Buffalo Pumps, Inc. is in the process of designing a new production facility and
ID: 2744233 • Letter: B
Question
Buffalo Pumps, Inc. is in the process of designing a new production facility and are currently considering whether to use a conventional induction motor or a more modern premium efficiency motor 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 kWh) and differ only in their efficiency ratings and cost. The conventional induction motor has an efficiency rating of 85% while the premium efficiency motor has a rating of 90%. Further, the induction motor costs only $15,000 per unit while the premium efficiency motor 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 (induction motor) and $1000 (premium efficiency motor). 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 Pumps 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 Pumps has an MARR of 12%, what are the savings per kWh they can expect if they choose the premium efficiency motor rather than the conventional induction motor?
(b) What number of operating hours would make the two motors equally economical?
Explanation / Answer
Cost of Conventional Motor
Year
Particulars
Present Value Factor@12%
Calculation
Calculated Amount
Discounted Value of Calculated Amount
0y
Initial Investment
1
$15,000*1
$ 15,000.00
$ 15,000.00
1y-20y
Operating Cost
7.46944
$.05perkwh*18.650kwh*12hrsper day*5 days per week*52 weeks*70% load/85% Efficiency
$ 2,395.98
$ 17,896.60
20y
Salvage Value
0.10367
$1,500*.10367
$ (1,500.00)
$ (155.51)
$ 32,741.10
Cost of Premium Efficiency Motor
Year
Particulars
Present Value Factor@12%
Calculation
Calculated Amount
Discounted Value of Calculated Amount
0y
Initial Investment
1
$17,500*1
$ 17,500.00
$ 17,500.00
1y-20y
Operating Cost
7.46944
$.05perkwh*18.650kwh*12hrsper day*5 days per week*52 weeks*70% load/90% Efficiency
$ 2,262.87
$ 16,902.35
20y
Salvage Value
0.10367
$1,000*.10367
$ (1,000.00)
$ (103.67)
$ 34,298.68
The conventional motor is better than premium efficiency motor as its cost is lower than Premium Efficiency Motor.
2).Let represent the number of hours of full load operation for two motars to economical break even.
The total cost may be equated as follows
Total cost=Initial cost + Facilitating Cost
15,000+($.05perkwh*18.650kwh*70% load/85% Efficiency)*X hrs=
17,500+($.05perkwh*18.650kwh*70% load/90% Efficiency)*X Hrs
=15000+0.76794*X hrs=17500+.725278*X hrs
2500=.04266Xhrs
X=58602 Hrs
So at 58602 hrs both motors are equally economical.
Cost of Conventional Motor
Year
Particulars
Present Value Factor@12%
Calculation
Calculated Amount
Discounted Value of Calculated Amount
0y
Initial Investment
1
$15,000*1
$ 15,000.00
$ 15,000.00
1y-20y
Operating Cost
7.46944
$.05perkwh*18.650kwh*12hrsper day*5 days per week*52 weeks*70% load/85% Efficiency
$ 2,395.98
$ 17,896.60
20y
Salvage Value
0.10367
$1,500*.10367
$ (1,500.00)
$ (155.51)
$ 32,741.10
Cost of Premium Efficiency Motor
Year
Particulars
Present Value Factor@12%
Calculation
Calculated Amount
Discounted Value of Calculated Amount
0y
Initial Investment
1
$17,500*1
$ 17,500.00
$ 17,500.00
1y-20y
Operating Cost
7.46944
$.05perkwh*18.650kwh*12hrsper day*5 days per week*52 weeks*70% load/90% Efficiency
$ 2,262.87
$ 16,902.35
20y
Salvage Value
0.10367
$1,000*.10367
$ (1,000.00)
$ (103.67)
$ 34,298.68
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