Question

There are 400kf/min of water being handled by a pump. The lift us firm 20-m deep well and the delivery velocity ks 15m/s. Find (a) the change in potential energy, (b) kinetic energy, (c) the required power of the pumping unit; g= 9.81m/s²

Answer 1

a)

$S=\frac Qv=0 .44\cdot 10^{-3}~m^2,$

$V=Sh=8.9\cdot 10^{-3}~m^3,$

$m=\rho V=8.9~kg,$

$U=mgh=1780~J,$

b)

$E=\frac{mv^2}2=1000~J,$

c)

$P=mgv=133~W.$