# A Novel Higher-Order Numerical Scheme for System of Nonlinear Load Flow Equations
## 3 Numerical Experiments
### Formation of Load Flow Equations
\right].\]  
For \(n=3\), the power flow Equations (28) and (29) for the unknowns, i.e., \(P_{2}\), \(P_{3}\), and \(Q_{2}\), can be written as  
\[P_{2}=V_{2}\sum_{k=1}^{3}Y_{2k}V_{k}\cos(\theta_{2k}+\delta_{k}-\delta_{2}), \tag{34}\]  
\[P_{3}=V_{3}\sum_{k=1}^{3}Y_{3k}V_{k}\cos(\theta_{3k}+\delta_{k}-\delta_{3}), \tag{35}\]  
\[Q_{2}=-V_{2}\sum_{k=1}^{3}Y_{2k}V_{k}\sin(\theta_{2k}+\delta_{k}-\delta_{2}). \tag{36}\]  
A \(3\times 3\) nonlinear system of equations is obtained by substituting the values in (34)-(36):  
\[P_{2}=23.478630V_{2}\cos(-2.03444+\delta_{2})+26.00005V_{2}^{2}+37.20818V_{2} \cos(-2.0344-\delta_{3}+\delta_{2})+4,\]  
Figure 2: Three-bus system. \[P_{3}=34.53209\cos(-1.89254+\delta_{3})+37.208184V_{2}\cos(2.034444788-\delta_{3}+ \delta_{2})+26.12160874,\]  
\[Q_{2}=23.478630V_{2}\sin(-2.03444+\delta_{2})+52.0000V_{2}^{2}+37.20818V_{2}\sin (-2.034444788-\delta_{3}+\delta_{2})+2.5.\]  
The required solution for the system of nonlinear equations for our problem is \(X=(-2.6964802629,-0.4988374911,0.9716777891)^{t}\) and the initial guess is taken as \(X^{(0)}=(0,0,1)^{t}\). Numerical results are shown in the following Table 5.  
As observed from Table 5, the newly proposed methods are relatively efficient in terms of error while we also observe mere difference in terms of computational time. **Example 3**. : _Consider a general four-bus power system as shown in the Figure 3; Bus 1 is considered as a slack bus with voltage magnitude \(|V_{1}|=1.04\) pu, \(\delta_{1}=0\). Bus 2 and Bus 3 are PQ buses, while Bus 4 is a slack bus. The magnitude of voltage is \(|V_{4}|=1.0\) pu, \(\delta_{4}=0\), \(P_{2}=0.5\) pu, \(Q_{2}=0.208\) pu and \(P_{3}=-1.0\) pu, \(Q_{3}=0.5\) pu._  
\(Y_{bus}\) matrix for this system is written as  
\[Y_{bus}=\left[ \begin{array}{cccc}Y_{11}&Y_{12}&Y_{13}&Y_{14}\\ Y_{21}&Y_{22}&Y_{23}&Y_{24}\\ Y_{31}&Y_{32}&Y_{33}&Y_{34}\\ Y_{41}&Y_{42}&Y_{43}&Y_{44}\end{array} \right],\]  
\[|Y_{bus}|=\left[ \begin{array}{cccc}9.486833&6.324555&3.162277&0\\ 6.324555&11.594807&2.107974&3.162277\\ 3.162277&2.107974&11.594807&6.324555\\ 0&3.162277&6.324555&9.486833\end{array} \right]\]  
and  
\[\left|Y_{\theta_{Bus}}\right|=\left[ \begin{array}{cccc}-71.565051&-68.42345 8&-68.423458&0\\ -68.423458&-71.568176&-68.440648&-68.423458\\ -68.423458&-68.440648&-71.568176&-68.423458\\ 0&-68.423458&-68.423458&-71.565051\end{array} \right].\] \begin{table}
\begin{tabular}{c c c c c} \hline _Cases_ & \(\mathbf{k}\) & \(\|\mathbf{X^{(k)}-X^{(k-1)}}\|_{\infty}\) & \(\|\mathbf{F(X^{(k)})}\|_{\infty}\) & \(\mathbf{t^{(k)}}\) \\ \hline \(HM_{1}\) & 1 & 4.7062458908 \((-2)\) & 2.9429422347 \((-6)\) & \\ \hline  & 2 & 5.3524284668 \((-8)\) & 1.8153562097 \((-36)\) & \\ \hline  & 3 & 3.35876206 \((-38)\) & 2.84164745 \((-47)\) & 1.987282 \\ \hline \(HM_{2}\) & 1 & 4.7062482460 \((-2)\) & 3.0040159229 \((-6)\) & \\ \hline  & 2 & 4.1730381388 \((-8)\) & 2.1707785791 \((-36)\) & \\ \hline  & 3 & 3.49657037 \((-38)\) & 2.84164745 \((-47)\) & 1.972989 \\ \hline \(HM_{3}\) & 1 & 4.7062549068 \((-2)\) & 8.0832420909 \((-6)\) & \\ \hline  & 2 & 9.0817922071 \((-8)\) & 3.6267951935 \((-34)\) & \\ \hline  & 3 & 6.02690441 \((-36)\) & 2.84164745 \((-47)\) & 1.963466 \\ \hline \(SF_{1}\) & 1 & 4.70624694 \((-2)\) & 1.46430173 \((-6)\) & \\ \hline  & 2 & 2.63918687 \((-8)\) & 2.26013119 \((-38)\) & \\ \hline  & 3 & 3.73850774 \((-40)\) & 2.84164745 \((-47)\) & 2.006422 \\ \hline \(SF_{2}\) & 1 & 4.70624886 \((-2)\) & 7.32584352 \((-7)\) & \\ \hline  & 2 & 2.71328333 \((-8)\) & 3.11954284 \((-38)\) & \\ \hline  & 3 & 7.52218608 \((-40)\) & 2.84164745 \((-47)\) & 2.003059 \\ \hline \end{tabular}
\end{table} 
Table 5: Comparison of seventh-order schemes for power flow problem. Thus, in this case, the \(Y_{Bus}\) admittance matrix is  
\[Y_{bus}=\left[ \begin{array}{cccc}3.0-j9.0&-2.0+j6.0&-1.0+j3.0&0\\ -2.0+j6.0&3.666-j11.0&-0.666+j2.0&-1.0+j3.0\\ -1.0+j3.0&-0.666+j2.0&3.666-j11.0&-2.0+j6.0\\ 0&-1.0+j3.0&-2.0+j6.0&3.0-j9.0\\ \end{array}