Answer to Question #222150 in Mechanics | Relativity for Kookie

Question

Find the power of engine that can fire water jet of density 1000 kg/m3
at the speed of 30m/s upward from the pipe of radius 1cm?

Answer 1

Explanations & Calculations

• Those given conditions are valid only at the pump outlet where the water is ejected at the speed of $\small 30\,ms^{-1}$ and beyond that these calculations may not seem to be effective.
• The power, as it is defined is = $\Large \frac{\Delta\text{work}}{\Delta\text{time}}$
• And the work done at the outlet is $\small W= \small \frac{1}{2}mv^2$ and lets assume that this work is done at a small duration of $\small t(s)$. Then the power needed is

$\qquad\qquad \begin{aligned} \small P&=\small \frac{W}{t}=\frac{1}{2}\Big(\frac{m}{t}\Big)v^2 \end{aligned}$

• That term within the parentheses is the rate of mass flow which can be deduced to be $\small Av\rho$ where A is the are of the pump.
• Then,

$\qquad\qquad \begin{aligned} \small P&=\small \frac{1}{2}(Av\rho)v^2\\ &=\small \frac{1}{2}Av^3\rho\\ &=\small \frac{1}{2}(\pi(0.01m)^2\times(30ms^{-1})^3\times1000kgm^{-3})\\ &=\small \bold{4241.15\,W\approx4.2kW} \end{aligned}$