Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 102: 2-71E

$\eta_{pump}=78.55%$

Here we are working with potential energy: The potential power is: $P_{p}=Q\Delta P$ The requirement of power is: $P_{p}=(15\frac{ft^3}{s}*1.2psi)*(\frac{1Btu}{5.404psift^3})*(\frac{1hp}{0.7068\frac{Btu}{s}})=4.713hp$ As the consume is $6hp$, the efficiency of the pump is: $\eta_{pump}=\frac{4.713hp}{6hp}=0.7855$ $\eta_{pump}=78.55%$