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Electrical Power & Machines βElectrical Engineering Deptβ Prepared By: Dr. Sahar Abd El Moneim Moussa Dr. Sahar Abd El Moneim Moussa 1 Dr. Sahar Abd El Moneim Moussa 2 Power Factor Correction ο 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= π π P Ρ² Q S Dr. Sahar Abd El Moneim Moussa 3 ο Accordingly the power factor can be defined as Cosine the angle between the voltage & the current in the a.c. circuit P.F= Cos ΞΈ I IC IR IL - IC V Ρ² IL Dr. Sahar Abd El Moneim Moussa Itotal 4 ο In inductive circuits ο There is lagging P.F ο In capacitive circuits ο there is leading P.F ο 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 ο To improve the P.F, some devices taking leading power should be connected in parallel to the load. ο This device draws a leading current & partly or completely neutralizes the lagging reactive component of the load current. Dr. Sahar Abd El Moneim Moussa 5 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: ο The active current of the motor is IL(active) = IL1 x Cos ο±1 = 50 x 0.6= 30 A Dr. Sahar Abd El Moneim Moussa 6 Dr. Sahar Abd El Moneim Moussa 7 πΌπΏ (πππ‘ππ£π) ο Motor new current IL2= πΆππ π2 = 30 0.9 = 33.3 π΄ ο IL1 (Reactive) = IL1 x Sin ο±1=50 x 0.8 = 40 A ο IL2(Reactive) = IL2 x Sin ο±2=33.3 x 0.435 =14.53 ο Current neutralized by the capacitor= IL1 - IL2 = 40-14.53 = 25.47 A ο Since the Capacitor is connected in parallel ο¦ π β΄ πΌπ = ππ = Vο¦ x 2 ο° f C πΌπ 25.47 β΄πΆ= = = 202 ππΉ πο¦ 2 π π 400 π 2 π π π π Dr. Sahar Abd El Moneim Moussa 8