The hallways of a university are lit by 40 fluorescent light

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.

Solution

Amount of electrical energy used every hour = 40*2*60 *60 J/h= 288000 J/h

No of kWh used in the entire day = 288000 * 24 /1000 = 6912 kWh

Cost of energy used for a day = 6912*7/100 = $ 483.84

Cost of energy for use in the entire week =$ 483.84 *7 = $ 3386.88 /week

No of kWh used if motion sensors are used= 288000*12*5/1000 = 17280 kWh /week

Cost of electricity when using motion sensors = 17280*7/100 = $ 1209.6 / week

Amount of Energy in kWh saved = 48384 - 17280 = 31104 kWh /week

Money saved= $ 3386.88 - $ 1209.6 = $ 2177.28 / week or $ 311.04 /day

Now the amount of money invested in installing the two sensors = $ 50 + $ 40 + $50 +$40 = $ 180

Value of investment after 1 day = $182.95

From this we can see that the pay back period is 1 day