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 93%. Either motor will be operated 4,000 hours per year at full load, and electricity costs $0.10 per kilowatt-hour (kWh). MARR= 15% per year, and neither motor will have a market value at the end of the eight-year study period. Most motors are operated at a fraction of their rated capacity (i.e., full load) in industrial applications. Suppose the average usage (load factor) of motors is expected to be 60%. Which motor should be recommended under this condition? Click the icon to view the interest and annuity table for discrete compounding when 15% per year. The PW of the motor with the efficiency of 90% is SD (Round to the nearest dollar) The PW of the motor with the efficiency of 93% is (Round to the nearest dollar.) Which motor should be chosen? Choose the correct answer below. 90% efficiency motor 93% efficiency motor O

Answer 1

cost per year is given by formula = machine capacity/Rate of efficiency (Rate of capacity) (cost per kwh) (Hours of operation) Therefore cost for 90% efficient motor = 30 hp/0.90 times 0.746 kw/hp times $ 0.10 kwh times 4000 hr/yr 90% = $ 9,946.67 per year Now cost for 93% efficient motor = 30 hp/0.93 times 0.746 kw/hp times $ 0.10 kwh times 4000 hr/yr c_93% = $ 9,625.81 per year \r\n pw = energy savings [(1 + i)^n - 1/i (1 + i)] Here i is value Therefore PW_90 % = -2200 - 9946.67 ((1 + 0.15)^8 - 1/0.15 (1 + 0.15)^8) Pw_90 % = -$ 46834 Pw_90 % = -3200 - 9625.81 ((1 + 0.15)^8 - 1/0.15 (1 + 0.15)^8) Pw_93 % = -$ 46394 since pw_93 % is higher Therefore this motor should be chosen that is motor with 93% efficiency should be chosen