A 100-hp (shaft output) motor that has an efficiency of 89.6 percent is worn out

Question

A 100-hp (shaft output) motor that has an efficiency of 89.6 percent is worn out and is to be replaced by a high-efficiency motor that has an efficiency of 96.6 percent. The motor operates 5,360 hours a year at a load factor of 0.66. Taking the cost of electricity to be $0.054/kWh, determine the amount of energy and money saved as a result of installing the high-efficiency motor instead of the standard motor. Energy Savings = kWh/year Cost Savings = $/year Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5,267 and $5,726, respectively. months

Explanation / Answer

Old Motor = 100 hp, 89.6%

Input powerOld = 100/89.6 = 116.28 hp (= 86.7 kW)

New Motor = 100 hp, 96.6%

Input powerNew = 100/89.6 = 103.52 hp (= 77.2 kW)

Energy Saving = LoadFactor*OperatingHours*( Input powerOld - Input powerNew)

= 0.66*5360*(86.7-77.2) = 33607.2 kWh

Cost Savings = Energy Saving*Cost/unit = 33607.2*0.054 = 1814.79 $

Price Difference = 5726-5267 = 459 $

Payback time (in months) = (459/1814.79)* 12 = 3.035 months

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