Difficulty: Easy
Correct Answer: work done on the wheel to the energy (head of water) actually supplied to the turbine
Explanation:
Introduction:
Hydraulic efficiency is a key performance indicator for water turbines. It compares how effectively the hydraulic energy delivered by the water is converted into useful work on the runner (wheel). For reaction turbines, both pressure and kinetic energy changes occur across the runner, so the proper definition must reference the energy actually available at the runner inlet.
Given Data / Assumptions:
Concept / Approach:
Let E_in be the specific hydraulic energy available at the runner entry plane (including appropriate pressure head and velocity head components relative to exit). The useful hydraulic work rate on the runner equals the runner torque times angular speed, which is the energy transfer predicted by Euler’s turbine equation. Hydraulic efficiency is therefore η_h = (work done on runner) / (hydraulic energy supplied to runner). This definition excludes mechanical losses (bearings, seals) and volumetric slip; those are captured by other efficiencies (mechanical, volumetric, overall).
Step-by-Step Solution:
Verification / Alternative check:
In textbooks, “hydraulic efficiency” explicitly omits mechanical losses and leakage; hence it cannot be “shaft power / water power at penstock.” That latter ratio corresponds to overall efficiency when combined with volumetric losses.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing hydraulic, mechanical, and overall efficiencies; using penstock head instead of runner-inlet available head; ignoring reaction effects through the runner.
Final Answer:
Discussion & Comments