Question

A fireman standing on a 10 m high ladder. operates a water hose with a round nozzle of diameter 2.74 inch. The lower end of the hose (10 m below the nozzle) is connected to the pump outlet of diameter 4.19 inch. The gauge pressure of the water at the pump is P(gauge) pump = P (abs) pump − Patm = 50.3 PSI = 346.806 kPa. Calculate the speed of the water jet emerging from the nozzle. Assume that water is an incompressible liquid of density 1000 kg/m3 and negligible viscosity. The acceleration of gravity is 9.8 m/s^2. Answer in units of m/s.

Answer 1

The speed of the water jet emerging from the nozzle can be calculated using Bernoulli's equation as follows:

V^2 = 2gP(gauge) pump/ρ

Substituting the given values in the above equation, we get

V^2 = 2*9.8*346.806*1000/1000

V = 166.905 m/s