Difficulty: Easy
Correct Answer: energy available at the impeller to the energy supplied to the pump by the prime mover
Explanation:
Introduction:
Pump efficiencies are categorized to isolate different loss mechanisms. Mechanical efficiency focuses on losses between the motor shaft and the impeller (bearings, seals, disc friction), distinct from hydraulic and volumetric effects inside the flow passages.
Given Data / Assumptions:
Concept / Approach:
By definition, mechanical efficiency eta_m = (power at impeller) / (power at shaft). It quantifies how effectively the mechanical drive delivers power to the impeller after subtracting mechanical losses. Overall efficiency multiplies mechanical and hydraulic efficiencies (and volumetric, if defined separately).
Step-by-Step Solution:
Identify numerator: impeller (available) power.Identify denominator: shaft (input) power from the prime mover.Form ratio: eta_m = P_impeller / P_shaft.
Verification / Alternative check:
Test data typically show eta_m > eta_h (hydraulic) for well-built pumps; overall eta_o = eta_m * eta_h * eta_v.
Why Other Options Are Wrong:
actual workdone / input: that is overall efficiency (if “actual workdone” is water output power).energy supplied to pump / energy at impeller: reciprocal of mechanical efficiency.manometric head / impeller energy per kN: relates to manometric (hydraulic) efficiency, not mechanical.
Common Pitfalls:
Mixing hydraulic efficiency with mechanical efficiency; they capture different losses.
Final Answer:
energy available at the impeller to the energy supplied to the pump by the prime mover
Discussion & Comments