*2.3. Electric Motor Model*

Electric motors used in sUAV applications exhibit high speed and high torque, as well as high power-to-weight ratios [29]. Assuming the power factor is equal to unity and the magnetic losses can be neglected, the output power of the motor is given by:

$$P\_P = (\mathcal{U}\_M - I\_M \mathcal{R}\_M)(I\_M - I\_{M,0}).\tag{8}$$

The angular velocity of the motor in revolutions per minute (RPM) can be expressed as:

$$N = (\mathbb{U}I\_M - R\_M I\_M) K\_{V\_{\prime}} \tag{9}$$

which should be equal to the RPM of propeller *N* = *<sup>v</sup> JdP* . From Equations (8) and (9), the motor current, *IM*, is:

$$I\_M = \frac{P\_P K\_V}{N} + I\_{M,0}.$$

The motor power and motor efficiency are given by, respectively,

$$P\_M = \mathcal{U}I\_M I\_{M\nu} \quad \eta\_M = \frac{P\_P}{P\_M}.$$

#### *2.4. Fuel Cell Model*

A PEMFC system is the primary power source for the sUAV. The total power generated by the fuel cell stack is calculated as:

$$P\_{\rm FC,total} = \pi\_{\rm FC} \mathcal{U}\_{\rm FC} I\_{\rm FC} \,\tag{10}$$

This power must cover the load demand *PFC*,*load* and the power required for auxiliaries [18], *Paux*,

$$P\_{\rm FC,total} = P\_{\rm FC,load} + P\_{\rm aux} \tag{11}$$

where *Paux* is the total power required for the compressor motor, the hydrogen circulation pump, the humidifier water circulation pump, the coolant pump, the cooling fan motor, and the bias power, *P*0. After simplifications, *Paux* could be written as a function of the fuel cell current [30],

$$P\_{\rm aux} = P\_0 + n\_{\rm FC} \kappa\_{\rm FC} I\_{\rm FC}.\tag{12}$$

The fuel cell current is a function of the current density and the fuel cell area,

$$I\_{\rm FC} = i\_{\rm FC} A\_{\rm FC\prime}$$

where *iFC* could be obtained by solving the equation,

$$
\mathcal{U}\_{FC} = \mathcal{U}\_{\text{rev}} - \mathcal{U}\_{\text{act}} - \mathcal{U}\_{\text{olum}} - \mathcal{U}\_{\text{conv}}.\tag{13}
$$

The reversible cell potential *Urev* is related to the molar specific Gibbs free energy Δ*gf* and the number of ions passed in the reaction *ne* [24],

$$
\mathcal{U}\_{rev} = \frac{\Delta \mathcal{g}\_f}{n\_c F}.
$$

The activation polarization *Uact* is a result of the energy required to initiate the reaction, which can be described by the semi-empirical Tafel equation [31–33],

$$
\mathcal{U}\_{act} = c\_0 + c\_1 \ln(i\_{FC})\_{\prime\prime}
$$

where *c*<sup>0</sup> and *c*<sup>1</sup> depend on temperature. When the current density is small, this equation can be modified [34] as:

$$
\mathcal{U}\_{\rm act} = c\_0 (1 - e^{-c\_1 i\_{F\mathcal{C}}}),\tag{14}
$$

where *<sup>c</sup>*<sup>0</sup> <sup>=</sup> <sup>−</sup>5.8 <sup>×</sup> <sup>10</sup>−4*T*¯ <sup>+</sup> 0.5736 and *<sup>c</sup>*<sup>1</sup> <sup>=</sup> *RT*¯ *neαFCF* .

The ohmic losses *Uohm* are due to the resistance to the flow of (i) ions in the membrane and in the catalyst layers and (ii) electrons through the electrodes [18],

$$
\hbar I\_{\rm olm} = i\_{\rm FC} \vec{\mathbb{R}}\_{\rm FC} \tag{15}
$$

where *R*˜ *FC* = *RFCAFC*.

The concentration polarization *Uconc*, is given by

$$dI\_{\rm conc} = d\_0 \mathfrak{e}^{d\_1 i\_{\rm FC}}.\tag{16}$$

With the parameters given in Appendix A, the polarization curve of a single cell is plotted in Figure 2. In reality, the current density could be controlled within a certain range. After excluding the very low current densities (*iFC* < 0.1 A/cm2), (13) could be linearized [34,35] as:

$$
\mathcal{U}\_{\rm FC} = \mathcal{U}\_{\rm CC} - \mathcal{R}\_{\rm FC} i\_{\rm FC} \,\tag{17}
$$

where *UOC* is the voltage at which the linearized curve crosses the y-axis, which should not be confused with *Urev*.

**Figure 2.** Polarization curve for a given PEMFC.

Unlike ground vehicles, the sUAV changes its orientation during the flight, which would change the inner resistance of the fuel cell by about five times [36] from horizontal to vertical. To this end, (17) is modified to account for this effect as:

$$
\mathcal{U}\_{\rm FC} = \mathcal{U}\_{\rm CC} - \mathcal{R}'\_{\rm FC} i\_{\rm FC} \,\tag{18}
$$

where *R*˜ *FC* <sup>=</sup> *<sup>R</sup>*˜ *FC*(<sup>1</sup> <sup>+</sup> *<sup>k</sup>*0*sin*(*k*1|*γ*|)). Combining Equation (18) with Equations (10)–(12) yields:

$$n\_{FC}R\_{FC}^{\prime}I\_{FC}^{2} - (n\_{FC}lL\_{OC} - n\_{FC} \kappa\_{aux})I\_{FC} + P\_{FC,load} + P\_0 = 0,\tag{19}$$

where *R FC* <sup>=</sup> *<sup>R</sup>*˜ *FC*/*AFC*. Overall, *IFC* can be expressed as:

$$I\_{\rm FC} = \frac{n\_{\rm FC}(\ell I\_{\rm OC} - \kappa\_{\rm aux}) - \left[n\_{\rm FC}^2(\ell I\_{\rm OC} - \kappa\_{\rm aux})^2 - 4n\_{\rm FC}R\_{\rm FC}'(P\_{\rm FC,Load} + P\_0)\right]^{\frac{1}{2}}}{2n\_{\rm FC}R\_{\rm FC}'}.\tag{20}$$
