Power transmission by pipeline — the maximum efficiency attainable for transmitting power through a pipeline occurs at what efficiency value?

Introduction / Context:
When power is transmitted hydraulically through a long pipeline to a turbine/nozzle, friction losses increase with velocity. There exists an operating condition that maximizes delivered power for a given supply head, which implies a specific efficiency at the optimum.

• Total supply head H (fixed).
• Head loss due to friction h_f increases with velocity squared.
• Delivered head at outlet H_d = H − h_f.
• Discharge Q ∝ velocity; delivered power P ∝ Q * H_d.

Concept / Approach:

Write P ∝ V * (H − k * V^2), where k aggregates friction factors. Differentiate dP/dV = 0 to find the optimum. The result shows h_f = H/3 at maximum power. Therefore efficiency η = H_d / H = (H − H/3) / H = 2/3 = 66.67%.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Substitute back into P(V) to confirm a maximum (second derivative negative). Classic textbooks report the same optimum: loss = one-third of head, efficiency = two-thirds.

Why Other Options Are Wrong:

25%, 33.3%, and 50% correspond to non-optimal operating points; 75% would require h_f < H/4, which does not satisfy the maximum-power condition.

Common Pitfalls (misconceptions, mistakes):

Maximizing efficiency instead of power; forgetting friction grows with V^2, not V.

Final Answer:

66.67%