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Ap p en d ix CT H R E E - P H A S E S Y S T E M SIn a three-phase system, there are two types of connections that are mostoften used: the wye (Y) connection and the delta (A) connection. Also,ther e are two phase sequences. In th e abc-sequence, the a-phase variableslead the 6-phase variables in time phase and the fr-phase variables lead thec-phase variables. In the acb-sequence, the c-phase variables lead the &-phasevariables.W y e C o n n e c t i o nTh e wye (Y) connection is illustrated in Fig. C -l. Balanced three -pha securrents are equal-amplitude sinusoidal currents displaced by 120. The instantaneous sum of balanced currents is zero; hence, a fourth wire is notneeded. In a Y connection, t he assigned negativ e-poten tial sides of th e windings are all connected to form what is called the neutral point, shown as nin Fig. C -l. Th e neu tral may be groun ded or left to float. A s mentioned,if the currents are balanced, their instantaneous sum is zero. If the neutralis floating, it is clear that the sum of the currents must be zero regardless oftheir form.

The voltages across each phase are referred to as line-to-neutral voltages,whereas the voltages between two of the three phases are the line-to-linevoltages. The phase currents are the line currents. We can relate the line-to-neutral and line-to-line voltages as

489

Electromechanical Motion Devices, Second Editionby Paul Krause, Oleg Wasynczuk and Steven PekarekCopyright 2012 Institute of Electrical and Electronics Engineers, Inc.

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490 APPENDIX C

Figure C-1: Wye connection.

Vab = Va-VbVbc = Vb-VcVC = VC- Va

(C-1)(C-2)(C-3)

If the system is balanced, the n we can express the stead y-sta te ph ase voltagesfor an abc sequence asVa = \/2Va COS 0J et

Vb = V2V scos(ujet-lTT)Vc = y/2Va cos (uet + 7r)

(C-4)(C-5)(C-6)

where the capital letters are used to denote steady-state conditions. For anabc sequence, the phase voltages may be written in phasor form as

Va = V s/W _Vb = VJ - 120vc = Vs/vxr

The line-to-line voltages may be expressed as

(C-7)(C-8)(C-9)

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THREE-PHASE SYSTEMS 491

Vab = K / 0 ~ Vsl - 120

= ^ ^ 7 3 0 (C-10)

Vbc = Vs/ ~ 120 - K /12 0

= > / 3 V r a / - 9 0 (C - l l )

K a = K / i 2 0 - n / o

= y ^ K / l S O 0 (C-12)Hence, the line-to-line voltages form a balanced three-phase set that is V3times the magnitude of the line-to-neutral voltages and shifted ahead in timephase by 30 for an abc sequence and shifted back by 30 for an acb sequence.For balanced steady-state conditions, we need to consider only one phasesince once we have determ ined th e variables associated w ith one of th e phaseswe can express the other phase variables by shifting the phase ahead or backby 120.D e l t a C o n n e c t i o nTh e A connection is illustrated in Fig. C-2. In this type of connection,the line-to-line voltages are the voltages across the phases, that is, va = i>a&,Vb Vba and so on. Th ere is no ne utra l connection. Th e line curren ts arethe sum of currents from two phases. For the connection shown in Fig. C-2,

lac la ~ tc (-loj

Ha = ib~ia (C-14)icb = ic-ib (c"15)

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492 APPENDIX C

- +Va b

Vca\Iba

T u i i*b V l

+ - >

Figure C-2: Delta connection.If, for example, the currents form a balanced abc sequence, then

la = Is/0 (C-16)h = Is/ - 120

The line currents becomeh

Lac

= I,/120

= / . /o^-

= Vsish

/s/120

-30

(C-17)(C-18)

(C-19)Thus, for an abc sequence, Iac is \/3 times the amplitude of Ia and shifted30 back in ph ase from it. Similarly, Iha is shifted back 30 from Ih and cbback 30 from Ic. In the case of the acb sequence, the line currents are shiftedahead of the phase currents.