In order to calculate the current necessary to remove the ice build-up, Joule heating, solar radiation, radiation of the conductor surface, convective heat transfer and melting of water have to be taken into account. Since Joule heating represents the dominant mechanism, convection and radiation are neglected. Specific thermal capacity (cFe, cAl, ci) is considered for each separate material. The necessary temperature that is needed to heat the conductor and the ice build-up from – 5 °C to 0 °C and to transform ice into liquid water is obtained by calculating the mass of the steel core, conductor, Al-stripes and ice.
Assuming that the ice build-up stopped at the time of a current increase, the time needed for the ice to melt is presented in a graph (picture on the right). The red line shows the result for the maximum current that is allowed by the HV equipment in the overhead power field (e.g. disconnector, break-switch, measuring transformers), i.e. 800 A. This line shows the shortest possible time for the removal of the ice from the conductor by increasing the current. In case of strong precipitation, when the quantity of freezing rain on the conductor is larger than the quantity of melted ice, the additional loading only increases and leads to the tearing of the conductor at a force of 86.4 kN and at an angle of incidence on the position of the OTLM device at 8.24°.