${\overline{H}}^{\left(h\right)}=\left[\begin{array}{llllll}\hfill & \hfill & {\overline{0}}_{2\left(m-1\right)\times 2\left(n-m+1\right)}\hfill & \hfill & \hfill & \hfill \\ ---\hfill & ---\hfill & --------\hfill & --\hfill & ---\hfill & ---\hfill \\ \frac{\partial {\mathrm{G}}_{r,m}^{\left(h\right)}}{\partial {\alpha }_m}\hfill & \frac{\partial {\mathrm{G}}_{r,m}^{\left(h\right)}}{\partial {\beta }_m}\hfill & .\hfill & .\hfill & 0\hfill & 0\hfill \\ \frac{\partial {\mathrm{G}}_{i,m}^{\left(h\right)}}{\partial {\alpha }_m}\hfill & \frac{\partial {\mathrm{G}}_{i,m}^{\left(h\right)}}{\partial {\beta }_m}\hfill & .\hfill & .\hfill & 0\hfill & 0\hfill \\ .\hfill & .\hfill & .\hfill & .\hfill & .\hfill & .\hfill \\ 0\hfill & 0\hfill & .\hfill & .\hfill & \frac{\partial {\mathrm{G}}_{r,n}^{\left(h\right)}}{\partial {\alpha }_n}\hfill & \frac{\partial {\mathrm{G}}_{r,n}^{\left(h\right)}}{\partial {\beta }_n}\hfill \\ 0\hfill & 0\hfill & .\hfill & .\hfill & \frac{\partial {\mathrm{G}}_{i,n}^{\left(h\right)}}{\partial {\alpha }_n}\hfill & \frac{\partial {\mathrm{G}}_{i,n}^{\left(h\right)}}{\partial {\beta }_n}\hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}\right]$

where h = 1, 5, 7, …, L. For h = 1 the entries are zero.

7.4.5 Harmonic Jacobian Entry Formulas Related to Line Currents

Figure 7.15 shows a typical nonlinear bus i of an n bus system. This ith bus is connected to three other buses, m, k, and l. The nonlinear device connected to bus i demands a complex load current ${\tilde{g}}_{\mathrm{load},i}^{\left(h\right)}$ (referred to the nonlinear bus i) at the hth harmonic frequency.

The total hth harmonic complex current leaving bus i via the lines is

${\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}={\tilde{I}}_k^{\left(h\right)}+{\tilde{I}}_{\ell }^{\left(h\right)}+{\tilde{I}}_m^{\left(h\right)}\text{,}$

$\begin{array}{l}{\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}=\left({\tilde{V}}_i^{\left(h\right)}-{\tilde{V}}_k^{\left(h\right)}\right){\tilde{y}}_{ik}^{\left(h\right)}+\left({\tilde{V}}_i^{\left(h\right)}-{\tilde{V}}_{\ell }^{\left(h\right)}\right){\tilde{y}}_{i\ell }^{\left(h\right)}+\left({\tilde{V}}_i^{\left(h\right)}-{\tilde{V}}_m^{\left(h\right)}\right){\tilde{y}}_{im}^{\left(h\right)}\text{,}\end{array}$

where ỹ_ij^(h) is the admittance of the line connecting buses i and j evaluated at the hth harmonic frequency. Rearranging terms gives

$\begin{array}{l}{\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}=\left({\tilde{y}}_{ik}^{\left(h\right)}+{\tilde{y}}_{i\ell }^{\left(h\right)}+{\tilde{y}}_{im}^{\left(h\right)}\right){\tilde{V}}_i^{\left(h\right)}-{\tilde{y}}_{ik}^{\left(h\right)}{\tilde{V}}_k^{\left(h\right)}-{\tilde{y}}_{i\ell }^{\left(h\right)}{\tilde{V}}_{\ell }^{\left(h\right)}-{\tilde{y}}_{im}^{\left(h\right)}{{\tilde{V}}_m}^{\left(h\right)}\text{,}\end{array}$

or

$\begin{array}{c}{\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}={\left({\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{ii}{\tilde{V}}_i^{\left(h\right)}+{\left({\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{ik}{\tilde{V}}_k^{\left(h\right)}+{\left({\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{i\ell }{\tilde{V}}_{\ell }^{\left(h\right)}+{\left({\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{im}{\tilde{V}}_m^{\left(h\right)}\text{,}\end{array}$

where ${\left({\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{ij}$ is the (i,j)th entry of the hth harmonic admittance matrix. Because k, l, and m are the only buses connected to bus i (see Fig. 7.15),

${\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}={\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right)}_{ii}{\overline{V}}_{\mathrm{bus}}^{\left(h\right)}\text{,}$

where ${\overline{V}}_{\mathrm{bus}}^{\left(h\right)}$ is a complex vector of the hth harmonic bus voltages. Therefore, the mismatch current for the hth harmonic is based on the Kirchhoff current law at bus i:

$\mathrm{\Delta }{\tilde{I}}_i^{\left(h\right)}={\tilde{I}}_{\mathrm{lines},i}^{\left(h\right)}+{\tilde{G}}_{\mathrm{load},i}^{\left(h\right)}=\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right){\overline{V}}_{\mathrm{bus}}^{\left(h\right)}+{\tilde{G}}_{\mathrm{load},i}^{\left(h\right)}\text{,}$

where ${\tilde{G}}_{\mathrm{load},i}^{\left(h\right)}$ is the hth harmonic nonlinear complex load current (for bus i), referred to the swing bus:

${\tilde{G}}_{\mathit{load},i}^{\left(h\right)}=G_{r,i}^{\left(h\right)}+j{G}_{i,i}^{\left(h\right)}$

In this equation, the second index i pertains to bus i and the first index i indicates imaginary. Breaking the mismatch current ΔĨ_i^(h) into a real part ΔI_r,i^(h) and an imaginary part ΔI_i,i^(h) we get

$\mathrm{\Delta }{I}_{r,i}^{\left(h\right)}=\mathrm{Real}\left\{\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right){\overline{V}}_{\mathrm{bus}}^{\left(h\right)}\right\}+G_{r,i}^{\left(h\right)}$

$\mathrm{\Delta }{I}_{i,i}^{\left(h\right)}=\mathrm{Imag}\left\{\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}^{\left(h\right)}\right){\overline{V}}_{\mathrm{bus}}^{\left(h\right)}\right\}+G_{i,i}^{\left(h\right)}\text{,}$

Therefore,

$\mathrm{\Delta }{I}_{r,i}^{\left(h\right)}=I_{r,i}^{\left(h\right)}+G_{r,i}^{\left(h\right)}={\displaystyle \sum _{j=1}^n{y}_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|cos\left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right)+G_{r,i}^{\left(h\right)}}\text{,}$

$\mathrm{\Delta }{I}_{i,i}^{\left(h\right)}=I_{i,i}^{\left(h\right)}+G_{i,i}^{\left(h\right)}={\displaystyle \sum _{j=1}^n{y}_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|sin\left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right)+G_{i,i}^{\left(h\right)}}\text{,}$

where y_ij^(h) and θ_ij^(h) are the magnitude and phase angle of the (i,j)th entry of the hth harmonic admittance matrix, respectively, and |${\tilde{V}}_j^{\left(h\right)}$| and δ_j^(h) are the hth harmonic voltage magnitude and phase angle of bus j with respect to swing bus.

The entries of the Jacobian matrix corresponding to line currents are submatrices ${\overline{Y}}^{\left(h,h\right)}$ (h = 1, 5, …, L). These matrices are defined as

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

Note that ${\overline{Y}}^{\left(h,j\right)}$ = 0 for h ≠ j. The zero terms are due to the fact that the (v-i) characteristics of the nonlinear devices are modeled in the Jacobian matrix using submatrices ${\overline{G}}^{\left(h,j\right)}$.

${\overline{Y}}^{\left(h,h\right)}=\left[\begin{array}{ccccc}\hfill \frac{\partial I_{r,1}^{\left(h\right)}}{\partial {\delta }_1^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{r,1}^{\left(h\right)}}{\partial V_1^{\left(h\right)}}\hfill & \hfill \cdots \hfill & \hfill \frac{\partial I_{r,1}^{\left(h\right)}}{\partial {\delta }_n^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{r,1}^{\left(h\right)}}{\partial V_n^{\left(h\right)}}\hfill \\ \hfill \frac{\partial I_{i,1}^{\left(h\right)}}{\partial {\delta }_1^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{i,1}^{\left(h\right)}}{\partial V_1^{\left(h\right)}}\hfill & \hfill \cdots \hfill & \hfill \frac{\partial I_{i,1}^{\left(h\right)}}{\partial {\delta }_n^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{i,1}^{\left(h\right)}}{\partial V_n^{\left(h\right)}}\hfill \\ \hfill \cdot \hfill & \hfill \cdot \hfill & \hfill \cdots \hfill & \hfill \cdot \hfill & \hfill \cdot \hfill \\ \hfill \frac{\partial I_{r,n}^{\left(h\right)}}{\partial {\delta }_1^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{r,n}^{\left(h\right)}}{\partial V_1^{\left(h\right)}}\hfill & \hfill \cdots \hfill & \hfill \frac{\partial I_{r,n}^{\left(h\right)}}{\partial {\delta }_n^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{r,n}^{\left(h\right)}}{\partial V_n^{\left(h\right)}}\hfill \\ \hfill \frac{\partial I_{i,n}^{\left(h\right)}}{\partial {\delta }_1^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{i,n}^{\left(h\right)}}{\partial V_1^{\left(h\right)}}\hfill & \hfill \cdots \hfill & \hfill \frac{\partial I_{i,n}^{\left(h\right)}}{\partial {\delta }_n^{\left(h\right)}}\hfill & \hfill \frac{\partial I_{i,n}^{\left(h\right)}}{\partial V_n^{\left(h\right)}}\hfill \end{array}\right]\text{,}$

For h = 1 only the last 2(n – m + 1) rows exist because fundamental current balance is not performed for linear load buses. Also for h = 1 the first two columns of the matrix do not exist (are zero) because $\left|{\tilde{V}}_1^{\left(1\right)}\right|$ and δ₁⁽¹⁾ are known and constant.

The entries of ${\overline{Y}}^{\left(h,h\right)}$ are computed from Eq. 7-119 by setting G_r,i^(h) and G_i,i^(h) to zero:

$\frac{\partial I_{r,i}^{\left(h\right)}}{\partial {\delta }_j^{\left(h\right)}}=-y_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|\sin \left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right),\mathrm{for}h=1,5,\dots ,L\text{,}$

$\frac{\partial I_{r,i}^{\left(h\right)}}{\partial \left|{\tilde{V}}_j^{\left(h\right)}\right|}=y_{ij}^{\left(h\right)}\cos \left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right),\mathrm{for}h=1,5,\dots ,L\text{,}$

$\frac{\partial I_{i,i}^{\left(h\right)}}{\partial {\delta }_j^{\left(h\right)}}=y_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|cos\left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right),\mathrm{fo}\mathrm{r}h=1,5,\dots ,L\text{,}$

$\frac{\partial I_{i,i}^{\left(h\right)}}{\partial \left|{\tilde{V}}_j^{\left(h\right)}\right|}=y_{ij}^{\left(h\right)}\sin \left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right),\mathrm{for}h=1,5,\dots ,L\text{,}$

where h and j denote harmonic order and bus number, respectively.

7.4.6 Newton-Based Harmonic Power Flow Algorithm

Based on the equations given, the harmonic power flow algorithm is the computation of bus voltage vector Ū for a given system configuration with linear and nonlinear loads. The Newton–Raphson approach (Eq. 7-107) will be used to force the mismatch vector (Eq. 7-101) to zero using the harmonic Jacobian matrix and obtaining appropriate correction terms ΔŪ. Therefore (ξ represents the iteration number),

$\mathrm{\Delta }{\overline{U}}^{\xi }={\overline{J}}^{-1}\mathrm{\Delta }\overline{M}\left({\overline{U}}^{\xi }\right)\text{,}$

${\overline{U}}^{\left(\xi +1\right)}={\overline{U}}^{\xi }-\mathrm{\Delta }{\overline{U}}^{\xi }\text{.}$

The solution procedure for the harmonic power flow algorithm is as follows (Fig. 7.16):

Step 1: Perform the fundamental load flow analysis (treating all nonlinear devices as linear loads) and compute an initial (approximate) value for the fundamental bus voltage magnitudes and phase angles. Make an initial guess for the harmonic bus voltage magnitudes and phase angles (e.g., 0.1 pu and 0 radians).

Step 2: Compute nonlinear device currents G_r,m^(h) and G_i,m^(h) (referred to the swing bus) for nonlinear loads.

Step 3: Evaluate $\mathrm{\Delta }\overline{M}\left(\overline{U}\right)$ using Eqs. 7-101, 7-102. If it is small enough then stop.

Step 4: Evaluate ${\overline{J}}^{\left(h\right)}$ (Eqs. 7-108 to 7-112 and 7-121) and calculate ΔŪ (Eqs. 7-107 and 7-123) using matrix inversion or forward/backward substitutions.

Step 5: Update Ū (Eq. 7-124).

Step 6: Update the total (fundamental plus harmonic) powers at nonlinear buses (P_j^t and Q_j^t) (Eq. 7-99, Eq. 7-100), and the specified total apparent power or total reactive power (Eq. 7-98), whichever is known.

Step 7: Go to Step 2.

7.4.7 Application Example 7.14: Computation of Harmonic Admittance Matrix

The harmonic power flow algorithm will be applied to the nonlinear power system as shown in Fig. E7.14.1 where all impedances are given in pu at 60 Hz and the base is S_base = 1 kVA. In order to simplify the problem, the harmonic (v–i) characteristic of the nonlinear load at bus 4 (referred to the voltage at bus 4) is given as

$\begin{array}{l}{\tilde{g}}_{\mathit{load},4}^{\left(5\right)}=g_{r,4}^{\left(5\right)}+j{g}_{i,4}^{\left(5\right)},\\ g_{r,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^3\cos \left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^2\cos \left(3{\delta }_4^{\left(5\right)}\right),\\ g_{i,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^3\sin \left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^2\sin \left(3{\delta }_4^{\left(5\right)}\right)\text{.}\end{array}$

It will be assumed that the voltage at the swing bus (bus 1) is $\left|{\tilde{V}}_1^{\left(1\right)}\right|$ = 1.00 pu and ${\delta }_1^{\left(1\right)}=0$ rad.

The swing bus can be represented by Fig. E7.14.2 at the fundamental frequency. For the 5th harmonic, Ẽ_an of Fig. E7.14.3 represents a short-circuit, and the synchronous machine impedance Z₁ is replaced by the (subtransient) impedance Z₂ ≈ jX⁻ = j0.0001 pu at 60 Hz (see Fig. E7.14.1). In this case the subtransient reactance has been chosen to be very small in order to approximate bus 1 as an ideal bus.

Compute the fundamental and 5th harmonic admittance matrices.

Solution to Application Example 7.14

The fundamental bus admittance matrix (X⁻ does not exist for fundamental quantities) is computed in Application Example 7.8. The 5th harmonic bus admittance matrix takes into account X⁻ and its entries are calculated as follows:

$\begin{array}{c}{y}_{11}^{\left(5\right)}=\frac{1}{0.01+j\left(0.02\cdot 5\right)}+\frac{1}{0.01+j\left(0.01\cdot 5\right)}+\frac{1}{j\left(0.0001\cdot 5\right)}\\ =2029.137\mathrm{p}\mathrm{u}\angle -1.5684\mathrm{rad}\end{array}$

$y_{12}^{\left(5\right)}=\frac{-1}{0.01+j\left(0.01\cdot 5\right)}=19.612\mathrm{p}\mathrm{u}\angle 1.7681\mathrm{rad}$

$y_{13}^{\left(5\right)}=0$

$y_{14}^{\left(5\right)}=\frac{-1}{0.01+j\left(0.02\cdot 5\right)}=9.9503\mathrm{p}\mathrm{u}\angle 1.6704\mathrm{rad}$

$y_{21}^{\left(5\right)}=y_{12}^{\left(5\right)}$

$\begin{array}{c}{y}_{22}^{\left(5\right)}=\frac{1}{0.01+j\left(0.01\cdot 5\right)}+\frac{1}{0.02+j\left(0.08\cdot 5\right)}\\ =22.084\mathrm{p}\mathrm{u}\angle -1.3900\mathrm{rad}\end{array}$

$y_{23}^{\left(5\right)}=\frac{-1}{0.02+j\left(0.08\cdot 5\right)}=2.4969\mathrm{p}\mathrm{u}\angle 1.62066\mathrm{rad}$

$y_{24}^{\left(5\right)}=0$

$y_{31}^{\left(5\right)}=y_{13}^{\left(5\right)}$

$y_{32}^{\left(5\right)}=y_{23}^{\left(5\right)}$

$\begin{array}{l}{y}_{33}^{\left(5\right)}=\frac{1}{0.02+j\left(0.08\cdot 5\right)}+\frac{1}{0.01+j\left(0.02\cdot 5\right)}\\ =12.445\mathrm{p}\mathrm{u}\angle –1.4811\mathrm{rad}\end{array}$

$y_{34}^{\left(5\right)}=\frac{-1}{0.01+j\left(0.02\cdot 5\right)}=9.9504\mathrm{p}\mathrm{u}\angle 1.6704\mathrm{rad}$

$y_{41}^{\left(5\right)}=y_{14}^{\left(5\right)}$

$y_{42}^{\left(5\right)}=y_{24}^{\left(5\right)}$

$y_{43}^{\left(5\right)}=y_{34}^{\left(5\right)}$

$\begin{array}{c}{y}_{44}^{\left(5\right)}=\frac{1}{0.01+j\left(0.02\cdot 5\right)}+\frac{1}{0.01+j\left(0.02\cdot 5\right)}\\ =19.901\mathrm{p}\mathrm{u}\angle -1.4711\mathrm{rad}\text{.}\end{array}$

Therefore,

${\overline{Y}}_{\mathrm{bus}}^{\left(5\right)}=\left[\begin{array}{cccc}\hfill 2029.137\mathrm{pu}\angle -1.5684\mathrm{rad}\hfill & \hfill 19.61\mathrm{pu}\angle 1.7681\mathrm{rad}\hfill & \hfill 0\hfill & \hfill 9.9503\mathrm{pu}\angle 1.6704\mathrm{rad}\hfill \\ \hfill 19.61\mathrm{pu}\angle 1.7681\mathrm{rad}\hfill & \hfill 22.0847\mathrm{pu}\angle -1.3900\mathrm{rad}\hfill & \hfill 2.4969\mathrm{pu}\angle 1.62066\mathrm{rad}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2.4969\mathrm{pu}\angle 1.62066\mathrm{rad}\hfill & \hfill 12.445\mathrm{pu}\angle -1.4811\mathrm{rad}\hfill & \hfill 9.9504\mathrm{pu}\angle 1.6704\mathrm{rad}\hfill \\ \hfill 9.9503\mathrm{pu}\angle 1.6704\mathrm{rad}\hfill & \hfill 0\hfill & \hfill 9.9504\mathrm{pu}\angle 1.6704\mathrm{rad}\hfill & \hfill 19.901\mathrm{pu}\angle -1.4711\mathrm{rad}\hfill \end{array}\right]\text{.}$

Note that: $-y_{ij}\angle {\theta }_{ij}=y_{ij}\angle \left({\theta }_{ij}+\pi \right)\text{.}$

7.4.8 Application Example 7.15: Computation of Nonlinear Load Harmonic Currents

For the system of Fig. E7.14.1, assume an initial value for the bus vector Ū⁰ and compute the fundamental and the 5th harmonic currents injected into the system by the nonlinear load. The computation of the nonlinear load harmonic currents is the initial step of the harmonic load flow algorithm.

Solution to Application Example 7.15

The initial step of harmonic load flow is as follows:

The real and imaginary nonlinear device currents $G_{r,4}^{\left(1\right)}$ and $G_{i,4}^{\left(1\right)}$ – referred to the swing bus – may be computed as follows, where the following powers are referred to the nonlinear bus 4:

$\left\{\begin{array}{c}\hfill P_4^{\left(1\right)}=V_4^{\left(1\right)}{I}_4^{\left(1\right)}cos\left({\delta }_4^{(1\_}-{\gamma }_4^{\left(1\right)}\right)\hfill \\ \hfill Q_4^{\left(1\right)}=V_4^{\left(1\right)}{I}_4^{\left(1\right)}sin\left({\delta }_4^{(1\_}-{\gamma }_4^{\left(1\right)}\right)\hfill \end{array}\right.$

I₄⁽¹⁾ and γ₄⁽¹⁾ = δ₄⁽¹⁾ – tan⁻¹(Q₄⁽¹⁾/P₄⁽¹⁾) are the magnitude and phase angle of the fundamental nonlinear device current, respectively. The diagram of Fig. E7.15.1 demonstrates the phasor relationships.

The real part of the fundamental of the nonlinear device current at bus 4 (referred to the swing bus) is

$G_{r,4}^{\left(1\right)}=I_4^{\left(1\right)}cos{\gamma }_4^{\left(1\right)}$

where

$I_4^{\left(1\right)}=\frac{P_4^{\left(1\right)}}{V_4^{\left(1\right)}cos\left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}\text{.}$

Thus

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

Correspondingly one obtains for the imaginary part of the fundamental of the nonlinear device current at bus 4 (referred to the swing bus):

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

The real and imaginary harmonic (5th) nonlinear device currents are given as (referred to bus 4)

$\left\{\begin{array}{c}\hfill g_{r,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^{3}cos\left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^{2}cos\left(3{\delta }_4^{\left(5\right)}\right)\hfill \\ \hfill g_{i,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^{3}sin\left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^{2}sin\left(3{\delta }_4^{\left(5\right)}\right)\hfill \end{array}\right.$

where δ₄⁽⁵⁾ is the angle between ${\tilde{V}}_4^{\left(5\right)}$ and ${\tilde{V}}_1^{\left(1\right)}$ the swing bus voltage, I₄⁽⁵⁾ is the fifth harmonic current magnitude

$I_4^{\left(5\right)}=\sqrt{{\left(g_{r,4}^{\left(5\right)}\right)}^2+{\left(g_{i,4}^{\left(5\right)}\right)}^2}$

and γ₄⁽⁵⁾ is the phase angle of Ĩ₄⁽⁵⁾ with respect to the swing bus voltage Ṽ₁⁽¹⁾.

Assuming bus 1 to be the swing bus $\left({\delta }_1^{\left(1\right)}=0\mathrm{radians},\left|{\mathit{V}}_1^{\left(1\right)}\right|=1.00\mathrm{p}\mathrm{u}\right)$, the solution procedure is as follows.

Step #1 of Harmonic Load Flow. Using the solution vector $\overline{x}$ of the fundamental power flow analysis (see Application Example 7.13) and assuming harmonic voltage magnitudes and phase angles of 0.1 pu and 0 radians, respectively, the bus vector Ū⁰ will be

$\begin{array}{c}{\overline{U}}^0=({\delta }_2^{\left(1\right)},\left|{\tilde{V}}_2^{\left(1\right)}\right|,{\delta }_3^{\left(1\right)},\left|{\tilde{V}}_3^{\left(1\right)}\right|,{\delta }_4^{\left(1\right)},\left|{\tilde{V}}_4^{\left(1\right)}\right|,{\delta }_1^{\left(5\right)},\left|{\tilde{V}}_1^{\left(5\right)}\right|,{\delta }_2^{\left(5\right)},\left|{\tilde{V}}_2^{\left(5\right)}\right|,\\ {\delta }_3^{\left(5\right)},\left|{\tilde{V}}_3^{\left(5\right)}\right|,{\delta }_4^{\left(5\right)},\left|{\tilde{V}}_4^{\left(5\right)}\right|,{\alpha }_4,{\beta }_4){}^t\end{array}$

$\begin{array}{c}{\overline{U}}^0=(-2.671\cdot {10}^{-4},0.9976,-2.810\cdot {10}^{-3},0.9964,\\ -3.414\cdot {10}^{-3},0.9960,0,0.1,0,0.1,0,0.1,0,0.1,0,0){}^t\end{array}$

where the angles are measured in radians.

Note that the nonlinear device variables (α₄ and β₄) are assumed to be zero because there are no device variables defined in this example for the nonlinear load at bus 4.

Step #2 of Harmonic Load Flow. Use Ū⁰ to compute the nonlinear device currents. With P₄⁽¹⁾ = 250 W corresponding to 0.25 pu, Q₄⁽¹⁾ = 100VAr corresponding to 0.100 pu, V₄⁽¹⁾ = 0.9960 pu, and δ₄⁽¹⁾ = –3.414 · 10⁻³ radians, one gets

${\varepsilon }_4^{\left(1\right)}={tan}^{-1}\left(\frac{Q_4^{\left(1\right)}}{P_4^{\left(1\right)}}\right)={tan}^{-1}\left(\frac{0.1}{0.25}\right)=0.3805\mathrm{rad}$

${\gamma }_4^{\left(1\right)}={\delta }_4^{\left(1\right)}-{\varepsilon }_4^{\left(1\right)}=\left(-0.003414-0.3805\right)=-0.38391\mathrm{rad}$

$\begin{array}{c}{G}_{r,4}^{\left(1\right)}=\frac{P_4^{\left(1\right)}cos\left({\gamma }_4^{\left(1\right)}\right)}{V_4^{\left(1\right)}cos\left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=\frac{0.25\cdot cos\left(-0.38391\right)}{0.9960\cdot cos\left(-0.003414+0.38391\right)}=0.2514\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}{G}_{i,4}^{\left(1\right)}=\frac{P_4^{\left(1\right)}sin\left({\gamma }_4^{\left(1\right)}\right)}{V_4^{\left(1\right)}cos\left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)}=\frac{0.25\cdot sin\left(-0.38391\right)}{0.9960\cdot cos\left(-0.003414+0.38391\right)}=-0.1015\mathrm{p}\mathrm{u}\text{.}\end{array}$

The fundamental magnitude of the nonlinear current is

$\begin{array}{c}{I}_4^{\left(1\right)}=\sqrt{{\left(G_{r,4}^{\left(1\right)}\right)}^2+{\left(G_{i,4}^{\left(1\right)}\right)}^2}=\sqrt{{\left(0.2514\right)}^2+{\left(-0.1015\right)}^2}\\ =0.2711\mathrm{p}\mathrm{u}\mathrm{with}{\gamma }_4^{\left(1\right)}=-0.38391\mathrm{rad}\text{.}\end{array}$

The fifth harmonic current components of the nonlinear load current at bus 4 are (referred to bus 4)

$\left\{\begin{array}{l}{g}_{r,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^{3}cos\left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^{2}cos\left(3{\delta }_4^{\left(5\right)}\right)\hfill \\ g_{i,4}^{\left(5\right)}=0.3{\left(V_4^{\left(1\right)}\right)}^{3}sin\left(3{\delta }_4^{\left(1\right)}\right)+0.3{\left(V_4^{\left(5\right)}\right)}^{2}sin\left(3{\delta }_4^{\left(5\right)}\right)\hfill \end{array}\right.\text{.}$

With V₄⁽¹⁾ = 0.9960 pu, δ₄⁽¹⁾ = –0.003414 rad, V₄⁽⁵⁾ = 0.1 pu, δ₄⁽⁵⁾ = 0 rad, one obtains

$\left\{\begin{array}{l}{g}_{r,4}^{\left(5\right)}=0.3{\left(0.9960\right)}^{3}cos\left(-3\cdot 0.003414\right)\\ +0.3{\left(0.1\right)}^{2}cos\left(3\cdot 0\right)=0.2994\mathrm{p}\mathrm{u}\\ g_{i,4}^{\left(5\right)}=0.3{\left(0.9960\right)}^{3}sin\left(-3\cdot 0.003414\right)\\ +0.3{\left(0.1\right)}^{2}sin\left(3\cdot 0\right)=-0.003036\mathrm{p}\mathrm{u}\end{array}\right.$

$\begin{array}{c}{I}_4^{\left(5\right)}=\sqrt{{\left(g_{r,4}^{\left(5\right)}\right)}^2+{\left(g_{i,4}^{\left(5\right)}\right)}^2}=\sqrt{{\left(0.2994\right)}^2+{\left(-0.003036\right)}^2}\\ =0.2994\mathrm{p}\mathrm{u}\end{array}$

${\varepsilon }_4^{\left(5\right)}={tan}^{-1}\left(\frac{\left|g_{i,4}^{\left(5\right)}\right|}{\left|g_{r,4}^{\left(5\right)}\right|}\right)=0.01014\mathrm{rad}\text{.}$

Therefore, the fifth harmonic currents at bus 4 referred to the swing bus are $G_{r,4}^{\left(5\right)}=I_4^{\left(5\right)}cos{\gamma }_4^{\left(5\right)}$ and $G_{i,4}^{\left(5\right)}=I_4^{\left(5\right)}sin{\gamma }_4^{\left(5\right)}$.

The phasor diagram for the 5th harmonic is shown in Fig. E7.15.2, where the angles are

${\delta }_4^{\left(5\right)}={\varepsilon }_4^{\left(5\right)}+{\gamma }_4^{\left(5\right)}$

or

${\gamma }_4^{\left(5\right)}={\delta }_4^{\left(5\right)}-{\varepsilon }_4^{\left(5\right)}=0-{\varepsilon }_4^{\left(5\right)}=-0.01014\mathrm{rad}$

resulting in $G_{r,4}^{\left(5\right)}$ = 0.2994 pu and $G_{i,4}^{\left(5\right)}$ = –0.003036 pu.

These nonlinear currents referred to the swing bus are about the same as those referred to bus 4 because the angle γ₄⁽⁵⁾ is very small for this first iteration.

Note that

$\left\{\begin{array}{c}{P}_4^{\left(1\right)}=I_4^{\left(1\right)}{V}_4^{\left(1\right)}\cos \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)\\ =0.2711\cdot 0.9960\cdot \cos \left(-0.003414+0.38391\right)\\ =0.250\mathrm{p}\mathrm{u}\\ Q_{\mathit{4}}^{\left(\mathit{1}\right)}=I_4^{\left(1\right)}{V}_4^{\left(1\right)}\sin \left({\delta }_4^{\left(1\right)}-{\gamma }_4^{\left(1\right)}\right)\\ =0.2711\cdot 0.9960\cdot \sin \left(-0.003414+0.38391\right)\\ =0.100\mathrm{p}\mathrm{u}\end{array}\right.$

are as expected. Correspondingly,

$\left\{\begin{array}{c}{P}_4^{\left(5\right)}=I_4^{\left(5\right)}{V}_4^{\left(5\right)}\cos \left({\delta }_4^{\left(5\right)}-{\gamma }_4^{\left(5\right)}\right)\\ =0.2994\cdot 0.1\cdot \cos \left(0+0.01014\right)=0.029994\mathrm{p}\mathrm{u}\\ Q_{\mathit{4}}^{\left(\mathit{5}\right)}=I_4^{\left(5\right)}{V}_4^{\left(5\right)}\sin \left({\delta }_4^{\left(5\right)}-{\gamma }_4^{\left(5\right)}\right)\\ =0.2994\cdot 0.1\cdot \sin \left(0+0.01014\right)=0.000304\mathrm{p}\mathrm{u}\text{.}\end{array}\right.$

Therefore, the total injected real and reactive powers at the nonlinear bus 4 are

$\left\{\begin{array}{c}{P}_4^t=P_4^{\left(1\right)}+P_4^{\left(5\right)}=0.250\mathrm{p}\mathrm{u}+0.02994\mathrm{p}\mathrm{u}\\ =0.27994\mathrm{p}\mathrm{u}\\ Q_{\mathit{4}}^t=Q_4^{\left(1\right)}+Q_4^{\left(5\right)}=0.100\mathrm{p}\mathrm{u}+0.000304\mathrm{p}\mathrm{u}\\ =0.100304\mathrm{p}\mathrm{u}\text{.}\end{array}\right.$

7.4.9 Application Example 7.16: Evaluation of Harmonic Mismatch Vector

Using the results of Application Example 7.15, evaluate the mismatch vector for the system of Fig. E7.14.1.

Solution to Application Example 7.16

Step #3 of Harmonic Load Flow. Evaluation of the mismatch vector

$\begin{array}{c}\mathrm{\Delta }{\overline{M}}^0={\left(\mathrm{\Delta }\overline{W},\mathrm{\Delta }{\overline{I}}^{\left(5\right)},\mathrm{\Delta }{\overline{I}}^{\left(1\right)}\right)}^t=(P_2^{\left(1\right)}+F_{r,2}^{\left(1\right)},Q_2^{\left(1\right)}+F_{i,2}^{\left(1\right)},\\ P_3^{\left(1\right)}+F_{r,3}^{\left(1\right)},Q_3^{\left(1\right)}+F_{i,3}^{\left(1\right)},P_4^t+F_{r,4}^{\left(1\right)}+F_{r,4}^{\left(5\right)},Q_4^t+F_{i,4}^{\left(1\right)}+F_{i,4}^{\left(5\right)},\\ I_{r,1}^{\left(5\right)},I_{i,1}^{\left(5\right)},I_{r,2}^{\left(5\right)},I_{i,2}^{\left(5\right)},I_{r,3}^{\left(5\right)},I_{i,3}^{\left(5\right)},I_{r,4}^{\left(5\right)}+G_{r,4}^{\left(5\right)},I_{i,4}^{\left(5\right)}+G_{i,4}^{\left(5\right)},\\ I_{r,4}^{\left(1\right)}+G_{r,4}^{\left(1\right)},I_{i,4}^{\left(1\right)}+G_{i,4}^{\left(1\right)})\text{.}\end{array}$

To compute these mismatches use equations Eqs. 7-80 and 7-81

$\mathrm{\Delta }{P}_i=P_i+F_{r,i}=P_i+{\displaystyle \sum _{j=1}^n{y}_{ij}\left|{\tilde{V}}_j\right|}\left|{\tilde{V}}_i\right|cos\left(-{\theta }_{ij}-{\delta }_j+{\delta }_i\right)$

$\mathrm{\Delta }{Q}_i=Q_i+F_{i,i}=Q_i+{\displaystyle \sum _{j=1}^n{y}_{ij}\left|{\tilde{V}}_j\right|}\left|{\tilde{V}}_i\right|sin\left(-{\theta }_{ij}-{\delta }_j+{\delta }_i\right)$

and Eqs. 7-119a,b

$\mathrm{\Delta }{I}_{r,i}^{\left(h\right)}=I_{r,i}^{\left(h\right)}+G_{r,i}^{\left(h\right)}={\displaystyle \sum _{j=1}^n{y}_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|}cos\left({\theta }_{ij}^{\left(h\right)}+{\delta }_{{}_j}^{\left(h\right)}\right)+G_{r,i}^{\left(h\right)}$

$\mathrm{\Delta }{I}_{i,i}^{\left(h\right)}=I_{i,i}^{\left(h\right)}+G_{i,i}^{\left(h\right)}={\displaystyle \sum _{j=1}^n{y}_{ij}^{\left(h\right)}\left|{\tilde{V}}_j^{\left(h\right)}\right|}sin\left({\theta }_{ij}^{\left(h\right)}+{\delta }_j^{\left(h\right)}\right)+G_{i,i}^{\left(h\right)}\text{.}$

Hence,

$\begin{array}{c}\mathrm{\Delta }{P}_2^{\left(1\right)}=P_2^{\left(1\right)}+F_{r.2}^{\left(1\right)}=P_2+y_{21}{V}_1{V}_{2}cos\left(-{\theta }_{21}-{\delta }_1+{\delta }_2\right)+\\ y_{22}{V}_2{V}_{2}cos\left(-{\theta }_{22}-{\delta }_2+{\delta }_2\right)+y_{23}{V}_3{V}_{2}cos\left(-{\theta }_{23}-{\delta }_3+{\delta }_2\right)\\ +y_{24}{V}_4{V}_{2}cos\left(-{\theta }_{24}-{\delta }_4+{\delta }_2\right)=-0.0170\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_2^{\left(1\right)}=Q_2^{\left(1\right)}+F_{i,2}^{\left(1\right)}=Q_2+y_{21}{V}_1{V}_{2}sin\left(-{\theta }_{21}-{\delta }_1+{\delta }_2\right)+\\ y_{22}{V}_2{V}_{2}sin\left(-{\theta }_{22}-{\delta }_2+{\delta }_2\right)+y_{23}{V}_3{V}_{2}sin\left(-{\theta }_{23}-{\delta }_3+{\delta }_2\right)\\ +y_{24}{V}_4{V}_{2}sin\left(-{\theta }_{24}-{\delta }_4+{\delta }_2\right)=0.083\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}\mathrm{\Delta }{P}_3^{\left(1\right)}=P_3^{\left(1\right)}+F_{r,3}^{\left(1\right)}=P_3+y_{31}{V}_1{V}_{3}cos\left(-{\theta }_{31}-{\delta }_1+{\delta }_3\right)+\\ y_{32}{V}_2{V}_{3}cos\left(-{\theta }_{32}-{\delta }_2+{\delta }_3\right)+y_{33}{V}_3{V}_{3}cos(-{\theta }_{33}-{\delta }_3\\ +{\delta }_3)+y_{34}{V}_4{V}_{3}cos\left(-{\theta }_{34}-{\delta }_4+{\delta }_3\right)=-0.00325\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_3^{\left(1\right)}=Q_3^{\left(1\right)}+F_{i,3}^{\left(1\right)}=Q_3+y_{31}{V}_1{V}_{3}sin\left(-{\theta }_{31}-{\delta }_1+{\delta }_3\right)+\\ y_{32}{V}_2{V}_{3}sin\left(-{\theta }_{32}-{\delta }_2+{\delta }_3\right)+y_{33}{V}_3{V}_{3}sin\left(-{\theta }_{33}-{\delta }_3+{\delta }_3\right)\\ +y_{34}{V}_4{V}_{3}sin\left(-{\theta }_{34}-{\delta }_4+{\delta }_3\right)=0.00192\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}\mathrm{\Delta }{P}_4^t=P_4^t+F_{r,4}^{\left(1\right)}+F_{r,4}^{\left(5\right)}\\ =P_4^t+{\displaystyle \sum _{j=1}^4{y}_{4j}{V}_j^{\left(1\right)}{V}_4^{\left(1\right)}}cos\left(-{\theta }_{4j}-{\delta }_j^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +{\displaystyle \sum _{j=1}^4{y}_{{}_{4j}}^{\left(5\right)}{V}_j^{\left(5\right)}{V}_4^{\left(5\right)}}cos\left(-{\theta }_{4j}^{\left(5\right)}-{\delta }_j^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)\\ =P_4^t+y_{41}{V}_1^{\left(1\right)}{V}_4^{\left(1\right)}cos\left(-{\theta }_{41}-{\delta }_1^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{43}{V}_3^{\left(1\right)}{V}_4^{\left(1\right)}cos\left(-{\theta }_{43}-{\delta }_3^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{44}{V}_4^{\left(1\right)}{V}_4^{\left(1\right)}cos\left(-{\theta }_{44}-{\delta }_4^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{41}^{\left(5\right)}{V}_1^{\left(5\right)}{V}_4^{\left(5\right)}cos\left(-{\theta }_{41}^{\left(5\right)}-{\delta }_1^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)\\ +y_{43}^{\left(5\right)}{V}_3^{\left(5\right)}{V}_4^{\left(5\right)}cos\left(-{\theta }_{43}^{\left(5\right)}-{\delta }_3^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)\\ +y_{44}^{\left(5\right)}{V}_4^{\left(5\right)}{V}_4^{\left(5\right)}cos\left(-{\theta }_{44}^{\left(5\right)}-{\delta }_4^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)=0.037\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_4^t=Q_4^t+F_{i,4}^{\left(1\right)}+F_{i,4}^{\left(5\right)}\\ =Q_4^t+{\displaystyle \sum _{j=1}^4{y}_{4j}{V}_j^{\left(1\right)}{V}_4^{\left(1\right)}}sin\left(-{\theta }_{4j}-{\delta }_j^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +{\displaystyle \sum _{j=1}^4{y}_{4j}^{\left(5\right)}}{V}_j^{\left(5\right)}{V}_4^{\left(5\right)}sin\left(-{\theta }_{4j}^{\left(5\right)}-{\delta }_j^{\left(5\right)}+{\delta }_4^{\left(1\right)}\right)\\ =Q_4^t+y_{41}{V}_1^{\left(1\right)}{V}_4^{\left(1\right)}sin\left(-{\theta }_{41}-{\delta }_1^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{43}{V}_3^{\left(1\right)}{V}_4^{\left(1\right)}sin\left(-{\theta }_{43}-{\delta }_3^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{44}{V}_4^{\left(1\right)}{V}_4^{\left(1\right)}sin\left(-{\theta }_{44}-{\delta }_4^{\left(1\right)}+{\delta }_4^{\left(1\right)}\right)\\ +y_{41}^{\left(5\right)}{V}_1^{\left(5\right)}{V}_4^{\left(5\right)}sin\left(-{\theta }_{41}^{\left(5\right)}-{\delta }_1^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)\\ +y_{43}^{\left(5\right)}{V}_3^{\left(5\right)}{V}_4^{\left(5\right)}sin\left(-{\theta }_{43}^{\left(5\right)}-{\delta }_3^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)\\ +y_{44}^{\left(5\right)}{V}_4^{\left(5\right)}{V}_4^{\left(5\right)}sin(-{\theta }_{44}^{\left(5\right)}-{\delta }_4^{\left(5\right)}+{\delta }_4^{\left(5\right)})=-0.188\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}{I}_{r,1}^{\left(5\right)}={\displaystyle \sum _{j=1}^4{y}_{1j}^{\left(5\right)}\left|{\tilde{V}}_j^{\left(5\right)}\right|cos\left({\theta }_{1j}^{\left(5\right)}+{\delta }_j^{\left(5\right)}\right)}\\ =y_{11}^{\left(5\right)}\left|{\tilde{V}}_1^{\left(5\right)}\right|cos\left({\theta }_{11}^{\left(5\right)}+{\delta }_1^{\left(5\right)}\right)\\ +y_{12}^{\left(5\right)}\left|{\tilde{V}}_2^{\left(5\right)}\right|cos\left({\theta }_{12}^{\left(5\right)}+{\delta }_2^{\left(5\right)}\right)\\ +y_{13}^{\left(5\right)}\left|{\tilde{V}}_3^{\left(5\right)}\right|cos\left({\theta }_{13}^{\left(5\right)}+{\delta }_3^{\left(5\right)}\right)\\ +y_{14}^{\left(5\right)}\left|{\tilde{V}}_4^{\left(5\right)}\right|cos\left({\theta }_{14}^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)=0.201\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}{I}_{i,1}^{\left(5\right)}={\displaystyle \sum _{j=1}^4{y}_{1j}^{\left(5\right)}\left|{\tilde{V}}_j^{\left(5\right)}\right|sin\left({\theta }_{1j}^{\left(5\right)}+{\delta }_j^{\left(5\right)}\right)}\\ =y_{11}^{\left(5\right)}\left|{\tilde{V}}_1^{\left(5\right)}\right|sin\left({\theta }_{11}^{\left(5\right)}+{\delta }_1^{\left(5\right)}\right)\\ +y_{12}^{\left(5\right)}\left|{\tilde{V}}_2^{\left(5\right)}\right|sin\left({\theta }_{12}^{\left(5\right)}+{\delta }_2^{\left(5\right)}\right)\\ +y_{13}^{\left(5\right)}\left|{\tilde{V}}_3^{\left(5\right)}\right|sin\left({\theta }_{13}^{\left(5\right)}+{\delta }_3^{\left(5\right)}\right)\\ +y_{14}^{\left(5\right)}\left|{\tilde{V}}_4^{\left(5\right)}\right|sin\left({\theta }_{14}^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)=-201.98\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}{I}_{r,2}^{\left(5\right)}={\displaystyle \sum _{j=1}^4{y}_{2j}^{\left(5\right)}\left|{\tilde{V}}_j^{\left(5\right)}\right|cos\left({\theta }_{2j}^{\left(5\right)}+{\delta }_j^{\left(5\right)}\right)}\\ =y_{21}^{\left(5\right)}\left|{\tilde{V}}_1^{\left(5\right)}\right|cos\left({\theta }_{21}^{\left(5\right)}+{\delta }_1^{\left(5\right)}\right)\\ +y_{22}^{\left(5\right)}\left|{\tilde{V}}_2^{\left(5\right)}\right|cos\left({\theta }_{22}^{\left(5\right)}+{\delta }_2^{\left(5\right)}\right)\\ +y_{23}^{\left(5\right)}\left|{\tilde{V}}_3^{\left(5\right)}\right|cos\left({\theta }_{23}^{\left(5\right)}+{\delta }_3^{\left(5\right)}\right)\\ +y_{24}^{\left(5\right)}\left|{\tilde{V}}_4^{\left(5\right)}\right|cos\left({\theta }_{24}^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)=0.000255\mathrm{p}\mathrm{u}\end{array}$

$\begin{array}{c}{I}_{i,2}^{\left(5\right)}={\displaystyle \sum _{j=1}^4{y}_{2j}^{\left(5\right)}\left|{\tilde{V}}_j^{\left(5\right)}\right|sin({\theta }_{2j}^{\left(5\right)}+{\delta }_j^{\left(5\right)}})\\ =y_{21}^{\left(5\right)}\left|{\tilde{V}}_1^{\left(5\right)}\right|sin\left({\theta }_{21}^{\left(5\right)}+{\delta }_1^{\left(5\right)}\right)\\ +y_{22}^{\left(5\right)}\left|{\tilde{V}}_2^{\left(5\right)}\right|sin\left({\theta }_{22}^{\left(5\right)}+{\delta }_2^{\left(5\right)}\right)+y_{23}^{\left(5\right)}\left|{\tilde{V}}_3^{\left(5\right)}\right|sin\left({\theta }_{23}^{\left(5\right)}+{\delta }_3^{\left(5\right)}\right)\\ +y_{24}^{\left(5\right)}\left|{\tilde{V}}_4^{\left(5\right)}\right|sin\left({\theta }_{24}^{\left(5\right)}+{\delta }_4^{\left(5\right)}\right)=-0.000620\mathrm{p}\mathrm{u}\end{array}$