Calculating Energy Savings from Lighting Retrofits

This is probably the simplest of energy savings calculations. The key here is to identify the lighting types in order to get the proper operating wattage.

If you have an incandescent, LED, halogen, or compact fluorescent lamp, you can just go off the lamp rating found on the lamp to determine the wattage. Also, for LED fixtures, you can just use the fixture rating.

However, if dealing with linear fluorescent, metal halide, induction, mercury vapor, or sodium (high and low-pressure) lighting, you will need both the lamp and ballast rating to determine fixture wattage, along with electrical system voltage. Ballasts are devices that regulate the flow of current into lamps and generally have a much longer life than the lamps.

On the ballast specification sheet, you will find the ballast input wattage when combined with lamps of different wattages and electrical system voltage. In the cover image, there is a ballast spec sheet that shows the lamp type in the first column, number of lamps in the second column, the system voltage in the 3rd column, and the input wattage in the fourth column. With the Universal B254PUNV-D ballast, if you install two standard F54T5HO lamps (any brand) in a building operating at 277V, the operating wattage would be 117W, despite the two 54W lamps having a combined wattage of 108W.

In addition to fixture wattage, you will also need the quantity of fixtures and the annual operating hours. If you have lighting controls, it could complicate or simplify the calculation depending on control type. Generally, assumptions are made based on business operating hours, with lighting controls factored in afterward.

So let’s go through a simple example. Bob’s Bakery with 10 4-lamp T8 fluorescent fixtures decides to replace the entire fixtures with new 40W LED fixtures. The bakery is open from 9:00 AM until 5:00 PM Mondays thru Friday and Bob arrives three hours earlier for prep and leaves two hours after closing. He also closes for 28 days per year for holiday and vacation. He also wants to apply to the local utility for energy efficiency incentives so he instructs the contractor to collect information from the old lamps and ballasts and take photos of them for their application. The contractor removes the fixtures and takes a photo of a T8 lamp installed which is a Philips F32T8 32W lamp and the ballast which is a General Electric GE432MAX-N/ULTRA ballast. Bob goes online and finds the ballast specification sheet here. He scrolls down and finds the F32T8 combination for 4 lamps. However, Bob is not sure what his system electrical voltage is, so he asks the contractor who informs him it is 120V. Bob goes back to the spec sheet and sees that the ballast input wattage would be 108W.

After collecting all the necessary information, Bob is ready to perform calculations to estimate the energy savings associated with his project.

Demand Savings (kW)
First he calculates the difference in wattage between the existing and new fixtures:

  • 108W (4L32T8) – 40W (LED) = 68W savings per fixture

Then he multiples that by the number of fixtures in his bakery:

  • 10 fixtures x 68W savings/fixture = 680W savings

These are called demand savings because it reduces the total electrical demand of the customer from the utility.

Annual Operating Hours (hr)
Now Bob must calculate his annual operating hours. First he determines the number of days he is open:

  • (52 weeks x 5 days per week) – 28 days holidays and vacation = 232 days open per year

Then he calculates the hours lights are on per day:

  • 8 opening hours + 5 hours for prep and closing = 13 hours/day

And multiplies that by number of days per year to get annual operating hours:

  • 232 days/year x 13 hours/day = 3,016 annual operating hours.

Annual Energy Savings (kWh)
Now Bob is ready to calculate his total annual energy savings, but first he must convert to kilowatts in order to match the standard his utility charges him in, kilowatt-hours, so first Bob divides 680W by 1000W/kW to get 0.68 kW.

Finally, he multiplies the demand savings by the annual operating hours to get his annual energy savings:

  • 0.68 kW x 3,016 hours = 2,051 kWh savings per year.

Annual Cost Savings
Bob is also curious what his cost savings will be from this upgrade considering his utility currently charges him $0.20 per kilowatt hour:

  • 2,051 kWh x $0.20/kWh = $410 cost savings per year

Bob is happy to know that not only is he reducing his impact on the environment, but also reducing his operating costs in doing so.