# Dynamics Analysis of a Double-Plunger-Type Turbine Overspeed Protection Mechanism
## 4 Analysis of the Trip Process  
### Trip Speed of Plungers  
When the rotation speed keeps rising to the preset tripping speed, there may be two kinds of movements of the inner and outer plungers, as follows. 1. The inner plunger moves first in the outer plunger cavity, and the outer plunger does not move at this time. Figure 6: Two different force conditions of the outer plunger. 2. The outer plunger drives the inner plunger to move both plungers together. Because the bottom of the inner plunger is limited by the outer plunger base's ability to move downward, there is no situation in which the outer plunger moves but the inner plunger remains in the same place. To confirm the actual motion for the given parameters, the spindle speed can be calculated by calculating the centrifugal force required to overcome the spring force at the initial position of the inner and outer plungers for the action to occur. The spindle speed is used as the unknown to solve the following two sets of equations. \[\left\{ \begin{array}{l}f_{2}+S_{2}=m\dot{\theta}_{2}{}^{2}\rho_{2}\\ F_{I2}=\dot{\theta}_{L2}\\ f_{2}=\mu F_{I2}\end{array} \right.\left\{ \begin{array}{l}f_{1}+S_{1}+S_{2}=m \dot{\theta}_{1}{}^{2}\rho_{1}\\ F_{I1}=m\ddot{\theta}L_{1}\\ f_{1}=\mu F_{I1}\end{array} \right. \tag{27}\]  
The spring forces \(S_{1}\) and \(S_{2}\) are calculated according to the following equation [12]:  
\[S_{1}=k_{1}(l_{1}+A_{1}X_{a}-B_{1}X_{h}) \tag{28}\]  
\[S_{2}=k_{2}(l_{2}+A_{2}Y_{a}-B_{2}Y_{h})\]  
\[A_{1}=1-\frac{\rho_{1}}{lg\varphi_{1}},\;\;B_{1}=\frac{\varphi_{1}}{\sin\varphi _{1}}-1,\;\;\varphi_{1}=\omega_{0}\sqrt{\frac{m_{A1}}{k_{1}}} \tag{29}\]  
\[A_{2}=1-\frac{\varphi_{2}}{lg\varphi_{2}},\;\;B_{1}=\frac{\varphi_{2}}{\sin \varphi_{2}}-1,\;\;\varphi_{2}=\omega_{0}\sqrt{\frac{m_{A2}}{k_{2}}}\]  
\(X_{a}\), \(X_{b}\), \(Y_{a}\), and \(Y_{b}\) are shown in Figure 7.  
For the problems of the centrifugal force on the spring in the rotating system, some scholars conducted relevant studies but did not propose a clear formula [13]. In this study, we followed the aforementioned research results to use Equations (28) and (29) and verified the equations as described in Section 5. Using the algorithm in Section 4.1 to solve the above equation, \(\dot{\theta}_{2}<\dot{\theta}_{1}\) can be obtained, which means that the action speed of the inner plunger is less than that of the outer plunger, and when the spindle speed keeps rising, the inner plunger first reaches the critical state  
Figure 7: Position description of \(X_{a}\), \(X_{b}\), \(Y_{a}\), and \(Y_{b}\). of imminent flying out, which is categorized as the first case mentioned above. The \(\dot{\theta}_{2}\) at this time is recorded as \(\omega_{2}\), which is the minimum speed to make the inner plunger action occur. Because the inner plunger is moved after causing the inner spring to be further compressed, the elastic force \(S_{1}\) of the inner spring is also increased, and the force situation is changed; thus, it is necessary to re-determine whether the outer plunger will move after the spring force of the inner spring is increased. When the inner plunger is moved, the inner spring force is a function of the radial displacement \(\rho_{2}\) of the inner plunger, denoted as \(S_{2}(\rho_{2})\). If the inner spring force is sufficient to overcome the frictional resistance and centrifugal force to make the outer plunger fly out, then there is  
\[S_{2}(\rho_{2})-f_{1}-m_{h1}\omega^{2}\rho_{1}>0 \tag{30}\]  
According to the calculation of the relevant parameters, Equation (30) does not hold when \(r_{0}<\rho_{2}<r_{0}+x_{max}\). Therefore, the outer plunger does not produce radial displacement during the movement of the inner plunger. When the inner plunger moves to the end, the inner wall of the outer plunger is in contact, and the inner and outer plungers become a whole.