For a centrifugal pump handling water under steady conditions, define the overall efficiency in practical engineering terms. Select the correct ratio representing overall efficiency.

Introduction:
Overall efficiency of a centrifugal pump quantifies how effectively input shaft power from the prime mover is converted into useful hydraulic power delivered to the fluid at the delivery outlet. It captures all internal losses collectively.

Given Data / Assumptions:

• Steady operation at a given duty point.
• Definition of useful output as water horsepower at delivery: ρ * g * Q * H_d.
• Input power measured at the pump shaft (after mechanical transmission losses).

Concept / Approach:
Overall efficiency η_o = (useful hydraulic output) / (input shaft power) = (ρ * g * Q * H_d) / P_in. It equals the product of volumetric, hydraulic, and mechanical efficiencies, but the simplest definition is output-over-input referred to the prime mover power supplied to the pump shaft.

Step-by-Step Solution:
1) Compute hydraulic output: P_out = ρ * g * Q * H_d.2) Measure input: P_in at shaft.3) Overall efficiency: η_o = P_out / P_in.4) Recognize that η_o = η_v * η_h * η_m, but the ratio definition remains output/input.

Verification / Alternative check:
Laboratory tests plotting pump curves routinely derive η_o from measured head–discharge and shaft power.

Why Other Options Are Wrong:
Energy at impeller vs energy supplied: describes manometric or hydraulic sub-efficiencies, not overall.

Energy supplied to pump divided by energy at impeller: inverted meaning.

Manometric head ratio per kN: partial metric, not overall definition.

Product of only two sub-efficiencies: omits mechanical losses.

Common Pitfalls:
Confusing overall efficiency with manometric or volumetric efficiency; ignoring mechanical losses in bearings and seals.

Final Answer:
actual workdone by the pump to the energy supplied to the pump by the prime mover