# Homogeneous Flow Model for Two-Phase Flow

The central assumption of the homogeneous flow model is that the two phases travel at equal velocities and mix well; therefore, they can be treated as if there is only one phase. This model works better for two-phase flow near the critical point, where the differences between the properties of the liquid and vapor are insignificant, or when the mass velocity of the two-phase flow is very high so that the flow regime is either bubbly or misty flow. In arriving at the homogeneous model for two-phase flow, area averaging is performed for both phases and the resultant governing equations become one-dimensional in nature. The density of the homogeneous mixture satisfies the relation $\frac{1}{\rho }=\frac{x}{{{\rho }_{v}}}+\frac{1-x}{{{\rho }_{\ell }}}\qquad\qquad(1)$

i.e., $\rho =\frac{{{\rho }_{v}}{{\rho }_{\ell }}}{{{\rho }_{\ell }}x+{{\rho }_{v}}(1-x)}\qquad\qquad(2)$

The mass flow rates of the liquid and vapor phases defined as (see Concepts and Notations for Two-Phase Flow) ${{\dot{m}}_{\ell }}={{\rho }_{\ell }}w{{A}_{\ell }}\qquad\qquad(3)$ ${{\dot{m}}_{v}}={{\rho }_{v}}w{{A}_{v}}\qquad\qquad(4)$

Substituting eqs. (3) and (4) into the from definition of void fraction $\alpha =\frac{\Delta z\int_{{{A}_{v}}}{dA}}{\Delta z\int_{A}{dA}}=\frac{{{A}_{v}}}{{{A}_{v}}+{{A}_{\ell }}}$

yields $\alpha =\frac{{{{\dot{m}}}_{v}}/{{\rho }_{v}}}{{{{\dot{m}}}_{v}}/{{\rho }_{v}}+{{{\dot{m}}}_{\ell }}/{{\rho }_{\ell }}}\qquad\qquad(5)$

Considering the definition of quality: $x=\frac{{{{\dot{m}}}_{v}}}{{{{\dot{m}}}_{\ell }}+{{{\dot{m}}}_{v}}}$

the void fraction becomes $\alpha =\frac{x}{x+(1-x){{\rho }_{v}}/{{\rho }_{\ell }}}\qquad\qquad(6)$

The total mass flux in the channel becomes $\dot{{m}''}=\frac{{{{\dot{m}}}_{\ell }}+{{{\dot{m}}}_{v}}}{A}=\frac{{{\rho }_{\ell }}w{{A}_{\ell }}+{{\rho }_{v}}w{{A}_{v}}}{A}=\rho w\qquad\qquad(7)$

The governing equations for the homogeneous model include continuity, momentum and energy equations, which are listed below: $A\frac{\partial \rho }{\partial t}+\frac{\partial }{\partial z}(\dot{{m}''}A)=0\qquad\qquad(8)$ $A\frac{\partial \dot{{m}''}}{\partial t}+\frac{\partial ({{{\dot{{m}''}}}^{2}}A/\rho )}{\partial z}=-\frac{\partial (pA)}{\partial z}-\rho g\cos \theta A-{{\tau }_{w}}P\qquad\qquad(9)$ $\frac{\partial }{\partial t}\left[ \rho \left( h+\frac{{{w}^{2}}}{2}+gz\cos \theta \right) \right]+\frac{1}{A}\frac{\partial }{\partial z}\left[ \rho wA\left( h+\frac{{{w}^{2}}}{2}+gz\cos \theta \right) \right]=\frac{P}{A}{{{q}''}_{w}}+{q}'''+\frac{\partial p}{\partial t}\qquad\qquad(10)$

where the angle brackets, “ $\left\langle {} \right\rangle$,” which represent the area average, have been dropped for ease of notation. P is perimeter, p is pressure, q''w is heat flux at the wall, and q''' is internal heat generation per unit volume within the fluid.

Substituting eq. (7) into eq. (10), the energy equation becomes $\frac{ \partial (\rho h) } {\partial t} + \frac{1}{A} \frac{\partial} {\partial z} (\dot{{m}''} A h) = \frac{P}{A} {{q''}_w} + q''' - \frac{1}{A} \frac{\partial }{ \partial z } (\frac { {{\dot{{m}''}}^3} A} { 2 {{\rho}^2} })- g \dot{{m}''} \cos \theta - \frac{\partial}{\partial t} (\frac{ {\dot{{m}''}}^2 } {2 \rho }) + \frac {\partial p} {\partial t} \qquad\qquad(11)$

For steady-state two-phase flow in a circular tube with constant cross-sectional area, the momentum eq. (9) reduces to $- \frac{{dp}}{{dz}} = \frac{{4{\tau _w}}}{D} + \frac{{\partial ({{\dot{{m}''}}^2}/\rho )}}{{\partial z}} + \rho g\cos \theta \qquad\qquad(12)$

The three terms on the right-hand side of eq. (12) represent pressure drops due to friction, dpF / dz, acceleration, dpa / dz, and gravity, dpg / dz. Thus, eq. (12) can be rewritten as $-\frac{dp}{dz}=-\frac{d{p_F}}{dz}-\frac{d{p_a}}{dz}-\frac{d{p_g}}{dz}\qquad\qquad(13)$

where the pressure drop due to friction, dpF / dz, must be obtained using appropriate correlations.

The energy equation (including mechanical and thermal energy) for steady-state two-phase flow in a circular tube with constant cross-sectional area can be obtained by simplifying eq. (11), $\frac{dh}{dz}=\frac{4{{{{q}''}}_{w}}}{\dot{{m}''}D}+\frac{{{q}'''}}{{\dot{{m}''}}}-\frac{{{{\dot{{m}''}}}^{2}}}{2}\frac{d}{dz}\left( \frac{1}{{{\rho }^{2}}} \right)-g\cos \theta \qquad\qquad(14)$

The enthalpy of the two-phase mixture can be expressed as $h={{h}_{\ell }}+x({{h}_{v}}-{{h}_{\ell }}).$ For a two-phase flow system with condensation or evaporation, where kinetic and potential energy as well as internal heat generation can be neglected, eq. (14) reduces to $\frac{dx}{dz}=\frac{4{{{{q}''}}_{w}}}{{\dot{m}}''D{{h}_{\ell v}}}\qquad\qquad(15)$

## References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA.