Pump Power Calculator

Overview

Pump power calculations determine the energy required to move fluid through a system. The three key power levels are hydraulic power (useful work on the fluid), shaft power (power delivered to the pump), and motor power (electrical input). Understanding these relationships is essential for pump selection, energy cost estimation, and system design.

Original source text preserved: "The ideal hydraulic power to drive a pump depends on the mass flow rate the liquid density the differential height - either it is the static lift from one height to an other or the total head loss component of the system - and can be calculated like Ph(kW) = q ρ g h / 3.6×106 = q p / 3.6×106." The source also states: "The shaft power - the power required transferred from the motor to the shaft of the pump - depends on the efficiency of the pump."

Key Formulas

The hydraulic power delivered to the fluid is:

$P_{hydraulic} = \frac{\rho \cdot g \cdot Q \cdot H}{1000}$

Accounting for pump efficiency, the required shaft power is:

$P_{shaft} = \frac{P_{hydraulic}}{\eta_{pump}}$

Accounting for motor efficiency, the electrical input power is:

$P_{motor} = \frac{P_{shaft}}{\eta_{motor}}$

Combining into a single expression:

$P_{motor} = \frac{\rho \cdot g \cdot Q \cdot H}{1000 \cdot \eta_{pump} \cdot \eta_{motor}}$

Hydraulic Horsepower - Imperial Units

The original source formula is preserved explicitly:

$P_h(hp) = \frac{q_{gpm} h_{ft} SG}{3960 \eta}$

where q_gpm is flow in US gallons per minute, h_ft is head in feet, SG is specific gravity, and efficiency is pump efficiency as a decimal. If efficiency is omitted, q_gpm h_ft SG / 3960 gives water horsepower.

Variables

Unit Converter

Example Calculations

SI Units

For water at 1000 kg/m³, flow rate 50 m³/h, total head 30 m, pump efficiency 75%, and motor efficiency 90%:

$P_{hydraulic} = \frac{1000 \cdot 9.81 \cdot (50/3600) \cdot 30}{1000} \approx 4.09\ kW$

$P_{motor} = \frac{4.09}{0.75 \cdot 0.90} \approx 6.06\ kW$

Imperial Units

For water at SG = 1.0, flow rate 500 US gpm, total head 100 ft, pump efficiency 70%, and motor efficiency 90%:

The imperial hydraulic horsepower formula from the source is:

$P_h(hp) = \frac{q_{gpm} h_{ft} SG}{3960 \eta}$

where q_gpm is flow in US gallons per minute, h_ft is head in feet, SG is specific gravity, and efficiency is pump efficiency as a decimal. Without the efficiency term, q_gpm h_ft SG / 3960 is water horsepower.

$Water\ hp = \frac{500 \cdot 100 \cdot 1.0}{3960} \approx 12.63\ hp$

$Motor\ hp = \frac{12.63}{0.70 \cdot 0.90} \approx 20.05\ hp$

Pump Shaft Power Charts

The original page includes metric and imperial shaft-power diagrams. The interactive chart below represents the same relationship for water at 70% pump efficiency.

Pump Power Formula Chart

The source pump_power.png image summarizes equivalent SI and Imperial pump power forms. This interactive representation compares hydraulic, shaft, and motor input power for the same duty point.

Pump Shaft Power Chart - Imperial Units

The original pump_shaft_power_imperial.png diagram is represented below using the water horsepower relationship at 70% pump efficiency.

Related Mobile Apps from The Engineering ToolBox

The source page included a related mobile-app section. In this migration the SI and Imperial pump power calculators are directly available on the page as browser-based tools for mobile and desktop use.

Typical Pump Efficiencies

Specific Speed and Efficiency Relationship

Restored Original Source Tables

The following tables are restored from the original source page to preserve the complete reference data.

The cached source table outside the diagrams contains site layout and calculator shell rows rather than engineering data. It is intentionally consolidated into the SI and Imperial calculators, the unit converter, and the metric/imperial shaft-power interactive charts above.

Original Source Images

The following original source images are preserved to avoid losing visual reference material. When an image contains chart or tabular data, its extracted values are represented in the page tables, calculators, or interactive charts; remaining images are retained as visual source references.

Pumps - shaft power vs. flow and lift diagram - Metric Units

Engineering Notes

• Unit consistency: Flow rate in m³/s is required in the SI formula. Convert from m³/h by dividing by 3600.
• Total dynamic head (H) includes static lift, friction losses, and any required discharge pressure converted to equivalent fluid column height.
• Efficiency varies with operating point: The listed efficiencies represent best efficiency point (BEP). Operating far from BEP significantly reduces actual efficiency.
• Specific gravity: For fluids other than water, use $\rho_{fluid}$ in place of water density. Alternatively, multiply water power by specific gravity (SG).
• VFD considerations: With variable frequency drives, motor efficiency may vary with speed. Refer to motor efficiency curves rather than a single nominal value.
• Viscous fluids: High-viscosity fluids reduce pump performance below catalog water-test values. Apply viscosity correction factors per HI (Hydraulic Institute) standards.
• Safety margins: It is common practice to select a motor rated 10–15% above calculated shaft power to handle transient overloads and future degradation.

References

• Original Source
• Hydraulic Institute, Engineering Data Book, pump application standards