Motor that has an efficiency of 87.5 percent is worn out and is to be replaced b

Question

Motor that has an efficiency of 87.5 percent is worn out and is to be replaced by a high-efficiency motor that has an efficiency of 95.8 percent. The motor operates 3, 960 hours a year at a load factor of 0.77. Taking the cost of electricity to be S0.061/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 = S/year Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5, 163 and $6, 052, respectively. 7 months

Explanation / Answer

1 hp = 746 W

Power output = 65 hp = 48490 W = 48.5 kW

Old power input = 48.5/0.875 = 55.428 kW

New power input = 48.5/0.958 = 50.626 kW

Energy = (Power)*(load-factor)*(time)

Old energy consumption = 55.428*0.77*3960 = 169011.05 kW-h

New energy consumption = 50.626*0.77*3960 = 154368.7992 kW-h

Hence energy saved = 169011.05 - 154368.7992 = 14642.2508 kW-h/year

Cost savings = Energy savings x unit cost of energy = 14642.2508 x $0.061 = $893.17 / year

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