Question

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?

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