# Application of Machine Learning for Fault Classification and Location in a Radial Distribution Grid
## 4 Case Studies, Results, and Discussion
### Unbalanced Load (Scenario II)
#### 4.2.1 Fault Detection and Classification (Scenario II)  
Like the balanced case, statistical features were extracted for 885 samples for each type of 11 faults and 526 non-fault samples. The detection and classification results are shown in the form of a confusion  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Components** & **Values** \\ \hline \multirow{3}{*}{Transformer I} & 12 MVA, 120 kV/25 kV, \(Y_{g}\)-D, \\  & Resistance 0.02 pu \\  & Leakage inductance 0.08 pu \\ \hline \multirow{3}{*}{Transformer II} & 12 MVA,25 kV/575 V, \(Y_{g}\)-D \\  & Resistance 0.02 pu \\  & Leakage inductance 0.08 pu \\ \hline \multirow{3}{*}{Unbalanced Load} & Phase A (0.5 MW, 0.48 Mvar) \\  & Phase B (2 MW, 0.65 Mvar) \\  & Phase C (3 MW, 1.2 Mvar) \\ \hline Frequency & 60 Hz \\ \hline Load variations & \(\pm\)5\% \\ \hline \multirow{3}{*}{Distribution line} & 30 km \\  & Positive sequence: 0.1153 Ohms/km, \(1.05\times 10^{-3}\) H/km \\  & Negative sequence: 0.413 Ohms/km, \(3.32\times 10^{-3}\) H/km \\ \hline Fault impedance & 0–15 \(\Omega\) \\ \hline \hline \end{tabular}
\end{table}
Table 4: Parameters of the distribution network for Scenario II.  
\begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline
**Fault** & **Techniques** & **RMSE** & **MAPE** & **R2** & **NSEC** & **WIA** & \
\begin{tabular}{c} **Training** \\ **Time (s)** \\ \end{tabular} &
\begin{tabular}{c} **Testing** \\ **Time (s)** \\ \end{tabular} \\ \hline \multirow{3}{*}{AG} & MLP & 0.0687 & 0.0077 & 1.000 & 0.9999 & 1.000 & 14.1718 & 0.2187 \\  & ELM & 0.0257 & 0.0008 & 1.000 & 1.000 & 1.000 & 0.8437 & 0.1406 \\  & MLP & 0.0673 & 0.0066 & 1.000 & 0.9999 & 1.000 & 9.6093 & 0.0156 \\  & ELM & 0.0411 & 0.0011 & 1.000 & 1.000 & 1.000 & 0.0937 & 0.03125 \\  & MLP & 0.0639 & 0.0060 & 1.000 & 0.9999 & 1.000 & 8.7031 & 0.0156 \\  & ELM & 0.0191 & 0.0006 & 1.000 & 1.000 & 1.000 & 0.0625 & 0.0625 \\  & MLP & 0.0720 & 0.0079 & 1.000 & 0.9999 & 1.000 & 13.1093 & \(-\)0.000 \\  & ELM & 0.0407 & 0.0011 & 1.000 & 1.0000 & 1.0562 & \(-\)0.000 \\  & MLP & 0.0781 & 0.0089 & 1.000 & 0.9999 & 1.000 & 8.5937 & 0.0313 \\  & ELM & 0.1369 & 0.0065 & 0.999 & 0.9999 & 0.999 & 0.1250 & \(-\)0.000 \\  & MLP & 0.0347 & 0.0026 & 1.000 & 1.0000 & 1.000 & 8.7813 & 0.03125 \\  & ELM & 0.0285 & 0.0019 & 1.000 & 1.0000 & 1.000 & 0.0781 & 0.01563 \\  & MLP & 0.0578 & 0.0054 & 1.000 & 1.0000 & 0.999 & 11.547 & 0.0312 \\  & ELM & 0.0870 & 0.0022 & 0.999 & 1.0000 & 1.000 & 0.0313 & \(-\)0.000 \\  & MLP & 0.0584 & 0.0046 & 1.000 & 0.9999 & 1.000 & 14.3281 & 0.1094 \\  & ELM & 0.0637 & 0.0021 & 1.000 & 0.9999 & 1.000 & 0.7187 & 0.0469 \\  & MLP & 0.0217 & 0.0029 & 1.000 & 1.0000 & 1.000 & 8.4688 & 0.01563 \\  & ELM & 0.0300 & 0.0013 & 1.000 & 1.0000 & 1.000 & 0.04687 & 0.04687 \\  & MLP & 0.0453 & 0.0046 & 1.000 & 1.0000 & 1.000 & 8.53125 & \(-\)0.000 \\  & ELM & 0.0208 & 0.0007 & 1.000 & 1.0000 & 1.000 & 0.12500 & \(-\)0.000 \\ \hline \hline \end{tabular}
\end{table}
Table 3: Performance measurements of MLP and ELM for locating faults in a distribution system with balanced load.  
matrix, as shown in Table 5. The overall performance in this case was 91.4%. It can be seen that the detection (no fault) and all types of faults except ABC and ABCG are classified 100%. The confusion between the ABC and ABCG is expected because the two faults are symmetrical and the fault currents are not affected by the fault impedance (i.e., the neutral current is zero).