View Single Post 11-17-2004, 03:26 PM #57 Cathar Thermophile   Join Date: Sep 2002 Location: Melbourne, Australia Posts: 2,538 Hmmm, maybe some of you more engineering types could help me out to understand as I may have the following wrong (being self-educated and all on this stuff): P=(Q/K)^2 seems to me to be simplification of Bernoulli's equation. However if we consider viscous head, being the energy that gets lost as heat (and absorbed into the liquid itself), then for turbulent flow the friction co-efficient of the viscous head term will fall away as the Reynold's number increases (which is proportional to the flow rate). So what we then have is an effect where P=(Q/K)^2 isn't really followed cleanly. For the more restrictive blocks I intuitively would have thought that they would deviate even further from the classic P=(Q/K)^2 equation. Now, quite true, Bernoulli's viscous equation consists of two primary terms, being the constant head loss due to inviscid flow (which is proportional to the flow rate squared), and the head loss due to viscous flow. The relative effect of viscous flow term diminishes as the flow rate increases, meaning that we end up with a curve that doesn't fit P=(Q/K)^2 any more. I threw out P = Q ^ 1.85 as a really quick and dirty approximation of this effect, perhaps only valid for within 0.5 to 2.0x the data point given at somewhere like OC.com. I measured various data points from 2.0 - 5.0LPM for the Storm/G4, and it didn't seem to me to quite follow a simple P=(Q/K)^2 curve. That could also just be due to measurement error too. Or perhaps one of you kind lads could tell me if viscous flow effects don't even apply here?  