The energy transmitted to the water by bomba is called water power - and can be calculated as
P whp = qh SG / (3960 μ) (1)
Where
P whp = Water power (HP)
Q = flow (gal/min)
H = Head (ft)
SG = 1 for specific gravity of water
μ = pump efficiency (decimal value)
Power can also be calculated as:
P whp = q dp / (1715 μ) (2)
Where
P whp = Water power (HP)
DP = Pressure provided (psi)
Example - Power required to pump water
20 gallons of water per minute They are high 20 feet . The necessary power (eg loss of friction in the pipe and efficiency = 1.0) can be calculated as
P whp = (20 gpm) (20 feet) (1) / (3960 (1.0))
= 0,10 cv
Power required to pump water to 60 o F with ideal pump efficiency 1.0:
| Power required to pump water (hp) | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| Volume flow (gpm) |
Height (feet) | ||||||||
| 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 50 | |
| 5 | 0,00631 | 0,0126 | 0,0189 | 0,0253 | 0,0316 | 0,0379 | 0,0442 | 0,0505 | 0,0631 |
| 10 | 0,0126 | 0,0253 | 0,0379 | 0,0505 | 0,0631 | 0,0758 | 0,0884 | 0,101 | 0,126 |
| 15 | 0,0189 | 0,0379 | 0,0568 | 0,0758 | 0,0947 | 0,114 | 0,133 | 0,152 | 0,189 |
| 20 | 0,0253 | 0,0505 | 0,0758 | 0,101 | 0,126 | 0,152 | 0,177 | 0,202 | 0,253 |
| 25 | 0,0316 | 0,0631 | 0,0947 | 0,126 | 0,158 | 0,189 | 0,221 | 0,253 | 0,316 |
| 30 | 0,0379 | 0,0758 | 0,114 | 0,152 | 0,189 | 0,227 | 0,265 | 0,303 | 0,379 |
| 35 | 0,0442 | 0,0884 | 0,133 | 0,177 | 0,221 | 0,265 | 0,309 | 0,354 | 0,442 |
| 40 | 0,0505 | 0,101 | 0,152 | 0,202 | 0,253 | 0,303 | 0,354 | 0,404 | 0,505 |
| 45 | 0,0568 | 0,114 | 0,170 | 0,227 | 0,284 | 0,341 | 0,398 | 0,455 | 0,568 |
| 50 | 0,0631 | 0,126 | 0,189 | 0,253 | 0,316 | 0,379 | 0,442 | 0,505 | 0,631 |
| 60 | 0,0758 | 0,152 | 0,227 | 0,303 | 0,379 | 0,455 | 0,530 | 0,606 | 0,758 |
| 70 | 0,0884 | 0,177 | 0,265 | 0,354 | 0,442 | 0,530 | 0,619 | 0,707 | 0,884 |
| 80 | 0,101 | 0,202 | 0,303 | 0,404 | 0,505 | 0,606 | 0,707 | 0,808 | 1,01 |
| 90 | 0,114 | 0,227 | 0,341 | 0,455 | 0,568 | 0,682 | 0,795 | 0,909 | 1,14 |
| 100 | 0,126 | 0,253 | 0,379 | 0,505 | 0,631 | 0,758 | 0,884 | 1,01 | 1,26 |
Observation! Individual bombs curves should always be used for exact calculations.
Energy consumption in metric units
Water consumption for water pumping can be expressed in metric units such as
P = qh ρ / (6116 10 3 m) (3)
Where
P = power (kW)
Q = flow (liter/min)
H = Head (m)
ρ = density (kg/m 3 ) (Water 1000 kg/m 3 )
μ = pump efficiency (decimal value)
Example - Power required to pump water
The power required to pump 100 l/min of water to an elevation of 10 m (Ex. Friction loss in the pipe and Efficiency = 1.0 ) can be calculated as
P = (100 litros/min) (10 m) (1000 kg/m 3 ) / (6116 10 3 (1,0))
= 0,16 kW
Source: https://www.engineeringtoolbox.com/pumping-water-horsepower-d_753.html
