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3.1.14Compressible Flow & Aerodynamics

Shock wave angle, deflection angle

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🇮🇳 Yeh note Hinglish mein padho →

WHAT are these two angles?

WHY decompose the velocity? (first principles)

The key trick: an oblique shock is just a normal shock for the velocity component perpendicular to it, with the tangential component unchanged.

Set up the geometry:

  • u1=V1sin⁡βu_1 = V_1\sin\beta — normal component upstream
  • w1=V1cos⁡βw_1 = V_1\cos\beta — tangential component (conserved: w2=w1w_2=w_1)
  • Behind the shock the flow has turned by θ\theta, so the normal component makes angle (β−θ)(\beta-\theta) with the downstream velocity: u2=V2sin⁡(β−θ),w2=V2cos⁡(β−θ)u_2 = V_2\sin(\beta-\theta), \qquad w_2 = V_2\cos(\beta-\theta)
Figure — Shock wave angle, deflection angle

HOW to derive the θ\theta–β\beta–MM relation

Step 1 — Normal Mach numbers. The normal component of the upstream Mach number is Mn1=M1sin⁡βM_{n1} = M_1\sin\beta Why this step? The perpendicular flow behaves like a normal shock, and a normal shock only "cares" about the normal Mach number.

Step 2 — Use the normal-shock density (continuity) ratio. For a normal shock the density ratio is the standard Rankine–Hugoniot result ρ2ρ1=(γ+1)Mn12(γ−1)Mn12+2\frac{\rho_2}{\rho_1}=\frac{(\gamma+1)M_{n1}^2}{(\gamma-1)M_{n1}^2+2} Why this step? Mass conservation across the shock face uses only the normal velocity: ρ1u1=ρ2u2\rho_1 u_1=\rho_2 u_2, so ρ2ρ1=u1u2\dfrac{\rho_2}{\rho_1}=\dfrac{u_1}{u_2}.

Step 3 — Express the velocity ratio with the angles. Divide normal by tangential on each side (tangential is equal): tan⁡β=u1w1,tan⁡(β−θ)=u2w2=u2w1\tan\beta=\frac{u_1}{w_1},\qquad \tan(\beta-\theta)=\frac{u_2}{w_2}=\frac{u_2}{w_1} Why this step? This turns velocity ratios into angle ratios, which is what we want. Dividing, tan⁡(β−θ)tan⁡β=u2u1=ρ1ρ2=(γ−1)Mn12+2(γ+1)Mn12\frac{\tan(\beta-\theta)}{\tan\beta}=\frac{u_2}{u_1}=\frac{\rho_1}{\rho_2}=\frac{(\gamma-1)M_{n1}^2+2}{(\gamma+1)M_{n1}^2}

Step 4 — Substitute Mn1=M1sin⁡βM_{n1}=M_1\sin\beta and simplify (using tan⁡=sin⁡/cos⁡\tan = \sin/\cos and trig identities). The standard compact form is:

Reading the relation (the 80/20 essentials)

Common mistakes (Steel-man them)

Recall Feynman: explain to a 12-year-old

Imagine running so fast that the "honk" of your warning can't reach the wall before you do. When supersonic air hits a wedge, it has no warning, so it crashes and bends all at once along a sharp line — that line is the shock. The slant of the line is the wave angle. How much the air's path bends is the turn angle. The neat trick: split the air's speed into "into the line" and "along the line." Only the "into the line" part gets squashed; the "along the line" part glides through unchanged. A small turn lets the line stay leaning and the air stays fast; turn too much and the air gives up — the shock pops off and floats in front like a curved cushion.

Flashcards

What does the wave angle β\beta measure?
The angle between the oblique shock and the upstream flow direction.
What does the deflection angle θ\theta measure?
The angle by which the streamline is turned across the shock (= wedge half-angle).
Why is tangential velocity unchanged across an oblique shock?
Inviscid shock pressure acts only perpendicular to its face, so there's no tangential force ⇒ tangential momentum (and velocity) is conserved.
What is the normal upstream Mach number?
Mn1=M1sin⁡βM_{n1}=M_1\sin\beta.
State the θ\theta–β\beta–MM relation.
tan⁡θ=2cot⁡β M12sin⁡2β−1M12(γ+cos⁡2β)+2\tan\theta = 2\cot\beta\,\dfrac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}.
What are the lower and upper limits of β\beta?
Lower = Mach angle μ=sin⁡−1(1/M1)\mu=\sin^{-1}(1/M_1); upper = 90∘90^\circ (normal shock). Both give θ=0\theta=0.
For θ<θmax⁡\theta<\theta_{\max} how many shock solutions exist?
Two — a weak shock (smaller β\beta, flow stays supersonic) and a strong shock (larger β\beta, flow subsonic).
What happens if the required θ>θmax⁡\theta>\theta_{\max}?
No attached oblique shock; a detached curved bow shock forms ahead of the body.
How do you get M2M_2 behind an oblique shock?
M2=Mn2/sin⁡(β−θ)M_2 = M_{n2}/\sin(\beta-\theta), where Mn2M_{n2} comes from the normal-shock relation applied to Mn1M_{n1}.
Setting θ=0\theta=0 in the relation gives what?
sin⁡β=1/M1\sin\beta = 1/M_1, i.e. β=\beta= Mach angle (a Mach wave).

Connections

  • Normal Shock Waves — the perpendicular-component limit (β=90∘\beta=90^\circ).
  • Mach Angle and Mach Waves — the θ→0\theta\to0 lower bound.
  • Rankine-Hugoniot Relations — density/pressure jumps used in Step 2.
  • Prandtl-Meyer Expansion — the opposite case (flow turning away, no shock).
  • Detached Bow Shock — what happens past θmax⁡\theta_{\max}.
  • Supersonic Wedge & Cone Flow — direct application of these angles.

Concept Map

cannot signal upstream

angle to upstream flow

flow turned by

equals

lower limit

upper limit 90 deg

decompose velocity

inviscid, no tangential force

normal acts like

Mn1 = M1 sin beta

combine with geometry

Supersonic flow hits wedge

Oblique shock forms

Wave angle beta

Deflection angle theta

Wedge half-angle

Mach angle mu = asin 1/M1

Normal shock

Normal and tangential components

Tangential velocity conserved

Normal shock analysis

Rankine-Hugoniot density ratio

theta-beta-M relation

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab supersonic flow ek wedge ya corner se takraata hai, toh flow ko pehle se "pata" nahi chalta ki aage obstacle hai — kyunki signal (sound) flow se tez upstream nahi ja sakta. Isliye flow dheere-dheere mudne ke bajaye ek patli shock ke across ekdum se mud jaata hai. Yeh shock incoming flow ke saath jo angle banaata hai use wave angle β\beta kehte hain, aur flow jitna mudta hai use deflection angle θ\theta (wedge ka half-angle) kehte hain. Yaad rakho: θ\theta hamesha β\beta se chhota hota hai.

Sabse important trick yeh hai: oblique shock asal me ek normal shock hota hai sirf us velocity component ke liye jo shock ke perpendicular hai. Jo component shock ke along (tangential) hai woh bilkul nahi badalta, kyunki inviscid shock par pressure sirf perpendicular lagta hai, koi shear force nahi. Isi liye hum Mn1=M1sin⁡βM_{n1}=M_1\sin\beta use karte hain aur saare normal-shock formulas isi normal Mach number par lagte hain.

Inko jodne wala main rishta hai θ\theta–β\beta–MM relation: tan⁡θ=2cot⁡β (M12sin⁡2β−1)/(M12(γ+cos⁡2β)+2)\tan\theta = 2\cot\beta\,(M_1^2\sin^2\beta-1)/(M_1^2(\gamma+\cos2\beta)+2). Iska matlab — agar tumhe M1M_1 aur θ\theta pata hai, toh β\beta nikal sakte ho, par aam taur par do answers milte hain: weak shock (chhota β\beta, flow supersonic hi rehta hai — real life external aero me yahi hota hai) aur strong shock (bada β\beta, flow subsonic ho jaata hai).

Ek aur cheez exam ke liye 80/20: har M1M_1 ke liye ek θmax\theta_{max} hota hai. Agar wedge ka angle isse zyada ho gaya, toh koi attached shock possible nahi — shock body se detach hokar aage curved bow shock ban jaata hai. Toh basic intuition: split the velocity, normal part squash hota hai, tangential glide karta hai, aur angles ka geometry sab kuch decide karta hai.

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Connections

Normal shock waves — Rankine-Hugoniot relations (all 5) — derivationsPhysics · 3.1.11Prandtl-Meyer expansion waves — isentropic, supersonic turningPhysics · 3.1.16Detached bow shockPhysics · 3.1.15