Question #222150
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?
1
Expert's answer
2021-08-02T08:40:23-0400

Explanations & Calculations


  • Those given conditions are valid only at the pump outlet where the water is ejected at the speed of 30ms1\small 30\,ms^{-1} and beyond that these calculations may not seem to be effective.
  • The power, as it is defined is = ΔworkΔtime\Large \frac{\Delta\text{work}}{\Delta\text{time}}
  • And the work done at the outlet is W=12mv2\small W= \small \frac{1}{2}mv^2 and lets assume that this work is done at a small duration of t(s)\small t(s). Then the power needed is

P=Wt=12(mt)v2\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 Avρ\small Av\rho where A is the are of the pump.
  • Then,

P=12(Avρ)v2=12Av3ρ=12(π(0.01m)2×(30ms1)3×1000kgm3)=4241.15W4.2kW\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}


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