Question

A 20kW rated motor has an average loading of 75% of rated power and has an efficiency of 80%. The motor has a peak loading of 100% of rated power. The motor operates 1000 hours a year. The cost of electricity is $0.10/kWh and $7/kW-month of demand. Assume the motor operates at periods of peak demand at the facility (i.e. the motor power impacts the billing demand). An energy efficiency upgrade reduces the average power consumed to an average loading of 50% of rated power and reduces the peak power to 75% of rated power, but increases the operating hours to 1500 hours per year. There is no change in the efficiency of the motor after the upgrade. The cost of the energy efficiency project is $10,000. What is the annual electricity cost of running the motor before and after the energy efficiency upgrade? What is the simple payback of the project?

Answer 1

	Before Energy Efficiency upgrade
	Peak Loading =75% of rated power=	20KW
	Effeciency	80%
	Power needed =20/0.8=	25	KW
	Number of hours in a year	1000
	Annual power consumption =25*1000	25000	KWh
	Cost of electricity =25000*$0.10=	$2,500
	Demand =$7*25*12	$2,100
	Total annual cost	$4,600
	After Energy Efficiency upgrade
	Peak Loading =100% of rated power=	15KW	(20*75%)
	Effeciency	80%
	Power needed =15/0.8=	18.75	KW
	Number of hours in a year	1500
	Annual power consumption =15*1500	22500	KWh
	Cost of electricity =22500*$0.10=	$2,250
	Demand =$7*18.75*12	$1,575
	Total annual cost	$3,825
	Annual Savings in cost=4600-3825=	$775
	Simple paybackin years=10000/775	12.90