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|
The power factor is given by :
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:
- 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.
- 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.
- 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.
- 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.
- 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
Methods of Improving Power Factor. The power factor may improve by using the following methods:
- Use of high power factor motors.
- Use of induction motors with phase advancers.
- Use of static capacitors.
- Use of capacitance boosters.
- Use of synchronous condensers.
POWER FACTOR IMPROVEMENT
Example 57. A 40-MVA, 11 kV 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.
Example 58. A 3-phase, 40 kW, 440 V, 50 Hz induction motor operates on full-load with efficiency of 90 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:
- Delta connected
Solution. Motor output = 40 kW
Supply voltage = 440 V
Efficiency of motor = 90%
Power factor 0.8 (lagging)
cos ɸ1 = 0.8
ɸ1 = cos ” 0.8 = 36.9°
tan ɸ1 = 0.75
Motor kVAR1 = P tan ɸ1 = 44.44 × 0.75 = 33.33
Power factor 0.9 (lagging)
Motor power input = 44.44 kW [As before]
[ Power is same as before since the capacitors
are loss free i.e., they do not absorb any power ]
cos ɸ2 = 0.9
ɸ2 = 25.8°
tan ɸ2 = 0.484
Motor kVAR2 = P tan ɸ2 = 44.44 × 0.484 = 21.5.
The difference in values of kVAR is due to the capacitors which supply leading kVAR to partially neutralize the lagging kVAR of the machine.
Leading kVAR supplied by capacitors is
= kVAR1 – kVAR2 = 33.33 – 21.5
= 11.83 ……… MQ or NT.
Since capacitors are loss-free, their kVAR is same as kVA
Total KVA rating of capacitors = 11.83 (Ans.)
(i) Delta connection. In delta-connection, voltage across each capacitor is 440 V.
Hence, capacitance of each capacitor = 50.35 µF. (Ans.)
(ii) Star-connection. In star-connection, voltage across each capacitor
Example 59. 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 there 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 of 0.9 lagging.
Solution. Supply voltage, EL = 400 V [Line-to-line]
Phase resistance, Rph = 10 Ω
Phase reactance, Xph = 7.5 Ω
When capacitors are connected:
Power factor, cos ɸ2 = 0.9
ɸ2 = 25.8°
tan ɸ2 = 0.4834
Since capacitors themselves do not absorb any power, power remains the same i.e., 10243 W even when capacitors are connected. The only thing that changes is the VAR.
Now VAR2 = P tan ɸ2 = 10243 × 0.4834 = 4951
Leading VAR supplied by three capacitors is
= VAR1 – VAR2 = 7682 – 4951 = 2731 = MQ or NT (Fig. 81)
For delta connection voltage across each capacitor is 400 V
Hence, capacitance of each capacitor = 18.10 µF. (Ans.)