Question

A plant owner is considering replacing each of the metal halide lamps in his facility with a 6-lamp T8 fixture. The plant has a total of 31 fixtures. Determine the annual energy savings for the proposed lighting upgrade (in kWh). Information about each light type is given below:

Metal Halides Each lamp is rated at 400 W with a ballast factor of 1.17. Lamp intensity is 22,000 lumens. Lamp lifetime is 20,000 hours. The cost of a replacement lamp is $14, and the time required to replace a bulb is 27 minutes. The cost of the fixture is $26.

T8 Fluorescent Each lamp is rated at 32 W with a ballast factor of 0.91. Lamp intensity is 3,000 lumens. Lamp lifetime is 30,000 hours. The cost of a replacement lamp is $2, and the time required to replace a bulb is 19 minutes. The cost of the fixture is $22.

Plant Information The plant operates for 7,967 hours per year. The labor rate is $15. The time required to retrofit an entire fixture is 38 minutes

Answer 1

HI,

The question is to calculate just the annual energy savings in kWh

So, we need to determine the number of KWs that are being saved by upgradation and then to multiply it with the number of hours per year that they are operating to get kWh savings

TO determine the number of kW, first we need to make sure that by replacing the system, the total luminous intensity should not change which means that the lighting of the room should be the same

Total Number of lumens emitted by the Metal Halides in the room should be same as the total number of lumens emitted by the replaced fluorescent lamp system.

Number of lumens emitted from each Metal halide lamp = 1.17X22000=25740 lumens

Number of lumens emitted from each fluorescent lamp = 0.91X3000=2730 lumens

SO, to maintain the same light intensity in the room, number of fluorescent lamps required for every Metal Halide lamp is given by N = 25740/2730=9.4 or approximately equal to 10 lamps

So each Metal halide lamp is to be replaced by 10 fluorescent lamps to get same light intensity

Now Wattage of one Metal halide lamp = 400W

Wattage of 10 fluorescent lamps = 32X10=320W

Hence saving in watts by new lighting system = 400-320=80W (for same light intensity)

Number of hour the plant operates = 7967 hrs per year

Hence savings in kWh per year = 80X7967/1000 =637kWh