## 2 Preliminary Design Based on Theoretical Calculation
### Thermal-Hydraulic Calculation of HE  
Heat transfer tubes are arranged as equilateral triangular layout in the present study, and the shell inner diameter is calculated by Equation (1) [34]. \[D_{s}=\left(1.1\sqrt{N_{t}}-1\right)\cdot s+3d_{o} \tag{1}\]  
where \(N_{t}\) is the number of tubes, \(s\) is the tube pitch for the equilateral triangular tube arrangement, and \(d_{o}\) is the out diameter of tubes. The heat transfer area of each (HEU) is given by:  
\[A_{j}=\frac{\dot{Q}_{j}}{k_{j}\Delta t_{\text{m},j}} \tag{2}\]  
Here, \(k_{j}\) represents the overall heat transfer coefficient of the \(j\)th HEU, \(\dot{Q}_{j}\) and \(\Delta t_{\text{m},j}\) are the heat transfer rate and the LMTD of the \(j\)th HEU, respectively. The overall heat transfer coefficient based on the external area of tubes is:  
\[k_{j}=\frac{\frac{d_{o}}{\frac{d_{o}}{h_{i,j}d_{i}}+\frac{d_{o}}{2\lambda}\ln \frac{d_{o}}{d_{i}}+\frac{1}{h_{i,j}}} \tag{3}\]  
where \(d_{i}\) and \(d_{o}\) are the inside and outside diameters of the tube, respectively; \(h_{i,j}\) and \(h_{o,j}\) are the heat transfer coefficients in the tube-side and shell-side, respectively; and \(\lambda\) is the heat conductivity of the tube-wall. \[d_{i}=d_{o}-2t \tag{4}\]  
\[t=\frac{Pd_{o}}{2([\sigma]E+PY)}+C_{1}+C_{2} \tag{5}\]  
where \(t\) is the tube thickness on the basis of national standard; \(P\) is the operating pressure of the tube-side; \([\sigma]\) is the allowable stress of the material under the operating temperature; \(E\) is the welding joint coefficient; \(Y\) is the pressure coefficient related to the material; \(C_{1}\) is the additional thickness due to thickness thinning; and \(C_{2}\) is the additional thickness due to corrosion. All parameters may find a value on the basis of a national standard. For the tube-side of each HEU, different correlations are proposed [35; 36]. The heat transfer coefficient is calculated by Gnielinski correlation [36] in this study, which is widely used in the heat transfer calculation of SCO\({}_{2}\), and listed as follows  
\[Nu_{t,j}=\frac{\frac{f_{j}}{8}\times(Re_{t,j}-1000)\times Pr_{j}}{12.7\sqrt{ \frac{f_{j}}{8}}\left(Pr_{t,j}^{2/3}-1\right)+1}\left[1+\left(\frac{d_{i}}{l_ {j}}\right)^{2/3}\right] \tag{6}\]  
The correlation is valuable in the range of 0.5 < \(Pr\) < 2000, 2300 < \(Re\) < 5 \(\times\) 10\({}^{5}\), where  
\[f_{j}=\left(1.81\times\lg Re_{j}-1.64\right)^{-2} \tag{7}\]  
For the shell-side of each unit, the heat transfer coefficient is calculated by the Graber-Rieger correlation [37], as follows:  
\[Nu_{s,j}=0.25+6.2x+(0.032x-0.007)P_{s,j}^{0.8-0.024x} \tag{8}\]  
Here, \(x\) (\(x=s/d_{o}\)) is the ratio of the tube pitch to tube outside diameter, and \(Pe_{j}\) is the Peclet number of each unit. \[Pe_{o,j}=Re_{o,j}Pr_{o,j} \tag{9}\]\[Re_{s,j}=\frac{\rho_{s,j}l_{s,j}D_{es}}{\mu_{s,j}} \tag{10}\]  
where \(d_{e}\) represents the hydraulic diameter of the shell side which can be expressed as follows:  
\[A_{j}=\frac{\dot{Q}_{j}}{k_{j}\Delta t_{\text{m},j}} \tag{2}\]0  
The pressure drop of the tube-side can be expressed as follows by neglecting the pressure drop for the elbow of single tube pass. \[A_{j}=\frac{\dot{Q}_{j}}{k_{j}\Delta t_{\text{m},j}} \tag{2}\]1  
where \(f_{t}\) is the tube-side friction factor, and \(l_{j}\) is the length of tubes of the \(j\)th unit. \(f_{t}\) is calculated by the Colebrook Equation [38] which is widely used in single-phase pressure drop calculation. \[\left\{ \begin{array}{l}f_{t,j}=\frac{64}{Re_{s,j}},Re\leq 2000\\ \frac{1}{\sqrt{f_{t,j}}}=1.74-2\lg\left(\frac{2\epsilon}{d_{i}}+\frac{18.7}{ Re_{s,j}\sqrt{f_{t,j}}}\right),Re>2000\end{array} \right. \tag{13}\]  
where \(\epsilon\) is the roughness of the tube-wall. The pressure drop of the shell-side can be calculated by the similar method such as the tube-side,  
\[\Delta P_{s}=\left(f_{s}\frac{l_{j}}{d_{e}}+1.5\right)\frac{\rho_{s,j}U_{s,j} ^{2}}{2} \tag{14}\]  
\(f_{s}\) is also calculated by the Colebrook correlation [38].