## 4 CO\({}_{2}\) Storage and Utilization  
To obtain a rough estimate for the amount of CO\({}_{2}\) that can be stored and utilized, we hypothetically consider a steam power plant where the steam at 540 \({}^{\circ}\)C (813 K) and 10 MPa expands through a turbine to become a saturated-vapor steam at 0.008 MPa (saturated temperature is 41.5 \({}^{\circ}\)C (314.5 K)). The steam discharged from the turbine enters a condenser and is condensed to a saturated liquid by a stream of CO\({}_{2}\) as the heat-removing fluid. Neglecting the turbine efficiency, the steam mass flow rate, \(\dot{m}_{st}\), is theoretically calculated from the steam work, \(\dot{W}_{t}\), as  
\[\dot{m}_{st}=\frac{\dot{W}_{t}}{(h_{st,in}-h_{st,out})} \tag{13}\]  
where \(h_{st,in}\) and \(h_{st,out}\) are the specific enthalpy of the steam at the turbine inlet and exit, respectively. The equation for the energy rate with isothermal heat released during the steam condensation is given as  
\[\dot{Q}_{st}=\dot{m}_{st}h_{fg}=\frac{\dot{W}_{t}h_{fg}}{(h_{st,in}-h_{st,out})} \tag{14}\]  
\begin{table}
\begin{tabular}{c c c c c} \hline
**Pstor** & **(MPa)** & \(\dot{m}_{f}\)**\(\alpha_{p}\)/\(\xi\)-K)** & (\(\lambda_{f}\))/(\(\dot{m}_{f}\)\(\alpha_{p}\)/(1/m)** \\ \hline  & Air & CO\({}_{2}\) & Air & CO\({}_{2}\) \\ \hline
1 & 101,884 & 107,702 & \(2.635\times 10^{-7}\) & \(1.55\times 10^{-7}\) \\
2 & 148,485 & 157,302 & \(1.797\times 10^{-7}\) & \(1.00\times 10^{-7}\) \\
3 & 183,906 & 195,623 & \(1.443\times 10^{-7}\) & \(7.68\times 10^{-8}\) \\
4 & 213,887 & 228,415 & \(1.234\times 10^{-7}\) & \(6.24\times 10^{-8}\) \\
5 & 240,350 & 258,122 & \(1.095\times 10^{-7}\) & \(5.26\times 10^{-8}\) \\ \hline \end{tabular}
\end{table}
Table 2: Heat capacity and exponential power of CO\({}_{2}\) and air for Equations (10) and (11) (cooling device at 0.1 MPa).  
\begin{table}
\begin{tabular}{c c c c} \hline
**Pstor** & \(\xi\)**47\%** & \multicolumn{2}{c}{**Condenser Area (WL\(\xi\)47\%)**} \\ \hline (MPa) & Air & CO\({}_{2}\) & Air & CO\({}_{2}\) \\ \hline
1 & 1 & 0.82 & WL & 0.82 WL \\
2 & 0.77 & 0.49 & 0.77 WL & 0.49 WL \\
3 & 0.63 & 0.37 & 0.63 WL & 0.37 WL \\
4 & 0.54 & 0.30 & 0.54 WL & 0.30 WL \\
5 & 0.51 & 0.27 & 0.51 WL & 0.27 WL \\ \hline \end{tabular}
\end{table}
Table 3: Condenser length, \(\xi\)_47\%_, and condenser area, WL\(\xi\)_47\%_, required for 47% of vapor content to be condensed.  
where \(h_{fg}\) is the latent heat of condensation of the steam. The heat absorbed by the cold steam of CO\({}_{2}\) is  
\[\dot{Q}_{CO2}=\dot{m}_{CO2}c_{p,CO2}(T_{CO2,in}-T_{CO2,out}) \tag{15}\]  
where \(\dot{m}_{CO2}\) is the mass flow rate for CO\({}_{2}\), \(c_{p,CO2}\) is the CO\({}_{2}\) specific heat, and \(T_{CO2,in}\) and \(T_{CO2,out}\) are the CO\({}_{2}\) temperatures at the condenser inlet and outlet, respectively. It is required that the energy released during the condensation must be absorbed by the CO\({}_{2}\); the CO\({}_{2}\) mass flow rate is calculated as  
\[\dot{m}_{CO2}=\frac{\dot{W}_{t}h_{fg}}{c_{p,CO2}(h_{st,in}-h_{st,out})(T_{CO2,in }-T_{CO2,out})} \tag{16}\]  
As an example, we use the condition of the steam given above (\(h_{st,in}\) = 3475.5 kJ/kg, \(h_{st,out}\) = 2577 kJ/kg, \(h_{fg}\) = 2403 kJ/kg). For CO\({}_{2}\) at 0.1 MPa and \(T_{CO2,in}\) = 278 K, the specific heat \(c_{p,CO2}\) = 0.85 kJ/kg-K, and assuming that \(T_{CO2,out}\) is equal to the steam saturated temperature at 0.008 MPa, \(T_{CO2,out}\) = 314.5 K. For a power plant of \(W_{t}\) = 1000 MW, the mass flow rate of the CO\({}_{2}\) is roughly calculated as 86,203 kg/s (744,7965 tons/day). It is noted that a large coal-fired power plant of 1000 MW generates approximately 30,000 tons of CO\({}_{2}\) per day [9]; thus, the amount of CO\({}_{2}\) emitted from such a large coal-power plant for 250 days can be stored and utilized.