Impulse Turbines — Maximum Hydraulic Efficiency What is the maximum hydraulic efficiency of an ideal impulse turbine (Pelton-type) in terms of the blade outlet angle phi (measured between relative exit velocity and the direction opposite to wheel motion)?

Introduction:
Hydraulic efficiency in an impulse turbine measures how much of the jet's kinetic energy is converted into useful work on the runner. The outlet blade angle controls the fraction of velocity effectively reversed, thus governing efficiency.

Given Data / Assumptions:

• Ideal conditions: smooth buckets, no shock at entry, no losses.
• Jet speed V, wheel speed u, outlet relative velocity leaves the bucket at angle phi to the negative direction of motion.
• Single jet; standard Pelton geometry.

Concept / Approach:
For maximum hydraulic efficiency, the optimal speed ratio u / V is 1 / 2. Substituting this into the efficiency expression based on whirl velocity change leads to a compact formula depending only on phi. As phi decreases (sharper deflection), efficiency approaches unity, but practical values use phi around 15 degrees to avoid jet interference.

Step-by-Step Solution:
Work per unit mass = Vw1 * u - Vw2 * u with ideal deflection relations.Set u / V = 1 / 2 for optimal speed.Simplify hydraulic efficiency to eta_h(max) = (1 + cos phi) / 2.

Verification / Alternative check:
For phi = 0 (complete reversal), eta_h(max) = 1; for phi = 90 degrees (no reversal), eta_h(max) = 0, matching physical intuition.

Why Other Options Are Wrong:
cos phi: overestimates or underestimates depending on phi; not derived from optimal speed ratio.1 - cos phi and (1 - cos phi) / 2: correspond to different energy partitions, not the Pelton optimum.

Common Pitfalls:
Mixing up relative and absolute velocity angles; phi is defined at outlet relative to the bucket.

Final Answer:
eta_h(max) = (1 + cos phi) / 2