Pump Power (HP)
Compute hydraulic and electric power (HP) of a pump given flow, head and efficiency.
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Hydraulic pump power in HP
To estimate the shaft power a water pump needs, use P (HP) = (Q · H) / (75 · η), where Q is in L/s, H (the manometric head) is in m and η stands for the pump efficiency, which sits around 0.6–0.8 on centrifugal pumps. That 75 is the factor that turns kgf·m/s into HP for water at 1000 kgf/m³. Take Q = 2 L/s, H = 20 m and η = 0.65: P = (2 × 20) / (75 × 0.65) ≈ 0.82 HP, which means you'd spec a 1 HP commercial motor.
Applications
This comes up when you size booster pumps, deep-well pumps, swimming-pool pumps, rural irrigation systems and reservoir transfer pumps on projects that have to meet NBR 12214 and NBR 5626. Round up to the next standard motor size (1/2, 3/4, 1, 1.5, 2 HP…), and keep in mind that H covers the geometric head together with the friction losses along the piping.
FAQ
Why divide by efficiency? No pump turns 100 % of its shaft power into hydraulic power, and η is there to account for the mechanical and hydraulic losses.
What is the manometric head? It's the geometric head plus the pressure you lose across the pipes, fittings and valves.
1 HP equals how many kW? Roughly 0.7457 kW (and 1 cv ≈ 0.7355 kW).
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