## 2 Materials and Methods for Coupled Thermal-Electrical Modeling of the Grounding Electrode
### Electrode-Soil Heat Exchange Model  
Consider a conductor in the form of a thin cylindrical wire with radius \(R_{1}\), length \(L\), electrical resistivity \(\rho_{c}\), and thermal conductivity \(k_{1}\), which is buried in a horizontal position at a depth p from the surface of a conductive soil with resistivity \(\rho_{s}\) and thermal conductivity \(k_{2}\). Figure 1 schematically illustrates the problem. The heat generated by the current \(I_{c}\) inside the conductor and the heat generated in the conductive medium by the current that leaks into the ground \(I_{s}\) from the exposed surface of the conductor is transmitted by conduction to the horizontal surface of the soil at temperature \(T_{s}\), and from there to the outside air at temperature \(T_{a}\) by free and/or forced convection with a convection coefficient \(h\). From a general perspective, we have a heat transfer problem with two distinct regions: the buried conductor and the semi-infinite external medium limited by the soil surface. In both media, there are heat sources, which are associated with the Joule effect due to the current passing through materials with non-zero resistivity. An initial formulation of the problem would consist of solving the heat equation with internal sources in both regions separately and applying the boundary conditions (BC) and initial conditions (IC), along with the equations for the interface between the two media, which is simply the surface of the buried conductor. A problem that arises is related to the choice of the coordinate system, as neither the Cartesian nor the cylindrical system fully adapts to this problem. The cylindrical system offers certain advantages, provided that the conductor is assumed to have infinite length. Indeed, it is easy to see that, in this case, a cylindrical system whose axis of symmetry aligns with the axis of the conductor leads to a one-dimensional heat conduction problem in the radial and angular variables. Furthermore, if the burial depth \(p\) is large compared to the radius of the conductor \(R_{1}\), it can be assumed with little error that there is no angular dependence in the temperature field, at least inside the conductor  
Figure 1: Schematic of a buried cylindrical conductor with electric and thermal interactions. and in areas of the soil close to it. A better approach to the problem could be achieved by modifying the system geometric configuration, taking advantage of the fact that, in our case, interest is focused on the transient solution over a very short time. In other words, the solution has barely evolved from the initial state. If we consider the soil as a cylindrical layer concentric with the conductor and with radius \(R_{2}=p\), the equations can be greatly simplified by considering the problem as axisymmetric and one-dimensional in the radial direction. This is equivalent to assuming that, at short times, both configurations are equivalent. Figure 2 aims to represent this simplification of the problem by showing a cross-sectional view of the conductor (shaded circle), surrounded by a cylindrical layer of soil with radius \(R_{2}\). The Joule power densities and \(\hat{\epsilon}_{2}\) generated in the conductor and the soil by the currents \(I_{c}\) and \(I_{s}\), respectively, are also shown. With these considerations, the equations to be solved are  
\[ \begin{array}{l}\frac{1}{\alpha_{1}}\left.\frac{\partial T_{1}(r,t)}{ \partial t}\right.=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial T_{ 1}}{\partial r})+\frac{\hat{\epsilon}_{1}}{k_{1}}\quad 0<r<R_{1}\\ \frac{1}{\alpha_{2}}\left.\frac{\partial T_{2}(r,t)}{\partial t}\right.=\frac {1}{r}\frac{\partial}{\partial r}(r\frac{\partial T_{2}}{\partial r})+\frac{ \hat{\epsilon}_{2}}{k_{2}}\quad R_{1}<r<R_{2}\\ \left.\frac{\partial T_{1}}{\partial r}\right|_{r=0}=0\\ T_{2}(r,t)\big{|}_{r=R_{2}}=T_{s}\\ T_{1}(r,0)=T_{i}\\ T_{1}(R_{1},t)=T_{2}(R_{1},t)\\ -k_{1}\left.\frac{\partial T_{1}}{\partial r}\right|_{r=R_{1}}=\left.-k_{2} \left.\frac{\partial T_{2}}{\partial r}\right|_{r=R_{1}}\end{array}