Question 3.8: Consider a distribution system with four buses, whose positi......

The primitive impedance or admittance matrices can be written by examination of the network. First consider the positive or negative sequence network of Figure 3.12a. The following steps illustrate the procedure:

1. The buildup is started with branches 01, 02, and 03 that are connected to the reference node, Figure 3.13a. The primitive impedance matrix can be simply written as

2. Next, add link 1–2, Figure 3.13b. As this link has no coupling with other branches of the system,

Z_{ej} = Z_{pj} \ – \ Z_{kj}
Z_{ee} = Z_{pk,pk} \ + \ Z_{pe} \ – \ Z_{ke}
p = 1,      k = 2.
Z_{e1} = Z_{11} \ – \ Z_{21} = 0.05
Z_{e2} = Z_{12} \ – \ Z_{22} = 0 – 0.2 = -0.2
Z_{e3} = Z_{13} \ – \ Z_{23} = 0
Z_{ee} = Z_{12, \ 12} \ + \ Z_{1e} \ – \ Z_{2e} = 0.04 + 0.05 + 0.2 = 0.29

The augmented matrix is

3. Eliminate the last row and last column by using Equation 3.76. This gives

\overline{Z}_{bus,modified} = \overline{Z}_{bus,primitive} \ – \ \frac{\overline{Z}_{je} \overline{Z}_{je}}{\overline{Z^\prime}_{ee}}       (3.76)

4. Add link 2–3, Figure 3.13c:
p = 2, k = 3 and there is no mutual coupling with other branches.
This gives
Z_{e1} = Z_{21} \ – \ Z_{31} = 0.345 – 0 = 0.345
Z_{e2} = Z_{22} \ – \ Z_{32} = 0.0621 – 0 = 0.0621
Z_{e3} = Z_{23} \ – \ Z_{33} = 0 – 0.05 = -0.05
Z_{ee} = Z_{23, 23} \ + \ Z_{2e} \ – \ Z_{3e} = 0.06 + 0.0621 – (-0.05) = 0.1721

The augmented matrix is

5. Eliminate last row and column. The modified matrix is

6. Add branch 3 to 4, p = 3, k = 4:

Z_{41} = Z_{31} = 0.01
Z_{42} = Z_{32} = 0.018
Z_{43} = Z_{33} = 0.0355
Z_{44} = Z_{34} \ + \ Z_{34, \ 34} = 0.0355 + 0.1 = 0.1355

The augmented matrix is

7. Add first parallel link 1–4, Figure 3.13d:
p = 1,    k = 4
Z_{e1} = Z_{1e} = 0.0345 – 0.01 = 0.0245
Z_{e2} = Z_{2e} = 0.0221 – 0.018 = 0.0041
Z_{e3} = Z_{3e} = 0.01 – 0.0355 = -0.0255
Z_{e4} = Z_{4e} = 0.01 – 0.1355 = 0.1255
Z_{ee} = 0.2 + 0.245 – (-0.1255) = 0.350

This gives the matrix

8. Eliminate last row and last column using Equation 3.76:

9. Finally, add second parallel link 1–4, Figure 3.13e:
Z_{e1} = Z_{1e} = 0.0328 – 0.0188 = 0.014
Z_{e2} = Z_{2e} = 0.0218 – 0.0195 = 0.0023
Z_{e3} = Z_{3e} = 0.0118 – 0.0264 = -0.0146
Z_{e4} = Z_{4e} = 0.0188 – 0.0905 = -0.0717
Z_{ee} = 0.2 + 0.014 – (-0.0717) = 0.2857
This gives

10. Eliminate last row and column:

\left|\begin{matrix} 0.0328 & 0.0218 & 0.0118  & 0.0188   \\ 0.0218 & 0.0397 & 0.0183 & 0.0195   \\ 0.0118 &  0.0183 & 0.0336 & 0.0264  \\ 0.0188 & 0.0195 & 0.0264 & 0.0905   \end{matrix} \right|
-\frac{ \left|\begin{matrix} 0.014 \\ 0.0023 \\ -0.0416 \\ -0.0717 \end{matrix} \right| \left|\begin{matrix} 0.0014 & 0.0023 & -0.0146 & -0.0717 \end{matrix} \right| }{0.2857}

This gives the final positive or negative sequence matrix as

Zero Sequence Impedance Matrix
The zero sequence impedance matrix is similarly formed, until the last parallel coupled line between buses 1 and 4 is added. The zero sequence impedance matrix, until the coupled branch is required to be added, is formed by a step-by-step procedure as outlined for the positive sequence matrix:

Add parallel coupled lines between buses 1 and 4, p = 1, k = 4.
The coupled primitive impedance matrix is

Its inverse is given by

Z^{-1}_{pr} = \left|\begin{matrix} 0.3 & 0.1   \\ 0.1 & 0.03  \end{matrix} \right|^{-1} =\left|\begin{matrix} 3.750 & -1.25   \\ -1.25 & 3.750  \end{matrix} \right|
Y_{pe,pe} = 3.75
Y_{pe,xy} = -1.25

p = 1, k = 4 coupled with 2-3. Thus,
Z_{x1} = 0.0184, \ \ Z_{x2} = 0.0123, \ \ Z_{x3} = 0.0098, \ \ Z_{x4} = 0.0132
Z_{y1} = 0.0132, \ \ Z_{y2} = 0.0314, \ \ Z_{y3} = 0.0523, \ \ Z_{y4} = 0.1567
This gives
Z_{e1} = Z_{11} – Z_{41} + \frac{Y_{pe,pe}(Z_{x1} \ – \ Z_{y1})}{Y_{pe,ep}}
= 0.0184 – 0.0132 + \frac{(-1.25)(0.0184) \ – \ (0.00132)}{3.75} = -0.0035

Similarly,

Z_{e2} = 0.0123 – 0.314 + (\frac{-1.25}{3.75}) (0.0123 – 0.314) = -0.0127
Z_{e3} = 0.0098 – 0.523 + (\frac{-1.25}{3.75})(0.0098 – 0.0523) = -0.0283
Z_{e4} = 0.0132 – 0.1567 + (\frac{-1.25}{3.75})(0.0132 – 0.1567) = -0.0957
Z_{ee} is given by
Z_{ee} = Z_{1e} \ – \ Z_{4e} + \frac{1 \ + \ (-1.25)(Z_{e1} \ – \ Z_{e4})}{3.75}
= 0.0035 – (-0.0957) + \frac{1 \ + \ (-25)(0.0035 \ – \ (-0.0957))}{3.75} = 0.3328

The modified impedance matrix is now

Finally, eliminate the last row and column:

\left|\begin{matrix} 0.0184 & 0.0123 & 0.0098  & 0.0132   \\ 0.0123 & 0.0670 & 0.0442 & 0.314   \\ 0.0098 &  0.0442 & 0.0806  & 0.0523  \\ 0.0132 & 0.0314 & 0.0523 & 0.1567   \end{matrix} \right|
-\frac{ \left|\begin{matrix} 0.0035 \\ -0.0127 \\ -0.0283 \\ -0.0957 \end{matrix} \right| \left|\begin{matrix} 0.0035 & -0.0127 & -0.0283 & -0.0957 \end{matrix} \right| }{0.3328}

This gives the final zero sequence bus impedance matrix:

Note that at each last row and column elimination, Z_{21} = Z_{12}, \ Z_{23} = Z_{32}, etc. Also, in the final matrix there are no negative elements. If a duplex reactor is modeled (Appendix C), some elements may be negative; the same is true, in some cases, for modeling of threewinding transformers. Other than that, there is no check on correct formation.