Wave drag — transonic and supersonic: Question bank

3.1.25 · D5 · Physics › Compressible Flow & Aerodynamics › Wave drag — transonic and supersonic

Pehle bank attempt karne se pehle, is page ki notation aur geometry ko solid kar lo. Neeche use hone wale har symbol ki definition pehle di gayi hai, aur chaar core geometric pictures draw ki gayi hain taaki tumhe unhe sirf text se imagine na karna pade.

Definition Is poore page par use hone wali Notation

== $M_{\infty}$ == — free-stream Mach number, yaani pure body ke kaafi aage ki undisturbed air ka Mach. Yeh aircraft ka flight Mach hai. Free stream ke liye hum poore page par $M_{\infty}$ use karte hain.
== $M_{1}$ == — ek specific shock mein enter karte waqt ka local Mach number. Yeh $M_{\infty}$ tab hi hota hai jab woh shock undisturbed flow mein baithta ho (jaise ek isolated wedge ki leading edge par). Jab bhi hum "jis shock ko dekh rahe hain uske bilkul upstream ka Mach" kehna chahte hain, hum $M_{1}$ likhte hain — taaki flight Mach se alag rahe.
== $C_{D}$ == (capital, subscript $D$ ) — poore body ka total drag coefficient: total drag force $\div (\frac{1}{2} ρ V^{2} \times reference area)$ . Yeh saare drag sources ko bundle karta hai (friction + wave + induced).
== $c_{d, w a v e}$ == (lower-case, per-unit-span) — ek 2-D airfoil section ka wave-drag coefficient, drag per unit span $\div (\frac{1}{2} ρ V^{2} \times chord)$ . Lower-case = "section" (2-D strip); capital = "whole 3-D body." $c_{d, w a v e}$ ek $C_{D}$ ka slice hai.
== $θ$ == — ek flow-turning angle: streamline kitne degrees deflect hoti hai. Ek wedge ke liye yeh wedge ke half-angle ke barabar hai.
== $θ_{ma x} (M_{1})$ == — sabse bada turn jo ek shock us $M_{1}$ par body se attached rehte hue kar sakta hai. Isse zyada turning maango toh shock detach ho jaata hai (bow shock ban jaata hai). Yeh $M_{1}$ dwara set ki gayi ek fixed ceiling hai.
== $d y_{t} / d x$ aur $d y_{c} / d x$ == — airfoil ki local surface slopes: $y_{t} (x)$ thickness distribution hai (chord line ke upar half-thickness) aur $y_{c} (x)$ camber (mean-line) shape hai. Inki slopes batati hain ki surface har station $x$ par flow mein kitni steeply tilt hoti hai; overbar $\overline{(\cdot)}$ ka matlab hai "chord ke saath average kiya hua."

Neeche chaar pictures woh hain jinhe yeh bank baar baar refer karta hai.

Figure — Wave drag — transonic and supersonic

Recall Teen "angles" jo sab

$M > 1$ ke paas rehte hain (figure s01 dekho)

Mach angle $μ = arcsin (1/ M_{\infty})$ — infinitesimally weak pressure pulses ka envelope (s01 mein teal cone).
Shock angle $β$ — ek finite-strength oblique shock ka lean (s01 mein orange line), $\theta$–$\beta$–$M$ relation se. Hamesha $β > μ$ .
Prandtl–Glauert factor $β_{P G} = M_{\infty}^{2} - 1$ — ek number hai, angle nahi, jo Ackeret theory mein appear karta hai. Yeh $M_{\infty} cos μ$ ke barabar hai: s01 dekho aur $cos μ = M_{\infty}^{2} - 1 / M_{\infty}$ seedha chhote Mach-angle triangle se padho. Shock angle ke saath ek hi letter — cruel tradition ke kaaran.

Recall Linked pages kya contribute karte hain (standalone summary)

Speed of sound and Mach number — pressure signals $a$ par travel karti hain; $M = V / a$ flight speed ko signal speed se compare karta hai.
Normal and oblique shock waves — ek shock ek thin, abrupt, irreversible compression hai; iske across entropy badhti hai.
Theta-beta-M relation and detached bow shocks — turn angle $θ$ , shock angle $β$ , upstream $M_{1}$ ko link karta hai; $θ_{ma x}$ aur detachment set karta hai.
Entropy and stagnation pressure loss — $p_{0, 2} / p_{0, 1} = e^{- Δ s / R} < 1$ : har shock stagnation pressure khota hai.
Critical and drag-divergence Mach number — $M_{cr}$ (pehla sonic point) aur $M_{dd}$ (steep drag rise) transonic regime ko delimit karte hain.
Whitcomb Area Rule and supercritical airfoils — length ke saath aircraft ka smooth total cross-sectional area transonic wave drag minimize karta hai (figure s04).

True or false — justify

True/false aur physical reason dono batao reveal karne se pehle.

Wave drag ke liye viscosity ka hona zaroori hai.

False. Wave drag entropy rise across a shock se paida hota hai (ek inviscid, adiabatic-but-irreversible process); yeh ek perfectly frictionless fluid mein bhi appear karta hai. Skin-friction drag alag viscous effect hai.

M_{\infty} < 1

par fly kar raha body kabhi wave drag experience nahi kar sakta.

False. Curved surface ke upar accelerate ho raha flow locally

M = 1

exceed kar sakta hai even jab free stream subsonic ho; woh local supersonic pocket ek shock par khatam hoti hai. Yahi exactly transonic drag-divergence mechanism hai.

Wave drag negative ho sakta hai agar airfoil ko cleverly shape diya jaaye.

False. Ackeret theory mein har wave-drag term ek square hai (

α^{2}

\overline{(d y_{c} / d x)^{2}}

\overline{(d y_{t} / d x)^{2}}

), isliye

c_{d, w a v e} \geq 0

hamesha. Squares encode karte hain ki "koi bhi deflection, kisi bhi sign mein, energy cost karti hai."

M_{\infty} > 1

par ek slender wedge ke liye, shock Mach cone ke saath lie karta hai.

False. Mach cone weak-signal envelope hai; ek real attached oblique shock ka

β > μ

hota hai kyunki finite turning ke liye finite compression chahiye (s01 mein orange vs teal compare karo). Yeh sirf zero-strength limit

θ \to 0

mein coincide karte hain.

Pure supersonic range mein

M_{\infty}

badhne par, wave drag badhta rehta hai.

False. Kyunki

c_{d, w a v e} \propto 1/ M_{\infty}^{2} - 1

, wave drag supersonic

M_{\infty}

badhne par actually girta hai (shocks zyada oblique aur per unit length kamzor ho jaate hain). Bada rise sirf transonic bump mein hota hai.

Drag-divergence Mach number, critical Mach number ke barabar hota hai.

False.

M_{cr}

woh jagah hai jahan flow pehli baar kaheen

M = 1

reach karta hai (abhi tak real drag nahi);

M_{dd} > M_{cr}

woh jagah hai jahan shock itna bada ho chuka ho ki total drag coefficient

C_{D}

steeply climb kare.

Ek thicker airfoil ka higher critical Mach number hota hai.

False. Thicker sections flow ko surface ke upar zyada accelerate karne par majboor karti hain, isliye local

M = 1

ek lower free-stream Mach par reach hota hai —

M_{cr}

lower hota hai, aur isliye fast wings thin hoti hain.

Wing ko sweep karna isliye help karta hai kyunki isse wing lambi ho jaati hai.

False. Sweep isliye help karta hai kyunki leading edge ke normal Mach component hi compressibility drive karta hai:

M_{n} = M_{\infty} cos Λ < M_{\infty}

(s03 dekho). Reduced effective Mach shock ko delay karta hai. Critical and drag-divergence Mach number dekho.

Ek shock ke across stagnation pressure constant rehta hai kyunki flow adiabatic hai.

False. Adiabatic total enthalpy

h_{0}

conserve karta hai, total pressure nahi. Entropy badhti hai, isliye

p_{0, 2} / p_{0, 1} = e^{- Δ s / R} < 1

— stagnation pressure hamesha girta hai.

Spot the error

Har statement mein ek flaw hai. Usse naam karo.

M_{\infty} = 2

par Mach angle

30°

hai, isliye

10°

wedge ka shock

30°

par baitha hai."

Error yeh hai ki Mach cone ko shock ke saath equate kar rahe hain (s01 dikhata hai ki woh alag hain). Wedge ka attached shock

β \approx 39.3°

par hai — zyada steep — kyunki use flow ko finite

10°

turn karna hai;

μ = 30°

sirf weak-disturbance envelope hai.

"Blunt noses wave drag minimize karne ke liye best hain kyunki woh shock ko spread out karte hain."

Ulta hai. Ek blunt body

θ_{ma x}

se zyada turning force karta hai, isliye shock detach hokar ek strong near-normal bow shock ban jaata hai (s02, right) — bade entropy jump ke saath — maximum wave drag. Slender bodies shocks ko attached aur weak rakhti hain (s02, left).

c_{d, w a v e} = \frac{4}{M _{\infty}^{2} - 1} (α + d y_{t} / d x + d y_{c} / d x)

Slopes aur

α

ko square kiya jaana chahiye, linearly sum nahi. Linear terms negative ho sakte hain aur galat thrust predict karte; squares guarantee karte hain ki

c_{d, w a v e} \geq 0

sin μ = 1/ M_{\infty}

use karte hue,

M_{\infty} = 1

par Mach angle

0°

hai, isliye cone body par collapse ho jaata hai."

Sign/limit error.

M_{\infty} = 1

par,

sin μ = 1

, isliye

μ = 90°

— "cone" flow ke perpendicular khade ek flat plane mein badal jaata hai. Angle sirf

M_{\infty} \to \infty

par

0°

ki taraf shrink karta hai.

"Prandtl–Glauert ka

β = M_{\infty}^{2} - 1

Ackeret formula mein shock angle hai."

Yeh ek compressibility factor hai (ek pure number jo

M_{\infty} cos μ

ke barabar hai), geometric shock angle

β

nahi. Yeh ek hi letter share karte hain lekin unrelated quantities hain — ek classic notation trap.

"Kyunki Ackeret ka

C_{p} = 2 θ / M_{\infty}^{2} - 1

θ

mein linear hai, ek downward-sloping surface negative

C_{p}

deta hai jo drag cancel karta hai."

Rearward-facing surface par negative

C_{p}

exactly wahi hai jo Prandtl–Meyer expansion deta hai, lekin yeh doosri taraf face karne wali surface par act karta hai, isliye force ko flow ke saath resolve karne par iska contribution drag mein phir bhi positive rehta hai. Drag terms squares mein end up hote hain.

Why questions

"Why" ka jawab mechanism ki ek ya do sentences mein do.

Wave drag critical Mach number ke upar hi kyun appear karta hai?

M_{cr}

se neeche poora flow subsonic hai, isliye compressions smooth aur reversible hain, koi shock nahi. Jab tak koi point

M = 1

reach nahi karta, shock form nahi ho sakta, aur shock hi entropy loss ka akela source hai jo wave drag banta hai.

Wake mein stagnation-pressure loss ka matlab body par force kyun hota hai?

Ek control-volume momentum balance momentum deficit (reduced

p_{0}

se) ko directly body par ek net rearward pressure force se tie karta hai — woh force hi wave drag hai. Lost stagnation pressure = lost push-back on the surface.

Transonic drag rise gentle subsonic drag ke comparison mein itna steep kyun hota hai?

M_{\infty} = 1

ke paas ek supersonic pocket suddenly form hoti hai aur uski terminating shock dono stagnation pressure lose karti hai aur iske peeche boundary layer thicken/separate hoti hai, isliye do drag sources ek saath ek narrow Mach band mein switch on hote hain.

Flat plate ke liye

L / D_{w a v e} = cot α

ek win se zyada warning kyun hai?

Chota

α

huge

L / D_{w a v e}

deta hai (efficient), lekin lift khud

α

ke saath scale karti hai, isliye almost koi lift generate nahi hoti — sirf efficiency par fly nahi kar sakte.

Area Rule sirf wing ke bajaay total cross-sectional area distribution par kyun focus karta hai?

Transonic wave drag chiefly is baat par depend karta hai ki aircraft ka combined cross-section apni length ke saath kitna smoothly grow aur shrink karta hai (s04), isliye wing ki wajah se ek bulge ko offset karna padta hai (jaise fuselage ko pinch karke) taaki area curve smooth rahe.

Linearized Ackeret pressure wahi

M_{\infty}^{2} - 1

kyun use karta hai jo Mach angle mein appear karta hai?

Small-deflection limit mein, real oblique waves Mach angle

μ = arcsin (1/ M_{\infty})

approach karti hain, aur

cos μ = M_{\infty}^{2} - 1 / M_{\infty}

geometry set karta hai (s01 se padho) — isliye weak-wave theory ko wahi

M_{\infty}^{2} - 1

factor virasat mein milta hai.

Bow shock ko

sin μ = 1/ M_{\infty}

se kabhi describe kyun nahi kiya ja sakta?

Ek bow shock ek finite, curved, near-normal compression hai jiska strength aur angle full shock relations se aate hain, jabki

sin μ = 1/ M_{\infty}

sirf vanishingly weak Mach waves ke envelope ko describe karta hai.

Edge cases

Har formula ko uski boundary tak push karo aur batao kya hota hai.

M_{\infty} = 1

par exactly Mach angle kya hai, aur "cone" kya ban jaata hai?

sin μ = 1/1 = 1

isliye

μ = 90°

; Mach cone flat hokar flow ke perpendicular khade ek plane mein badal jaata hai — signals barely body se aage nahi nikal paate. (Yeh woh value hai jis par neeche wala mnemonic point karta hai — $0°$ nahi.)

M_{\infty} \to 1^{+}

par

c_{d, w a v e} = 4 (\dots) / M_{\infty}^{2} - 1

ka kya hota hai?

Denominator

M_{\infty}^{2} - 1 \to 0

, isliye linearized theory predict karti hai ki

c_{d, w a v e} \to \infty

. Yeh sonic ke paas thin, small-perturbation assumption ka breakdown hai; real transonic drag large lekin finite hai aur nonlinear theory ki zaroorat hai.

M_{\infty} \to \infty

par Mach cone ka kya hota hai?

sin μ = 1/ M_{\infty} \to 0

, isliye

μ \to 0°

— cone body ke around tightly wrap ho jaata hai aur disturbance region surface ke saath ek thin sliver mein shrink ho jaata hai.

Ek wedge jiska half-angle

θ_{ma x} (M_{1})

se exceed karta hai, shock kya karta hai?

Us upstream Mach

M_{1}

par koi attached oblique-shock solution exist nahi karta, isliye shock detach ho jaata hai aur ek curved bow shock ke roop mein stand off karta hai (s02, right) — centreline ke paas yeh locally ek normal shock hai jisme maximum entropy jump aur drag hota hai.

Exactly zero lift (

α = 0

) aur zero camber par, kya wave drag zero hai?

Nahi. Thickness term

\overline{(d y_{t} / d x)^{2}}

survive karta hai, nonzero "volume" wave drag deta hai — finite thickness ki body ko abhi bhi air ko side mein push karna hoga, waves form karke.

θ \to 0^{+}

limit mein,

θ

–

β

–

M

relation ka weak-shock root

β

ke liye kya deta hai?

Yeh Mach angle ki taraf tend karta hai:

β \to μ = arcsin (1/ M_{1})

. Yeh zero-strength limit single case hai jahan ek real shock angle Mach angle ke barabar hota hai.

Agar wing ko poore

90°

sweep kiya jaaye (theoretical limit), toh sweep component

M_{n}

kya hoga?

M_{n} = M_{\infty} cos 90° = 0

, isliye normal Mach vanish ho jaata hai aur leading edge ke saath compressibility effects disappear ho jaate hain — ek idealized extreme jo dikhata hai ki sweep shock kyun delay karta hai (s03). Speed of sound and Mach number dekho.