## 2 Analytic Model Development
### The Thermal Modeling  
In Equation (5), \(Q\) represents the discharge heating power and the thermal energy loss. The discharge heating power can be represented by Ohm heating easily. The thermal energy loss through the wall and the electrode is modeled based on the lumped capacitance method.  
The heat transfer process in the PSA is shown in Figure 3. Firstly, by forced convection, the heat transfers from the chamber gas to the shell inner wall. Secondly, by heat conduction through the electrodes and shell wall, the heat transfers from the shell inner wall to the shell outer wall. Thirdly, by natural convection, the heat transfers from the shell outer wall to the external air.  
Similar to the electric circuit, the heat transfer process can be described as a thermal circuit, as shown in Figure 4. \(T_{\rm c}\), \(T_{\rm w,in}\), \(T_{\rm w,out}\), and \(T_{\infty}\) indicate the temperature of chamber gas, shell inner wall, shell outer wall, and external air, respectively. \(R_{\rm in}\), \(R_{\rm e}\), \(R_{\rm w}\), and \(R_{\rm out}\) represent the thermal resistance of force convection, electrode conduction, shell conduction, and natural convection. \(C_{\rm w}\) represents the thermal capacitance of the shell.  
Based on the theory of engineering thermal transmission, the thermal resistance of force convection, electrode conduction, and natural convection is given as follows.  
\[\left\{\begin{array}{l}R_{\rm in}\ =\dfrac{1}{h_{\rm in}A_{\rm c,in}}\\ R_{\rm out}=\dfrac{1}{h_{\rm out}A_{\rm c,out}}\\ R_{\rm e}=\dfrac{0.5l_{\rm e}}{\kappa_{\rm e}A_{\rm e}}\end{array}\right. \tag{16}\]  
The structure of the shell wall is complex, and is regarded as a combination of three simple structures, as shown in Figure 5. The upper structure of the shell is seen as a single wall. The middle structure of the shell is seen as a single hollow cylinder wall. The subjacent structure of the shell is also seen as a single wall. The thermal resistance of the three structures is given as follows.  
\[\left\{\begin{array}{l}R_{\rm up}\ =\dfrac{h_{3}}{\kappa_{\rm up}\dfrac{ \pi}{4}\big{(}d_{1}^{2}-d_{3}^{2}\big{)}}\\ R_{\rm out}=R_{\rm mid}=\dfrac{\ln\dfrac{d_{2}}{d_{1}}}{2\pi\kappa_{\rm mid} h_{1}}\\ R_{\rm sub}=\dfrac{h_{2}}{\kappa_{\rm sub}\dfrac{\pi}{4}d_{1}^{2}}\end{array}\right. \tag{17}\]  
The total thermal resistance of the shell wall is calculated as Equation (18).  
\[R_{\rm w}=\dfrac{1}{1/R_{\rm up}\ +1/R_{\rm mid}+1/R_{\rm sub}} \tag{18}\]  
With all of the circuit components defined, the differential function obtained from the circuit shown in Figure 4 is used to solve for the wall temperature.  
\[\dfrac{dT_{\rm w,in}}{dt}\ =\dfrac{1}{C_{\rm w}}\bigg{(}\dfrac{T_{\rm c}-T_{\rm w,in}}{R_{\rm in}}-\dfrac{T_{\rm w,in}-T_{\infty}}{R_{\rm out}+R_{\rm e}R_{\rm w }/(R_{\rm e}+R_{\rm w})}\bigg{)} \tag{19}\]  
Figure 4: Electrical representation of the thermal heat transfer process.  
Figure 3: The schematic diagram of the heat transfer process.  
Hence, combined with the discharge heating source, the parameter \(Q\) in Equation (10) is given.  
\[\dot{Q}=Q_{\mathrm{arc}}\,-\,\frac{T_{\mathrm{c}}-T_{\mathrm{w,in}}}{R_{\mathrm{ in}}}=0.5I(t)^{2}R(t)-\frac{T_{\mathrm{c}}-T_{\mathrm{w,in}}}{R_{\mathrm{in}}} \tag{20}\]  
It must be pointed out that the energy calculated directly using Ohm heating theory is not the heating energy. The sheath energy loss and radiation energy loss must be taken into consideration. Based on the analysis of [PERSON] [19], a heating efficiency of 50% is adopted.