# Page History

## Key

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 No Pin Description Pin Type Value/Unit Instruction 1 Get sending end current magnitude of wire j O A (RMS) transformerID/ImagFromjwhere j is 1, 2 or 3 2 Get receiving end current magnitude of wire j O A (RMS) transformerID/ImagTojwhere j is 1, 2 or 3 3 Get sending end current angle of wire j O Degree transformerID/IangFromjwhere j is 1, 2 or 3 4 Get receiving end current angle of wire j O Degree transformerID/IangTojwhere j is 1, 2 or 3 5 Set/Get tap position I/O Integer between [min_tap, max_tap] transformerID/tap_jwhere j is 1, 2 or 3

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# Model Equations

This multiphase transformer is modeled based on the primitive nodal admittance matrix Yprim [1],[2].

Yprim = A N B ZB-1 BT NT AT  matrix dimension:  np*m x np*m, np = number of phases, m= number of windings

Y1 = B ZB-1 BT   ;   Yw = N Y1 NT  ;   Yprim = A Yw AT

Y1 is the ground-referenced nodal admittance matrix on a 1 volt base. Matrix dimension: np*m x np*m

N is the incidence matrix whose non-zero elements are the inverse of the numbers of turns in the windings. This matrix represents the effect of the ideal transformers shown to obtain actual windings voltages. Matrix dimension: 2*np*m x np*m

B is the incidence matrix whose elements are either 1,-1 or 0. It relates currents in the short circuit reference frame where the first winding is assumed shorted to the currents in the nodal admittance reference frame on a 1 volt base. Matrix dimension: np*m x np

A is the incidence matrix whose non-zero elements are generally either 1 and -1, that relates the winding currents to the actual terminal currents. Matrix dimension: nc x 2*np*m, nc = number of terminal currents

ZB is the short circuit impedance matrix. Matrix dimension: np*(m-1) x np*(m-1)

Yw is the winding admittance matrix. Matrix dimension: 2*np*m x 2*np*m

# Examples

1) A single-phase 2W transformer with the following data:  7.2/0.12 kV, 25 kVA, X = 20%, R=1.1%

In this case np = 1, m = 2.

ZB in pu = 0.011+0.02i, ZB in 1V base = (ZB in pu)*12/25 kVA = 4.4e-7 + 8e-7i.  ZB-1= 527.831e3 - 959.692e3i

Y1 = B ZB-1 BTB is a matrix [np*m=2 x np=1]

B =

 1 -1

Y1  =

 527.831e3 - 959.692e3i -527.831e3+959.692e3i -527.831e3+959.692e3i 527.831e3 - 959.692e3i

N is a matrix  [2*np*m=4 x np*m=2]

N

 1 /7200 0 -1 /7200 0 0 1/120 0 -1/120

Yw = N Y1 NT  =

 0.0102-0.0185i -0.0102+0.0185i -0.6109+1.1108i 0.6109-1.1108i -0.0102+0.0185i 0.0102-0.0185i 0.6109-1.1108i -0.6109+1.1108i -0.6109+1.1108i 0.6109-1.1108i 36.6549-66.6453i -36.6549+66.6453i 0.6109-1.1108i -0.6109+1.1108i -36.6549+66.6453i 36.6549-66.6453i

To generate matrix A is necessary to define the number of terminal currents in the model. In this case there are 2 terminal currents (see the red currents in the figure above) so nc=2 and A matrix is [nc=2 x 2*np*m=4]

A =

 1 0 0 0 0 0 1 0

Finally the matrix Yprim is calculated

Yprim = A Yw AT  =

 0.0101-0.0185i -0.6105+1.1100i -0.6105+1.1100i 36.6007-66.5467i

Below it can be seen how to add this single-phase transformer in the excel file. The total resistance was divided equally between the 2 windings (RW1 =  RW2 = 0.011 pu/2 = 0.0055 pu). Note that the voltages voltages must be added as phase to phase voltages even though the model is single-phase (according to the table above)

2) A three-phase 2W transformer with the following data:  12.47/0.208 kV (wye/delta), 75 kVA, X = 20%, R=1.1%

In this case np = 3, m = 2.

ZB in pu = 0.011+0.02i, ZB in 1V base = (ZB in pu)*12/75 kVA = 1.4667e-7 + 2.6667e-7i.  ZB-1= 158.349e3 - 287.907e3i

Y1 = B ZB-1 BT ;  B is a matrix [np*m=6 x np=3]

B =

 1 0 0 -1 0 0 0 1 0 0 -1 0 0 0 1 0 0 -1

Y1  =

 158.349e3 - 287.907e3i -158.349e3 + 287.907e3i 0 0 0 0 -158.349e3 + 287.907e3i 158.349e3 - 287.907e3i 0 0 0 0 0 0 158.349e3 - 287.907e3i -158.349e3 + 287.907e3i 0 0 0 0 -158.349e3 + 287.907e3i 158.349e3 - 287.907e3i 0 0 0 0 0 0 158.349e3 - 287.907e3i -158.349e3 + 287.907e3i 0 0 0 0 -158.349e3 + 287.907e3i 158.349e3 - 287.907e3i

N is a matrix  [2*np*m=12 x np*m=6]

N

 1 /12470 0 0 0 0 0 -1 /12470 0 0 0 0 0 0 1/(208*sqrt(3)) 0 0 0 0 0 -1/(208*sqrt(3)) 0 0 0 0 0 0 1 /12470 0 0 0 0 0 -1 /12470 0 0 0 0 0 0 1/(208*sqrt(3)) 0 0 0 0 0 -1/(208*sqrt(3)) 0 0 0 0 0 0 1 /12470 0 0 0 0 0 -1 /12470 0 0 0 0 0 0 1/(208*sqrt(3)) 0 0 0 0 0 -1/(208*sqrt(3))

Yw = N Y1 NT  =

 0.0102-0.0185i -0.0102+0.0185i -0.3525+0.6409i 0.3525-0.6409i 0 0 0 0 0 0 0 0 -0.0102+0.0185i 0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i 0 0 0 0 0 0 0 0 -0.3525+0.6409i 0.3525-0.6409i 12.2002-22.1823i -12.2002+22.1823i 0 0 0 0 0 0 0 0 0.3525-0.6409i -0.3525+0.6409i -12.2002+22.1823i 12.2002-22.1823i 0 0 0 0 0 0 0 0 0 0 0 0 0.0102-0.0185i -0.0102+0.0185i -0.3525+0.6409i 0.3525-0.6409i 0 0 0 0 0 0 0 0 -0.0102+0.0185i 0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i 0 0 0 0 0 0 0 0 -0.3525+0.6409i 0.3525-0.6409i 12.2002-22.1823i -12.2002+22.1823i 0 0 0 0 0 0 0 0 0.3525-0.6409i -0.3525+0.6409i -12.2002+22.1823i 12.2002-22.1823i 0 0 0 0 0 0 0 0 0 0 0 0 0.0102-0.0185i -0.0102+0.0185i -0.3525+0.6409i 0.3525-0.6409i 0 0 0 0 0 0 0 0 -0.0102+0.0185i 0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i 0 0 0 0 0 0 0 0 -0.3525+0.6409i 0.3525-0.6409i 12.2002-22.1823i -12.2002+22.1823i 0 0 0 0 0 0 0 0 0.3525-0.6409i -0.3525+0.6409i -12.2002+22.1823i 12.2002-22.1823i

To generate matrix A is necessary to define the number of terminal currents in the model. In this case there are 6 terminal currents (see figure above) so nc=6 and A matrix is [nc=6 x 2*np*m=12]

A =

 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0

Yprim = A Yw AT  =

 0.0101-0.0185i 0 0 -0.3524+0.6408i 0.3524-0.6408i 0 0 0.0101-0.0185i 0 0 -0.3524+0.6408i 0.3524-0.6408i 0 0 0.0101-0.0185i 0.3524-0.6408i 0 -0.3524+0.6408i -0.3524+0.6408i 0 0.3524-0.6408i 24.4004-44.3645i -12.2002+22.1822i -12.2002+22.1822i 0.3524-0.6408i -0.3524+0.6408i 0 -12.2002+22.1822i 24.4004-44.3645i -12.2002+22.1822i 0 0.3524-0.6408i -0.3524+0.6408i -12.2002+22.1822i -12.2002+22.1822i 24.4004-44.3645i

The following image shows how to add this component in the excel file.