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?

Question

Engineering

Jasmine Gregory

21 March, 13:30

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?

+1

Answers (1)

Know the Answer?

Sierra Mendez · Answer 1

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