Power required to drive a centrifugal pump — correct formula Select the correct expression for input power P (watts) in terms of Hm (manometric head, m), w (specific weight, N/m^3), Q (discharge, m^3/s), and ηo (overall efficiency).

Introduction:
To size a motor for a centrifugal pump, we must relate the hydraulic power imparted to the fluid to the electrical/mechanical input power, accounting for overall efficiency (volumetric, hydraulic, mechanical losses lumped together).

Given Data / Assumptions:

• Incompressible fluid (water) with specific weight w (N/m^3).
• Flow rate Q (m^3/s) against manometric head Hm (m).
• Overall efficiency ηo includes all losses.

Concept / Approach:
Hydraulic output power (water power) equals w * Q * Hm. Since not all input power is converted to useful hydraulic power, input power must be higher by a factor 1/ηo. Thus P_input = (w * Q * Hm) / ηo.

Step-by-Step Solution:

Verification / Alternative check:
Dimensionally, w (N/m^3) * Q (m^3/s) * Hm (m) gives Nm/s = W, confirming correctness.

Why Other Options Are Wrong:

• wQHm: omits losses; gives water power, not input.
• (wQHm)ηo: would be less than water power, nonphysical for ηo < 1.
• (ρQHm)/ηo: uses density instead of specific weight; missing factor g.
• (wQ)/(Hmηo): incorrect placement of head.

Common Pitfalls:
Mixing ρ with w; forgetting to divide by overall efficiency; confusing manometric head with static head only.

Final Answer: