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29 December, 22:13

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high - efficiency motor that has an efficiency of 95.4 percent. the motor operates 4368 hours a year at a load factor of 0.75. taking the cost of electricity to be $0.12/kwh, determine the amount of energy and money saved as a result of installing the high-efficiency motor instead of the standard motor. also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5449 and $5520, respectively.

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  1. 30 December, 01:31
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    Convert the shaft ouput from HP to kW

    Shaft output = 75HP = 55.93kW

    1st: Finding for the power consumption based on 55.93kW output

    Power consumption (Old) = 55.93kW /.910 = 61.46kW

    Power consumption (New) = 55.93kW /.954 = 58.63kW

    2nd: Total power used in kWh:

    Power Used = Power consumption * load factor * Hours:

    Power (Old) = 61.46kW *.75 * 4368 = 201343 kWh

    Power (New) = 58.63kW *.75 * 4368 = 192072 kWh

    Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

    3rd: Calculating for the price:

    Price = kW-Hr * $/kWh

    Price (Old) = 201343kWh * $0.08/kWh = $16107.44

    Price (New) = 192072 kWh * $0.08/kWh = $15365.76

    Cost saved = $16107.44 - $15365.76 = $741.68/yr

    4th: Setting up the cost equation:

    Cost over time, F (t) = Motor_Cost + (Price * Number of Years, t)

    Cost (Old) = 5449 + 16107.44*t

    Cost (New) = 5520 + 15365.76*t

    Equate the two to find for t when they cost equally:

    5449 + 16107.44*t = 5520 + 15365.76*t

    16107.44*t = 15365.76*t + 71

    16107.44*t - 15365.76*t = 71

    741.68*t = 71

    t = 71 / 741.68 =.095 years = 35 days

    So the payback period is after 35 days.
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