Equivalent depth for deep burial
The calculation of temperatures of deep tunnels as well as deeply buried cables/ducts/pipes (e.g. from directional drilling) involves special considerations regarding their thermal inertia.
- The depth can easily reach hundreds of meters (e.g. a tunnel between two valleys in a mountainous area).
- The initial temperature of rocky soils at the core of mountains can reach unexpected temperature values (more than 50°C).
- The heating source can be the combination of cable losses, thermal effects from other circulating fluids (in multipurpose structures), but also water-cooling and forced ventilation with cold air renewal.
The calculation of the equivalent depth of a deep tunnel is based on the IEEE transactions on power delivery paper 'Ampacity Calculation for Deeply Installed Cables' by E. Dorison, G.J. Anders, and F. Lesur, dated 2010. The installation of cables in shared tunnels (e.g. railway tunnels) can lead to large diameters, because some tunnel-boring machines are capable of excavating a diameter greater than eight meters. Therefore, some of the assumptions regarding the depth and the diameter may no longer be valid and a more accurate, although somewhat more complex, equation is used, which is described in section
IV. Case of Deep Tunnels.
Cableizer does also use the exact equation for deeply buried cables/ducts/pipes and never the approximative solution described in section III. Equivalent Depth of Deeeply Buried Cables:$$\frac{1}{2}e^{\frac{1}{2}\left[\ln{\left(4\tau_L\delta_{soil}\right)}-0.5772+\operatorname{expi}{\left(-\frac{L_{cm}^2}{\tau_L\delta_{soil}}\right)}\right]}$$
With unlimited computation power, using the above approximation is not really necessary, especially as this solution does not work for low values of transient load periods $\tau_L$. With a large tunnel diameter and relatively shallow burial, the approximate solution does also result in a thermal resistance to ambient $T_{4iii}$ that is not on the safe side as compared to the exact solution.
Formulae
$\frac{D_o}{2} cosh\left(\frac{-\operatorname{expi}\left(\frac{-{D_o}^2}{16\tau_L \delta_{soil}}\right)+\operatorname{expi}\left(\frac{-{L_{cm}}^2}{\tau_L \delta_{soil}}\right)}{2}\right)$