example symmcomp

6
Consider a thevenin source supplying a delta/wye-grounded transformer, and assume a SLG fault at the wye-grounded side of the transformer For simplicity, we’ll neglect all resistances, and assume j1 ohm reactive impedance for each element, in all sequence networks. The symmetrical component networks for this fault condition are connected in series as seen from the fault location, as follows: At the fault location, Current at the fault location, on the faulted phase Current at the fault location, on the unfaulted phases: XS XT SLG Fault 1 pu / wye- gnd Sourc e j1 p u j1 pu 1 pu j1 pu j1 pu -ve seq j1 pu j1 pu 0-seq If1 If2 If0 IS1 IS2 IS0 +ve seq

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Example of application of symmetrical components

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Page 1: Example SymmComp

Consider a thevenin source supplying a delta/wye-grounded transformer, and assume a SLG fault at the wye-grounded side of the transformer

For simplicity, we’ll neglect all resistances, and assume j1 ohm reactive impedance for each element, in all sequence networks. The symmetrical component networks for this fault condition are connected in series as seen from the fault location, as follows:

At the fault location,

Current at the fault location, on the faulted phase

Current at the fault location, on the unfaulted phases:

XS XT

SLG Fault1 pu

/ wye-gndSource

j1 pu j1  pu1 pu

j1  pu j1  pu-ve seq

j1  pu j1  pu0-seq

If1

If2

If0

IS1

IS2

IS0

+ve seq

Page 2: Example SymmComp

Next, consider current seen on the source side of transformer terminals (i.e. on the side of the delta-connected winding):

Consider if the result is reasonable.

Consider the following winding configuration, where the voltage on the delta side lags that on the wye-connected side by 30 degrees. With a fault on Phase a on the wye-side, current is expected to flow between phases a & b on the delta side. However current in Phase ‘c’ ought to be zero!!!

Va

Vc

Vb

va

vc

vb

Page 3: Example SymmComp

Similarly, if on the other hand the voltage on the delta connected side were to lead that on the wye-connected side by 30 degrees, as shown below:

Then a fault on Phase a on the wye-side should result in current between phases a & c on the delta side, and no current in Phase ‘b’!!!

**** What’s wrong? ***

ANSWER: We have not taken into account the phase shift resulting from the delta-wye transformation.

Let’s say we consider the case where the voltage on the delta-connected side lags the voltage on the wye-connected side by 30 degrees.

Then, in the positive sequence mode, Va_delta_pos = Va-wye_pos * e-j30

whereas in the negative sequence mode, Va_delta_neg = Va-wye_neg * e+j30

Hence

Finally,

Va

Vc

Vb

vavc

vb

Page 4: Example SymmComp

Thus, as expected, Ia = -Ib, and Ic = 0

Page 5: Example SymmComp

In this case, let’s check for energy balance:

Three phase power at the sending end

Three-phase power at the receiving end