I have a question regarding the power consumption of a saw-tooth impeller at the turbulent regime. I read many places that the power consumption of the impeller is given by: P=Po*rho*w^3*D^5 (Po – power number, rho – fluid density, w – rpm, D – impeller diameter). But if I try to derive this formula I get the following result: P=F*v (F- force, v- linear velocity) F=0.5*rho*v^2*Cd*A (Cd – drag coefficient, A – saw tooth cross section). v=w*D/2 Substituting: P=Po*rho*w^3*D^3*A. And if h is the height of the tooth and b is the width of the tooth I get: P=Po*rho*w^3*D^3*b*h. According to this in the case of saw-tooth impeller power is related to the impeller diameter by the 3rd power only and not by 5th power as I often see. Can you please clarify what is the correct answer?
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Dave Dickey Forum Moderator 320 Posts
Re: Can you help me sort out the power consumption of the impeller?27 November 2012 at 4:38pmFor geometric similarity, which is a basic assumption for the power number, both the tooth height (h) and blade width (b) are some fraction of the impeller diameter. So, if h = 0.1 D and b = 0.2 D, or any other proportions of the impeller diameter, those factors become part of the power number and your expression involves impeller diameter (D) to the 5th power. Power number for any impeller is always some function of geometry factors such as blade width, height, angle, shape, etc.