4.1. Upward co-current annular flow

Suppose that the gas and liquid phases flow upward in a vertical pipe of radius \(R~(=D/2)\) as shown in Fig. 4.1(a). The momentum balance in the two phases can be written as

(4.1)\[\begin{split}\begin{split} &- \alpha_{G} \left. \frac{dp}{dz} \right|_{TP} - \tau_{i} \frac{Pe_{i}}{A} - \alpha_{G} \rho_{G} g = 0 \\ &- \left. \frac{dp}{dz} \right|_{TP} A_{L} + \tau_{i} Pe_{i} - \rho_{L} g A_{L} - \tau_{W} Pe_{W} = 0 \end{split}\end{split}\]

where

(4.2)\[\begin{split}\begin{split} &A = \pi R^{2} \\ &A_{G} = \pi ( R - \delta )^{2} \\ &A_{L} = A - A_{G} \\ &Pe_{W} = 2 \pi R \\ &Pe_{i} = 2 \pi (R - \delta) = 2 \pi R \sqrt{\alpha_{G}} \end{split}\end{split}\]

and \(\delta\) is the mean liquid film thickness. We employ the following expressions for the interfacial and wall frictions:

(4.3)\[\begin{split}\begin{split} &\tau_{i} = f_{i} \frac{\rho_{G}}{2} (u_{G} - u_{L})^{2} \\ &\tau_{W} = f_{W} \frac{\rho_{L}}{2} u_{L}^{2} \end{split}\end{split}\]

Thus,

(4.4)\[\begin{split}\begin{split} &- \alpha_{G} \left. \frac{dp}{dz} \right|_{TP} - f_{i} \frac{\rho_{G}}{2} (u_{G} - u_{L})^{2} \frac{2 \pi R \sqrt{\alpha_{G}}}{A} - \alpha_{G} \rho_{G} g = 0 \\ &- \alpha_{L} \left. \frac{dp}{dz} \right|_{TP} + f_{i} \frac{\rho_{G}}{2} (u_{G} - u_{L})^{2} \frac{2 \pi R \sqrt{\alpha_{G}}}{A} - \alpha_{L} \rho_{L} g - f_{W} \frac{\rho_{L}}{2} u_{L}^{2} \frac{Pe_{W}}{A} = 0 \end{split}\end{split}\]

For a given set of the volume flow rates,

(4.5)\[\begin{split}\begin{split} &u_{G} = \frac{Q_{G}}{A_{G}} \\ &u_{L} = \frac{Q_{L}}{A_{L}} \\ \end{split}\end{split}\]

Therefore, we can obtain \(- \left. dp/dz \right|_{TP}\) and \(\alpha_{G}\) by solving the momentum equations, provided that the friction factors are given.

Fig. 4.1 Simple model of annular flow.

Being similar to single-phase flows, the wall friction factor is often given by the following functional form:

(4.6)\[ f_{W} = \frac{a}{Re_{L}^{n}}\]

where \(Re_{L}\) is the liquid Reynolds number defined by

(4.7)\[ Re_{L} = \frac{\rho_{L} j_{L} D}{\mu_{L}}\]

For example, Wallis [Wal70] used the following expression:

(4.8)\[\begin{split} f_{W} = \left\{ \begin{array}{ll} \frac{16}{Re_{L}} &\text{for}~Re_{L} \le 2300 \\ \frac{0.079}{Re_{L}} & \text{otherwise} \end{array} \right.\end{split}\]

The interfacial friction factor is considered to be a function of \(\delta\), e.g., [Wal69]

(4.9)\[ f_{i} = 0.005 \left( 1 + 150 \frac{\delta}{R} \right)\]

When \(\delta \ll R\), \(\alpha \sim 1 - 2 \delta / R\); therefore,

(4.10)\[ f_{i} = 0.005 \left( 1 + 75 (1 - \alpha_{G}) \right)\]

From Eq. (4.1),

(4.11)\[ \tau_{i} = \frac{R}{2} \sqrt{\alpha_{G}} \left( - \left. \frac{dp}{dz} \right|_{TP} - \rho_{G} g \right)\]

By summing the two in Eq. (4.1), we obtain the global balance equation:

(4.12)\[ \tau_{W} = \frac{R}{2} \left( - \left. \frac{dp}{dz} \right|_{TP} - \rho_{m} g \right)\]

where \(\rho_{m} = \alpha_{G} \rho_{G} + \alpha_{L} \rho_{L}\). Eliminating the pressure drop yields

(4.13)\[ \tau_{i} = \sqrt{\alpha_{G}} \tau_{W} + \frac{R}{2} \sqrt{\alpha_{G}} \left( \rho_{m} - \rho_{G} \right) g\]

or

(4.14)\[ \tau_{i} = \sqrt{\alpha_{G}} \tau_{W} + \frac{R}{2} \sqrt{\alpha_{G}} \alpha_{L} \Delta \rho g\]

where \(\Delta \rho = \rho_{L} - \rho_{G}\). The liquid film is, thus, suspended by the interfacial friction. Using (4.3) and evaluating \(\tau_{W} \sim \mu_{L} u_{L} / \delta\) give

(4.15)\[ f_{i} \frac{\rho_{G}}{2} (u_{G} - u_{L})^{2} = \sqrt{\alpha_{G}} \mu_{L} \frac{u_{L}}{\delta} + \frac{R}{2} \sqrt{\alpha_{G}} \alpha_{L} \Delta \rho g\]

(4.16)\[ u_{G} - j = \left[ \frac{2 D \sqrt{\alpha_{G}}}{\alpha_{L} \delta f_{i}} \left( 1 + \alpha_{L}^{2} \frac{\Delta \rho g D \delta}{4 \mu_{L} j_{L}} \right) \right]^{1/2} \left( \frac{\mu_{L} j_{L}}{\rho_{G} D} \right)^{1/2} + ( u_{L} - j )\]

This can be the basis of the drift velocity as shown in Table 2.1.