Hydraulic power from a nozzle: if the total head at a pipe nozzle is 37.5 m and the discharge is 1 m^3/s, estimate the power output (in horsepower).

Introduction / Context:
Power available from a jet/nozzle is the rate at which potential head is converted to kinetic energy. For water systems, hydraulic power can be computed directly from discharge and head, then converted to horsepower (HP) for conventional rating comparisons.

Given Data / Assumptions:

• Total head H = 37.5 m of water at the nozzle.
• Discharge Q = 1 m^3/s (1 cumec).
• Water density rho ≈ 1000 kg/m^3; g ≈ 9.81 m/s^2.
• Neglect losses at the nozzle (ideal conversion).
• Use 1 HP ≈ 735.5 W (metric horsepower) for comparison in many civil/mech contexts.

Concept / Approach:

Hydraulic power P equals rho * g * Q * H. This is the mechanical power theoretically obtainable from the head and discharge.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Using 1 HP (imperial) = 746 W gives 367,875/746 ≈ 493 HP, still closest to 500 H.P. among the given choices.

Why Other Options Are Wrong:

400 and 450 H.P. underestimate the computed power; 550 H.P. overestimates. 500 H.P. is the nearest correct estimate with standard conversions.

Common Pitfalls (misconceptions, mistakes):

Mixing metric and imperial horsepower without noting small differences; omitting density or g; forgetting that losses reduce usable power (here neglected by assumption).

Final Answer:

500 H.P.