Hydraulic Machines — Optimum Vane Speed for Maximum Efficiency A jet of water strikes the center of a smooth curved vane that moves in the jet direction at uniform speed. For maximum efficiency of energy transfer to the moving vane, the vane (bucket) speed should be what fraction of the jet speed?

Introduction:
When a free jet impinges on a moving vane, useful work is done through the change of whirl (streamwise) velocity. There exists an optimum vane speed that maximizes the fraction of jet kinetic energy converted into work. This classical result is widely applied in Pelton buckets and other jet–vane systems.

Given Data / Assumptions:

• Jet speed V, vane speed u along jet direction.
• Jet hits the center of a smooth, lossless curved vane and exits without shock.
• Relative exit speed equals relative inlet speed (ideal case).

Concept / Approach:
Power delivered is proportional to the product of vane speed and change in whirl component. For an ideal deflection, optimizing with respect to u yields the condition u = V / 2 for maximum hydraulic efficiency of the moving surface.

Step-by-Step Solution:
Work per unit mass = u * (V_w1 − V_w2).Ideal turning with no losses and no shock gives V_w2 proportional to (V − u).Differentiate work with respect to u and set derivative to zero.The optimum occurs at u = V / 2.

Verification / Alternative check:
Substitute u = V / 2 into the hydraulic efficiency expression to obtain the known maximum value for an ideal jet–vane pair.

Why Other Options Are Wrong:
one-third, two-third, three-fourth: do not satisfy the optimum from the derivative of the power expression; they either underutilize or overrun the jet.

Common Pitfalls:
Confusing fixed-vane formulas (where u = 0) with moving-vane optimization.

Final Answer:
one-half