Ebook Electric circuit analysis: Part 2

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Part 2 book "Electric circuit analysis" includes content: Sinusoidal steady state in three phase circuits, dynamic circuits with periodic inputs – analysis by fourier series, first order RL circuits; first order RC circuits; series and parallel RLC circuits, analysis of dynamic circuits by laplace transforms, magnetically coupled circuits.

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Nội dung Text: Ebook Electric circuit analysis: Part 2

8.1 Chapter 8 SinusoidalSteady- StateinThree-Phase Circuits CHAPTEROBJECTIVES • To define and explain balanced and unbalanced three-phase systems and to extend the concepts of active power, reactive power and complex power to such systems. • To develop and employ relations between line and phase quantities in Y-connected and D-connected three-phase circuits. • To emphasize constraints on line quantities in three-phase three-wire and four-wire systems. • To develop a systematic procedure to analyse sinusoidal steady-state in three-phase balanced circuits using Y-equivalent and single-phase equivalent circuits. • To show how to use phasors, phasor equivalent circuits and phasor diagrams for solving three-phase circuits under sinusoidal steady-state condition. • To introduce and illustrate the problem of neutral-shift voltage in four-wire systems. • To illustrate circuit analysis procedures for unbalanced three-phase circuit analysis through solved examples. • To introduce and explain symmetrical components in detail and use it in analysis of unbalanced circuits. • To develop power relations using sequence components. IntroductIon Electrical power generation, transmission and distribution employ a particular system of three sinusoidal voltages/currents with certain symmetry in voltage/current magnitude and relative phase. This system of three sinusoidal voltage/current waveforms is called a three-phase voltage/current and
8.2 SinusoidalSteady-StateinThree-PhaseCircuits a circuit employing a three-phase voltage/current system is called a three-phase circuit. A three-phase circuit, like any other circuit, will have to go through a transient period subsequent to the application of source functions before it can settle down to a sinusoidal steady-state condition. We focus on analysis of three-phase circuits under sinusoidal steady-state condition in this chapter. No new concepts are required for analysis of three-phase circuits under sinusoidal steady-state conditions. It requires only the application of all the concepts we developed in the last two chapters. But before we take up the analysis of three-phase circuits, let us see why electrical engineers thought up three-phase system in the first place. 8.1 three-Phase system Versus sIngle-Phase system Industrial heaters are used in various material processing industries for a variety of heating applications. Typically, heater boxes that contain heating elements powered by AC supply, transfer the heat developed to a gaseous (usually air) or liquid medium which conducts the heat to the material that is to be heated up. Resistor elements made of some specially constructed wire are laid out uniformly inside the heater box to develop uniform heating of the contact surface. Many such heating elements are finally connected in parallel to form the heater resistive load. Consider such a heater which, after all parallel connection of sub-elements, is finally represented by three equal resistors of R each. These three resistors may be connected in parallel and supplied from a single-phase AC source or they may be supplied individually from different AC sources – for instance, for redundancy and reliability. 3r + I VP Є0º R – 3r 3I – 3r – + r 3r R R R + VP Є0º I R R VP Є0º + 3r – r VP Є0º I 3r 3I (a) (b) Fig. 8.1-1 a) Three equal resistors powered from a single-phase supply (b) Three equal ( resistorssuppliedfromthreeidenticalsources The circuit in Fig. 8.1-1(a) shows the three heater elements tied in parallel and supplied from a single-phase AC supply. The rms value of source is specified as VP. Let r be the resistance of the connecting cable and 3I be the phasor current in the cable. r is usually small compared to R and hence we can take the total circuit resistance as R/3 ignoring the contribution 2r. Therefore, circuit current magnitude ≈ 3VP/R A rms. Therefore, magnitude of I is ≈VP/R A rms. The total power loss in the cable is ≈18r (VP/R)2 W. The circuit in Fig. 8.1-1(b) shows the same three heater elements supplied from same single- phase supply – but with sources shown as three identical sources. Since the current in the connecting cable is only one-third of the earlier value we will choose a cable that has one-third the area of cross section that the cable in the circuit in Fig. 8.1-1(a) had. This implies that the cables in the circuit in Fig. 8.1-1(b) have a resistance of 3r each. Thus, the total power loss in the circuit in Fig. 8.1-1(b) is ≈6 × 3r × (VP /R)2 = 18r (VP /R)2 W again.
Three-PhaseSystemVersusSingle-PhaseSystem 8.3 R IR + 3r I VP Є0º R + 3r – In VP Є0º R N 3I – – – r VP Є α R N R – + + – VP Є0º R B VP Єβ Y 3r IY IB R 3r + + VP Є0º 3r I I VP 3r In = (1 + 1Єα + Єβ ) R (a) (b) Fig. 8.1-2 Fromsingle-phasecircuittothree-phasecircuit Next, we imagine that the negative polarity terminals of all the three sources are joined together to form a common node and that that a single cable connects this node to a similarly formed node on the load side as shown in the circuit of Fig. 8.1-2(a). The return currents from the three load branches flow through this cable. Then, this cable must have three times the cross-sectional area of the other cables in circuit in Fig. 8.1-2(a). Therefore, its resistance will be r. The power loss in four cables put together will again be 18r (VP /R)2 W. Non-uniform heating will result if the three branches get different powers delivered to them. Therefore, we wish to keep the power dissipated in the three resistors equal. Therefore, the three sources must have the same rms voltage. But that does not mean that they should be in-phase too. We are not particularly worried about the relative phase of currents flowing in the three branches of a heater resistance. However, is there any advantage in making the phase angles non zero? We assume in circuit of Fig. 8.1-2(b) that the phase of source connected at the terminal identified Y is at an angle a with respect to the source connected at terminal marked as R. Similarly, the phase of source connected at the terminal identified as B is at an angle b with respect to the source connected at terminal marked as R. Moreover, we assume that now we are using an extra-thick cable between node N to a similar node on the load side such that it is virtually a resistance-free connection. Now, the currents flowing out from the sources have different phase; but same magnitude of VP/(R + 3r) A rms. Note that we use ‘R’ as a terminal marking and ‘R’ as symbol and value for a resistance. The cable connecting the node N to a similar node on the load side will have a current In in it. VP V I n = (1 + 1∠a + 1∠b ) A rms = [(1 + cos a + cos b ) + j (sin a + sin b )] P A rms. R + 3r R + 3r Why do not we make the magnitude of this current zero so that the cable power loss will become zero in the corresponding cable even if we reduce the size of that cable? We try to find a (a, b) pair such that magnitude of In is zero. a and b should be such that (sina + sinb) = 0 and (cosa + cosb) = -1 simultaneously. First equation implies that b = -a and with this constraint the second equation results in a = ±120°. Therefore, (a, b) = (-120°, 120°) or (120°, -120°) are the two choices for (a, b) pair such that the current In = 0. If we make one of these two choices, the total power delivered to the resistors remains the same; however, the total cable loss becomes ≈9r (VP /R)2 W, i.e., only 50% of the cable loss in the case of single-phase supply.
8.4 SinusoidalSteady-StateinThree-PhaseCircuits If a cable (or any resistance for that matter) carries only zero current in a circuit, then, that cable can be removed without affecting the behaviour of any circuit variable in the circuit. This implies that we can remove the cable connecting the common point N of the sources to the common point in the load without affecting the load currents or load power. This is possible since the instantaneous currents in the three lines add up to zero at all t leaving only zero current in the common point link. Let us verify this with a = -120° b = 120°. 2VP 2VP 2VP iR (t ) = cos w t A, iY (t ) = cos(w t − 120°) A, iB (t ) = cos(w t + 120°) A R + 3r R + 3r R + 3r 2VP ∴ iR (t ) + iY (t ) + iB (t ) = [cos w t + cos(w t − 120°) + cos(w t + 120°)] R + 3r 2VP = [cos w t + {−0.5 cos w t + 0.866 sin w t} + {−0.5 cos w t − 0.866 sin w t}] = 0 for all t R + 3r Hence, no wire is needed between the common point of sources and common point of load. Therefore, we need only three wires of 3r W each in circuit of Fig. 8.1-2(b) instead of two wires of r W each as in circuit in Fig. 8.1-1(a). Cross-sectional area of wires in circuit of Fig. 8.1-1(a) will be three times that of the wires in circuit of Fig. 8.1-2(b). Therefore, the amount of copper (volume or weight) we use in wiring up circuit in Fig. 8.1-2(b) is only 50% of the copper that we use in wiring up circuit in Fig. 8.1-1(a). A set of three sinusoidal quantities, all at the same frequency, with equal peak (and hencerms)valuesandshiftedsuccessivelyby120° inphaseisdefinedasaBalanced Three-Phase Quantity.Therefore,ifx1(t),x2(t)andx3(t)isathree-phaseset,then, x1(t)=Xmcoswt,x2(t)=Xmcos(wt-120°)andx3(t)=Xmcos(wt+ 120°)or x1(t)=Xmcoswt,x2(t)=Xmcos(wt+ 120°)andx3(t)=Xmcos(wt- 120°) andx1(t)+ x2(t)+ x3(t)=0forallt. Eachlimborbranchofathree-phasesystem(sourceorload)istermedasaphase.R,Y andBareusedtodesignatethelineterminalsofathree-phasesourceorload. Ifthepeakvaluesareunequaland/orsuccessivephaseshiftsaredifferentfrom120°, thesetwillbecalledanUnbalanced Three-Phase Quantity. Abalancedthree-phasesystemrequiresonly50%ofthecopperandincursonly50%of thepowerlosstotransmitagivenamountofpowertoloadcomparedtoasingle-phase system. The sub elements of an industrial heater designed for three-phase operation are sometimes arranged in such a way that adjacent sub-elements belong to different phases. That is, if three sub-elements are adjacent, the first one will be a part of R that gets connected to R-line, the second one will be part of R that gets connected to Y-line and the third one will be part of R that gets connected to B-line. Then, the cycle repeats. This is, obviously, done on purpose. We discuss the sum of instantaneous power delivered by the three phases of a three-phase system under balanced operation to appreciate the purpose involved.
Three-PhaseSystemVersusSingle-PhaseSystem 8.5 Let us assume that the impedance per branch (i.e., phase) is Z = Z∠q and that Y-phase voltage lags R-phase by 120° and B-phase voltage lags Y-phase voltage by 120°. Then, the instantaneous currents in three lines can be expressed as iR (t ) = I m cos(w t − q ), iY (t ) = I m cos(w t − 120° − q ), and iB (t ) = I m cos(w t + 120° − q ) where Im = Vm/Z. Also, vRN (t ) = Vm cos(w t ), vYN (t ) = Vm cos(w t − 120°), and VBN (t ) = Vm cos(w t + 120°). where Vm = 2VP . Now, the instantaneous powers delivered by the sources in each phase are expressed as pR (t ) = Vm I m cos(w t ) cos(w t − q ) = 0.5Vm I m [cos q + cos q cos 2w t ] + 0.5Vm I m [sin q sin 2w t ] W. pY (t ) = Vm I m cos(w t − 120°) cos(w t − 120° − q ) = 0.5Vm I m cos q + cos q cos( 2w t − 240°)] + 0.5Vm I m [sin q sin( 2w t − 240°)] W. pB (t ) = Vm I m cos(w t + 120°) cos(w t + 120° − q ) = 0.5Vm I m [cos q + cos q cos( 2w t + 240°)] + 0.5Vm I m [sin q sin( 2w t + 240°)] W. The sum of instantaneous powers can now be expressed as pR (t ) + pY (t ) + pB (t ) = 1.5Vm I m cos q + 0.5Vm I m cos q {cos 2w t + cos( 2w t − 240°) + cos( 2w t + 240°)} + 0.5Vm I m sin q {sin 2w t + sin( 2w t − 240°) + sin( 2w t + 240°)}] W. A phase angle of -240° is same as +120° and a phase angle of +240° phase is same as –120°. Therefore, the three terms in the sums enclosed by curly brackets in the preceding equation form balanced three-phase sets. Sum of a three-phase balanced set is zero for all time instants. Therefore, VP pR (t ) + pY (t ) + pB (t ) = 1.5Vm I m cos q = 3 cos q W. Z Ifthepowerdissipatedinthethreebranchesofabalancedthree-phaseloadactson thesame physical system,then,abalancedthree-phasesourcedeliversactivepowerto theloadwithzeropowerpulsation.Thedouble-frequencypowerpulsationinthecaseof asingle-phaseACsupplycanneverbereducedtozero. Some instances in which the total active power from three branches of a three-phase load acts on the same physical system are industrial heaters and three-phase motors. In a three-phase heater, power dissipated in three branches will heat up the same body. In a three-phase motor, the torque developed by the branch windings act on the same mechanical shaft. This leads to steady heating and steady temperature in a heater and speed without fluctuations in the case of a motor. The next point of superiority of three-phase system over single-phase system has to be accepted as a statement of fact. It concerns matters beyond the scope of a book on circuits. It is the fact that Electrical Machines and Power Systems Components are more efficiently designed in three-phase
8.6 SinusoidalSteady-StateinThree-PhaseCircuits version rather than in single-phase version in all senses of the word ‘efficiency’ – efficiency in material utilisation, efficiency in power utilisation etc. Hence, we conclude: Abalancedthree-phasesystemcomprisingbalancedthree-phasesourcesandbalanced three-phaseloadsissuperiortoasingle-phasesystemthankstothefollowingfacts: (i) Powerlossintransmissionsystemislowerinthree-phasesystem. (ii) Copperutilisationissuperiorinthree-phasesystem. iii) Powerdeliveredtoabalancedthree-phaseloadbyabalancedthree-phasesupplyis ( freeofpulsation. iv) Electricalequipmentdesignedforthree-phaseoperationismoreefficientthantheir ( single-phasecounterparts. 8.2 three-Phase sources and three-Phase Power Three-phase sources and loads can take two forms – star connected or delta connected. The star connection is also referred to as Y-connection. The delta connection is referred to as D-connection and mesh connection too. 8.2.1 the y-connected source Two options to construct a three-phase Y-connected source are shown in Fig. 8.2-1. In Fig. 8.2-1(a) the terminal identified by Y has a voltage phasor that lags the voltage phasor of terminal identified as R by 120°. Further, the terminal identified as B has a voltage phasor that leads the R-line by 120° – i.e., B-line voltage lags R-line voltage by 240°. Thus, similarly located points on the waveform will appear first in R-line, then in Y-line after one-third of a cycle period and finally in B-line after another one-third of a cycle period. See the waveforms in Fig. 8.2-1(a). Thus, the sequence of appearance of peaks will be …–RYB–RYB–… A three-phase source that has this sequence of appearance of similarly located waveform points is called a positive sequence three-phase source. Note that the sequence can also be written as …–YBR–YBR–… or BRY–BRY–… The source in Fig. 8.2-1(b) illustrates a negative phase sequence source. In this case, the voltage of B-line lags that of R-line by 120° and voltage of Y-line lags that of R-line by 240°. Thus, the order of appearance of similarly located waveform points in time will be …–RBY–RBY–… (equivalently, …–BYR–BYR–… or …–YRB–YRB–… too) in this case. There are loads that cannot distinguish between the two sequences – a three-phase industrial heater is one. There are loads that can distinguish between the sequences – a three-phase induction motor is one. If it rotates in clockwise direction for positive sequence applied voltage, it will rotate in counter- clockwise direction for negative sequence applied voltage. So will a three-phase synchronous motor. The three-phase sources we consider from this point onwards will be positive sequence sources by default. The voltage VP used in Fig. 8.2-1 is an rms value. We will specify three-phase source in rms values in this chapter. The Y-connected source has a common point at which the three phase-sources join. This point is called the ‘neutral’ point of the source. Phases in the context of three-phase source or load refer to the three branches that constitute the three-phase system. Lines refer to the three terminals – R, Y and B –
Three-PhaseSourcesandThree-PhasePower 8.7 R + R + VP Є0º – VP Є0º N – – – VP Є–120º – – VP Є120º VP Є–120º VP Є120º + + + + Y Y B B (V) 2 VP (V) 2 VP RN YN BN YN RN BN ωt ωt 2π /3 4π /3 2π 2π /3 4π /3 2π – 2 VP – 2V (a) P (b) Fig. 8.2-1 a)ApositivesequenceY-connectedthree-phasesource(b)Anegativesequence ( Y-connectedthree-phasesource that come out of the three-phase source or load and are available for external connections. Thus, in the case of Y-connected source, the phase-source is connected between a Line and the Neutral. Therefore, the phase-source voltage is same as Line-to-Neutral voltage in the case of Y-connected source. As a matter of common practice, phase-source voltage is shortened to phase voltage. Beginners often get confused between phase voltage (i.e., phase-source voltage or phase-load voltage) and line-to-neutral voltage and conclude that they are the same. They are not the same always. Phase voltage is the voltage across phase source or phase load. Phase voltages exist for any three-phase source or load. Line-to- neutral voltage will exist only if there is a neutral point – i.e., only for Y-connected source or load. Though line-to-neutral voltage does not exist unless it is a Y-connected system, line-to-line voltages exist for all three-phase systems. VRY is the phasor voltage of the R-line with respect to the Y-line. VYB and VBR are similarly defined. Line-to-line voltage is usually shortened to line voltage. Similarly, phase current is the phase-source current or phase-load current flowing in one of the branches that constitute the three-phase source or load as per passive sign convention. Line current is the current that flows out of or into one of the three terminals R, Y or B. Line current is usually taken to be flowing out of a source line terminal and flowing into a load line terminal. Refer to Fig. 8.2-2 that shows all the phase and line quantities in a Y-connected source. VRN, VYN and VBN are the phase voltages and IRN, IYN and IBN are the phase currents. There is a neutral point in an Y-connected source. Hence, line-to-neutral voltages exist and they are the same as phase voltages. Line currents are IR, IY and IB and are assumed to flow out of the source towards the load. In a Y-connected source, the line currents defined in this manner are equal to negative of phase currents defined as per passive sign convention. Since the R-line is same as positive terminal of the source that feeds the R-line in a Y-connected source, this source may be referred to as the R-phase source. Similarly for other two sources too.
8.8 SinusoidalSteady-StateinThree-PhaseCircuits IRN IR VBN – VYN VRN – + R VRN + VBR 30º 90º VRY VP Є0º – VRY 30º N 90º VRN – – VP Є–120º VBR VP VP Є120º – 90º + + 30º IY VBN VYN + Y VYN IBN IYN VYB 3 VP = VL – IB VYB + B Fig. 8.2-2 AY-connectedsourcewithallphaseandlinequantitiesidentified We start at the Y-line terminal and traverse through the two sources, reach R-line terminal and move to Y-terminal by falling through VRY. The KVL equation we get is VRY = VRN – VYN. Refer to the phasor diagram shown in Fig. 8.2-2. VRN is taken as reference and a horizontal line of length VP (the rms value of phase sources) to suitable scale is drawn to represent it. The remaining two phase voltages – VYN and VBN – are at –120° and –240° positions as shown. Now, the equation VRY = VRN – VYN is implemented by rotating VYN by 180°, moving a copy of VRN to the tip of –VYN and completing the triangle. The result is the first line voltage VRY. We observe by projecting the lengths of VRN and –VYN on to VRY that magnitude of VRY is 2 cos30° VP = √3 VP. We represent the magnitude of line voltage by VL. Hence, VL = √3 VP. We observe further that the first line voltage, VRY, leads the first phase voltage VRN by 30°. The remaining two line voltages are also obtained by using equations VYB = VYN – VBN and VBR = VBN – VRN. Obviously, the three line voltages form a three-phase balanced set of voltages. ThelinevoltagesinabalancedY-connectedsourceformabalancedthree-phasesetof voltagesthatare√3timesinmagnitudeand30°aheadinphasewithrespecttophase voltages. The line currents in Y-connected source need not be cophasal with the phase voltages. That depends on the load. Assuming that the load is a balanced one, the line currents themselves will form a three- phase balanced set of phasors that are displaced in phase by a phase-lag angle q with respect to respective phase voltages. Thus, IR = IL∠-q, IY = IL∠-q - 120° and IB = IL∠-q + 120°, where IL is the rms value of line current. These current phasors are shown in relation to phase voltage phasors in Fig. 8.2-3(b). Also shown are the phase positions of line voltages with respect to phase voltages in Fig. 8.2-3(a). We make the following observations from these phasor diagrams. 1. The line current lags the phase-source voltage by q and it lags behind the corresponding line voltage by (30° + q). 2. Each source contributes an active power of VPIL cosq W. Hence, total three-phase active power delivered by the three-phase source = 3 VPIL cosq = √3 VLIL cosq W where q is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., -IRN in the case of R-line) lags behind the phase voltage.
Three-PhaseSourcesandThree-PhasePower 8.9 VBN 90º VRY VBR 30º VBN 30º IB 90º VRN VP θ 90º VRN 30º θ VP VYN IY θ 3 VP = VL 120º IR VYB VYN (a) (b) Fig. 8.2-3 a)PhasordiagramofphaseandlinevoltagesofaY-connectedsource(b)Phasor ( diagramoflinecurrentsandphasevoltagesinY-connectedsource 3. Each source contributes a reactive power of VPIL sinq W. Hence, total three-phase reactive power delivered by the three-phase source = 3 VPIL sinq = √3 VLIL sinq VArs, where q is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., -IRN in the case of R-line) lags behind the phase voltage. Equivalently, q is the angle by which the current delivered by a phase source lags behind the phase voltage. 4. Therefore, the complex power delivered by the three-phase source S = √3 VLIL∠q VA. 8.2.2 the D-connected source The generators in a power system are almost invariably Y-connected. However, power systems use Y-D transformers and a D-connected three-phase source can be used to model the secondary side of such a transformer. A D-connected three-phase source and the phasor diagram of its voltages and currents are shown in Fig. 8.2-4. We assume that the load on the source is balanced. VRY is taken as the reference phasor for phasor diagrams. Note that phase voltage and line voltage are the same for this source and that there is no neutral point and hence no line-to-neutral voltage can be defined for this source. Applying KCL at the three corners of delta, we get the following equations. I BR − I RY = I R − I YB + I RY = I Y I YB − I BR = I B This system of equations has no unique solution since we can add a constant to IRY, IYB and IBR without affecting these equations. Such a constant added to the three currents will represent a circulating current within the delta-loop. Such a circulating current is possible only since three ideal sources are connected in a loop. However, in practice, all the sources in the delta will have some small impedance or other which will force the condition that (IRY + IYB + IBR)Z = 0 where Z is the small source impedance of the sources. Assuming that the circulating current component in delta is zero, let the current -IRY be IP∠-q. Then, the remaining two phase currents will be -IYB = IP∠-q -120° and -IBR = IP∠-q + 120°, where IP is the magnitude of phase currents.
8.10 SinusoidalSteady-StateinThree-PhaseCircuits Now, IR can be constructed by IR = -IRY – (-IBR). This is shown in the phasor diagram in Fig. 8.2-4. The resulting IR = √3 IP ∠-(30° + q ) A rms. IR IBR VBR R VBR – IRY VL ∠120º + + VRY – – – IBR IYB VL Є0º + VRY VYB IY VL VL Є–120º Y θ – IRY IB – IYB IL 30º B VYB IR Fig. 8.2-4 Adelta-connectedsourceanditsphasordiagrams Hence, the line current magnitude is IL = √3 IP in a delta-connected three-phase source. The line currents delivered by a balanced D-connected source form a balanced three- phase set of currents that are √3 times in magnitude and 30° behind in phase with respecttophasecurrentsdeliveredbythephase-sources. We make the following observations from these phasor diagrams. 1. The delivered phase current lags the phase-source voltage by q and line current lags behind the corresponding line voltage by (30° + q). 2. Each source contributes an active power of VLIP cosq W. Hence, total three-phase active power delivered by the three-phase source = 3 VLIP cosq = √3 VLIL cosq W, where q is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., -IRY in the case of RY- phase source) lags behind the phase voltage. 3. Each source contributes a reactive power of VLIP sinq W. Hence, total three-phase reactive power delivered by the three-phase source = 3 VLIP sinq = √3 VLIL sinq VArs where q is the angle by which the ‘negative of phase current defined as per passive sign convention’ (i.e., -IRY in the case of RY-phase source) lags behind the phase voltage. Equivalently, q is the angle by which the current delivered by a phase-source lags behind the phase voltage. 4. Therefore, the complex power delivered by the three-phase source S = √3 VLIL∠q VA. Therefore, we state the following conclusion: Thethree-phasecomplexpowerdeliveredbyathree-phasesourceisgivenby S=√3VLIL∠q=[√3VLILcosq+ j√3VLILsinq]VA whereqistheanglebywhichthecurrentphasordelivered byaphasesourcelagsbehind itsvoltagephasor.ThisrelationshipisindependentofwhetherthesourceisY-connected orD-connected.Itshouldbenotedthatqis nottheanglebetweenlinevoltagephasor andlinecurrentphasor.
AnalysisofBalancedThree-PhaseCircuits 8.11 Electrical Power Engineering adopts a convention to specify a three-phase source by specifying its line-to-line voltage in rms value. We follow this convention unless stated otherwise in the remaining portions of this chapter. Thus a 400 V, 50 Hz source means a source that has a line-to-line voltage magnitude of 400 V rms at 50 Hz. If it is Y-connected its phase voltage will be 400/√3 V rms and peak of its phase voltage will be 400√2/√3 V. 8.3 analysIs of Balanced three-Phase cIrcuIts Balanced three-phase circuits comprising a balanced three-phase source and a balanced three-phase load can be of four varieties, viz., –Y–Y, Y–D, D–Y or D–D. We develop a common analysis procedure for all the possible configurations. 8.3.1 equivalence Between a y-connected source and a D-connected source Let VRY = VL∠0°, VYB = VL∠-120° and VBR = VL∠120° be the three line voltages observed in a three-phase circuit and let IR = IL∠-(q + 30°), IY = IL∠-(q + 150°) and IB = IL∠-(q -90°) be the observed line currents where the observed phase lag of first line current is expressed as 30° plus some angle designated by q. If the source is within a black box and we are to guess whether it is a Y-connected source or a D-connected source, we will find that we will not be able to resolve the matter using the observed line voltage phasors and line current phasors. This is so because both configurations shown in Fig. 8.3-1 will result in these observed line voltages and line currents. IR = IL ∠–( θ +30°) IBR IRN IR = IL ∠–(θ +30°) R VBR IL – – IRY = – ∠– θ + VL R+ VL ∠120° + 3 VRN ∠–30° + VRY 3 VRY = VL∠0º – – N VL VL ∠0º VL – – ∠–150º VBR = V ∠120º – ∠90º L IYB 3 + 3 – IY + + IY VYB VYN VBN + Y VL ∠–120° IBN IYN VYB = V ∠–120º Y L IB – IB + B B Fig. 8.3-1 llustratingequivalencebetweenaY-connectedsourceandaD-connectedsource I Hence,foreverysetoflinevoltagesandlinecurrentsobservedinathree-phasecircuit, thereexistadelta-connectedsourceandastar-connectedsourcewhichwilljustifythe observedlinequantitiesandwhichareindistinguishablefromoutside. Therefore, if a balanced three-phase source is really D-connected, it may be replaced by an equivalent Y-connected source for purposes of analysis. The details of phase-source currents in the real delta-connected source can be obtained after the circuit problem is solved using the star equivalent source.
8.12 SinusoidalSteady-StateinThree-PhaseCircuits 8.3.2 equivalence Between a y-connected load and a D-connected load But there is nothing new in this equivalence – we already know that any Y-connected set of impedances can be transformed into a D-connected impedance and vice versa. Therefore, any given three-phase balanced D-connected impedance can be converted into a three-phase balanced Y-connected impedance such that the terminal voltage and current behaviour remain unaffected. The individual branch currents in the branches of delta in the real delta-connected load can be obtained after line quantities have been obtained by using its equivalent Y-connected model. Three- phase symmetry and 1/√3 factor connecting line currents and phase currents in a delta-connected system can be used for this purpose. The impedance to be used in Y-connected load is 1/3 times the impedance present in D-connected load. See Fig. 8.3-2. IR = IL ∠–(θ + 30°) IBR R IRN IR = IL ∠–(θ + 30°) VBR IL IRY = ∠–θ VL + ∠–30º + – VL ∠120° 3 VRN = R Z∠ θ 3 Z ∠θ VRY = V ∠0º VRY = VL∠0º L 3 Z∠ θ Z∠ θ – N VL VYN = ∠–150º IYB VL – – VBN = ∠90º 3 VBR = VL∠120º VYB IY 3 – I + + Z ∠θ Z ∠θ Y VL ∠–120° + Y Y IBN 3 3 IYN VYB = V ∠–120º IB L – IB + B B Fig. 8.3-2 EquivalentY-connectedloadforaD-connectedload 8.3.3 the single-Phase equivalent circuit for a Balanced three-Phase circuit Any balanced three-phase circuit can be equivalenced to a Y–Y balanced three-phase circuit by replacing D-connected sources and loads (if any) by their Y-connected equivalents developed in the last two subsections. The equivalence is valid as far as line voltages, line currents and complex power flow are concerned. Thus, analysis of balanced three-phase circuits reduces to a three-step procedure. The D-connected sources and loads are replaced by their Y-connected equivalents in the first step. The resulting balanced Y-connected circuit is solved for line voltages and line currents everywhere in the second step. The relevant line voltage values and line current values are used to determine the phase-source voltages and currents in the case of those D-connected sources that were replaced by star equivalents earlier and phase-load voltages and currents in the case of those D-connected loads that were replaced by star equivalents earlier. This will constitute the third step. The difference between a real Y-connected source/load and Y-connected equivalent of D-connected source/load is that the neutral point of the first is accessible for connection, whereas the neutral point of the second is an inaccessible node. However, the neutral connection in a balanced three-phase circuit does not carry any current (even if the neutral link has non-zero impedance) and hence the voltage difference between source neutral and load neutral will remain zero whether we connect the neutral points together or not. Therefore, the circuit solution in a balanced Y-connected circuit is independent of neutral connection.
AnalysisofBalancedThree-PhaseCircuits 8.13 If all the neutral points in a balanced three-phase circuit are at the same potential, then, each phase- source appears directly across the net impedance connected from the respective source line terminal (R or Y or B) to the load neutral point. Thus, the circuit may be solved phase by phase independent of each other. But then, why should we solve all the three phases at all? We expect the circuit solution to be balanced set of three-phase phasors in any case. Hence, we can determine the solution for Y-line and B-line if we know the solution for phasors in R-line by using three-phase symmetry. Therefore, we need to determine only the R-line phasors. The equivalent circuit that we employ to solve for R-line phasors is called the single-phase equivalent circuit for a balanced three-phase circuit. This equivalent circuit will contain the R-line phase-source in series with all impedances that come between R-line terminal of the source and neutral point of load. The neutral points of source and load are shorted to form the reference node in a single-phase equivalent circuit even if the neutral points are connected through impedance in the actual circuit. This is so since the neutral connection in a balanced circuit will not carry any current. Note that if a three-phase circuit is star connected everywhere, the use of ‘R-phase’ instead of ‘R-line’ does not create any confusion. Such confusion can arise in the case of delta connection. A delta-connected source has only R-line and no R-phase. However, we have now decided that all our circuits will be converted to star equivalents. Thus, use of ‘R-phase’ instead of R-line is permitted. example: 8.3-1 A balanced Y-connected load with phase impedance of 7.9 + j5.7 W at 50 Hz is supplied from a 400V, 50 Hz balanced Y-connected source through connection impedance of 0.1 + j0.3 W in each line. (i) Find the line current and load voltage. (ii) Find the active and reactive power delivered by the source and delivered to the load. Solution R 0.1 + j0.3 Ω It is a Y–Y connection already. The single-phase equi- + + IR valent circuit required for solving the problem is shown 7.9 + j5.7 Ω VL – 230.9 Є0º – in Fig. 8.3-3. V rms A three-phase source is specified by specifying line- NS,NL to-line voltage rms value. Thus, the source voltage in single-phase equivalent must be 400/√3 = 230.9 V rms Fig. 8.3-3 S ingle-phaseequivalent forExample8.3-1 in magnitude. The R-phase source voltage is taken as the reference phasor with an angle of 0°. Solving the circuit in Fig. 8.3-3, we get, 230.9∠0° 230.9∠0° 230.9∠0° IR = = = = 23.09∠ − 36.9° A rms. (0.1 + j 0.3) + (7.9 + j 5.7) 8 + j6 10 ∠36.9° VL = 23.09∠ − 36.9° × ( 7.9 + j 5.7) = 23.09∠ − 36.9° × 9.74 ∠35.8° = 224.9∠ − 1.1° V rms. Active power delivered by source = 3 × 230 × 23.09 × cos(0 − ( −36.9°)) = 12.75 kW Reactive power delivered by source = 3 × 230 × 23.09 × sin(36.9°) = 9.56 kVAr Active power delivered to load = 3 × 224.9 × 23.09 × cos( −1.1° − ( −36.9°)) = 12.64 kW Reactive power delivered to load = 3 × 224.9 × 23.09 × sin( −1.1° − ( −36.9°)) = 9.11 kVAr Magnitude of line voltage across load = 3 × 224.9 = 389.56 V rms
8.14 SinusoidalSteady-StateinThree-PhaseCircuits Complete Phasor Solution Phase voltages at source: 230∠0°, 230∠-120°, 230∠120° (V rms) The first line voltage, i.e., VRY, is known to lead the first phase voltage, i.e., VRN, by 30°. Therefore, line voltages at source terminals: 400∠30°, 400∠-90°, 400∠150° (V rms) Phase voltages at load: 224.9∠-1.1°, 230∠-121.1°, 230∠118.9° (V rms) Line voltages at load terminals: 389.56∠28.9°, 389.56∠-91.1°, 389.56∠148.9° (V rms) Line currents: 23.09∠-36.9°, 23.09∠-156.9°, 23.09∠83.1° (A rms) Complex power delivered by source = 3 × 400 × 23.09∠(0 − (36.9°)) = 16 ∠36.9° kVA Complex power in load = 3 × 389.56 × 23.09∠( −1.1 − (36.9°)) = 15.54 ∠35.8° kVA example: 8.3-2 A balanced D-connected load with phase impedance of 23.7 + j17.1 W at 50 Hz is supplied from a 400 V, 50 Hz balanced Y-connected source through connection impedance of 0.1 + j0.3 W in each line. (i) Find the line current and load voltage. (ii) Find the load-branch currents. (iii) Find the active and reactive power delivered by the source and delivered to the load. Solution The circuit with all the relevant phasors identified is shown in Fig. 8.3-4. IR 0.1 + j0.3 Ω IBR + R IRY 230.9∠0° 23.7 + j17.1 Ω – V rms 230.9∠120° N 23.7 + j17.1 Ω IYB – – 230.9∠–120° V rms + + V rms IY 0.1 + j0.3 Ω 23.7 + j17.1 Ω Y IB 0.1 + j0.3 Ω B Fig. 8.3-4 Y-connectedsourcedeliveringpowertoaD-connectedloadthroughconnection A impedance The delta-connected impedance is transformed into an Y-connected one by star-delta transformation equations. Since the delta impedances are equal, the impedance in star connection will be one-third of delta branch impedance, i.e., 7.9 + j5.7 W. The circuit after this transformation is shown in Fig. 8.3-5. Now, the single-phase equivalent of this circuit can be identified as the same as in Example: 8.3-1 and hence the solution is same as in that example. \Line current (R) = 23.09∠-36.9° A rms, Load phase voltage (R) = 224.9∠-1.10 V rms and Load line voltage (RY) = 389.56∠28.9° V rms. Load active power can be found as 3 × 224.9 × 23.09 × cos(36.9°-1.1°) or as √3 × 389.56 × 23.09 × cos(36.9°-1.1°). The value is 12.64 kW. Similarly, the load reactive power is 3 × 224.9 × 23.09 × cos(36.9°-1.1°) or √3 × 389.56 × 23.09 × cos(36.9°-1.1°). The value is 9.11 kVAr.
AnalysisofBalancedThree-PhaseCircuits 8.15 + R IR 0.1 + j0.3 Ω 230.9∠0° V rms – 7.9 + j5.7 Ω N 230.9∠120° – – 230.9∠–120° V rms V rms 7.9 + j5.7 Ω 7.9 + j5.7 Ω + + IY 0.1 + j0.3 Ω Y IB 0.1 + j0.3 Ω B Fig. 8.3-5 TheY-connectedequivalentofcircuitinFig.8.3-4 Branch Currents in Delta-connected Load Applying KCL at the corners of delta in R-line and Y-line in Fig. 8.3-4, we get IRY – IBR = IR and IYB – IRY = IY Multiplying the second equation by –1 and adding the result to the first equation, we get 2IRY – (IBR + IYB) = IR - IY But (IRY + IBR + IYB) = 0 since we expect these three currents to be a three-phase balanced set. Therefore, -(IBR + IYB) = IRY. Therefore, 2IRY + IRY = IR - IY ⇒ IRY = (IR - IY)/3. We know that IR = 23.09∠-36.9° A rms. IR, IY and IB form a balanced three-phase set of phasors. Therefore, IY = 23.09∠-156.9° A rms. \3 IRY = (23.09∠-36.9° - 23.09∠-156.9°) A rms. This is the difference between two phasors with 120° between them. We have determined the result of this kind of difference operation before. The result will have a magnitude that is √3 × 23.09 and will lead the first quantity in the difference by 30°. \ 3IRY = 40∠-6.9° and IRY = 13.33 ∠-6.9° A rms. This can also be obtained by observing that RY line voltage at load is 389.56∠28.9° V rms and IRY is the current drawn by 23.7 + j17.1W (29.225∠35.8° W) from this voltage. Therefore, IRY = 389.56∠28.9° ÷ 29.225∠35.8° = 13.33∠-6.9° A rms as before. Now, the remaining two branch currents can be obtained by using three-phase symmetry as IYB = 13.33∠-126.9° A rms and IBR = 13.33∠-113.1° A rms example: 8.3-3 A delta-connected source with an open-circuit voltage of 200 V rms at 50 Hz is used to deliver power to a star-connected load with phase impedance of 5.9 + j7.7 W. Each phase source in the D-connected source has internal impedance of 0.3 + j0.9 W in series with it. (i) Find the phase voltage and line voltage across the load. (ii) Find the line currents and delta branch currents. (iii) Find the active power and reactive power delivered to the load and consumed by the internal impedance of the source. Solution The three-phase circuit to be analysed is shown in Fig. 8.3-6.
8.16 SinusoidalSteady-StateinThree-PhaseCircuits 200∠120° V rms – I1 R IR + 0.3 + j0.9 Ω I3 0.3 + j0.9 Ω – 5.9 + j7.7 Ω I2 + + 200∠0° V rms 5.9 + j7.7 Ω 5.9 + j7.7 Ω 200∠–120° – IY V rms 0.3 + j0.9 Ω Y IB B Fig. 8.3-6 Delta-connected source with source impedance supplying a star-connected A load The open-circuit line voltage across source terminals is given to be 200 V rms. The balanced delta connection shown in Fig. 8.3-6 will not have any current in delta when the terminals are left open since the loop voltage inside delta is zero. Therefore, the source voltages must themselves be 200 V rms. A straightforward determination of Y-equivalent of delta connected source is not possible in this case. We transform the ‘voltage source in series with impedance combinations’ into ‘current source in parallel with impedance combinations’ by using source transformation theorem. This is shown in Fig. 8.3-7. 200∠120° V rms – R R 200∠120° I1 + 0.3 + j0.9 0.3 + j0.9 Ω 0.3 + j0.9 Ω A rms 0.3 + j0.9 Ω I3 – I2 + 200∠0° + 200∠0° V rms 0.3 + j0.9 200∠–120° – A rms V rms 200∠–120° 0.3 + j0.9 Ω 0.3 + j0.9 0.3 + j0.9 Ω Y A rms Y B B Fig. 8.3-7 pplying source transformation theorem on the delta-connected source in A Fig.8.3-6 We take the RY line quantity as the reference phasor in this example. Now, the delta-connected current source has a star equivalent and the delta-connected source impedance also has a star equivalent. Combining the current source in star equivalent and the impedance in star equivalent into a voltage source in series with impedance, we get (200/√3)∠-30° = 115.5∠-30° V rms in series with (0.3 + j0.9)/3 =0.1 + j0.3 W in R-phase of the final star equivalent. The complete star equivalent of the source along with the load is shown in Fig. 8.3-8. The single-phase equivalent of this circuit is shown in Fig. 8.3-9. (i) The line current (R) is 115.5∠-30° V rms ÷ 6 + j8 W = 11.55 ∠-83.13° A rms and the phase voltage (RN) across load is 11.55∠-83.13° A rms × (5.9 + j7.7) W = 11.55∠-83.13° × 9.7∠52.54° =112∠-30.6° V rms.
AnalysisofUnbalancedThree-PhaseCircuits 8.17 + R 115.5∠–30° 0.1 + j0.3 Ω IR V rms – 5.9 + j7.7 Ω N 115.5∠90° – – 115.5∠–150° V rms V rms 5.9 + j7.7 Ω 5.9 + j7.7 Ω + + IY 0.1 + j0.3 Ω Y 0.1 + j0.3 Ω IB B Fig. 8.3-8 ThestarequivalentofthecircuitinFig.8.3-6 Hence, line voltage (RY) across the load = 112√3∠(-30.6° + 30°) = 194∠-0.6° V rms. (ii) The line current (R) is 115.5∠-30° ÷ 6 + j8 = 11.55 ∠-83.13° A rms. Refer to Fig. 8.3-6. The current 115.5∠–30° + R marked as I1 = [200∠0° - line voltage (RY) across V rms 0.1 + j0.3 Ω I R – 5.9 + j7.7 Ω load] ÷ (0.3 + j0.9) = [200∠0° - 194∠-0.6°] ÷ 0.9487 ∠71.57° = (6 + j2) ÷ 0.9487 ∠71.57° = Fig. 8.3-9 ingle-phaseequivalent S 6.68∠-51.13° A rms. Then, I2 will be 6.68∠- ofcircuitinFig.8.3-6 171.13° A rms and I3 will be 6.68∠68.87° A rms (iii) The active power delivered to the load = √3 × 194 × 11.55 × cos(-30°-(-83.13°)) = 2.33 kW. The reactive power delivered to load = √3 × 194 × 11.55 × sin(-30°-(-83.13°)) = 3.11 kVAr. The internal impedance of the source is 0.3 + j0.9 W and it carries an rms current of 6.68 A. Therefore, the active power consumed by the internal impedance of the delta-connected source = 3 × 6.682 × 0.3 = 40.2 W. The reactive power consumed by this impedance = 3 × 6.682 × 0.9 = 120.6 VAr. 8.4 analysIs of unBalanced three-Phase cIrcuIts An unbalanced three-phase circuit is one that contains at least one source or load that does not possess three-phase symmetry. A source with the three source-function magnitudes unequal and/or the phase displacements different from 120° can make a circuit unbalanced. Similarly, a three-phase load with unequal phase impedances can make a circuit unbalanced. The single-phase equivalent circuit technique of analysis does not work for unbalanced three- phase circuits. General circuit analysis techniques like mesh analysis or nodal analysis will have to be employed for analysing such circuits. 8.4.1 unbalanced y–y circuit Y–Y connection is typically used in the last mile in a power system, i.e., in the low-tension (LT) distribution system. The primary distribution system usually runs at 11 kV line voltage level. Distribution transformers that are rated for 11 kV/400 V and are D-connected in the primary side and Y-connected in the secondary side are used to step down the primary distribution voltage to the LT distribution level. Such transformers are located at the load centre location of the area that a particular transformer is expected to serve. The neutral point of secondary side of a distribution transformer is usually earthed solidly.
8.18 SinusoidalSteady-StateinThree-PhaseCircuits 4-wire LT lines emanating from the secondary of transformer distribute power to various subareas of the service area. Individual consumers are provided service by means of service drops tapped from these lines. Both single-phase loads and three-phase loads are provided power from these LT lines at various points. Thus, the load on the LT line will be unbalanced in general. Even if the single phase loads are to get distributed equally among three lines (R, Y and B), only the transformer secondary will perceive a balanced load and various sections of the LT line will see varying degree of unbalance in load. This is due to the spatially dispersed nature of loads. Thus, the neutral wire in an LT distribution line will invariably carry current – that too, different currents in different locations – even if the aggregate load is balanced. The system in which the neutral wire is available for connecting single-phase loads and three-phase loads that require a neutral tie-up is called a four-wire system. The system in which neutral wire is not available for load connection is called a three-wire system. This could be Y-Y system with neutrals isolated, Y-D, D-Y or D-D system. The exposed metal parts of all electrical equipment at the consumer location are tied together and earthed locally. This earth point is made available as the third pin of all three-pin power points installed in the consumer electrical system. If the earthing resistances at the transformer neutral and at the consumer location are negligibly small, then, the earth-pin at all consumer locations will be at the same potential as the transformer secondary neutral. However, the neutral-pin at the consumer locations will differ in potential from the transformer neutral potential due to the neutral wire drop. This difference will be more if there is unbalance in the load and neutral is carrying heavy currents. Moreover, this potential difference between the earth-pin and neutral-pin will be different for different consumers due to spatially distributed nature of line load. This problem of neutral – earth voltage difference is called the neutral- shift problem. Electronic equipment in general, and, computing equipment in particular, are sensitive to this neutral- shift voltage and often malfunction when it exceeds certain pre-specified levels. Damage to components can also take place due differences in neutral-shift potential that exist among various power points when different components of an interconnected computing system are powered from different power points. example: 8.4-1 A balanced three-phase source of 400 V 50 Hz supplies an unbalanced load through an LT 4-wire line , as in Fig. 8.4-1. Each wire in the line has impedance of 0.1 + j0.3 W. (a) Find (i) phase and line voltages at load location and neutral-shift voltage, (ii) wire currents, (iii) power and reactive power delivered to the load, (iv) three-phase apparent power and power factor. (b) Repeat (a) assuming that the neutral wire is a thick conductor with zero impedance. (c) Repeat (a) assuming that the neutral wire is open. + IR 0.1 + j0.3 Ω R 230.9∠0° 9.9 – j0.3 Ω – V rms G V 230.9∠120° – N – IN V rms 230.9∠–120° 0.1 + j0.3 Ω 7.9 – j6.3 Ω + + V rms IY 0.1 + j0.3 Ω 7.9 + j5.7 Ω Y IB 0.1 + j0.3 Ω B Fig. 8.4-1 Unbalancedthree-phasecircuitinExample8.4-1
AnalysisofUnbalancedThree-PhaseCircuits 8.19 Solution (a) We solve the problem by finding out the phasor voltage at load neutral with respect to the ground (earth) node G by using the KCL equation at load neutral. The sum of currents flowing away from that node is set to zero. V − 230.9∠0° V V − 230.9∠ − 120° V − 230.9∠120° + + + = 0, 10 0.1 + j 0.3 8 + j6 8 − j6 1 1 1 1  230.9∠0° 230.9∠ − 120° 230.9∠120° i.e., V  + + + = + + .  10 0.1 + j 0.3 8 + j 6 8 − j 6   10 8 + j6 8 − j6 Solving for V, we get, V = -2.31-j5.49 = 5.96∠-112.8° V rms. Now, we can find wire currents as 230.9∠0° − 5.96 ∠ − 112.8° IR = = 23.33∠1.35° A rms 10 230.9∠ − 120° − 5.96 ∠ − 112.8° IY = = 22.5∠ − 157.06° A rms 8 + j6 230.9∠120° − 5.96 ∠ − 112.8° IY = = 23.45∠155.71° A rms 8 + j6 I N = I R + I Y + I B = 18.83∠175.65° A rms The load phase voltages can be determined by multiplying line current phasors by phasor impedance of each phase. VRN = 23.33∠1.35° × (9.9 − j 0.3) = 231.04 ∠ − 0.39° V rms VYN = 22.5∠ − 157.06° × (7.9 + j 5.7) = 219.18∠ − 121.25° V rms VBN = 23.45∠ − 155.71° × (7.9 + j 6.3) = 237∠ − 123.65° V rms The line voltages across the load can be determined in terms of the phase voltage phasors now. VRY = VRN − VYN = 231.04 ∠ − 0.39° − 219.18∠ − 121.25° = 391.63 ∠28.33°V rms VYB = VYN − VBN = 219.18∠ − 121.25° − 237∠ − 123.65° = 398.32 ∠ − 90.8°V rms VBR = VBN − VRN = 237∠123.65° − 231.04 ∠ − 0.39° = 400.2 ∠147.94°V rms The active power and reactive power delivered to the load is determined by adding up the corresponding power delivered to each phase of the load. P = PR + PY + PB = 231.04 × 23.33 × cos ( −0.39° − 1.35°) + 231.04 × 22.5 × cos (121.25° − ( −157.06°)) 1 + 237 × 23.45 × cos (123.65° − ( −155.71°)) = 5387.1W + 3999W + 4345.9 kW Q = QR + QY + QB = 231.04 × 23.33 × sin ( −0.39° − 1.35°) + 231.04 × 22.5 × sin (121.25° − ( −157.06°)) + 237 × 23.45 × sin (123.65° − ( −155.71°)) = 163.25VAr + 2885.3VAr − 3465.7VAr = −0.744 kVAr
8.20 SinusoidalSteady-StateinThree-PhaseCircuits Two different definitions of apparent power and power factor are possible in the case of unbalanced three-phase circuits. The first one is using the total complex power, S, delivered to the load. In this definition, the magnitude of S is taken as apparent power and ratio of active power to the magnitude of S is taken as average power factor of the circuit. The apparent power in the load in the circuit in this example then is (13.732 + 0.7442)°.5 = 13.75 kVA and power factor is 0.9985 lead. The second definition accepts the sum of apparent powers of load phases as the three-phase apparent power. Power factor is taken as the ratio of active power to apparent power.. The apparent power in the load in the circuit in this example then is 15.88 kVA and power factor is 0.865. We cannot specify it as a lead or lag power factor in this case. The second definition is more appropriate in the sense that power factor and apparent power are expected to give us an indication as to the utilisation efficiency of current in carrying active power. The first method of calculating apparent power and power factor will hide the fact that underutilization of current is taking place when one phase takes leading reactive power and another phase takes lagging reactive power. The phases may consume large reactive power and yet the complex power may come out with only real part or with small imaginary part. Therefore, the definition of apparent power and power factor based on apparent power of individual phases of load is a more meaningful one. (b) The neutral-shift voltage in this case is zero since the source neutral and load neutral are tied together by a zero impedance link. The circuit becomes a collection of three single-phase circuits sharing a common point at neutral. Therefore, the line currents can be obtained as 230.9∠0° IR = = 23.09∠0° A rms 10 230.9∠ − 120° IY = = 23.09∠ − 156.87° A rms 8 + j6 230.9∠120° IB = = 23.09∠156.87° A rms 8 − j6 I N = I R + I Y + I B = −19.38 A rms The load phase voltages can be determined by multiplying line current phasors by phasor impedance of each phase. VRN = 23.09∠0° × (9.9 − j 0.3) = 228.7∠ − 1.74° V rms VYN = 23.09∠ − 156.87° × (7.9 + j 5.7) = 224.93∠ − 121.06° V rms VBN = 23.09∠ − 156.87° × (7.9 − j 6.3) = 233.31∠118.3° V rms The line voltages across the load can be determined in terms of the phase voltage phasors now. VRY = VRN − VYN = 228.7∠ − 1.74° − 224.93∠ − 121.06° = 391.51 ∠28.32°V rms m VYB = VYN − VBN = 224.93∠ − 121.06° − 233.31∠118.3° = 398.15 ∠ − 90.8°V rms VBR = VBN − VRN = 233.31∠118.3° − 228.7∠ − 1.74° = 400.2 ∠147.95°V rms VRY = VYB − VBR = 391.51∠28.32° + 398.15∠ − 90.8° + 400.2 ∠147.95° = 344.65 + j185.73 − 5.56 − j 398.11 − 339.204 + j 212.37 ≈ 0 within round-off error n The active power and reactive power delivered to the load are determined by adding up the corresponding power delivered to each phase of the load.