## Motion Sensors to Reduce Costs

The lighting needs of a storage room are being met by 6 fluorescent light fixtures, each fixture containing four lamps rated at 60 W each. All the lamps are on during operating hours of the facility, which are 6 AM to 6 PM 365 days a year. The storage room is actually used for an average of 3 hours a day. If the price of electricity is $0.08/kWh, determine the amount of energy and money that will be saved as a result of installing motion sensors. Also, determine the simple pay-back period if the purchase price of the sensor is $32 and it takes 1 hour to install it at a cost of $40.

##
__
__**Hint**

**Hint**

Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours)

##
__
__**Hint 2**

**Hint 2**

Cost Savings = (Energy Savings)(Unit cost of energy)

*Assumptions*: The electrical energy consumed by the ballasts is negligible.

The plant operates 12 hours a day, and currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9×365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become

Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours)

= (24 lamps)(60 W/lamp)(3285 hours/year)

=

**4730 kWh/year**
Cost Savings = (Energy Savings)(Unit cost of energy)

= (4730 kWh/year)($0.08/kWh)

=

**$378/year**
The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor.

Implementation Cost = Material + Labor = $32 + $40 = $72

Solving for the payback period:

$$$Payback\:Period=\frac{Implementation\:Cost}{Annual\:Cost\:Savings}$$$

$$$=\frac{72}{378}=0.19\:year\:(2.3\:months)$$$

Energy Savings = 4730 kWh/year

Cost Savings = $378/year

Payback Period = 0.19 year (2.3 months) or ~2 months