Empirical sizing – Lea’s formula for economical pipe diameter for pumped flow For a discharge Q (in m^3/s) to be pumped in a transmission main, Lea suggested an empirical relation for the economical internal diameter D (in m). Which of the following expressions represents Lea’s formula?

Introduction / Context:Selecting an economical diameter for a pumped main balances capital cost (larger pipe is costlier) against operating cost (smaller pipe increases headloss and pumping energy). Empirical formulas like Lea’s give a quick first estimate before refined hydraulic and life-cycle analyses.

Given Data / Assumptions:

• D in metres, Q in m^3/s (cumecs).
• Continuous pumped transmission (not intermittent distribution line sizing).
• Typical cost and energy price assumptions embedded in the empirical constant.

Concept / Approach:For turbulent flow, headloss varies roughly with Q^2/D^5 (for Darcy–Weisbach with fixed roughness). Pumping power therefore rises rapidly as diameter decreases, while pipe capital cost increases with diameter. Minimizing total annualized cost yields a diameter proportional to the square root of discharge: D ∝ Q^0.5. Lea’s constant 1.22 reflects specific economic assumptions in traditional practice.

Step-by-Step Solution:Identify the scaling → economical D varies as Q^0.5 (square root dependence).Among the options, only one includes Q^0.5 → D = 1.22 * Q^0.5.Select the expression that matches Lea’s suggested value.

Verification / Alternative check:Back-of-the-envelope checks: a tenfold increase in discharge increases the economical diameter by about √10 ≈ 3.16 times, which aligns with energy–capital trade-offs found in practice.

Why Other Options Are Wrong:

• Linear relations (D ∝ Q) significantly oversize small flows and undersize large flows.
• Different constants multiplying Q (without exponent) ignore the correct scaling law.

Common Pitfalls:Treating empirical results as final; after obtaining a first estimate, perform full headloss, NPSH, transients, and life-cycle cost analyses.

Final Answer:D = 1.22 * Q^0.5