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# Power Factor Correction using Capacitor Bank

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Power Factor
The ratio of the true power of an AC Circuit to the apparent power; the ratio of R/Z; equal to the cosine of the phase angle between applied voltage and circuit current.

The ratio of true power to apparent power in an AC Circuit is called the power factor and can be expressed as follows:

$PF=\frac{True\text{ Power P}}{Apparent\text{ Power S}}$

It is also defined as the ratio of resistance to the impedance (series circuit):

$PF=\frac{\operatorname{Resistance}\text{ R}}{\operatorname{Impedance}\text{ Z}}$

Since R/Z is the cosine of the impedance angle, we can say that the power factor is equal to the cosine θ:

$PF=\cos (\theta )$

Because the impedance of a circuit is equal to or greater than the resistance, the power factor is a decimal fraction that has a value between 0 and 1.

When θ=±90o, for a purely reactive circuit, PF is 0 because

$Cos({{90}^{o}})=0$

At the other extreme, when θ=0o, for a purely resistive circuit, PF is 1 because

$Cos({{0}^{o}})=1$

The power factor is a dimensionless quantity because it is the ratio of two numbers with the same dimensions.

Many industrial and residential electrical loads are inductive in nature that is they operate at a lagging power factor. The consequence of a lagging power factor is that a large apparent power S is required in comparison to that required for a similar load operating at unity power factor.

This is easily seen ointhe following power triangle.

In above figure, an apparent power S is required to produce a power, P, for a load operating at a power factor angle of θ. However, If the same power, P, is maintained while the power factor angle of θ is brought closer to 0°, the hypotenuse of the power triangle becomes smaller. The apparent Power S needed to supply a given true Power, P, is the minimum. A lower apparent power S is desirable since for a given voltage, a lower line current, and greater efficiency results.

Although it is not possible to change the inductive nature of a load itself, it is possible to connect a capacitive load in parallel with the inductive load to cancel some inductive reactive power, thereby decreasing the apparent power.

The addition of a capacitor in parallel with an inductive load so as to reduce the apparent power while not altering the voltage or current to the original load is known as power factor correction. A power factor correction circuit and associated power triangle are shown in the following Fig.

In reference to the power triangle, the parallel capacitor supplies a reactive power, QC, which cancels some of the original reactive power, Q1, leaving a net inductive power Q2 .Accordingly, the apparent power is decreased from S1 to S2. Notice that if an overall power factor of unity is desired, the capacitor is chosen so that its reactive power equals the reactive power of the inductive load. In such a case;

$P=S~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~at\text{ }Cos\theta =1$

## Example of Power factor correction

A load operating at a lagging power factor of 0.7 dissipates 2 KW when connected to a 220 V, 60 Hz power line. What value of capacitance is needed to correct the power factor to 0.9?

Solution

Referring to the given data and above mentioned figure,

${{\theta }_{1}}=Co{{s}^{-1}}\left( 0.7 \right)={{45.6}^{{}^\circ }}$

${{\theta }_{2}}=Co{{s}^{-1}}\left( 0.9 \right)={{26}^{{}^\circ }}$

Then,

${{Q}_{1}}=Ptan{{\theta }_{1}}=2000*tan({{45.6}^{{}^\circ }})=2040\text{ }~vars$

${{Q}_{2}}=Ptan{{\theta }_{2}}=2000*tan({{26}^{{}^\circ }})=975~\text{ }vars$

And

${{Q}_{C}}={{Q}_{1}}-{{Q}_{2}}=1065~\text{ }vars$

From $~~~~{{Q}_{C}}=\frac{{{V}^{2}}}{{{X}_{c}}}$ ,

${{X}_{C}}=\frac{{{V}^{2}}}{{{Q}_{C}}}=\frac{{{200}^{2}}}{1065}=\frac{4.84*{{10}^{4}}}{1.065*{{10}^{3}}}=45.4~\Omega$

And

$C=\frac{1}{\omega {{X}_{C}}}=\frac{1}{2*\pi *f*45.4}=58.3~\mu F$

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