Solution to Application Example 7.17

Step #4 of Harmonic Load Flow. The matrix equation shown below makes it easy to identify the entries of the Jacobian matrix:

${\overline{J}}^0=\begin{array}{c}\hfill \hfill \\ \hfill \left|\begin{array}{ccccccccccccccccc}\hfill \underset{\downarrow }{\underbrace{{\delta }_2^{\left(1\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_2^{\left(1\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_3^{\left(1\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_3^{\left(1\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_4^{\left(1\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_4^{\left(1\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_1^{\left(5\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_1^{\left(5\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_2^{\left(5\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_2^{\left(5\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_3^{\left(5\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_3^{\left(5\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{{\delta }_4^{\left(5\right)}}}\hfill & \hfill \underset{\downarrow }{\underbrace{\left|{\tilde{V}}_4^{\left(5\right)}\right|}}\hfill & \hfill \underset{\downarrow }{\underbrace{\alpha }}\hfill & \hfill \underset{\downarrow }{\underbrace{\beta }}\hfill & \hfill \hfill \\ \hfill J_{1,1}^0\hfill & \hfill J_{1,2}^0\hfill & \hfill J_{1,3}^0\hfill & \hfill J_{1,4}^0\hfill & \hfill J_{1,5}^0\hfill & \hfill J_{1,6}^0\hfill & \hfill J_{1,7}^0\hfill & \hfill J_{1,8}^0\hfill & \hfill J_{1,9}^0\hfill & \hfill J_{1,10}^0\hfill & \hfill J_{1,11}^0\hfill & \hfill J_{1,12}^0\hfill & \hfill J_{1,13}^0\hfill & \hfill J_{1,14}^0\hfill & \hfill J_{1,15}^0\hfill & \hfill J_{1,16}^0\hfill \\ \hfill J_{2,1}^0\hfill & \hfill J_{2,2}^0\hfill & \hfill J_{2,3}^0\hfill & \hfill J_{2,4}^0\hfill & \hfill J_{2,5}^0\hfill & \hfill J_{2,6}^0\hfill & \hfill J_{2,7}^0\hfill & \hfill J_{2,8}^0\hfill & \hfill J_{2,9}^0\hfill & \hfill J_{2,10}^0\hfill & \hfill J_{2,11}^0\hfill & \hfill J_{2,12}^0\hfill & \hfill J_{2,13}^0\hfill & \hfill J_{2,14}^0\hfill & \hfill J_{2,15}^0\hfill & \hfill J_{2,16}^0\hfill \\ \hfill J_{3,1}^0\hfill & \hfill J_{3,2}^0\hfill & \hfill J_{3,3}^0\hfill & \hfill J_{3,4}^0\hfill & \hfill J_{3,5}^0\hfill & \hfill J_{3,6}^0\hfill & \hfill J_{3,7}^0\hfill & \hfill J_{3,8}^0\hfill & \hfill J_{3,9}^0\hfill & \hfill J_{3,10}^0\hfill & \hfill J_{3,11}^0\hfill & \hfill J_{3,12}^0\hfill & \hfill J_{3,13}^0\hfill & \hfill J_{3,14}^0\hfill & \hfill J_{3,15}^0\hfill & \hfill J_{3,16}^0\hfill \\ \hfill J_{4,1}^0\hfill & \hfill J_{4,2}^0\hfill & \hfill J_{4,3}^0\hfill & \hfill J_{4,4}^0\hfill & \hfill J_{4,5}^0\hfill & \hfill J_{4,6}^0\hfill & \hfill J_{4,7}^0\hfill & \hfill J_{4,8}^0\hfill & \hfill J_{4,9}^0\hfill & \hfill J_{4,10}^0\hfill & \hfill J_{4,11}^0\hfill & \hfill J_{4,12}^0\hfill & \hfill J_{4,13}^0\hfill & \hfill J_{4,14}^0\hfill & \hfill J_{4,15}^0\hfill & \hfill J_{4,16}^0\hfill \\ \hfill J_{5,1}^0\hfill & \hfill J_{5,2}^0\hfill & \hfill J_{5,3}^0\hfill & \hfill J_{5,4}^0\hfill & \hfill J_{5,5}^0\hfill & \hfill J_{5,6}^0\hfill & \hfill J_{5,7}^0\hfill & \hfill J_{5,8}^0\hfill & \hfill J_{5,9}^0\hfill & \hfill J_{5,10}^0\hfill & \hfill J_{5,11}^0\hfill & \hfill J_{5,12}^0\hfill & \hfill J_{5,13}^0\hfill & \hfill J_{5,14}^0\hfill & \hfill J_{5,15}^0\hfill & \hfill J_{5,16}^0\hfill \\ \hfill J_{6,1}^0\hfill & \hfill J_{6,2}^0\hfill & \hfill J_{6,3}^0\hfill & \hfill J_{6,4}^0\hfill & \hfill J_{6,5}^0\hfill & \hfill J_{6,6}^0\hfill & \hfill J_{6,7}^0\hfill & \hfill J_{6,8}^0\hfill & \hfill J_{6,9}^0\hfill & \hfill J_{6,10}^0\hfill & \hfill J_{6,11}^0\hfill & \hfill J_{6,12}^0\hfill & \hfill J_{6,13}^0\hfill & \hfill J_{6,14}^0\hfill & \hfill J_{6,15}^0\hfill & \hfill J_{6,16}^0\hfill \\ \hfill J_{7,1}^0\hfill & \hfill J_{7,2}^0\hfill & \hfill J_{7,3}^0\hfill & \hfill J_{7,4}^0\hfill & \hfill J_{7,5}^0\hfill & \hfill J_{7,6}^0\hfill & \hfill J_{7,7}^0\hfill & \hfill J_{7,8}^0\hfill & \hfill J_{7,9}^0\hfill & \hfill J_{7,10}^0\hfill & \hfill J_{7,11}^0\hfill & \hfill J_{7,12}^0\hfill & \hfill J_{7,13}^0\hfill & \hfill J_{7,14}^0\hfill & \hfill J_{7,15}^0\hfill & \hfill J_{7,16}^0\hfill \\ \hfill J_{8,1}^0\hfill & \hfill J_{8,2}^0\hfill & \hfill J_{8,3}^0\hfill & \hfill J_{8,4}^0\hfill & \hfill J_{8,5}^0\hfill & \hfill J_{8,6}^0\hfill & \hfill J_{8,7}^0\hfill & \hfill J_{8,8}^0\hfill & \hfill J_{8,9}^0\hfill & \hfill J_{8,10}^0\hfill & \hfill J_{8,11}^0\hfill & \hfill J_{8,12}^0\hfill & \hfill J_{8,13}^0\hfill & \hfill J_{8,14}^0\hfill & \hfill J_{8,15}^0\hfill & \hfill J_{8,16}^0\hfill \\ \hfill J_{9,1}^0\hfill & \hfill J_{9,2}^0\hfill & \hfill J_{9,3}^0\hfill & \hfill J_{9,4}^0\hfill & \hfill J_{9,5}^0\hfill & \hfill J_{9,6}^0\hfill & \hfill J_{9,7}^0\hfill & \hfill J_{9,8}^0\hfill & \hfill J_{9,9}^0\hfill & \hfill J_{9,10}^0\hfill & \hfill J_{9,11}^0\hfill & \hfill J_{9,12}^0\hfill & \hfill J_{9,13}^0\hfill & \hfill J_{9,14}^0\hfill & \hfill J_{9,15}^0\hfill & \hfill J_{9,16}^0\hfill \\ \hfill J_{10,1}^0\hfill & \hfill J_{10,2}^0\hfill & \hfill J_{10,3}^0\hfill & \hfill J_{10,4}^0\hfill & \hfill J_{10,5}^0\hfill & \hfill J_{10,6}^0\hfill & \hfill J_{10,7}^0\hfill & \hfill J_{10,8}^0\hfill & \hfill J_{10,9}^0\hfill & \hfill J_{10,10}^0\hfill & \hfill J_{10,11}^0\hfill & \hfill J_{10,12}^0\hfill & \hfill J_{10,13}^0\hfill & \hfill J_{10,14}^0\hfill & \hfill J_{10,15}^0\hfill & \hfill J_{10,16}^0\hfill \\ \hfill J_{11,1}^0\hfill & \hfill J_{11,2}^0\hfill & \hfill J_{11,3}^0\hfill & \hfill J_{11,4}^0\hfill & \hfill J_{11,5}^0\hfill & \hfill J_{11,6}^0\hfill & \hfill J_{11,7}^0\hfill & \hfill J_{11,8}^0\hfill & \hfill J_{11,9}^0\hfill & \hfill J_{11,10}^0\hfill & \hfill J_{11,11}^0\hfill & \hfill J_{11,12}^0\hfill & \hfill J_{11,13}^0\hfill & \hfill J_{11,14}^0\hfill & \hfill J_{11,15}^0\hfill & \hfill J_{11,16}^0\hfill \\ \hfill J_{12,1}^0\hfill & \hfill J_{12,2}^0\hfill & \hfill J_{12,3}^0\hfill & \hfill J_{12,4}^0\hfill & \hfill J_{12,5}^0\hfill & \hfill J_{12,6}^0\hfill & \hfill J_{12,7}^0\hfill & \hfill J_{12,8}^0\hfill & \hfill J_{12,9}^0\hfill & \hfill J_{12,10}^0\hfill & \hfill J_{12,11}^0\hfill & \hfill J_{12,12}^0\hfill & \hfill J_{12,13}^0\hfill & \hfill J_{12,14}^0\hfill & \hfill J_{12,15}^0\hfill & \hfill J_{12,16}^0\hfill \\ \hfill J_{13,1}^0\hfill & \hfill J_{13,2}^0\hfill & \hfill J_{13,3}^0\hfill & \hfill J_{13,4}^0\hfill & \hfill J_{13,5}^0\hfill & \hfill J_{13,6}^0\hfill & \hfill J_{13,7}^0\hfill & \hfill J_{13,8}^0\hfill & \hfill J_{13,9}^0\hfill & \hfill J_{13,10}^0\hfill & \hfill J_{13,11}^0\hfill & \hfill J_{13,12}^0\hfill & \hfill J_{13,13}^0\hfill & \hfill J_{13,14}^0\hfill & \hfill J_{13,15}^0\hfill & \hfill J_{13,16}^0\hfill \\ \hfill J_{14,1}^0\hfill & \hfill J_{14,2}^0\hfill & \hfill J_{14,3}^0\hfill & \hfill J_{14,4}^0\hfill & \hfill J_{14,5}^0\hfill & \hfill J_{14,6}^0\hfill & \hfill J_{14,7}^0\hfill & \hfill J_{14,8}^0\hfill & \hfill J_{14,9}^0\hfill & \hfill J_{14,10}^0\hfill & \hfill J_{14,11}^0\hfill & \hfill J_{14,12}^0\hfill & \hfill J_{14,13}^0\hfill & \hfill J_{14,14}^0\hfill & \hfill J_{14,15}^0\hfill & \hfill J_{14,16}^0\hfill \\ \hfill J_{15,1}^0\hfill & \hfill J_{15,2}^0\hfill & \hfill J_{15,3}^0\hfill & \hfill J_{15,4}^0\hfill & \hfill J_{15,5}^0\hfill & \hfill J_{15,6}^0\hfill & \hfill J_{15,7}^0\hfill & \hfill J_{15,8}^0\hfill & \hfill J_{15,9}^0\hfill & \hfill J_{15,10}^0\hfill & \hfill J_{15,11}^0\hfill & \hfill J_{15,12}^0\hfill & \hfill J_{15,13}^0\hfill & \hfill J_{15,14}^0\hfill & \hfill J_{15,15}^0\hfill & \hfill J_{15,16}^0\hfill \\ \hfill J_{16,1}^0\hfill & \hfill J_{16,2}^0\hfill & \hfill J_{16,3}^0\hfill & \hfill J_{16,4}^0\hfill & \hfill J_{16,5}^0\hfill & \hfill J_{16,6}^0\hfill & \hfill J_{16,7}^0\hfill & \hfill J_{16,8}^0\hfill & \hfill J_{16,9}^0\hfill & \hfill J_{16,10}^0\hfill & \hfill J_{16,11}^0\hfill & \hfill J_{16,12}^0\hfill & \hfill J_{16,13}^0\hfill & \hfill J_{16,14}^0\hfill & \hfill J_{16,15}^0\hfill & \hfill J_{16,16}^0\hfill \end{array}\right|\hfill \end{array}\begin{array}{c}\hfill \hfill \\ \hfill \leftarrow \mathrm{\Delta }{P}_2^{\left(1\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{Q}_2^{\left(1\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{P}_3^{\left(1\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{Q}_3^{\left(1\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{P}_4^t\hfill \\ \hfill \leftarrow \mathrm{\Delta }{Q}_4^t\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{r,1}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{i,1}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{r,2}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{i,2}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{r,3}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{i,3}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{r,4}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{i,4}^{\left(5\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{r,4}^{\left(1\right)}\hfill \\ \hfill \leftarrow \mathrm{\Delta }{I}_{i,4}^{\left(1\right)}\hfill \end{array}$

The top row, which is not part of the Jacobian, lists the variables:

${\delta }_2^{\left(1\right)},\left|{\tilde{V}}_2^{\left(1\right)}\right|,{\delta }_3^{\left(1\right)},\left|{\tilde{V}}_3^{\left(1\right)}\right|,{\delta }_4^{\left(1\right)},\left|{\tilde{V}}_4^{\left(1\right)}\right|,{\delta }_1^{\left(5\right)},\left|{\tilde{V}}_1^{\left(5\right)}\right|,{\delta }_2^{\left(5\right)},\left|{\tilde{V}}_2^{\left(5\right)}\right|,{\delta }_3^{\left(5\right)},\left|{\tilde{V}}_3^{\left(5\right)}\right|,{\delta }_4^{\left(5\right)},\left|{\tilde{V}}_4^{\left(5\right)}\right|,\alpha ,\beta$

and the right-hand side column, which is not part of the Jacobian, lists the mismatch quantities that must be satisfied:

${\left[\mathrm{\Delta }{P}_2^{\left(1\right)},\mathrm{\Delta }{Q}_2^{\left(1\right)},\mathrm{\Delta }{P}_3^{\left(1\right)},\mathrm{\Delta }{Q}_3^{\left(1\right)},\mathrm{\Delta }{P}_4^t,\mathrm{\Delta }{Q}_4^t,\mathrm{\Delta }{I}_{r,1}^{\left(5\right)},\mathrm{\Delta }{I}_{i,1}^{\left(5\right)},\mathrm{\Delta }{I}_{r,2}^{\left(5\right)},\mathrm{\Delta }{I}_{i,2}^{\left(5\right)},\mathrm{\Delta }{I}_{r,3}^{\left(5\right)},\mathrm{\Delta }{I}_{i,3}^{\left(5\right)},\mathrm{\Delta }{I}_{r,4}^{\left(5\right)},\mathrm{\Delta }{I}_{i,4}^{\left(5\right)},\mathrm{\Delta }{I}_{r,4}^{\left(1\right)},\mathrm{\Delta }{I}_{i,4}^{\left(1\right)}\right]}^t\text{.}$

These two identifications help in defining the partial derivatives of the Jacobian. For example,

$J_{3,5}^0=\frac{\partial \left(\mathrm{\Delta }{P}_3^{\left(1\right)}\right)}{\partial \left({\delta }_4^{\left(1\right)}\right)},J_{3,10}^0=\frac{\partial \left(\mathrm{\Delta }{P}_3^{\left(1\right)}\right)}{\partial \left(\mathrm{\Delta }{V}_2^{\left(5\right)}\right)}\text{,}$

$\begin{array}{c}{J}_{8,5}^0=\frac{\partial \left(\mathrm{\Delta }{I}_{i,1}^{\left(5\right)}\right)}{\partial \left({\delta }_4^{\left(1\right)}\right)},J_{8,10}^0=\frac{\partial \left(\mathrm{\Delta }{I}_{i,1}^{\left(5\right)}\right)}{\partial \left(\mathrm{\Delta }{V}_2^{\left(5\right)}\right)},\\ J_{9,11}^0=\frac{\partial \left(\mathrm{\Delta }{I}_{r,2}^{\left(5\right)}\right)}{\partial \left({\delta }_3^{\left(5\right)}\right)},\mathrm{and}{J}_{12,16}^0=\frac{\partial \left(\mathrm{\Delta }{I}_{i,3}^{\left(5\right)}\right)}{\partial \left(\beta \right)}\text{.}\end{array}$

In the solution below submatrices are defined, which is an alternative way of defining the Jacobian entries.

Step #4 of Harmonic Load Flow (continued). Evaluate the Jacobian matrix $\overline{J}$ and compute ΔŪ⁰. The Jacobian matrix has the following form (${\overline{0}}_{n\times m}$ denotes an n by m zero submatrix with zero entries):

$\overline{J}=\left|\begin{array}{ccc}\hfill {\overline{J}}^{\left(1\right)}\hfill & \hfill {\overline{J}}^{\left(5\right)}\hfill & \hfill {\overline{0}}_{6,2}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill {\overline{0}}_{6,2}\hfill \\ \hfill {\overline{G}}^{(5,1)}\hfill & \hfill {\overline{Y}}^{(5,5)}+{\overline{G}}^{(5,5)}\hfill & \hfill {\overline{H}}^{\left(5\right)}\hfill \\ \hfill {\overline{Y}}^{(1,1)}+{\overline{G}}^{(1,1)}\hfill & \hfill {\overline{G}}^{(1,5)}\hfill & \hfill {\overline{H}}^{\left(1\right)}\hfill \end{array}\right|\text{.}$

The Jacobian submatrix for the fundamental is

${\overline{J}}^{\left(1\right)}=\left|\begin{array}{cccccc}\hfill 61.68\hfill & \hfill 52.932\hfill & \hfill -11.764\hfill & \hfill -2.944\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -52.94\hfill & \hfill 61.945\hfill & \hfill 2.944\hfill & \hfill -11.764\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -11.764\hfill & \hfill -2.944\hfill & \hfill 51.772\hfill & \hfill 22.917\hfill & \hfill -40.0076\hfill & \hfill -19.98\hfill \\ \hfill 2.944\hfill & \hfill -11.764\hfill & \hfill -22.926\hfill & \hfill 51.773\hfill & \hfill 19.982\hfill & \hfill -40.0076\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -40.0076\hfill & \hfill -19.982\hfill & \hfill 80.015\hfill & \hfill 39.961\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 19.982\hfill & \hfill -40.0076\hfill & \hfill -39.963\hfill & \hfill 80\hfill \end{array}\right|\text{.}$

The Jacobian submatrix for the 5th harmonic is

${\overline{J}}^{\left(5\right)}=\left|\begin{array}{cccccccc}\hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill {\overline{0}}_{4\times 8}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill -0.099\hfill & \hfill -0.099\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.099\hfill & \hfill -0.09\hfill & \hfill 0.198\hfill & \hfill 0.198\hfill \\ \hfill 0.010\hfill & \hfill -0.099\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.010\hfill & \hfill -0.99\hfill & \hfill -0.020\hfill & \hfill 1.98\hfill \end{array}\right|\text{.}$

Also

${\overline{Y}}^{(1,1)}=\left|\begin{array}{cccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -39.889\hfill & \hfill -19.89\hfill & \hfill 79.81\hfill & \hfill 39.73\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -19.81\hfill & \hfill 40.06\hfill & \hfill 39.56\hfill & \hfill -80.14\hfill \end{array}\right|$

${\overline{Y}}^{(5,5)}=\left|\begin{array}{cccccccc}\hfill 202.91\hfill & \hfill 4.84\hfill & \hfill -1.92\hfill & \hfill -3.85\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill \\ \hfill 0.48\hfill & \hfill -2029.1\hfill & \hfill -0.39\hfill & \hfill 19.23\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.10\hfill & \hfill 9.90\hfill \\ \hfill -1.92\hfill & \hfill -3.846\hfill & \hfill 2.17\hfill & \hfill 3.97\hfill & \hfill -0.25\hfill & \hfill -0.13\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -0.39\hfill & \hfill 19.23\hfill & \hfill 0.40\hfill & \hfill -21.73\hfill & \hfill -0.01\hfill & \hfill 2.45\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -0.25\hfill & \hfill -0.13\hfill & \hfill 1.24\hfill & \hfill 1.12\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -0.01\hfill & \hfill 2.49\hfill & \hfill 0.11\hfill & \hfill -12.40\hfill & \hfill -0.10\hfill & \hfill 9.90\hfill \\ \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 1.98\hfill & \hfill 1.98\hfill \\ \hfill -0.01\hfill & \hfill 9.90\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.01\hfill & \hfill 9.90\hfill & \hfill 0.20\hfill & \hfill -19.80\hfill \end{array}\right|\text{.}$

In this example, ${\overline{H}}^{\left(1\right)}$ and ${\overline{H}}^{\left(5\right)}$ are zero submatrices since there are no nonlinear device variables defined for the system of Fig. E7.14.1. However, setting ${\overline{H}}^{\left(1\right)}$ and ${\overline{H}}^{\left(5\right)}$ to zero introduces two columns of zeros in the Jacobian matrix, which makes it singular. In order to avoid this situation, ${\overline{H}}^{\left(1\right)}$ and ${\overline{H}}^{\left(5\right)}$ are set to random matrices. Note that the solution does not depend on the values of these submatrices.

In order to compute ${\overline{G}}^{(1,1)}$ and ${\overline{G}}^{(1,5)}$, we need to differentiate $G_{r,4}^{\left(1\right)}$ and $G_{i,4}^{\left(1\right)}$ as follows:

$\frac{\partial G_{r,4}^{\left(1\right)}}{\partial {\delta }_4^{\left(1\right)}}=\frac{P_4^{\left(1\right)}\cos {\gamma }_4^{\left(1\right)}\tan \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}{V_4^{\left(1\right)}\cos \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=0.00167$

$\frac{\partial G_{r,4}^{\left(1\right)}}{\partial V_4^{\left(1\right)}}=-\frac{P_4^{\left(1\right)}\cos {\gamma }_4^{\left(1\right)}}{{\left(V_4^{\left(1\right)}\right)}^2\cos \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=-0.252$

$\frac{\partial G_{i,4}^{\left(1\right)}}{\partial {\delta }_4^{\left(1\right)}}=\frac{P_4^{\left(1\right)}\sin {\gamma }_4^{\left(1\right)}\tan \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}{V_4^{\left(1\right)}\cos \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=-0.0000112$

$\frac{\partial G_{i,4}^{\left(1\right)}}{\partial V_4^{\left(1\right)}}=-\frac{P_4^{\left(1\right)}\sin {\gamma }_4^{\left(1\right)}}{{\left(V_4^{\left(1\right)}\right)}^2\cos \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=0.00169$

The partial derivatives of $G_{r,4}^{\left(1\right)}$ and $G_{i,4}^{\left(1\right)}$ with respect to V₄⁽⁵⁾ and δ₄⁽⁵⁾ are zero. Therefore, ${\overline{G}}^{(1,5)}$ is a 2 × 8 submatrix of zeros and

${\overline{G}}^{(1,1)}=-\left|\begin{array}{ccc}\hfill \hfill & \hfill 0.10026\hfill & \hfill -0.2517\hfill \\ \hfill {\overline{0}}_{2\times 4}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill -0.0405\hfill & \hfill 0.25415\hfill \end{array}\right|\text{.}$

In order to compute ${\overline{G}}^{(5,1)}$ and ${\overline{G}}^{(5,5)}$, we need to differentiate ${\overline{G}}_{r,4}^{\left(5\right)}$ and ${\overline{G}}_{i,4}^{\left(5\right)}$ as follows:

$\frac{\partial G_{r,4}^{\left(5\right)}}{\partial {\delta }_4^{\left(1\right)}}=-0.9{\left(V_4^{\left(1\right)}\right)}^{3}sin\left(3{\delta }_4^{\left(1\right)}\right)=0.00896$

$\frac{\partial G_{r,4}^{\left(5\right)}}{\partial V_4^{\left(1\right)}}=0.9{\left(V_4^{\left(1\right)}\right)}^{2}cos\left(3{\delta }_4^{\left(1\right)}\right)=0.893$

$\frac{\partial G_{i,4}^{\left(5\right)}}{\partial {\delta }_4^{\left(1\right)}}=0.9{\left(V_4^{\left(1\right)}\right)}^{3}cos\left(3{\delta }_4^{\left(1\right)}\right)=0.8894$

$\frac{\partial G_{i,4}^{\left(5\right)}}{\partial V_4^{\left(1\right)}}=0.9{\left(V_4^{\left(1\right)}\right)}^{2}sin\left(3{\delta }_4^{\left(1\right)}\right)=-0.009\text{.}$

Also

$\frac{\partial G_{r,4}^{\left(5\right)}}{\partial {\delta }_4^{\left(5\right)}}=-0.9{\left(V_4^{\left(5\right)}\right)}^{2}sin\left(3{\delta }_4^{\left(5\right)}\right)=0.00$

$\frac{\partial G_{r,4}^{\left(5\right)}}{\partial V_4^{\left(5\right)}}=0.6V_4^{\left(5\right)}cos\left(3{\delta }_4^{\left(5\right)}\right)=0.06$

$\frac{\partial G_{i,4}^{\left(5\right)}}{\partial {\delta }_4^{\left(5\right)}}=0.9{\left(V_4^{\left(5\right)}\right)}^{2}cos\left(3{\delta }_4^{\left(5\right)}\right)=0.009$

$\frac{\partial G_{i,4}^{\left(5\right)}}{\partial V_4^{\left(5\right)}}=0.6V_4^{\left(5\right)}sin\left(3{\delta }_4^{\left(5\right)}\right)=0.00\text{.}$

Therefore,

$\begin{array}{c}{\overline{G}}^{\left(5.1\right)}=\left|\begin{array}{cccc}\hfill \hfill & \hfill {\overline{0}}_{6\times 8}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 0.00896\hfill & \hfill 0.893\hfill \\ \hfill {\overline{0}}_{2\times 6}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 0.8894\hfill & \hfill -0.009\hfill \end{array}\right|,\\ {\overline{G}}^{\left(5.5\right)}=\left|\begin{array}{cccc}\hfill \hfill & \hfill {\overline{0}}_{6\times 8}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 0.06\hfill \\ \hfill {\overline{0}}_{2\times 6}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 0.009\hfill & \hfill 0\hfill \end{array}\right|\text{.}\end{array}$

The Jacobian matrix is

$\left|\begin{array}{cccccccccccccccc}\hfill -61.57\hfill & \hfill 52.71\hfill & \hfill -11.70\hfill & \hfill -2.90\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -52.79\hfill & \hfill 61.52\hfill & \hfill 2.89\hfill & \hfill -11.74\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -11.69\hfill & \hfill -2.69\hfill & \hfill 51.37\hfill & \hfill 22.86\hfill & \hfill -39.71\hfill & \hfill -19.90\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2.95\hfill & \hfill -11.72\hfill & \hfill -22.77\hfill & \hfill 51.58\hfill & \hfill 19.82\hfill & \hfill -39.87\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -39.66\hfill & \hfill -19.94\hfill & \hfill 79.45\hfill & \hfill 39.59\hfill & \hfill -0.099\hfill & \hfill -0.099\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.099\hfill & \hfill -0.099\hfill & \hfill 0.198\hfill & \hfill 0.198\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 19.86\hfill & \hfill -39.83\hfill & \hfill -39.93\hfill & \hfill 79.58\hfill & \hfill 0.010\hfill & \hfill -0.99\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.010\hfill & \hfill -0.99\hfill & \hfill -0.020\hfill & \hfill 1.98\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 202.9\hfill & \hfill 4.84\hfill & \hfill -1.92\hfill & \hfill -3.85\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.48\hfill & \hfill -2029\hfill & \hfill -0.39\hfill & \hfill 19.23\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.10\hfill & \hfill 9.90\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1.92\hfill & \hfill -3.846\hfill & \hfill 2.17\hfill & \hfill 3.97\hfill & \hfill -0.25\hfill & \hfill -0.13\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.39\hfill & \hfill 19.23\hfill & \hfill 0.40\hfill & \hfill -21.73\hfill & \hfill -0.01\hfill & \hfill 2.45\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.25\hfill & \hfill -0.13\hfill & \hfill 1.24\hfill & \hfill 1.12\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.01\hfill & \hfill 2.49\hfill & \hfill 0.11\hfill & \hfill -12.40\hfill & \hfill -0.10\hfill & \hfill 9.90\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.009\hfill & \hfill 0.893\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.99\hfill & \hfill -0.99\hfill & \hfill 1.98\hfill & \hfill 2.04\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.889\hfill & \hfill -0.009\hfill & \hfill -0.01\hfill & \hfill 9.90\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.01\hfill & \hfill 9.90\hfill & \hfill 0.201\hfill & \hfill -19.80\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -39.89\hfill & \hfill -19.89\hfill & \hfill 79.91\hfill & \hfill 39.48\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -19.81\hfill & \hfill 40.06\hfill & \hfill 39.81\hfill & \hfill -80.04\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right|$

7.4.11 Application Example 7.18: Computation of the Correction Bus Vector and Convergence of Harmonic Power Flow

Using the results of Application Examples 7.15 to 7.17, evaluate the correction bus vector ΔŪ⁰ for the system of Fig. E7.14.1. Update the mismatch vector and comment on the convergence of the harmonic load flow.

Solution to Application Example 7.18

Step #5 of Harmonic Load Flow. Evaluate ΔŪ⁰ and update ΔŪ⁰.

The correction bus vector ΔŪ⁰ is computed as follows:

$\mathrm{\Delta }{\overline{U}}^0={\overline{\left(J^0\right)}}^{-1}\left[\begin{array}{l}3.43472E-6\\ -1.33365E-6\\ -7.62939E-6\\ -3.81470E-6\\ 2.99421E-2\\ 2.97265E-4\\ 4.217029E-6\\ -200.0\\ -4.09782E-8\\ 2.384186E-7\\ -1.93715E-7\\ -5.960464E-8\\ 0.29935\\ -3.03515E-3\\ -1.78814E-7\\ 8.86619E-7\end{array}\right]=\left[\begin{array}{c}1.20815E-6\\ -8.59742E-7\\ 9.552948E-6\\ 1.54111E-6\\ 1.184274E-5\\ 2.3572E-6\\ 1.46588E-3\\ 0.10000228\\ 2.38763E-2\\ 0.1005407\\ 0.2042002\\ 0.1021379\\ 0.249194\\ 0.1027635\\ --\\ --\end{array}\right]\text{.}$

The updated bus vector is

$\begin{array}{c}{\overline{U}}^1={\overline{U}}^0-\mathrm{\Delta }{\overline{U}}^0=(-2.68332E-4,0.99759,-2.80029E-3,0.99640,-3.4257E\\ -3,0.99594,-1.465844E-3,-2.28E-6,-2.38763E-2,-5.407E-4,\\ -0.2042,-2.138E-3,-0.24919,-2.7635E-3,0,0){}^t\text{.}\end{array}$

Note that all harmonic voltage magnitudes are negative. To prevent negative magnitudes, the corresponding phase angles are shifted by π radians; therefore,

$\begin{array}{c}{\overline{U}}^1=(-2.68332E-4,0.99759,-2.80029E-3,\\ 0.99640,-3.4257E-3,0.99594,3.14013,2.28E-6,\\ 3.11772,5.407E-4,2.93738,2.138E-3,2.89239,2.7635E-3,0,0){}^t\text{.}\end{array}$

The new (updated) nonlinear device currents are $G_{r,4}^{\left(1\right)}$ = 0.256629, $G_{i,4}^{\left(1\right)}$ = –0.1011142, $G_{r,4}^{\left(5\right)}$ = 0.2965726 and $G_{i,4}^{\left(5\right)}$ = –2.59613E – 3. The new (updated) mismatch vector is

$\begin{array}{c}\mathrm{\Delta }{\overline{W}}^1=(5.342E-6,-3.800E-7,5.722E-6,\\ 3.815E-6,-1.403E-3,4.332E-4,-2.269E-3,\\ -3.300E-2,-2.677E-3,6.475E-3,-1.021E-3,\\ -2.105E-3,0.3023,3.015E-2,-6.264E-4,-1.473E-4){}^t\text{.}\end{array}$

This mismatch vector is not small enough and the solution has not converged. However, the iterative procedure will converge in eight iterations. The final solution is (all magnitudes are in percentage of the base values and all angles are in degrees):

Fundamental power flow output solution (Fig. E7.14.1):

Harmonic power flow output solution for harmonic number 5, frequency = 300 Hz (Fig. E7.14.1):

Total current/power:

Bus voltage summary:

7.5.1 Decoupled Harmonic Power Flow

The Newton-based harmonic power flow of Section 7.4 is very accurate because harmonic couplings are included at all frequencies. However, it might encounter convergence problems for large power systems with many nonlinear loads. For such applications, a decoupled approach is more appropriate.

At the fundamental frequency, the conventional power flow algorithm (Section 7.3) is used to model the system. At higher frequencies, harmonic couplings are neglected (to simplify the procedure, to reduce computing time, and to limit memory storage requirements) and the distorted power system is modeled by the harmonic admittance matrix. The general model of a linear load consisting of a resistance in parallel with a reactance is utilized [7,8], and nonlinear loads are modeled with decoupled harmonic current sources. The magnitudes and phase angles of harmonic current sources depend on the nature of the nonlinear loads and can be computed using analytical models or may be estimated based on measured nonsinusoidal current waveforms. This simple approach of nonlinear load modeling is very practical for most industrial loads.

The hth harmonic load admittance at bus i $\left(y_i^{\left(h\right)}\right)$, shunt capacitor admittance at bus i $\left(y_{ci}^{\left(h\right)}\right)$, and feeder admittance between buses i and i + 1 $\left(y_{i,i+1}^{\left(h\right)}\right)\text{,}$ are [7,8,16,18,19]

$y_i^{\left(h\right)}=\frac{P}{\left|V_i^{\left(1\right)}\right|{}^2}-j\frac{Q}{h\left|V_i^{\left(1\right)}\right|{}^2}\text{,}$

$y_{ci}^{\left(h\right)}=h{y}_{ci}^{\left(1\right)}\text{,}$

$y_{i,i+1}^{\left(h\right)}=\frac{1}{R_{i,i+1}+jh{X}_{i,i+1}}\text{,}$

where P_i and Q_i are the fundamental active and reactive load powers at bus i, $\left(y_{ci}^{\left(1\right)}\right)$ is the fundamental frequency admittance of the capacitor bank connected to bus i, and the superscript h indicates harmonic orders.

Nonlinear loads are modeled as decoupled harmonic current sources that inject harmonic currents into the system. The fundamental and the hth harmonic currents of the nonlinear load installed at bus m with fundamental real power P_m and fundamental reactive power Q_m are [8,16,19]

$I_m^{\left(1\right)}=\left[P_m+j{Q}_m)/V_m^{\left(1\right)}\right]*\text{,}$

$I_m^{\left(h\right)}=C\left(h\right)I_m^{\left(1\right)}\text{,}$

where C(h) is the ratio of the hth harmonic current to its fundamental value. A conventional (fundamental) Newton–Raphson algorithm (Section 7.3) is used to solve the system at the fundamental frequency and harmonic voltages are calculated by solving the following decoupled load flow equation (derived from the node equations) [16,17,19]

$Y^{\left(h\right)}{V}^{\left(h\right)}=I^{\left(h\right)}\text{.}$

The flowchart of the decoupled harmonic power flow (DHPF) based on Eqs. 7-125 to 7-130 is shown in Fig. 7.17. This simple and fast approach for the calculation of harmonic voltages and currents provides acceptable solutions as compared with Newton-based HPF (Section 7.4), which considers harmonic couplings [5,6].

7.5.2 Fast Harmonic Power Flow

This method employs the equivalent current injection transformation and the forward and backward sweep techniques to solve the harmonic load flow problem [17]. For bus i, the complex power S_i is

$S_i=\left(P_i+j{Q}_i\right),i=1,\dots ,n\text{,}$

where n is the total number of buses. The equivalent current injection (I_i^k) at the kth iteration is

$I_i^k=I_i^r\left(V_i^k\right)+j{I}_i^i\left(V_i^k\right)=\left(\frac{P_i+j{Q}_i}{V_i^k}\right)*\text{,}$

where V_i^k is the voltage of bus i at the kth iteration.

For fundamental load flow, the equivalent current injection needs to be iteratively transformed (e.g., Eq. 7-132 is used at the end of each iteration to compute the equivalent current injection (I_i^k) from the updated voltage (V_i^k), active (P_i), and reactive (Q_i) powers). However, this transformation is not required for harmonic currents since they have been obtained in a suitable form based on harmonic analysis (which represents nonlinear loads with harmonic current sources).

Figure 7.18 shows part of a lateral distribution system and the associated harmonic currents. The relationships between branch currents and harmonic currents are

$B_{jk}^{\left(h\right)}=-I_k^{\left(h\right)}\text{,}$

${\mathrm{B}}_{jl}^{\left(h\right)}=-I_l^{\left(h\right)}\text{,}$

${\mathrm{B}}_{ij}^{\left(h\right)}=B_{jk}^{\left(h\right)}+B_{j\ell }^{\left(h\right)}-I_j^{\left(h\right)}\text{,}$

where B and I are branch and injection (harmonic) currents, respectively.

It can be seen that the relationship between branch currents and harmonic (injection) currents of a radial feeder can be calculated by summing the injection currents from the receiving bus toward the sending bus of the feeder. The general form of the harmonic currents can be expressed as

$B_{ij}^{\left(h\right)}=-I_j^{\left(h\right)}+{\displaystyle \sum _{\ell =1}^n{B}_{j\ell }^{\left(h\right)}}\text{.}$

On the other hand, the relationship between branch currents and bus voltages can be shown to be

$\begin{array}{l}{V}_j^{\left(h\right)}=V_i^{\left(h\right)}-B_{ij}^{\left(h\right)}{Z}_{ij}^{\left(h\right)}\\ V_k^{\left(h\right)}=V_j^{\left(h\right)}-B_{jk}^{\left(h\right)}{Z}_{jk}^{\left(h\right)}\\ V_i^{\left(h\right)}=V_j^{\left(h\right)}-B_{jl}^{\left(h\right)}{Z}_{jl}^{\left(h\right)}\text{,}\end{array}$

where Z^(h) is the equivalent impedance of the line section at the hth harmonic frequency.

It is evident for a radial distribution system that if branch currents have been calculated, then the bus voltages can be easily determined from

$V_j^{\left(h\right)}=V_i^{\left(h\right)}-Z_{ij}^{\left(h\right)}{B}_{ij}^{\left(h\right)}\text{.}$

The bus voltages can be obtained by calculating the voltage from the sending bus forward to the receiving bus of the feeder. Power system components are adjusted for each harmonic order.

7.5.3 Modified Fast Decoupled Harmonic Power Flow

This method extends and generalizes the conventional fast decoupled (fundamental) power flow algorithm to accommodate harmonic sources [20]. The reduced and decoupled features of the fast decoupled algorithm is explored for the harmonic power flow application. A mathematical nonlinear model is integrated into an existing fast decoupled power flow program. This method allows separate modeling of nonlinear loads and subsequently integrating them in the main algorithm without substantially increasing computation time.

The hth harmonic bus injection current vector is defined as

$I^{\left(h\right)}=Y^{\left(h\right)}{V}^{\left(h\right)}\text{,}$

where Y^(h) is the hth harmonic admittance matrix. The injected current at any bus p is given by

$I_p^{\left(h\right)}={\displaystyle \sum _{i=1}^n{Y}_{p,i}^{\left(h\right)}{V}_i^{\left(h\right)}}\text{.}$

This equation provides the change of voltage magnitude and phase angle for each harmonic as a result of the injected currents.

The problem is then solved by the Newton–Raphson method. The current injected by a nonlinear load is assumed to be constant throughout the iteration and is used to modify the solution of voltage amplitudes and angles without reinverting the Jacobian matrix. At the end of each iteration, the device parameters are adjusted to reflect the new voltage magnitudes and phase angles. The harmonic currents and injected power are then reevaluated and the process is repeated until convergence is achieved.

7.5.4 Fuzzy Harmonic Power Flow

All elements of Eq. 7-139 are complex valued. Therefore, it can be reformulated as [13]

$\left[{\psi }^{\left(h\right)}\right]\left[\begin{array}{c}\hfill V_r^{\left(h\right)}\hfill \\ \hfill V_i^{\left(h\right)}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill I_r^{\left(h\right)}\hfill \\ \hfill I_i^{\left(h\right)}\hfill \end{array}\right]\text{,}$

where V_r^(h), V_i^(h), I_r^(h), and I_i^(h) are the real and imaginary parts of the hth harmonic voltage and current, respectively. ψ^(h) is a 2n × 2n matrix restructured from Y^(h) and n is the bus number.

Fuzzy set theory is employed to represent the injection harmonic currents. For an n bus distribution system the solution is obtained by solving a set of 4n × 4n linear equations. The fuzzy solution for both harmonic magnitudes and phase angles can be individually expressed by two explicit linear equations. The solutions for any harmonic level can be easily obtained from these equations.

Harmonic sources are represented as (voltage-independent) current sources. Because harmonic penetration is a steady-state phenomenon, monitoring harmonics requires a long period of time for accurate measurements. It can be shown that the value of the harmonic currents at a bus is most likely near the center of a monitored range. Therefore, it is suitable to model the harmonic current source with a triangle fuzzy number, denoted by Ĩ_h. With a proper fuzzy model, the new fuzzy harmonic power flow equations can be expressed in the following form [13]:

$\left[{\psi }^{\left(h\right)}\right]\left[\begin{array}{c}\hfill {\tilde{V}}_r^{\left(h\right)}\hfill \\ \hfill {\tilde{V}}_i^{\left(h\right)}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill {\tilde{I}}_r^{\left(h\right)}\hfill \\ \hfill {\tilde{I}}_i^{\left(h\right)}\hfill \end{array}\right]\text{,}$

Note that ψ^(h) is a constant whereas the harmonic voltages ${\tilde{V}}_i^{\left(h\right)}$ and ${\tilde{V}}_i^{\left(h\right)}$ are fuzzy numbers.

An arbitrary fuzzy number can be represented by an ordered pair of functions (u_min (r), u_max (r)), 0 ≤ r ≤ 1, which meet the following conditions:

• u_min(r) is a bounded left continuous nondecreasing function over [0,1].

• u_max(r) is a bounded right continuous nonincreasing function over [0,1].

• u_man(r) ≤ u_max(r), 0 ≤ r ≤ 1.

The fuzzy harmonic power flow can then be expressed as follows [13]:

$\left[\begin{array}{cc}\hfill {\psi }_{11}\hfill & \hfill {\psi }_{12}\hfill \\ \hfill {\psi }_{21}\hfill & \hfill {\psi }_{22}\hfill \end{array}\right]\left[\begin{array}{c}\hfill V_{r\min }^{\left(h\right)}\left(r\right)\hfill \\ \hfill V_{i\min }^{\left(h\right)}\left(r\right)\hfill \\ \hfill -V_{r\max }^{\left(h\right)}\left(r\right)\hfill \\ \hfill -V_{i\max }^{\left(h\right)}\left(r\right)\hfill \end{array}\right]=\left[\begin{array}{c}\hfill I_{r\min }^{\left(h\right)}\left(r\right)\hfill \\ \hfill I_{i\min }^{\left(h\right)}\left(r\right)\hfill \\ \hfill -I_{r\max }^{\left(h\right)}\left(r\right)\hfill \\ \hfill -I_{i\max }^{\left(h\right)}\left(r\right)\hfill \end{array}\right]\text{,}$