Question

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 \(\$ 0.12 / \mathrm{kWh}\), 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.

Short answer

Additionally, what is the simple payback period for the high-efficiency motor? Answer: By installing the high-efficiency motor instead of the standard motor, a total of 11,257 kWh of energy and $1,350.84 in electricity costs can be saved annually. The simple payback period for the high-efficiency motor is approximately 19 days.

Step-by-step solution

Calculate the power consumption of both motors

To calculate the power of the standard motor and the high-efficiency motor, we will use: Power (kW) = (Horsepower * 0.746) / Efficiency For the standard motor: Power_standard = (75 * 0.746) / 0.91 = 61.494 kW For the high-efficiency motor: Power_high_efficiency = (75 * 0.746) / 0.954 = 58.576 kW

Calculate the yearly energy consumption for each motor

Given the operating hours and load factor, we can calculate the yearly energy consumption for each motor: Energy_consumption (kWh) = Power (kW) * Operating hours * Load factor For the standard motor: Energy_standard = 61.494 kW * 4368 hours * 0.75 = 203,597 kWh For the high-efficiency motor: Energy_high_efficiency = 58.576 kW * 4368 hours * 0.75 = 192,340 kWh

Calculate energy and money saved by installing the high-efficiency motor

We can calculate the energy saved by subtracting the energy consumption of the high-efficiency motor from the standard motor's energy consumption: Energy_saved = Energy_standard - Energy_high_efficiency = 203,597 kWh - 192,340 kWh = 11,257 kWh The cost of electricity is given as \(\$ 0.12 / \mathrm{kWh}\). Therefore, we can calculate the money saved: Money_saved = Energy_saved * Cost_of_electricity = 11,257 kWh * \$0.12/kWh = \$1,350.84

Calculate the simple payback period

The simple payback period is calculated by dividing the additional investment required to purchase the high-efficiency motor by the money saved per year. Additional_investment = High_efficiency_motor_price - Standard_motor_price = \$5520 - \$5449 = \$71 Simple_payback_period = Additional_investment / Money_saved_per_year = \$71 / \$1,350.84 = 0.0526 years or approximately 19 days. In conclusion, by installing the high-efficiency motor instead of the standard motor, a total of 11,257 kWh of energy and \$1,350.84 in electricity costs can be saved annually. The simple payback period for the high-efficiency motor is approximately 19 days.