## 2 Fuel Cell  
Among the various types of fuel cells, proton exchange membrane fuel cells (PEMFCs) have attracted much attention in recent years since they can operate at relatively low temperatures with a fast start-up [14]. The process of generating electricity from a PEM fuel cell is linked to chemical reactions that lead to electron transfer. These chemical reactions can be divided into two categories: oxidation (emission of electrons) and reduction (absorption of electrons). In the PEM fuel cells, an acidic polymer membrane is used as the electrolyte and the chemical reactions that produce electricity take place at the electrodes based on the following reactions [15]:  
\[\text{Oxidation: H}_{2}\to 2\text{H}^{+}+2\text{e}^{-}\]  
\[\text{Reduction: 4H}^{+}+\text{O}_{2}+4\text{e}^{-}\to 2\text{H}_{2}\text{O}\]To achieve the overall reaction, the first equation (oxidation reaction) must be doubled and then summed up with the second one (reduction reaction), as the number of electrons is the same on both sides of the equations:  
Oxidation: 2H2 4H+ 4e-  
Reduction: 4H+ O2 + 4e- 2H2O  
2H2 + O2 2H2O + energy  
According to the fact that water (H2O) is a liquid for temperatures below 100 \({}^{\circ}\)C and the chemical equation of the PEM fuel cell, the thermodynamic voltage equation would change for two different temperature ranges to Equation (1) [16]. In fact, \(P_{H2O}\) exhibits the water activity between anode and electrolyte. \[E_{n}=\left\{ \begin{array}{cc}E_{0}+(T-T_{0})\frac{\Delta S_{0}}{2F}+\frac{ RT}{2F}\ln\left(P_{H_{2}}\frac{P_{2}^{\frac{1}{2}}}{P_{H_{2}O}}\right)&T\leq 100\ ^{\circ}\text{C}\\ \\ E_{0}+(T-T_{0})\frac{\Delta S_{0}}{2F}+\frac{RT}{2F}\ln\left(\frac{P_{H_{2}}P_{ 2}^{\frac{1}{2}}}{P_{H_{2}O}}\right)&T>100\ ^{\circ}\text{C}\end{array} \right. \tag{1}\]  
where \(F\) is the Faraday constant, \(z\) is the number of electrons transferred (\(z=2\)), \(E_{n}\) is the thermodynamic voltage, \(E_{0}\) is the standard electrode voltage, \(T_{0}\) is the temperature at standard condition (298 K), and \(\Delta S_{0}\) exhibits the entropy of the chemical reaction (\(-44.43\) J/mol K). The parameters of \(P_{H_{2}}\), \(P_{O_{2}}\), and \(P_{H_{2}O}\) denote the partial pressure of H2, O2, and H2O in atm, respectively [17]. By substituting the aforementioned values in Equation (1), the Nernst equation can be obtained as follows [18]:  
\[E_{n}=\left\{ \begin{array}{cc}1.229+(T-298)\frac{-44.43}{2F}+\frac{RT}{2F}\ln \left(P_{H_{2}}\frac{P_{O_{2}}^{\frac{1}{2}}}{P_{H_{2}O}}\right)&T\leq 100\ ^{\circ}\text{C}\\ \\ 1.229+(T-298)\frac{-44.43}{2F}+\frac{RT}{2F}\ln\left(\frac{P_{H_{2}}\frac{P_{O _{2}}^{\frac{1}{2}}}{P_{H_{2}O}}}{P_{H_{2}O}}\right)&T>100\ ^{\circ}\text{C}\end{array} \right. \tag{2}\]  
Assuming that the fuel contains \(x\%\) hydrogen and the input air contains \(y\%\) oxygen and \(w\%\) water vapor, the partial pressures can be expressed based on the hydrogen and oxygen utilization as follows:  
\[\left\{ \begin{array}{cc}P_{H_{2}}=\left(1-U_{fH_{2}}\right)x\%P_{fuel}\\ \\ P_{H_{2}O}=\left(w+2y\%U_{fO_{2}}\right)P_{air}\\ \\ P_{O_{2}}=\left(1-U_{fO_{2}}\right)y\%_{o}P_{air}\end{array} \right. \tag{3}\]  
where \(U_{fH_{2}}\) and \(U_{fO_{2}}\) are the values of hydrogen and oxygen utilization, respectively; these parameters can be calculated in terms of flow rate at the inlet, cell current, and pressures as follows [19]:  
\[U_{f_{H_{2}}}=\frac{6000\times R\times T\times N\times i_{fc}}{2\times F\times P _{fuel}\times V_{lpmf}\times x\%} \tag{4}\]  
\[U_{f_{O_{2}}}=\frac{6000\times R\times T\times N\times i_{fc}}{4\times F\times P _{air}\times V_{lpma}\times y\%} \tag{5}\]  
where \(P_{fuel}\) and \(P_{air}\) are absolute supply pressures of hydrogen and air, respectively. The parameters \(V_{lpmf}\) and \(V_{lpma}\) are the flow rates of hydrogen and air at the inlet in L/min.