= Solution
Let $\chi_1(\mathbf x,t)$ equal one in phase one and zero in phase two; its interface value is immaterial for area integrals. In the absence of phase change or interfacial <mass> transfer, the distributional phase <mass conservation> is $\partial_t(\chi_1\rho)+\nabla\cdot(\chi_1\rho\mathbf u)=0$. For a cross-section $S$ of area $A$, define
$$
\boxed{\alpha=\frac1A\int_S\chi_1\,dA,\qquad
\rho_1=\frac1{A\alpha}\int_S\chi_1\rho\,dA,\qquad
u_1=\frac{\int_S\chi_1\rho u_x\,dA}{A\alpha\rho_1}.}
$$
Here the symbol $u_x$ in the integral denotes the axial velocity component, not a derivative. Integrating through an arbitrary axial slice and using zero normal flux at the rigid pipe wall gives
$$
\boxed{\partial_t(\rho_1\alpha)+\partial_x(\rho_1\alpha u_1)=0.}
$$
The velocity is <mass> weighted, so the averaged <mass flux> factors exactly; an unweighted phase mean would not in general have this property. This is the <cross-section averaged phase mass balance>.
For phase two use $\chi_2=1-\chi_1$, <volume fraction> $1-\alpha$, and
$$
\rho_2=\frac{\int_S\chi_2\rho\,dA}{A(1-\alpha)},\qquad
u_2=\frac{\int_S\chi_2\rho u_x\,dA}{A(1-\alpha)\rho_2}.
$$
Then $\boxed{\partial_t[\rho_2(1-\alpha)]+\partial_x[\rho_2(1-\alpha)u_2]=0}$. Mean quantities in a phase that is absent are unnecessary; the products giving conserved masses and fluxes remain the meaningful variables.
Back to article page
