## 3 Results and Discussion
### Calculating the CO\({}_{2}\) Uptake in a Concrete Structure  
Applicable factors are considered when calculating the CO\({}_{2}\) uptake of a concrete structure. The calculations were supposed to provide the carbonation depth at a given time. It is also assumed that the maximum amount of CaO that can be carbonated in OPC is approximately 75% [19; 20; 21]. If the data for the cement paste or cement used are known, the amount of CO\({}_{2}\) uptake is calculated using Equation (18). \[\text{CO}_{2}\text{ uptake}=0.75\times\text{C}\times\text{CaO}\times(\text{MCO} _{2}/\text{MCaO})\times(\text{K}\surd(\text{time in years})\text{ m}\]  
where C = amount of OPC in the concrete kg per m\({}^{3}\). CaO = amount of CaO in cement (wt.%). \(\text{MCO}_{2}\) = Molar weight of CO\({}_{2}\). MCaO = Molar weight of CaO  
### Calculating the CO\({}_{2}\) Uptake in a Concrete Structure  
Calculate the 50-years amount of CO\({}_{2}\) captured by a block of concrete slab (binder is OPC with 10 wt.% \(\gamma\)-C\({}_{2}\)S), erected with the following dimensions, 2 m wide, 3 m height, and 0.2 m thickness, standing in the open (un-sheltered). Using Equation (18), the estimated carbonation depth was calculated from the corrected k values in Tables 6-8, while also considering that k\({}_{1}\) is a combined correction factor for the strength and environment:  
K = k\({}_{1}\) (exposed, (Table 6)) \(\times\) k\({}_{2}\) (outdoors, (Table 7)) \(\times\) k\({}_{3}\) (type of cement, (Table 8)). The values of k\({}_{1}\), k\({}_{2}\), and k\({}_{3}\) were k\({}_{1}\) = 5 mm\(\sqrt{t}\)(years), k\({}_{2}\) = 0.9, and k\({}_{3}\) = 1.05. Then, the carbonation depth for 50 years, x1 = (5 mm/\(\sqrt{50}\) years) \(\times\) (0.9) \(\times\) (1.05). The carbonation depth = 33.41 mm in 50 years. The carbonation depth on both sides of the slab was 33.41 \(\times\) 2 = 66.82 mm. \begin{table}
\begin{tabular}{c c c} \hline \hline
**No** & **Surface Class** & **Correction Factor** \\ \hline
1 & Indoor house concrete & k \(\times\) 0.7 \\ \hline
2 & Outdoors house concrete & k \(\times\) 0.9 \\ \hline
3 & Infrastructure concrete when painted & k \(\times\) 1.0 \\ \hline \hline \end{tabular}
\end{table} 
Table 7: Correction factors for surface treatment. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Amount** & **\textless{}10** & **10–20** & **20–30** & **30–40** & **40–60** & **60–80** \\
**wt.\%** & & & & & \\ \hline Limestone & & K \(\times\) 1.05 & K \(\times\) 1.10 & & & \\ \hline Silica fume & K \(\times\) 1.05 & K \(\times\) 1.10 & & & & \\ \hline Fly ash & & K \(\times\) 1.05 & & K \(\times\) 1.10 & & \\ \hline GBFS & 1.05 & K \(\times\) 1.10 & K \(\times\) 1.15 & K \(\times\) 1.20 & K \(\times\) 1.25 & K \(\times\) 1.30 \\ \hline \hline \end{tabular}
\end{table} 
Table 8: Suggested correction factors dependent on the type of cement. Surface area of slab = 2 m \(\times\) 3 m = 6 m\({}^{2}\). Therefore, the carbonated volume after 50 years = (66.82 mm \(\times\) 1 m/1000 mm) \(\times\) 6 m\({}^{2}\) = 0.4 m\({}^{3}\) of carbonated concrete was used. In mixing 0.4 m\({}^{3}\) of concrete, using the mixing ratios of cement mortar,  
Cement: sand: water = 100:300:50. Amount of cement in 0.4 m concrete = 100/(100 + 300 + 50)  
0.4 \(\times\) 100/450 = 0.089 m\({}^{3}\) cement. The specific gravity of OPC (Table 1) was 3.15; therefore, the mass of the OPC mixture was 0.089 m\({}^{3}\)\(\times\) 3150 Kg/m\({}^{3}\) = 280.35 kg of cement used (and carbonated).