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Pin Description

Pin Type

Value/Unit

Instruction

1

Get sending end current magnitude of wire j

O

A (RMS)

transformerID/ImagFromj

where j is 1, 2 or 3

2

Get receiving end current magnitude of wire j

O

A (RMS)

transformerID/ImagToj

where j is 1, 2 or 3

3

Get sending end current angle of wire j

O

Degree

transformerID/IangFromj

where j is 1, 2 or 3

4

Get receiving end current angle of wire j

O

Degree

transformerID/IangToj

where j is 1, 2 or 3

5

Set/Get tap position

I/O

Integer between [min_tap, max_tap]

transformerID/tap_j

where j is 1, 2 or 3

...

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.

Image Added

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+959.692e3i527.831e3 - 959.692e3i

N is a matrix  [2*np*m=4 x np*m=2] 

N

1 /72000
-1 /72000
01/120
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+1.1108i0.6109-1.1108i 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 =  

1000
0010

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)

Image Added


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.


Image Added

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 =

100
-100
010
0-10
001
00-1

Y1  =  

158.349e3 - 287.907e3i-158.349e3 + 287.907e3i0000
-158.349e3 + 287.907e3i158.349e3 - 287.907e3i0000
00158.349e3 - 287.907e3i-158.349e3 + 287.907e3i00
00-158.349e3 + 287.907e3i158.349e3 - 287.907e3i00
0000158.349e3 - 287.907e3i-158.349e3 + 287.907e3i
0000-158.349e3 + 287.907e3i158.349e3 - 287.907e3i

N is a matrix  [2*np*m=12 x np*m=6] 

N

1 /1247000000
-1 /1247000000
01/(208*sqrt(3))0000
0-1/(208*sqrt(3))0000
001 /12470000
00-1 /12470000
0001/(208*sqrt(3))00
000-1/(208*sqrt(3))00
00001 /124700
0000-1 /124700
000001/(208*sqrt(3))
00000-1/(208*sqrt(3))

Yw = N Y1 NT  =   

0.0102-0.0185i -0.0102+0.0185i-0.3525+0.6409i0.3525-0.6409i00000000
   -0.0102+0.0185i  0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i00000000
-0.3525+0.6409i0.3525-0.6409i12.2002-22.1823i-12.2002+22.1823i00000000
0.3525-0.6409i-0.3525+0.6409i-12.2002+22.1823i12.2002-22.1823i00000000
00000.0102-0.0185i -0.0102+0.0185i-0.3525+0.6409i0.3525-0.6409i0000
0000   -0.0102+0.0185i  0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i0000
0000-0.3525+0.6409i0.3525-0.6409i12.2002-22.1823i-12.2002+22.1823i0000
00000.3525-0.6409i-0.3525+0.6409i-12.2002+22.1823i12.2002-22.1823i0000
000000000.0102-0.0185i -0.0102+0.0185i-0.3525+0.6409i0.3525-0.6409i
00000000-0.0102+0.0185i0.0102-0.0185i 0.3525-0.6409i -0.3525+0.6409i
00000000-0.3525+0.6409i0.3525-0.6409i12.2002-22.1823i-12.2002+22.1823i
000000000.3525-0.6409i -0.3525+0.6409i-12.2002+22.1823i12.2002-22.1823i

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 =  

100000000000
000010000000
000000001000
001000000001
000100100000
000000010010

Yprim = A Yw AT  =  

0.0101-0.0185i 00-0.3524+0.6408i0.3524-0.6408i0
00.0101-0.0185i 00-0.3524+0.6408i0.3524-0.6408i
000.0101-0.0185i 0.3524-0.6408i0-0.3524+0.6408i
-0.3524+0.6408i00.3524-0.6408i24.4004-44.3645i-12.2002+22.1822i-12.2002+22.1822i
0.3524-0.6408i-0.3524+0.6408i0-12.2002+22.1822i24.4004-44.3645i-12.2002+22.1822i
00.3524-0.6408i-0.3524+0.6408i-12.2002+22.1822i-12.2002+22.1822i24.4004-44.3645i

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

Image Added


3) Multiple transformers in the same model

See the Transformer page in phasor08_IEEE13.xls file in demo PHASOR-08.

References

[1] Roger C. Dugan, "A Perspective on Transformer Modeling for Distribution Systems Analysis". 2003 IEEE Power Engineering Society General Meeting. DOI: 10.1109/PES.2003.1267146

[2] Roger C. Dugan and Surya Santoso, "An Example of 3-phase Transformer Modeling for Distribution Systems Analysis". 2003 IEEE PES Transmission and Distribution Conference and Exposition. DOI: 10.1109/TDC.2003.1335084