Answer to Question #102219 in Mechanics | Relativity for Nyx

Question

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.

Answer 1

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

$$\frac{v_p}{v_n}=\frac{A_n}{A_p}=\frac{r_n^2}{r_p^2}$$

Applying Bernoulli’s equation,

$$0.5\rho(v_n^2-v_p^2)=P_n-P_p-\rho gh$$

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

Let

$$R=\frac{2}{\rho}P_p^{gauge}-2gh$$

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

Then,

$$R=v_n^2-v_p^2=v_n^2-\frac{r_n^4}{r_p^4}v_n^2$$

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

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