## 3 FEM Simulation and Experimental Validation  
### Electrical-Thermal Balance Equations for the FEM Simulation  
This section presents the implementation of the \(q_{\text{total}}^{{}^{\prime\prime}}(x)\) model into the FEM code. The IR lamp converts electrical energy into IR radiation energy using the tungsten filaments inside the lamp. During the energy conversion, the filament temperature increases to 2500-3000 K and the radiant heat energy was emitted into the surrounding space; the lamp should satisfy the electrical-thermal energy balance. First, the conservation of energy charge [35] is considered:  
\[\
abla\cdot(K_{\text{e}}\
abla V)=0\ \ \ \text{where}\ \mathbf{I}=-K_{\text{e}} \
abla V. \tag{16}\]  
\(\mathbf{I}\) is the electric current, \(K_{\text{e}}\) is the electrical conductivity, and \(V\) is the electrical potential. A weak form for the FEM can be derived as below:  
\[\int_{\Omega}[\
abla N]^{T}K_{\text{e}}[\
abla N]d\Omega[V]=\mathcal{F}_{ \text{elec}^{\prime}}^{\text{ext}}\ \text{where}\ \mathcal{F}_{\text{elec}}^{\text{ext}}=-\int_{\Gamma}[N]^{T}\mathbf{I}\cdot\mathbf{nd}\Gamma, \tag{17}\]  
where \(d\Omega\) is the volume integral term and \(d\Gamma\) represents the surface integral term. \(\mathbf{n}\) is the outward unit vector normal to the surface and \([N]\) is the shape function of the FEM, \(\mathcal{F}_{\text{elec}}^{\text{ext}}=-\int_{\Gamma}[N]^{T}\mathbf{I}\cdot\mathbf{nd}\Gamma\), which denotes an external force as the boundary condition of the electrical potential. By solving Equation (17), the electrical current vector, \(\mathbf{I}\), can be obtained by substituting \(\mathbf{I}\) into [PERSON]'s law [36], leading to thermal energy conversion:  
\[\dot{Q}_{\text{res}}=(\mathbf{I}\cdot\mathbf{I})R. \tag{18}\]  
\(R\) is the electrical resistance of the filament and \(\dot{Q}_{\text{res}}\) is the time rate of the thermal energy converted from electrical energy. The thermal balance [37] of the lamp was considered:  
\[\
abla\cdot(K_{\text{th}}\
abla T)+\dot{Q}_{\text{res}}=\rho\dot{T} \tag{19}\]  
where \(K_{\text{th}}\) is the thermal conductivity, \(\rho\) is the density, \(C\) is the thermal capacity, and \(T\) is the absolute temperature. A weak form of Equation (19) can be derived under static condition (\(\dot{T}=0\)) as follows:  
\[\int_{\Omega}\
abla\delta TK_{\text{th}}\
abla Td\Omega=\int_{\Gamma}\delta T \mathbf{q}_{\text{out}}\cdot\mathbf{nd}\Gamma+\int_{\Omega}\delta T\dot{Q}_{\text{res }}d\Omega \tag{20}\]  
Figure 5: Modeling of the IR energy distribution: (**a**) Focal length 20 mm; (**b**) Focal length 40 mm. \(\mathbf{q_{\text{out}}}\cdot\mathbf{n}=-\big{\{}h(T-T_{\infty})+\sigma\varepsilon\big{(}T^{4 }-T_{\infty}^{4}\big{)}\big{\}}\) includes the energy loss through convection (\(h(T-T_{\infty})\)) and radiation \(\big{(}\sigma\varepsilon\big{(}T^{4}-T_{\infty}^{4}\big{)}\big{)}\) to the surrounding air; \(h\) is the heat convection coefficient, \(T_{\infty}\) is the ambient temperature, \(\sigma\) is the Stefan-Boltzmann constant, and \(\varepsilon\) denotes the emissivity of the filament. The radiation energy \(\big{(}\sigma\varepsilon\big{(}T^{4}-T_{\infty}^{4}\big{)}\big{)}\) is considered as the IR energy emitted from the lamp. By solving the electrical-thermal energy balance of the lamp in Equation (20), the emitted IR power, \(I_{0}\), was calculated; \(I_{0}\) can be converted to the IR distribution using Equation (12). Next, the material heated by the IR energy should be modeled. The heated material is based on the same thermal balance in Equation (19). The differences are that the heated material considers a transient state (\(\dot{T}\
eq 0\)) and IR heating condition (\(q_{\text{total}}^{\prime\prime}\)).