Power transmission through a pipeline: at what fraction of the supplied head should the friction head loss be for maximum power at the outlet?

Introduction / Context:
When fluid power is delivered through a long pipe to a turbine or nozzle, frictional head loss reduces the net head at the outlet. The power delivered P is maximized at a particular balance between losses and available head.

Given Data / Assumptions:

• Total supply head = H.
• Friction head loss in pipe = h_f; outlet head = H − h_f.
• Flow rate Q depends on head and losses (for a given pipe, Q ∝ √(H − h_f) when losses scale with Q^2).

Concept / Approach:

Power at outlet is P = ρ g Q (H − h_f). For turbulent pipe flow with Darcy–Weisbach losses, h_f ∝ Q^2. Writing h_f = k Q^2, we obtain P = ρ g (H Q − k Q^3). Differentiating with respect to Q and setting dP/dQ = 0 gives H − 3 k Q^2 = 0 ⇒ h_f = k Q^2 = H/3. Thus, maximum power occurs when friction loss is one-third of the total head.

Step-by-Step Solution:

Verification / Alternative check:

At optimum, outlet head is 2H/3, a known result quoted in standard hydraulics texts.

Why Other Options Are Wrong:

(a) and (c) do not satisfy the optimization; (d) leaves zero usable head; (e) is impossible in real pipes.

Common Pitfalls:

Maximizing efficiency rather than absolute power; overlooking that maximizing P allows significant but optimal loss.

Final Answer:

one-third of the total head supplied