Required horsepower (hp) to pump 1 cubic foot of water per minute (ft3/min) with efficiency 85% - is indicated in the diagram below:
According the diagram above 1 hp is required to lift 1 ft3/min of water 600 ft. Required power to pump 10 ft3/min can be calculated as
(10 ft3/min) (1 hp) / (1 ft3/min) = 10 hp
The brake horsepower is the amount of real horsepower going to the pump, not the horsepower used by the motor. In the metric system kilowatts (kW) is used.
Due to hydraulic, mechanical and volumetric losses in a pump or turbine the actual horsepower available for work on or from the fluid is less than the total horsepower supplied.
The brake horse power - bhp - for a pump or fan can be expressed as:
Pbhp = ( γ Q h / 33000 ) / η (2)
Pbhp = brake horse power (horsepower, hp)
Q = volume flow rate (ft3/min, cfm)
η = overall efficiency
The brake horse power - bhp - for a turbine can be expressed as:
Pbhp = η ( γ Q h / 33000 ) (2b)
The input horsepower to the electrical motor for a pump or fan can be expressed as:
Php_el = Pbhp /ηe (3)
Php_el = (m h g / 33000) / (ηηe) (3b)
Php_el = input power to the electrical motor
ηe = mechanical efficiency of the electrical motor
Horsepower can be converted to other common units as:
Transforming horsepower to kW:
PkW = 0.746 Php (4)
PkW = power (kW)
Together with the equations above it's possible to express (4) in many common combinations - such as:
PkW = 0.746 (m h / 33000) / η ηe (5)
The horsepower required to pump 50 lbm/min water a head of 10 ft with a pump with overall efficiency 0.7 - can be calculated with eq. (3) as
Php = ((50 lbm/min)(10 ft) (32 ft/s2) / 33000) / 0.8
= 0.6 hp
The power in kW can be calculated as
PkW = 0.746 (0.6 hp)
= 0.45 kW
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