Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 106: 2-119E

$P_{p}=11.32kW$

Here we are working with potential energy: The potential work is: $W_{p}=mgh$ And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$ The potential power needed is: $P_{p}=1000\frac{kg}{m^3}*(200\frac{gal}{min}*\frac{1m^3}{264.172gal}*\frac{1min}{60s})*9.81\frac{m}{s^2}*[(200ft-(-100ft))*\frac{1m}{3.28ft}]$ $P_{p}=11.32kW$