Power requirement for pumping: motor horsepower expression Let ω be the unit weight of water (N/m^3), Q the discharge (m^3/s), H the total head (m), and η the overall pump–motor efficiency (decimal). What is the required motor horsepower (HP)?

Introduction / Context:
Determining pump power is central to water-supply and irrigation design. The hydraulic power equals the product of unit weight, discharge, and head. Accounting for efficiency and converting watts to horsepower yields the motor rating required for continuous operation at the design point.

Given Data / Assumptions:

• ω in N/m^3, Q in m^3/s, H in m.
• η is the overall efficiency (pump + motor + transmission) expressed as a fraction (e.g., 0.7).
• 1 horsepower (mechanical) ≈ 746 W.

Concept / Approach:

Hydraulic power P_h = ω * Q * H in watts. The shaft power required is P_shaft = P_h / η. Converting to horsepower uses division by 746.

Step-by-Step Solution:

Verification / Alternative check:

Using example values (ω = 9810 N/m^3, Q = 0.05 m^3/s, H = 20 m, η = 0.7) gives HP ≈ (98100.0520)/(0.7*746) ≈ 18.8 HP, consistent with handbook estimates.

Why Other Options Are Wrong:

• (ω * Q * H)/(η * 75): 75 is for metric horsepower (kgf·m/s); the problem defines ω in N/m^3 and requests HP, so 746 is appropriate.
• Multiplying by η: Would underestimate the motor size; efficiency losses must be divided out.
• Omitting η: Ignores real losses.
• Missing ω: Loses unit-weight term, not dimensionally correct.

Common Pitfalls:

• Confusing N/m^3 with kgf/m^3; ensure consistent unit system.
• Using pump efficiency only and forgetting motor/drive losses.

Final Answer:

(ω * Q * H) / (η * 746).