Type a line (100 m3/h 50 m eff 70 %, 400 l/min 6 bar, 15 kW 50 m eff 70 % → flow) or fill in the fields: the line and the fields stay in step.
Enter the flow rate of a pump and its head or differential pressure, and the calculator gives the hydraulic (useful) power, P = rho x g x Q x H = Q x dp. Add the pump efficiency and it gives the shaft power; add the motor efficiency as well and it gives the motor input power. Both efficiencies are your own figures. Head and differential pressure are converted into each other with the density of the liquid. In reverse, it works out the flow from a power and a head or the head from a power and a flow (a shaft power needs the pump efficiency, a motor input power both efficiencies), or the pump efficiency from a flow, a head and a measured shaft power. The working is shown with your numbers, and the result is a line you can copy.
Arithmetic only: no typical efficiencies, no NPSH, no motor sizing margin and no recommendations. The duty point is where the pump curve meets the system curve, the efficiency comes from the pump curve of the maker, and the motor is chosen by its maker and the rules of the class society.
P = rho x g x Q x H, with g = 9.80665 m/s2, Q in m3/s and H in m. 100 m3/h of sea water at 1025 kg/m3 to a head of 50 m: 1025 x 9.80665 x 0.02778 x 50 = 13961 W = 13.96 kW. From a differential pressure, P = Q x dp: 400 l/min at 6 bar is 0.006667 m3/s x 600000 Pa = 4000 W.
Shaft power = hydraulic power / pump eff; motor input power = shaft power / motor eff. With 70 % and 90 % entered as examples: 13.96 / 0.70 = 19.94 kW and 19.94 / 0.90 = 22.16 kW.
dp = rho x g x H: 50 m of sea water is 5.026 bar, of fresh water at 15 °C 4.899 bar. In reverse, the flow is Q = hydraulic power / (rho x g x H), the head H = hydraulic power / (rho x g x Q), and the pump eff = hydraulic power / shaft power.
| Flow, m3/h | Head, m | dp, bar (fresh water) | Power, kW (fresh water 999.1 kg/m3) | Power, kW (sea water 1025 kg/m3) |
|---|---|---|---|---|
| 10 | 20 | 1.960 | 0.54 | 0.56 |
| 25 | 30 | 2.939 | 2.04 | 2.09 |
| 50 | 30 | 2.939 | 4.08 | 4.19 |
| 100 | 50 | 4.899 | 13.61 | 13.96 |
| 200 | 40 | 3.919 | 21.77 | 22.34 |
| 300 | 80 | 7.838 | 65.32 | 67.01 |
| 500 | 25 | 2.449 | 34.02 | 34.90 |
| 1000 | 20 | 1.960 | 54.43 | 55.84 |
The figures are the arithmetic P = rho x g x Q x H with no efficiency: the power the pump gives to the liquid. Fresh water is 999.1 kg/m3 at 15 °C, sea water a typical 1025 kg/m3. The shaft power is this figure divided by the pump efficiency read from the pump curve.
The working is the definition of the hydraulic (useful) power of a pump, P = rho x g x Q x H = Q x dp, and of the shaft power and motor input power through the efficiencies. Standard gravity, g = 9.80665 m/s2, is exact by definition. The unit factors are exact definitions: 1 ft = 0.3048 m, the US gallon 3.785411784 l, the Imperial gallon 4.54609 l, 1 bar = 100000 Pa; 1 psi = 6894.757 Pa and 1 hp (550 ft lbf/s) = 745.7 W follow from the exact pound, foot, inch and g, 1 PS (metric horsepower) = 735.49875 W from g alone, and each is checked against two independent references. The density picks — fresh water at 15 °C, 999.1 kg/m3, and sea water, a typical 1025 kg/m3 (about 1020 to 1029 kg/m3 at the surface) — are each checked against two independent references; the density from the data sheet of the liquid or a measurement always comes first.
P = rho x g x Q x H, with the density rho in kg/m3, g = 9.80665 m/s2, the flow Q in m3/s and the head H in m. 100 m3/h is 100 / 3600 = 0.02778 m3/s; at 50 m with sea water of 1025 kg/m3, P = 1025 x 9.80665 x 0.02778 x 50 = 13961 W, or 13.96 kW. From a differential pressure it is P = Q x dp: 400 l/min = 0.006667 m3/s at 6 bar = 600000 Pa gives 4000 W, or 4 kW.
The hydraulic (useful) power is what the pump gives to the liquid. The shaft power is what the pump takes from its shaft: the hydraulic power divided by the pump efficiency. The motor input power is what the motor takes from the supply: the shaft power divided by the motor efficiency. With 13.96 kW, a pump efficiency of 70 % and a motor efficiency of 90 %, entered as examples, the shaft power is 13.96 / 0.70 = 19.94 kW and the motor input power 19.94 / 0.90 = 22.16 kW. The 70 % and 90 % are examples, not typical figures.
dp = rho x g x H. 50 m of sea water at 1025 kg/m3 is 1025 x 9.80665 x 50 = 502591 Pa, or 5.026 bar; 50 m of fresh water at 999.1 kg/m3 (15 °C) is 4.899 bar. The same head gives a higher pressure in a denser liquid. Back the other way, H = dp / (rho x g): 6 bar of fresh water at 999.1 kg/m3 is 61.24 m.
This page gives neither. The pump efficiency is read from the pump curve of the maker at the duty point, and the motor efficiency from the motor rating plate or the data of the motor maker. The motor itself is chosen by its maker and the rules of the class society; this page gives no sizing margin.
In reverse, Q = shaft power x pump efficiency / (rho x g x H). 15 kW at 50 m with sea water of 1025 kg/m3 and a pump efficiency of 70 %, entered as an example, gives a hydraulic power of 10.5 kW and Q = 10500 / (1025 x 9.80665 x 50) = 0.02089 m3/s, or 75.21 m3/h. The real duty point is where the pump curve meets the system curve; the arithmetic only says what that power can lift to that head at that efficiency.
For a given head, yes: the power rises in step with the density, so sea water of 1025 kg/m3 takes 1025 / 999.1 = 1.026 times the power of fresh water of 999.1 kg/m3 at 15 °C, 2.6 % more. For a given differential pressure, no: P = Q x dp does not contain the density, which only sets the head that the pressure stands for.
The calculator works out the power from the figures you enter. When a pump, an impeller, a seal or the motor for it is needed, attach photographs of the pump and motor plates, and Vessel Core checks the details before an offer goes back.
