Answer:
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
