## 6 Thermodynamic Efficiency: From System to Integration in Buildings  
Whatever the system, open or closed, it is important to answer the issue related to the thermodynamic efficiency limit of the heat storage. Basically, the designer of such system must know the thermodynamics limit to evaluate the enhancement possibilities of its prototype. Let us first define \(Q^{+}\) as the supplied energy per mole of salt to the storage system and \(Q^{-}\) as the recovered energy per mole of salt. For the sake of universality, the efficiency of the system \(\eta\) is defined as:  
\[\eta=\frac{Q^{-}}{Q^{+}} \tag{14}\]  
This efficiency can also be found as _COP_ in the literature [3]. However, the definition of _COP_ may vary from one author to the other: an example is the definition of _COP_ given by [10] which is completely different from the one given by [18]. The maximum reachable efficiency \(\eta_{max}\) is calculated with the maximum \(Q^{-}\), called \(Q^{-}_{max}\), and, of course, the minimum \(Q^{+}_{min}\). The assumptions used to evaluate the **theoretical limits** of the energy efficiency are:  
* Heat losses in the system are not taken into account. * Sensible heat of materials and parts of the reactor are neglected. Figure 6: Prototype energy density versus material energy density. The energy released or absorbed by the reaction is approximated by the standard enthalpy of reaction. The equilibrium drop influence is neglected (see [18]). Only total hydration/dehydration processes are under investigation. Given the previous assumptions, if the heat of condensation is not recovered and if the heat of evaporation is energy-free (Energy-free means that no additional energy is required for producing water vapor), the trivial efficiency is \(\eta_{max}=1\) (see Figure 7--Q\({}_{max}^{-}=Q_{min}^{+}=\Delta_{r}H^{0}\)). Considering the previous assumptions, if non-free heat of evaporation, the maximum efficiency becomes (see Figure 8):  
\[\eta_{max}=\frac{Q_{max}}{Q_{min}^{+}}=\frac{\Delta_{r}H^{0}}{\Delta_{r}H^{0}+ Q_{evap}}=\frac{1}{1+\frac{Q_{evap}}{\Delta_{r}H^{0}}} \tag{15}\]  
The quantity of energy required to evaporate liquid water per mole of salt is evaluated with:  
\[Q_{evap}=\
u\times M_{H_{2}O}\times L_{\
u} \tag{16}\]  
where \(L_{\
u}\) is the water heat of vaporization in standard conditions taken as 2456 kJ kg\({}^{-1}\) and \(M_{H_{2}O}\) the molar mass of water equal to 18 g mol\({}^{-1}\). Considering data from Table 1, and considering that \(\Delta_{r}H^{0}/\
u\) is varying between 55.1 kJ mol\({}^{-1}\) and 67.8 kJ mol\({}^{-1}\), the ratio \(Q_{evap}/\Delta_{r}H^{0}\) can be straightforwardly evaluated as:  
\[0.65\leq\frac{Q_{evap}}{\Delta_{r}H^{0}}\leq 0.80 \tag{17}\]  
Figure 8: System integration: heat of condensation is not recovered and heat of evaporation is not free: \(\eta_{max}=\frac{1}{1+\frac{Q_{evap}}{\Delta_{r}H^{0}}}\). Figure 7: System integration: heat of condensation is not recovered, and heat of evaporation is energy–free: \(\eta_{max}=1\). Introducing Equation (17) into Equation (15), the energy efficiency of the system becomes:  
\[0.55\leq\eta_{max}\leq 0.6 \tag{18}\]  
The previous equations also show that energy required for evaporation is, at least, 65% **of the available energy of reaction**! And the maximum energy efficiency of the integrated system \(\eta_{max}\) is 60% only! Let us now consider the case with a total recover of condensation heat and energy-free heat of evaporation. Then, the energy efficiency becomes (see Figure 9):  
\[\eta_{max}=\frac{Q_{max}^{-}}{Q_{min}^{+}}=\frac{\Delta_{r}H^{0}+Q_{cond}}{ \Delta_{r}H^{0}}=1+\frac{Q_{cond}}{\Delta_{r}H^{0}} \tag{19}\]  
The quantity of energy (per mole of salt) recoverable from condensation is evaluated by:  
\[Q_{cond}=\
u\times M_{H2O}\times L_{v} \tag{20}\]  
Starting from Equation (19) and introducing Equation (20), the energy efficiency of the system becomes:  
Figure 6:0  
The maximum theoretical energy efficiency of the system is about 1.8, whatever the salt!