Factor for apportioning the thermal capacitance of the insulation (dielectric) for calculations of partial transients for long durations.
If changes in load current and system voltage occur at the same time, then an additional transient temperature rise due to the dielectric loss has to be calculated. For cables at voltages up to and including 275 kV, it is sufficient to assume that half of the dielectric loss is produced at the conductor and the other half at the insulation screen or sheath. The cable thermal circuit is derived by the method given with the Van Wormer coefficient computed from equations for long- and short-duration transients, respectively. For paper-insulated cables operating at voltages higher than 275 kV, the dielectric loss is an important fraction of the total loss and the Van Wormer coefficient is calculated by the third equation. In practical calculations for all voltage levels for which dielectric losses are important, half of the dielectric loss is added to the conductor loss and half to the sheath loss; therefore, the loss coefficients ($1+\lambda_1$) and ($1+\lambda_1+\lambda_2$) are used to evaluate thermal resistances and capacitances are set equal to 2.
| $\frac{1}{2\ln\left(\frac{D_i}{d_{ct}}\right)}-\frac{1}{\left(\frac{D_i}{d_{ct}}\right)^2-1}$ | long-term transients |
| $\frac{1}{\ln\left(\frac{D_i}{d_{ct}}\right)}-\frac{1}{\frac{D_i}{d_{ct}}-1}$ | short-term transients |
| $\frac{\left(\frac{D_i}{d_{ct}}\right)^2 \ln\left(\frac{D_i}{d_{ct}}\right)-{\ln\left(\frac{D_i}{d_{ct}}\right)}^2-\frac{\left(\frac{D_i}{d_{ct}}\right)^2-1}{2}}{\left(\left(\frac{D_i}{d_{ct}}\right)^2-1\right) {\ln\left(\frac{D_i}{d_{ct}}\right)}^2}$ | dielectric loss for paper-insulated cables $U_o$ > 275 kV |
