A motor with a 200-horsepower output is needed in the factory for intermittent u

Question

A motor with a 200-horsepower output is needed in the factory for intermittent use. A Graybar motor costs 7000, and has an electrical efficiency of 89 percent. A blueball motor costs 6000 and has an 85 percent efficiency. Neither motor would have any salvage value, since the cost to remove it would equal its scrap value. The annual maintenance cost for either motor is estimated at 300 per year. Electric power costs 0.0072/kWh (1 hp = 0.746 kW). If a 10% interest rate is used in calculations, what is the minimum number of hours the higher initial cost Graybar motor must be used each year to justify its purchase?

Explanation / Answer

Difference in cost is $1,000. Interest is 10%. So we need to save $100 per year to just pay the interest on the difference. It would be nice to know when we can apply the energy savings; monthly or yearly? But as long as we pay interest on the difference in price as often as we apply the energy savings, the numbers work out the same. 89% efficiency 200*0.746 = 149.2 kW is what we need x*0.89 = 149.2 KW x = 149.2/.89 = 167.6404494 this is kW electricity needed to generate 149.2 kW motion 85% efficienct x = 149.2/0.85 = 175.5294118 difference = 7.888962327 kW electric company charges for energy. Power times time = energy kW is power h = hours = time Each hour the more efficent motor saves us 0.0072 $/kWh*7.89kW = $0.0568 $100/$0.0568/h = 1,760.5 hr each year (or about 34 hours per week).

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