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

Use $\rho Av=\text{const}$ (the mass flow rate). Why: the fluid is compressible — $\rho$ changes — so the volume form $Av$ is not conserved. Only mass can never be created or destroyed, so only $\rho Av$ is guaranteed constant. Dropping $\rho$ here would silently assume ${\rho }_1={\rho }_2$, which is false for steam.

It speeds up. Rule: $v\propto 1/A$ for incompressible flow, so smaller area $\Rightarrow$ larger speed. Look again at the s01 figure — the skinny slab is longer, i.e. faster.

$Q=Av\Rightarrow v=\frac{Q}{A}=\frac{3\times 10^{-3}}{6\times 10^{-4}}=5\text{m/s}$. Units: $\frac{{\text{m}}^3/\text{s}}{{\text{m}}^2}=\text{m/s}$ ✓ — a speed, as expected. (Areas already in m², so no conversion needed.)

Incompressible, so $A_1{v}_1=A_2{v}_2$ with $A=\pi r^2$. This is a ratio problem — we only need $A_1/A_2$, so the units of $r$ (cm or m) cancel and the $\pi$ cancels: $v_2=v_1\frac{r_1^2}{r_2^2}=1.5\times \frac{6^2}{2^2}=1.5\times 9=13.5\text{m/s}.$ Why the cm are safe here: both radii sit in the ratio $(r_1/r_2{)}^2$, so as long as $r_1$ and $r_2$ use the same unit the unit disappears — no conversion to metres is required for a ratio. Why $r^2$: area grows as radius squared; the radius shrank by factor 3, so area shrank by $3^2=9$, so speed grew ×9.

Density changes → use mass form: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.2\times 0.20\times 8}{0.80\times 0.20}=\frac{1.92}{0.16}=12\text{m/s}.$ Why it speeds up: same mass per second, but each kilogram now occupies more volume (lower density), so it must move faster to clear the same area.

This produces a real $Q$, so we convert the area to SI first: $A=2{\text{cm}}^2=2\times 10^{-4}{\text{m}}^2$. $Q=Av=(2\times 10^{-4})(4)=8\times 10^{-4}{\text{m}}^3/\text{s}$. $t=\frac{V}{Q}=\frac{0.024}{8\times 10^{-4}}=30\text{s}.$ Why $t=V/Q$: continuity says the same volume-per-second $Q$ that leaves the tap must arrive in the tank (no water vanishes). So the tap delivers $Q$ cubic metres each second at a steady rate; the total volume $V$ divided by that per-second rate is exactly the number of seconds needed — rate × time = volume, rearranged.

Volume conservation across the junction: what enters per second = what leaves per second. This is a ratio-style balance — every area is in cm² on both sides, so the cm² cancel and we get a clean speed in m/s without converting to m²: $Q_0=2Q_b\Rightarrow A_0{v}_0=2A_b{v}_b.$ $v_b=\frac{A_0{v}_0}{2A_b}=\frac{10\times 3}{2\times 4}=\frac{30}{8}=3.75\text{m/s}.$ Why the cm² are safe: $A_0$ and $A_b$ appear as the ratio $A_0/A_b$, so their common unit disappears — only the numerical size ratio matters here. Why the factor 2: continuity is about the whole flow, so both branches together must carry the full incoming $Q$. Look at s02 — the incoming volume-per-second fans out into two streams.

$A_1{v}_1=A_2{v}_2$ gives $\frac{A_1}{A_2}=\frac{v_2}{v_1}=3$. The upstream (slow) section has three times the area of the downstream (fast) section — wide-slow, narrow-fast, exactly reversed from the speed ratio. Why the ratio flips: in $A_1{v}_1=A_2{v}_2$ the product $Av$ is fixed, so $A$ and $v$ are inversely tied — if $v$ multiplies by 3, $A$ must divide by 3 to keep the product constant. That is why the area ratio ($3$) equals the reciprocal of the speed change, i.e. it is the speed ratio $v_2/v_1$ read the other way round. No units are needed since only a ratio is asked.

Two things change, so keep the full mass form: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.0\times 0.30\times 20}{1.5\times 0.10}=\frac{6.0}{0.15}=40\text{m/s}.$ Why: the area shrank ×3 (pushes speed up) while density rose ×1.5 (pulls speed down). Net factor $=\frac{3}{1.5}=2$, so $v$ doubles from 20 to 40.

The general junction rule is total in = sum of totals out: $A_0{v}_0=A_1{v}_1+A_2{v}_2={\sum }_i{A}_i{v}_i.$ Incoming: $Q_0=A_0{v}_0=10\times 3=30$ (cm²·m/s). Branch 1 takes $Q_1=A_1{v}_1=4\times 2=8$. So branch 2 must carry the rest: $Q_2=Q_0-Q_1=30-8=22,v_2=\frac{Q_2}{A_2}=\frac{22}{3}\approx 7.33\text{m/s}.$ Why the sum, not a factor 2: the "$2$" in L3.1 was a shortcut valid only because the two branches were identical. In general each branch carries whatever $Av$ it carries, and continuity just demands they add up to the incoming flow. The equal-branch case is the special instance $A_1{v}_1=A_2{v}_2$. (All $Q$ here are in consistent cm²·m/s units, so the subtraction is legitimate; only the speed $v_2$ needs the shared cm² to cancel, which it does.)

Step 0 (which form?): water is incompressible ($\rho$ constant), so the volume form $A_1{v}_1=A_2{v}_2$ is legal — we do not need $\rho Av$ for the speeds. Step 1 (continuity, a ratio): the areas appear as $A_1/A_2$, so the cm² cancel: $v_2=v_1\frac{A_1}{A_2}=2\times \frac{8}{2}=8\text{m/s}.$ Step 2 (Bernoulli, now in SI): speeds are in m/s and $\rho$ in kg/m³, so pressure comes out in Pa: $p_1-p_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\frac{1}{2}(1000)(8^2-2^2)=500\times (64-4)=30000\text{Pa}.$ Why continuity first: Bernoulli needs both speeds; continuity is the only tool that hands you $v_2$ from geometry. This chained logic is exactly how a Venturi meter works — narrow spot goes fast, pressure drops, and you read the drop off a gauge.

Continuity between the free surface (section 1) and the hole (section 2), incompressible water: $v_1=v_2\frac{A_2}{A_1}=6\times \frac{5\times 10^{-4}}{1.0}=3\times 10^{-3}\text{m/s}=3\text{mm/s}.$ Why so tiny: the surface is 2000× wider than the hole, so it creeps down 2000× slower than the jet shoots out — which is why we usually treat the tank surface as stationary in energy problems.

$Q=A_1{v}_1=12\times 1=12$ (in cm²·m/s — same for every section since incompressible; the cm² cancel when we divide back out for each speed). $v_2=\frac{Q}{A_2}=\frac{12}{4}=3\text{m/s},v_3=\frac{Q}{A_3}=\frac{12}{6}=2\text{m/s}.$ Fastest where narrowest → section 2 ($3\text{m/s}$). Notice $v_3<v_2$: widening from 4 to 6 cm² slows the flow back down. Continuity is local — each section's speed is set by its own area.

(a) Incompressible: $v_2=v_1\frac{A_1}{A_2}=30\times \frac{0.40}{0.20}=60\text{m/s}.$ (b) Full mass form: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.2\times 0.40\times 30}{1.6\times 0.20}=\frac{14.4}{0.32}=45\text{m/s}.$ (c) The incompressible guess (60) over-estimates. When the gas is squeezed denser, each kilogram now packs into less volume, so you need less speed to move the same mass through — the true speed (45) is lower. The area ratio alone (×2) is partly "absorbed" by the density rise (×1.33): net speed factor $=\frac{2}{1.33}=1.5$, giving $30\times 1.5=45$. See Incompressible vs compressible flow.

All quantities are already in SI (m², m³, m³/s), so no conversions are needed. (a) Time to empty. The same $Q$ leaves every second, so total volume ÷ rate gives the time: $t=\frac{V}{Q}=\frac{0.90}{1.5\times 10^{-3}}=600\text{s}=10\text{min}.$ Why $t=V/Q$: continuity guarantees the volume leaving per second is exactly $Q$ (no water vanishes), so emptying $0.90{\text{m}}^3$ at $1.5\times 10^{-3}{\text{m}}^3$ each second takes $V/Q$ seconds — rate × time = volume, rearranged. (b) Exit pipe area. The jet is that same $Q$ squeezed through the small pipe, so $Q=A_{\text{exit}}v$: $A_{\text{exit}}=\frac{Q}{v}=\frac{1.5\times 10^{-3}}{3}=5\times 10^{-4}{\text{m}}^2=5{\text{cm}}^2.$ (c) Surface drop speed. The surface (area $A_1$) also passes the same $Q$: $Q=A_1{v}_{\text{surface}}$, so $v_{\text{surface}}=\frac{Q}{A_1}=\frac{1.5\times 10^{-3}}{0.50}=3\times 10^{-3}\text{m/s}=3\text{mm/s}.$ Why $Q$ links all three parts: the same volume-per-second empties the tank, feeds the jet, and lowers the surface — continuity ties the big slow surface to the small fast jet through one shared number $Q$.

Mass flow rate $\rho Av$ — never the volume form $Av$.

Total in = sum of totals out: $A_0{v}_0={\sum }_i{A}_i{v}_i$. Equal branches each get half; unequal branches each carry their own $A_i{v}_i$.

Bernoulli needs both speeds; continuity is what converts the area geometry into the second speed.

Ratio-only problems (speed from area ratio) let cm² cancel; any problem yielding a real $Q$, time, or pressure needs SI (m²).

$v$ mean in the signed continuity statement? Flow reversed relative to the chosen positive direction — fluid crossing that section backwards.