## 2 Materials and Methods
### Equations of Motion and Their Solutions
\begin{array}{l}\frac{d}{dt}\frac{\partial T}{\partial\varphi}-\frac{\partial T }{\partial\varphi}=Q_{\varphi u},\\ \frac{d}{dt}\frac{\partial T}{\partial\varphi}-\frac{\partial T}{\partial \varphi}=Q_{\varphi d},\end{array}  \tag{1}\]  
where \(T\) is the kinetic energy of the system, and \(Q_{\varphi u}\) and \(Q_{\varphi d}\) are the generalized forces in cases of upward and downward movement of the piston, respectively. To do this, it is necessary to make expressions of the kinetic energy of the system: the device motor. The kinetic energy of the system consists of two components: the kinetic energy of the rotary motion of the engine rotor and the kinetic energies of the translational energy of the piston with the slide at r << lcr:  
\[T=\frac{1}{2}J_{mr}\dot{\varphi}^{2}+\frac{1}{2}m\dot{v}^{2}, \tag{2}\]  
where \(J_{mr}=\frac{1}{2}M_{mr}R^{2}\) is the inertia moment of the engine rotor, and \(v\) is the speed of the piston with the slide. To determine the velocity \(v\), it is necessary to take into account some geometric properties of the device. If the influence of the final length of the connecting rod is neglected, the path traveled by the piston can be related to the angle \(\varphi\) of rotation of the crank by the following relation:  
\[x=r_{0}(1-\cos\varphi). \tag{3}\]  
Piston and slide speed are as follows:  
\[v=\frac{dx}{dt}=r_{0}\sin\varphi\dot{\varphi}. \tag{4}\]  
Substituting this expression in (2), we obtain the kinetic energy of the system in the form  
\[T=\frac{1}{2}\Big{(}J_{mr}+mr_{0}^{2}\sin^{2}\varphi\Big{)}\dot{\varphi}^{2}. \tag{5}\]  
Considering that \(M_{mr}>>m\),  
\[T=\frac{1}{2}J_{mr}\dot{\varphi}^{2}. \tag{6}\]  
The crank of the mechanism is driven by the dynamic torque of the motor \(M\) and, as a result of the translational movement of the slide, creates the resulting \(F\) of the pressure forces \(F_{1}\), friction \(F_{2}\) and \(F_{3}\) acting on the piston. For the proposed mechanism, the DC motor with rectilinear characteristic has a dynamic torque as follows:  
\[M=a-b\dot{\varphi}, \tag{7}\]  
There are relatively few studies dedicated to the calculation of viscous fluid flow through plate holes. For example, articles [47; 48] study the flow of a viscous compressible medium through drilled (porous) plate holes, including the influence of the entrance effect. The mass flow rate through the holes depends on the pressure difference on the plate, depending on the relevant parameters such as the diameter of the hole, the thickness of the plate, and the properties of the medium. The force of interaction between the piston and the fluid, \(F\), includes the pressure force, buoyant force, and gravitational force \(F_{1}\), the friction force acting on the outer lateral surface of the piston \(F_{2}\), and the friction force acting on the surfaces of the piston's holes \(F_{3}\)We determine the projection of the pressure force, buoyant force, and gravity force acting on the piston. For this purpose, let us consider a flow tube with two cross-sections: the first coincides with the cross-section of the cylinder and the lower boundary of the piston, and the second coincides with the cross-sections of the holes and the clearance between the piston and the cylinder, as well as the upper boundary of the piston. Assuming the fluid is incompressible and has low viscosity, we apply the continuity equation and Bernoulli's equation to these sections. By solving the resulting system of equations simultaneously, we find the pressure difference of the fluid between the upper and lower surfaces of the piston:  
\[\Delta p=p_{2}-p_{1}=-\frac{\rho}{2}\Bigg{[}\frac{1}{n_{2}^{2}}\bigg{(}\frac{ R_{c}}{R_{2}}\bigg{)}^{4}-1\Bigg{]}v^{2}+\rho g_{x}h+\frac{mg_{x}}{\pi\big{(}R_{1}^{ 2}-nR_{2}^{2}\big{)}}. \tag{8}\]  
Here, \(g_{x}\) is the projection of gravitational acceleration onto the direction of fluid flow through the piston holes and the gaps between the piston and cylinder: \(g_{x}=-g\) when the fluid moves downward, and \(g_{x}=g\) when the fluid moves upward.