Difficulty: Medium
Correct Answer: No
Explanation:
Introduction / Context:
Water-jet propulsion produces thrust by accelerating a mass of water astern. Propulsive efficiency compares useful power (thrust * ship speed) to the rate of kinetic energy imparted to the jet plus losses. Maximum efficiency is achieved when the velocity increase of the water over the ship speed is minimized for a given thrust (large mass flow, small velocity rise).
Given Data / Assumptions:
Concept / Approach:
Propulsive efficiency for a momentum jet can be expressed (in simplified form) as η_p = 2 * V_s / (V_s + V_j,rel), where V_s is ship speed and V_j,rel is jet speed relative to the ship. This monotonically increases as V_j,rel approaches V_s from above; it does not peak at 2 V_s. The closer the jet exit speed is to the ship speed (for sufficient mass flow to produce required thrust), the higher the efficiency.
Step-by-Step Solution:
Let relative jet speed be V_j,rel.Compute η_p trend: as V_j,rel decreases toward V_s, η_p increases toward a maximum (approaching 1 in the ideal limit).If V_j,rel = 2 V_s, η_p = 2 V_s / (3 V_s) = 2/3, which is below the maximum achievable trend.Therefore the statement claiming the maximum at twice the ship speed is false.
Verification / Alternative check:
Momentum theory of propulsors yields the same conclusion: for fixed thrust, raising jet velocity rather than mass flow reduces efficiency due to higher kinetic energy losses in the slipstream.
Why Other Options Are Wrong:
Answers asserting conditional yes are incorrect; maximum does not occur at 2 V_s under typical idealizations, regardless of depth or Froude number.
Common Pitfalls:
Confusing intake angle effects (mainly affecting losses) with the optimal jet/ship velocity ratio; assuming “faster jet is better,” which reduces efficiency.
Final Answer:
No
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