3.3. Frictional pressure drop: Lockhart-Martinelli correlation

3.3.1. Separate flow model

Following Lockhart and Martinelli [LM49], we assume that the two-phase pressure drop can be expressed by

(3.17)\[\begin{split} - \left. \frac{dp}{dz} \right|_{TP} = \lambda_{k} \frac{\rho_{k} v_{k}^{2}}{2D} \end{split}\]

where the friction factors of both phases are given in the Blasius form:

(3.18)\[ \lambda_{k} = C_{k} Re_{k}^{n_{k}} \]

The Reynolds number of the phase \(k\) is defined by

(3.19)\[ Re_{k} = \frac{\rho_{k} j_{k} D}{\mu_{k}}\]

If the phase \(k\) flows alone at the mass flux \(G_{k} = \alpha_{k} \rho_{k} v_{k} = \rho_{k} j_{k}\), the pressure drop may be expressed as

(3.20)\[\begin{split} - \left. \frac{dp}{dz} \right|_{k} = \lambda_{k} \frac{\rho_{k} j_{k}^{2}}{2D} \end{split}\]

since the mean velocity is given by \(G_{k} / \rho_{k} = j_{k}\).

Note

\(- | dp/dz |_{L}\) and \(- | dp/dz |_{L0}\) are different quantities!

The square root of the ratio of the liquid-phase pressure drop to the gas-phase pressure drop is the Lockhart-Martinelli parameter:

(3.21)\[ X^{2} = \frac{\left. dp/dz \right|_{L}}{\left. dp/dz \right|_{G}} = \frac{\lambda_{L} \rho_{L} j_{L}^{2}/2D}{\lambda_{G} \rho_{G} j_{G}^{2}/2D} = \frac{C_{L}}{C_{G}} \frac{\mu_{G}^{n_{G}}}{\mu_{L}^{n_{L}}} \frac{\rho_{L}^{1+n_{L}}}{\rho_{G}^{1+n_{G}}} \frac{j_{L}^{2+n_{L}}}{j_{G}^{2+n_{G}}} \frac{D^{n_{L}}}{D^{n_{G}}}\]

The last expression seems complex, but if we take \(n_{L} = n_{G} = 0\) and \(C_{L} = C_{G}\) under an assumption that both phases are in fully turbulent conditions we obtain

(3.22)\[ X^{2} = \frac{\rho_{L} j_{L}^{2}}{\rho_{G} j_{G}^{2}} = \frac{\rho_{G} G_{L}}{\rho_{L} G_{G}} = \frac{\rho_{G}}{\rho_{L}} \frac{1 - x}{x}\]

Writing the two-phase pressure drop as a product of the pressure drop of the phase \(k\) and a two-phase multiplier \(\phi_{k}\), we have

(3.23)\[\begin{split} - \left. \frac{dp}{dz} \right|_{TP} = - \left. \frac{dp}{dz} \right|_{k} \phi_{k}^{2} \end{split}\]

Therefore,

(3.24)\[ \frac{\phi_{G}^{2}}{\phi_{L}^{2}} = \frac{\left. dp/dz \right|_{L}}{\left. dp/dz \right|_{G}} = X^{2}\]

By obtaining \(\phi\) as a function of \(X\), we can calculate the two-phase pressure drop since \(X\) is determined by the flow condition.

3.3.2. Theoretical basis

Chisholm [Chi67] gave a theoretical basis for an empirical fit of \(\phi (X)\) as follows. The force balance in each phase is given by

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

where \(Pe_{k}\) is the perimeter of the phase \(k\) (\(Pe_{G} + Pe_{L} = \pi D\)), \(Pe_{i}\) is the perimeter of the gas-liquid interface, \(A_{k}\) is the cross sectional area of the phase \(k\), \(\tau_{k}\) is the wall shear stress at the contact between the phase \(k\) and the pipe wall, and \(\tau_{i}\) is the interfacial shear stress.

Summing the two equations yields

\(- \left. \frac{dP}{dz} \right|_{TP} = \tau_{G} \frac{Pe_{G}}{A} + \tau_{L} \frac{Pe_{L}}{A} = \frac{4 \tau_{G}}{4A/Pe_{G}} + \frac{4 \tau_{L}}{4A/Pe_{L}} = \frac{4 \tau_{G}}{D_{G}} + \frac{4 \tau_{L}}{D_{L}}\)

where \(D_{k}\) is the hydraulic equivalent diameter.

Factorizing the pressure gradient and interfacial friction terms gives

(3.26)\[\begin{split}\begin{split} &- \left. \frac{dp}{dz} \right|_{TP} \left\{ 1 - \frac{\tau_{i} Pe_{i}}{- A_{G} \left. dp/dz \right|_{TP} } \right\} = \frac{\tau_{G} Pe_{G}}{A_{G}} \\ &- \left. \frac{dp}{dz} \right|_{TP} \left\{ 1 + \frac{\tau_{i} Pe_{i}}{- A_{L} \left. dp/dz \right|_{TP}} \right\} = \frac{\tau_{L} Pe_{L}}{A_{L}} \end{split}\end{split}\]

The ratio of the interfacial friction to the pressure drop is denoted by

(3.27)\[ S_{R} = \frac{\tau_{i} Pe_{i}}{- A_{G} \left. dp/dz \right|_{TP}}\]

and the wall shear stresses are given by the following constitutive equations:

(3.28)\[ \tau_{k} = \frac{f_{k} \rho_{k} v_{k}^{2}}{2}\]

where \(f_{k}\) is Fanning’s friction coefficient for the phase \(k\). Therefore,

(3.29)\[\begin{split}\begin{split} &- \left. \frac{dp}{dz} \right|_{TP} \left\{ 1 - S_{R} \right\} = f_{G} \frac{Pe_{G}}{A_{G}} \frac{\rho_{G} v_{G}^{2}}{2} \\ &- \left. \frac{dp}{dz} \right|_{TP} \left\{ 1 + \frac{A_{G}}{A_{L}} S_{R} \right\} = f_{L} \frac{Pe_{L}}{A_{L}} \frac{\rho_{L} v_{L}^{2}}{2} \end{split}\end{split}\]

Dividing the second equation with the first one and using the definition

(3.30)\[ Z^{2} = \frac{1 + S_{R} A_{G} / A_{L}}{1 - S_{R}}\]

yield

(3.31)\[ Z^{2} = \frac{f_{L}}{f_{G}} \frac{Pe_{L} A_{G}}{Pe_{G} A_{L}} \frac{\rho_{L} v_{L}^{2}}{\rho_{G} v_{G}^{2}}\]

Therefore, the velocity ratio is given by

(3.32)\[ K = \frac{v_{G}}{v_{L}} = \frac{1}{Z} \sqrt{ \frac{f_{L}}{f_{G}} \frac{Pe_{L} A_{G}}{Pe_{G} A_{L}} \frac{\rho_{L}}{\rho_{G}} }\]

For the cases in which the two phases flow alone, the liquid pressure drop can be written as

(3.33)\[ - \left. \frac{dp}{dz} \right|_{L} = Pe_{L}' f_{L}' \frac{\rho_{L} j_{L}^{2}}{2} = Pe_{L}' f'_{L} \frac{A_{L}^{2}}{A^{2}} \frac{\rho_{L} v_{L}^{2}}{2}\]

where \(f'_{L}\) is the friction factor for the liquid phase flowing alone and \(D = 4 A_{L} / Pe'\). The two-phase multiplier is calculated from this equation and Eq. (3.29) as

(3.34)\[ \phi_{L}^{2} = \frac{\left. dp/dz \right|_{TP}}{\left. dp/dz \right|_{L}} = \frac{( 1 + A_{G}/A_{L} )^{2}}{ 1 + S_{R} A_{G} / A_{L} } \frac{f_{L}}{f'_{L}} \frac{Pe_{L}}{Pe'} \]

Rearranging the first factor in the third equation yields

(3.35)\[ \phi_{L}^{2} = \frac{f_{L}}{f'_{L}} \frac{Pe_{L}}{Pe'} \left(1 + \frac{A_{G}}{A_{L}} \right) \left( 1 + \frac{A_{G}}{A_{L} Z^{2}} \right)\]

Consider the limiting case where both phases are in turbulent conditions; the friction factors are the same constant and the phase distributions are uniform, and there is no velocity slip, yielding

(3.36)\[ Z^{2} = \frac{\rho_{L}}{\rho_{G}}\]

Using Eq. (3.22), we obtain

(3.37)\[ \frac{X}{Z} = \frac{1 - x}{x} = \frac{A_{L}}{A_{G}} \]

Thus,

(3.38)\[ \phi_{L}^{2} = 1 + \frac{C}{X} + \frac{1}{X^{2}}\]

where the coefficient \(C\) depends on the flow state as shown in Table 3.1, and Fig. 3.2 shows the model for each flow conditions.

Table 3.1 Chisholm parameter

Liquid/Gas	Turbulent/Turbulent	Laminar/Turbulent	Turbulent/Laminar	Laminar/Laminar
\(C\)	20	12	10	5

Fig. 3.2 Chisholm model for two-phase multiplier. solid line: \(\phi_{L}\), broken line: \(\phi_{G}\).