Condition for maximum horsepower at a nozzle fed by a long pipeline For maximum power delivered by a nozzle at the end of a pipe, the total head supplied must be equal to how many times the head lost due to pipe friction?

Introduction / Context:
Power transmission by water through pipelines is limited by frictional head losses. There exists an optimal operating condition where the product of discharge and available head at the nozzle is maximized, yielding maximum horsepower (HP) at the outlet.

Given Data / Assumptions:

• Total head at the pipe entrance: H.
• Head loss due to friction along the pipe: h_f.
• Head available at the nozzle: H_o = H − h_f.
• Other losses (minor losses) neglected for the classical result.

Concept / Approach:

Delivered power P ∝ Q * H_o. For a given pipe, h_f ∝ V^2 and Q ∝ V (V is mean velocity). Thus P ∝ V * (H − k V^2). Maximizing this with respect to V provides the optimum balance between flow and head loss.

Step-by-Step Solution:

Verification / Alternative check:

At this optimum, the efficiency of transmission is η_max = (H − h_f)/H = (H − H/3)/H = 2/3 (a well-known benchmark value).

Why Other Options Are Wrong:

Equating H to 1×, 2×, or 4× h_f does not satisfy the optimum condition from calculus; those settings yield less outlet power.

Common Pitfalls:

Maximizing head alone or flow alone instead of their product; ignoring that friction rise with V^2 shifts the optimum.

Final Answer:

Thrice the head loss (3×)