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

3.1.5 · D4 · Physics › Compressible Flow & Aerodynamics › Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivat

Ye problems ek ladder ki tarah upar chadhte hain: pehle sirf sign rule ko pehchanno, phir numbers plug karo, phir competing effects analyse karo, phir ek nozzle synthesise karo, aur finally subtle edge cases master karo. Har problem ka full solution ek collapsible callout ke andar chhupa hua hai — pehle khud try karo, phir reveal karo.

Aapko sirf ek tool chahiye: area–velocity relation:

Neeche kai problems differential continuity relation bhi use karte hain, toh aage kaam mein laane se pehle batate hain ki wo kahaan se aata hai:

Master equation ko ek sign machine ki tarah padho:

$(M^{2} - 1) < 0$ (subsonic): $d A$ aur $d V$ ke signs opposite hote hain.
$(M^{2} - 1) > 0$ (supersonic): $d A$ aur $d V$ ke signs same hote hain.
$(M^{2} - 1) = 0$ (sonic): $d A = 0$ , ek throat.

Figure s01 — the sign machine. Neeche ka curve $(M^{2} - 1)$ ko $M$ ke against plot karta hai. Dekho kahan curve zero cross karta hai (red dashed line $M = 1$ par): uske left mein shaded region negative hai (subsonic, opposite signs), uske right mein positive hai (supersonic, same signs). Yeh ek single crossing is poore page ki puri kahani hai.

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

Level 1 — Recognition

Exercise 1.1

Ek gas $M = 0.7$ par flow kar rahi hai. Aap chahte ho ki wo slow down ho ( $d V < 0$ ). Duct converge hona chahiye ya diverge?

Recall Solution

WHAT: Hume $d A$ ka sign chahiye jab $d V < 0$ subsonic speed par. WHY: Equation dono ko $(M^{2} - 1)$ ke sign ke through tie karti hai. $M = 0.7 \Rightarrow M^{2} - 1 = 0.49 - 1 = - 0.51 < 0$ . Toh $\frac{d A}{A} = (- 0.51) \frac{d V}{V}$ . Jab $d V / V < 0$ , product $(- 0.51) (negative) = positive$ hoga, toh $d A > 0$ . Answer: duct diverge hona chahiye. (Subsonic gas tab slow hoti hai jab pipe wide hoti hai — bilkul paani ki tarah.)

Exercise 1.2

Har regime ko match karo ki area ka kya hota hai jab flow accelerate hoti hai ( $d V > 0$ ): (a) $M = 0.3$ , (b) $M = 1$ , (c) $M = 3$ .

Recall Solution

$(M^{2} - 1)$ ka sign decide karta hai, $d V / V > 0$ ke saath:

(a) $M = 0.3$ : $M^{2} - 1 = - 0.91 < 0 \Rightarrow d A < 0$ → converging.
(b) $M = 1$ : $M^{2} - 1 = 0 \Rightarrow d A = 0$ → throat (stationary area).
(c) $M = 3$ : $M^{2} - 1 = 8 > 0 \Rightarrow d A > 0$ → diverging.

Level 2 — Application

Exercise 2.1

Ek supersonic flow $M = 2.5$ par speed mein $2%$ gain karti hai ( $d V / V = 0.02$ ). Fractional area change $d A / A$ (percent mein) nikalo.

Recall Solution

WHAT: Master equation mein numbers plug karo. WHY: $M$ aur $d V / V$ dono given hain, toh $d A / A$ ek direct product hai. $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V} = (2. 5^{2} - 1) (0.02) = (6.25 - 1) (0.02) = 5.25 \times 0.02 = 0.105.$ Answer: $d A / A = 0.105 = 10.5%$ — area ko 10.5% increase karna padega sirf 2% speed gain ke liye. Supersonic acceleration area ke maamle mein "expensive" hoti hai.

Exercise 2.2

Ek subsonic flow $M = 0.4$ par ek aisi section se guzarti hai jahan area $6%$ shrink hoti hai ( $d A / A = - 0.06$ ). Fractional speed change $d V / V$ kya hoga?

Recall Solution

WHAT: Equation ko $d V / V$ ke liye solve karo. WHY: Ab $d A / A$ input hai; rearrange karo. $\frac{d V}{V} = \frac{d A / A}{M ^{2} - 1} = \frac{- 0.06}{0. 4 ^{2} - 1} = \frac{- 0.06}{0.16 - 1} = \frac{- 0.06}{- 0.84} = 0.0714 \dots$ Answer: $d V / V \approx 0.0714 = 7.14%$ — flow speed up hoti hai lagbhag 7.1% se. (Subsonic gas ko squeeze karo → wo accelerate karti hai, jaise expected.)

Exercise 2.3

$M = 1.2$ par area $3%$ increase hoti hai. $d V / V$ nikalo.

Recall Solution

WHAT: Master equation ko unknown $d V / V$ ke liye solve karo, given $M$ aur $d A / A$ . WHY: Yahan area change input hai, toh hum $(M^{2} - 1)$ se divide karke $d V / V$ isolate karte hain — same rearrangement jaise Ex 2.2 mein, lekin ab flow supersonic hai toh bracket positive hai. $M^{2} - 1 = 1.44 - 1 = 0.44$ . $\frac{d V}{V} = \frac{d A / A}{M ^{2} - 1} = \frac{0.03}{0.44} = 0.0682 = 6.82% .$ Answer: $d V / V \approx 6.82%$ . Supersonic, area badhti hai aur speed badhti hai (same sign) — sign machine ke saath consistent.

Level 3 — Analysis

Exercise 3.1

Density relation $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V}$ use karke (isentropic combination se), $M = 2$ par $d V / V = + 0.01$ ke liye fractional density drop aur fractional speed rise ki magnitude compare karo. Kaun sa effect dominate karta hai, aur kyun wo area ko grow karne par majboor karta hai?

Recall Solution

WHAT: Dono fractional changes compute karo. WHY: Page ke upar bataya hua continuity relation $\frac{d ρ}{ρ} + \frac{d A}{A} + \frac{d V}{V} = 0$ teen effects ko balance karta hai; density/velocity mein se jo bhi win kare woh $d A$ ka sign decide karta hai. $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V} = - (4) (0.01) = - 0.04 (4% drop) .$ Speed rise sirf $+ 0.01$ (1%) hai. Toh density speed rise se 4 guna tez girती hai. Plug into continuity: $\frac{d A}{A} = - \frac{d ρ}{ρ} - \frac{d V}{V} = - (- 0.04) - 0.01 = 0.04 - 0.01 = 0.03.$ Answer: $d A / A = + 0.03$ (3% area increase). Kyunki density speed se tez thin ho rahi hai, product $ρ V$ actually girta hai; $ρ A V$ constant rakhne ke liye area ko grow karna hoga. Yahi physical reason hai ki supersonic ducts diverge karte hain.

Exercise 3.2

Dikhao ki exactly $M = 1$ par density-drop rate aur speed-rise rate magnitude mein equal hote hain, aur isliye $ρ V$ momentarily stationary hota hai. $d A$ ke liye iska kya matlab hai?

Recall Solution

$M = 1$ par: $\frac{d ρ}{ρ} = - (1) \frac{d V}{V} = - \frac{d V}{V}$ . Toh fractional density drop exactly cancel karta hai fractional speed rise ko: $\frac{d ρ}{ρ} + \frac{d V}{V} = 0 \Rightarrow d (ρ V) / (ρ V) = 0.$ Continuity se $\frac{d A}{A} = - (\frac{d ρ}{ρ} + \frac{d V}{V}) = 0$ . Answer: $d A = 0$ — area stationary hai. Yeh throat hai: exactly woh crossover point jahan density aur velocity effects balance karte hain. Sonic flow sirf wahan ho sakti hai jahan $d A = 0$ .

Figure s02 — the hourglass. Neeche ka nozzle profile duct half-width (area ke proportional) ko position ke against dikhata hai. Flow arrow ko left se right trace karo: walls subsonic region mein squeeze in karti hain (green label), red dashed throat ( $M = 1$ , $d A = 0$ ) par minimum tak pinch hoti hain, phir supersonic region mein flare open hoti hain. Notice karo ki throat exactly wahan hai jahan dono halves closest hain — yahi woh $d A = 0$ point hai jo Exercise 3.2 ne prove kiya.

Level 4 — Synthesis

Exercise 4.1

Design reasoning: ek rocket ko combustion gas ko slow $M = 0.2$ chamber flow se lekar $M = 3$ exit tak accelerate karna hai, hamesha accelerating ( $d V > 0$ ). Duct ke required shape ko region by region describe karo, aur har transition ko $(M^{2} - 1)$ ke sign se justify karo.

Recall Solution

$(M^{2} - 1)$ ko trace karo jab $M$ badhta hai, hamesha $d V > 0$ demand karte hue:

Region	$M$	$M^{2} - 1$	$d V > 0$ ke liye chahiye	Shape
Chamber → throat	$0.2 \to 1$	negative	$d A < 0$	Converging
Throat	$1$	zero	$d A = 0$	Minimum area
Throat → exit	$1 \to 3$	positive	$d A > 0$	Diverging
Answer: ek converging–diverging (de Laval) nozzle: ye throat tak narrow hota hai jahan $M = 1$ hota hai, phir $M = 3$ tak reach karne ke liye flare open hota hai. Har transition $(M^{2} - 1)$ ke sign se force hoti hai — aap $M = 1$ sirf throat ke alawa aur kahan se pass nahi kar sakte, kyunki sirf wahan $d A = 0$ hota hai. Dekho De Laval Nozzle and Choked Flow.

Exercise 4.2

Ek sanity estimate ke taur par ki supersonic bell kitni sharply flare hoti hai, flow ka ek chota step lo $M = 3$ par jahan $d V / V = 0.10$ (10% speed gain). Kaun sa area change result hota hai, aur kya ye confirm karta hai ki nozzle flaring karti rehti hai?

Recall Solution

WHAT: $M = 3$ par supersonic step ke liye $d A / A$ nikalo given $d V / V = 0.10$ . WHY: Master equation directly $M$ aur $d V / V$ se $d A / A$ deti hai; high Mach par evaluate karne se pata chalta hai ki bracket $(M^{2} - 1)$ kitna bada hota hai, jo batata hai ki bell kitni tezi se kholni hogi. $M = 3$ par: $M^{2} - 1 = 9 - 1 = 8$ . $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V} = 8 \times 0.10 = 0.80 = 80% .$ Answer: 10% speed gain ke liye 80% area increase. Bracket $(M^{2} - 1)$ high Mach par bahut bada hota hai, toh bell ko bahut tezi se flare karna padta hai — exactly isliye real supersonic nozzles mein wide exit bells hoti hain. Yeh continual divergence confirm karta hai.

Level 5 — Mastery

Exercise 5.1

Ek duct converging hai ( $d A < 0$ ) lekin flow measure karne par decelerating ( $d V < 0$ ) nikli. Aap Mach regime ke baare mein kya conclude kar sakte ho? Sign logic use karke fully explain karo, aur physical setting identify karo jahan ye hota hai.

Recall Solution

WHAT: Dono $d A < 0$ aur $d V < 0$ hain → unke signs same hain. WHY: Same-sign $d A$ aur $d V$ ke liye $(M^{2} - 1) > 0$ chahiye. $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}, ( neg ) = (M^{2} - 1) (neg) \Rightarrow (M^{2} - 1) > 0.$ Answer: flow supersonic honi chahiye ( $M > 1$ ). Ek converging supersonic duct flow ko decelerate karta hai — yeh ek supersonic diffuser hai (supersonic engine intakes aur wind-tunnel diffusers mein use hota hai). Subsonically wahi shrinking duct accelerate karta; sign flip yahi toh poora point hai.

Exercise 5.2

Ek choked de Laval nozzle ke throat par ( $d A = 0$ ), flow sonic hai ( $M = 1$ ). Equation deti hai $0 = (M^{2} - 1) d V / V = 0 \cdot d V / V$ . Explain karo ki $d V / V$ yahan zero hone par force kyun nahi hota, aur ye indeterminate form exactly kyun ek smooth subsonic→supersonic transition allow karta hai.

Recall Solution

WHAT: $M = 1$ par bracket $(M^{2} - 1) = 0$ hai, toh equation padhti hai $0 = 0 \cdot (d V / V)$ . WHY: $d V / V$ ki koi bhi value $0 = 0$ satisfy karti hai. Equation throat par speed change par koi constraint nahi lagati. Physics: kyunki $d V / V$ unconstrained hai, flow throat ke through accelerate ( $d V > 0$ ) karte reh sakti hai chahe area momentarily stationary ho. Agar $(M^{2} - 1)$ nonzero hota, toh $d A = 0$ force karta $d V = 0$ (flow na speed up na slow down) — ek dead point. Sirf $M = 1$ par indeterminate $0 \cdot (d V / V)$ ek "gate" kholata hai jisse velocity smoothly sound se neeche se sound se upar slide kar sake. Answer: $d V / V$ sonic throat par indeterminate hai (zero nahi); yahi indeterminacy exactly woh mathematical door hai jisse flow subsonic se supersonic mein pass hoti hai.

Exercise 5.3

Degenerate check: constant area duct ( $d A = 0$ ) mein subsonic flow ( $M < 1$ , sonic nahi) ke liye relation kya predict karta hai? Kya acceleration possible hai?

Recall Solution

WHAT: $d A = 0$ impose karo $M < 1$ ke saath (toh $M^{2} - 1 \neq = 0$ ) aur master equation ko $d V / V$ ke liye solve karo. WHY: Area change ko zero set karne se pata chalta hai ki equation speed ke liye kya allow karta hai; nonzero bracket ke saath product sirf tab vanish ho sakta hai jab $d V / V$ khud vanish ho — Ex 5.2 ke sonic throat ki opposite situation, jahan bracket zero tha. $0 = (M^{2} - 1) \frac{d V}{V}, (M^{2} - 1) \neq = 0 \Rightarrow \frac{d V}{V} = 0.$ Answer: $d V = 0$ — constant-area subsonic duct mein koi speed change nahi hoti (friction/heat ignore karte hue). Area change ke bina acceleration impossible hai. Yeh sonic throat ka opposite hai: bracket nonzero hone ke saath, $d A = 0$ strictly force karta hai $d V = 0$ . Isse highlight hota hai ki throat ka special behaviour purely $(M^{2} - 1) = 0$ ki wajah se hai, sirf $d A = 0$ ki wajah se nahi.

Recall Master recall — the sign machine

Area–velocity relation ek sign rule hai ::: $d A / A = (M^{2} - 1) d V / V$ ; $(M^{2} - 1)$ ka sign area aur velocity changes ko link karta hai. Subsonic accelerate ke liye chahiye ::: converging duct ( $d A < 0$ ). Supersonic accelerate ke liye chahiye ::: diverging duct ( $d A > 0$ ). Constant-area subsonic duct deta hai ::: $d V = 0$ (koi acceleration nahi). Converging supersonic duct karta hai ::: flow ko decelerate (supersonic diffuser). Sonic throat subsonic→supersonic allow karta hai kyunki ::: $0 \cdot (d V / V)$ indeterminate hai, $d V / V$ unconstrained rehta hai.