Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and the water flow rate is 16 m3/min. Take the density of water as 1000 kg/m3 and the momentum-flux correction factor as unity.

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Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, [tex]Q=16 m^{3}/min[/tex]

Diameter d= 8cm= 0.08 m

Assumptions

  • The flow is jet flow hence momentum-flux correction factor is unity
  • Gravitational force is not considered
  • The flow is steady, frictionless and incompressible
  • Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

[tex]V=\frac {Q}{A}[/tex] where V is the exit velocity and A is area

Area, [tex]A=\frac {\pi d^{2}{4}[/tex] where d is the diameter

By substitution

[tex]V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min[/tex]

To convert v to m/s from m/s, we simply divide it by 60 hence

[tex]V=\frac {3183.098862  m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s[/tex]

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