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Pharos University EE-385. Electrical Power & Machines “Electrical Engineering Dept” Prepared By: Dr. Sahar Abd El Moneim Moussa. Power Factor Correction. Power Factor Correction.
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Pharos UniversityEE-385 Electrical Power & Machines “Electrical Engineering Dept” Prepared By: Dr. SaharAbd El MoneimMoussa Dr. Sahar Abd El Moneim Moussa
Power Factor Correction Dr. Sahar Abd El Moneim Moussa
Power Factor Correction • The power factor is the ratio between the active power of the load(P in watt) to the total power of the load (S in VA) P.F= P Ѳ Q S Dr. Sahar Abd El Moneim Moussa
Accordingly the power factor can be defined as Cosine the angle between the voltage & the current in the a.c. circuit P.F= Cos θ I IC IR V Ѳ IL - IC Itotal IL Dr. Sahar Abd El Moneim Moussa
In inductive circuits There is lagging P.F • In capacitive circuits there is leading P.F • Most of loads are inductive in nature & have low lagging P.F which is highly undesirable as it causes an increase in current resulting in additional losses of active power in all elements of the power system • To improve the P.F, some devices taking leading power should be connected in parallel to the load. • This device draws a leading current & partly or completely neutralizes the lagging reactive component of the load current. Dr. Sahar Abd El Moneim Moussa
Example 2: A single phase 400 V, 50 Hz motor takes a supplying current 50 A at a power factor 0.6 lagging. The motor p.f. has been improved to 0.9 lagging by connecting a capacitor in parallel. Calculate the capacity of the capacitor. Solution: • The active current of the motor is IL(active) = IL1 x Cos 1 = 50 x 0.6= 30 A Dr. Sahar Abd El Moneim Moussa
Motor new current IL2= • IL1 (Reactive) = IL1 x Sin 1=50 x 0.8 = 40 A • IL2(Reactive) = IL2x Sin 2=33.3 x 0.435 =14.53 • Current neutralized by the capacitor= IL1 - IL2 = 40-14.53 = 25.47 A • Since the Capacitor is connected in parallel = V x 2 f C Dr. Sahar Abd El Moneim Moussa