I don’t understand something about fittings tables and equivalent lengths (Le). Since pressure drop increases with velocity, and velocity increases as pipe (or duct) diameter (D) decreases, then why should Le/D yield larger and larger pressure drop as D increases? (I am using Le = Le/D * D) Obviously, you can see, the larger D is…the larger Le becomes…
Isn’t this paradoxical? Is there some unspoken assumption that the Le/D method is just being conservative to approximate what the losses are across fittings and bends?
From Marks’ Standard Handbook For Mechanical Engineers, pg 4-23,4:
The fundamental equation as previously given on a unit weight bases[sic], assuming the pipe horizontal, is
(vbar dvbar/g) + v dp + dF = 0
The friction term dF includes not only losses due to frictional flow along the pipe but also those due to fittings, valves, etc., as well as losses occasioned by any enlargement or contraction of the pipe as, for instance, the loss occuring when a fluid passes from a pipe into a tank. For long straight pipes of uniform diameter, dF is approximately equal to 2f’ [vbar[sup]2[/sup] dL/(gD)].
<…snip…>
The coefficient of friction f is not a constant but is a function of the dimensionless expression μ/(ρ vbar a) or μv/(vbar a), which is the reciprocal of the Reynolds number. McAdams and Sherwood formulate the expression
f’ = 0.0054 + 0.375[μv/(vbar d)]
[end quotation]
(Note that vbar is mean velocity, and v is specific volume, as vCode won’t allow for an overline.)
So the answer is that both increaing the diameter simultaneously decreases the “friction”[sup]1[/sup] on the flow AND increases the Reynolds number (ratio of interia to viscous forces). What this tells us is that when going to a larger diameter, inertial effects dominate rather than “frictional”[sup]1[/sup] viscous forces, and give the counterintuitiave result that going to a substantiall larger tube creates more resistance. :eek: This makes a little more intuitive sense when you realize that the larger diameter tube will result in the fluid slowing down, which means it puts a backpressure on the fluid prior to the opening, kind of like how, how LA drivers coming back from Vegas will be jetting along just fine with four lanes of interstate, but as soon as they hit San Bernardino County and get eight lanes of traffic they’ll slow to a crawl.
It’s been a few years since I’ve worked with hydraulics, and our systems (off-road construction handling equipment) were small enough and were essentially hydrostatic that we didn’t worry about pressure drop due to flow rates, so perhaps someone can offer a little more practical advice on the matter.
Thank you, though, for the opportunity to use my Marks’ twice in one day. I don’t think it’s seen this much use since I needed to prop up my keyboard.
Stranger
Although it is standard terminology, I hate, hate, hate the use of the term “friction” with regard to fluids, as the effect is not frictional in a strict continuum mechanics sense. But, it is the accepted jargon. so,
