Equivalent Length of a Fitting

Leq=KDfL_{eq} = \frac{K D}{f}

Worked example: 6 m equivalent on 100 mm pipe at f 0.025 → K = 1.5 — press Try an example to run it live, then adjust anything.

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Equivalent Length of a Fitting explained

KDfLeq

Set Darcy–Weisbach equal to the minor-loss equation — f(L/D)(v²/2g) = K(v²/2g) — and the velocity heads cancel, leaving Leq=KD/fL_{\mathrm{eq}} = KD/f. The point is bookkeeping: rather than tracking two different loss equations, a designer converts every valve and fitting into a phantom length of pipe, adds it to the measured run, and computes one friction loss for the whole circuit. A 90° elbow (K = 0.9) on 4 in pipe with f = 0.02 is worth 0.9 × 0.333/0.02 = 15 ft of straight pipe.

Because f cancels out of nothing here, the equivalent length depends on the pipe's friction factor — which is why the old rule "an elbow equals 30 diameters" is only true near f = 0.03. On smooth plastic with f = 0.018 the same elbow is worth 50 diameters. Fire-protection and refrigeration codes finesse this by publishing fixed equivalent-length tables for a nominated schedule and material; those tables are fine inside their intended context and misleading outside it.

Equivalent Length of a Fitting formula

Leq=KDfL_{eq} = \frac{K D}{f}
Where
  • LeqL_{eq}= Equivalent length (m)
  • KK= Resistance coefficient
  • DD= Inside diameter (mm)
  • ff= Darcy friction factor

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