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

3.1.5 · D3 · Physics › Compressible Flow & Aerodynamics › Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivat

Yeh page ek hi equation ko $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ har us case mein push karta hai jahan yeh apply ho sakti hai. Numbers chhune se pehle, hum ek map banate hain saare scenario types ka taaki kuch bhi surprise na kare. Agar koi symbol yahaan unfamiliar lage, toh parent derivation aur prerequisite notes revisit karo.

Definition Yeh relation kab apply hoti hai? (assumptions)

Equation $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ teen physical laws ko combine karke derive ki gayi thi, aur har ek ke saath ek condition aati hai. Yeh relation tabhi valid hai jab flow:

Steady ho — time ke saath koi cheez fixed point par nahi badlti (isliye mass flow $ρ A V$ sach mein constant hai, continuity equation se).
Quasi one-dimensional ho — gas essentially ek hi axis ke along duct mein chalti hai, isliye $A$ , $V$ , $ρ$ ek cross-section par ek hi value lete hain.
Isentropic aur compressible ho — adiabatic (andar/bahar koi heat nahi) aur reversible (koi friction/shock nahi), taaki isentropic link $d p = a^{2} d ρ$ hold kare aur density change ho sake.
Inviscid ho — koi viscosity nahi, isliye Euler's momentum equation $d p = - ρ V d V$ bina kisi friction term ke apply hoti hai. Agar inmen se koi bhi toot jaaye (ek shock, heat addition, strong wall friction), toh yeh relation as written hold nahi karti.

Scenario matrix

Is equation ke har problem ka ek in cells mein se kisi ek mein fall hona zaroori hai. Do inputs jo matter karte hain woh hain (a) Mach regime (jo $M^{2} - 1$ ka sign fix karta hai) aur (b) hum flow ko kya karna chahte hain (speed up ya slow down). Saath hi edge/limit cases bhi hain. "Example" column niche ke matching worked example ko point karta hai (cell C1 ↔ Example 1, aur aise hi).

#	Cell (scenario class)	$M^{2} - 1$ sign	Kya poochha gaya	Example
C1	Subsonic, accelerate karna hai	negative	shape?	Example 1
C2	Subsonic, decelerate karna hai (diffuser)	negative	shape?	Example 2
C3	Supersonic, accelerate karna hai	positive	shape + magnitude	Example 3
C4	Supersonic, decelerate karna hai	positive	shape?	Example 4
C5	Exactly sonic $M = 1$ (degenerate: multiplier $(M^{2} - 1) = 0$ )	zero	$d V$ kya karta hai?	Example 5
C6	Limit $M \to 0$ (incompressible check)	$\to - 1$	water-pipe law recover karo	Example 6
C7	Real-world word problem (rocket bell sizing)	positive	magnitude	Example 7
C8	Exam twist: $d A / A$ diya, required $M$ nikalo	$M$ ke liye solve	inverse problem	Example 8
C9	Sign/geometry sanity: throat par area kis taraf jaata hai?	mixed	figure reasoning	Example 9

Ab hum har cell ko hit karte hain.

Sign padhna — ek picture jo dimaag mein rakhni chahiye

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

Figure 1 kaise padhen. Horizontal axis Mach number $M$ hai, $0$ (bilkul still) se $3$ (sound ki speed se teen guna) tak chalti hai. Vertical axis multiplier $M^{2} - 1$ ki value hai — woh quantity jo hamare equation mein $d V / V$ ke aage baithti hai. Red curve yahi multiplier hai. Isse left se right trace karo: yeh $M = 0$ par $- 1$ se shuru hoti hai, chadhti hai, exactly $M = 1$ ki dashed vertical line par zero se guzarti hai (red dot), aur phir positive territory mein tezi se upar jaati hai. Horizontal axis ke niche multiplier negative hai (subsonic region, left par labelled); uske upar multiplier positive hai (supersonic region, right par labelled). Woh ek zero-crossing — jahan curve axis ko pierce karti hai — de Laval nozzle ki poori physics hai.

Worked examples — matrix ke har cell ke liye ek

(Niche har " $%$ " ek chote local $d V / V$ ki jagah hai; upar differentials wali caution dekhein — answers local approximations hain.)

Worked example Example 1 — C1: subsonic, speed badhani hai

$M = 0.40$ par air ko accelerate karna hai. Duct converge karega ya diverge, aur $2%$ speed gain ke liye kitna (fractionally)?

Forecast: subsonic → mera andaza hai converge karega (garden hose ki tarah).

Multiplier nikalo: $M^{2} - 1 = 0.4 0^{2} - 1 = 0.16 - 1 = - 0.84$ . Yeh step kyun? Is number ka sign duct shape decide karta hai aur iska size magnitude set karta hai — isliye yeh hamesha pehli cheez pin down karte hain.
Hum chahte hain $d V / V = + 0.02$ . Plug in karo: $\frac{d A}{A} = (- 0.84) (0.02) = - 0.0168$ . Yeh step kyun? Equation ab fully determined hai; target speed change substitute karne se required area change mil jaata hai.
Negative $d A / A$ matlab area shrink hogi → converging. Magnitude $1.68%$ .

Verify: Sign check — subsonic aur speeding up mein oppose hona chahiye ( $d A < 0$ jabki $d V > 0$ ). ✅ Units: dono sides dimensionless. ✅ Converging inlet se match karta hai.

Worked example Example 2 — C2: subsonic diffuser, SLOW DOWN karna hai

$M = 0.60$ par air ek diffuser mein enter karti hai jो ise $5%$ slow kare (pressure recover kare). Kaisi shape?

Forecast: subsonic flow slow karna — shayad duct bahar khulega.

Multiplier: $M^{2} - 1 = 0.36 - 1 = - 0.64$ . Yeh step kyun? Hamesha pehla move — multiplier ka sign hi shape decide karta hai, isliye hum isse pehle nikalte hain.
Ab $d V / V = - 0.05$ (ek decrease). $\frac{d A}{A} = (- 0.64) (- 0.05) = + 0.032$ . Yeh step kyun? Yahaan target slow-down hai, isliye $d V / V$ negative hai; negative $\times$ negative $=$ positive, aur isi tarah decelerating case ka outcome Example 1 ke baraaks ho jaata hai.
Positive $d A / A$ → area $3.2%$ badhi → diverging.

Verify: Subsonic diffuser sach mein ek choda hota pipe hota hai; flow slow karne se pressure badh ta hai. Sign logic continuity ke saath consistent hai: slower + fatter se $ρ A V$ fixed rehta hai. ✅

Worked example Example 3 — C3: supersonic, speed badhani hai

$M = 2.5$ par supersonic stream ko $4%$ accelerate kiya jaata hai. $d A / A$ aur shape nikalo.

Forecast: supersonic mein aur tez jaane ke liye open up karna padta hai — isliye area badhegi, aur shayad bahut zyada.

Multiplier: $M^{2} - 1 = 6.25 - 1 = 5.25$ . Yeh step kyun? Hum sign fix karte hain (positive → supersonic behaviour) aur note karte hain ki iska size bada hai, jo pehle hi batata hai ki area change badi hogi.
$\frac{d A}{A} = (5.25) (0.04) = 0.21$ . Yeh step kyun? Multiplier jaanne ke baad, target $d V / V = + 0.04$ substitute karne se required area change directly mil jaati hai.
$d A / A = + 0.21$ → area $21%$ grow karni chahiye → strongly diverging.

Verify: Same sign ( $d A > 0$ , $d V > 0$ ) jaisa $M > 1$ ke liye zaroori hai. ✅ Sirf $4%$ speed gain ke liye $21%$ zyada area chahiye — isliye rocket bells enormously flare karte hain (Example 7 dekho).

Worked example Example 4 — C4: supersonic, SLOW DOWN karna hai

$M = 1.8$ par supersonic air ko $3%$ decelerate kiya jaata hai. Kaisi shape?

Forecast: supersonic "ulta duniya" hai; slow down karne se area shrink honi chahiye.

Multiplier: $M^{2} - 1 = 3.24 - 1 = 2.24$ . Yeh step kyun? Hamesha ki tarah, multiplier ka sign duct shape fix karta hai, isliye hum pehle compute karte hain; yahaan positive confirm karta hai ki hum supersonic regime mein hain.
$d V / V = - 0.03$ : $\frac{d A}{A} = (2.24) (- 0.03) = - 0.0672$ . Yeh step kyun? Target slow-down hai, isliye $d V / V$ negative hai, aur yeh positive multiplier se seedha through carry hota hai aur ek negative area change deta hai.
$d A / A = - 0.0672$ → area $6.72%$ shrink hogi → converging.

Verify: Supersonic flow mein $d A$ aur $d V$ ek sign share karte hain; dono yahaan negative gaye. ✅ Yeh ek supersonic diffuser hai — supersonic flow slow karne ke liye converge karta hai, Example 2 ka mirror image.

Worked example Example 5 — C5: degenerate sonic point

$M = 1$ Flow exactly sonic hai, $M = 1$ , ek throat par jahan $d A = 0$ hai. $d V$ kya kar sakta hai?

Forecast: kuch khaas hona chahiye — yeh crossover hai.

Multiplier: $M^{2} - 1 = 1 - 1 = 0$ . Yeh step kyun? $M = 1$ par poora right-hand side zero se scale ho jaata hai, isliye hum ise pehle compute karte hain taaki woh collapse samajh aaye.
Equation ban jaati hai $\frac{d A}{A} = (0) \frac{d V}{V} = 0$ , yaani $0 = 0$ . Yeh step kyun? Multiplier ke zero hone se relation kisi bhi value of $d V / V$ ke liye hold karti hai — yeh ab speed ko constrain nahi karti, jo is degenerate feature ki khaasiyat hai.
Interpretation: $d V / V$ is equation ke dwara $M = 1$ par unconstrained hai. Velocity throat ke across smoothly badhti reh sakti hai even though $d A = 0$ wahan hai — aise hi flow sub→supersonic transition karti hai.

Verify: $(1 - 1) \cdot x = 0$ har $x$ ke liye true hai, isliye equation $d V$ par koi limit nahi lagaati. ✅ Parent note ke " $0 \cdot (d V / V)$ indeterminate" remark ke saath consistent. Sonic conditions ke liye ek throat chahiye, lekin ek throat sonic force nahi karta (choking pressure ratio par depend karta hai).

Worked example Example 6 — C6: limit

$M \to 0$ water-pipe law recover karta hai Dikhao ki jab $M \to 0$ (almost incompressible), relation familiar continuity law ban jaati hai liquid ke liye.

Forecast: bahut low speed par, density barely change hoti hai, isliye yeh reduce ho kar "area up ⇒ speed down" banana chahiye.

$M \to 0$ lo: $M^{2} - 1 \to - 1$ . Yeh step kyun? Hum multiplier ka leading behaviour chahte hain jab flow sound speed se bahut neeche hai, isliye hum $M$ ko uske extreme par bhejte hain.
Equation → $\frac{d A}{A} = (- 1) \frac{d V}{V}$ , yaani $\frac{d A}{A} + \frac{d V}{V} = 0$ . Yeh step kyun? Limiting multiplier substitute karne se compressibility term hat jaati hai aur sirf pure area–speed trade-off bachta hai.
Yeh exactly $A V = constant$ (differentiated) hai — incompressible continuity result jismein $ρ$ frozen hai.

Verify: Numerically $M = 0.05$ par: multiplier $= 0.0025 - 1 = - 0.9975 \approx - 1$ . ✅ Isliye $1%$ area drop $\approx 1%$ speed rise deta hai — garden-hose intuition, limit ke roop mein recover ki gayi.

Worked example Example 7 — C7: real-world rocket bell

Ek rocket exhaust ko throat par $M = 1$ se exit par $M = 3$ tak accelerate karna hai. Equation ko locally exit par use karte hue, roughly ek final $2%$ speed increase ke saath kitni area change hoti hai?

Forecast: far supersonic → area change bahut badi honi chahiye.

$M = 3$ par: $M^{2} - 1 = 9 - 1 = 8$ . Yeh step kyun? Hum multiplier ko local exit Mach number par evaluate karte hain kyunki equation ek point relation hai — iska coefficient $M$ ke nozzle ke along badhne par badal ta hai.
$\frac{d A}{A} = (8) (0.02) = 0.16$ . Yeh step kyun? Local multiplier fix ho jaane ke baad, desired $d V / V = + 0.02$ substitute karne se exactly wahan exit par required area change mil jaati hai.
Sirf $2%$ speed gain ke liye $16%$ area increase chahiye — isliye wide flaring bell hoti hai.

Verify: $8 \times 0.02 = 0.16$ . ✅ Bada $M$ ⇒ bada multiplier ⇒ steeper flare, jo real nozzle geometry se match karta hai (bells exit ke paas sabse tezi se chodi hoti hain).

Worked example Example 8 — C8: exam twist,

$M$ ke liye solve karo Ek duct segment mein $d A / A = + 0.09$ measure hota hai jabki flow $d V / V = + 0.03$ se accelerate hoti hai. Local Mach number kya hai, aur flow sub- ya supersonic hai?

Forecast: area aur speed dono saath badhein → supersonic hona chahiye; $M$ nikaalte hain.

Rearrange karo: $M^{2} - 1 = \frac{d A / A}{d V / V} = \frac{0.09}{0.03} = 3$ . Yeh step kyun? Equation multiplier mein linear hai, isliye dono measured fractional changes ko divide karne se $(M^{2} - 1)$ isolate ho jaata hai — usual plug-in ka inverse.
$M^{2} = 4 \Rightarrow M = 2$ . Yeh step kyun? $1$ add karo aur positive square root lo, kyunki Mach number ek positive physical speed ratio hai.
$M = 2 > 1$ → supersonic, $d A$ aur $d V$ ka same sign share karna consistent hai.

Verify: Back-substitute karo: $(2^{2} - 1) (0.03) = 3 \times 0.03 = 0.09 = d A / A$ . ✅ Dono fractional changes $M > 1$ par positive — correct regime.

Worked example Example 9 — C9: throat par geometry sanity

Figure 2 dekho. Converging side ( $M < 1$ ) par, diverging side ( $M > 1$ ) par, aur throat par, batao ki ek accelerating de Laval nozzle mein downstream move karte hue $d A$ negative, positive, ya zero hai.

Forecast: shrink, phir flat, phir grow.

Converging side: subsonic accelerating → Example 1 logic se $d A < 0$ (area shrink ho rahi hai). Yeh step kyun? Hum subsonic sign result reuse karte hain upstream half ko label karne ke liye bina recompute kiye — physics C1 jaisi hi hai.
Throat: $M = 1$ → $d A = 0$ (area minimum, momentarily flat). Yeh step kyun? $M = 1$ par multiplier vanish ho jaata hai (Example 5), jo $d A = 0$ force karta hai; geometrically yeh minimum-area point hai.
Diverging side: supersonic accelerating → Example 3 logic se $d A > 0$ (area grow ho rahi hai). Yeh step kyun? Throat ke baad flow supersonic hai, isliye C3 sign result apply hota hai aur accelerate karte rehne ke liye area badhni chahiye.

Verify: Figure 2 mein nozzle outline ek single narrowest red point (throat) tak pinch hoti hai phir flare karti hai — exactly wahi shape jo hamare teen signs predict karte hain. ✅

Figure 2 kaise padhen. Black curves cross-section mein dekhe gaye de Laval nozzle ki top aur bottom walls hain; horizontal axis nozzle ke along travel ki gayi distance hai aur vertical axis duct half-width hai (har wall centre-line se kitni door hai). Gas left se right flow karti hai (black arrow, labelled $d V > 0$ ). Walls left par andar ki taraf pinch karti hain — yeh converging, subsonic region hai jahan $d A < 0$ hai. Woh sabse narrow gap par pahunchi red vertical line par, throat, jahan $M = 1$ aur $d A = 0$ hai (red dot ise centre-line par mark karta hai). Throat ke baad walls bahar ki taraf flare karti hain — diverging, supersonic region jahan $d A > 0$ hai. Teen regions ko left-to-right padhne se exactly wahi shrink → flat → grow pattern reproduce hota hai jo Example 9 mein hai.

Recall Quick self-test

Supersonic flow, area $6%$ shrink karti hai — kya yeh speed up ho rahi hai ya slow down? ::: Slow down. $M > 1$ matlab $d A$ aur $d V$ ek sign share karte hain, isliye $d A < 0 \Rightarrow d V < 0$ (Example 4 logic). Tum $d A / A = - 0.005$ aur $d V / V = + 0.005$ measure karte ho. Yeh $M$ ke baare mein kya bataata hai? ::: Ratio deta hai $M^{2} - 1 = - 1$ , yaani $M^{2} \to 0$ , isliye $M \to 0$ : flow essentially incompressible hai (bahut low Mach). Yeh Example 6 ka limiting water-pipe case hai, aur yeh batata hai ki local Mach number negligibly small hai, na ki speed literally zero hai.