Answer to Question #102988 in Mechanics | Relativity for Deandra

Question

A fireman standing on a 13 m high ladder
operates a water hose with a round nozzle of
diameter 2.53 inch. The lower end of the hose
(13 m below the nozzle) is connected to the
pump outlet of diameter 3.61 inch. The gauge
pressure of the water at the pump is
P
(gauge)
pump = P
(abs)
pump − Patm
= 31.5 PSI = 217.185 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

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,

$$0.5ρ(v_ n^2 ​ −v _ p ^2 ​ )=P _ n ​ −P _ p ​ −ρgh$$

$$P _ n ​ −P _ p ​ =P _ p ^{gauge} ​ =217.185\ kPa$$

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

$$R=\frac{2}{1000}217185-2(9.8)(13)=179.57\frac{m^2}{s^2}$$

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

$$179.57=v_n^2-\frac{2.53^4}{3.61^4}v_n^2$$

$$v_n=15.3839\frac{m}{s}$$