# 10.8 Flow rate and its relation to velocity  (Page 2/6)

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$\begin{array}{c}{Q}_{1}={Q}_{2}\\ {A}_{1}{\overline{v}}_{1}={A}_{2}{\overline{v}}_{2}\end{array}\right\}\text{.}$

This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases.

Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion.

## Calculating fluid speed: speed increases when a tube narrows

A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle.

Strategy

We can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and 2 for the nozzle.

Solution for (a)

First, we solve $Q=A\overline{v}$ for ${v}_{1}$ and note that the cross-sectional area is $A={\mathrm{\pi r}}^{2}$ , yielding

${\overline{v}}_{1}=\frac{Q}{{A}_{1}}=\frac{Q}{{\mathrm{\pi r}}_{1}^{2}}\text{.}$

Substituting known values and making appropriate unit conversions yields

${\overline{v}}_{1}=\frac{\left(0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}\text{L/s}\right)\left({\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}/\text{L}\right)}{\pi \left(9\text{.}\text{00}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}=1\text{.}\text{96}\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$

Solution for (b)

We could repeat this calculation to find the speed in the nozzle ${\overline{v}}_{2}$ , but we will use the equation of continuity to give a somewhat different insight. Using the equation which states

${A}_{1}{\overline{v}}_{1}={A}_{2}{\overline{v}}_{2}\text{,}$

solving for ${\overline{v}}_{2}$ and substituting ${\mathrm{\pi r}}^{2}$ for the cross-sectional area yields

${\overline{v}}_{2}=\frac{{A}_{1}}{{A}_{2}}{\overline{v}}_{1}=\frac{{\mathrm{\pi r}}_{1}^{2}}{{\mathrm{\pi r}}_{2}^{2}}{\overline{v}}_{1}=\frac{{r}_{{1}^{2}}}{{r}_{{2}^{2}}}{\overline{v}}_{1}\text{.}$

Substituting known values,

${\overline{v}}_{2}=\frac{\left(0\text{.}\text{900}\phantom{\rule{0.25em}{0ex}}\text{cm}{\right)}^{2}}{\left(0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{cm}{\right)}^{2}}1\text{.}\text{96}\phantom{\rule{0.25em}{0ex}}\text{m/s}=\text{25}\text{.}\text{5 m/s}\text{.}$

Discussion

A speed of 1.96 m/s is about right for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube.

The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.

In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes

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