## 3 Modeling
### Slug Flow Pressure Model in Inclined Pipe  
The pressure drop of the slug unit can be calculated by the Section 3.1 slug unit pressure model, which is crucial for the slug flow pressure in the pipeline. The slug flow in the pipe is formed by the movement of the slug unit at different times. By calculating the pressure drop of the slug unit at different times, the movement state of the slug flow in a pipe can be divided, and the slug flow pressure model is established. The liquid slug movement in the inclined pipe is roughly divided into four stages, as presented in Figure 5.  
Figure 4: Pressure drop of slug flow unit. At \(t_{1}\), the liquid slug area of the slug unit at the elbow just enters the upward inclined pipe section, and there is no liquid backflow in the inclined pipe as presented in Figure 5a. The slug head speed \(V_{\rm s}\) is obtained as follows. \[V_{s}=V_{\rm LS}+V_{\rm GS} \tag{37}\]  
where \(V_{\rm LS}\) and \(V_{\rm GS}\) are the superficial velocities of liquid and gas phases, respectively. According to the mass conservation of the liquid phase, the liquid volume in the liquid film area is \(\rm AV_{\rm LS}/\it f\), which can be converted into the length of the liquid film area \(\rm AV_{\rm LS}/\it f_{\rm s}\)). The length of the liquid film-free area is calculated as follows. \[L_{\rm g}=\frac{L_{\rm s}V_{\rm s}-\rm AV_{\rm LS}V_{\rm t}/\it f\alpha_{\rm s} }{V_{\rm t}-V_{\rm s}} \tag{38}\]  
The number of slug in the inclined pipe is obtained,  
\[\rm n_{1}=\left[\frac{L-L_{g}}{L_{\rm u}}\right] \tag{39}\]  
where [] signifies rounding. \begin{table}
\begin{tabular}{c c} \hline \hline
**Parameter** & **Expression** \\ \hline Gas phase wetting perimeter \(S_{\rm G}\) & \(S_{\rm G}=\pi\rm D-S_{\rm L}\left(m\right)\) \\ Liquid phase wetting perimeter \(S_{\rm L}\) & \(S_{\rm L}=\arccos(1-2h_{\rm L}/\rm D)/\rm D,\left(m\right)\) \\ Cross sectional area of pipe A & \(\rm A=\pi\rm D^{2}/4\), \(\rm(m^{2})\) \\ Gas phase cross-sectional area \(A_{\rm G}\) & \(A_{\rm G}=\pi(1-a_{\rm L})^{2}/4\), \(\rm(m^{2})\) \\ Liquid phase cross-sectional area \(A_{\rm L}\) & \(A_{\rm L}=\pi\rm A_{\rm L}\rm D^{2}/4\), \(\rm(m^{2})\) \\ Shear force of liquid film and wall \(\tau_{\rm f}\) & \(\tau_{\rm f}=f_{\rm f}\rho_{\rm f}V_{\rm t}^{2}/2\) \\ Shear force of gas phase and wall \(\tau_{\rm G}\) & \(\tau_{\rm G}=f_{\rm G}\rho_{\rm G}V_{\rm G}^{2}/2\) \\ Shear stress in the liquid slug area \(\tau_{\rm s}\) & \(\tau_{\rm s}=f_{\rm g}\rho_{\rm s}V_{\rm s}^{2}/2\) \\ Friction coefficient of liquid film and wall \(f_{\rm L}\) & \(f_{\rm f}=\zeta_{\rm L}\left(D_{\rm L}\left(V_{\rm f}\right)/m\right)^{-n}\) \\ Friction coefficient of gas phase and wall \(f_{\rm G}\) & \(f_{\rm G}=C_{\rm G}\left(D_{\rm G}V_{\rm f}/f_{\rm G}\right)^{-m}\) \\ Friction factor between gas and phase in liquid slug area \(f_{\rm s}\) & \(f_{\rm s}=C_{\rm G}\left(D\zeta_{\rm s}/\it f_{\rm s}\right)^{-m}\) \\ Hydraulic diameter of liquid phase \(\rm D_{\rm L}\) & \(D_{\rm L}=4A_{\rm L}/S_{\rm L}\) \\ \hline \end{tabular} For laminar flow, \(\rm C_{\rm L}\), \(\rm C_{\rm G}\), \(m\), and \(n\) are 1. For turbulent flow, \(\rm C_{\rm L}\) and \(\rm C_{\rm G}\) are 0.046, \(m\) and \(n\) are 0.2.  
\end{table} 
Table 2: Parameter calculation method. Ignoring the pressure drop caused by the gas in the pipeline in the liquid film-free area, the pressure at point A at \(t_{1}\) is obtained. \[p_{A}^{t_{1}}=p_{1}+(n_{1}+1)Dp_{\rm u}+\big{(}L-(n_{1}+1)L_{\rm u}-L_{\rm g}+L_ {\rm f}\big{)}\frac{Dp_{\rm f}}{L_{\rm f}} \tag{40}\]  
At \(t_{2}\), the liquid slug area at the elbow completely enters the upward inclined pipeline, and part of the liquid film area also enters this inclined passage. The first slug in the inclined pipe is about to dissipate at the top of the pipe as shown in Figure 5b. Comparing the position of the liquid slug at \(t_{1}\) and \(t_{2}\), the expression for \(t_{2}\) is:  
\[t_{2}=t_{1}+(L_{\rm g}+L_{\rm s})/V_{t} \tag{41}\]  
At this time, the number of slugs in the inclined pipe is obtained.