Is it better to oversize the motor for a little extra power, “just in case?” Consider this. Utilities often impose demand charges for poor power factor when a motor is seri- ously under-utilized. They also may subject cyclical power users to demand charges based on peak usage. The way it usually works is that one episode of high usage raises the kW/hr cost of electricity for the entire billing period. This imposes severe penalties for start- ing large motors across the line. Identifying oversized motors can help many users reduce peak demand charges.

The letter designations for locked- rotor kV•a / hp as measured at full voltage and rated frequency are shown in the table. To calculate the range of inrush current (locked rotor amps) for a motor, determine its NEMA code letter from the name- plate and solve the following equa- tion for the corresponding kV•a/hp values shown in the table:

CLR= CL x hp x 1000/(1.732 x V)

where CLR is locked-rotor current,

CL is kV•a/hp from the table,

hp is the motor horsepower, and

V is the voltage.

For a 40-hp motor with letter G designation (5.6 - 6.3 kVA/hp), the locked-rotor current will be 281 to 316 a:

CLR= 5.6 x 40 x 1000/(1.732 x 460)

= 281 a

CLR= 6.3 x 40 x 1000/(1.732 x 460)

=316 a

Based on the previous example, substituting a 25-hp replacement motor (code letter F) for the over- sized 40-hp model not only decreases full-load current from 37 to 30 a, but also drops starting current from 299 to 188 a. These factors help reduce wear and tear on motor starters, contacts, and other parts from unnecessarily high in- rush currents. The substitution also improves power factor appreciably.

Power factor and efficiency

Power factor, PF, goes hand-in-hand with efficiency, E, so it’s no surprise that the power factor for the 3-phase, 40 hp motor in our example is 0.76 when driving the 25.1 hp load:

E= Wout/Win

= 746 x hp/(1.732 x V x A x PF)

=746 x 25.1/(1.732 x 460 x 37 x 0.76)

=18,725/22,404

=0.836, or 83.6% efficiency

By comparison, the replace- ment 25-hp motor operates very efficiently:

E= 746 x hp/(1.732 x V x a x PF)

=746 x 25.1/(1.732 x 460 x 30 x 0.84)

=18,725/20,077

=0.933, or 93.3% efficiency

The power factor can be measured directly with a power factor meter. If a power factor meter is not available, but a watt meter is, the following equation can be used to calculate the power factor of a 3-phase motor:

PF= Win/V-ain

=22404/(1.732 x 460 x 37

= 22404/29479

=0.76

The original 40-hp motor was operating at only about 84% efficiency — nearly 10% below the 93% efficiency of the correctly sized premium-efficient replacement motor. According to the MotorMasterPlus software available from the U.S. Department of Energy (DOE), if the motor operated 8760 hr/yr (that’s 24/7), the original motor would have used more than 196,000 kW•hr/year, versus 176,000 kW•hr/yr for the 25-hp premium efficient model. At \$0.07/kW•h, that’s a savings of more \$1400/yr, every year. Where do you want to spend your money?

Thomas H. Bishop, P.E., is a technical support specialist with the Electrical Apparatus Service Association (EASA), St. Louis, an international trade association of more than 2400 firms that sell and service electrical, electronic, and mechanical apparatus. Call EASA at (314) 993-2220; or visit www.easa.com.