Question #102219

A fireman standing on an 11 m high ladder

operates a water hose with a round nozzle of

diameter 2.1 inch. The lower end of the hose

(11 m below the nozzle) is connected to the

pump outlet of diameter 4.1 inch. The gauge

pressure of the water at the pump is

P(gauge)pump = P(abs)pump − Patm= 55 PSI = 379.212 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/s2. Answer in units of m/s.

operates a water hose with a round nozzle of

diameter 2.1 inch. The lower end of the hose

(11 m below the nozzle) is connected to the

pump outlet of diameter 4.1 inch. The gauge

pressure of the water at the pump is

P(gauge)pump = P(abs)pump − Patm= 55 PSI = 379.212 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/s2. Answer in units of m/s.

Expert's answer

By the continuity equation "v A = const" the ratio of the speed of the water jet emerging

from the nozzle to the speed of water in the hose is

Applying Bernoulli’s equation,

"P_n-P_p=P_p^{gauge}=379.212\\ kPa"

Let

"R=\\frac{2}{1000}379212-2(9.8)(11)=542.824\\frac{m^2}{s^2}"

Then,

"542.824=v_n^2-\\frac{2.1^4}{4.1^4}v_n^2"

"v_n=24.1442\\frac{m}{s}"

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