Hydraulic resistance of pipelines. An interesting alternative method of calculation Chernikina A.V.

While solving another problem, partly related to the calculations of hydraulic resistance, I once again encountered the problem of a “stepwise” function when transitioning from one flow regime to another. It was precisely these “steps” that often confused my algorithm for determining the hydraulic resistance of a complex branched hydraulic system.

To study the problem, I sketched out a small example in MathCad….

Accordingly, I calculated the dimensionless coefficient of hydraulic friction resistance for five modes (Laminar, Blasius smooth-wall resistance zone,
Konakov smooth-wall resistance zone, Altshul subquadratic resistance zone, Shifrinson quadratic resistance zone)

Calculation of the dimensionless coefficient of hydraulic resistance

Well, in fact, we observe a typical picture of break points of a function and its “continuity”…

1-3 range

Of course, the task of smoothing out these transitions is not particularly difficult, but…. I remembered that somewhere I saw a formula that….

[1] Chernikin A.V. Generalization of the calculation of the hydraulic resistance coefficient of pipelines // Science and technology of hydrocarbons. M.: 1998. No. 1. pp. 21–23.

λ=0.11[(68/Re+k/D+(1904/Re)^14)/(115·(1904/Re)^10+1)]^0.25

where: k – equivalent roughness of the inner pipe wall (average height of the projections), m.

Vyacheslav Leonidovich performed verification calculations and found that the above formula is the most universal in a wide range of Reynolds numbers! The values obtained using this formula are extremely close to the values

functions λ=64/Re for a laminar flow zone in the range 10

functions λ=0.11·(68/Re+k/D)0.25 for the zone of turbulent flow at Re>4500;

in the range 1500 < Re < 4500, according to the analysis, there is a transition zone.

Let's check it in practice…

The result was pleasantly surprising…

There are no more problems with continuity, questions remain regarding the difference in values in the second range, but this, I think, is a separate topic….

And so, my algorithm worked as it should, which is what was required) I hope A.V. Chernikin’s method. will be useful for colleagues)