**PartB: HPFA with non-linear loads and D-STATCOM device.**


$$
\begin{bmatrix}
\mathbf{V\_a} \\
\mathbf{V\_b} \\
\mathbf{V\_c}
\end{bmatrix}^\mathrm{h} = \begin{bmatrix}
\mathbf{0} \\
\mathbf{0} \\
\mathbf{0}
\end{bmatrix} \tag{26}
$$

22. Find the net harmonic current matrix at all the buses with the harmonic current matrix drawn by the linear loads and the harmonic current injection matrix of non-linear loads and the D-STATCOM device. The current matrix drawn by the linear loads at all the buses is zero for the first iteration as the harmonic voltage at all the buses is zero for the first iteration. This is illustrated with the sample section as shown in Figure 6. The net harmonic current matrix at bus-j is given by Equation (27), and the harmonic current matrix in branch-ij is given by Equation (28):

$$\left[\mathrm{I\_{abc}}\right]^{\mathrm{h}}\_{\mathrm{j}} = -\left[\mathrm{IS\_{abc}}\right]^{\mathrm{h}}\_{\mathrm{j}} - \left[\mathrm{ID\_{abc}}\right]^{\mathrm{h}}\_{\mathrm{j}} + \left[\mathrm{IL\_{abc}}\right]^{\mathrm{h}}\_{\mathrm{j}}\tag{27}$$

$$\left[\mathbf{I}\_{\rm abc}\right]\_{\rm ij}^{\rm h} = \left[\mathbf{I}\_{\rm abc}\right]\_{\rm j}^{\rm h} \tag{28}$$

where:

[Iabc] h j : Harmonic current matrix at bus-j for harmonic order-h;

[Iabc] h ij : Harmonic current matrix in branch-ij for harmonic order-h;

[ILabc] h j : Harmonic current matrix drawn by linear load at bus-j for harmonic order-h;

[ISabc] h j : Harmonic current injection matrix by non-linear load at bus-j for harmonic order-h;

[IDabc] h j : Harmonic current injection matrix by D-STATCOM device at bus-j for harmonic order-h.

Likewise, the harmonic currents in all branches are to be obtained by moving up to the substation as explained in step 3 to step 4 in PartA for FPFA.

23. Then, start finding the harmonic voltages at all buses downstream from the substation bus with Equation (29) as explained in step 5 in PartA:

$$
\begin{aligned}
\left[\begin{array}{c}\mathbf{V\_{a}}\\ \mathbf{V\_{b}}\\ \mathbf{V\_{c}}\end{array}\right]\_{\mathbf{j}}^{\mathbf{h}} &= \left[\begin{array}{c}\mathbf{V\_{a}}\\ \mathbf{V\_{b}}\\ \mathbf{V\_{c}}\end{array}\right]\_{\mathbf{j}}^{\mathbf{h}} - \left[\begin{array}{c}\mathbf{Z\_{a}} & \mathbf{Z\_{ab}} & \mathbf{Z\_{ac}}\\ \mathbf{Z\_{ba}} & \mathbf{Z\_{bb}} & \mathbf{Z\_{bc}}\\ \mathbf{Z\_{ca}} & \mathbf{Z\_{cb}} & \mathbf{Z\_{cc}}\end{array}\right]\_{\mathbf{j}}^{\mathbf{h}} \cdot \left[\begin{array}{c}\mathbf{I\_{a}}\\ \mathbf{I\_{b}}\\ \mathbf{I\_{c}}\end{array}\right]\_{\mathbf{j}}^{\mathbf{h}}\end{aligned}\tag{29}
$$


$$\begin{aligned} \left[\begin{array}{c} \text{SLoss}\_{\mathsf{a}} \\ \text{SLoss}\_{\mathsf{b}} \\ \text{SLoss}\_{\mathsf{c}} \end{array}\right]\_{\mathsf{ij}}^{\mathsf{h}} &= \left[\begin{array}{c} \left(\mathsf{V}\_{\mathsf{a}}\right)\_{\mathsf{i}}\cdot\left(\mathsf{I}\_{\mathsf{a}}\right)\_{\mathsf{ij}}^{\*} \\ \left(\mathsf{V}\_{\mathsf{b}}\right)\_{\mathsf{i}}\cdot\left(\mathsf{I}\_{\mathsf{b}}\right)\_{\mathsf{ij}}^{\*} \\ \left(\mathsf{V}\_{\mathsf{c}}\right)\_{\mathsf{i}}\cdot\left(\mathsf{I}\_{\mathsf{c}}\right)\_{\mathsf{ij}}^{\*} \end{array}\right]^{\mathsf{h}} - \left[\begin{array}{c} \left(\mathsf{V}\_{\mathsf{a}}\right)\_{\mathsf{j}}\cdot\left(\mathsf{I}\_{\mathsf{a}}\right)\_{\mathsf{ji}}^{\*} \\ \left(\mathsf{V}\_{\mathsf{b}}\right)\_{\mathsf{j}}\cdot\left(\mathsf{I}\_{\mathsf{b}}\right)\_{\mathsf{ji}}^{\*} \\ \left(\mathsf{V}\_{\mathsf{c}}\right)\_{\mathsf{j}}\cdot\left(\mathsf{I}\_{\mathsf{c}}\right)\_{\mathsf{ji}}^{\*} \end{array}\right]^{\mathsf{h}} \end{aligned} \tag{30}$$

$$\left[\text{TS\\_loss}\_{\text{abc}}\right]^{\text{h}} = \sum\_{\text{br}=1}^{\text{Nbr}} \left[\text{SLoss}\_{\text{abc}}\right]^{\text{h}}\_{\text{br}} \tag{31}$$


$$\left[\text{Total\\_loss}\right] = \sum\_{\text{h=h\_0}}^{\text{h\_m}} \sum\_{\text{br}=1}^{\text{Nbr}} \left[\left[\text{SLoss}\_{\text{abc}}\right]\_{\text{br}}^{\text{h}}\right] \tag{32}$$

28. The total r.m.s voltage at bus-i, say, phase '*a*', is calculated as:

$$\mathbf{(V\_a)\_i} = \sqrt{\left| \left( \mathbf{V\_a)\_i^1} \right|^2 + \sum\_{\mathbf{h=h\_o}}^{\mathbf{h\_m}} \left| \left( \mathbf{V\_a)\_i^h} \right|^2 \right. \tag{33}$$

29. The total harmonic distortion at every bus is calculated using Equation (34):

$$(\text{THD})\_{\text{i}}^{\text{a}} = \frac{\sqrt{\sum\_{\text{h}=\text{h}\_{\text{o}}}^{\text{h}\_{\text{m}}} \left| (\mathbf{V}\_{\text{a}})\_{\text{i}}^{\text{h}} \right|^{2}}{\left| (\mathbf{V}\_{\text{a}})\_{\text{i}}^{\text{1}} \right|} \tag{34}$$

where: h<sup>o</sup> : Minimum harmonic order;

h<sup>m</sup> : Maximum harmonic order;

br : Branch number;

Nbr : Total number of branches.

**Figure 4.** A simple URDN three busses.

**Figure 5.** A simple URDN with two buses with D-STATCOM placed at bus-j.

**Figure 6.** Sample section of two buses for HPFA.
