Equivalent diameter for multiple identical parallel pipes: To replace one pipe of diameter D by n parallel pipes each of diameter d (same length and headloss), the relationship is best given by:

Introduction / Context:
Designers often split a large pipeline into several smaller parallel lines to reduce cost or to ease installation. An equivalent diameter relation helps preserve total discharge under the same headloss and length when using the Darcy–Weisbach equation in the turbulent regime with roughly constant friction factor.

Given Data / Assumptions:

• Same length L for all pipes.
• Same headloss h_f for the single pipe and the set of n parallel pipes.
• Turbulent flow with approximately constant friction factor f (fully rough assumption) so f does not vary strongly with Reynolds number.

Concept / Approach:
For Darcy–Weisbach, h_f ∝ (L/D) * V^2. For a given h_f and L, V ∝ √D. Discharge Q = (π D^2/4) V ∝ D^2 * √D = D^(5/2). For n identical parallel pipes, total Q is the sum of n equal branch discharges, each proportional to d^(5/2). Hence D^(5/2) = n d^(5/2).

Step-by-Step Solution:
Single pipe capacity: Q_single ∝ D^(5/2).Each branch: q_branch ∝ d^(5/2).Total with n branches: Q_total = n q_branch ∝ n d^(5/2).Equate capacities: D^(5/2) = n d^(5/2).

Verification / Alternative check:
If laminar flow applied, Q ∝ D^4, yielding D^4 = n d^4 (option (c)); but for most water-distribution designs (turbulent), the 5/2 power rule is used—validating option (b).

Why Other Options Are Wrong:
D^2 and D^3 relations do not follow from Darcy–Weisbach with constant f.D^4 applies to laminar Hagen–Poiseuille flow, not typical design turbulence.

Common Pitfalls:

• Using laminar equivalence in turbulent contexts.
• Ignoring unequal distribution if branch minor losses differ (not assumed here).

Final Answer:
D^(5/2) = n d^(5/2)