For maximum power transmission through a nozzle at the end of a pipeline (with pipeline friction head loss h_f and available head H at the pipe entrance), the condition for maximum power is:

Introduction / Context:
When a nozzle is supplied by a long pipeline from a reservoir, the jet power at the nozzle depends on the jet velocity (hence residual head) and on losses in the pipe. There exists an optimal balance between velocity gain and friction loss that maximizes power delivered by the jet.

Given Data / Assumptions:

• Total available head at pipe entrance = H.
• Head loss in pipeline due to friction = h_f (minor losses neglected for simplicity).
• Nozzle converts remaining head (H − h_f) to velocity head at the exit.

Concept / Approach:
Jet power P ∝ ρ g Q (H − h_f). For a given pipe, discharge Q increases with head difference but also increases h_f (typically h_f ∝ Q^2 for turbulent flow with constant f). Maximizing P with respect to Q (or h_f) yields the classical result h_f = H/3.

Step-by-Step Solution:
Let h_f = K Q^2 (turbulent, Darcy–Weisbach with f constant → K lumps geometry and properties).Power at nozzle: P = ρ g Q (H − K Q^2).Differentiate: dP/dQ = ρ g (H − 3K Q^2) = 0 → K Q^2 = H/3 → h_f = K Q^2 = H/3.Thus maximum power occurs when pipe friction loss equals one-third of total head.

Verification / Alternative check:
Substituting h_f = H/3 gives nozzle head = 2H/3 and shows that one-third of H is spent in friction while two-thirds appears as jet head, a well-known design proportion for penstocks and fire nozzles (neglecting minors).

Why Other Options Are Wrong:
H/2 or H/4 do not satisfy the optimum condition from differentiation.2H/3 assigns excessive friction, leaving insufficient energy for the jet.

Common Pitfalls:

• Including minor losses without adjusting the optimum; more complete models change the constant slightly but keep the one-third rule as a strong guideline.
• Maximizing velocity rather than power; the objectives differ.

Final Answer:
h_f = H/3