Note: This is the first part of the PF Correction tutorial series.

In industrial plants, induction motors comprises majority of the load. Without power factor correction, the nominal industrial plant overall power factor ranges from about 0.70 to 0.80 lagging.

A low power factor means that the **I ^{2}R** losses is high. Some electrical utility companies also penalize plant with low power factors as this leads to a high system loss. On the hand, plant with high power factor are given incentive discounts on the billing which will make any investment on a power factor correction equipment viable.

To illustrate this better, let us provide a real world example.

Plant AMonthly average

Monthly energy consumption : 125 000 kwhs

Power Factor : 0.75 Lagging

Utilization Voltage : 400V, 3 phase

**What is the size of capacitor bank needed to raise the power factor to 0.95 lagging?**

**Solution**:

Number of hours per month : 720 h

The formula of KVAR in terms or the kwh and PF is

KVAR = [kwh/720] * [ (1/PF)^{2}- 1 ]^{0.5}

Let us calculate the reactive power of the plant at 0.75 PF and at 0.95 PF. The difference of the two KVARs shall be the size of the capacitor bank.

**At 0.75 PF**

KVAR_{0.75}= [125 000/720] * [ (1/0.75)^{2}- 1 ]^{0.5}

KVAR_{0.75}= 153.10 kVAR

**At 0.95 PF**

KVAR_{0.95}= [125 000/720] * [ (1/0.95)^{2}- 1 ]^{0.5}

KVAR_{0.95}= 57.06 kVAR

KVAR = KVARsay 100 kVAR_{0.75}- KVAR_{0.95}

KVAR = 153.10 - 57.06

KVAR = 96.04 kVAR

The best combination for the capacitor bank may be

4 x 25kVAR or

5 x 20kVAR

The higher the number of steps the better the PF regulation will be when it is used with a PF controller. This however has a cost implication as the latter will be most costly than the former. A 20kVAR unit does not cost much less than a 25VAR unit. Proper selection of combination is desirable when designing PF correction system.

Next time - we shall be calculating the savings for using power factor correction. Will it pay for itself? Will find that out on the next tutorial.