Pelton wheel: optimal speed ratio for maximum efficiency For a Pelton turbine operating at design head and discharge, the maximum efficiency occurs at which speed ratio (runner peripheral speed / jet velocity)?

Introduction / Context:
Impulse turbines like the Pelton wheel extract energy from a high-velocity jet striking double-cupped buckets. The runner speed relative to jet speed—the speed ratio—controls the exit relative velocity and thus the work extracted per unit jet momentum.

Given Data / Assumptions:

• Single-jet Pelton at design condition (nozzle coefficient and bucket losses typical).
• Jet deflection near 165–170 degrees with some relative speed loss in the buckets.

Concept / Approach:
The ideal analysis (neglecting losses) suggests maximum efficiency near a speed ratio of 0.5. Accounting for practical losses and incomplete flow deflection, empirical practice places the optimum slightly lower, about 0.46.

Step-by-Step Solution:
Let jet velocity be V_jet and runner peripheral speed u.Power ∝ mass flow * (V_jet - u) * (velocity change along jet direction).Differentiate efficiency with respect to u and set derivative to zero (including loss coefficients).Solution yields u / V_jet ≈ 0.46 for typical Pelton buckets.

Verification / Alternative check:
Design manuals commonly recommend 0.43–0.48 as the practical optimum; 0.46 is a standard textbook value.

Why Other Options Are Wrong:
0.26 and 0.36 under-speed the runner and waste jet kinetic energy; 0.56 and 0.66 over-speed the wheel, reducing relative jet deflection and work.

Common Pitfalls:
Using the ideal 0.5 without accounting for bucket losses; confusing specific speed trends of reaction turbines with Pelton settings.

Final Answer:
0.46