Optimal area ratio for maximum power transmission through a nozzle on a long pipe Given a pipe of length L, diameter D, and Darcy friction factor f, what is the optimal area ratio (a/A) of nozzle area to pipe area for maximum jet power?

Introduction / Context:When a nozzle is attached to a long pipeline, increasing the nozzle diameter raises discharge but also increases pipe velocity and friction loss, reducing available head at the nozzle. There is a specific area ratio that maximizes the transmitted power P = ρ Q V_jet^2 / 2.

Given Data / Assumptions:

• Steady, incompressible flow.
• Single straight pipe of length L and diameter D with Darcy friction factor f.
• Negligible minor losses except friction and nozzle conversion to jet.

Concept / Approach:Balancing available head with friction loss in the pipe yields a relation between pipe velocity and nozzle area. Maximizing jet power with respect to nozzle area leads to a classical optimum: (8 f L / D) * (a/A)^2 = 1, hence a/A = 1 / √(8 f L / D).

Step-by-Step Solution:

Verification / Alternative check:This optimum is equivalent to the condition that the pipe friction head equals one-third of the supply head at maximum power for the end-nozzle configuration.

Why Other Options Are Wrong:Other forms invert or misplace parameters, leading to non-optimal or dimensionally inconsistent relationships. The presence of the 8 f L / D group is critical and must be in the denominator under the square root.

Common Pitfalls:Confusing Darcy and Fanning friction factors (factor of 4 difference); ignoring minor losses when they are significant, which would shift the exact optimum.

Final Answer:a/A = 1 / √(8 f L / D)