sinotech.org – Power factor can be improved by installing capacitors correcting power factor in electric distribution systems / electrical installations in factories / industries. Capacitors act as reactive power generation and therefore will reduce the amount of reactive power, as well apparent power generated by the utility.
An example showing the improvement of power factor by installing capacitors is shown below:
Example 1. A chemical plant installed a transformer 1500 KVA . The plant needs in the first 1160 KVA with power factor 0.70. The percentage of transformer loading about 78 percent (1160/1500 = 77.3 percent). To improve the power factor and to avoid penalties by electricity suppliers, the plant adds approximately 410 kVAr the motor load. This improves the power factor up to 0.89 and this can reduces the KVA that required becomes 913 KVA , which is the sum vector of KVAR and KW. 1500 kVA transformer and then only under load 60 % from the capacity. So that the plant will be able to increase the load on the transformer in the future
Example 2. A group of incandescent lamps with a voltage of 220V/58 W, coupled with a fluorescent lamp 12,, 11 W, there are 30 pieces of incandescent and fluorescent lamp. Power factor measured by cos alpha1 = 0.5. Calculate the apparent power of the load and the magnitude of current I1 before the compensation, if desired work factors becomes cos alpha2 = 0.9. Calculate the magnitude of current I2 (after compensation).
1. The amount of the combined light
PG = (58 W x 18) + (11 W x 12) = 1176 watts = 1.176 KW
Cos phi 1 => PG/S1 > S1 => Pg / Cos phi 1 => 1.176 KW/ 0.5 => 2.352 kVA.
I1 = S1 / U = 2.352 V = 10.69 kVA/220 ampere (A) -> before compensation
2. The amount of power after compensation (cos phi = 0.9)
S2 = PG / Cos phi2 = 1.176 kW / 0.9 = 1.306 kVA
Then I2 = S2 / U = 1.306 V = 5.94 kVA/220 A -> after compensation