Example 12.2 from College Physics, 12.1 Flow Rate and Its Relation to Velocity

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.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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 $Q = A \bar{v}$ for $v_{1}$ and note that the cross-sectional area is $A = {πr}^{2}$ , yielding
${\bar{v}}_{1} = \frac{Q}{A_{1}} = \frac{Q}{{πr}_{1}^{2}} .$
12.8

Substituting known values and making appropriate unit conversions yields

${\bar{v}}_{1} = \frac{(0.500 L/s) (10^{- 3} m^{3} / L)}{π (9 . 00 \times 10^{- 3} m)^{2}} = 1 . 96 m/s .$
12.9
Solution for (b)
We could repeat this calculation to find the speed in the nozzle ${\bar{v}}_{2}$ , but we will use the equation of continuity to give a somewhat different insight. Using the equation which states
$A_{1} {\bar{v}}_{1} = A_{2} {\bar{v}}_{2},$
12.10

solving for ${\bar{v}}_{2}$ and substituting ${πr}^{2}$ for the cross-sectional area yields
${\bar{v}}_{2} = \frac{A_{1}}{A_{2}} {\bar{v}}_{1} = \frac{{πr}_{1}^{2}}{{πr}_{2}^{2}} {\bar{v}}_{1} = \frac{r_{1^{2}}}{r_{2^{2}}} {\bar{v}}_{1} .$
12.11

Substituting known values,

${\bar{v}}_{2} = \frac{(0.900 cm)^{2}}{(0.250 cm)^{2}} 1.96 m/s = 25 . 5 m/s .$
12.12
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.

0 of 4 steps open

How did it go?

Strategy

Solution for (a)

Solution for (b)

Discussion

Strategy

Solution for (a)

Solution for (b)

Discussion