# Study on the Leakage and Inter-Stage Pressure Drop Characteristics of Two-Stage Finger Seal
## 3 Analysis Method of Two-Stage Finger Seal Based on FEM/CFD Coupling
### CFD Model of Two-Stage Finger Seal
#### 3.3.2 Calculation of Resistance Coefficients of Porous Media  
In this paper, the leakage flow through the finger stick bundles is converted into the seepage in porous media. Compared to the flow in a non-porous medium, the flow in a porous medium is subject to additional resistance forces, and the effect of the additional resistance forces can be reflected by adding an additional source term to the fluid momentum equation. According to the theory of non-Darcian porosity model, the additional source term can be described as follows [20]. \[S_{i}=-a_{i}\mu u_{i}-\frac{1}{2}b_{i}\rho|u|u_{i} \tag{1}\]  
where \(S\) is the additional source term; \(a\) and \(b\) are viscous resistance coefficient and inertial resistance coefficient; \(\mu\) is the dynamic viscosity; \(\rho\) is the fluid density; \(u\) is the superficial velocity; the subscript \(i\) represents the spatial coordinate direction. For each 1 SFS in 2 SFS, the calculation methods of resistance coefficients of porous media are the same, therefore, this paper only introduces the calculation method of resistance coefficients of porous media for a 1 SFS. As shown in Figure 6, for the porous medium equivalent to finger foot bundle, the flow resistance in the \(s_{1}\) direction (radial direction of 1 SFS) is much smaller than that in the other two directions (\(z_{1}\) and \(n_{1}\)) (axial and circumferential directions of 1 SFS); for the porous medium equivalent to finger stick stem bundle, the flow resistance in the \(s_{2}\) direction (direction from the foot upper circle to the finger base circle along the finger interstice) is also much smaller than that in the other two directions (\(z_{2}\) and \(n_{2}\)) (axial direction of 1 SFS and direction perpendicular to direction \(s_{2}\) in the axial end face of 1 SFS). Therefore, the two porous media have anisotropic structural characteristics and the resistance coefficients of porous media are different in different directions. In addition, the flow resistance of the fluid in the finger foot bundle is greater than that of the fluid in the finger stick stem bundle, so the resistance coefficients of  
Figure 5: Sample mesh view. the porous medium equivalent to finger foot bundle is different from that of the porous medium equivalent to finger stick stem bundle. In Ref. [19], the authors regarded the whole finger stick bundle as a porous medium, and the equations of resistance coefficients were derived by analyzing the flow resistance forces in the rough contact clearance between adjacent finger sticks and the finger interstices. In order to better model the side leakage flow, the finger stick bundle is equivalent to two porous media in this paper. Considering that the derivation process of the equations of resistance coefficients of the two porous media is similar to that shown in Ref. [19], the equations of resistance coefficients of each porous medium will be given directly instead of the derivation process in order to reduce the length of paper. For the porous medium equivalent to finger foot bundle, the resistance coefficients (\(a_{z1}\), \(a_{n1}\), \(a_{s1}\), \(b_{z1}\), \(b_{n1}\), and \(b_{s1}\)) are calculated by  
\[\left\{ \begin{array}{l}a_{z1}=\frac{12(n-1)(\pi\sigma_{-}ms)\beta_{z1}}{\phi \mu^{2}mL_{z1}}\\ b_{z1}=\frac{29(n-1)\beta_{z1}^{2}}{L_{z1}}\\ a_{n1}=\frac{24(\pi\sigma_{-}ms)\beta_{z1}}{\phi\mu^{2}L_{z1}}\\ b_{n1}=\frac{5.8m\beta_{z1}^{2}}{L_{z1}}\\ a_{s1}=\frac{32\delta_{z1}\delta_{z1}}{L_{z1}\delta_{z1}^{2}}\\ b_{s1}=0\end{array} \right. \tag{2}\]  
For the porous medium equivalent to finger stick stem bundle, the resistance coefficients (\(a_{z2}\), \(a_{n2}\), \(a_{s2}\)\(b_{z2}\), \(b_{n2}\), and \(b_{s2}\)) are calculated by  
\[\left\{ \begin{array}{l}a_{z2}=\frac{6(n-1)(f_{n-s})\beta_{z2}}{\phi\mu^{2}L_ {z2}^{2}}\\ b_{z2}=\frac{2.9(n-1)\beta_{z2}^{2}}{L_{z2}}\\ a_{n2}=\frac{12m(f_{n-s})\beta_{z2}}{\phi\mu^{2}L_{z2}}\\ b_{n2}=\frac{5.8m\beta_{z2}^{2}}{L_{z2}}\\ a_{s2}=\frac{32\delta_{z1}f_{s2}}{L_{z2}\delta_{z}^{2}}\\ b_{s2}=0\end{array} \right.