Question

A special-purpose 30-horsepower electric motor has an efficiency of 90%. Its purchase and installation price is $2,200. A second 30-horsepower high-efficiency motor can be purchased for $3,200, and its efficiency is 92%. Either motor will be operated 4,000 hours per year at full load, and electricity costs $0.10 per kilowatt-hour (kWh). MARR = 12% per year, and neither motor will have a market value at the end of the eight-year study period. a. Which motor should be chosen? b. For an incremental investment of $1,000 in the more efficient motor, what is the PW of the energy savings over the eight-year period? Recall that 1 hp = 0.746 kW.

Answer 1

a. Energy consumed by 90% efficient motor in one year = 30 * 0.746 * 4000 / 0.90 = 99466.67 Kwh

Cost of energy consumed by 90% efficient motor = 99466.67 Kwh * 0.10 = 9946.67

Energy consumed by 92% efficient motor in one year = 30 * 0.746 * 4000 / 0.92 = 97304.35 Kwh

Cost of energy consumed by 92% efficient motor = 97304.35 Kwh * 0.10 = 9730.43

EUAC of 90% efficient motor = 2200 * (A/P, 12%,8) + 9946.67

= 2200 * 0.201303 + 9946.67

= 10389.54

EUAC of 92% efficient motor = 3200 * (A/P, 12%,8) + 9730.43

= 3200 * 0.201303 + 9730.43

= 10374.60

As annual cost of 92% efficient motor is less, it should be selected

b.

energy savings per year = 9946.67 - 9730.43 = 216.24

PW of energy savings = 216.24 * (P/A, 12%,8) = 216.40 * 4.967640 = 1074.20