Q: A hose pipe has a diameter of 2.5 cm and its required to direct a jet of water to a height of at least 40 m. Find the minimum power of the pump needed for this hose.

Sol: Volume of water ejected per sec is

$\large A v = \pi \times (\frac{d}{2})^2 \times \sqrt{2 g h}$ ; ( v = √(2gh))

Mass ejected per sec is $\large m = \frac{\pi d^2}{4}\times \sqrt{2 g h}\times \rho Kg/s $

Kinetic energy of water leaving hose/sec

$\large = \frac{1}{2}m v^2 = \frac{\pi d^2}{8}\times (2gh)^{3/2}\times \rho $

$\large = \frac{3.14 \times (2.5 \times 10^{-2})^2}{8}\times (2 \times 9.8 \times 40)^{3/2}\times 1000 $

= 21.5 KJ