Question 2.16: Cost Savings Associated with High- Efficiency Motors A 60-hp...

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

\begin{array}{l}\dot{W}_{\text {electric in, standard }}=\dot{W}_{\text {shaft }} / \eta_{\mathrm{st}}=(\text { Rated power })(\text { Load factor }) / \eta_{\mathrm{st}} \\\dot{W}_{\text {electric in, efficient }}=\dot{W}_{\text {shaft }} / \eta_{\mathrm{eff}}=(\text { Rated power })(\text { Load factor }) / \eta_{\text {eff }}\end{array}

Power savings =\dot{W}_{\text {electric in, standard }}-\dot{W}_{\text {electric in, efficient }}

\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=(\text { Rated power })(\text { Load factor })\left(1 / \eta_{\mathrm{st}}-1 / \eta_{\mathrm{eff}}\right)

where \eta_{\mathrm{st}} is the efficiency of the standard motor, and \eta_{\mathrm{eff}} 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

\begin{aligned}\text { Energy savings } &=(\text { Power savings })(\text { Operating hours }) \\&=(\text { Rated power })\left(\text { Operating hours)(Load factor) }\left(1 / \eta_{\mathrm{st}}-1 \eta_{\mathrm{eff}}\right)\right.\\&=(60 \mathrm{hp})(0.7457 \mathrm{~kW} / \mathrm{hp})(3500 \mathrm{~h} / \mathrm{year})(1)(1 / 0.89-1 / 0.932) \\&=7929 \mathrm{kWh} / \text { year } \\\text { Cost savings } &=(\text { Energy savings })(\text { Unit cost of energy }) \\&=(7929 \mathrm{kWh} / \text { year })(\$ 0.08 / \mathrm{kWh}) \\&=\$ \mathbf{6 3 4 / y e a r}\end{aligned}

Also,

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

This gives a simple payback period of

Simple payback period =\frac{\text { Excess initial cost }}{\text { Annual cost savings }}=\frac{\$ 640}{\text { \$634/year }}=1.01 year

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.