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 = $7000-$6000 = $1,000.

Interest = 10%

So it is required to save $100 per year to just pay the interest on the difference.

89% efficiency, i.e. 200×0.746 = 149.2 kW is what is required

Thus,

(x)×(0.89) = 149.2 KW

x = 149.2/.89 = 167.64 kW electricity is needed to generate 149.2 kW motion

85% efficient implies:

x = 149.2/0.85 = 175.52

Thus, net difference = 7.88 kW

Electric company charges for energy.

Each hour the more efficient 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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