Pool Pump Electricity Cost Calculator
Give the pump's power one of three ways, with run hours, season length and your rate. The comparison runs the same pump at a reduced speed for proportionally longer, so the same volume of water is filtered each day.
Pump Input Power
1,491 W
Energy Per Day
11.93 kWh
Cost Per Day
$2.15
Cost Per Month
$65.37
Cost Per Season
$386.57
Reduced Speed: Hours Per Day
16 h
Reduced Speed: Input Power
186 W
Reduced Speed: Energy Per Day
2.98 kWh
Reduced Speed: Cost Per Season
$96.64
Season Savings at Reduced Speed
$289.93
How it works
A pump's cost is its electrical input power times the hours it runs times your rate. A motor's horsepower is its shaft output, so the electricity it draws is the horsepower in watts (745.7 W per hp) divided by the motor's efficiency. If you have a nameplate instead, input power is volts × amps × power factor; if you have measured watts, enter them directly.
Centrifugal pumps follow the affinity laws: flow is proportional to speed and power to the cube of speed. Halving the speed halves the flow, so filtering the same water takes twice as long, but the pump draws only one-eighth the power. The energy for the same water is the speed squared: one-quarter at half speed.
The comparison applies those laws to the same pump at the reduced speed you enter. If moving the same water at that speed would take more than 24 hours a day, the reduced-speed results read "Not possible within 24 h" and the full-speed costs are still shown. A monthly figure uses an average month of 30.44 days; the season figure uses the days you enter.
Formula
input W (hp) = hp × 745.7 ÷ motor efficiency input W (nameplate) = volts × amps × power factor kWh per day = input W × hours ÷ 1,000 cost per day = kWh per day × rate ÷ 100 cost per month = cost per day × 30.44; cost per season = cost per day × days reduced speed s: hours = hours ÷ s; W = input W × s³; kWh per day = kWh × s² season savings = season cost × (1 − s²)
Example
A 1.5 hp motor at 75 % efficiency draws 1.5 × 745.7 ÷ 0.75 = 1,491 W. Running 8 hours a day it uses 11.93 kWh, which at 18 ¢ per kWh is $2.15 a day, $65.37 a month and $386.57 over a 180-day season.
At 50 % speed the same water takes 16 hours a day at 186 W, 2.98 kWh a day: $96.64 a season, $289.93 less.
Assumptions and limitations
- The affinity laws are ideal. Real pool systems have static head, filter and heater pressure drops, and drive and motor losses that change with speed, so variable-speed savings at very low speeds are smaller than the cube law predicts. ENERGY STAR states the ideal case: "reducing pump speed by one-half allows the pump to use just one-eighth as much energy", meaning per hour of running, which is power.
- Horsepower is converted with the mechanical horsepower of 745.7 W (the 746 W often quoted is the rounded electrical horsepower); the motor efficiency default is a typical value. A pump's actual draw depends on how heavily it is loaded, so a measured wattage is more reliable than horsepower or nameplate amps.
- Run hours, rate and season length are held constant; the 18 ¢ rate default is illustrative. Tiered and time-of-use tariffs are not modelled.
- Whether the reduced flow still meets the pool's filtration, chlorination, heater and skimmer requirements is not checked; those depend on the equipment manufacturer's instructions and local health or building codes.
Frequently asked questions
Why does slowing the pump save so much?
Power falls with the cube of speed but run time only rises in proportion to it, so the energy for the same water falls with the square of speed. At 75 % speed the energy is 0.75² = 56 % of full speed; at 50 %, 25 %.
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