Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 99: 2-31E

$\tau=787.82lbf*ft$

First we calculate the angular velocity: $\omega=3000rpm*(\frac{2\pi rad}{1rev})*(\frac{1min}{60s})=314.16\frac{rad}{s}$ And the power is: $450hp*(\frac{550\frac{lb*ft}{s}}{1hp})=247500\frac{lb*ft}{s}$ Now the torque is: $\tau=\frac{W}{\omega}=\frac{247500\frac{lb*ft}{s}}{314.16\frac{rad}{s}}=787.82lbf*ft$