Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle): Question bank

3.1.5 · D5 · Physics › Compressible Flow & Aerodynamics › Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivat

Figure — Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle)

True or false — justify karo

Kisi bhi duct ko squeeze karna hamesha gas ko speed up karta hai.

False. Sirf jab

M < 1

ho (subsonic): tab

M^{2} - 1 < 0

, isliye

d A < 0

d V > 0

forced hota hai. Jab

M > 1

hota hai sign flip ho jaata hai aur squeeze karna gas ko slow karta hai.

Agar area badhta hai, toh flow hamesha slow ho jaati hai.

False. Yeh subsonic ki aadat hai. Supersonic flow mein

M^{2} - 1 > 0

, isliye

d A > 0

ke saath

d V > 0

jaata hai — ek chauda hota duct gas ko accelerate karta hai.

De Laval nozzle ke throat par flow guaranteed sonic (

M = 1

) hoti hai.

False. Throat par

d A = 0

sirf

M = 1

ko allow karta hai; yeh ek subsonic maximum-speed point bhi allow karta hai. Sonic conditions ke liye itna bada pressure ratio chahiye ki flow "choke" ho — dekho De Laval Nozzle and Choked Flow.

Area–velocity relation sirf mass conservation se derive ki gayi thi.

False. Iske liye teen laws chahiye: continuity, Euler's momentum equation, aur isentropic sound-speed relation

d p = a^{2} d ρ

Ek incompressible liquid ke liye wahi equation

M

ki jagah zero rakhke apply hoti hai.

Partly true, lekin sahi se padho.

M = 0

set karne par

d A / A = - d V / V

milta hai, yaani

ρ A V =

const jahan

ρ

constant hai — woh jaani-pehchani "narrow pipe = faster" rule. Full relation iske reduce hoti hai sirf isliye kyunki incompressible flow

M \to 0

limit hai.

Supersonic jaane ke liye duct ko aur aur zyada converge karte rehte hain.

False.

M = 1

par factor

(M^{2} - 1) = 0

hota hai, isliye

d A = 0

; aur converge karna (

d A < 0

) supersonic flow mein

d V < 0

force kar dega, jo acceleration ko forbid karta hai. Throat ke baad diverging switch karna padega.

Ek gas bina kisi pressure drop ke accelerate kar sakti hai.

False. Euler deta hai

d p = - ρ V d V

, isliye

d V > 0

maangta hai

d p < 0

. Gas speed khareedne ke liye pressure kharach karta hai.

Equation mein density ka koi zikr nahi, isliye density yahaan irrelevant hai.

False. Density iske andar chupi hai. Poora sign-flip

d ρ / ρ = - M^{2} (d V / V)

se aata hai: supersonically density speed se tezi se girta hai, aur yahi wajah hai ki area badhna padta hai.

Spot the error

"Kyunki

\overset{m}{˙} = ρ A V

constant hai, aur hum

A

squeeze karte hain, toh

V

hamesha compensate karne ke liye badhega."

Error: yeh maanta hai ki

ρ

fixed rehta hai. Compressible flow mein

ρ

bhi change hota hai. Supersonically

ρ

itna tezi se girta hai ki

V

gir sakti hai jabki

ρ A V

constant rehta hai.

"Throat par

d A = 0

hai, aur equation deta hai

0 = (M^{2} - 1) d V / V

, isliye

d V

zero hona chahiye."

Error: jab

M = 1

hota hai toh bracket bhi zero hota hai, jisse yeh

0 = 0 \cdot (d V / V)

ban jaata hai — indeterminate.

d V

unconstrained hai, aur yahi wajah hai ki flow subsonic se supersonic mein smoothly slide karta hai.

"Ek converging–diverging nozzle isliye diverge karta hai kyunki gas ko thermally expand hone ke liye jagah chahiye."

Error: cause mix ho gaya. Divergence continuity + $(M^{2} - 1)$ ke sign se demand ki jaati hai supersonic flow mein, thermal expansion se nahi. Geometry equation follow karta hai, heating argument nahi.

M < 1

ke liye term

(M^{2} - 1)

negative hai, isliye flow kabhi accelerate nahi ho sakti."

Error: negative factor acceleration forbid nahi karta; yeh sirf

d V > 0

d A < 0

ke saath pair karta hai. Subsonic flow theek se accelerate hoti hai — duct converge karo.

a^{2} = (\partial p / \partial ρ)_{s}

ka matlab hai ki speed of sound depend karti hai ki pressure time ke saath kitni tezi se change hoti hai."

Error: yeh constant entropy par density ke respect mein derivative hai, time ke nahi. Yeh measure karta hai ki compress hone par pressure kitni stiffly respond karta hai — dekho Speed of Sound in a Gas.

"Hum isentropic assumption drop kar sakte hain; relation kisi bhi flow ke liye hold karta hai."

Error: Step 3 mein

d p = a^{2} d ρ

use hota hai, jo sirf isentropic changes ke liye valid hai. Friction ya heat addition ke saath,

d ρ

d p

se alag relation hota hai aur yeh exact form toot jaata hai.

Why questions

Hum

ρ A V =

const ko directly differentiate karne ki jagah log-differentiate kyun karte hain?

Ek product ka derivative messy hota hai, lekin

d (ln ρ + ln A + ln V) = d ρ / ρ + d A / A + d V / V = 0

seedha fractional changes deta hai — wohi language jismein yeh poora topic bolta hai.

M

is equation mein ek "switch" ki tarah kyun kaam karta hai?

Kyunki

(M^{2} - 1)

ka sign

M = 1

par flip hota hai: neeche negative, upar positive. Yeh ek sign change hi reverse kar deta hai ki "accelerate" ka matlab converge hai ya diverge.

Sonic flow sirf throat par kyun honi chahiye?

M = 1

(M^{2} - 1) = 0

hota hai, jo

d A = 0

force karta hai. Jis point par

d A = 0

ho woh area extremum hai — ek accelerating nozzle mein woh minimum hota hai, throat.

Supersonic acceleration area mein "expensive" kyun hai?

M > 1

ke saath,

(M^{2} - 1) > 1

, isliye choti

d V / V

ke liye badi

d A / A

chahiye.

M = 2

par,

1%

speed gain ke liye

3%

zyada area chahiye — isliye rocket bells itne wide flare karti hain.

d ρ / ρ = - M^{2} (d V / V)

mein density term minus sign kyun carry karta hai?

Kyunki tezi se flow karna matlab lower pressure (Euler), aur lower pressure matlab lower density (sound-speed relation). Speed up, density down — opposite signs.

Yahan use hue Euler equation mein koi gravity term kyun nahi hai?

Nozzle ko horizontal treat kiya jaata hai aur gas halki hai, isliye pressure force per unit mass kisi bhi weight term ko utterly dominate karta hai; use negligible drop kiya jaata hai.

Ek purely converging nozzle kabhi supersonic exhaust produce kyun nahi kar sakti?

Jo zyada se zyada hoga woh hai

M = 1

apni exit throat par.

M = 1

se exceed karne ke liye

d A > 0

chahiye, jo ek sirf-converging duct kabhi provide nahi karta — diverging section add karna padega.

Edge cases

M = 0

(stagnant limit) par relation kya kehta hai?

d A / A = - d V / V

: pure incompressible behaviour. Density koi role nahi khelti aur flow simple pipe continuity follow karta hai.

Jab

M \to \infty

(hypersonic) hota hai toh kya hota hai?

(M^{2} - 1) \to M^{2}

, isliye

d A / A \approx M^{2} d V / V

— enormous area increases se sirf tiny speed gains milte hain. Nozzles impractically large ho jaati hain; doosri physics (real-gas, dissociation) bhi intervene karti hai.

Bilkul

M = 1

par,

d A

ki value kya hai?

Zero.

d A = A (M^{2} - 1) (d V / V) = A \cdot 0 \cdot (d V / V) = 0

regardless of

d V

. Area momentarily stationary hai — ek genuine extremum.

Agar flow subsonic duct mein decelerate (

d V < 0

) karti hai, toh shape kaisi hai?

M < 1

M^{2} - 1 < 0

milta hai;

d V < 0

ke saath

d A > 0

milta hai — diverging duct (ek subsonic diffuser). Everyday nozzles yahaan ulte run karte hain.

Kya

d A = 0

kisi aisi jagah ho sakta hai jo throat nahi hai aur sonic nahi hai?

Haan — agar

d V = 0

locally hai (flow momentarily na speed up ho rahi na slow) toh equation kisi bhi

M

ke liye satisfy hoti hai.

d A = 0

signal deta hai "yahaan koi acceleration nahi," jiska sonic flow sirf ek tarika hai.

Agar ek diverging duct subsonic flow carry kare toh kya hoga?

M < 1

aur

d A > 0

d V < 0

force karte hain: gas decelerate karta hai aur pressure badhta hai. Yeh ek subsonic diffuser hai, stagnation pressure recover karta hai — nozzle ka ulta.

Recall Ek-sentence summary

$(M^{2} - 1)$ ka ek akela sign poori kahani hai: yeh decide karta hai ki "speed up" ka matlab squeeze hai ya spread, sonic flow ko sirf throat par rehne pe majboor karta hai, aur de Laval nozzle ki hourglass shape dictate karta hai.