7.3.3 Fundamental Jacobian Entry Formulas

Figure 7.9 shows a typical i bus of an n-bus system. This i bus is connected to three other buses m, k, and l. The load connected to bus i demands a complex power S_load,i.

The total complex (apparent) power leaving bus i via the lines to buses k, l, m is

$S_{\mathrm{lines},i}={\tilde{V}}_i{\left[{\tilde{I}}_k+{\tilde{I}}_l+{\tilde{I}}_m\right]}^{*}\text{,}$

where* denotes the complex conjugate and ỹ_ij are the line admittances:

$S_{\mathrm{lines},i}={\tilde{V}}_i{\left[\left({\tilde{V}}_i-{\tilde{V}}_k\right){\tilde{y}}_{ki}+\left({\tilde{V}}_i-{\tilde{V}}_l\right){\tilde{y}}_{li}+\left({\tilde{V}}_i-{\tilde{V}}_m\right){\tilde{y}}_{mi})\right]}^{*}\text{,}$

or

$S_{\mathrm{lines},i}={\tilde{V}}_i{\left[{\tilde{V}}_i\left({\tilde{y}}_{ki}+{\tilde{y}}_{li}+{\tilde{y}}_{mi}\right)-{\tilde{y}}_{ki}{\tilde{V}}_k-{\tilde{y}}_{li}{\tilde{V}}_l-{\tilde{y}}_{mi}{\tilde{V}}_m\right]}^{*}\text{.}$

With the entries of the admittance matrix

$\begin{array}{c}{y}_{ii}={\left({\overline{Y}}_{\mathrm{bus}}\right)}_{ii}={\tilde{y}}_{ki}+{\tilde{y}}_{li}+{\tilde{y}}_{mi},\\ y_{ki}={\left({\overline{Y}}_{\mathrm{bus}}\right)}_{ki}=-{\tilde{y}}_{ki},\\ y_{li}={\left({\overline{Y}}_{\mathrm{bus}}\right)}_{li}=-{\tilde{y}}_{li},\\ y_{mi}={\left({\overline{Y}}_{\mathrm{bus}}\right)}_{mi}=-{\tilde{y}}_{mi}\text{,}\end{array}$

it follows

$\begin{array}{l}{S}_{\mathrm{lines},i}=[{\left({\overline{Y}}_{\mathrm{bus}}\right)}_{ii}{\tilde{V}}_i+{\left({\overline{Y}}_{\mathrm{bus}}\right)}_{ki}{\tilde{V}}_k+\\ {\left({\overline{Y}}_{\mathrm{bus}}\right)}_{li}{\tilde{V}}_l+{\left({\overline{Y}}_{\mathrm{bus}}\right)}_{mi}{\tilde{V}}_m]*{\tilde{V}}_i\text{.}\end{array}$

Using the notation

${\overline{V}}_{\mathrm{bus}}=\left[\begin{array}{c}\hfill {\tilde{V}}_i\hfill \\ \hfill {\tilde{V}}_k\hfill \\ \hfill {\tilde{V}}_l\hfill \\ \hfill {\tilde{V}}_m\hfill \end{array}\right]\text{,}$

one can write in a condensed manner

$S_{\mathrm{lines},i}={\left[\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}\right){\overline{V}}_{\mathrm{bus}}\right]}^{*}{\tilde{V}}_i\text{.}$

Therefore, for the PQ buses we have the mismatch powers

$\mathrm{\Delta }{W}_i=S_{\mathrm{load},i}+S_{\mathrm{lines},i}$

$=S_{\mathrm{load},i}+{\left[\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}\right){\overline{V}}_{\mathrm{bus}}\right]}^{*}{\tilde{V}}_i\text{,}$

where S_load,i is the load (demand) complex (apparent) power at bus i (S_load,i = P_i + jQ_i). Breaking the mismatch power ΔW_i into a real part ΔP_i and an imaginary part ΔQ_i, we have the real and reactive mismatch powers, respectively:

$\mathrm{\Delta }{P}_i=P_i+\mathrm{Real}\left\{{\left[\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}\right){\overline{V}}_{\mathrm{bus}}\right]}^{*}{\tilde{V}}_i\right\}$

and

$\mathrm{\Delta }{Q}_i=Q_i+\mathrm{Imag}\left\{{\left[\left(\mathrm{row}i\mathrm{of}{\overline{Y}}_{\mathrm{bus}}\right){\overline{V}}_{\mathrm{bus}}\right]}^{*}{\tilde{V}}_i\right\}$

with P_i and Q_i being the real and reactive load (demand) powers at bus i.

7.3.4 Newton–Raphson Power Flow Algorithm

Based on the equations given, the conventional fundamental power flow algorithm is the computation of the bus voltage vector for a given system configuration and linear loads. The Newton–Raphson approach (Eq. 7-64) will be used to force the mismatch power to zero (Eq. 7-57). The iterative solution procedure for the fundamental Newton–Raphson power flow algorithm is as follows (Fig. 7.10):

• Step 1: Construct the admittance matrix ${\overline{Y}}_{\mathrm{bus}}$ (Eq. 7-41).

• Step 2: Make an initial guess for $\overline{x}$ (e.g., 1.0 pu volts and zero radians for all bus voltages, Eq. 7-42).

• Step 3: Evaluate mismatch power ($\mathrm{\Delta }\overline{W}\left(\overline{x}\right)$, Eq. 7-56). If it is small enough, then stop.

• Step 4: Establish Jacobian $\overline{J}$ (Eqs. 7-30 and 7-82) and calculate $\mathrm{\Delta }{\overline{x}}^{\xi }$ (Eq. 7-62) using either matrix inversion (for problems with a few buses) or upper (lower) triangular factorization combined with backward/forward substitution.

• Step 5: Update the bus vector voltage (Eq. 7-63) and go to Step 3.

It is recommended to use upper (lower) triangular factorization combined with backward/forward substitutions rather than matrix inversion in computing the solution because an iterative method is used where the coefficient matrix is updated from iteration to iteration.

7.3.5 Application Example 7.8: Computation of Fundamental Admittance Matrix

A simple (few buses) power system (Fig. E7.8.1) is used to illustrate in detail the Newton–Raphson load flow analysis. All impedances are in pu and the base apparent power is S_base = 1 kVA. Compute the fundamental admittance matrix for this system.

Solution to Application Example 7.8

For this example y₁₄ = –ỹ₁₄, where ỹ₁₄ is the admittance between bus 1 and bus 4 and y₁₄ is an entry of the bus admittance matrix Y_bus. Note the minus sign.

$\begin{array}{cc}{y}_{14}& =-{\tilde{y}}_{14}=\frac{-1}{0.01+j0.02}=\frac{-1}{0.02236\angle 1.1071\mathrm{rad}}\\ & =\frac{1\angle \pm 3.1415\mathrm{rad}}{0.02236\angle 1.1071\mathrm{rad}}=44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\end{array}$

$\begin{array}{cc}{y}_{12}& =-{\tilde{y}}_{12}=\frac{-1}{0.01+j0.01}=\frac{-1}{0.01414\angle 0.7854}\\ & =\frac{1\angle \pm 3.1415\mathrm{rad}}{0.01414\angle 0.7854}=70.72\mathrm{p}\mathrm{u}\angle 2.356\mathrm{rad}\end{array}$

$\begin{array}{l}{y}_{11}=-\left\{{\tilde{y}}_{14}+{\tilde{y}}_{12}\right\}=-\{44.72\cos \left(2.034\right)+70.72\cos \left(2.356\right)+j[44.72\sin \left(2.034\right)\\ +70.72\sin \left(2.356\right)]\}=-\{-19.98-49.997+j[40.008+50.02]\}\\ =\left(69.98-j90.03\right)=114.03\mathrm{p}\mathrm{u}\angle -0.910\mathrm{rad}\end{array}$

$y_{13}=-{\tilde{y}}_{13}=0$ (because there is no direct line between buses 1 and 3)

$\begin{array}{c}{y}_{23}=-{\tilde{y}}_{23}=\frac{-1}{0.02+j0.08}=\frac{1\angle \pm 3.1415\mathrm{rad}}{0.082462\angle 1.3258\mathrm{rad}}\\ =12.127\mathrm{pu}\angle 1.816\mathrm{rad}\end{array}$

$y_{24}=-{\tilde{y}}_{42}=0$

$\begin{array}{c}{y}_{34}=-{\tilde{y}}_{34}=\frac{-1}{0.01+j0.02}=\frac{1\angle \pm 3.1415\mathrm{rad}}{0.02236\angle 1.1071\mathrm{rad}}\\ =44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\end{array}$

$\begin{array}{l}{y}_{22}=-\left({\tilde{y}}_{12}+{\tilde{y}}_{23}\right)=-\{70.72\cos \left(2.356\right)+\\ 12.127\cos \left(1.816\right)+j(70.72\sin \left(2.356\right)+\\ 12.127\sin \left(1.816\right)]\}=-\{-49.997-2.994+\\ j\left(50.102+11.764\right)\}=-\{-52.941+\\ j61.862\}=81.42\mathrm{p}\mathrm{u}\angle -0.863\mathrm{rad}\end{array}$

$\begin{array}{l}{y}_{33}=-\left({\tilde{y}}_{23}+{\tilde{y}}_{34}\right)=-\{12.127\cos \left(1.816\right)+\\ 44.72\cos \left(2.034\right)+j[12.127\sin \left(1.8161\right)+\\ 44.72\sin \left(2.034\right)]\}=-\{-2.944-19.982+j(11.764+\\ 4.008)\}=-\left\{-22.93+j51.772\right\}=56.62\mathrm{p}\mathrm{u}\angle –1.154\end{array}$

$\begin{array}{l}{y}_{44}=-\left({\tilde{y}}_{14}+{\tilde{y}}_{34}\right)=-\{44.72\cos \left(2.034\right)+\\ 44.72\cos \left(2.034\right)+j(44.72\sin \left(2.034\right)+\\ 44.72\sin \left(2.034\right)]\}=-\{-19.982-19.982+j(40.008+\\ 40.008)\}=\left\{-39.964+j80.016\right\}=89.44\mathrm{p}\mathrm{u}\angle -1.1076\mathrm{rad}\text{.}\end{array}$

The bus admittance matrix is now

${\overline{Y}}_{\mathrm{bus}}=\left[\begin{array}{cccc}\hfill 114.03\mathrm{p}\mathrm{u}\angle -0.910\mathrm{rad}\hfill & \hfill 70.72\mathrm{p}\mathrm{u}\angle 2.356\mathrm{rad}\hfill & \hfill 0\hfill & \hfill 44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\hfill \\ \hfill 70.72\mathrm{p}\mathrm{u}\angle 2.356\mathrm{rad}\hfill & \hfill 81.42\mathrm{p}\mathrm{u}\angle -0.863\mathrm{rad}\hfill & \hfill 12.127\mathrm{p}\mathrm{u}\angle 1.816\mathrm{rad}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 12.127\mathrm{p}\mathrm{u}\angle 1.816\mathrm{rad}\hfill & \hfill 56.62\mathrm{p}\mathrm{u}\angle -1.154\mathrm{rad}\hfill & \hfill 44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\hfill \\ \hfill 44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\hfill & \hfill 0\hfill & \hfill 44.72\mathrm{p}\mathrm{u}\angle 2.034\mathrm{rad}\hfill & \hfill 89.44\mathrm{p}\mathrm{u}\angle -1.1076\mathrm{rad}\hfill \end{array}\right]\text{.}$

Note that ${\overline{Y}}_{\mathit{bus}}$ is singular; that is, ${\left({\overline{Y}}_{\mathit{bus}}\right)}^{-1}$ cannot be found, because there is no admittance connected to the ground bus. This is unimportant for the Newton-Raphson approach because for forming of the Jacobian the entries of ${\overline{Y}}_{\mathit{bus}}$ are used and ${\overline{Y}}_{\mathit{bus}}$ is not inverted.

Comment about Singularity

• Considering

$\left[{\overline{I}}_{\mathrm{bus}}\right]={\left[{\overline{Y}}_{\mathrm{bus}}\right]}^{-1}\left[{\overline{V}}_{\mathrm{bus}}\right]\text{,}$

without any connection (admittance) from any bus to ground (Fig. 7.11), ${\overline{Y}}_{\mathrm{bus}}^{-1}$ is singular. However, if ${\overline{Y}}_{\mathrm{bus}}$ contains load information (admittance to ground) then it will not be singular.

• Considering

$\left[\mathrm{\Delta }\overline{x}\right]={\left[\overline{J}\right]}^{-1}\left[\mathrm{\Delta }\overline{W}\right]\text{,}$

if [$\overline{J}$] contains load information (P₃, Q₃, Fig. 7.12), then [$\overline{J}$] is not singular.

7.3.6 Application Example 7.9: Evaluation of Fundamental Mismatch Vector

For the system of Fig. E7.8.1, assume bus 1 is the swing bus (δ₁ = 0 and |${\tilde{V}}_1$| = 1.0 pu). Then we can make an initial guess for bus voltage vector ${\overline{x}}^0$ = (δ₂, |${\tilde{V}}_2$|, δ₃, |${\tilde{V}}_3$|, δ₄, |${\tilde{V}}_4$|)^t = (0, 1, 0, 1, 0, 1)^t. Note, the superscript 0 means initial guess. Compute the mismatch power vector for this initial condition.

Solution to Application Example 7.9

${\overline{X}}^o={\left({\delta }_2,\left|{\tilde{V}}_2\right|,{\delta }_3,\left|{\tilde{V}}_3\right|,{\delta }_4,\left|{\tilde{V}}_4\right|\right)}^t={\left(0,1,0,1,0,1\right)}^t$

$\begin{array}{l}\mathrm{\Delta }{\overline{W}}^0=(P_2+F_{r,2},Q_2+F_{i,2},P_3+F_{r,3},Q_3\\ +F_{i,3},P_4+F_{r,4},Q_4+F_{i,4}){}^t\end{array}$

$\begin{array}{ll}\mathrm{\Delta }{P}_2& =P_2+F_{r,2}=P_2+{\displaystyle \sum _{j=1}^4{y}_{2j}{V}_j{V}_2}cos\left(-{\theta }_{2j}-{\delta }_j+{\delta }_2\right)\\ & =P_2+y_{21}{V}_1{V}_{2}cos\left(-{\theta }_{21}-{\delta }_1+{\delta }_2\right)\\ & +y_{22}{V}_2{V}_{2}cos\left(-{\theta }_{22}-{\delta }_2+{\delta }_2\right)\\ & +y_{23}{V}_3{V}_{2}cos\left(-{\theta }_{23}-{\delta }_3+{\delta }_2\right)\\ & +y_{24}{V}_4{V}_{2}cos\left(-{\theta }_{24}-{\delta }_4+{\delta }_2\right)\end{array}$

$\begin{array}{c}\mathrm{\Delta }{P}_2=P_2+F_{r,2}=0.10+\{70.72\cdot 1\cdot 1\cdot cos(-2.356-\\ 0+0)+81.42\cdot 1\cdot 1\cdot cos\left(0.863-0+0\right)+12.127\cdot \\ 1\cdot 1\cdot cos\left(-1.816-0+0\right)+0\}=0.10-49.996+\\ 52.936-2.942=0.0978\end{array}$

$\begin{array}{cc}\mathrm{\Delta }{Q}_2& =Q_2+F_{i,2}=Q_2+\sum _{j=1}^4{y}_{2j}{V}_j{V}_{2}sin\left(-{\theta }_{2j}-{\delta }_j+{\delta }_2\right)\\ & =Q_2+y_{21}{V}_1{V}_{2}sin\left(-{\theta }_{21}-{\delta }_1+{\delta }_2\right)\\ & +y_{22}{V}_2{V}_{2}sin\left(-{\theta }_{22}-{\delta }_2+{\delta }_2\right)\\ & +y_{23}{V}_3{V}_{2}sin\left(-{\theta }_{23}-{\delta }_3+{\delta }_2\right)\\ & +y_{24}{V}_4{V}_{2}sin\left(-{\theta }_{24}-{\delta }_4+{\delta }_2\right)\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_2=Q_2+F_{i,2}=0.10+\{70.72\cdot 1\cdot \sin (-2.356-0\\ +0)+81.42\cdot 1\cdot \sin \left(0.863-0+0\right)+12.127\cdot \\ 1\cdot 1\cdot \sin \left(-1.816-0+0\right)+0\}=0.10-50.016+\\ 61.86-11.57=0.374\end{array}$

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

$\begin{array}{c}\mathrm{\Delta }{P}_3=P_3+F_{r,3}=0.0+0.0+\{12.127\cdot 1\cdot 1\cdot \cos (-1.816-0+0)\\ +56.62\cdot 1\cdot 1\cdot \cos (1.154-0+0)+44.72\cdot 1\cdot 1\cdot \cos (-2.034-0+0)\}\\ =0.0+0.0-2.944+22.92-19.98=-0.0056\end{array}$

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

$\begin{array}{c}\mathrm{\Delta }{Q}_3=Q_3+F_{i,3}=0.0+0.0+\{12.127\cdot 1\cdot 1\cdot \\ \sin \left(-1.816-0+0\right)+56.62\cdot 1\cdot 1\cdot \sin \left(1.154-0+0\right)\\ +44.72\cdot 1\cdot 1\cdot \sin \left(-2.034-0+0\right)\}=0.0+0.0-\\ 11.764+51.77-40.008=-0.00166\end{array}$

$\begin{array}{c}\mathrm{\Delta }{P}_4=P_4+F_{r,4}=P_4+y_{41}{V}_1{V}_4\cos \left(-{\theta }_{41}-{\delta }_1+{\delta }_4\right)+\\ y_{42}{V}_2{V}_4\cos \left(-{\theta }_{42}-{\delta }_2+{\delta }_4\right)+y_{43}{V}_3{V}_4\cos \left(-{\theta }_{43}-{\delta }_3+{\delta }_4\right)\\ +y_{44}{V}_4{V}_4\cos \left(-{\theta }_{44}-{\delta }_4+{\delta }_4\right)\end{array}$

$\begin{array}{c}\mathrm{\Delta }{P}_4=P_4+F_{r,4}=0.25+\{44.72\cdot 1\cdot 1\cdot \cos (-2.034-\\ 0+0)+0+44.72\cdot 1\cdot 1\cdot \cos \left(-2.034-0+0\right)+\\ 89.44\cdot 1\cdot 1\cdot \cos \left(1.1076-0+0\right)\}=0.25-\\ 19.98+0-19.98+39.96=0.2527\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_4=Q_4+F_{i,4}=Q_4+y_{41}{V}_1{V}_4\sin \left(-{\theta }_{41}-{\delta }_1+{\delta }_4\right)+\\ y_{42}{V}_2{V}_{4}sin\left(-{\theta }_{42}-{\delta }_2+{\delta }_4\right)+y_{43}{V}_3{V}_4\sin (-{\theta }_{43}-\\ {\delta }_3+{\delta }_4)+y_{44}{V}_4{V}_4\sin \left(-{\theta }_{44}-{\delta }_4+{\delta }_4\right)\end{array}$

$\begin{array}{c}\mathrm{\Delta }{Q}_4=Q_4+F_{i,4}=0.1+\{44.72\cdot 1\cdot 1\cdot \sin (-2.034-\\ 0+0)+0+44.72\cdot 1\cdot 1\cdot \sin \left(-2.034-0+0\right)+\\ 89.44\cdot 1\cdot 1\cdot \sin \left(1.1076-0+0\right)\}=0.1-\\ 40.0076+0-40.0076+80.016=0.1004\text{.}\end{array}$

The mismatch power vector is now

$\begin{array}{c}\mathrm{\Delta }{\overline{W}}^0=(0.0978,0.374,-0.0056,\\ -0.00166,0.2527,0.1004){}^t\text{.}\end{array}$

7.3.7 Application Example 7.10: Evaluation of Fundamental Jacobian Matrix

For the system of Fig. E7.8.1, compute the Jacobian matrix.

Solution to Application Example 7.10

$\begin{array}{ll}{J}_{11}& =\frac{\partial \mathrm{\Delta }{P}_2}{\partial {\delta }_2}=-\sum _{\begin{array}{l}j=1\\ \ne 2\end{array}}^4{y}_{2j}{V}_2{V}_j\sin \left({\delta }_2-{\delta }_j-{\theta }_{2j}\right)\\ & =-y_{21}{V}_2{V}_1\sin \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)\\ & -y_{23}{V}_2{V}_3\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\\ & -y_{24}{V}_2{V}_4\sin \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)\end{array}$

$\begin{array}{l}{J}_{11}=\frac{\partial \mathrm{\Delta }{P}_2}{\partial {\delta }_2}=-70.72\cdot 1\cdot 1\cdot \sin \left(0-0-2.356\right)-12.127\cdot 1\cdot 1\cdot \sin \left(0-0-1.816\right)-0\\ =61.68\end{array}$

$\begin{array}{ll}{J}_{12}& =\frac{\partial \mathrm{\Delta }{P}_2}{\partial V_2}=\sum _{\begin{array}{l}j=1\\ \ne 2\end{array}}^4{y}_{2j}{V}_j\cos \left({\delta }_2-{\delta }_j-{\theta }_{2j}\right)+2y_{22}{V}_2\cos \left(-{\theta }_{22}\right)\hfill \\ & =y_{21}{V}_1\cos \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+y_{23}{V}_3\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & +y_{24}{V}_4\cos \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)+2V_2{y}_{22}\cos \left(-{\theta }_{22}\right)\hfill \end{array}$

$\begin{array}{ll}{J}_{12}& =\frac{\partial \mathrm{\Delta }{P}_2}{\partial V_2}=70.72\cdot 1\cdot \cos \left(0-0-2.356\right)\hfill \\ & +12.127\cdot 1\cdot \cos \left(0-0-1.816\right)+0+2\cdot 1\cdot 81.42\cos \left(0.863\right)\hfill \\ & =-49.996-2.944+0+105.872=52.932\hfill \end{array}$

$\begin{array}{ll}{J}_{13}& =\frac{\partial \mathrm{\Delta }{P}_2}{\partial {\delta }_3}=y_{23}{V}_1{V}_3\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & =12.127\cdot 1\cdot 1\cdot \sin \left(0-0-1.816\right)=-11.764\hfill \end{array}$

$\begin{array}{ll}{J}_{14}& =\frac{\partial \mathrm{\Delta }{P}_2}{\partial V_3}=y_{23}{V}_2\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & =12.127\cdot 1\cdot \cos \left(0-0-1.816\right)=-2.944\end{array}$

$J_{15}=\frac{\partial \mathrm{\Delta }{P}_2}{\partial {\delta }_4}=y_{24}{V}_1{V}_5\sin \left({\delta }_2-{\delta }_5-{\theta }_{25}\right)=0$

$J_{16}=\frac{\partial \mathrm{\Delta }{P}_2}{\partial V_4}=y_{24}{V}_2\cos \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)=0$

$\begin{array}{ll}{J}_{21}& =\frac{\partial \mathrm{\Delta }{Q}_2}{\partial {\delta }_2}={\displaystyle \sum _{\begin{array}{l}j=1\\ \ne 2\end{array}}^4{y}_{2j}{V}_2{V}_j\cos \left({\delta }_2-{\delta }_j-{\theta }_{2j}\right)}\hfill \\ & =y_{21}{V}_2{V}_1\cos \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)\hfill \\ & +y_{23}{V}_2{V}_{3}cos\left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & +y_{24}{V}_2{V}_4\cos \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)\hfill \end{array}$

$\begin{array}{ll}{J}_{21}& =\frac{\partial \mathrm{\Delta }{Q}_2}{\partial {\delta }_2}=70.72\cdot 1\cdot 1\cdot \cos \left(0-0-2.3560\right)\hfill \\ & +12.127\cdot 1\cdot 1\cdot \cos \left(0-0-1.816\right)+0\hfill \\ & =-49.996-2.944+0=-52.94\hfill \end{array}$

$\begin{array}{ll}{J}_{22}& =\frac{\partial \mathrm{\Delta }{Q}_2}{\partial V_2}={\displaystyle \sum _{\begin{array}{l}j=1\\ \ne 2\end{array}}^4{y}_{2\mathrm{j}}{\mathit{V}}_{\mathit{j}}\sin \left({\delta }_2-{\delta }_j-{\theta }_{2j}\right)}+2V_2{y}_{22}\sin \left(-{\theta }_{22}\right)\hfill \\ & =y_{21}{V}_1\sin \left({\delta }_2-{\delta }_1-{\theta }_{21}\right)+y_{23}{V}_3\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & +y_{24}{V}_4\sin \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)+2V_2{y}_{22}\sin \left(-{\theta }_{22}\right)\hfill \end{array}$

$\begin{array}{ll}{J}_{22}& =\frac{\partial \mathrm{\Delta }{Q}_2}{\partial V_2}=70.72\cdot 1\cdot \sin \left(0-0-2.356\right)\hfill \\ & +12.127\cdot 1\cdot \sin \left(0-0-1.816\right)+0+2\cdot 1\cdot 81.42\sin \left(0.863\right)\hfill \\ & =-50.016-11.764+0+123.725=61.945\hfill \end{array}$

$\begin{array}{ll}{J}_{23}& =\frac{\partial \mathrm{\Delta }{Q}_2}{\partial {\delta }_3}=-y_{23}{V}_2{V}_3\cos \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\hfill \\ & =-12.127\cdot 1\cdot 1\cdot \cos \left(0-0-1.816\right)\hfill \\ & =2.944\hfill \end{array}$

$\begin{array}{l}{J}_{24}=\frac{\partial \mathrm{\Delta }{Q}_2}{\partial V_3}=y_{23}{V}_2\sin \left({\delta }_2-{\delta }_3-{\theta }_{23}\right)\\ =12.127\cdot 1\cdot \sin \left(0-0-1.816\right)=-11.764\hfill \end{array}$

$J_{25}=\frac{\partial \mathrm{\Delta }{Q}_2}{\partial {\delta }_4}=-y_{24}{V}_2{V}_4\cos \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)=0$

$J_{26}=\frac{\partial \mathrm{\Delta }{Q}_2}{\partial V_4}=y_{24}{V}_2\sin \left({\delta }_2-{\delta }_4-{\theta }_{24}\right)=0$

$\begin{array}{l}{J}_{31}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial {\delta }_2}=y_{32}{V}_3{V}_{2}sin\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\\ =12.127\cdot 1\cdot 1\cdot sin\left(0-0-1.816\right)=-11.764\end{array}$

$\begin{array}{l}{J}_{32}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial V_2}=y_{32}{V}_{3}cos\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\\ =12.127\cdot 1\cdot cos\left(0-0-1.816\right)=-2.944\end{array}$

$J_{33}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial {\delta }_3}=-{\displaystyle \sum _{\begin{array}{c}\hfill j=1\hfill \\ \hfill \ne 3\hfill \end{array}}^4{y}_{3j}{V}_3{V}_{j}sin\left({\delta }_3-{\delta }_j-{\theta }_{3j}\right)}$

$\begin{array}{ll}{J}_{33}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial {\delta }_3}=& -y_{31}{V}_3{V}_{1}sin\left({\delta }_3-{\delta }_1-{\theta }_{31}\right)\hfill \\ & -y_{32}{V}_3{V}_{2}sin\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\hfill \\ & -y_{34}{V}_3{V}_{4}sin\left({\delta }_3-{\delta }_4-{\theta }_{34}\right)\hfill \\ & =0-12.127\cdot 1\cdot 1\cdot sin\left(0-0-1.816\right)\hfill \\ & -44.72\cdot 1\cdot 1\cdot sin\left(0-0-2.034\right)=51.772\hfill \end{array}$

$\begin{array}{ll}{J}_{34}& =\frac{\partial \mathrm{\Delta }{P}_3}{\partial V_3}={\displaystyle \sum _{\begin{array}{c}\hfill j=1\hfill \\ \hfill \ne 3\hfill \end{array}}^4{y}_{3j}{V}_{j}cos\left({\delta }_3-{\delta }_j-{\theta }_{3j}\right)}+2V_3{y}_{33}cos\left(-{\theta }_{33}\right)\hfill \\ & =y_{31}{V}_{1}cos\left({\delta }_3-{\delta }_1-{\theta }_{31}\right)+y_{32}{V}_{2}cos\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\hfill \\ & +y_{34}{V}_{4}cos\left({\delta }_3-{\delta }_4-{\theta }_{34}\right)+2V_3{y}_{33}cos\left(-{\theta }_{33}\right)\hfill \\ & =0+12.127\cdot 1\cdot cos\left(0-0-1.186\right)+44.72\cdot 1\cdot \hfill \\ & cos\left(0-0-2.034\right)+2\cdot 1\cdot 56.62cos\left(1.154\right)\hfill \\ & =0-2.944-19.982+45.843=22.917\hfill \end{array}$

$\begin{array}{l}{J}_{35}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial {\mathrm{\Delta }}_4}=y_{34}{V}_3{V}_{4}sin\left({\delta }_3-{\delta }_4-{\theta }_{34}\right)\\ =44.72\cdot 1\cdot 1\cdot sin\left(0-0-2.034\right)=-40.0076\end{array}$

$\begin{array}{l}{J}_{36}=\frac{\partial \mathrm{\Delta }{P}_3}{\partial V_4}=y_{34}{V}_{3}cos\left({\delta }_3-{\delta }_4-{\theta }_{34}\right)\\ =44.72\cdot 1\cdot cos\left(0-0-2.034\right)=-19.98\end{array}$

$\begin{array}{c}{J}_{41}=\frac{\partial \mathrm{\Delta }{Q}_3}{\partial {\delta }_2}=-y_{32}{V}_3{V}_{2}cos\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\\ =-12.127\cdot 1\cdot 1\cdot cos\left(0-0-1.816\right)=2.944\end{array}$

$\begin{array}{c}{J}_{42}=\frac{\partial \mathrm{\Delta }{Q}_3}{\partial V_2}=y_{32}{V}_{3}sin\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\\ =12.127\cdot 1\cdot sin\left(0-0-1.816\right)=-11.764\end{array}$

$J_{43}=\frac{\partial \mathrm{\Delta }{Q}_3}{\partial {\delta }_3}={\displaystyle \sum _{\begin{array}{c}\hfill j=1\hfill \\ \hfill \ne 3\hfill \end{array}}^4{y}_{3j}{V}_3{V}_{j}cos\left({\delta }_3-{\delta }_j-{\theta }_{3j}\right)}$

$\begin{array}{c}{J}_{43}=\frac{\partial \mathrm{\Delta }{Q}_3}{\partial {\delta }_3}=y_{31}{V}_3{V}_{1}cos\left({\delta }_3-{\delta }_1-{\theta }_{31}\right)\\ +y_{32}{V}_3{V}_{2}cos\left({\delta }_3-{\delta }_2-{\theta }_{32}\right)\hfill \\ +y_{34}{V}_3{V}_{4}cos\left({\delta }_3-{\delta }_4-{\theta }_{34}\right)\hfill \end{array}$