# Constant Force Control of Centrifugal Pump Housing Robot Grinding Based on Pneumatic Servo System
## 4 Modeling of the Pneumatic Servo System
### Mass Flow Equation of the Pneumatic Servo Valve  
The flow process of the gas passing through the valve port of the servo valve is relatively complex. Generally, the flow process of the gas passing through the valve port of a proportional servo valve is considered to be the one-dimensional isentropic flow of an ideal gas. According to the Sanville flow formula [35], the mass flow formula of a proportional valve port is  
\[q_{m}=\mu A_{t}P_{u}\sqrt{\frac{2}{RT}}\Phi\left(\frac{P_{d}}{P_{u}}\right) \tag{12}\]  
\[\Phi\left(\frac{P_{d}}{P_{u}}\right)= \begin{cases}\sqrt{\frac{k}{k-1}}\left[ \left(\frac{P_{d}}{P_{u}}\right)^{\frac{2}{k}}-\left(\frac{P_{d}}{P_{u}} \right)^{\frac{k+1}{k}}\right]&0.528<\frac{P_{d}}{P_{u}}\leq 1\\ \sqrt{\frac{k}{k+1}\left(\frac{2}{k+1}\right)^{\frac{2}{k-1}}}&0\leq\frac{P_{ d}}{P_{u}}\leq 0.528\end{cases} \]  
The opening area of the servo valve is  
\[A_{t}=\omega x_{v} \tag{13}\]  
where \(x_{v}\) is the valve core displacement, m, and \(\omega\) is the area gradient of the valve port (\(\omega=\pi D\)), m.  
Equation (13) can be substituted into Equation (12) to get:  
\[q_{m}=\mu\pi Dx_{v}P_{u}\sqrt{\frac{2}{RT}}\Phi\left(\frac{P_{d}}{P_{u}}\right) \tag{14}\]  
Then, the inlet flow and outlet flow of the valve are, respectively:  
\[\left\{ \begin{array}{l}q_{m_{1}}=\mu\pi Dx_{v}P_{p}\sqrt{\frac{2}{RT}}\Phi \left(\frac{P_{1}}{P_{p}}\right)\\ q_{m_{2}}=\mu\pi Dx_{v}P_{2}\sqrt{\frac{2}{RT}}\Phi\left(\frac{P_{1}}{P_{2}} \right)\end{array} \right. \tag{15}\]  
where \(P_{p}\) is the air source pressure, Pa; \(P_{1}\) is the pressure without a rod chamber, Pa; \(P_{2}\) is the pressure with a rod chamber, Pa; and \(P_{k}\) is the atmospheric pressure, Pa. Since the displacement of the proportional valve spool is linear with the input voltage \(u\)[11], \(x_{v}=K_{u}u\), \(K_{u}\) is the gain coefficient of the proportional valve. Then, we get  
\[\left\{ \begin{array}{l}q_{m_{1}}=\mu\pi DK_{u}uP_{p}\sqrt{\frac{2}{RT}}\Phi \left(\frac{P_{1}}{P_{p}}\right)\\ q_{m_{2}}=\mu\pi DK_{u}uP_{2}\sqrt{\frac{2}{RT}}\Phi\left(\frac{P_{1}}{P_{2}} \right)\end{array} \right. \tag{16}\]  
It can be seen from the above formula that the actual flow at the valve port is related to the pressure at the air outlet and the control voltage of the electric servo valve, so the flow formula can be simplified as  
\[q_{m}=f(u,P) \tag{17}\]  
where \(u\) is the control voltage of the servo valve. Equation (16) can be linearized near the zero position to obtain  
\[\left\{ \begin{array}{l}q_{m_{1}}=K_{1}\Delta u-K_{2}\Delta p_{1}\\ q_{m_{2}}=K_{3}\Delta u+K_{4}\Delta p_{2}\end{array} \right. \tag{18}\]  
\[K_{1}=\left. \begin{array}{l}\frac{\partial q_{m_{1}}}{\partial u}\bigg{|}_{u }=0\\ p_{1}=P_{10}\end{array} \right.,K_{2}=\left. \begin{array}{l}\frac{\partial q_{m_ {1}}}{\partial p_{1}}\bigg{|}_{u}=0\\ p_{1}=P_{10}\end{array} \right.,K_{3}=\left. \begin{array}{l}\frac{\partial q_{ m_{2}}}{\partial u}\bigg{|}_{u}=0\\ p_{2}=P_{20}\end{array} \right.,K_{4}=\left. \begin{array}{l}\frac{\partial q_ {m_{2}}}{\partial p_{2}}\bigg{|}_{u}=0\\ p_{2}=P_{20}\end{array} \right.\]  
where  
\[K_{1}=\left. \begin{array}{l}\frac{\partial q_{m_{1}}}{\partial u}\bigg{|}_{u }=0\\ p_{1}=P_{10}\end{array} \right.=\mu\tau DK_{u}P_{p}\sqrt{\frac{2}{RT}}\Phi\left( \frac{P_{1}}{P_{p}}\right)\bigg{|}_{u}=\mu\tau DK_{u}P_{p}\sqrt{\frac{2}{RT}} \sqrt{\frac{k}{k-1}}\Bigg{[}\left(\frac{P_{1}}{P_{p}}\right)^{\frac{2}{k}}- \left(\frac{P_{1}}{P_{p}}\right)^{\frac{k+1}{k}}\Bigg{]},\\ p_{1}=P_{10}\end{array}\]  
\[K_{2}=\left. \begin{array}{l}\frac{\partial q_{m_{1}}}{\partial p_{1}}\bigg{|}_{u }=0\\ p_{1}=P_{10}\end{array} \right.=0,\]  
\[K_{3}=\left. \begin{array}{l}\frac{\partial q_{m_{2}}}{\partial u}\bigg{|}_{u }=0\\ p_{2}=P_{20}\end{array}