Pump Power Formula
Calculate pump power using P = ρgQH/η, relating fluid density, flow rate, head, and pump efficiency to required power input.
The Formula
The pump power formula calculates the electrical or mechanical power required to drive a pump. It accounts for the fluid properties, the desired flow rate, the height (head) the fluid must be raised, and the pump's efficiency.
This is the input power required. The useful (hydraulic) power delivered to the fluid is P_hydraulic = ρgQH, and the actual input is higher because no pump is 100% efficient.
Variables
| Symbol | Meaning |
|---|---|
| P | Input power required (in watts, W) |
| ρ | Fluid density (in kg/m³; water ≈ 1000 kg/m³) |
| g | Gravitational acceleration (9.81 m/s²) |
| Q | Volumetric flow rate (in m³/s) |
| H | Total head — the height equivalent the pump must overcome (in meters) |
| η | Pump efficiency (dimensionless, between 0 and 1; typically 0.6 to 0.85) |
Example 1
A pump must deliver water (ρ = 1000 kg/m³) at a flow rate of 0.05 m³/s to a height of 20 m. The pump efficiency is 75%. What power is needed?
P = ρgQH / η
P = 1000 × 9.81 × 0.05 × 20 / 0.75
P = 9810 / 0.75
P = 13,080 W ≈ 13.1 kW (about 17.5 horsepower)
Example 2
A small fountain pump moves 500 liters per hour of water to a height of 3 m. Pump efficiency is 60%. What power is required?
Convert flow rate: 500 L/hr = 500/3600 L/s = 0.139 L/s = 0.000139 m³/s
P = ρgQH / η = 1000 × 9.81 × 0.000139 × 3 / 0.60
P = 4.09 / 0.60
P ≈ 6.8 W (a very small pump — typical for a garden fountain)
When to Use It
The pump power formula is essential in hydraulic and mechanical engineering.
- Sizing pumps for water supply and irrigation systems
- Designing cooling water circuits in industrial plants
- HVAC system design (chilled water and hot water loops)
- Calculating energy costs for pumping operations
- Wastewater treatment plant design
- Fire suppression system engineering
Remember: total head H includes not just the static lift height but also friction losses in pipes and fittings. In real systems, friction losses can be significant and must be calculated separately.