We have an idea in mind for a mixer/paddle that would be inserted in a process stream. Any ideas on sizing the motor to drive it? Any ideas on figuring the hydraulic loads or drag? We're at the very beginning of this.
Motor horsepower will be a function of spindle rpm and torque on the mixing blade shaft. In turn torque on the shaft will be a function of the resisting force on the mixing blade from the fluid. The resisting force will be a function of fluid viscocity,fluid speed relative to mixing blade speed and mass flow displaced by mixing blade.
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?
I did not carried out your horsepower calculation of 500HP but for 900 rpm(which appears fast), I would say that your are in the ball park. The 500 horsepower is at the end of the shaft and does not take into account Hp losses thru gear train.
What are you mixing and have you done any shear rate evaluation of fluid.
It's an inline mixer mixing pulp stock (8% by weight) and chemicals. We have other mixers running at 1800 rpm. We're experimenting with some new geometries.