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

3.1.5 · D2 · Physics › Compressible Flow & Aerodynamics › Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivat

Step 0 — Woh picture jiske baare mein hum baat kar rahe hain

KYA HAI. Ek pipe (ek "duct") jiska cross-section apni length ke saath badalta hai. Gas left se right ki taraf flow karti hai. Ek slice par pipe ki area $A$ hai, gas $V$ speed se chal rahi hai, density $ρ$ hai (gas kitni tightly packed hai), aur pressure $p$ hai (gas kitna zyada bahar ki taraf push karti hai).

KYUN. Kisi bhi equation se pehle, humein agree karna hoga ki letters picture mein kahan point kar rahe hain. Neeche har symbol is drawing par ek label hai.

PICTURE.

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

Kisi symbol ke aage chhota letter $d$ — jaise $d V$ mein — matlab hai "thoda sa change us cheez mein jab hum ek baal-bhar downstream jaate hain." Toh $d V$ ek chhoti speed increase hai, $d A$ ek chhota area change. Hum fractional changes jaise $d V / V$ (" $V$ kitne fraction se badha?") chase karte hain kyunki woh dimensionless hote hain aur saaf saaf add hote hain.

Step 1 — Mass conserved hai: $ρ A V = const$

KYA HAI. Har second mein har slice cross karne wale mass ko gino. Woh amount, $\overset{m}{˙} = ρ A V$ , har slice par identical hai.

KYUN. Steady flow mein, gas pipe ke beech mein pile up nahi ho sakti aur na hi bahar leak ho sakti. Jo bhi mass ek second mein ek chunk ke left mein enter karta hai, woh right mein nikalna chahiye — warna mass appear ya vanish ho jaata. Yeh Continuity Equation (Compressible Flow) hai.

PICTURE. Do slices, patli aur moti. Patli slice mein dots dense aur fast hain; moti slice mein woh spread aur slow ho jaate hain — lekin har ek ke through count per second barabar hai.

Ab clever move. Ek product $=$ constant ko differentiate karna awkward hai. Toh pehle natural log lo — logs multiplication ko addition mein convert karte hain:

$ln ρ + ln A + ln V = const$

LOG KYUN? Kyunki $d ( ln X ) = \frac{d X}{X}$ — log ka derivative automatically fractional change hai. Aur fractional change is poore topic ki language hai. Differentiate karne par (ek chhota step downstream, toh constant ka change $0$ hai):

$\boxed{\ \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V} = 0\ }\tag{1}$

Step 2 — Flow ke saath Newton's law: Euler's equation

KYA HAI. Gas ke ek tiny slab ko dekho. Peechhe ka pressure aage push karta hai; aage ka pressure peeche push karta hai. Agar aage pressure kam hai, toh ek net forward push hai, aur slab accelerate karta hai.

KYUN. Yeh sirf $F = ma$ hai jo ek fluid slab par apply hua hai bina friction aur gravity ke — yeh Euler's Equation for Steady Flow hai. Ek horizontal inviscid gas mein pressure hi ek maatra force hai.

PICTURE. Ek slab jisme peeche se bada red push hai, aage se chhota red push hai, aur ek green acceleration arrow hai.

Slab ke liye force aur acceleration balance karne par:

$\underbrace{dp}_{\text{pressure change}} + \underbrace{\rho\, V\, dV}_{\text{mass}\times\text{accel term}} = 0 \quad\Longrightarrow\quad \frac{dp}{\rho} = -V\,dV \tag{2}$

Step 3 — Jab pressure change hota hai tab density kya karti hai: speed of sound

KYA HAI. Gas ko thoda squeeze karo (raise $p$ ) aur yeh thodi denser ho jaati hai (raise $ρ$ ). Us response ki stiffness — kitna pressure har unit density change par jump karta hai — $a^{2}$ nikalta hai, sound speed ka square.

KYUN. Ek sound wave hi ek chhoti travelling pressure pulse hai. Yeh kitni fast jaati hai yeh is baat par depend karta hai ki gas compress hone par kitna stiffly push back karti hai. Stiffer gas ⇒ faster sound. Formally yeh isentropic ("no heat lost, reversible") derivative hai — dekho Isentropic Flow Relations.

PICTURE. Ek chhoti compression pulse pipe ke saath chal rahi hai; ek spring cartoon "stiffness = $a^{2}$ " dikhata hai.

Step 4 — Pressure bahar karo: density ko $M$ ke terms mein

KYA HAI. Equations (2) aur (3) dono $d p$ mention karte hain. $d p$ eliminate karo taaki density directly speed ke terms mein mile.

KYUN. Hamara final goal sirf $A$ aur $V$ chahta hai. Pressure aur density middlemen hain; hum unhe ek ek karke chase karte hain. Pehle $d p$ , yahan; phir $d ρ$ , Step 5 mein.

PICTURE. Ek flow diagram: $(2)$ aur $(3)$ merge hote hain, $d p$ cancel ho jaata hai, aur $M^{2}$ wala ek relation nikalta hai.

(2) ko (3) mein daalo:

$d ρ = \frac{d p}{a ^{2}} = \frac{- ρ V d V}{a ^{2}}$

Fractional change paane ke liye dono sides ko $ρ$ se divide karo, aur speeds group karo:

$\frac{d ρ}{ρ} = - \frac{V ^{2}}{a ^{2}} \frac{d V}{V}$

Ab magic substitution. Kyunki $M = V / a$ , hume milta hai $\frac{V ^{2}}{a ^{2}} = M^{2}$ :

$\boxed{\ \frac{d\rho}{\rho} = -M^2\,\frac{dV}{V}\ }\tag{4}$

Step 5 — Combine karke area–velocity relation nikalo

KYA HAI. Density result (4) ko mass equation (1) mein substitute karo. Density gayab ho jaati hai; sirf $A$ aur $V$ rehte hain.

KYUN. (1) ne teen fractional changes ko link kiya tha. Ab hum (4) se $d ρ / ρ$ jaante hain, toh hum ise trade kar sakte hain aur sirf woh ek link reh jaata hai jo hum chahte the.

PICTURE. Equation (1) mein $d ρ / ρ$ box $- M^{2} d V / V$ se replace ho jaata hai, phir $d V / V$ terms merge ho jaate hain.

(1) se shuru karo aur density term ke liye (4) daalo:

$yeh hai d ρ / ρ - M^{2} \frac{d V}{V} + \frac{d A}{A} + \frac{d V}{V} = 0$

Dono $d V / V$ terms ko right mein move karo aur collect karo:

$\frac{d A}{A} = M^{2} \frac{d V}{V} - \frac{d V}{V} = (M^{2} - 1) \frac{d V}{V}$

Step 6 — Switch padhna: teeno regimes

KYA HAI. Goal fix karo "gas ko accelerate karo," yani $d V > 0$ . Tab $(M^{2} - 1)$ ka sign $d A$ ka sign force karta hai. Teeno cases walk karo.

KYUN. Ek relation tab hi samajh aata hai jab aapne har case cover kiya ho. Switch exactly teen signs le sakta hai, aur har ek alag pipe shape maangta hai.

PICTURE. Teen ducts side by side — converging, throat, diverging — har ek apne regime aur sign ke saath labelled.

Regime	$M^{2} - 1$	$d V > 0$ ke liye hume chahiye	Shape
Subsonic $M < 1$	negative	$d A < 0$	converging
Sonic $M = 1$	zero	$d A = 0$	throat (min area)
Supersonic $M > 1$	positive	$d A > 0$	diverging

Step 7 — Degenerate case $M = 1$ : sonic flow sirf ek throat par kyun rehta hai

KYA HAI. $M = 1$ par switch $(M^{2} - 1)$ exactly $0$ hai, toh equation padhti hai $d A / A = 0$ . Area momentarily stationary hai — ek minimum.

KYUN. Yeh De Laval Nozzle and Choked Flow ka linchpin hai. Yeh kehta hai ki gas sirf wahan sonic ho sakti hai jahan pipe ka narrow hona ruk jaata hai aur wide hona shuru hota hai — throat.

PICTURE. Throat par zoom: area curve flat ( $d A = 0$ ) neck par, $M$ smoothly $1$ ke through jaata hua.

$At M = 1 : \frac{d A}{A} = = 0 (1 - 1) \frac{d V}{V} = 0 ⟹ d A = 0$

Worked examples (har ek aapke dimag mein ek mini-picture)

Ek-picture summary

Recall Poore walkthrough ki Feynman retelling

Ek pipe mein gas ki picture karo. Rule ek: koi gas create ya lost nahi hoti, toh crowding × width × speed poore saath fixed rehta hai (Step 1). Rule do: gas tab hi speed up hoti hai jab aage pressure kam ho — yeh low-pressure side ki taraf girrti hai (Step 2). Rule teen: gas ko squeeze karo aur yeh denser ho jaati hai, aur us push-back ki stiffness sound speed squared hai (Step 3). Inhe chain karo: rules do aur teen se, jab gas speed up hoti hai uski crowding girrti hai, aur yeh utna zyada girrti hai jitni gas pehle se sound se compare karke fast ho — yahi $M^{2}$ hai (Step 4). Ise rule ek mein daalo aur pressure aur density middlemen cancel ho jaate hain, ek clean sentence bacha ke: $d A / A = (M^{2} - 1) d V / V$ (Step 5). Bracket $(M^{2} - 1)$ ek switch hai. Slow gas: speed up karne ke liye squeeze karo. Sound-jaisi-fast gas: switch zero hai, toh pipe apni sabse narrow jagah par honi chahiye — throat (Step 7). Sound se faster: switch positive flip karta hai, toh speed up karte rehne ke liye pipe ko khola chahiye (Step 6). Squeeze karo, ek neck par pinch karo, phir wide flare karo — woh hourglass ek de Laval nozzle hai, aur ab aap jaante ho exactly kyun.