Pump Power Calculator
Calculate the shaft power required to drive a pump given flow rate, total head, efficiency, and fluid type. Ideal for engineers sizing pumps or estimating energy consumption in water supply, HVAC, and process systems.
About this calculator
Pump power is derived from the hydraulic power needed to move fluid against a pressure head, divided by the pump efficiency. The hydraulic (water) power is P_hydraulic = ρ × g × Q × H, where ρ is fluid density (kg/m³), g = 9.81 m/s², Q is volumetric flow rate (m³/s), and H is total head (m). Dividing by efficiency η (expressed as a decimal) gives the shaft power input: P_shaft = (ρ × g × Q × H) / η. In this calculator flow is entered in L/min (converted by dividing by 60,000 to get m³/s) and the result is expressed in kilowatts. Water density is assumed as 1,000 kg/m³; a lighter fluid option uses 800 kg/m³. The formula becomes: P (kW) = (Q/60000 × H × 9.81 × ρ) / (η/100) / 1000.
How to use
A water pump delivers a flow rate of 500 L/min against a total head of 30 m at 75% efficiency. Q = 500/60,000 = 0.00833 m³/s. P = (0.00833 × 30 × 9.81 × 1000) / (75/100) / 1000. Numerator: 0.00833 × 30 × 9810 = 2452 W. Divide by 0.75: 3,269 W. Divide by 1000: 3.27 kW. Enter flowRate = 500, totalHead = 30, efficiency = 75, and select water to confirm a required shaft power of approximately 3.27 kW.
Frequently asked questions
What is total head in a pump power calculation?
Total head H is the total energy per unit weight that the pump must add to the fluid, expressed in metres. It is the sum of static head (height difference between suction and discharge), pressure head (difference in system pressures converted to metres of fluid), and velocity head (kinetic energy difference), plus friction losses in pipes, fittings, and valves. Engineers calculate total head using the extended Bernoulli equation applied between the inlet and outlet of the pumping system. Using an accurate total head value is critical because power scales linearly with H — underestimating it leads to an undersized, overloaded pump motor.
How does pump efficiency affect the required motor power?
Pump efficiency η represents how much of the shaft input power is actually converted into useful hydraulic power. A pump with 60% efficiency requires 67% more shaft power than a theoretically perfect (100% efficient) pump for the same duty. Motor size, and therefore energy costs, scale inversely with efficiency: halving the efficiency doubles the required power. Efficiency typically peaks at the Best Efficiency Point (BEP) on the pump curve and falls off at both lower and higher flow rates. Always select a pump that operates near its BEP under design conditions to minimise energy consumption and wear.
Why does fluid density matter when calculating pump power?
Hydraulic power is proportional to fluid density (P = ρgQH), so a denser fluid requires proportionally more power to pump at the same flow rate and head. Water at ~1,000 kg/m³ is the reference; a light oil at ~800 kg/m³ requires only 80% of the power needed for water under identical conditions, while a heavy brine at 1,200 kg/m³ would require 20% more. When pumping fluids other than water it is important to input the correct density (or select the appropriate fluid option) to avoid undersizing the motor, which can lead to overheating and premature failure.