Question

A fireman standing on a 9.1 m high ladder operates a water hose with a round nozzle of diameter 1.8 inch. The lower end of the hose (9.1 m below the nozzle) is connected to the pump outlet of diameter 3.1 inch. The gauge pressure of the water at the pump is(gauge) Plamp-P,tin 50.6 PSI 348.875 kPa (abs) pump atmCalculate the speed of the water jet emerging from the nozzle. Assume that water is an incompressible liquid of density 1000 kg/m^3 and negligible viscosity. The acceleration of gravity is 9.8 m/s^2 . Answer in units of m/s.

Answer 1

Answer:

The speed of the water jet emerging from the nozzle is 24.21 m/s.

Explanation:

Given that,

Height = 9.1 m

Diameter =1.8 inch

Gauge pressure = 348.875 kPa

We need to calculate the speed of the water jet emerging from the nozzle

Using Bernoulli's equation

$\dfrac{1}{2}\rho(v_{n}^2-v_{p}^2)=P_{gauge}-\rho gh$

$(v_{n}^2-v_{p}^2)=(\dfrac{2}{\rho})P_{gauge}-2gh$

$v_{n}^2-(\dfrac{A_{n}}{A_{p}})^2v_{n}^2=(\dfrac{2}{\rho})P_{gauge}-2gh$

$v_{n}^2-(\dfrac{r_{n}}{r_{p}})^4v_{n}^2=(\dfrac{2}{\rho})P_{gauge}-2gh$

$v_{n}^2=\dfrac{(\dfrac{2}{\rho})P_{gauge}-2gh}{1-(\dfrac{r_{n}}{r_{p}})^4}$

$v_{n}=\sqrt{\dfrac{(\dfrac{2}{\rho})P_{gauge}-2gh}{1-(\dfrac{r_{n}}{r_{p}})^4}}$

Put the value into the formula

$v_{n}=\sqrt{\dfrac{\dfrac{2}{1000}\times348.875\times10^{3}-2\times9.8\times9.1}{1-\dfrac{(0.9)^4}{(1.55)^4}}}$

$v_{n}=24.21\ m/s$

Hence, The speed of the water jet emerging from the nozzle is 24.21 m/s.