\sqrt{\gamma R T}$ and $\rho u^2 = \gamma p M^2$ > [!intuition] > The trick that makes everything collapse to functions of $M$ alone: speed of sound $a=\sqrt{\gamma R T}$, so $u=Ma$, and > $$\rho u^2 = \rho (Ma)^2 = \rho M^2 \gamma R T = \gamma M^2 (\rho R T) = \gamma p M^2.$$ > So $\rho u^2 = \gamma p M^2$. This single identity is the engine of the whole derivation. --- ## 3. Relation 1 — Velocity / density ratio (the Prandtl-style result) We derive the density (= velocity) ratio first because the others lean on it. **Step A.** Rewrite momentum using $\rho u^2=\gamma p M^2$: $$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{$\star$}$$ *Why this step?* It turns the momentum equation into a clean pressure ratio in Mach numbers. **Step B.** Energy with total enthalpy. Since $h=\frac{a^2}{\gamma-1}$ and $u=Ma$: $$\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{$\dagger$}$$ *Why this step?* Total enthalpy constant ⇒ $a^2(1+\frac{\gamma-1}{2}M^2)$ is the same on both sides. **Step C.** Mass + ideal gas: $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{M_1 a_1}{M_2 a_2}.$$ Also from ideal gas and ($\star$): $\dfrac{\rho_2}{\rho_1}=\dfrac{p_2}{p_1}\dfrac{T_1}{T_2}=\dfrac{p_2/p_1}{a_2^2/a_1^2}$. Set the two density expressions equal and use ($\star$),($\dagger$). After simplification (algebra below) you get the famous **Mach-number relation** which then gives every ratio. --- ## 4. Relation 2 — Downstream Mach number $M_2$ Combine ($\star$) and the mass/energy results. The cleanest route: substitute $\frac{a_2^2}{a_1^2}$ from ($\dagger$) and $\frac{u_1}{u_2}=\frac{\rho_2}{\rho_1}$ into mass conservation written as $$\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}}.$$ Equating to $\dfrac{p_2/p_1}{a_2^2/a_1^2}$ and grinding through (a quadratic that factors, the trivial root $M_2=M_1$ being the no-shock case) leaves the surviving root: > [!formula] Downstream Mach number > $$\boxed{\,M_2^2=\dfrac{1+\dfrac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\dfrac{\gamma-1}{2}}\,}$$ > For $M_1>1$ this gives $M_2<1$ (subsonic). For $M_1\to\infty$, $M_2^2\to\frac{\gamma-1}{2\gamma}$ (e.g. $\approx0.378$ for $\gamma=1.4$). > [!intuition] > There are always two algebraic solutions: $M_2=M_1$ (no shock, the flow just continues) and the genuine shock. The Second Law (entropy must rise) selects the shock branch and forbids the reverse (expansion shock $M_1<1\to M_2>1$). **Shocks are always compressive.** --- ## 5. Relation 3 — Pressure ratio Put the $M_2^2$ expression into ($\star$). After algebra the $1+\gamma M_2^2$ in the denominator simplifies beautifully: > [!formula] Static pressure ratio > $$\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}$$ > *Sanity:* at $M_1=1$, $p_2/p_1=1$ (infinitely weak shock). For $M_1\to\infty$ it grows without bound — strong shocks pile up huge pressure. --- ## 6. Relation 4 — Density (velocity) ratio From mass $\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}$, substitute the results above: > [!formula] Density / velocity ratio > $$\boxed{\,\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}=\dfrac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2}\,}$$ > [!intuition] **Density saturates.** As $M_1\to\infty$, $\rho_2/\rho_1\to\frac{\gamma+1}{\gamma-1}$ ($=6$ for $\gamma=1.4$). No matter how strong the shock, you cannot compress a perfect gas more than 6×. The extra energy goes into **temperature**, not density. --- ## 7. Relation 5 — Temperature ratio Ideal gas: $\dfrac{T_2}{T_1}=\dfrac{p_2/p_1}{\rho_2/\rho_1}$. Multiply Relations 3 and 4: > [!formula] Static temperature ratio > $$\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}\,}$$ > Unlike density, temperature grows like $M_1^2$ for large $M_1$ — this is why re-entry vehicles roast. --- ## 8. Bonus — Total (stagnation) pressure ratio & entropy Total enthalpy is conserved ⇒ $T_{01}=T_{02}$, so $T_{0}$ is **unchanged**. But entropy rises, so $p_0$ **drops**: $$\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 is the price (drag, inefficiency) of crossing a shock. --- ## 9. Worked examples > [!example] (a) Air, $\gamma=1.4$, $M_1=2$ > - $M_2^2=\dfrac{1+0.2(4)}{1.4(4)-0.2}=\dfrac{1.8}{5.4}=0.3333\Rightarrow M_2=0.577$. *Why:* plug $M_1^2=4$. > - $p_2/p_1=1+\frac{2.8}{2.4}(4-1)=1+1.1667(3)=4.50$. *Why:* Relation 3. > - $\rho_2/\rho_1=\frac{2.4(4)}{0.4(4)+2}=\frac{9.6}{3.6}=2.667$. *Why:* Relation 4. > - $T_2/T_1=4.50/2.667=1.687$. *Why:* ideal gas $T=p/\rho$ ratio. > [!example] (b) Strong-shock limit, $M_1=10$, $\gamma=1.4$ > - $\rho_2/\rho_1=\frac{2.4(100)}{0.4(100)+2}=\frac{240}{42}=5.71$ (near the cap 6). *Why:* density saturates. > - $T_2/T_1=\frac{[2.8(100)-0.4][0.4(100)+2]}{2.4^2(100)}=\frac{279.6\times42}{576}=20.4$. *Why:* energy dumped into heat. > - Notice $T$ blows up but $\rho$ barely moves past example (a) — the steel-man insight made quantitative. --- ## 10. Common mistakes (Steel-man + fix) > [!mistake] "Total pressure is conserved because the flow is adiabatic." > *Why it feels right:* adiabatic ⇒ no heat ⇒ "energy conserved" ⇒ feels like $p_0$ should stay. **Fix:** adiabatic conserves total **enthalpy/temperature** ($T_0$), not total **pressure**. The shock is *irreversible* (internal viscosity/conduction), entropy rises, so $p_0$ **falls**. Adiabatic ≠ isentropic. > [!mistake] "A shock can speed a subsonic flow up to supersonic." > *Why it feels right:* the math gives two roots, one looks like the reverse. **Fix:** that root would *lower* entropy — forbidden by the 2nd Law. Shocks only go supersonic→subsonic (compression). Expansion shocks don't exist. > [!mistake] Using $\rho_2/\rho_1=p_2/p_1$ ("pressure and density rise together equally"). > *Why it feels right:* both increase across the shock. **Fix:** it's not isothermal! $T$ also jumps, so $\rho_2/\rho_1=(p_2/p_1)(T_1/T_2)$. Density saturates at $\frac{\gamma+1}{\gamma-1}$ while pressure is unbounded. --- ## 11. Recall > [!recall]- Active recall — cover the answers > - What three conservation laws give the shock relations? ⇒ mass, momentum, energy. > - What stays constant across a shock, and what drops? ⇒ $T_0$ (total temp) constant; $p_0$ drops. > - Limit of $\rho_2/\rho_1$ as $M_1\to\infty$ for $\gamma=1.4$? ⇒ 6. > - Why is $M_2<1$ guaranteed for $M_1>1$? ⇒ entropy must rise; the subsonic root is the physical one. > [!recall]- Feynman — explain to a 12-year-old > Picture cars on a highway suddenly hitting a traffic jam. Going fast, they have no warning (no honk reaches them because they're faster than sound), so they slam together into a thin packed wall. After that wall the cars move slowly, are squished close together (high density), and get hot from bumping (high temperature). That packed wall is a **shock**. You can squish the cars only so much (density limit), but the bumping heat keeps growing the faster they came in. And some "orderliness" is lost forever in the crash — that's the lost total pressure. > [!mnemonic] > **"My Mom Eats Pancakes Daily, Temperature Sky-high"** → > **M**ass, **M**omentum, **E**nergy give → **P**ressure, **D**ensity, **T**emperature ratios. > And remember **"$T_0$ 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]] (selects the shock branch) - [[Oblique shock waves]] (normal shock = oblique with deflection 0) - [[Rayleigh & Fanno flow]] (heat/friction-driven Mach changes) - [[Stagnation properties]] #flashcards/physics What assumptions underlie the Rankine–Hugoniot relations? ::: Steady, 1-D, adiabatic, no body forces/external work, calorically perfect gas; viscosity/conduction only inside the thin shock. The identity that reduces everything to functions of M ::: $\rho u^2=\gamma p M^2$ (since $u=M\sqrt{\gamma RT}$). Downstream Mach number formula ::: $M_2^2=\dfrac{1+\frac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\frac{\gamma-1}{2}}$. Static pressure ratio across a normal shock ::: $\dfrac{p_2}{p_1}=1+\dfrac{2\gamma}{\gamma+1}(M_1^2-1)$. 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 ::: $\dfrac{T_2}{T_1}=\dfrac{[2\gamma M_1^2-(\gamma-1)][(\gamma-1)M_1^2+2]}{(\gamma+1)^2M_1^2}$. Maximum density ratio for γ=1.4 ::: 6 (limit of $\frac{\gamma+1}{\gamma-1}$ as $M_1\to\infty$). What is conserved vs lost across a shock? ::: Total temperature $T_0$ conserved (adiabatic); total pressure $p_0$ decreases (entropy rises). Why no expansion shocks? ::: They'd lower entropy, violating the 2nd Law; shocks must be compressive. For M1=2, γ=1.4: 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, normal shock ek bahut patli "deewar" hai jo supersonic flow ke saamne perpendicular khadi rehti hai. Jab air sound se tez aa rahi hoti hai, downstream tak koi pressure signal pahunch hi nahi paata (kyunki flow signal se bhi tez hai), isliye nature ek achanak compression wall bana deti hai — wahi shock hai. Iske aar-paar flow supersonic se subsonic ho jaata hai, aur pressure, temperature, density sab achanak badh jaate hain. > > Saari Rankine–Hugoniot relations sirf teen cheezon se nikalti hain: mass, momentum aur energy conservation. Ek magic identity yaad rakho — $\rho u^2 = \gamma p M^2$ — bas isse sab kuch sirf upstream Mach number $M_1$ ke function ban jaate hain. Phir nikalta hai $M_2$ (hamesha 1 se kam jab $M_1>1$), $p_2/p_1$, $\rho_2/\rho_1$, $T_2/T_1$. > > Sabse important intuition do hain. Pehla: density ek limit pe ruk jaati hai — chahe shock kitna bhi strong ho, $\rho_2/\rho_1$ zyada se zyada 6 ho sakta hai ($\gamma=1.4$ ke liye), lekin temperature unbounded badhta hai. Isiliye re-entry vehicles itne garam hote hain. Doosra: total temperature $T_0$ same rehta hai (adiabatic hai), par total pressure $p_0$ girta hai kyunki shock irreversible hai — entropy badhti hai. Yaad rakho: "**T0 stays, p0 pays**". Aur expansion shock (subsonic se supersonic) kabhi nahi hota — wo 2nd law todega. ![[audio/3.1.11-Normal-shock-waves-—-Rankine-Hugoniot-relations-(all-5)-—-derivations.mp3]]