Question

A fireman standing on a 14 m high ladderoperates a water hose with a round nozzle ofdiameter 2.65 inch. The lower end of the hose(14 m below the nozzle) is connected to thepump outlet of diameter 3.49 inch. The gaugepressure of the water at the pump isCalculate the speed of the water jet emerg-ing from the nozzle. Assume that water is anincompressible liquid of density 1000 kg/m3and negligible viscosity. The acceleration ofgravity is 9.8 m/s?Answer in units of m/s.

Answer 1

Given data,

The height, H = 14 m

The diameter, D = 2.65 inch

The gauge pressure, P = 317.84 kPa

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

Using Bernoulli's equation,

`\begin{gathered} (1)/(2)\rho(v^2_n-v^2_p)=P_{gauge\text{ }}-\rho gh \\ (v^2_n-v^2_p)=((2)/(\rho))P_(gauge)-2gh \\ v^2_n-((A_n)/(A_p))^2v^2_n=((2)/(\rho))P_(gauge)-2gh \\ v^2_n-(\frac{r_n}{r_p_{}})^4v^2_n=((2)/(\rho))P_(gauge)-2gh \end{gathered}`

Further solved as,

`\begin{gathered} v_n=\sqrt[]{(((2)/(\rho))P_(gauge)-2gh)/(1-((r_n)/(r_p))^4)} \\ v_n=\sqrt[]{\frac{((2)/(1000))*317.84*10^3-2*9.8*14}{1-(\frac{1.325_{}}{1.745_{}})^4}} \\ v_n=\sqrt[]{(635-274.4)/(0.667)} \\ v_n=\sqrt[]{540.62} \end{gathered}`

Thus, the speed of the water jet is

`v=23.25\text{ m/s}`