# Fluids Question for a M.E. **URL:** \ **Category:** Factual Questions\ **Created:** [March 4, 2005, 11:34am UTC](https://boards.straightdope.com/t/fluids-question-for-a-m-e/292884 "2005-03-04T11:34:29Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [March 4, 2005, 11:34am UTC](https://boards.straightdope.com/t/fluids-question-for-a-m-e/292884/1 "2005-03-04T11:34:29Z")
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? Thanks, - Jinx 😕 ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [March 4, 2005, 10:51pm UTC](https://boards.straightdope.com/t/fluids-question-for-a-m-e/292884/2 "2005-03-04T22:51:35Z")
Hmm…I guess I stumped the best of 'em! We engineers are trained to use the Le/D method, but I never stopped to think about why it _shouldn’t_ work! - Jinx 🆒 ---
**Author:** ![chaparralv8](https://avatars.discourse-cdn.com/v4/letter/c/e274bd/32.png) [@chaparralv8](https://boards.straightdope.com/u/chaparralv8)\ **Post date:** [March 4, 2005, 11:47pm UTC](https://boards.straightdope.com/t/fluids-question-for-a-m-e/292884/3 "2005-03-04T23:47:03Z")
It’s pressure LOSS you’re concerned about here. What happens when the pipe opens back out? ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [March 4, 2005, 11:55pm UTC](https://boards.straightdope.com/t/fluids-question-for-a-m-e/292884/4 "2005-03-04T23:55:32Z")
> [@Jinx](#): > > 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? > > Thanks, > > - Jinx 😕 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 2_f’_ [_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, ☹