There are some complex formulas (differential equations) for calculating the actual force of a moving fluid against an obstruction. However, a simple but rough estimate that is commonly used for water is Force (lbs) = Area (ft²) x K x Velocity² (ft/sec) where K is a drag coefficient. For a rectangular flat obstruction, K can be between 1.8 and 2.0.

Considering a 4 ft diameter, 2 ft wide paddle wheel: for a single paddle 2 ft x 2 ft, that will be in contact with the water for one half revolution, (a paddle wheel that is half submerged), the maximum area would be 4 square ft. At 70 RPM the velocity of the center of the paddle would be 70 x 2 x pi = 440 ft/min = 7.3 ft/sec. Using k=2.0, the force of the water against the paddle is approximately 4 x 2.0 x 7.3 = 58.4 lb force. The radius of the center of the paddle would be 1 ft, so the torque would be 1 x 58.4 = 58.4 lb-ft.

The above estimate is for a single blade of the paddle wheel that is half submerged. To calculate an estimate for the actual wheel the number of paddles and the depth they are submerged would have to be known. Depending on the depth the wheel is submerged, trigonometric functions may be necessary to complete the calculations. Also, an estimate of the maximum torque could be arrived at geometrically.

Editor's Note: Gene Vogel provided the answer on behalf of our Motor & Drive experts Tom Bishop and Chuck Yung, technical support specialists, Electrical Apparatus Service Association.