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

Before anything else, one reminder so every symbol is earned:

The scenario matrix

Continuity problems all obey one master line — mass in per second = mass out per second — but they look different depending on what changes and what is held fixed. Here is every class:

Cell	What changes	Held fixed	Conserved quantity	Watch out for
A Narrowing	$A$ shrinks	$ρ$	$A v$	speed goes up
B Widening	$A$ grows	$ρ$	$A v$	speed goes down
C Radius given	$r$ (not $A$ )	$ρ$	$A v$ , with $A \propto r^{2}$	square the ratio
D Splitting / merging	one pipe → many	$ρ$	total $A v$	sum the branches
E Compressible gas	$ρ$ changes	$A$ (here)	$ρ A v$ only	do not cancel $ρ$
F Word problem (fill/drain)	time from rate	$ρ$	$Q$	$t = V / Q$
G Degenerate: $A \to 0$	window closes	$ρ$	$\overset{m}{˙} = ρ A v$	$v \to \infty$ limit
H Exam twist: solve for area	$A_{2}$ unknown	$ρ$ , mixed units	$A v$	convert cm↔m first
I Compressible split	one gas pipe → many, $ρ$ changes	(nothing)	total $ρ A v$	sum mass flows

The 9 examples below hit cells A–I in order. Forecast each answer before reading the steps — guessing first is how the intuition sticks.

Examples

Cell A — Narrowing pipe (speed up)

Figure s01 below draws exactly this pipe — compare the thin blue arrow ( $v_{1}$ ) in the wide section with the fat orange arrow ( $v_{2}$ ) in the throat: same flow, faster where narrow.

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

Figure s01 — Cell A: a wide pipe (blue, area $A_{1}$ , slow $v_{1}$ ) narrowing into a throat (orange, area $A_{2}$ , fast $v_{2}$ ). The two arrows show the speed jump that $A_{1} v_{1} = A_{2} v_{2}$ forces.

Cell B — Widening pipe (slow down)

Cell C — Given radius, not area (square the ratio)

Figure s02 contrasts the wrong straight line ( $r$ ) with the true parabola ( $r^{2}$ ); the red dot marks where the radius halves and the area falls to a quarter.

Figure s02 — Cell C: area ratio vs radius ratio. The dashed gray line is the naive (wrong) $r$ ; the blue curve is the true $A \propto r^{2}$ . At the red dot, radius halved $\Rightarrow$ area $1/4$ $\Rightarrow$ speed $\times 4$ .

Cell D — Splitting into branches (sum the flows)

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

Worked example E: air compressed in a duct

Air enters a duct at $ρ_{1} = 1.0 kg/m^{3}$ , $A_{1} = 0.4 m^{2}$ , $v_{1} = 20 m/s$ . Downstream it is compressed to $ρ_{2} = 2.5 kg/m^{3}$ in a wider section $A_{2} = 0.8 m^{2}$ . Find $v_{2}$ .

Forecast: area doubled and density $\times 2.5$ — both slow it. Guess well below $20$ .

Because $ρ$ changes, only mass flow is conserved (see Incompressible vs compressible flow): $ρ_{1} A_{1} v_{1} = ρ_{2} A_{2} v_{2}$ . Why this step? Cancelling $ρ$ here would be the classic error — $A v$ is not constant for gases.
$v_{2} = \frac{ρ _{1} A _{1} v _{1}}{ρ _{2} A _{2}} = \frac{1.0 \times 0.4 \times 20}{2.5 \times 0.8} = \frac{8}{2} = 4 m/s .$ Why this step? Denser + wider both shrink the required speed to move the same mass.

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 is not conserved, exactly as warned.

Cell F — Word problem (fill a tank)

Cell G — Degenerate limit: the pipe nearly closes

Worked example G: thumb on the hose

Water flows in at $A_{1} = 1 \times 1 0^{- 4} m^{2}$ , $v_{1} = 4 m/s$ . You press your thumb so the exit shrinks to $A_{2}$ . (a) Find $v_{2}$ when $A_{2} = 1 \times 1 0^{- 5} m^{2}$ . (b) What happens to $v_{2}$ as $A_{2} \to 0$ ?

Forecast: (a) area cut $\times 10$ → speed $\times 10$ = $40 m/s$ . (b) as the gap closes, speed blows up.

Incompressible: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = \frac{A _{1} v _{1}}{A _{2}}$ . Why this step? Same volume/sec through a tiny window.
(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) Take the limit: $A_{2} \to 0^{+} lim v_{2} = A_{2} \to 0^{+} lim \frac{A _{1} v _{1}}{A _{2}} = + \infty.$ Why this step? Dividing a fixed positive flow by a shrinking area sends speed to infinity — the mathematical face of "squeeze harder, jet faster." In reality friction and the fluid's limits cap it, but the trend is real (this is your hose-jet).

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 $\to 0^{+}$ , so $\to + \infty$ ✓.

Figure s03 plots $v_{2}$ against the shrinking exit area: the orange dot is part (a), and the curve rockets upward toward the vertical axis — the visual meaning of the limit in part (b).

Figure s03 — Cell G: exit speed $v_{2}$ vs exit area $A_{2}$ . As $A_{2}$ shrinks toward zero, $v_{2} = A_{1} v_{1} / A_{2}$ climbs without bound; the orange dot marks the $A_{2} = 1 \times 1 0^{- 5}$ m² answer of $40$ m/s.

Cell H — Exam twist: solve for area, mixed units

Worked example H: what exit area gives a target speed?

Water enters at $A_{1} = 20 cm^{2}$ , $v_{1} = 1.5 m/s$ . You want the exit speed to be exactly $v_{2} = 6 m/s$ . What exit area $A_{2}$ (in $cm^{2}$ ) is needed? This is the design idea behind a Venturi meter.

Forecast: speed must quadruple, so area must shrink to a quarter → $5 cm^{2}$ .

Convert: $A_{1} = 20 cm^{2} = 20 \times 1 0^{- 4} m^{2} = 2 \times 1 0^{- 3} m^{2}$ . Why this step? $1 cm^{2} = 1 0^{- 4} m^{2}$ ; mixing cm² with m/s silently corrupts the answer.
Rearrange $A_{1} v_{1} = A_{2} v_{2}$ for the unknown area: $A_{2} = \frac{A _{1} v _{1}}{v _{2}} = \frac{( 2 \times 1 0 ^{- 3} ) ( 1.5 )}{6} = \frac{3 \times 1 0 ^{- 3}}{6} = 5 \times 1 0^{- 4} m^{2} .$ Why this step? Continuity is one equation; solve for whichever symbol is unknown.
Convert back: $5 \times 1 0^{- 4} m^{2} = 5 cm^{2}$ .

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 that also splits (D + E combined)

Worked example I: hot gas manifold

A gas enters a single duct at $ρ_{1} = 1.2 kg/m^{3}$ , $A_{1} = 0.5 m^{2}$ , $v_{1} = 10 m/s$ . It splits into two outlet ducts. Outlet 1: $ρ = 0.8 kg/m^{3}$ , $A = 0.5 m^{2}$ , $v = 6 m/s$ . Find the mass flow rate $\overset{m}{˙}_{2}$ that must leave through outlet 2.

Forecast: whatever mass comes in must leave; outlet 1 can't carry it all, so outlet 2 takes the remainder.

Because density changes across the manifold, conserve mass flow, not volume, and sum the branches: $ρ_{1} A_{1} v_{1} = \overset{m}{˙}_{out, 1} + \overset{m}{˙}_{out, 2}$ . Why this step? This fuses cell D (split → sum) with cell E (gas → keep $ρ$ ). Volume flow $A v$ is meaningless to sum here.
Mass in: $\overset{m}{˙}_{in} = ρ_{1} A_{1} v_{1} = 1.2 \times 0.5 \times 10 = 6 kg/s$ .
Mass out of outlet 1: $\overset{m}{˙}_{out, 1} = 0.8 \times 0.5 \times 6 = 2.4 kg/s$ .
Remainder through outlet 2: $\overset{m}{˙}_{out, 2} = 6 - 2.4 = 3.6 kg/s$ . Why this step? The junction stores nothing; total mass out equals 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 Which conserved quantity for a compressing gas?

Mass flow rate $\overset{m}{˙} = ρ A v$ — not $A v$ , because $ρ$ changes.

If radius halves (incompressible), speed does what?

Becomes

4 \times

, since

v \propto 1/ r^{2}

One pipe splits into two branches — what is conserved at the junction?

Total volume flow

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

(mass can't accumulate).

When a compressible gas splits, what do you sum across branches?

The mass flows:

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

As the exit area

A_{2} \to 0

at fixed input, what does

v_{2}

approach?

+ \infty

(speed

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

blows up).

Convert

20 cm^{2}

m^{2}

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

Time to fill volume

V

at volume flow rate

Q

t = V / Q

Connections

2.2.12 Continuity equation — derivation (conservation of mass), ρAv = const (Hinglish) — parent derivation.
Bernoulli's equation — uses these speeds to get pressures.
Volume flow rate and discharge — the $Q = A v$ used in cells B, D, F.
Incompressible vs compressible flow — why cells E and I keep $ρ$ .
Venturi meter — the design in cell H.
Streamlines and stream tubes — the junction picture in cells D and I.
Conservation of mass — the law behind every cell.