Engineering · Mechanical Engineering · Fluid Power
Pump Power Calculator
Calculates the hydraulic and shaft power required to pump a fluid given flow rate, head, fluid density, and pump efficiency.
Calculator
Formula
\rho is the fluid density (kg/m³), g is gravitational acceleration (9.81 m/s²), Q is the volumetric flow rate (m³/s), H is the total dynamic head (m), and \eta is the pump efficiency as a decimal (0 < \eta \leq 1). P_{\text{hydraulic}} is the useful power delivered to the fluid (W), and P_{\text{shaft}} is the actual mechanical power that must be supplied to the pump shaft (W).
Source: Munson, Young & Okiishi — Fundamentals of Fluid Mechanics, 8th Edition; ISO 9906: Rotodynamic pumps — Hydraulic performance acceptance tests.
How it works
When a pump moves fluid through a system, it must overcome the total dynamic head — a combined measure of elevation change, pressure difference, and velocity head losses along the pipe network. The power actually transferred to the fluid is called hydraulic power (also known as water power), and it depends directly on the fluid's density, the gravitational constant, the volumetric flow rate, and the total head. This is the theoretical minimum power the system requires.
In practice, no pump is perfectly efficient. Friction in bearings, turbulence in the impeller, recirculation, and mechanical seal losses all consume additional energy. The ratio of hydraulic power output to shaft power input defines pump efficiency (η). The shaft power — sometimes called brake power or brake horsepower (BHP) — is therefore always greater than the hydraulic power, calculated as Pshaft = Phydraulic / η. The two core formulas are: Phydraulic = ρ · g · Q · H and Pshaft = Phydraulic / η. For water at 20 °C, density is approximately 998 kg/m³, often rounded to 1000 kg/m³ for engineering estimates.
This calculator is widely used in municipal water supply design, HVAC chilled water loops, chemical process plant pumping circuits, irrigation systems, oil and gas pipelines, and wastewater treatment. Engineers use shaft power to select the correct drive motor — typically choosing a motor rated at least 10–15% above the calculated shaft power to accommodate start-up surges and operating margin. Results are provided in watts, kilowatts, and horsepower to suit both SI and imperial conventions.
Worked example
Problem: A centrifugal pump circulates water (density = 1000 kg/m³) at a flow rate of 0.05 m³/s against a total dynamic head of 20 m. The pump has an efficiency of 75%. Find the hydraulic power and the required shaft power.
Step 1 — Calculate hydraulic power:
Phydraulic = ρ × g × Q × H
Phydraulic = 1000 × 9.81 × 0.05 × 20
Phydraulic = 9,810 W = 9.81 kW
Step 2 — Calculate shaft power:
Pshaft = Phydraulic / η = 9,810 / 0.75
Pshaft = 13,080 W = 13.08 kW ≈ 17.54 hp
Step 3 — Motor selection: Adding a 15% safety margin, the engineer would specify a motor of at least 15.0 kW (the next standard motor size above 13.08 × 1.15 = 15.04 kW). This ensures adequate capacity under variable demand and prevents motor overload.
Limitations & notes
This calculator assumes steady-state, incompressible flow and uses a single average efficiency value for the pump. In reality, pump efficiency varies across the operating range and peaks at the Best Efficiency Point (BEP); operating far from the BEP significantly increases actual power consumption and can cause cavitation. The total dynamic head input must account for all system losses — including pipe friction (Darcy-Weisbach), minor losses (bends, valves, fittings), static elevation, and pressure difference between inlet and outlet — failing to include any component will underestimate shaft power. For compressible fluids (gases, steam), this incompressible model is not valid and a compressor or fan power formula should be used instead. The calculator does not include motor efficiency; to find the electrical input power consumed from the supply, divide shaft power by the motor efficiency as well. Fluid viscosity effects on pump performance curves are also not captured here — for highly viscous fluids, consult Hydraulic Institute viscosity correction standards.
Frequently asked questions
What is the difference between hydraulic power and shaft power in a pump?
Hydraulic power (water power) is the useful power actually transferred to the fluid — it represents the ideal minimum energy needed to move the fluid against the system head. Shaft power (brake power) is the actual mechanical power delivered to the pump shaft by the motor, which is always higher than hydraulic power because real pumps have internal losses such as impeller friction, recirculation, and bearing drag. The ratio of hydraulic to shaft power is the pump efficiency.
What is a typical pump efficiency value to use?
Pump efficiency varies widely by type, size, and operating point. Small centrifugal pumps (under 1 kW) may only achieve 40–60%, while large industrial centrifugal pumps can reach 85–92% at their Best Efficiency Point (BEP). For initial sizing estimates, 70–80% is commonly assumed for medium-duty centrifugal pumps. Always verify efficiency from the manufacturer's pump curve at the intended operating flow rate and head.
How do I calculate total dynamic head for a pumping system?
Total Dynamic Head (TDH) is the sum of: (1) static head — the vertical elevation difference between source and destination, (2) pressure head — the difference in pressure between suction and discharge converted to meters of fluid, (3) velocity head — usually small and often neglected, and (4) friction losses — calculated using the Darcy-Weisbach equation for pipe friction plus minor loss coefficients for fittings, bends, and valves. Accurate TDH calculation is critical because pump power scales directly with it.
Can I use this calculator for fluids other than water?
Yes. Simply enter the correct fluid density for your liquid. For example, seawater is approximately 1025 kg/m³, light crude oil around 850 kg/m³, sulfuric acid at 93% concentration about 1835 kg/m³, and glycol solutions vary with concentration. Be aware that highly viscous fluids require additional corrections to the pump curve beyond what this calculator provides.
How do I convert the shaft power result to electrical input power?
To find the electrical power drawn from the supply, divide the shaft power by the combined efficiency of the motor and any variable speed drive (VSD). For example, if shaft power is 13.08 kW, the motor efficiency is 92%, and there is no VSD, the electrical input power is 13.08 / 0.92 ≈ 14.22 kW. Including the motor efficiency gives you the true energy consumption figure needed for electricity cost calculations and energy audits.
Last updated: 2025-01-15 · Formula verified against primary sources.