Buffalo Pumps, Inc. is in the process of designing a new production facility and
ID: 2743302 • 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. 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? What number of operating hours would make the two motors equally economical?Explanation / Answer
a.)
b.)
Ignore depriciation impact as this is non-cash expense has no impact expense on cash flow and its impact on output will have no effect.
Conventional motor(85%) Premium efficiency motor(90%) Incremental cost Cost 15000 17500 2500 Depriciation on 20 years salvage value 675 825 150 Operations for hrs 60 60 0 Price negotiated per hour 0.05 0.05 0 Total output = 18.650*Capacity*price per hour*operation 47.5575 50.355 2.7975 Cash Flow Generated =Total output * PVIFA 323.2534947 342.2684062 19.014911 PVIFA based on MARR basis = (1/1+MARR)*(1+MARR)^Expected project life-1) 6.797108652Related Questions
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