Equivalent parallel pipes — replacing one pipe by three identical pipes A single pipe of length L and diameter D is to be replaced by three pipes of the same material, each of length L and equal diameter d (d < D), connected in parallel between the same reservoirs. To carry the same total discharge under the same head loss, the relation between d and D is:

Introduction / Context:
When multiple equal pipes are laid in parallel between the same heads, each pipe experiences the same head loss, and the total discharge is the sum of individual discharges. For turbulent flow using Darcy–Weisbach with the same roughness and length, discharge scales with diameter to the power 2.5 for a given head loss. This yields a clean diameter relation.

Given Data / Assumptions:

• Three identical pipes in parallel, each length L, diameter d.
• Original single pipe: length L, diameter D.
• Same material → same friction factor regime.
• Same head loss between reservoirs.

Concept / Approach:
For Darcy–Weisbach, with head loss h_f fixed and length L, velocity V ∝ sqrt(h_f * D / (f L)). Discharge Q = A V ∝ (D^2) * sqrt(D) = D^(2.5). For n equal parallel pipes at the same head loss, Q_total = n * k * d^(2.5), where k is a constant for given h_f and L. Equate to the original single-pipe discharge k * D^(2.5).

Step-by-Step Solution:
1) Write Q_single = k * D^(2.5).2) Write Q_parallel = 3 * k * d^(2.5).3) Equate: D^(2.5) = 3 * d^(2.5).4) Solve: d = D / 3^(2/5).

Verification / Alternative check:
Numerically, 3^(2/5) ≈ 1.552 → d ≈ 0.644 D, a plausible reduction when splitting into three parallel branches.

Why Other Options Are Wrong:
Exponents 1/4, 1/2, 3/5 do not arise from the D^(2.5) scaling; they correspond to incorrect flow–diameter relationships.

Common Pitfalls:
Using laminar (D^4) or Hazen–Williams exponents indiscriminately; forgetting that each parallel pipe has the same length and head loss.

Final Answer:
d = D / 3^(2/5)