What we have is basically a paddle wheel in a 14" process line. The paddles are flat bars (A=0.139 ft²) turning at 900 rpm (what we're thinking so far). With a 6.3" radius, that gives us a tip speed of 49 fps. If the drag force on a paddle equals Cd*(1/2*ñ*V²*A), then I get 675 lb per blade (Cd=2). Eight blades is 5400 lb of drag so that's over 2800 ft·lb of torque required. If torque = hp*5250/rpm, I'll need a 500 hp motor!
Is this line of reasoning right? Am I overlooking anything else?
Patrick