Cost Savings Associated with High-Efficiency Motors A 60-hp electric motor (a motor that delivers 60 hp of shaft power at full load) that has an efficiency of 89.0 percent is worn out and is to be replaced by a 93.2 percent efficient high-efficiency motor (Fig. 2–61). The motor operates 3500 hours

Question

Cost Savings Associated with High-Efficiency Motors A 60-hp electric motor (a motor that delivers 60 hp of shaft power at full load) that has an efficiency of 89.0 percent is worn out and is to be replaced by a 93.2 percent efficient high-efficiency motor (Fig. 2–61). The motor operates 3500 hours a year at full load. Taking the unit cost of electricity to be $0.08/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 $4520 and $5160, respectively.

Accepted Answer

A worn-out standard motor is to be replaced by a high-efficiency one. The amount of electrical energy and money saved as well as the simple payback period are to be determined.
Assumptions The load factor of the motor remains constant at 1 (full load) when operating.
Analysis The electric power drawn by each motor and their difference can be expressed as

[latex]\begin{aligned}\dot{W}_{ ext {electric in, standard }} &=\dot{W}_{ ext {shaft }} / \eta_{ ext {st }}=( ext { Rated power })( ext { Load factor }) / \eta_{ ext {st }} \\dot{W}_{ ext {electric in, efficient }} &=\dot{W}_{ ext {shaft }} / \eta_{ ext {eff }}=( ext { Rated power })( ext { Load factor }) / \eta_{ ext {eff }} \ ext { Power savings } &=\dot{W}_{ ext {electric in, standard }}-\dot{W}_{ ext {electric in, efficient }} \&=( ext { Rated power })( ext { Load factor })\left(1 / \eta_{ ext {st }}-1 / \eta_{ ext {eff }} ight)\end{aligned}[/latex]

where [latex]\eta _{st}[/latex] is the efficiency of the standard motor, and [latex]\eta _{eff}[/latex] is the efficiency of the comparable high-efficiency motor. Then the annual energy and cost savings associated with the installation of the high-efficiency motor become

[latex]\begin{aligned} ext { Energy savings } &=( ext { Power savings })( ext { Operating hours }) \&=( ext { Rated power })( ext { Operating hours })( ext { Load factor })\left(1 / \eta_{ st }-1 / \eta_{ ext {eff }} ight) \&=(60 hp )(0.7457 kW / hp )(3500 h / ext { year })(1)(1 / 0.89-1 / 0.93 .2) \&=7929 kWh / ext { year } \ ext { Cost savings } &=( ext { Energy savings })( ext { Unit cost of energy }) \&=(7929 kWh / ext { year })(\$ 0.08 / kWh ) \&=\$ 634 / ext { year }\end{aligned}[/latex]

Also,

Excess initial cost = Purchase price differential = $5160 - $4520 = $640

This gives a simple payback period of

[latex] ext { Simple payback period }=\frac{ ext { Excess initial cost }}{ ext { Annual cost savings }}=\frac{\$ 640}{\$ 634 / ext { year }}=1.01 ext { year } [/latex]

Discussion Note that the high-efficiency motor pays for its price differential within about one year from the electrical energy it saves. Considering that the service life of electric motors is several years, the purchase of the higher efficiency motor is definitely indicated in this case.

Question 2.17: Cost Savings Associated with High-Efficiency Motors A 60-hp ...

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