USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

POWER FACTOR IMPROVEMENT

The power factors of different loads are as follows:

Type of load                                     Power factor (range)

Heating and lighting                                0.95 to unity

Motor loads                                                0.5 to 0.9

Single-phase motors                               as low as 0.4

Electric welding units                                0.2 to 0.3

 

4

In each case, the kVA is directly proportional to current.

The higher current due to poor power factor affects the system and lead to the following undesirable results:

  1. Rating of alternators and transformers are proportional to their output current hence inversely proportional to power factor, therefore large generators and transformers are required to deliver same load but at a low power factor.
  2. When a load having a low lagging power is switched on, there is a large voltage drop in the supply voltage because of the increased voltage drop in the supply lines and transformers. This drop in voltage adversely affects the starting torque of motors and necessitates expensive voltage stabilizing equipment for keeping the consumer’s voltage fluctuations within the statutory limits.
  3. The cross-sectional area of the bus-bar, and the contact surface of switch gear is required to be enlarged for the same power to be delivered but at a low power factor.
  4. To transmit same power at low power factor, more current will have to be carried by the transmission line or the distributor or cable. If the current density in the line is to be kept constant the size of the conductor will have to be increased. Thus more copper is required to deliver the same load but at a low power factor.
  5. Copper losses are proportional to the square of the current hence inversely proportional to the square of the power factor i.e. more copper losses at low power factor and hence poor efficiency.

Thus we find that the capital cost of transformers, alternators, distributors, transmission lines etc. increase with the low power factor.

Causes of Low Power Factor:

(i)                           All A.C. motors (except overexcited synchronous motors and certain types of commutators motors) and transformers operate at lagging power factor.

(ii)                        Due to typical characteristic of the arc, arc lamps operate at low power factor.

(iii)                      When there is increase in supply voltage, which usually occurs during low load periods (such as lunch hours, night hours etc.) the magnetizing current of inductive reactances increases and power factor of the electrical plant as a whole decreases.

(iv)                      Arc and induction furnaces etc., operate at a very low lagging power factor.

(v)                         Due to improper maintenance and repairs of motors the power factor at which motors operate fall.

Methods of Improving Power Factor. The power factor may improve by using the following methods:

  1. Use of high power factor motors.
  2. Use of induction motors with phase advancers.
  3. Use of static capacitors.
  4. Use of capacitance boosters.
  5. Use of synchronous condensers.

Example 1. A 40-MV A, 11 k V 3-phase alternator supplies full-load at a lagging power factor of 0.6. Find the percentage increase in earning capacity if the power factor is increased to 0.9.

Solution. Rating of the alternator = 40 MV A

Supply voltage = 11 kV

Initial power factor = 0.6 (lagging)

Final power factor after increase = 0.9 (lagging)

Percentage increase in earning capacity:

The earning capacity is proportional to the power (in MW or kW) supplied by the alternator.

MW supplied at 0.6 lagging = 40 × 0.6 = 24

MW supplied at 0.9 lagging = 40 × 0.9 = 36

Increase in MW = 36 – 24 = 12

The increase in earning capacity is proportional to 12.

5

Example 2. A 3-phase, 40 kW, 440 V, 50 Hz induction motor operates on full-load with an efficiency of90 per cent and at a power factor of 0.8 lagging. Calculate the total kVA rating of capacitors required to raise the full-load power factor to 0.9 lagging. What will be capacitance per phase if the capacitances are:

(i)                           Delta connected

(ii)                        Star-connected.

Solution. Motor output = 40 kW

Supply voltage =440 V

Efficiency of motor =90%

6

Example 3. Three impedance coils each having a resistance of 10 Ω and a reactance of 7.5 Ω are connected in star to a 400 V, 3-phase 50 Hz supply. Calculate:

(i)                           Power factor

(ii)                        Line current, and

(iii)                      Power supplied.

If three capacitors, each of the same capacitance, are connected in delta to the same supply so as to from parallel circuit with the above impedance coils, calculate the capacitance of each capacitor to obtain resultant power factor 0.9lagging.

Solution. Supply voltage, EL = 400 V [Line-to-line]

15