Equivalent single pipe for a compound (series) pipe system Three pipes in series have diameters d1, d2, d3 and lengths l1, l2, l3. They are to be replaced by one equivalent pipe of uniform diameter d and total length l = l1 + l2 + l3, carrying the same discharge with the same total head loss (same roughness, same f). The equivalent diameter d is given by:

Introduction / Context:
Designers often replace a series of pipes of different diameters with a single “equivalent” pipe to simplify calculations. For turbulent flow modeled with the Darcy–Weisbach relation and equal friction factor, the total head loss must be preserved at a given discharge to determine the equivalent diameter.

Given Data / Assumptions:

• Pipes in series: lengths l1, l2, l3; diameters d1, d2, d3.
• Equivalent single pipe: length l = l1 + l2 + l3; diameter d.
• Same discharge Q through all series segments.
• Same friction factor f and same fluid; use Darcy–Weisbach head loss.

Concept / Approach:
For a segment i: h_fi = f * (li/di) * (vi^2 / (2g)). With Q constant, vi = 4Q/(π di^2). Thus h_fi ∝ li * Q^2 / di^5. Total head loss is the sum over segments. The equivalent pipe must satisfy h_f(eq) = Σ h_fi at the same Q, yielding the classical 1/d^5 relationship.

Step-by-Step Solution:
Write h_f(total) = K * [ (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5) ] * Q^2, where K collects constants.For the equivalent pipe: h_f(eq) = K * (l / d^5) * Q^2.Equate: l / d^5 = (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5).Solve for d: d = [ l / ( Σ li / di^5 ) ]^(1/5).

Verification / Alternative check:
Dimensional reasoning: as diameters increase, head loss reduces rapidly (power 5); combining smaller pipes increases the equivalent resistance, hence decreases equivalent d, consistent with the derived formula.

Why Other Options Are Wrong:

• Power 4 relationships (options with d^4) arise in laminar (Hagen–Poiseuille) conditions, not in this Darcy–Weisbach turbulent-equivalent assumption.
• Option using weighted sum l1 d1^5 etc. in the numerator inverts the correct dependence.
• Although option (a) is an intermediate relation, the asked “size of equivalent pipe” requires the explicit expression for d, given in option (c).

Common Pitfalls:

• Mixing laminar and turbulent formulas; ensure consistency of friction model and exponent.

Final Answer:
d = [ l / ( (l1 / d1^5) + (l2 / d2^5) + (l3 / d3^5) ) ]^(1/5)