# Pneumatic Rotary Actuator Position Servo System Based on ADE-PD Control
## 3 Modeling the System
### Mass Flow Continuity Equation  
According to the law of mass flow, the mass flow of the gas in chamber \(\mathbf{a}\) and \(\mathbf{b}\) is equal to the mass change of inflow and outflow of gas per unit time, respectively, which can be written as  
\[\left\{ \begin{array}{ll}\dot{M_{a}}=\frac{\mathrm{d}M_{b}}{\mathrm{d}t}= \frac{\mathrm{d}(\rho_{a}V_{a})}{\mathrm{d}t}=\rho_{a}\frac{\mathrm{d}V_{a}}{ \mathrm{d}t}+V_{a}\frac{\mathrm{d}\rho_{a}}{\mathrm{d}t}\\ -\dot{M_{b}}=\frac{\mathrm{d}M_{b}}{\mathrm{d}t}=\frac{\mathrm{d}(\rho_{b}V_{b} )}{\mathrm{d}t}=\rho_{b}\frac{\mathrm{d}V_{b}}{\mathrm{d}t}+V_{b}\frac{ \mathrm{d}\rho_{b}}{\mathrm{d}t}\end{array} \right. \tag{12}\]  
where \(M_{a}\) and \(M_{b}\) are masses of gas in chamber \(\mathbf{a}\) and \(\mathbf{b}\); \(\rho_{a}\) and \(\rho_{b}\) are densities of gas in chamber \(\mathbf{a}\) and \(\mathbf{b}\); and \(V_{a}\) and \(V_{b}\) are volumes of gas in chamber \(\mathbf{a}\) and \(\mathbf{b}\). If gas state equation \(\rho=\frac{p}{RT}\) is substituted into Equation (12), we can obtain Equation (13). \[\left\{ \begin{array}{ll}\dot{M_{a}}=\frac{1}{RT_{s}}\Big{(}p_{a}\frac{ \mathrm{d}V_{a}}{\mathrm{d}t}+V_{a}\frac{\mathrm{d}p_{a}}{\mathrm{d}t}-\frac{p _{a}V_{a}}{T_{a}}\frac{\mathrm{d}T_{a}}{\mathrm{d}t}\Big{)}\\ -\dot{M_{b}}=\frac{1}{RT_{b}}\Big{(}p_{b}\frac{\mathrm{d}V_{b}}{\mathrm{d}t}+V_ {b}\frac{\mathrm{d}p_{b}}{\mathrm{d}t}-\frac{p_{b}V_{b}}{\mathrm{d}t}\frac{ \mathrm{d}T_{b}}{\mathrm{d}t}\Big{)}\end{array} \right. \tag{13}\]  
According to the previous hypotheses, the temperature \(T\) and initial temperature \(T_{0}\) in the whole process of the system satisfy the isentropic condition [31]:  
\[T=T_{0}\bigg{(}\frac{p}{p_{0}}\bigg{)}^{\frac{k-1}{k}} \tag{14}\]  
where \(p_{0}\) is the initial pressure. We can get the derivative of Equation (14). \[\frac{\mathrm{d}T}{\mathrm{d}t}=\frac{k-1}{k}\frac{T_{0}}{p_{0}}\bigg{(}\frac{p}{p_ {0}}\bigg{)}^{-\frac{1}{k}}\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{k-1}{k}\frac{T_ {0}}{p}\bigg{(}\frac{p}{p_{0}}\bigg{)}^{\frac{k-1}{k}}\frac{\mathrm{d}p}{\mathrm{ d}t}=\frac{k-1}{k}\frac{T}{p}\frac{\mathrm{d}p}{\mathrm{d}t} \tag{15}\]  
Substituting Equation (15) into (13) yields  
\[\left\{ \begin{array}{c}\dot{M}_{a}=\frac{1}{RT_{a}k}\Big{[}V_{a}\frac{ \mathrm{d}p_{a}}{\mathrm{d}t}+kp_{a}\frac{\mathrm{d}V_{a}}{\mathrm{d}t}\Big{]} \\ -\dot{M}_{b}=\frac{1}{RT_{b}k}\Big{[}(V_{b}\frac{\mathrm{d}p_{b}}{\mathrm{d}t} +kp_{b}\frac{\mathrm{d}V_{b}}{\mathrm{d}t}\Big{]}\end{array} \right.. \tag{16}\]  
When the piston of pneumatic rotary actuator works near position \(i\), it can be considered to make small changes near the position [27]. By linearizing Equation (16), we obtain the increment of mass flow:  
\[\left\{ \begin{array}{c}\Delta\dot{M}_{a}=\frac{1}{RT_{a}k}\Big{[}(V_{ai}) \frac{\mathrm{d}(\Delta p_{a})}{\mathrm{d}t}+(kp_{ai})\frac{\mathrm{d}(\Delta V _{a})}{\mathrm{d}t}\Big{]}\\ -\Delta\dot{M}_{b}=\frac{1}{RT_{b}k}\Big{[}(V_{bi})\frac{\mathrm{d}(\Delta p_{ b})}{\mathrm{d}t}+(kp_{bi})\frac{\mathrm{d}(\Delta V_{b})}{\mathrm{d}t}\Big{]} \end{array} \right.. \tag{17}\]  
where \(T_{ai}\) stands for temperature of chamber \(\mathbf{a}\) in working position \(i\), \(V_{ai}\) and \(V_{bi}\) are volumes of gas of chamber \(\mathbf{a}\) and \(\mathbf{b}\) in working position \(i\). After setting the initial position as an intermediate position, we take the total working volume of two chambers as \(2V_{0}\), then the initial values are as follows:  
\[T_{ai}=T_{bi}=T_{s}V_{ai}=V_{bi}=V_{0},\ p_{ai}=p_{bi}=p_{i}\]  
where \(p_{i}\) stands for pressure of chambers in initial position \(i\). The displacement of the actuator piston is represented by \(y\), and the effective area of the piston is represented by \(A\). The rotation angle of the actuator is represented by \(\theta\), and the pitch diameter of the pinion is \(d_{f}\).