Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 108: 2-132

a) $8.3\frac{L}{s}$

Here we are working with potential energy: The potential work is: $W_{p}=mgh$ And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$ Solving for $Q$: $Q=\frac{P_{p}}{\rho gh}=\frac{2000W}{(0.820\frac{kg}{L}*\frac{1000L}{1m^3})*9.81\frac{m}{s^2}30m}=0.0083\frac{m^3}{s}*\frac{1000L}{1m^3}=8.3\frac{L}{s}$