Continuity equation — derivation (conservation of mass), ρAv = const: Worked examples

2.2.12 · D3 · Physics › Fluid Mechanics › Continuity equation — derivation (conservation of mass), ρAv

Kuch bhi shuru karne se pehle, ek reminder taaki har symbol samajh aaye:

Scenario matrix

Continuity problems sab ek hi master line follow karte hain — mass per second andar = mass per second bahar — lekin dikhte alag-alag hain depending on kya change ho raha hai aur kya fixed hai. Yeh rahi har class:

Cell	Kya change hota hai	Fixed kya hai	Conserved quantity	Dhyan rakho
A Narrowing	$A$ kam hoti hai	$ρ$	$A v$	speed badhti hai
B Widening	$A$ badhti hai	$ρ$	$A v$	speed ghatti hai
C Radius diya hai	$r$ (na ki $A$ )	$ρ$	$A v$ , jahan $A \propto r^{2}$	ratio square karo
D Splitting / merging	ek pipe → kaafi	$ρ$	kul $A v$	branches sum karo
E Compressible gas	$ρ$ change hoti hai	$A$ (yahan)	sirf $ρ A v$	$ρ$ cancel mat karo
F Word problem (fill/drain)	rate se time	$ρ$	$Q$	$t = V / Q$
G Degenerate: $A \to 0$	window band hoti hai	$ρ$	$\overset{m}{˙} = ρ A v$	$v \to \infty$ limit
H Exam twist: area ke liye solve karo	$A_{2}$ unknown	$ρ$ , mixed units	$A v$	pehle cm↔m convert karo
I Compressible split	ek gas pipe → kaafi, $ρ$ change hoti hai	(kuch nahi)	kul $ρ A v$	mass flows sum karo

Niche ke 9 examples cells A–I ko order mein cover karte hain. Steps padhne se pehle har answer forecast karo — pehle guess karna hi intuition pakki karta hai.

Examples

Cell A — Narrowing pipe (speed up)

Figure s01 niche exactly yahi pipe draw karta hai — wide section mein thin blue arrow ( $v_{1}$ ) aur throat mein fat orange arrow ( $v_{2}$ ) compare karo: same flow, narrow mein faster.

Figure — Continuity equation — derivation (conservation of mass), ρAv = const

Figure s01 — Cell A: ek wide pipe (blue, area $A_{1}$ , slow $v_{1}$ ) narrow hokar throat mein (orange, area $A_{2}$ , fast $v_{2}$ ). Dono arrows woh speed jump dikhate hain jo $A_{1} v_{1} = A_{2} v_{2}$ force karta hai.

Cell B — Widening pipe (slow down)

Cell C — Radius diya hai, area nahi (ratio square karo)

Figure s02 wrong straight line ( $r$ ) aur sahi parabola ( $r^{2}$ ) ko contrast karta hai; red dot woh jagah mark karta hai jahan radius half hoti hai aur area quarter reh jaata hai.

Figure s02 — Cell C: area ratio vs radius ratio. Dashed gray line naive (galat) $r$ hai; blue curve sahi $A \propto r^{2}$ hai. Red dot par, radius half hua $\Rightarrow$ area $1/4$ $\Rightarrow$ speed $\times 4$ .

Cell D — Branches mein split (flows sum karo)

Worked example D: ek pipe do ko feed karti hai

Ek main pipe water $Q_{in} = 1.2 \times 1 0^{- 2} m^{3} / s$ par carry karti hai. Yeh do identical branches mein split hoti hai, har ek area $A = 3 \times 1 0^{- 3} m^{2}$ ka. Agar dono branches same speed par chalen, toh woh speed find karo.

Forecast: har branch half flow handle karegi. Guess speed $= \frac{0.006}{0.003} = 2 m/s$ .

Mass (yahan volume, kyunki $ρ$ const hai) junction par pile up nahi kar sakta: $Q_{in} = Q_{1} + Q_{2}$ . Yeh step kyun? Junction par conservation of mass ka matlab sirf yeh hai ki jo aata hai woh jaata hai, ab outlets mein share hoke (dekho Streamlines and stream tubes).
Equal branches → har ek $Q_{1} = Q_{2} = \frac{Q _{in}}{2} = 6 \times 1 0^{- 3} m^{3} / s$ carry karti hai.
$v = \frac{Q _{1}}{A} = \frac{6 \times 1 0 ^{- 3}}{3 \times 1 0 ^{- 3}} = 2 m/s .$ Yeh step kyun? $Q = A v \Rightarrow v = Q / A$ per branch.

Verify: total out $= 2 \times (A v) = 2 \times (3 \times 1 0^{- 3} \times 2) = 1.2 \times 1 0^{- 2} m^{3} / s = Q_{in}$ ✓.

Cell E — Compressible gas ( $ρ$ rakho!)

Worked example E: duct mein compress hua air

Air ek duct mein $ρ_{1} = 1.0 kg/m^{3}$ , $A_{1} = 0.4 m^{2}$ , $v_{1} = 20 m/s$ par enter karti hai. Downstream ise wider section $A_{2} = 0.8 m^{2}$ mein $ρ_{2} = 2.5 kg/m^{3}$ tak compress kiya jaata hai. $v_{2}$ find karo.

Forecast: area double aur density $\times 2.5$ — dono ise slow karenge. Guess $20$ se kaafi neeche.

Kyunki $ρ$ change hota hai, sirf mass flow conserve hota hai (dekho Incompressible vs compressible flow): $ρ_{1} A_{1} v_{1} = ρ_{2} A_{2} v_{2}$ . Yeh step kyun? Yahan $ρ$ cancel karna classic error hoga — gases ke liye $A v$ constant nahi hota.
$v_{2} = \frac{ρ _{1} A _{1} v _{1}}{ρ _{2} A _{2}} = \frac{1.0 \times 0.4 \times 20}{2.5 \times 0.8} = \frac{8}{2} = 4 m/s .$ Yeh step kyun? Denser + wider dono us speed ko shrink karte hain jo same mass move karne ke liye chahiye.

Verify: $\overset{m}{˙}_{in} = 1.0 \times 0.4 \times 20 = 8 kg/s$ ; $\overset{m}{˙}_{out} = 2.5 \times 0.8 \times 4 = 8 kg/s$ ✓. Note $A_{1} v_{1} = 8 \neq = A_{2} v_{2} = 3.2$ — volume flow conserve nahi hota, bilkul jaise warning thi.

Cell F — Word problem (tank bharna)

Cell G — Degenerate limit: pipe lagbhag band hoti hai

Worked example G: hose par thumb

Water $A_{1} = 1 \times 1 0^{- 4} m^{2}$ , $v_{1} = 4 m/s$ par andar flow karta hai. Tum apna thumb dabate ho toh exit $A_{2}$ tak shrink hota hai. (a) $v_{2}$ find karo jab $A_{2} = 1 \times 1 0^{- 5} m^{2}$ . (b) $v_{2}$ ka kya hota hai jab $A_{2} \to 0$ ?

Forecast: (a) area $\times 10$ se cut → speed $\times 10$ = $40 m/s$ . (b) jaise gap band hoti hai, speed blow up hoti hai.

Incompressible: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = \frac{A _{1} v _{1}}{A _{2}}$ . Yeh step kyun? Same volume/sec ek tiny window se.
(a) $v_{2} = \frac{( 1 \times 1 0 ^{- 4} ) ( 4 )}{1 \times 1 0 ^{- 5}} = \frac{4 \times 1 0 ^{- 4}}{1 \times 1 0 ^{- 5}} = 40 m/s .$
(b) Limit lo: $A_{2} \to 0^{+} lim v_{2} = A_{2} \to 0^{+} lim \frac{A _{1} v _{1}}{A _{2}} = + \infty.$ Yeh step kyun? Ek fixed positive flow ko shrinking area se divide karna speed ko infinity ki taraf le jaata hai — "zyada dabaao, jet faster" ka mathematical chehra. Reality mein friction aur fluid ki limits ise cap karti hain, lekin trend real hai (yeh tumhari hose-jet hai).

Verify: $A_{2} v_{2} = 1 \times 1 0^{- 5} \times 40 = 4 \times 1 0^{- 4} = A_{1} v_{1}$ ✓. Limit sign: numerator $> 0$ , denominator $\to 0^{+}$ , toh $\to + \infty$ ✓.

Figure s03 $v_{2}$ ko shrinking exit area ke against plot karta hai: orange dot part (a) hai, aur curve vertical axis ki taraf rocket karta hai — part (b) ki limit ka visual meaning.

Figure s03 — Cell G: exit speed $v_{2}$ vs exit area $A_{2}$ . Jaise $A_{2}$ zero ki taraf shrink hota hai, $v_{2} = A_{1} v_{1} / A_{2}$ bina bound ke chadhta hai; orange dot $A_{2} = 1 \times 1 0^{- 5}$ m² wala $40$ m/s answer mark karta hai.

Cell H — Exam twist: area ke liye solve karo, mixed units

Worked example H: kaun sa exit area target speed deta hai?

Water $A_{1} = 20 cm^{2}$ , $v_{1} = 1.5 m/s$ par enter karta hai. Tum chahte ho exit speed exactly $v_{2} = 6 m/s$ ho. Kaun sa exit area $A_{2}$ (in $cm^{2}$ ) chahiye? Yeh Venturi meter ke peeche ka design idea hai.

Forecast: speed four times karni hai, toh area ek-chauthai tak shrink hona chahiye → $5 cm^{2}$ .

Convert karo: $A_{1} = 20 cm^{2} = 20 \times 1 0^{- 4} m^{2} = 2 \times 1 0^{- 3} m^{2}$ . Yeh step kyun? $1 cm^{2} = 1 0^{- 4} m^{2}$ ; cm² ko m/s ke saath mix karna silently answer corrupt kar deta hai.
$A_{1} v_{1} = A_{2} v_{2}$ ko unknown area ke liye rearrange karo: $A_{2} = \frac{A _{1} v _{1}}{v _{2}} = \frac{( 2 \times 1 0 ^{- 3} ) ( 1.5 )}{6} = \frac{3 \times 1 0 ^{- 3}}{6} = 5 \times 1 0^{- 4} m^{2} .$ Yeh step kyun? Continuity ek equation hai; jo bhi symbol unknown ho uske liye solve karo.
Wapas convert karo: $5 \times 1 0^{- 4} m^{2} = 5 cm^{2}$ .

Verify: $A_{1} v_{1} = 2 \times 1 0^{- 3} \times 1.5 = 3 \times 1 0^{- 3}$ ; $A_{2} v_{2} = 5 \times 1 0^{- 4} \times 6 = 3 \times 1 0^{- 3}$ ✓. Area ratio $20/5 = 4$ = speed ratio $6/1.5 = 4$ ✓.

Cell I — Compressible fluid jo split bhi karta hai (D + E combined)

Worked example I: hot gas manifold

Ek gas single duct mein $ρ_{1} = 1.2 kg/m^{3}$ , $A_{1} = 0.5 m^{2}$ , $v_{1} = 10 m/s$ par enter karti hai. Yeh do outlet ducts mein split hoti hai. Outlet 1: $ρ = 0.8 kg/m^{3}$ , $A = 0.5 m^{2}$ , $v = 6 m/s$ . Mass flow rate $\overset{m}{˙}_{2}$ find karo jo outlet 2 se nikalna chahiye.

Forecast: jo mass andar aata hai woh bahar jaana chahiye; outlet 1 sab carry nahi kar sakta, toh outlet 2 remainder leta hai.

Kyunki density manifold ke across change hoti hai, mass flow conserve karo, volume nahi, aur branches sum karo: $ρ_{1} A_{1} v_{1} = \overset{m}{˙}_{out, 1} + \overset{m}{˙}_{out, 2}$ . Yeh step kyun? Yeh cell D (split → sum) aur cell E (gas → $ρ$ rakho) ko fuse karta hai. $A v$ volume flow yahan sum karna meaningless hai.
Mass in: $\overset{m}{˙}_{in} = ρ_{1} A_{1} v_{1} = 1.2 \times 0.5 \times 10 = 6 kg/s$ .
Outlet 1 se mass out: $\overset{m}{˙}_{out, 1} = 0.8 \times 0.5 \times 6 = 2.4 kg/s$ .
Outlet 2 se remainder: $\overset{m}{˙}_{out, 2} = 6 - 2.4 = 3.6 kg/s$ . Yeh step kyun? Junction kuch store nahi karta; total mass out = total mass in.

Verify: $\overset{m}{˙}_{out, 1} + \overset{m}{˙}_{out, 2} = 2.4 + 3.6 = 6 kg/s = \overset{m}{˙}_{in}$ ✓. Units $kg/m^{3} \cdot m^{2} \cdot m/s = kg/s$ ✓.

Active recall

Recall Compressing gas ke liye kaun si conserved quantity?

Mass flow rate $\overset{m}{˙} = ρ A v$ — $A v$ nahi, kyunki $ρ$ change hota hai.

Agar radius half ho (incompressible), speed kya karti hai?

4 \times

ho jaati hai, kyunki

v \propto 1/ r^{2}

Ek pipe do branches mein split hoti hai — junction par kya conserved hota hai?

Total volume flow

Q_{in} = Q_{1} + Q_{2}

(mass accumulate nahi ho sakta).

Jab ek compressible gas split hoti hai, branches mein kya sum karte ho?

Mass flows:

ρ_{1} A_{1} v_{1} = \overset{m}{˙}_{out, 1} + \overset{m}{˙}_{out, 2}

Fixed input par exit area

A_{2} \to 0

hone par

v_{2}

kya approach karta hai?

+ \infty

(speed

= A_{1} v_{1} / A_{2}

blow up hoti hai).

20 cm^{2}

m^{2}

mein convert karo.

20 \times 1 0^{- 4} = 2 \times 1 0^{- 3} m^{2}

Volume

V

ko volume flow rate

Q

par bharne ka time?

t = V / Q

Connections

2.2.12 Continuity equation — derivation (conservation of mass), ρAv = const (Hinglish) — parent derivation.
Bernoulli's equation — in speeds se pressures nikalta hai.
Volume flow rate and discharge — $Q = A v$ jo cells B, D, F mein use hota hai.
Incompressible vs compressible flow — kyun cells E aur I mein $ρ$ rakhte hain.
Venturi meter — cell H mein design.
Streamlines and stream tubes — cells D aur I mein junction picture.
Conservation of mass — har cell ke peeche ka law.