Calculate what your pool pump costs to run per day, month and year from its watts, horsepower or amps, plus upgrade savings and budget runtime.
Pool Pump Energy Cost Formula
The cost of running a pool pump comes from three numbers: the input power of the pump, how long it runs, and what you pay for electricity.
kWh/day = W × h / 1000
Cost/day = kWh/day × R
Cost/month = Cost/day × 30.44
Use measured real electrical input watts when available. Horsepower or AC current and voltage require additional operating assumptions: DOE motor-load guidance, NIST conversion factors and ABB motor-nameplate guidance explain these distinctions.
W = shaft mechanical hp × 745.699872 / (motor-and-drive efficiency / 100)
Single-phase W = RMS volts × RMS amps × true power factor. For a balanced three-phase supply, W = √3 × RMS line-to-line volts × RMS line amps × true power factor. Do not apply motor efficiency again to real electrical input watts.
Variables:
- W is the electrical input power of the pump in watts
- h is the runtime in hours per day
- R is your own electricity rate in USD per kWh; the $0.18 examples are hypothetical, not a current national-average claim
- HP is actual mechanical shaft output; rated motor horsepower applies only at full rated shaft load
- Efficiency is the motor-and-drive efficiency at the same operating load. It is separate from power factor, pump hydraulic efficiency and service factor; no universal 60% value is assumed
- A and V are RMS operating current and voltage. True power factor is required for AC real power; line-to-line voltage and balanced line current are used for three phase
The calculator’s cost mode applies these formulas directly and reports cost per hour, day, month, and year, along with the kWh used. The comparison mode runs the daily cost formula for two pumps and subtracts the results, then divides the new pump’s price by the annual savings to get a payback period. The budget mode inverts the monthly cost formula to solve for hours: it divides your monthly budget by 30.44 times the rate times the pump’s kilowatts, which tells you how many hours per day you can run before exceeding your budget.
Illustrative Pump Power and Energy Charges
The first table uses hypothetical electrical input powers, an illustrative $0.18/kWh rate and an 8-hour daily schedule. These scenarios are not a horsepower-to-watts lookup or equipment measurements. Use actual input power at the chosen operating speed and load.
| Scenario | Illustrative input watts | Cost per hour | Cost per month (8 h/day) |
|---|---|---|---|
| A | ~1,000 | $0.18 | $43.83 |
| B | ~1,350 | $0.24 | $59.18 |
| C | ~1,800 | $0.32 | $78.90 |
| D | ~2,300 | $0.41 | $100.82 |
| E | ~3,100 | $0.56 | $135.88 |
The second table is an ideal affinity-law illustration for the same pump and impeller under similar operating conditions and constant pump, motor and drive efficiencies. Flow scales with speed and power with speed cubed. For equal pumped volume, longer runtime makes energy scale with speed squared. Actual operating points depend on system head and losses; static head or changing efficiency can invalidate this simple scaling. Use operating measurements for cost comparisons. See Pentair’s affinity-law explanation and KSB’s similarity and efficiency qualifications.
| Motor speed | % of full speed | Power draw (1,800 W pump) | Energy to move the same water |
|---|---|---|---|
| 3,450 RPM | 100% | 1,800 W (100%) | 100% |
| 2,588 RPM | 75% | 759 W (42%) | 56% |
| 1,725 RPM | 50% | 225 W (12.5%) | 25% |
| 1,150 RPM | 33% | 67 W (3.7%) | 11% |
Example Problems
Example 1: Monthly cost of a single-speed pump. An example pump has 1,800 watts of real electrical input and runs 8 hours per day. Electricity costs $0.18 per kWh. Daily energy is 1,800 × 8 / 1000 = 14.4 kWh, so the daily cost is 14.4 × 0.18 = $2.59. Over an average month of 30.44 days that is 438.3 kWh and $78.90. Over a full year it comes to about 5,260 kWh and $946.81.
Example 2: Upgrade savings and payback. The same 1,800 watt pump running 8 hours per day is replaced by a variable-speed pump that runs 16 hours per day at 300 watts. At $0.18 per kWh the old pump costs $946.81 per year while the new schedule uses 4.8 kWh per day and costs $315.60 per year. The upgrade saves $631.20 per year, roughly a 67% reduction. If the new pump costs $1,200 installed, the payback period is 1,200 / 631.20, or about 1.9 years.
Frequently Asked Questions
How many hours per day should a pool pump run?
Runtime must satisfy equipment instructions, water-treatment needs and applicable local requirements. Volume divided by flow gives a nominal turnover time, but one theoretical turnover does not mean every part of the water was filtered or establish a safe daily schedule. The budget result is a financial allowance, not a filtration recommendation. Confirm that any proposed schedule meets the pool’s circulation and treatment requirements. See CDC operating guidance for manufacturer-directed recirculation and separate water-quality monitoring.
How do I find my pump’s actual wattage?
Prefer a correctly obtained real-input power measurement at the operating speed and load, or manufacturer operating data. A pump display may report an estimate rather than a meter measurement. Nameplate horsepower and full-load current describe rated conditions, not necessarily actual use. Single-phase volts × amps is apparent power unless true power factor is included. The HP method requires shaft output and motor-and-drive efficiency at the same load; it does not infer either from a nominal pump rating.
Why does a variable-speed pump cost so much less to run?
The ideal affinity model predicts lower power at lower speed, but actual energy savings depend on the pump, plumbing, static head, efficiency and the schedule needed for the required treatment performance. Compare real input watts and valid operating schedules. This calculator does not infer a universal savings percentage from a pump’s speed or horsepower rating.
