Question

Q 20 w.p.e:

3-20. A motor is to be used to lift water from a depth of 10 m & discharging it ground level at a speed 10 m/s. If 2 kg of water is lifted per second & the efficiency of motor is 40%, minimum power of the motor should be nearly :
(A) 1.5 HP (B) 1 HP (C) 0.67 HP (D) 2 HP

Answer 1

Dear student

$$\begin{gathered} Workdonepersecond=mgh+\frac{1}{2}m{v}^2 \\ W=2\times 9.8\times 10+\frac{1}{2}\times 2\times {10}^2=196+100=296 \\ efficiency=\frac{P_{output}}{P_{input}} \\ 0.4=\frac{296}{P} \\ P=\frac{296}{0.4}=740watt \\ 1Hp=746watt \\ sopowerisapproximately1Hp \\ Regards \end{gathered}$$