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Power-Factor Correction by Synchronous Motors

Author: E.E. Kimberly

As explained in Chapter 19, the current taken by a synchronous motor may be made to lead the applied voltage by over-exciting the field. Thus, a synchronous motor may be made to provide a load having a component that simulates a condenser and provides some capacitive correction to any associated load of lagging power factor. Fig. 25-5 shows the power factor of the load of Fig. 25-2 (a) corrected to cos 62 by a synchronous motor.

Fig. 25-5. Power-Factor Correction by Synchronous Motor

Most synchronous motors are provided with a field winding of sufficient size to permit excitation to a power factor of 0.8 leading, when carrying the full rated shaft load. Thus, if a motor is rated at 100 hp with an efficiency of 92% at full load, the power input at 100% power factor would be


When the motor is excited to a leading power factor of 0.8, the total kv-a input would be


and the reactive kv-a would be


Power-factor correction obtained by over-excitation of synchronous motors is very low in cost and is highly attractive.

Example 25-1. - A consumer has an average monthly energy consumption of 140,000 kw-hr. The average maximum demand is 670 kw. The average power factor is 0.60. The plant works 576 hours per month. Calculate the kv-a of static condensers required to avoid a power-factor penalty under Schedule C of Chapter 24.

Solution. - The average hourly values are:


To avoid penalty the power factor must be raised to 0.75. The allowable kv-a values are:


The number of kv-a in static condensers required to avoid a power-factor penalty is


The vector diagram is shown in Fig. 25-6,

Fig. 25-6. Vector Diagram of Example 25-1

Example 25-2. - The static condenser of the preceding example would cost $3000 installed. Under Schedule C of Chapter 24, what would be the saving in the power bill in the first year?

Solution. - Inasmuch as the power factor is to be corrected to the minimum of 0.75 allowable without penalty, the billing demand will be equal to the measured demand of 670 kw. The charges are as follows:

Demand Charges:

10 · $2.50-$ 25.00

30 · 2.10= 63.00

160 · 1.80= 288.00

200 · 1.60= 320.00

(670-400) · 1.50= 405.00

Total = $1101.00

Last Update: 2010-10-06