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Jack went to his local hardware store and purchased a plug-in power meter. He immediately went back to the office and monitored the energy use of an idling coffee maker for a few days. He concluded that, on average, an idling coffee maker consumes 150 watts per hour.
Assuming the cost of electricity is $0.06/kWh, calculate the annual savings that would result from eliminating 5 idling coffee makers for 50 hours each week.
Remember: 1000 watts = 1 kW
$ per year
Can you suggest an alternative to idling coffee makers? (Choose the most appropriate option)
Apart from the direct energy savings, can you think of any other benefits that would result form eliminating idling coffee makers? (Select all that apply)
Let’s assume that you could eliminate 100,000 kWh of electricity use during the cooling season in your facility. That would save $6,000 of direct energy costs at $0.06 per kWh.
Assume your cooling plant COP is 5.0. Other savings that might occur are:
After a quick after-hours survey of offices in your building, you have determined that there are 350 – 2 lamp T8 fixtures “left on” by office occupants.
Assume 60 watts per fixture and $0.06 per kWh. The hourly cost of unnecessary operation is:
$ per hour
Calculate your answer to two decimals.
From the figure in the previous question ($1.26 per hour), determine the annual costs based on leaving the lights on for an extra 12 hours per day, 5 days per week and 50 weeks per year.