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Pump power

Pump input power equals hydraulic power raised by the pump divided by pump efficiency.

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Гидравлика

Формула

$$P = \frac{\rho g Q H}{\eta}$$
equation-flow Pump power balance

Block relation from flow and head to shaft power demand.

Flow and head determine hydraulic and shaft power requirements.

Обозначения

$P$
Pump shaft power, W
$\rho$
Fluid density, kg/m^3
$g$
Gravitational acceleration, m/s^2
$Q$
Volumetric flow rate, m^3/s
$H$
Total head added by the pump, m
$\eta$
Pump efficiency, -

Условия применения

  • Steady operation at a fixed operating point.
  • Known total head includes both static and friction head for intended route.
  • Efficiency is expressed as decimal fraction.

Ограничения

  • Efficiency varies with flow and head; using a single value may oversimplify.
  • Cavitation, NPSH, and mechanical losses outside the pump are not included explicitly.
  • Transient starts/stops and surge events need additional dynamic modeling.

Подробное объяснение

Hydraulic power is ρgQH; dividing by efficiency gives mechanical input needed for that delivered head.

Как пользоваться формулой

  1. Determine Q and total head H.
  2. Choose efficiency η at the design point.
  3. Compute P and add a safety margin in motor selection.

Историческая справка

Pump power relations are core to pump stations design since the early development of water-supply networks.

Пример

ρ=1000 kg/m^3, g=9.81, Q=0.05 m^3/s, H=35 m, η=0.75 => P=1000·9.81·0.05·35/0.75 = 22.9 kW.

Частая ошибка

Confusing hydraulic power with shaft power can under-size motors.

Практика

Задачи с решением

Compute required pump power

Условие. ρ=1000 kg/m^3, Q=0.03 m^3/s, H=25 m, η=0.80.

Решение. P = 1000·9.81·0.03·25/0.80 = 9.2 kW.

Ответ. P = 9.2 kW.

Find resulting head from given power

Условие. P=5 kW, ρ=980 kg/m^3, Q=0.02 m^3/s, η=0.70, g=9.81.

Решение. H = Pη/(ρ g Q) = 5000·0.7/(980·9.81·0.02) = 1.82 m.

Ответ. H ≈ 1.82 m.

Дополнительные источники

  • Stewart, R. H. (2020). Pumping Station Hydraulics.
  • Karassik, I. J., et al. (2001). Pump Handbook.

Связанные формулы

Инженерия

Pressure head

$h_p = \frac{p}{\rho g}$

Pressure head converts absolute pressure to an equivalent water column height, useful for balancing energies in flow systems.

Инженерия

Darcy–Weisbach head loss

$h_f = f\frac{L}{D_h}\frac{v^2}{2g}$

The Darcy–Weisbach equation estimates major head loss in fully developed pipe flow using friction factor and geometric ratio.

Инженерия

Bernoulli equation (basic)

$\frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2$

The basic Bernoulli equation links pressure head, velocity head, and elevation head along a streamline for frictionless, steady, incompressible flow.