Find the useful power used in pumping `3425 m^(3)` of water per hour from a well 8 m deep to the surface, supposing `40%` of the horse power during pumping is wasted. What is the horse power of the engine ?

Explanation

Here `V = 342 m^(3) , d = 10^(3) kg m^(-3) , h = 8m,`
`g = 9 . 8 ms^(-2) , t = 1 " hour " = 60 xx 60 s`
Utilised power `= (mgh)/(t)`
`60% P = (mph)/(t) = (Vdgh)/(t)`
`= (3425 xx 10^(3) xx 9.8 xx 8)/(60 xx 60)`
`(60)/(100) xx P = (3425 xx 10^(3) xx 9.8 xx 8)/( 60 xx 60)`
`p = (3425 xx 10^(3) xx 9.8 xx 8 xx 100)/( 60 xx 60 xx 60) = 1243.14` watts
= 1.666 h.p [ 1hp = 746 watt]