# Increasing Compressed Gas Energy Storage Density Using CO\({}_{2}\)-N\({}_{2}\) Gas Mixture
## 4 Model  
A schematic of the model is shown in Figure 4. A vessel is initially charged, to a known initial pressure, with a CO2/N2 mixture to some known mole fraction. A liquid, mineral oil in which both N2 and CO2 are immiscible, is pumped into the vessels at a constant flow rate. As the liquid column rises inside the vessel, the gaseous mixture is compressed, and its pressure rises. When the partial pressure of CO2 equals, and subsequently exceeds its saturation pressure, CO2 starts to condense, forming liquid condensate on top of the compression liquid column. Compression is terminated when the volume of the gaseous mixture reaches less than 5% of the total volume of the vessel or until the pressure reaches 140 bar. Figure 3: Within the same pressure limits, a non-condensable gas system (**a**) will utilize less compression liquid volume than a system with condensable gas (**b**). For simplicity, the compression process is assumed to be isothermal. This is justified because energy is stored incrementally at a slow rate. This allows sufficient time for heat to flow through the conductive walls of the pressure vessel to sustain isothermal compression. Furthermore, the analysis being presented in this paper aims to find a bounded solution for the maximum possible increase in energy density that results from using the CO\({}_{2}\)/N\({}_{2}\) mixture instead of air. The CO\({}_{2}\) and N\({}_{2}\) are modeled as real gases using the Redlich-Kwong equation of state. At any time, the volume of the compression liquid in the vessels is:  
\[V_{CL}=Q\ \Delta t \tag{2}\]  
And the volume of the gaseous mixture is:  
\[V_{G}=V_{V}-V_{CL}-V_{CO_{2}(L)} \tag{3}\]  
The change in the volume of the gaseous mixture in the vessel causes the pressure to rise in the vessel. The relationship between the pressure and the volume of the compressed gaseous mixture is modeled by the Redlich-Kwong equation of state:  
\[P=\frac{RT}{V_{m}-b}-\frac{a}{\sqrt{T}\ V_{m}(V_{m}+b)} \tag{4}\]  
where:  
\[a=0.427480\ \frac{R^{2}T_{c}^{2}}{P_{c}} \tag{5}\]  
\[b=0.086640\ \frac{RT_{c}}{P_{c}} \tag{6}\]  
\[V_{m}=\frac{V_{G}}{N_{G}} \tag{7}\]  
For a mixture of gases, \(a\) and \(b\) in Equation (4) are replaced by \(a_{G}\) and \(b_{G}\) that are calculated from each species' individual \(a\) and \(b\) as follows:  
\[a_{ij}=\sqrt{a_{i}a_{j}} \tag{8}\]  
\[a_{G}=\sum\limits_{i}\sum\limits_{j}y_{i}y_{j}a_{ij} \tag{9}\]  
Figure 3:0  
Figure 3:1 Schematic of the modeled system. When the partial pressure of CO2 reaches its saturation pressure at the tank temperature (which is constant), CO2 condenses and its mole fraction in the gaseous mixture becomes:  
Figure 3:2  
The number of moles of gaseous CO2 is then:  
Figure 3:3  
The number of moles of CO2 in liquid phase is calculated by applying the conservation of mass:  
Figure 3:4  
When condensation is taking place, \(N_{CO_{2}(G)}\) and \(N_{CO_{2}(L)}\) are both changing with time. At the beginning of each time step, the volume (or molar volume) of the gas mixture and its pressure are unknown. To solve for them, the solution must be iterative which is computationally more expensive that an explicit solution. Besides, an iterative solution is further complicated by the discontinuity in the solution that is caused by phase change. Instead of driving the solution procedure by marching in time and starting from the change of volume, a computationally simpler approach is to solve for volume by marching in pressure. This methodology leads to a more tractable and explicit solution. The steps are detailed below and schematically in Figure 3:5  
Figure 3:6 Solution scheme for one step in pressure. We increment the total pressure every step. Because the total pressure is known, a determination is made as to whether condensation is occurring or not. If no condensation is occurring, then the total number of moles in the gaseous phase is known and equals its initial value. Equation (4) can then be solved for molar volume. A direct solution of Equation (4) is cumbersome and requires a trial and error procedure.