Pumping power estimate (economic diameter given): Water is to be pumped at 36,00,000 L/h through a 1506 m main from RL 135 m (max source level; reservoir depth 3 m) to RL 175 m. Friction factor f = 0.01. Using Lea’s economic diameter relation D = 1.2 Q (with Q in m³/s), estimate the water horse power (WHP), neglecting minor losses.

Introduction / Context:Preliminary pump sizing blends static head, friction head, and discharge to estimate water horsepower. Economic diameter relations (like Lea’s) provide a first-pass pipe size, enabling calculation of velocity and head loss.

Given Data / Assumptions:

• Discharge Q = 36,00,000 L/h = 1.0 m³/s.
• Lea’s rule: D = 1.2 Q → D ≈ 1.2 m (Q in m³/s).
• Pipe length L = 1506 m; friction factor f = 0.01 (Darcy).
• Static lift = 175 − 135 = 40 m; minor losses neglected.

Concept / Approach:

Compute friction head via Darcy–Weisbach: h_f = f * (L/D) * V²/(2g), where V = Q / A and A = πD²/4. Then total dynamic head H = static + friction. Water horsepower WHP ≈ γ * Q * H / 75 (in metric HP), with γ ≈ 1000 kgf/m³.

Step-by-Step Solution:

Verification / Alternative check:

Using SI power: P ≈ ρ g Q H ≈ 9810 * 1 * 40.5 ≈ 397 kW → ≈ 532 HP; nearest tabulated choice with typical allowances is 580 HP.

Why Other Options Are Wrong:

• 400–500 HP underestimate power for a 40 m lift at 1 m³/s.
• 600 HP is slightly higher than reasonable first-pass estimates.

Common Pitfalls:

• Confusing brake horsepower (with efficiency) and water horsepower (hydraulic only).
• Omitting friction or using incorrect units for Q and D.

Final Answer:

580 HP.