Energy Savings Calculator

How to use the energy savings calculator

Enter the old device power in watts, the new device power in watts, how many hours a day it runs, and your electricity rate, and you get the savings from the swap: the dollars saved a year, the saving a month, and the kilowatt-hours saved over the year. The defaults price the classic upgrade, a 100 W incandescent bulb replaced by a 15 W LED on five hours a day, which comes to 155.1 kWh and about $27.15 a year. The rate is the input that swings the answer most. The tool defaults to 17.5 cents per kWh, while the Energy Information Administration reported a US residential average of 18.11 cents year to date for 2026 and 18.44 cents in May 2026, so use the figure from your own bill.

The saving comes entirely from the gap in wattage, so getting both numbers right is what makes the answer real. The old wattage is whatever the device draws today: the nameplate on an incandescent bulb, the sticker on an old fridge, the label on a window air conditioner. The new wattage is what the replacement pulls, and LED packaging lists it plainly while Energy Star appliance tags give a yearly kWh figure you can work back from by dividing by the hours it runs. A plug-in power meter beats both, because a nameplate usually states a maximum rather than a typical draw. If the two devices do the same job at the same brightness, the same cooling, or the same throughput, then the difference between them is pure saving.

Hours per day is the multiplier that decides whether a swap is worth doing at all. The same 85 watts saved is worth $22.47 a year on a fitting that runs 4 hours a day and $134.85 on one that runs continuously — six times as much for the identical purchase. Be honest about real runtime rather than how long the device is plugged in: a refrigerator compressor cycles rather than running flat out, a living-room lamp works a few hours an evening, and a basement bulb almost never. Set the hours to match the device you are actually pricing, and if you are unsure, run the calculation twice at the low and high ends of your guess to see how much the uncertainty is worth.

Use the tool to compare candidates before you buy, not just to confirm a swap you have already made. Run the always-on, high-hours loads first — an old refrigerator, a porch or hallway light that never goes off, a pool pump, a well pump — because those return the most for the watts saved. Then check the rarely-used items and you will usually find they are not worth the trouble. Once you have the yearly saving, the payback period follows in one step: divide the extra purchase cost over the cheaper, thirstier option by the annual saving and you get the years to break even. Anything shorter than the expected life of the device puts you ahead, which is why runtime rather than the wattage gap alone decides where to spend first.

The formula

The saving is the power you no longer draw, turned into energy and then into money. Subtract the new wattage from the old to get the watts saved, divide by 1,000 to reach kilowatts so the units match the kilowatt-hours your utility bills, multiply by daily hours for the energy saved each day, then by 365 days and your rate for the yearly figure. Nothing else enters the calculation, because both devices are assumed to be doing the same job.

The shape of that expression is worth noticing, because it explains every counter-intuitive result the tool produces. Saving is proportional to watts and to hours in equal measure, so 20 watts saved on something that runs all day beats 200 watts saved on something that runs half an hour. It is also floored at zero: if the replacement draws as much as the original or more, there is nothing to save, and the tool will not invent a negative saving to flatter a bad upgrade.

watts saved = old watts − new watts   (floored at zero)
daily kWh saved = (watts saved ÷ 1000) × hours per day
annual saving = daily kWh saved × 365 × rate
payback in years = extra purchase price ÷ annual saving

Worked example with the defaults, a 100 W incandescent replaced by a 15 W LED running five hours a day at $0.175 per kWh: 100 − 15 = 85 W saved; 85 ÷ 1000 × 5 = 0.425 kWh saved a day; 0.425 × 365 = 155.1 kWh a year; 155.1 × $0.175 = $27.15 a year, and the monthly figure the tool shows is 0.425 × 30 × $0.175 = $2.23. Push the same 85 watts to a device that never switches off and the arithmetic gives 85 ÷ 1000 × 24 × 365 × $0.175 = $130.31 a year, nearly five times as much for an identical wattage gap.

Payback follows from that annual figure in a single division. An LED costing two dollars more than the incandescent it replaces but saving $27.15 a year pays for itself in about 27 days. A more efficient appliance costing $150 more than a basic model and saving $40 a year breaks even in 3.75 years, which is worth doing if the appliance will plausibly last ten. The comparison that fails is the pricey efficient model on a low-hours load: the same $150 premium against a $5 a year saving needs 30 years, which no appliance survives.

Two honest limits on the number. First, it is running cost only, so it ignores the purchase price, installation, disposal of the old unit, and any rebate, and it is not a bill prediction. Second, nameplate wattage is often a maximum rather than a typical draw, and many appliances cycle rather than running flat out — a refrigerator compressor is a clear example, which is why entering its nameplate watts at 24 hours a day overstates the result badly. Use a plug-in power meter or an Energy Star annual kWh figure divided by real running hours when the number has to be defensible.

Frequently asked questions