## 2 Technical Solution
### Carnot Battery Integration
#### 2.3.2 Carnot Battery with Solid Particles: Scenario IV  
For this scenario, a fluidized bed storage system was developed, aligning with the technical solution implemented in the solar thermal plant. As in the previous case, surplus electricity will power an electric resistance, which in this instance will heat an air stream. The selection of the electric resistance is based on the aspects discussed in the previous section, as well as the high efficiency of the resistance-air system, which can reach up to 100% [55].  
Once the air reaches the desired temperature, it will pass through the tubes of a heat exchanger, transferring most of its heat to the particles in a fluidized bed surrounding the tubes. This process reduces the air temperature. To improve the system's efficiency, the exiting air, which remains at a relatively high temperature, will be recovered and used both to fluidize and preheat the particles [56]. This heat exchange system is similar to the one used in [57].  
The power of the electric resistance responsible for heating the air was determined using the same methodology as in the previous section, resulting in a resistance of 15 MWe.  
Next, the required air flow rate for both propulsion and particle fluidization was calculated. For this calculation, a target air temperature of 670 \({}^{\circ}\)C was set, and the following energy balance equation was applied:  
\[P_{EH}\cdot\eta_{EH}=\dot{m}_{air}\cdot\overline{C_{P_{air}}}\cdot\Delta T \tag{1}\]  
where PEH is the selected resistance power; \(\eta_{\text{EH}}\) is the efficiency of the selected resistance, fixed at 97.5% according to Table 9; \(\dot{m}_{\text{air}}\) is the air mass flow rate; Cp is the average specific heat of air [58]; and \(\Delta\)T is the temperature difference the air experiences during heating.  
After calculating the required flow rate and considering that pressure loss in such beds is low [21; 59], a pressure drop of 0.3 bar was assumed. It was determined that the air propulsion element must be a centrifugal compressor, based on the required pressure and flow rate [60].  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Area** & **Typical Efficiencies (\%)** \\ \hline Electric transformer & 97.5 [2] \\ Electric resistance & 97.5 [2] \\ Electricity-to-heat & 95.1 \\ \hline \hline \end{tabular}
\end{table}
Table 8: Typical efficiencies for charging process in Scenario II.  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Area** & **Typical Efficiencies (\%)** \\ \hline Electric transformer & 97.5 [2] \\ Electric resistance & 97.5 [2] \\ Heat exchanger & 96.0 [44] \\ Electricity-to-heat & 95.1 \\ \hline \hline \end{tabular}
\end{table}
Table 9: Typical efficiencies of charging process in Scenario IV.  
The compressor power was calculated using (2), assuming air as an ideal gas and an adiabatic compressor [61]:  
\[P=\frac{R\cdot T_{1}\cdot\gamma}{\eta\cdot(\gamma-1)}\cdot\left[\left(\frac{P_{2} }{P_{1}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]\cdot\dot{n}_{air} \tag{2}\]  
where P is the resulting compressor power in kW; T\({}_{1}\) is the air temperature at the compressor inlet, 943 K in this case; \(\gamma\) is the specific heat ratio; R is the gas constant for air; \(\eta\) is the compressor's isentropic efficiency, assuming an average value of 81% [60]; and P\({}_{1}\) and P\({}_{2}\) are the inlet and outlet pressures at the compressor, respectively. It is assumed that air enters at atmospheric pressure (1 bar) and exits at 1.3 bar due to the pressure drop.  
Using these values, the final compressor power was calculated to be 1.9 MW.  
Efficiencies involved in the entire charging process are critical. These include transformer efficiency, resistance efficiency, and overall conversion efficiency from electricity to heat. A summary of these efficiencies is presented in Table 9.