Power-Factor Correction Using Synchronous Motor

For a constant load, the power factor of a synchronous motor can be varied from a leading value to a lagging value by adjusting the DC field excitation (Figure bellow). Field excitation can be adjusted so that PF = 1 (Figure a). With a constant load on the motor, when the field excitation is increased, the counter EMF (VG) increases. The result is a change in phase between stator current (I) and terminal voltage (Vt), so that the motor operates at a leading power factor (Figure b). Vp in Figure  is the voltage drop in the stator winding’s due to the impedance of the windings and is 90o out of phase with the stator current. If we reduce field excitation, the motor will operate at a lagging power factor (Figure c). Note that torque angle, a, also varies as field excitation is adjusted to change power factor.

Synchronous motors are used to accommodate large loads and to improve the power factor of transformers in large industrial complexes.

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. bellow shows the power factor of the load  corrected to cos 62 by a synchronous motor

 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 . - 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  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 

 Vector Diagram of Example

Example 2. - The static condenser of the preceding example would cost $3000 installed. Wwhat 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