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Creating a 4 lines 3 phases net in
a 3 lines 3 phases net using a Zn-transformer?

About operation mode of a Zn - transformer and its design.

Typical application schemas  of a Zn-transformer is shown in the Fig.1. The Zn-transformer has 6 equal windings which are connected in zig - zag.

If the 3 phase load is symetrical  ( linear load complex impedances ZA = ZB = ZC ) then the there is no current in the neutlal line ( In = 0 ) and the Zn-transformer runs in the no-load operation mode.

If the 3 phase load is symetrical  ( l ZA = ZB = ZC) but non-linear (typical for some rectifier) then the currents IA, IB and Ic can have the following form:

iA = I1m sin(ωt)           +I3m sin(3ωt)+I5m sin(5ωt)          +I7m sin(7ωt)             +I9m sin(9ωt)+...
iB = I1m sin(ωt+120°) +I3m sin(3ωt) +I5m sin(5ωt+120°)+I7m sin(7ωt+120°)   +I9m sin(9ωt)+...
iC = I1m sin(ωt+2400°) +I3m sin(3ωt)+I5m sin(5ωt+240°)+I7m sin(7ωt+2400°)+I9m sin(9ωt)+...

and the current in the neutral line is:

in = iA + iB + iC = 3 *I3m sin(3ωt) + 3*I9m sin(9ωt) + ...

where I3m and I9m are max. values of the 3. and 9. current harmonics.

For this operation mode the curents through the Zn transformer are:

Ia = ib = ic = in/3 = I3m sin(3ωt) + I9m sin(0ωt) + ...

Note that the Zn-transformer works in this operation mode as a very powerful filter of th 4., 9., ... current harmonicd. 

Assune the rms values of the 3. and 9. harmoc in all 3 phases of a 3x400V, 50Hz,  three phase net are:
I3rms=33A
I9rms = 10A
then:
UA= UB = UC = 230 and the zig & zag voltage is 230/2/cos(30°) = 133V

Run theLlarge Transformer èrogram and set the input as follows in the Fig.2:

Rermsrks:

  • Due to the fact that the transformer is driven by 50Hz you have to set a small  50Hz current.

  • In order to set the 3. and 9. current harmonics you have to set Circuit = 41

  • Do not forget to encrease the eddy current factpr RacRdc

  • Note that the turns of the zig and zag windings have to be equal.

Normally the worst case for the Zn-transformer is unsymetrical operation mode when only one phase is on load:

iA = I1m sin(ωt)  + I3m sin(3ωt)+I5m sin(5ωt)  + I7m sin(7ωt)             +I9m sin(9ωt)+...
iB = 0
iC = 0

and the current in the neutral line is:

in = iA  = I1m sin(ωt)  + I3m sin(3ωt) + I9m sin(9ωt) + ...

For this operation mode the curents through the Zn transformer are:

Ia = ib = ic = in/3 = (I1m/3)  sin(ωt)  + (I3m/3)  sin(3ωt)+(I5m/3)  sin(5ωt)  + (I7m/3)  sin(7ωt) ...

 Assune the rms values of the current harmonics in the phase A in a 3x400V, 50Hz net are:
Î1rms=90A
I3rms=33A
I5rms = 21A
I7rms=15A

then:
UA= UB = UC = 230 and the zig & zag voltage is 230/2/cos(30°) = 133V

Run theLlarge Transformer èrogram and set the input as follows in the Fig.4:

 

Rermsrks:

  • In order to set the 3.  current harmonics you have to set Circuit = 41

  • Do not forget to encrease the eddy current factpr RacRdc

  • Note that the turns of the zig and zag windings have to be equal.

Finally note that you can use the Small Transformers Program if there are no current harmonics:

iA = I1m sin(ωt)

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