Buffalo Wire Works is in the process of designing a new production facility and are currently considering which new motor to use for their testing equipment. Both motors have the same specifications (25 HP, 230 V, 60 Hz, 1800 rpm), provide the same power output

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Buffalo Wire Works is in the process of designing a new production facility and are currently considering which new motor to use for their testing equipment. Both motors have the same specifications (25 HP, 230 V, 60 Hz, 1800 rpm), provide the same power output (0.746 kW/HP or 18.650 kW) and differ only in their efficiency ratings and cost. The first motor (Motor A) has an efficiency rating of 85% while the second motor option (Motor B) has a rating of 90%. Further, Motor A costs only $15,000 per unit while Motor B costs $17,500. Both have an expected lifecycle of 20 years. At the end of their useful lives the motors will have salvage values of $1500 (Motor A) and $1000 (Motor B). The testing facility is expected to operate in support of their new extended shift which operates 12 hours per day, 5 days a week, year round. Buffalo Wire Works have negotiated a price of $0.05 per kWh with the local energy provider and plans to operate the testing station, and thus the motor, at 70% load.

(a) If Buffalo Wire Works has an MARR of 12%, what are the savings per kWh they can expect if they choose Motor B rather than Motor A?

(b) [15 points] What number of operating hours would make the two motors equally economical?

Answer : A) $1522.25    B) 623 HOURS

Explanation:

Given data :

Motor A : efficiency = 85% . cost = $15000. life = 20 years. salvage value = $1500. efficiency at 70% work load = 56%

Motor B : efficiency = 90%. cost = $1750. life = 20 years. salvage value = $1000. efficiency at 70% work load = 63%

A) Calculate  the savings per kWh they can expect if they choose Motor B rather than Motor A

i) = Total power ( Kwh ) * ( difference in efficiencies at 70% work load for motor A and motor B )

= 18.650 * ( 63% - 56% ) = 1.3055 kwh

ii) Discounting factor for MARR of 12% for 20 years  = 7.4737

iii) calculation of new extended shift hours

= 12 hours * 5 days * 52 weeks = 3120 hours

to ascertain the amount of savings PER Kwh if Motor B is chosen rather than motor A

= 18650 Kwh * 7% * 7.4737 * $0.005 per kwh * 3120 hours = $1522.25

B)  What number of operating hours would make the two motors equally economical

To get this right we have to calculate the NPV  for both machine A and machine B

assuming effective cost of motor A = effective cost of motor B

also assume of hours = p

Motor A

cost for year 1 = $15000

discounting factor at year 1 = 1

Testing cost at 20 years = 7.4737 * $0.05 * 18.650 kwh * p * 85% * 70% = 4.1467 p

discounting factor at 20 years on testing cost = 7.4737

salvage value as on 20 years = ( 0.1040 * $1500 ) = $156

discounting factor at 20 years on salvage value = 0.1040

Total value for motor A = cost for year 1 + testing cost at 20 years - salvage value on 20 years = $15000 + 4.1467 p - $156

Motor B

we apply the same discounting factor values as with motor A

Cost for year 1 = $17500

testing cost at 20 years = 7.4737 * $0,05 * 18.650 kwh * p * 90% * 70% = 4.3907 P

salvage value as on 20 years = 0.1040 * $1000 = $104

Total value for motor B = $17500 + 4.3907 p - $104

equate the Total vale of motor A  to the total value of motor B to find p

=  $15000 + 4.1467 p - $156 = $17500 + 4.3907 p - 104

therefore p = 622.69 ≈ 623 hours

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explanation: