Question

The hallways of a university are lit by 40 fluorescent lights, each containing two lamps rated at 60 W each. The lights are on all day and night during the year, but the building is only used from 7 a.m. to 7 p.m. 5 days per week during the year. If the price of electricity is 7 cents/kWh, estimate the amount of energy and money that could be saved by installing two motion sensors that turn off the lights when the building is not being used. Then, estimate the simple payback and the payback with a 6% effective interest rate if the price of a sensor is $50 and the cost of installation is $40.

Answer 1

simple payback period = 1.9 years

discounted payback period = 2.08 years

Step-by-step explanation:

the total cost of electricity per year = [(2 x 60 x 40 x 24 x 365) / 1,000] x $0.07 = $2,943.36

consumption for the actual time needed, assuming that the university is open all year long = [(2 x 60 x 40 x 12 x 5 x 52) / 1,000] x $0.07 = $1,048.32

savings resulting from installing motion sensors = $2,943.36 - $1,048.32 = $1,895.04

cost of installing the 40 sensors = ($50 + $40) x 40 sensors = $3,600

simple payback period = $3,600 / $1,895.04 = 1.9 years

discounted cash flows:

$1,895.04 / 1.06 = $1,787.77

$1,895.04 / 1.06² = $1,686.58

$1,895.04 / 1.06³ = $1,591.11

discounted payback period = 2 years + $125.65/$1,591.11 = 2.08 years