A nozzle with a radius of 0.250 cm

Gerry Palme

Gerry Palme

Answered question

2022-06-18

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 2 marks (b) in the nozzle.

Answer & Explanation

star233

star233

Skilled2022-09-14Added 403 answers

Solution for (a)
First, we solve Q=Av¯ for v¯1 and note that the cross-sectional area is A=πr2, yielding
v¯1=QA1=Qπr12.
Substituting known values and making appropriate unit conversions yields v¯1=(0.500 L/s)(103 m3/L)π(9.00×103 m)2=1.96 m/s
 

Solution for (b)
We could repeat this calculation to find the speed in the nozzle v¯2, but we will use the equation of continuity to give a somewhat different insight. Using the equation which states
A1v¯1=A2v¯2
Solving for v¯2 and substituting πr2 for the cross-sectional area yields
v¯2=A1A2v¯1=πr12πr22v¯1=r12r22v¯1
Substituting known values,
v¯2=(0.900 cm)2(0.250 cm)21.96 m/s=25.5 m/s

 

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.

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