n electric motor is rated at 10 horsepower (HP) and costs $800. Its full-load ef

Question

n electric motor is rated at 10 horsepower (HP) and costs $800. Its full-load efficiency is specified to be 85%. A newly designed, high- efficiency motor of the same size has an effi- ciency of 90%, but costs $1,200. It is estimated that the motors will operate at a rated 10-HP output for 1,500 hours a year, and the cost of energy will be $0.07 per kilowatt-hour. Each motor is expected to have a 15-year life. At the end of 15 years, the first motor will have a sal- vage value of $50, and the second motor will have a salvage value of $100. Consider the MARR to be 8%. (Note: 1 HP 0.7457 kW.) 5.42 A (a) Determine which motor should be in stalled, based on the PW criterion. (b) What if the motors operated 2,500 hours a year instead of 1,500 hours a year? Would the motor selected in (a) still be the choice?

Explanation / Answer

PART-1)

HP is required to produce 10 HP:

Motor A: X1 = 10/0.85 = 11.765 HP

Motor B: X2 = 10/0.90 = 11.11 HP

Cost of Annual energy computation:

Motor A: 11.765 * 0.7457 * 1,500 * 0.07 = 921.18

Motor B: 11.111 * 0.7457 * 1500 * 0.07 = 869.97

Present equivalent cost computation:

PE (8%) A = -800 - 921.18(P/A, 8%, 15) +100(P/F, 15) = -800- (921.18 * 8.5595) + (100 * 0.3152) = -8716

PE (8%) B = -1200 - 869.97(P/A, 8%, 15) = -1200 - (869.97 * 8.5595) = -8646.51

Motor B is preferred

PART-2)

When 2,500 operating hours we have:

PE (8%) A = -800-1535.26(P/A, 8%, 15) +50(P/F, 8%, 15) +100(P/F, 8%, 15)

= - 800 - (1535.26 * 8.5595) + (50*0.3152) + (100*0.3152)

= -13,894

PE (8%) B = -1200-1449.97(P/A, 8%, 15) = -1200 - 1449.97 * 8.5595 = -13,611

Thus Motor B will be preferred and still remains the choice

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