EE201 Lecture 39 P 1 Power Factor and Power Factor Correction Power Factor pf Average Power Pave pf S Apparent Power Pave VeffIeff pf cos v i Power factor angle pfa Terminology pf lagging 0 v i 180o pf leading 0 i v 180o EE201 Lecture 39 I P 2 I 1 k V 0 4 H 1 k V a b a Parallel RL circuit illustrating lagging pf b Parallel RC circuit illustrating leading pf P P P pf S P jQ P2 Q2 Solving for Q 1 Q P 1 2 pf Q 0 pf lagging Q 0 pf leading 1 F EE201 Lecture 39 P 3 Comparison of systems with same average power but different apparent power S1 S2 Q1 Q2 P1 P2 S1 low pf large Q Large losses S2 high pf small Q Small losses Conclusion System with S1 is inefficient EE201 Lecture 39 P 4 Example Suppose the motor absorbs 50kW of average power with pf 0 8 lagging The terminal voltage is Vm 230 V The frequency of operation is 60 Hz or 120 rad s For the first part of the example the capacitor is not connected to the motor In part c the capacitor is connected to the motor to improve the power factor a Find the complex power delivered to the motor b Find Is Vs and the power delivered by the source which might represent the power delivered by the local electric company c Correct pf to 0 95 lagging of the combined motor capacitor load d Compute the power delivered by the source to the combined motor capacitor load Rline 0 5 Is Vs Vm Im Motor 50 kW IC C EE201 Lecture 39 P 5 Solution Part a Find the complex power delivered to the motor Pav 50 kW Sm pf Sm 0 8 Sm 50 0 8 62 5 kVA VmIm Where Vm Vm and Im Im Compute Sm Lagging means the current phase lags behind the voltage phase i e v i 0 Consider the diagram below which shows the current phasor Im lags behind the voltage phasor i e the current phasor makes an angle of 36 87 cos 1 0 8 Sfrom the voltage phasor Vm m VmI m I m Hence Sm 36 87 0 36 87 0 36 87 Im Vm EE201 Lecture 39 P 6 Compute complex power Sm By definition Sm Sm Sm VmIm Sm 62 5 36 87 kVA 50 j37 5 kVA Pav jQ Part b Find Is Vs and the power delivered by the source Is Sm 62 5 103 271 74 A Vm 230 Is 271 74 36 87 217 4 j163 A From KVL Vs RlineIs Vm 0 5 217 4 j163 230 338 7 j81 5 348 4 13 53 V Ss VsI s 348 4 13 53 271 74 36 87 VA 94 66 23 34 86 92 j37 5kVA EE201 Lecture 39 P 7 Part c Correct the power factor to 0 95 lagging of the combined motor capacitor load Smnew 50 cos 1 0 95 52 63 18 195 50 j16 43 kVA Recall that Smold 50 j37 5 kVA Find the capacitor value to reduce reactive power jQold jQnew j21 07 kvar j reactive power of capacitor jQC Therefore jQC j21 07 kvar VmI C Vm YC j Vm j C Vm 2 It follows that C QC Vm 2 21 07 103 1 057 10 3 F 120 2302 EE201 Lecture 39 P 8 Part d Compute the new power delivered by the source Smnew 50 j16 43 kVA Smnew Isnew V m 50 j16 43 230 103 217 j71 43 A From KVL and Ohm s law Vsnew 0 5Isnew Vm 338 5 j35 72 V Ssnew Vsnew Isnew 76 j16 48 kVA Hence the new average power delivered by the source is 76 kW as opposed to 86 9 kW without power factor correction With this power factor correction there is a reduction of 86 9 76 10 9 kW of power loss in the line connecting the source to the load EE201 Lecture 39 P 9 Example Motor A 50 kW pf 0 8 Motor B 5 kW pf 0 7 R 230V 60Hz 100kVA A B B CA CB CB a What are the generator current and the pf for the combined load of 1 large and 3 small motors b Compute the number of small motors that can be run simultaneously with the large motor without exceeding the gen rating c Repeat b assuming the pf of all motors equals 0 9 by adding parallel capacitors d Compute capacitances required in c EE201 Lecture 39 P 10 Solution part a For each type of motor the reactive power is Qa Pa 1 1 1 50 1 37 5kVAR 2 2 pf 0 8 Qb Pb 1 1 1 5 1 5 1kVAR 2 2 pf 0 7 The complex power supplied by the generator is found from conservation principle S gen S a 3S b S gen Pa jQa 3 Pb jQb S gen 50 j 37 5 3 5 j 5 1 S gen 65 j 52 8 83 7 39 1 kVA Below 100kVA EE201 Igen Lecture 39 P 11 83 7kVA 230V pf combined 364A P 2 P Q 2 65 2 65 52 8 2 0 78 Solution part b S gen Pa jQa n Pb jQb S gen 50 j 37 5 n 5 j 5 1 The apparent power is S gen 50 5n 37 5 5 1n kVA 2 2 S gen 100kVA Roots of equation n1 5 3 n2 22 5 5 small motors can be run simultaneously with large motor EE201 Lecture 39 P 12 Solution part c Recalculate reactive power numbers 50 Motor A Qa Motor B Qb 5 1 1 24 2kVA 2 0 9 1 1 2 4kVA 0 9 2 Calculate complex power for n small motors S gen 50 5n 24 2 2 4n 100kVA 2 2 Roots n1 8 n2 28 8 small motors can be run with 1 large motor EE201 Lecture 39 P 13 Solution part d The reactive power supplied by a capacitor is QC VCIC VC CVC The difference between reactive power in part b and part c is Qa 37 5 24 2 13 3kVAR For the large motor and Qb 5 1 2 4 2 7 kVAR For the small motors EE201 Lecture 39 Therefore Qa CaVca 2 Qa 2 60 Ca 230 13 3kVAR 2 Ca 6 66 10 4 F Qb CbVcb 2 Qb 2 60 Cb 230 2 7 kVAR 2 Cb 1 34 10 4 F P 14
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