# Water flows through a fire hose of diameter 6.35 centimeters at a rate of 0.012 m3/s. the fire hose ends in a nozzle of inner diameter 2.2 centimeters. what is the velocity with which the water exits the nozzle?

Jun 22, 2014

$31.6 m {s}^{-} 1$

This solution assumes an incompressible fluid, so the volumetric flow rate is constant at all points in the hose and nozzle (this is because the density is constant and the mass flow rate must be the same at all points).

In order to calculate the of velocity of the water in the nozzle we need to know the cross-sectional area of the nozzle.
A_N=(πD_N^2)/4=(π(0.022)^2)/4=3.8*10^-4 m^2

Now we can use the cross-sectional area with the volumetric flow rate to calculate the velocity of the water in the nozzle:
$\frac{\mathrm{dV}}{\mathrm{dt}} = A \frac{\mathrm{ds}}{\mathrm{dt}}$

Where $\frac{\mathrm{dV}}{\mathrm{dt}}$ is the volumetric flow rate and $\frac{\mathrm{ds}}{\mathrm{dt}}$ is the rate of change of displacement, i.e. the velocity.

$v = \frac{\left(\mathrm{dV}\right) / \left(\mathrm{dt}\right)}{A} = \frac{0.012}{3.8 \cdot {10}^{-} 4} = 31.6 m {s}^{-} 1$

That is just over 70 mph!