# Fast Prediction of Characteristics in Wound Rotor Synchronous Condenser Using Subdomain Modeling
## 2 Subdomain Modeling
### General Solutions  
The infinite permeability assumption in the core leads to tangential-direction boundary conditions.  
\[\left.\frac{\partial A_{z}^{p}}{\partial\theta}\right|_{\theta=\theta_{p}}=\left. \frac{\partial A_{z}^{p}}{\partial\theta}\right|_{\theta=\theta_{p}+\alpha} \tag{7}\]  
\[\left.\frac{\partial A_{z}^{q}}{\partial\theta}\right|_{\theta=\theta_{q}}= \left.\frac{\partial A_{z}^{q}}{\partial\theta}\right|_{\theta=\theta_{q}+\gamma} \tag{8}\]  
\[\left.\frac{\partial A_{z}^{i}}{\partial\theta}\right|_{\theta=\theta_{i}}= \left.\frac{\partial A_{z}^{i}}{\partial\theta}\right|_{\theta=\theta_{i}+\beta} \tag{9}\]  
\[\left.\frac{\partial A_{z}^{i}}{\partial\theta}\right|_{\theta=\theta_{j}}= \left.\frac{\partial A_{z}^{j}}{\partial\theta}\right|_{\theta=\theta_{j}+\delta} \tag{10}\]  
Applying these four boundary conditions to Equations (1)-(5), general solutions of PDEs are expressed as a sum of homogeneous and particular solutions.  
\[A_{z}^{p}=\mathbf{A_{0}^{p}}+\ln(r)\ \mathbf{B_{0}^{p}}-\frac{\mu_{0}}{4}r^{2}J_{0}^{p}+ \sum\limits_{h=1,2}^{\infty}\Big{(}r^{h\frac{\pi}{\alpha}}\mathbf{C_{h}^{p}}+r^{- h\frac{\pi}{\alpha}}\mathbf{D_{h}^{p}}\Big{)}\cos\Big{(}h\frac{\pi r}{\alpha}( \theta-\theta_{p})\Big{)} \tag{11}\]  
\[A_{z}^{q}=\mathbf{A_{0}^{q}}+\ln(r)\ \mathbf{B_{0}^{q}}+\sum\limits_{g=1,2}^{\infty} \Big{(}r^{g\frac{\pi}{\gamma}}\mathbf{C_{g}^{q}}+r^{-g\frac{\pi}{\gamma}}\mathbf{D_{g }^{q}}\Big{)}\cos\Big{(}g\frac{\pi}{\gamma}\big{(}\theta-\theta_{q}\big{)} \Big{)} \tag{12}\]  
\[A_{z}^{I}=\sum\limits_{n=1,2}^{\infty}\Big{(}r^{n}\mathbf{A_{n}^{I}}+r^{-n}\mathbf{B_{ n}^{I}}\Big{)}\sin(n\theta)+\Big{(}r^{n}\mathbf{C_{n}^{I}}+r^{-n}\mathbf{D_{n}^{I}} \Big{)}\cos(n\theta) \tag{13}\]  
\[A_{z}^{i}=\mathbf{A_{0}^{i}}+\ln(r)\ \mathbf{B_{0}^{i}}+\sum\limits_{k=1,2}^{\infty} \Big{(}r^{k\frac{\pi}{\beta}}\mathbf{C_{k}^{i}}+r^{-k\frac{\pi}{\gamma}}\mathbf{D_{k}^ {i}}\Big{)}\cos\Big{(}k\frac{\pi}{\beta}(\theta-\theta_{i})\Big{)} \tag{14}\]  
\[A_{z}^{j}=\mathbf{A_{0}^{j}}+\ln(r)\ \mathbf{B_{0}^{j}}-\frac{\mu_{0}}{4}r^{2}J_{0}^{j}+ \tag{15}\]  
In total, there are 20 coefficients in (11)-(15). Thus, to determine a unique solution requires us to derive 20 corresponding equations. The continuity of the vector potential radial components results in the following boundary conditions:  
\[r=R_{1}\rightarrow\frac{\partial A_{z}^{p}}{\partial r}=0 \tag{16}\]  
\[r=R_{2}\rightarrow\left\{\begin{array}{l}\frac{\partial A_{z}^{t}}{ \partial r}=\frac{\partial A_{z}^{q}}{\partial b}\\ A_{z}^{q}=A_{z}^{r}\end{array}\right. \tag{17}\]  
\[r=R_{3}\rightarrow\left\{\begin{array}{l}A_{z}^{i}=A_{z}^{i}\\ \frac{\partial A_{z}^{i}}{\partial r}=\sum\limits_{p=1,2}^{N_{r}}\frac{\partial A _{z}^{p}}{\partial r}\end{array}\right. \tag{18}\]  
\[r=R_{4}\rightarrow\left\{\begin{array}{l}\frac{\partial A_{z}^{i}}{\partial r }=\sum\limits_{i=1,2}^{N_{r}}\frac{\partial A_{z}^{i}}{\partial r}\\ A_{z}^{i}=A_{z}^{i}\end{array}\right. \tag{19}\]  
\[r=R_{5}\rightarrow\left\{\begin{array}{l}A_{z}^{i}=A_{z}^{i}\\ \frac{\partial A_{z}^{i}}{\partial r}=\frac{\partial A_{z}^{i}}{\partial r} \end{array}\right. \tag{20}\]\[r=R_{6}\rightarrow\frac{\partial A_{z}^{i}}{\partial r}=0 \tag{21}\]  
By employing integrals \(sni(n,\theta_{i},\beta)\), \(rni(n,\theta_{i},\beta)\), \(gkni(k,n,\theta_{i},\beta)\), \(fkni(k,n,\theta_{i},\beta)\), and \(Fmk(m,k,\beta,\delta)\) in [19] to shorten equations, 20 equations are given in Appendices A and B.