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Chapter 1. Three-Phase System 1 1.1: Review of Single-Phase System The Sinusoidal voltage v1(t) = Vm sin wt i v1 v2 Load AC generator 2 1.1: Review of Single-Phase System The Sinusoidal voltage v(t) = Vm sin wt where Vm = the amplitude of the sinusoid w = the angular frequency in radian/s t = time v(t) Vm wt -Vm 3 v(t) Vm t -Vm 2 T w 1 f T w 2f The angular frequency in radians per second 4 A more general expression for the sinusoid (as shown in the figure): v2(t) = Vm sin (wt + q) where q is the phase v(t) Vm q V1 = Vm sin wt wt -Vm V2 = Vm sin wt + q) 5 A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. We can transform a sinusoid from sine to cosine form or vice versa using this relationship: sin (ωt ± 180o) = - sin ωt cos (ωt ± 180o) = - cos ωt sin (ωt ± 90o) = ± cos ωt cos (ωt ± 90o) = + sin ωt 6 Sinusoids are easily expressed in terms of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoid. v(t) = Vm cos (ωt + θ) V Vmq Time domain Phasor domain Time domain Phasor domain Vm cos( wt q ) Vm q Vm sin( wt q ) Vm q 90 o I m cos( wt q ) I m q I m sin( wt q ) I m q 90o 7 Time domain v(t) v2(t) = Vm sin (wt + q) V1 = Vm sin wt Vm v1(t) = Vm sinwt q wt -Vm V2 = Vm sin wt + q) Phasor domain V2 θ V1 Vm 0 or V1 Vrms0 V2 Vmq or V2 Vrmsq V1 8 1.1.1: Instantaneous and Average Power The instantaneous power is the power at any instant of time. p(t) = v(t) i(t) Where v(t) = Vm cos (wt + qv) i(t) = Im cos (wt + qi) Using the trigonometric identity, gives 1 1 p( t ) Vm I m cos( q v qi ) Vm I m cos( 2wt q v qi ) 2 2 9 The average power is the average of the instantaneous power over one period. 1 P T P T 0 p (t ) dt 1 Vm I m cos( q v q i ) 2 p(t) 1 Vm I m 2 1 Vm I m cos( q v q i ) 2 t 10 The effective value is the root mean square (rms) of the periodic signal. The average power in terms of the rms values is P Vrms I rms cos( qv qi ) Where Vrms I rms Vm 2 Im 2 11 1.1.2: Apparent Power, Reactive Power and Power Factor The apparent power is the product of the rms values of voltage and current. S Vrms I rms The reactive power is a measure of the energy exchange between the source and the load reactive part. Q Vrms I rms sin( qv qi ) 12 The power factor is the cosine of the phase difference between voltage and current. P Power factor cos( qv qi ) S The complex power: P jQ Vrms I rms qv qi 13 1.2: Three-Phase System In a three phase system the source consists of three sinusoidal voltages. For a balanced source, the three sources have equal magnitudes and are phase displaced from one another by 120 electrical degrees. A three-phase system is superior economically and advantage, and for an operating of view, to a singlephase system. In a balanced three phase system the power delivered to the load is constant at all times, whereas in a single-phase system the power pulsates with time. 14 1.3: Generation of Three-Phase Three separate windings or coils with terminals R-R’, Y-Y’ and B-B’ are physically placed 120o apart around the stator. R Stator Y R B B’ N Y’ Rotor Y S B R’ 15 V or v is generally represented a voltage, but to differentiate the emf voltage of generator from voltage drop in a circuit, it is convenient to use e or E for induced (emf) voltage. 16 v(t) vR vY vB wt The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin wt eY = EmY sin (wt -120o) eB = EmB sin (wt -240o) = EmBsin (wt +120o) 17 Three-phase AC generator IR VR ER ZR IN EB VB EY IY VY ZB ZY IB Three-phase 18 Load Phase voltage The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin ωt eY = EmY sin (ωt -120o) eB = EmB sin (ωt -240o) = EmBsin (ωt +120o) In phasor domain: 120o ER = ERrms 0o EY = EYrms -120o EB = EBrms 120o 0o -120o ERrms = EYrms = EBrms = Ep Magnitude of phase voltage 19 Three-phase AC generator Line voltage IR ERY ER VR ZR IN EB VB EY IY VY ZB ZY IB ERY = ER - EY Three-phase Load Line voltage ERY = Ep 0o - Ep = EL ERY -EY 30o = 1.732Ep = √3 Ep -120o 30o 30o 120o 0o -120o ERY = ER - EY 21 Three-phase AC generator Line voltage IR VR ER ZR IN EB VB EY IY IB VY ZB ZY EYB EYB = EY - EB Three-phase Load Line voltage EYB = Ep -120o - Ep -90o = 1.732Ep = √3 Ep = EL 120o 120o 0o -90o -90o -120o -EB EYB = EY - EB EYB Three-phase AC generator Line voltage IR ER EBR VR ZR IN EB VB EY IY VY ZB ZY IB EBR = EB - ER Three-phase Load Line voltage EBR = Ep 120o - Ep = EL EBR 120o 150o = 1.732Ep = √3 Ep 0o 0o 150o 150o -ER -120o For star connected supply, EL= √3 Ep EBR = EB - ER 25 Phase voltages ER = Ep 0o EY = Ep -120o EB = Ep 120o It can be seen that the phase voltage ER is reference. 120o 0o Line voltages -120o ERY = EL 30o EYB = EL -90o EBR = EL 150o 26 Phase voltages ER = Ep -30o EY = Ep -150o EB = Ep Or we can take the line voltage ERY as reference. 90o Line voltages ERY = EL 0o EYB = EL -120o EBR = EL 120o 27 Three-phase AC generator Delta connected Three-Phase supply IR ERY VR ZR ER EB VB EY VY ZB IY ZY IB ERY = ER = Ep 0o Three-phase Load Three-phase AC generator Delta connected Three-Phase supply IR VR ZR ER EB EBR VB EY ZB IY IB VY ZY EYB For delta connected supply, EL= Ep Three-phase Load Connection in Three Phase System 4-wire system (neutral line with impedance) 3-wire system (no neutral line ) 4-wire system (neutral line without impedance) 3-wire system (no neutral line ), delta connected load Star-Connected Balanced Loads a) 4-wire system b) 3-wire system Delta-Connected Balanced Loads a) 3-wire system 30 Three-phase AC generator 4-wire system (neutral line with impedance) IR VR ER IN ZN EB EY ZR VN IY VB VY ZB ZY IB Voltage drop across neutral impedance: VN = INZN 1.1 Three-phase Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KCL at star point IR VR ER IN ZN EB EY ZR VN VB VY ZB IY ZY IB IR + IY + IB= IN 1.2 Three-phase Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KVL on R-phase loop IR VR ER IN ZN EB EY ZR VN IY VB VY ZB ZY IB Three-phase 33 Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KVL on R-phase loop IR VR ER IN ER – VR – VN = 0 ZR ZN VN ER – IRZR – VN = 0 Thus IR = ER – VN ZR 1.3 Three-phase 34 Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KVL on Y-phase loop IR VR ER IN ZN EB EY ZR VN IY VB VY ZB ZY IB Three-phase 35 Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KVL on Y-phase loop EY – VY – VN = 0 Thus IY = EY – IYZY – VN = 0 IN EY EY – VN 1.4 ZY ZN VN IY VY ZY Three-phase 36 Load Three-phase AC generator 4-wire system (neutral line with impedance) Applying KVL on B-phase loop EB – VB – VN = 0 Thus IB = EB – IBZB – VN = 0 IN EB – VN 1.5 ZB ZN EB VN VB ZB IB Three-phase 37 Load 4-wire system (neutral line with impedance) Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1: IR + IY + IB= IN ER – VN EY – VN + ZR + ZY ZR + EY ZY + EB ZB VN = ZB ER – VN + EY – VN + ZR ZR ZY ZY ER EB – VN = VN ZN EB – VN VN = ZB ZB 1 ZN + ZN 1 ZR + 1 ZY + 1 ZB 38 4-wire system (neutral line with impedance) ER VN + ZR ZR + ZY + ZN + EB ZB 1.6 = 1 ER EY EY ZY + 1 + ZR EB ZB = VN 1 1 + ZY ZB 1 ZN + 1 ZR + 1 ZY + 1 ZB 39 4-wire system (neutral line with impedance) VN is the voltage drop across neutral line impedance or the potential different between load star point and supply star point of three-phase system. ER VN + ZR EY EB + ZY ZB 1.6 = 1 ZN + 1 ZR + 1 ZY + 1 ZB We have to determine the value of VN in order to find the values of currents and voltages of star connected loads of 40 three-phase system. Example IR EL = 415 volt ER VR ZR = 5 Ω IN ZN =10 Ω EB EY VN IY VB ZY= 2 Ω ZB = 10 Ω IB Find the line currents IR ,IY and IB. Also find the neutral current IN. Three-phase Load Three-phase AC generator 3-wire system (no neutral line ) IR VR ER IN ZN EB EY ZR VN IY VB VY ZB ZY IB Three-phase Load Three-phase AC generator 3-wire system (no neutral line ) IR VR ER ZR VN EB VB EY VY ZB IY ZY IB No neutral line = open circuit , ZN = ∞ Three-phase 43 Load 3-wire system (no neutral line ) ER VN = EY + ZR 1 ZN EB + ZY 1 + ZR ZB + 1 + ZY 1.6 1 ZB ZN = ∞ 1 ∞ = 0 44 3-wire system (no neutral line ) ER VN = ZR + EY EB + ZY 1 ZR ZB + 1 ZY + 1.7 1 ZB 45 Example IR EL = 415 volt ER EB EY VN IY VR ZR = 5 Ω VB ZY= 2 Ω ZB = 10 Ω IB Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB. Three-phase Load 3-wire system (no neutral line ),delta connected load Three-phase AC generator IR VR ER EB ZR VB EY IY VY ZB ZY IB Three-phase Load 3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir ER VBR EB Ib EY VRY ZBR ZRY ZYB Iy IY IB VYB Three-phase Load 3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir ERY =VRY ER EBR =VBR EY EB VRY VBR ZBR ZRY Ib ZYB Iy IY IB EYB =VYB VYB Three-phase Load 3-wire system (no neutral line ),delta connected load Phase currents Ir = Iy = Ib = VRY = ERY = EL ZRY ZRY ZRY VYB EYB EL = = ZYB ZYB ZYB VBR EBR EL ZBR = ZBR = 30o -90o 150o ZBR 50 3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir Line currents IR = Ir - Ib E=B EL 30o ERY =VRY ER o E 150 E - BRL =VBR EZ Y BR ZRY = ZYB -90o VBR ZBR ZRY Ib ZYB Iy IY IY = Iy - Ir EL VRY - EL 30EoYB =VYB VYB IB ZRY Three-phase Load 3-wire system (no neutral line ),delta connected load Three-phase AC generator IR Ir Line currents IB = E=B Ib - Iy EL ZBR ERY =VRY ER 150o o E -90 E - BRL =VBR EZ Y YB VRY VBR ZBR ZRY Ib ZYB Iy IY EYB =VYB VYB IB Three-phase Load Star to delta conversion + ZYZB + ZBZR ZRY = ZRZY ZB ZR ZBR ZRY + ZYZB + ZBZR ZYB = ZRZY ZR ZY ZB + ZYZB + ZBZR ZBR = ZRZY ZY ZYB Example Use star-delta conversion. IR EL = 415 volt ER EB EY VN IY VR ZR = 5 Ω VB ZY= 2 Ω ZB = 10 Ω IB Find the line currents IR ,IY and IB . Three-phase Load Three-phase AC generator 4-wire system (neutral line without impedance) IR VR ER IN ZN = 0 Ω EB EY ZR VN VB IR VY ZB ZY IB VN = INZN = IN(0) = 0 volt Three-phase 55 Load 4-wire system (neutral line without impedance) For 4-wire three-phase system, VN is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become, IR = IY = IB = ER – VN ZR EY – VN 1.3 1.4 ZY EB – VN 1.5 ZB 56 Example IR EL = 415 volt ER VR ZR = 5 Ω IN EB EY VN IY VB ZY= 2 Ω ZB = 10 Ω IB Find the line currents IR ,IY and IB . Also find the neutral current IN. Three-phase Load v(t) vR vY vB wt The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin wt eY = EmY sin (wt -120o) eB = EmB sin (wt -240o) = EmBsin (wt +120o) 58 1.4: Phase sequences RYB and RBY VB (a) RYB or positive sequence VR VR ( rms ) 0o w 120o 120o VY VY ( rms ) 120 o VR -120o VB VB ( rms ) 240o VB ( rms ) 120o VY VR leads VY, which in turn leads VB. This sequence is produced when the rotor rotates in the counterclockwise direction. 59 (b) RBY or negative sequence VR VR ( rms ) 0o VY w VB VB ( rms ) 120o 120o 120o VR -120o VY VY ( rms) 240o VY ( rms) 120o VB VR leads VB, which in turn leads VY. This sequence is produced when the rotor rotates in the clockwise direction. 60 1.5: Connection in Three Phase System 1.5.1: Star Connection a) Three wire system R ZR ZY Z B Y B 61 Star Connection b) Four wire system R VRN V BN ZR V YN ZY Z B Y N B 62 Wye connection of Load R R 2 Z Z1 Y Z1 Y Load Z3 B B N N Z2 Z3 Load 63 1.5.2: Delta Connection R R Y Y B B 64 Delta connection of load R Load R Zc Z c Zb Y B Za Y Zb Za B Load 65 1.6: Balanced Load Connection in 3-Phase System 66 Example Wye-Connected Balanced Loads b) Three wire system IR EL = 415 volt ER EB EY VN IY VR ZR = 20 Ω ZY= 20 Ω VB ZB = 20 Ω IB Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB. Three-phase 67 Load Wye-Connected Balanced Loads b) Three wire system VN = = 0 volt VR = ER VY = EY VB = EB 68 Example 1.6.1: Wye-Connected Balanced Loads a) Four wire system IR EL = 415 volt ER VR ZR = 20 Ω IN EB EY VN IY ZY= 20 Ω VB ZB = 20 Ω IB Find the line currents IR ,IY and IB . Also find the neutral current IN. Three-phase 69 Load 1.6.1: Wye-Connected Balanced Loads a) Four wire system IR R IN IR IY IB Z1 VRN IN N V YN V BN Z2 IY Z Y B IB VRN Vphasa 0 VYN Vphasa 120 VBN Vphasa 120 where Vphasa VRN VYN VBN 3 For balanced load system, IN = 0 and Z1 = Z2 = Z3 VRN 0 o IR Z1 VYN 120o IY Z2 VBN 120o IB Z3 70 Wye-Connected Balanced Loads b) Three wire system IR R IR IY IB 0 Z1 VBR VRY S IY Z2 Z 3 Y VYB B IB VRS Vphasa 0 VYS Vphasa 120 VBS Vphasa 240 VRS 0 o IR Z1 VYS 120o IY Z2 VBS 120o IB Z3 where Vphasa VRS VYS VBS 71 1.6.2: Delta-Connected Balanced Loads IR Phase currents: R I YB VYB 120o Z2 I BR VBR 120o Z3 Z VBR I RY Z VRY I BR I RY VRY 0 o Z1 IY Y VYB B I YB Z IB where I RY I YB I BR I phasa and I R I Y I B I line Line currents: I R I RY I BR I Y I YB I RY I B I BR I YB 72 1.7: Unbalanced Loads 73 1.7.1: Wye-Connected Unbalanced Loads Four wire system IR IN IR IY IB R Z1 VRN For unbalanced load system, IN 0 and Z1 Z2 Z3 IN N V YN V BN IY Z2 Z 3 VRN 0 o IR Z1 Y B IB VRN Vphasa 0 VYN Vphasa 120 VBN Vphasa 120 VYN 120o IY Z2 VBN 120o IB Z3 74 1.7.2: Delta-Connected Unbalanced Loads Phase currents: IR I RY VRY 0 o Z1 I YB VYB 120o Z2 I BR VBR 120o Z3 R I RY VBR Z Z VRY I BR IY Y VYB I YB Z B IB Line currents: VRN Vphasa 0 I R I RY I BR VYN Vphasa 120 I Y I YB I RY VBN Vphasa 120 I B I BR I YB 75 1.8 Power in a Three Phase System 76 Power Calculation The three phase power is equal the sum of the phase powers P = PR + PY + PB If the load is balanced: P = 3 Pphase = 3 Vphase Iphase cos θ 77 1.8.1: Wye connection system: I phase = I L and VLL 3 V phase Real Power, P = 3 Vphase Iphase cos θ 3 VLL I L cos q Watt Reactive power, Q = 3 Vphase Iphase sin θ 3 VLLI L sin q VAR Apparent power, S = 3 Vphase Iphase 3 VLLI L VA or S = P + jQ 78 1.8.2: Delta connection system I L 3 I phase VLL= Vphase P = 3 Vphase Iphase cos θ 3 VLL I L cos q Watt 79 1.9: Three phase power measurement 80 Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters. 81 Three Phase Circuit Four wire system, Each phase measured separately PA IA Phase A VAN Phase B W A V W A VBN Phase C PB IB V PC IC W A VCN V Neutral (N) 82 watt-meter connection W Current coil (low impedance) voltage coil (high impedance) 83 a) Four wire system Example WR IR ER EL = 415 volt IN EB EY VR WY VN ZR = 5 ZY = 10 VB ZB = 20 30o Ω 90o Ω 45o Ω IY IB WB Find the three-phase total power, PT. Three-phase Load b) Three wire system Example WR IR ER EL = 415 volt VR ZR = 5 EB EY WY ZY = 10 VN ZB = 20 30o Ω 90o Ω 45o Ω IY IB WB Find the three-phase total power, PT. Three-phase Load b) Three wire system Example WR IR ER EL = 415 volt EB EY WY VN VR ZR = 5 VB ZB = 20 ZY = 10 30o Ω 90o Ω 45o Ω IY IB WB Find the three-phase total power, PT. Three-phase Load Three Phase Circuit Three wire system, The three phase power is the sum of the two wattP meters reading I Phase A AB A A VAB = VA - VB W V Phase B VCB = VC - VB PT PAB PCB IC A V W Phase C PCB 87 PT PAB PCB Proving: The three phase power (3-wire system) is the sum of the two watt-meters reading IA Phase A Instantaneous power: A VAN pA = v A i A PA W V Phase B IB A VBN pB = v B i B Phase C pC = v C i C PB W V IC PC A VCN W V Neutral (N) pT = pA + pB + pC = vA iA + vB iB +vC iC = vA iA + vB iB +vC iC = vA iA + vB (-iA -iC) +vCiC 88 Proving: The three phase power (3-wire system) is the sum of the two watt-meters reading IA Phase A Instantaneous power: pT = vA iA + vB (-iA –iC) +vCiC PA A IA VAN Phase B PAB W A V W IB VAB = VA - VB Phase A A PB V W Phase B = (vA – vB )iA + (vC – vB )iC VBN Phase C VCB = VC - VB = vAB iA + vCBiC Neutral (N) IC A V IC V W PC A W VCN V Phase C PCB pT = pAB + pCB PT PAB PCB 89 Power measurement In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters. In a three-wire system (three phases without neutral) the power is measured using only two single phase watt-meters. The watt-meters are supplied by the line current and the line-to-line voltage. 90