Lecture 06: Power (continued)Vijay Singh∗February 5, 2003AbstractContinuation of the discussion on power in R-L-C circuits.1 ExampleVS+-L1L2I1I2250 <0oV(rms)ISFigure 1: The circuit for the example.Given• Load L1(8 kW at pF of 0.8) leading• Load L2(20 kVA at pF of 0.6) lagging• L1and L2are in parallel with each other and the voltage across them is 2506(0) (rms).Find the power factor (pF) of the total load.1. For load L1:S1= P1+ jQ1θ1= cos−1(0.8) = 36.87◦|S1| =P1cos θ1=8 kW0.8= 10 kVA|Q1| = 6 kVAR∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky, Lex-ington, KY, USA. E-mail: [email protected] Lecture presented on January 31, 2003. Typeset in LATEX1P1 = 8 kWQ1S1theta1Figure 2: Complex power diagram for load L1.2. Similarly for load L2:S2= P2+ jQ2|S2| = 20 kVAθ2= cos−1(0.6)P2 = 12 kWQ2S2theta2Figure 3: Complex Power diagram for load L2.|P2| = |S2| cos θ2= 20(0.6) = 12 kW|Q2| = 20 sin θ2= 16kVAR3. Complex Power:S1= 8000 − j6000S2= 12000 + j16000Therefore, total power isS = S1+ S2= (20000 + j10000) VABut,S = VsI∗sTherefore,I∗s=20000 + j100002506(0)= 80 + j40 A2Is= 80 − j40 = 896(−26.57◦)VStheta3ISFigure 4: Illustration for θ3.Thus, θ3= 26.57◦.Power factor of the combined load iscos(θv− θi) = cos(26.57◦)lagging.pF = 0.8944 laggingthat is, current lags the voltage.4.|S| = Apparent Power= |(20000 + j10000)| = 22.36 kVAMagnitude of current supplied is|80 − j40| = 89.89 A5. Average power lost in the line is:I2sR = (89.89)2(0.05) = 404 wattsThus the power company must supply a real power of 20000 + 404 = 20404 watts of whichonly 20000 W is useful.Also it must supply an apparent power of (22360 + . . . . . . ) VA.6. If we put a capacitor in parallel to the two loads such that the power in the capacitor is−10 kVAR, then the net power factor will be 1.We want Q = −10000 VAR.But,Q =(Veff)21/(ωC)or,1ωC=250210000= 6.25Ω310 kVARtheta422.36 kVA20 kWFigure 5: How the capacitor will work.L1L2Figure 6: How the capacitor is connected to the load.C =1ω(6.25)=1276.99 × 6.25= 4.24.4 µFω = 2π(60) = 376.99With the corrective capacitor in place,Apparent power = |S| = 20 kVAIs=20000250= 80 AAverage power lost in the line is:I2sR = (80)2(0.05) = 320 WThus, the addition of capacitor has reduced the line loss from 404 W to 320
View Full Document
Unlocking...