Oblique shock waves — θ-β-M relation

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WHAT is an oblique shock, physically?

The key trick (WHY it works): decompose the upstream velocity into components normal and tangential to the shock.

• Normal component $u_{n1} = V_1\sin\beta$ → behaves like a normal shock with effective Mach number $M_{n1} = M_1\sin\beta$
• Tangential component $w = V_1\cos\beta$ → unchanged across the shock.

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HOW to derive the θ-β-M relation from scratch

We build everything from the velocity triangle. Upstream the flow is along the wall (angle $\beta$ to shock). Downstream it is deflected by $\theta$, so it makes angle $(\beta-\theta)$ with the shock.

Step 1 — Normal components from geometry. $u_{n1} = V_1\sin\beta, \qquad u_{n2} = V_2\sin(\beta-\theta)$ Why this step? The downstream flow makes angle $(\beta-\theta)$ with the shock, so its normal component uses that angle.

Step 2 — Tangential component is conserved. $w = V_1\cos\beta = V_2\cos(\beta-\theta)$ Why this step? No tangential force (proved above), so the same $w$ appears on both sides.

Step 3 — Form the ratio (eliminate $V$). Divide the normal equations and use Step 2:

= \frac{\cos\beta}{\cos(\beta-\theta)}\cdot\frac{\sin(\beta-\theta)}{\sin\beta} = \frac{\tan(\beta-\theta)}{\tan\beta}$$ *Why this step?* We replaced $V_2/V_1$ using the tangential identity $V_2 = V_1\cos\beta/\cos(\beta-\theta)$. Now everything is in angles. **Step 4 — Use the normal-shock density (velocity) ratio.** Across a normal shock the mass conservation + Rankine–Hugoniot give $$\frac{u_{n2}}{u_{n1}} = \frac{\rho_1}{\rho_2} = \frac{(\gamma-1)M_{n1}^2 + 2}{(\gamma+1)M_{n1}^2}, \qquad M_{n1}=M_1\sin\beta$$ *Why this step?* The normal direction obeys exactly the normal-shock laws — that was the whole point of decomposing. **Step 5 — Equate Steps 3 and 4:** $$\frac{\tan(\beta-\theta)}{\tan\beta} = \frac{(\gamma-1)M_1^2\sin^2\beta + 2}{(\gamma+1)M_1^2\sin^2\beta}$$ **Step 6 — Solve for $\tan\theta$** (algebra using $\tan(\beta-\theta)$ expansion) to get the standard explicit form: > [!formula] The θ-β-M relation > $$\boxed{\;\tan\theta = 2\cot\beta\,\dfrac{M_1^2\sin^2\beta - 1}{M_1^2(\gamma + \cos 2\beta) + 2}\;}$$ > - $\theta$: flow deflection, $\beta$: shock angle, $M_1$: upstream Mach, $\gamma$: ratio of specific heats ($1.4$ for air). > - Reduces to a **normal shock** at $\beta = 90^\circ$ (then $\tan\theta = 0 \Rightarrow \theta=0$, no deflection). > - Requires $\sin\beta > 1/M_1$ for a real shock (i.e. $M_{n1}>1$). At $\beta=\mu=\arcsin(1/M_1)$, $\theta=0$ — that's a **Mach wave**. --- ## Reading the curve (the 80/20 of using it) For a fixed $M_1$, plotting $\theta$ vs $\beta$ gives a **hump**: - For each $\theta < \theta_{max}$ there are **TWO** solutions: - **Weak shock** (smaller $\beta$): downstream usually still supersonic — *nature normally picks this*. - **Strong shock** (larger $\beta$): downstream subsonic. - $\theta_{max}$ = peak of the hump. If the wedge demands $\theta > \theta_{max}$, **no attached oblique shock exists** → the shock **detaches** and becomes a curved **bow shock** standing off the body. > [!example] Worked Example 1 — find $\beta$ for a wedge > Air ($\gamma=1.4$), $M_1 = 3$, half-angle wedge $\theta = 20^\circ$. Find the weak-shock angle. > > **Why:** the wedge forces $\theta=20^\circ$; we invert θ-β-M for $\beta$. > Plug into $\tan 20^\circ = 2\cot\beta\,\dfrac{9\sin^2\beta-1}{9(1.4+\cos2\beta)+2}$ and solve numerically. > **Result:** $\beta \approx 37.8^\circ$ (weak). *Why this root?* It's the smaller of the two; the strong one is $\approx 82^\circ$. > Then $M_{n1}=3\sin37.8^\circ = 1.84$ → use normal-shock tables for $p_2/p_1$, $M_{n2}$, and finally $M_2 = M_{n2}/\sin(\beta-\theta)$. > [!example] Worked Example 2 — Mach wave limit > $M_1 = 2$, $\theta \to 0$. **Why:** an infinitesimally weak disturbance. > Setting numerator $M_1^2\sin^2\beta-1=0$ gives $\sin\beta = 1/M_1 = 0.5 \Rightarrow \beta=30^\circ$. > *Interpretation:* this is the **Mach angle** $\mu=\arcsin(1/M)$ — the limiting weakest oblique "shock". > [!example] Worked Example 3 — downstream Mach number > From Example 1: $M_{n1}=1.84 \Rightarrow M_{n2}\approx 0.608$ (normal-shock relation). > $$M_2 = \frac{M_{n2}}{\sin(\beta-\theta)} = \frac{0.608}{\sin(37.8^\circ-20^\circ)} = \frac{0.608}{\sin17.8^\circ}\approx 1.99.$$ > *Why divide by $\sin(\beta-\theta)$?* Because $u_{n2}=V_2\sin(\beta-\theta)$, so the Mach version inverts that geometry. Note $M_2>1$: this weak shock stays supersonic. --- ## Common mistakes (Steel-manned) > [!mistake] "Use $M_1$ in the normal-shock formula directly." > **Why it feels right:** you're crossing a shock, and normal-shock formulas use upstream Mach. **The fix:** only the **normal component** $M_{n1}=M_1\sin\beta$ crosses the shock like a normal shock. The full $M_1$ overstates the compression. > [!mistake] "Downstream flow is always subsonic after a shock." > **Why it feels right:** true for normal shocks. **The fix:** for the **weak** oblique shock $M_2$ is often still **supersonic** — only $M_{n2}<1$ is guaranteed. > [!mistake] "There's one shock angle per deflection." > **Why it feels right:** geometry seems unique. **The fix:** the θ-β curve is double-valued — **weak and strong** roots. Specify which by context (free flight → weak). > [!mistake] "Bigger wedge → bigger shock, always." > **Why it feels right:** more turning = more compression. **The fix:** only up to $\theta_{max}$. Beyond it, the shock **detaches** (bow shock) and the attached-shock formula has no real solution. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine you're skating fast and you ram into a slanted wall. The part of your motion *straight into* the wall gets stopped hard (that's the "shock"), but the part *sliding along* the wall keeps going the same. Because the wall is tilted, you don't stop fully — you just get bent sideways a bit and slowed down. If you make the wall tilt steeper than your skates can handle, you can't follow it at all and a big curved cushion of squished air forms in front. The θ-β-M rule is just the math that says: "for this speed and this turn angle, here's exactly how tilted the shock must be." > [!mnemonic] Remember the geometry > **"Normal Normally, Tangent Tags Along."** > The **normal** component does **N**ormal-shock physics; the **tangential** component is conserved (tags along unchanged). And **"Weak is what nature seeks."** > [!recall] Active recall checkpoints > - What stays constant across an oblique shock? *(tangential velocity)* > - Which Mach number enters the normal-shock laws? *($M_{n1}=M_1\sin\beta$)* > - What happens when $\theta>\theta_{max}$? *(detached bow shock)* --- ### #flashcards/physics What two velocity components do we decompose the flow into at an oblique shock? ::: Normal (to shock) and tangential (along shock). Why is the tangential velocity unchanged across an oblique shock? ::: No tangential pressure force exists along the shock face, so tangential momentum is conserved. What effective Mach number governs the normal-shock physics? ::: $M_{n1}=M_1\sin\beta$. State the θ-β-M relation. ::: $\tan\theta = 2\cot\beta\,\dfrac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$. What does $\beta=90^\circ$ reduce the oblique shock to? ::: A normal shock (deflection $\theta=0$). What is the minimum $\beta$ for a real oblique shock? ::: The Mach angle $\beta=\arcsin(1/M_1)$, where $M_{n1}=1$ and $\theta=0$. For a given $\theta<\theta_{max}$, how many shock-angle solutions exist? ::: Two — a weak (smaller $\beta$) and a strong (larger $\beta$) shock. Which solution does nature usually pick in external flow? ::: The weak shock. Is downstream flow subsonic for a weak oblique shock? ::: Usually no — $M_2$ stays supersonic, though $M_{n2}<1$ always. What happens when wedge angle exceeds $\theta_{max}$? ::: No attached oblique shock; a detached curved bow shock forms ahead of the body. How do you recover $M_2$ after finding $M_{n2}$? ::: $M_2=M_{n2}/\sin(\beta-\theta)$. --- ### Connections - [[Normal Shock Waves]] — the building block ($M_{n1}$ closure). - [[Mach Angle and Mach Waves]] — the $\theta\to0$ limit. - [[Prandtl-Meyer Expansion]] — the opposite case (flow turning away, expansion fan). - [[Detached Bow Shocks]] — what happens past $\theta_{max}$. - [[Rankine-Hugoniot Relations]] — source of the density/velocity ratio in Step 4. - [[Supersonic Wedge and Cone Flow]] — direct engineering application. ## 🖼️ Concept Map ```mermaid flowchart TD M1[Supersonic flow M1] -->|turns into itself| WEDGE[Wedge compression corner] WEDGE -->|deflects flow by theta| OBS[Oblique shock at angle beta] OBS -->|decompose velocity| NORM[Normal component un1] OBS -->|decompose velocity| TANG[Tangential component w] NORM -->|acts like| NS[Normal shock Mn1 = M1 sin beta] TANG -->|no tangential force| CONS[w conserved across shock] NS -->|geometry plus Rankine-Hugoniot| REL[theta-beta-M relation] CONS -->|tan ratio identity| REL REL -->|gives shock angle beta| DOWN[Downstream properties] DOWN -->|yields| PROPS[Pressure temperature M2] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, oblique shock ka core idea bahut simple hai: jab supersonic flow ko ek wedge ya corner force karta hai turn karne ke liye, toh flow smoothly mud nahi sakta — kyunki information upstream travel nahi kar sakti (flow toh sound se fast hai). Isliye ek tilted shock ban jaata hai jo angle $\beta$ pe baithta hai, aur flow $\theta$ angle se mud jaata hai. Sabse important trick: incoming velocity ko shock ke **normal** aur **tangential** components mein toड़ do. Normal part bilkul ek normal shock ki tarah behave karta hai (effective Mach $M_{n1}=M_1\sin\beta$), aur tangential part **change hi nahi hota** — kyunki shock ke along koi pressure force nahi hai. > > Yahi se θ-β-M relation nikalta hai. Velocity triangle banao, normal aur tangential conserve/transform karo, aur normal-shock ki density ratio plug karo — bas formula aa jaata hai: $\tan\theta = 2\cot\beta\,(M_1^2\sin^2\beta-1)/(M_1^2(\gamma+\cos2\beta)+2)$. Iska matlab: agar tumhe wedge angle $\theta$ aur flight Mach $M_1$ pata hai, toh shock angle $\beta$ nikal sakte ho, aur uske baad saari downstream properties (pressure, temperature, $M_2$). > > Ek bahut important baat: ek hi $\theta$ ke liye **do** answers aate hain — **weak shock** (chhota $\beta$, flow usually supersonic rehta hai) aur **strong shock** (bada $\beta$, flow subsonic). Real life external flow mein nature weak waala choose karta hai. Aur agar wedge angle $\theta_{max}$ se zyada ho gaya, toh attached shock possible hi nahi — shock detach ho ke ek curved **bow shock** ban jaata hai body ke aage. Yeh concept missile noses, supersonic inlets aur fighter jet design mein direct kaam aata hai. ![[audio/3.1.13-Oblique-shock-waves-—-θ-β-M-relation.mp3]]