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2 November, 21:46

A special-purpose 30-horsepower electric motor has an efficiency of 9090 %. Its purchase and installation price is $2 comma 2002,200. A second 30-horsepower high-efficiency motor can be purchased for $3 comma 2003,200 , and its efficiency is 9393 %. Either motor will be operated 4 comma 0004,000 hours per year at full load, and electricity costs $0.100.10 per kilowatt-hour (kWh). MARRequals=1515 % 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 6060 %. Which motor should be recommended under this condition?

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  1. 2 November, 22:11
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    Motor 1 recommended

    Given Information:

    Two 30-horsepower motors

    Consumption = 4,000 hours per year

    Electricity cost = $0.10 per kWh

    Load factor of motors = 60 %

    MARR = 15 %

    Study period = 8 years

    Motor 1:

    Efficiency = 90 %

    Purchase + Installation cost = $2,200

    Motor 2:

    Efficiency = 93%

    Purchase + Installation cost = $3,200

    Required information:

    Which motor should be recommended under these conditions = ?

    Solution:

    Electricity cost = Rate of electricity*Units consumed*Load factor

    Electricity cost: Motor 1

    Units consumed = (30 hp/0.90) * (0.746 kW/hp) * (4000 h/year)

    Units consumed = 99,466.66 kWh/year

    Electricity cost = (99,466.66 kWh/year) * (0.10/kWh) * (0.60)

    Electricity cost = $5,968 per year

    Electricity cost: Motor 2

    Units consumed = (30 hp/0.93) * (0.746 kW/hp) * (4000 h/year)

    Units consumed = 96,258 kWh/year

    Electricity cost = (96,258 kWh/year) * (0.10/kWh) * (0.60)

    Electricity cost = $5,775 per year

    Present Worth Analysis: Motor 1

    PW = - $2,200 - $5,968 (P/A, 15%, 8)

    where (P/A, 15%,8) = Uniform series present worth at 15% MARR and n = 8 years

    P/A = (1 + i) ⁿ - 1 / i * (1 + i) ⁿ

    P/A = (1 + 0.15) ⁿ - 1 / 0.15 (1 + 0.15) ⁿ

    P/A = 4.487

    PW = - $2,200 - $5,968 (4.487)

    PW = - $28,978

    Present Worth Analysis: Motor 2

    PW = - $3,200 - $5,775 (P/A, 15%, 8)

    P/A = 4.487

    PW = - $3,200 - $5,775 (4.487)

    PW = - $29,112

    Recommendation:

    In present worth analysis, we select the alternative which has largest PW value i. e more positive or least negative.

    Since the PW of motor 1 (90% efficiency) is greater (less negative) than the motor 2 (93% efficiency).

    Therefore, Motor 1 recommended
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