# Answer to Question #102988 in Mechanics | Relativity for Deandra

Question #102988
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
.
1
2020-02-17T09:07:58-0500

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\u03c1(v_\nn^2\n\u200b\t\n \u2212v _\np\n^2\n\u200b\t\n )=P _\nn\n\u200b\t\n \u2212P _\np\n\u200b\t\n \u2212\u03c1gh"

"P _\nn\n\u200b\t\n \u2212P _\np\n\u200b\t\n =P _\np\n^{gauge}\n\u200b\t\n =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}"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!