# The Stationary Thermal Field in a Multilayer Elliptic Cylinder
## 3 Calculation Example--Electric Conductor with Double Insulation
\begin{array}{ll}I_{2}(m=2)&=\frac{\delta_{1}c\sinh(2\eta_{1})}{4\lambda_{3}}\cdot h (m=2,n=0)-\\ &-\frac{\delta_{1}c\cosh(2\eta_{2}-\eta_{1})\sinh[2(\eta_{2}-\eta_{3})]}{4\lambda_ {3}}\cdot h(m=2,n=2)+\\ &+\frac{\delta_{1}c\lambda_{2}\sinh[2(\eta_{2}-\eta_{1})]\cosh[2(\eta_{2}- \eta_{3})]}{4\lambda_{1}\lambda_{3}}\cdot h(m=2,n=2)+\\ &+\frac{\pi\tau_{5}\Gamma_{2}\alpha}{8\lambda_{1}\lambda_{3}}\cosh[2(\eta_{2} -\eta_{1})]\cosh[2(\eta_{2}-\eta_{3})]-\\ &-\frac{\pi\tau_{5}\Gamma_{2}\alpha}{8\lambda_{1}\lambda_{3}^{2}}\sinh[2(\eta _{2}-\eta_{1})]\sinh[2(\eta_{2}-\eta_{3})],\end{array}  \tag{27}\]  
\[ \begin{array}{ll}I_{3}(n)=r(n)\cdot h(m=0,n),\end{array}  \tag{28}\]  
\[ \begin{array}{ll}I_{4}=&\frac{g_{1}c\sinh(2\eta_{1})}{4\lambda_{3}}\cdot h(m =n=0)-\frac{\delta_{1}c\cosh[2(\eta_{2}-\eta_{1})]\sinh[2(\eta_{2}-\eta_{3})] }{4\lambda_{1}}\cdot h(m=2,n=0)+\\ &+\frac{g_{1}c\lambda_{2}\sinh[2(\eta_{2}-\eta_{1})]}{4\lambda_{1}\lambda_{3}} \cosh[2(\eta_{2}-\eta_{3})]\cdot h(m=2,n=0)+\\ &-\frac{\pi(\eta_{2}-\eta_{1})g_{1}c^{2}\alpha}{2\lambda_{2}\lambda_{3}}\sinh (2\eta_{1})-\frac{\pi(\eta_{2}-\eta_{3})g_{1}c^{2}\alpha}{2\lambda_{3}^{2}} \sinh(2\eta_{1}),\end{array}  \tag{29}\]  
\[ \begin{array}{ll}h(m,n)=\int\limits_{0}^{2\pi}\frac{\cos(m\psi)\cos(n\psi)d \psi}{\sqrt{\cosh^{2}(\eta_{3})-\cos^{2}(\psi)}},\end{array}  \tag{30}\]  
\[ \begin{array}{ll}r(n)=\frac{n}{c}[p(n)\sinh(n\eta_{3})+q(n)\cosh(n\eta_{3}) ].\end{array}  \tag{31}\]  
where \(p(n)\) and \(q(n)\) are determined by the Formulas (19) and (20). After the numerical calculation of integrals (30), the system of Equations (22) and (23) should be solved with respect to the coefficients \(A_{n}\) and the constant \(A_{0}\). In Formulas (8), (15) and (16), the ambient temperature \(T_{a}\) was taken into account (according to formula (1)). In this way, the sought field distributions were obtained. The calculations were performed for the case of an aluminum core of twisted bundles with a total active cross-sectional area of \(S_{1}=270\) mm\({}^{2}\) (the aluminum fills 90% of the area of the \(\pi a_{1}b_{1}\) ellipse). Materials for the inner and outer insulation were, respectively, PVC and rubber. The following set of data was adopted:  
\[ \begin{array}{ll}b_{1}=0.00690988\text{ m},&a_{1}=2b_{1},&b_{2}=0.0114796 \text{ m},&a_{2}=0.0165837\text{ m},&b_{3}=0.0158993\text{ m},\\ a_{3}=0.0199005\text{ m},&\lambda_{1}=180\text{ W/(mK)},&\lambda_{2}=0.17\text{ W/(mK)},&\lambda_{3}=0.16\text{ W/(mK)},&\alpha=14.9\\ \text{W/(m${}^{2}$K)},&\rho(70\text{ }^{\circ}\text{C})=3.39592\times 10^{-8}\text{ }\Omega\text{m},&T_{max}=70\text{ }^{\circ}\text{C},&T_{a}=25\text{ }^{\circ}\text{C},&k_{s}=1.02,&k_{n}=1.02.\end{array}  \tag{32}\]  
Using the data set (32), additional parameters were calculated: abscissa of the ellipse focus \(c=\sqrt{a_{1}^{2}-b_{1}^{2}}\) and the equations for the perimeters of individual ellipses, \(\eta_{i}=\text{arcsinh}h(b_{i}/c)\) for \(i=1,2,3\)[29]. For the determination of the field distributions (8), (15) and (16), a computer program was developed in the Mathematica package [39]. The program computes numerically the integrals \(h\) in (24)-(30), solves iteratively the system of Equations (22) and (23) and sums the truncated series (8), (15) and (16). The above-mentioned series turned out to be strong converging. During summing of more than 12 terms, the computation results varied at the sixth decimal point at different points on the conductor. Therefore, only 12 terms were summed in series (8), (15) and (16). After completing the computation, the program visualized the results. The key parameter of the conductor is current-carrying capacity (or ampacity). This parameter is defined as the maximum amount of current the insulation can withstand before it heats up above its maximum operating temperature. The most sensitive point of insulation is the contact with the conductor (\(\eta=\eta_{1}\)) at the point closest to the center of the heat source (\(\psi=\pi/2\)).