Question

3. Consider the investment project given in the below table:

An electric motor is rated at 10 horsepower (HP) and costs $1,200. Its full-load efficiency is specified to be 85%. A newly designed high-efficiency motor of the same size has an efficiency of 90%, but it costs $1,600. It is estimated that the motors will operate at a rated 10 HP output for 2,000 hours/year, and the cost of energy will be $0.09/kilowatt- hour. Each motor is expected to have a 15-year life. At the end of the 15 years, the first motor will have a salvage value of $50 and the second motor will have a salvage value of $100. Consider the MARR to be 8%. (1 HP = 0.7457 kW)

Use NPW to determine which motor should be installed.

Answer 1

		10 HP   Rated Motor
	I	Initial Cost	$1,200
		Efficiency of motor	85%
		Power Consumption=10/0.85	11.7647059	HP
		Power Consumption in KW	8.77	(11.76471*0.7457)
		Number of hours per year	2000
		Total energy consumption per year	17545.8824	KWh	(8.77*2000)
		Annual Energy Cost =17545.88*$0.09	$1,579.13
	Rate	Discount Rate =MARR=	8%
	Nper	Number of years	15
	Pmt	Annual Energy Cost	$1,579.13
	Fv	Salvage Value	$50
	PV	Present value of annual   costs	$13,500.76	(Using PV function of excel with Rate=8%, Nper=15, Pmt=-1579.13, Fv=50)
	NPW=PV+I	Net Present worth of Costs	$14,700.76
		High Efficiency Motor
	I	Initial Cost	$1,600
		Efficiency of motor	90%
		Power Consumption=10/0.9	11.1111111	HP
		Power Consumption in KW	8.29	(11.1111*0.7457)
		Number of hours per year	2000
		Total energy consumption per year	16571.1111	KWh	(8.29*2000)
		Annual Energy Cost =16571.11*$0.09	$1,491.40
	Rate	Discount Rate =MARR=	8%
	Nper	Number of years	15
	Pmt	Annual Energy Cost	$1,491.40
	Fv	Salvage Value	$100
	PV	Present value of annual   costs	$12,734.08	(Using PV function of excel with Rate=8%, Nper=15, Pmt=-1491.40, Fv=100)
	NPW=PV+I	Net Present worth of Costs	$14,334.08
		High efficiency Motor has lower NPW of costs
		Hence,High efficiency Motor should be installed