Lighting Efficiency
What is the goal of lighting efficiency? Provide the amount of light needed at a minimum wattage. Before we get into how to achieve the goal of lighting efficiency, we need to familiarize ourselves with some basic terminology in lighting:
 Lumens (lm)
–The amount of light given off is measured in lumens, also referred as brightness.
–One lumen is the equivalent of the light given off by one candle in one square foot.
 Watts
–The amount of power input to a light bulb to produce light is measured in watts
 Energy/Electricity Consumption
–is measured by KWh (kilowatts hour):
KWh = Watts × hours of operation/1000 ...................................................................equ (1)
How to Measure Efficiency of lamps?
`````````````````````````````````````````````````````equ (2)
Unit of lamp efficiency is lumen/watt.
Note the above equation can be rearranged into:
Lumen output = Lamp efficiency X watts of power input ........................equ (3)
`````````````````````````````````````````````equ (4)
Applications of Lighting Efficiency
Now let's look at a couple of applications of the lighting efficiency concept.
Example #1
The hallway of the Physics Department at Michigan State University has 20, 72watts energysaving halogen light bulbs whose lamp efficiency is 20.7 lumen/watt. The university is considering changing these light bulbs into compact fluorescent light (CFL) bulbs whose lamp efficiency is 60 lumen/watt.
1. What watts of the CFL bulbs should the university buy, assuming each energysaving halogen light bulb is replaced with a CFL bulb?
2. Assuming these light bulbs are on 24 hours a day, 7 days a week for 50 week of a year, how much energy/electricity will be saved in a year by switching to CFLs?
SOLUTION:
1. Since each halogen lightbulb is replaced with a CFL bulb, the lumen output from a CFL bulb should be the same as that from a halogen lightbulb.
Based on equ (3);
For a halogen lightbulb, lumen output = 20.7 lumen/watts x 72 watts = 1490.4 lumens
Then based on equ (4);
So the university should buy 25watts CFLs, these CFLs will produce the same amount of light (i.e. lumens) as compared to existing lightbulbs. One can see as we use lamps with higher efficiency, for the same amount of light produced, we can use a much lower watts lamps.
2. Let's first figure out electricity consumed for both bulbs in a year, the difference is the energy saving.
Based on equ (1);
When using halogen lightbulbs, electricity/energy consumption (kWh) = 20 (bulbs) x 72 watts x (24 hr/day x 7 days/week x 50 weeks/year) /1000 = 12,096 kWh
When using CFLs, electricity/energy consumption (kWh) = 20 (bulbs) x 25 watts x (24 hr/day x 7 days/week x 50 weeks/year) /1000 = 4200 kWh
Electricity saving = 12,096 kWh  4200 kWh =7,896 kWh.
Example #2
The dormitory supervisor of a major university is considering switching 60 watts energysaving halogen lightbulbs with 9W Light Emitting Diode (LED) replacement bulbs. Halogen efficiency is 13 lumens/watt and LED efficiency is 90 lumens/watt. There are a total of 400 such lightbulbs in the dorm. Assume each lightbulb burns an average of 4 hours per day for 300 days/year. The university is paying 8 cents per kWh for electricity.
1. Is the light output from LED bulbs comparable to halogen bulbs?
2. How much electricity will be saved in a year by switching to LEDs?
3. How much ($) will university save in electricity bill in a year by switching to LEDs?
SOLUTION:
1. Let's calculate the lumen output for both bulbs.
For halogen bulbs, Lumen output = Lamp efficiency X watts of power input = 13 (lumens/watt) x 60 watt = 780 lumens
For LEDs, Lumen output = Lamp efficiency X watts of power input = 90 (lumens/watt) x 9 watt = 810 lumens
Yes, the light output from LED bulbs are comparable to (actually greater than) halogen bulbs.
2. Let's calculate electricity consumption for both bulbs in a year, the difference in the saving.
For halogen bulbs, electricity consumption (kWh) = 400 (bulbs) x 60 watt x (4 hr/day x 300 days/yr)/1000 = 28,800 kWh
For LEDs, electricity consumption (kWh) = 400 (bulbs) x 9 watt x (4 hr/day x 300 days/yr)/1000 = 4,320 kWh
Electricity saving = 28,800 kWh  4,320 kWh = 24,480 kWh
3. Savings in electricity bill ($) $0.08 kWh x 24,480 kWh = $1,958.40
Have I Grasped the Key Concepts Here?
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