<p> Notes 6: Symmetrical Components 2</p><p>6.1 Symmetrical components: examples</p><p>These examples were adapted from [1].</p><p>Example 1: Compute sequence components of the following balanced a-b-c sequence line-to-neutral voltages.</p><p>Van   2770  V  V   277 120 abc  bn    Vcn   277120 </p><p>1 Solution:</p><p>Example 2: Compute the sequence components for a balanced Y-load that has phase b opened.</p><p>2 Ia=10/_0 </p><p>Ic=10/_120 </p><p>ZY ZY </p><p>In </p><p>ZY Ib=0 </p><p>Fig. 1: Balanced Y load with open phase b</p><p>Implication: Zero-sequence component results from unbalanced load.</p><p>3 Example 3: A 12.47kV feeder provides service to an unbalanced delta-connected load consuming the following power: . Phase ab: 1500 kVA, 0.95 lagging . Phase bc: 1000 kVA, 0.85 lagging . Phase ca: 950 kVA, 0.9 lagging Give the following expressions: a.The currents in each phase. b. The matrix K, where K is defined in the equation below.</p><p>Ia  Iab  I   K I   b   bc  Ic  Ica  c.Are the line currents balanced? Solution: The situation is shown in Fig. 2.</p><p>Ia </p><p>Vab Iab </p><p>Ib Vca Z∆2 </p><p>Z∆3 Z∆1 Ibc Vbc Ica Ic </p><p>Fig. 2</p><p>4 With line-line voltages given as 7200v, it is easy to see that we can obtain the phase * currents using, for example, Iab=(Sab/Vab) . We will assume that Vab is the reference. * *  S  1500 103 cos1 0.95   ab    Iab        Vab   72000  * 1500 103 18.19      208.3 18.19  72000  * *  S  1000 103 cos1 0.85   bc    Ibc        Vbc   7200 120  * 1000 103 31.788      138.89 151.788  7200 120  * *  S   950 103 cos1 0.90   ca    I ca        Vca   7200120  *  950 103 25.84      131.94145.84  7200 120  Notice that the phase currents are most definitely unbalanced.</p><p>Now what are the line currents? Consider Fig. 2 and note that we may relate the line currents to the phase currents using KCL:</p><p>5 Ia=Iab-Ica=(1)Iab+(0)Ibc+(-1)Ica Ib=-Iab+Ibc=(-1)Iab+(1)Ibc+(0)Ica Ic=-Ibc+Ica=(0)Iab+(-1)Ibc+(1)Ica</p><p>Writing in matrix form, we have:</p><p>I a   1 0 1I ab  I  I   1 1 0 I  abc  b    bc  I c   0 1 1 I ca  So the desired matrix is  1 0 1   K  1 1 0   0 1 1  Now let’s use it to obtain the line currents:</p><p>6 I a   1 0 1 208.3 18.19  I  I   1 1 0 138.89 151.788 abc  b     I c   0 1 1  131.94145.84   208.3 18.19 131.94145.84      208.3 18.19 138.89 151.788 138.89 151.788 131.94145.84  337.1  24.37     320.3 168.81  140.484.56  Note that line currents are also unbalanced.</p><p>Note an interesting thing about the phase currents and the line currents in the above.</p><p>Even though they both are unbalanced, they both add to 0 (do the math to see).</p><p>One can observe immediately that the line currents must add to zero, by considering the delta configuration as a single node. </p><p>7 In fact, we can use the delta-Y transformation to change Fig. 2 into Fig. 3.</p><p>Ia </p><p>Vab </p><p>Ib ZY1 Vca ZY2 </p><p>ZY3 </p><p>Vbc Ic </p><p>Fig. 3</p><p>To remind you from Nilsson’s book, </p><p>Z 1Z 2 ZY1  Z 1  Z 2  Z 3</p><p>Z 2 Z 3 ZY 2  Z 1  Z 2  Z 3</p><p>Z 1Z 3 ZY 3  Z 1  Z 2  Z 3</p><p>We see by applying KCL at the center node, that the 3 currents must sum to 0.</p><p>This will always be true for line currents feeding a delta connection.</p><p>8 However, try to add up the phase currents. You will find that they are not zero. </p><p>(Question: How can there be zero sequence components in the phase currents but not in the line currents? Answer: At any junction in the delta connection, the entire zero sequence current flowing in one phase also flows in the other phase, i.e., none of it “escapes” to the line. This mean that zero sequence currents circulate in the delta.)</p><p>Conclusion: Line currents into a delta or into an ungrounded Y always sum to zero. </p><p>Recall: Anytime you can show that the three phasors add to zero, the zero-sequence components will all be zero.</p><p>Implication: Line currents into a delta or into an ungrounded Y never have zero- sequence components (i.e., the zero</p><p>9 sequence components for line currents into a delta or into an ungrounded Y).</p><p>What about . Unbalanced currents into a grounded-Y? . Unbalanced line-to-line voltages?</p><p>10</p>