## 2 Working Principle and Modeling
### Diaphrag Mechanical Model
\begin{table}
\begin{tabular}{c c c c c} \hline \hline \multicolumn{5}{c}{**Grid Independence Test of the RD**} \\ \hline
**Number** & **Domain Unit** & **Edge Unit** & **Displacement** &
\begin{tabular}{c} **Deviation Rate Compared** \\ **to the Next Grid** \\ \end{tabular} \\ \hline
1 & 4050 & 432 & 3.9572 mm & 2.48\% \\
2 & 6136 & 527 & 4.0552 mm & 2.14\% \\
3 & 12,437 & 796 & 4.1419 mm & 1.09\% \\
4 & 40,262 & 1414 & 4.1869 mm & - \\ \hline \hline \multicolumn{5}{c}{Grid Independence Test of the CD} \\ \hline Number & Domain Unit & Edge Unit & Displacement &
\begin{tabular}{c} Deviation Rate Compared \\ to the Next Grid \\ \end{tabular} \\ \hline
5 & 2633 & 347 & 1.1599 mm & 2.73\% \\
6 & 3954 & 421 & 1.1916 mm & 2.24\% \\
7 & 11,737 & 695 & 1.2183 mm & 0.84\% \\
8 & 41,785 & 1287 & 1.2285 mm & - \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Grid independence tests of finite element simulations. Figure 9: Simulation results of the RD part model using the free triangle grid: (**a**) the displacement cloud diagram. (**b**) The stress cloud diagram. 4.78 MPa. When the displacement of the CD is 1.2 mm, the maximum stress is 5.42 MPa. It is less than the tensile strength. To investigate the elastic characteristics and volume deformation of the diaphragm under the force conditions, finite element simulations were conducted on the diaphragms. Under the two-dimensional axial symmetry, the expression of the diaphragm volume deformation V is  
\[V=\int z\mathrm{d}\sigma \tag{21}\]  
Figure 11: Simulation results of the RD part model with the quadrilateral grid: (**a**) The displacement cloud diagram. (**b**) The stress cloud diagram. Figure 12: Displacement and stress cloud diagrams. (**a**) Displacement cloud diagram of the RD; (**b**) stress cloud diagram of the RD; (**c**) displacement cloud diagram of the CD; and (**d**) stress cloud diagram of the CD. Figure 10: Quadrilateral grid of the RD part. where z is the displacement in the z-direction, and \(\sigma\) is the area of the projection in the z-direction. The relationships between the displacement, volume, and force of the diaphraggrams obtained through the simulations and the least squares polynomial fittings are shown in Figure 13.  
When the displacement range of the RD is from \(-\)1 mm to 4 mm, and the displacement range of the CD is from \(-\)0.12 mm to 1.2 mm, the volume deformation of the diaphragms increases with the increase in the diaphragm displacements, and the root mean square errors are 0.011 and 0.0064, respectively. Therefore, Equations (13) and (14) can be expressed as linear equations, and their expressions are  
\[V_{1}=k_{\rm{v1}}x_{1} \tag{22}\]  
\[V_{2}=k_{\rm{v2}}x_{2} \tag{23}\]  
where k\({}_{\rm{v1}}\) and k\({}_{\rm{v2}}\) are the elastic slopes of the RD and the CD, respectively. In the displacement range, k\({}_{\rm{v1}}\) and k\({}_{\rm{v2}}\) are obtained by least squares polynomial fittings with values of 0.1336 and 0.5722, respectively. Therefore, combining with Equations (12), (15), (22), and (23), k\({}_{\rm{p1}}\) is 4.2829. In the same displacement range, the elastic forces of the diaphragms increase with the increase in the diaphragm displacements, and the root mean square errors are 0.6809 and 0.228, respectively. The expressions of the least squares polynomial fittings are  
\[F_{\rm{1k}}=k_{\rm{1}}x_{\rm{1}}+b_{\rm{1}} \tag{24}\]  
Figure 9:0  
where k\({}_{\rm{1}}\), b\({}_{\rm{1}}\), k\({}_{\rm{2}}\), and b\({}_{\rm{2}}\) are 2.182, 0.052, 1.13, and 0.018, respectively. Because the fitted intercepts are close to 0 compared to the magnitudes of the diaphragm elastic forces, they can be ignored.