Motors & Drives Information

Many professionals who work with AC motors consider their operational behavior a mystery. This month I'll try to clear up some of the questions and address typical concerns.

Question

I have a 20hp motor running at full load producing 20hp of torque and the motor is rated as 90% efficient. Does that mean that the motor actually pulls 16.4 kW in order to produce 14.9 kW worth of work? If that is correct then how does a power factor of 0.60 versus a power factor of 0.90 affect the motor?

Discussion Group Answers

Input power to the motor increases as the motor efficiency falls. Power factor will not affect the input power, but it will affect the input current. If the motor input power is 16.4 kW at a power factor of 1, it will be the same with a power factor of 0.6. The current however will increase as the power factor drops due to the addition of reactive currents, harmonic currents, or both. In the case of an induction motor fed from a standard supply, the additional current is inductive and does not add to kilowatt usage.

If my understanding is correct, then what someone pays for is the true power and not the kVA and electric companies just have an additional charge for the increased current being used to deliver the power as a result of a low power factor.

Standard electrical meters measure true kilowatt hours and electrical utilities bill for each of these hours. Utilities may have surcharges designed to minimize line losses due to poor power factor and reduce maximum demand. Surcharges vary by region. Some utilities use thermal meters and bill based on maximum demand (Amps). In such cases, charges are calculated based on the highest current draw in a three-month period. Thus, a single high current draw for 30 minutes would be used to as the basis for the entire 3-month power bill. Some utilities charge extra when your maximum demand coincides with the maximum load on the power distributor. Such fees provide a major incentive to draw a constant load 24/7.
Efficiency is dependent on motor loading. If motor loading is 50%, for example, the efficiency will drop according to the manufacturer-provided efficiency versus loading curve. At low loads, the motor becomes less efficient. If the motor shaft load is equal to zero, the motor efficiency is equal to zero.

The power factor is dependent on the motor load. It decreases rapidly with smaller loads. The magnetizing branch of the motor significantly influences the power factor decrease for small motor loading. In many cases, power factor versus motor loading curves are available from manufacturers.
Power factor issues relate only to AC machines. Windings represent an inductive load, and this produces a lagging power factor. (A pf = 1 is perfect; this means that, for example, that 0.9 is better than 0.6.) Because circuits must be capable of supplying this unwanted current, cables and switchgear must be sized up to carry it. Improve power factor by installing power factor correction capacitors. They produce a leading current that offsets the lagging current produced by the motor windings. Electricity tariffs often have power factor penalty clauses in the billing structure.
Clarifications

In the question, 20hp of torque is misleading. The motor could produce from 30 lb-ft of torque to 120 lb-ft of torque, depending on the base speed of the motor. A standard 4-pole motor (1750rpm base speed) would produce about 60 lb-ft of torque.

Application of AC motors requires an understanding of both the power demand by the motor and the duty cycle demand by the motor. Users have a tendency to oversize motors for a given application. Motors are designed to operate best at their nameplate rating.

However, it is also important to consider that the cables feeding power to the motor are an important factor. Losses in any cables should be kept below 5% voltage loss in all cases, and as close to 3% as possible. Whenever possible, power factor correction should be applied at the motor to reduce distribution system losses due to reactive currents. In cases where the motor loading varies widely, the changes in power factor will not typically result in additional utility penalties as long as peak power levels are below maximum demand levels.

Motor manufacturers have improved efficiency by increasing the cross sectional dimension of the wire used in the windings. Since this reduces the I²R losses, less heat is lost in the motor. The downside to this improvement is that inrush currents are greater than with many older motors. In addition, since a lower slip occurs in the induction motor, the actual speed of the motor, when operated across the line, is greater. This increase in speed may have a negative effect in the application.

In applications where pulse width modulated (PWM) AC drives control the motors, input power factors are almost unity under all motor loading conditions. Although some concerns exist regarding harmonic currents, additional distribution system loading is generally less when compared to the current when the motor is operated across the line.

# posted by Medical Information : 3:42 AM