Dave Dickey Forum Moderator 245 Posts
Re: Agitator efficiency equation10 May 2007 at 1:29pmLet's start with the real answers to the questions:
There is no "formula" that will predict the "efficiency" of an impeller, especially for solids suspension and dissolution. With respect to the last sentence, tank and impeller geometry are very important in solids suspension. Impeller diameter (D) with respect to tank diameter (T) can be a critical variable. Small impellers (D/T less than 0.25) and large impellers (D/T more than 0.45) are much less effective than intermediate size impellers of any type.
In one respect, all impellers are 100% efficient. All of the power delivered to the fluid eventually becomes molecular motion (heat) regardless of the impeller type.
The most complete (easy-to-read) reference on solids suspension is by Corpstein, et. al., "The High-Efficiency Road to Liquid-Solid Agitation," Chemical Engineering, October 1994, pp. 138-144.
The simplest answer to the general question is that for the same amount of torque (horsepower divided by rotational speed - hp/rpm) the propeller or a hydrofoil impeller will be more effective than a turbine impeller. Remember torque, not horsepower, is the critical factor and axial flow is usually better than any mixed or radial flow. More detail is a consulting project.