Question

What is the change in speed (in m/s) due to the Bernoulli effect as water goes into a 3.19 cm-diameter nozzle from a 9.26 cm-diameter fire hose while carrying a flow of 40.9 L/s?

Answer 1

Using continuity equation we know that Volume flow rate will remain same in both nozzle and hose. So

Q = Volume flow rate = V*A

A = Area of pipe = pi*d^2/4

V = Speed of water = Q/A = 4Q/(pi*d^2)

Given values are: Q = 40.9 L/s = 40.9*10^-3 m^3/s

(Since 1 L = 1*10^-3 m^3)

Now speed of water in nozzle will be:

V1 = 4*40.9*10^-3/(pi*0.0319^2) = 51.17 m/s

And speed of water in hose will be:

V2 = 4*40.9*10^-3/(pi*0.0926^2) = 6.07 m/s

So change in speed will be:

dV = V1 - V2 = 51.17 - 6.07

dV = 45.1 m/s

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