6-29. Two electric motors are being considered to drive a centrifugal pump. One

Question

6-29. Two electric motors are being considered to drive a centrifugal pump. One of the motors delivering 60 horsepower (output) to the pumping operation. It is expected that the motors will be in use 800 hours per must be selected. Each motor is capable of year. The following data are available: (6.5) Motor A Motor B Capital investment Electrical efficiency Annual maintenance Useful life $1,200 0.92 $160 3 years $1,000 0.80 $100 6 years a. If electricity costs $0.07 per kilowatt-hour, which motor should be selected if the MARR is 8% per year? Recall that 1 hp 0.746 kW. Assume repeatability b. What is the basic trade-off being made in this problem?

Explanation / Answer

By applying the annual worth for motor A, we get AW(8%) = R-E-CR(i%) R = Revenue E = Expense CR = Capital recovery Awa(8%) = -1200(A/P,8%,3)-$160 - 60hp/0.92*(0.746kW/hp)*(800 hrs/yr)*(0.07kWh) Aw=-1200*0.388-160-((60/0.92)*0.746*800*0.07) -3350.12 Awb(8%) = -1000(A/P,8%,6)-$100 - 60hp/0.80*(0.746kW/hp)*(800 hrs/yr)*(0.07kWh) AW = - 1000*0.2163-100-((60/0.80)*0.746*800*0.07) -3449.5 Motor A should be selected as it has lower Annual worth The basic trade off being made in this problem is the cost involved in the maintenance of the machine and the one which has lower annual worth is selected

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