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Chapter 12 Three-Phase Circuits Power Factor Correction Medium Voltage Metal Enclosed Capacitor Banks A balanced three-phase circuit Three-Phase Voltages vaa 2 V cos t vbb 2 V cos( t 120) vcc 2 V cos( t 240) (a) The three windings on a cylindrical drum used to obtain three-phase voltages (b) Balanced three-phase voltages Three-Phase Voltages Generator with six terminals Three-Phase Balanced Voltages Vaa V 0 Vbb V 120 Vcc V 240 V 120 Phase sequence or phase rotation is abc Positive Phase Sequence neutral terminal Va V 0 Vc V 120 Vb V 240 V 120 Phase sequence or phase rotation is acb Negative Phase Sequence Two Common Methods of Connection phase voltage (a) Y-connected sources (b) -connected sources Phase and Line Voltages Vab Va Vb V p0 V p 120 V p V p ( 0.5 j 0.866) The line-to-line voltage Vab of the Y-connected source 3V p30 Similarly Vbc 3V p 90 Vca 3Vp 210 The Y-to-Y Circuit A four-wire Y-to-Y circuit The Y-to-Y Circuit (cont.) Four - wire Va Vb Vc I aA , IbB , and I cC ZA ZB ZC InN IaA IbB IcC The average power delivered by the three-phase source to the three-phase load P PA PB PC When ZA = ZB = ZC the load is said to be balanced Va V 0 Vb V 120 Vc V 120 I aA , IbB , and I cC Z A Z ZB Z ZC Z V V V I aA , IbB ( 120), and I cC ( 120) Z Z Z The Y-to-Y Circuit(cont.) There is no current in the wire connecting the neutral node of the source to the neutral node of the load. InN IaA IbB I cC 0 The average power delivered to the load is P PA PB PC V V V V cos( ) V cos( ) V cos( ) Z Z Z V2 3 cos( ) Z The Y-to-Y Circuit (cont.) A three-wire Y-to-Y circuit The Y-to-Y Circuit (cont.) Three - wire We need to solve for VNn Va VNn Vb VNn Vc VNn 0 ZA ZB ZC V 0 VNn V 120 VNn V 120 VNn ZA ZB ZC Solve for VNn (V 120)Z AZC V 120Z AZ B V 0Z B ZC VNn Z AZC Z AZ B Z B ZC Va VNn Vb VNn Vc VNn I aA , IbB , and I cC ZA ZB ZC The Y-to-Y Circuit (cont.) When the circuit is balanced i.e. ZA = ZB = ZC (V 120) ZZ V 120ZZ V 0ZZ VNn ZZ ZZ ZZ 0 The average power delivered to the load is P PA PB PC V2 3 cos( ) Z The Y-to-Y Circuit (cont.) Transmission lines A three-wire Y-to-Y circuit with line impedances The Y-to-Y Circuit (cont.) The analysis of balanced Y-Y circuits is simpler than the analysis of unbalanced Y-Y circuits. VNn = 0. It is not necessary to solve for VNn . The line currents have equal magnitudes and differ in phase by 120 degree. Equal power is absorbed by each impedance. Per-phase equivalent circuit Example 12.4-1 S = ? Va 1100 Vrms Vb 110 120 Vrms Vc 110120 Vrms Z A 50 j80 Z B j50 ZC 100 j 25 Unbalanced 4-wire I aA Va 1100 Z A 50 j80 IbB Vb 110 120 2.2150 A rms ZB j50 I cC 1.16 58 A rms Vc 110120 1.07 106 A rms ZC 100 j 25 Example 12.4-1 (cont.) S A I*aAVa 68 j109 VA S B I*bB Vb j 242 VA SC I*cC Vc 114 j 28 VA The total complex power delivered to the three-phase load is S S A S B SC 182 j379 VA Example 12.4-2 S = ? Balanced 4-wire Va 1100 Vrms Z A 50 j80 Vb 110 120 Vrms Z B 50 j80 Vc 110120 Vrms ZC 50 j80 I aA Va 1100 Z A 50 j80 1.16 58 A rms S A I*aAVa 68 j109 VA The total complex power delivered to the three-phase load is S 3S A 204 j326 VA Also IbB 1.16 177 Arms , IcC 1.1662 Arms SB 68 j109 VA SC Example 12.4-3 S = ? Va 1100 Vrms Vb 110 120 Vrms Vc 110120 Vrms Z A 50 j80 Z B j50 ZC 100 j 25 Unbalanced 3-wire Determine VNn (110 120)Z AZC 110120Z AZ B 1100Z B ZC VNn Z AZC Z AZ B Z B ZC 56 151 Vrms Va VNn Vb VNn Vc VNn I aA , IbB , and I cC ZA ZB ZC Example 12.4-3 (cont.) IaA 1.71 48, IbB 2.453, and IcC 1.1979 S A I*aA Va I*aA (I aAZ A ) 146 j 234 VA S B I*bB Vb I*bB (I bB Z B ) j 94 VA SC I*cC Vc I*cC (I cC ZC ) 141 j 35 VA The total complex power delivered to the three-phase load is S S A S B SC 287 j364 VA Example 12.4-4 S = ? Balanced 3-wire Va 1100 Vrms Z A 50 j80 Vb 110 120 Vrms Z B 50 j80 Vc 110120 Vrms ZC 50 j80 I aA Va 1100 Z A 50 j80 1.16 58 A rms S A I*aAVa 68 j109 VA The total complex power delivered to the three-phase load is S 3S A 204 j326 VA Example 12.4-5 PLoad = ? PLine = ? PSource = ? Balanced 3-wire Va 1000 I aA ( ) Z A 50 j (377)(0.045) per-phase equivalent circuit 1.894 18.7 A The phase voltage at the load is VAN ( ) (40 j(377)(0.04))IaA ( ) 812 V Example 12.4-5 (cont.) The power delivered by the source is Vm I m Pa cos(V I ) 2 (100)(1.894) cos(18.7) 89.7 W 2 The power delivered to the load is I m2 (1.894)2 PA Re( Z A ) 40 71.7 W 2 2 The power lost in the line is I m2 (1.894)2 PaA Re( Z Line ) 10 17.9 W 2 2 Line loss 20 % The -Connected Source and Load Vab 1200 Vrms Vbc 120.1 121 Vrms Vca 120.2121 Vrms Circulating current ( Vab Vbc Vca ) I 3.75 A 1 Unacceptable Total resistance around the loop Therefore the sources connection is seldom used in practice. The -Y and Y- Transformation Z1Z3 ZA Z1 Z 2 Z3 ZA ZB ZC Z1Z 2 ZC Z1 Z 2 Z3 Z A Z B Z B ZC Z A ZC Z1 ZB Z3 Z1 Z2 Z3 ZB Z1 Z 2 Z3 Z2 Z A Z B Z B ZC Z A ZC Z2 ZA Z A Z B Z B ZC Z A ZC Z3 ZC Example The Y- Circuits I aA I AB ICA IbB I BC I AB I cC ICA I BC where I AB VAB Z3 I BC VBC Z1 ICA VCA Z2 The Y- Circuits (cont.) IaA I AB ICA I cos j sin I cos( 120) j sin( 120) 3I ( 30) or IaA 3 I I L 3I p Example 12.6-1 IP = ? IL = ? 220 30 Vrms 3 220 Vb 150 Vrms 3 220 Vc 90 Vrms 3 Va The -connected load is balanced with Z 1050 VAB Va Vb 2200 Vrms VBC Vb Vc 220 120 Vrms VCA Vc Va 220 240 Vrms The line currents are I AB I BC ICA VAB 2250 A rms Z VBC 22 70 A rms Z VCA 22 190 A rms Z IaA I AB ICA 22 320, IbB 22 3 100, IcC 22 3 220 The Balanced Three-Phase Circuits Y-to- circuit equivalent Y-to-Y circuit Z ZY 3 per-phase equivalent circuit Example 12.7-1 IP = ? Va 1100 Vrms Vb 110 120 Vrms Vc 110120 Vrms Z L 10 j5 Z 75 j 225 Z ZY 25 j 75 3 Va I aA 1.26 66 A rms Z L ZY Example 12.7-1 (cont.) IbB 1.26 186 Arms and IcC 1.26 54 Arms The voltages in the per-phase equivalent circuit are VAN I aAZY 99.65 Vrms VBN 99.6 115 Vrms VCN 99.6125 Vrms The line-to-line voltages are VAB VAN VBN 17235 Vrms VBC VBN VCN 172 85 Vrms I AB VAB 0.727 36 A rms Z I BC VBC 0.727 156 A rms Z VCA VCN VAN 172155 Vrms ICA VCA 0.72784 A rms Z Instantaneous and Average Power in BTP Circuits One advantage of three-phase power is the smooth flow of energy to the load. vab V cos t , vbc V cos( t 120), The instantaneous power and vca V cos( t 240) 2 vab vbc2 vca2 (1 cos 2 t ) 2 p (t ) cos t R R R 2 V2 1 cos 2 t 1 cos 2( t 120) 1 cos 2( t 240) 2R 3V 2 V 2 cos 2 t cos(2 t 240) cos(2 t 480) 2R 2R 0 3V 2 2R Instantaneous and Average Power in BTP Circuits The total average power delivered to the balanced Y-connected load is IaA I L AI , VAN VP AV Phase A 3VL I L cos( ) 3 3 VL I L cos( ) 3VP I L cos( ) PY 3PA 3VP I L cos( AV AI ) Instantaneous and Average Power in BTP Circuits The total average power delivered to the balanced -connected load is 3VP I L cos( ) 3 3( 3VP ) L cos( ) I P 3PAB 3VAB I AB cos( ) Example 12.8-1 P = ? Va 1100 Vrms Vb 110 120 Vrms Vc 110120 Vrms Z L 10 j5 Z 75 j 225 Va I aA 1.26 66 A rms Z L ZY VAN IaAZY 99.65 Vrms P 3(99.6)(1.26) cos(5 ( 66)) 122.6 W Two-Wattmeter Power Measurement cc = current coil vc = voltage coil W1 read P1 VAB I A cos1 W2 read P2 VCB I C cos 2 For balanced load with abc phase sequence 1 a 30 and 2 a 30 a is the angle between phase current and phase voltage of phase a Two-Wattmeter Power Measurement(cont.) P P1 P2 2VL I L cos cos30 3VL I L cos To determine the power factor angle P1 P2 VL I L 2cos cos30 P1 P2 VL I L (2sin sin30) P1 P2 VL I L 2 cos cos 30 3 P1 P2 VL I L ( 2sin sin 30) tan P1 P2 tan 3 P1 P2 P1 P2 or tan 3 P1 P2 1 Example 12.9-1 P = ? Z 1045 line-to-line voltage = 220Vrms The phase voltage 220 VA 30 3 The line current VA 220 30 IA 12.7 75 and IB 12.7 195 Z 10 345 P1 VAC I A cos1 2698 W P2 VBC I B cos 2 723 W P P1 P2 3421 W Electrodynamic Wattmeter Digital Power Meter pf Meter VAR Meter Summary Three-Phase voltages The Y-to-Y Circuits The -Connected Source and Load The Y-to- Circuits Balanced Three-Phase Circuits Instantaneous and Average Power in Bal. 3 Load Two-Wattmeter Power Measurement