Power Factor Correction Calculation

Power Factor Correction Calculation

Let the inductive load operating at lagging p.f. cos phase advancers and taking a current of I₁ from the supply. It is required to improve p.f. of this circuit. For this purpose, a capacitor is connected in parallel with the load which takes a leading current for compensating lagging reactive component of load. This is shown in Fig. 7.22.1.

The capacitor current I₂ leads the voltage V by 90°. The supply current I is phasor sum of Ii and I₂ and its phase angle is ϕ which is less than ϕ. As ϕ < ϕ cos ϕ  > cos ϕ the p.f. of the circuit is improved.

From the phasor diagram we have,

I sin ϕ = I₁ sin ϕ  - I₂

Capacitor current, I₂ = I₁ sin ϕ – I sin ϕ

The capacitive reactance of capacitor is given by,

Capacitance required to improve p.f., C = I_C / ωV

Let us understand the power factor improvement with the help of power triangle shown in the Fig. 7.22.2 For p.f. cos ϕ, the power triangle is OXZ while for improved p.f. cos ϕ, the power triangle is OXY. The active part of power remains unchanged. But the lagging kVAR of load is reduced by capacitor to enhance the p.f. from cos ϕ ' to cos ϕ. Leading kVAR supplied by capacitor = YZ

= XZ - XY = kVAR₁ - kVAR

= OX tan ϕ ' – OX tan ϕ

= OX(tan ϕ ' – tan ϕ)

= kW (tan ϕ ' - tan ϕ)

With the knowledge of leading kVAR supplied by capacitor, other results can be obtained.

Example 7.22.1 A 3 phase, 50 Hz, 3300 V star connected induction motor develops 250 hp (186.5 kW), the power factor being 0.707 lagging and the efficiency 0.86. Three capacitors in delta are connected across the supply terminals and power factor raised to 0.9 lagging. Calculate

i) The kVAR rating of the capacitor bank

ii) The capacitance of each unit

Solution :

Example 7.22.2 A 3 phase, 50 Hz, 3000 V motor develops 447.6 kW, the p.f. being 0.75 lagging and the efficiency 0.93. A bank of capacitors is connected in delta across the supply terminals and p.f. raised to 0.95 lagging. Each of the capacitance units is built of five similar 600 V capacitors. Determine the capacitance of each capacitor.

Solution :

p.f. before improvement = cos ϕ ' = 0.75 lag  ϕ ' = cos^-1 0.75 = 41.40°

tan ϕ ' = tan 41.40° = 0.8819

p.f. after improvement = cos ϕ = 0.95 lag  ϕ = cos^-1 0.95 = 18.19°

tan ϕ = tan 18.19° = 0.3286

kW = Motor input = Motor output / Efficiency = 447.6 / 0.93 = 481.29 kW.