Question

A firehose must be able to shoot water to the top of a building 35 m tall when aimed straight up. Water enters this hose at a steady rate of 0.5 m^3/s and shoots out a round nozzle. A) what is the maximum diameter this nozzle can have? B) if the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

Answer 1

(a)

Rate of Volume of liquid flowing \((\mathrm{V} / \mathrm{t})=\operatorname{Area}(\mathrm{A}) *\) Velocity \((\mathrm{v})\).

$$ \begin{array}{l} \frac{V}{t}=\mathrm{Av} \\ 0.5=\left(\frac{\pi D^{2}}{4}\right)(\sqrt{2 \mathrm{gh}}) \\ 0.5=\left(\frac{\pi D^{2}}{4}\right)(\sqrt{2 \times 9.8 \times 35}) \end{array} $$

by solving the above eqation we get \(D=0.156 \mathrm{~m}\).

(b)

If Diameter is doubled the maximum height reached by the water can be calculated using equation of continuity.

$$ \begin{array}{c} A_{1} \cdot V_{1}=A_{2} \cdot V_{2} \\ \left(\frac{\pi D_{1}^{2}}{4}\right)\left(\sqrt{2 g h_{1}}\right)=\left(\frac{\pi D_{2}^{2}}{4}\right)\left(\sqrt{2 g h_{2}}\right) \\ \Rightarrow D_{1}^{2}\left(\sqrt{h_{1}}\right)=D_{2}^{2}\left(\sqrt{h_{2}}\right) \\ D_{1}^{2} \times \sqrt{35_{\square}}=4 D_{1}^{2} \sqrt{h_{2}} \\ \therefore \mathrm{h}_{2}=2.19 \mathrm{~m} . \end{array} $$