= Solution
Define $\chi_1$ to be one in <phase> 1 and zero in <phase> 2, including its dependence on both transverse coordinates even though the averaged fields depend only on $x,t$. With no interphase <interphase mass transfer>[mass transfer], its distributional <phase>-mass equation is $\partial_t(\rho\chi_1)+\nabla\cdot(\rho\mathbf v\chi_1)=0$. At a pipe <cross section> $\Sigma$ of area $A$, set
$$
\boxed{\alpha=\frac1A\int_\Sigma\chi_1\,dA,\quad
\rho_1=\frac1{A\alpha}\int_\Sigma\rho\chi_1\,dA,\quad
u_1=\frac{\int_\Sigma\rho v_x\chi_1\,dA}{A\alpha\rho_1}.}
$$
Thus $\rho_1$ is an intrinsic <phase> <density> and $u_1$ is the <density-weighted mean> axial <velocity>. This weighting is necessary when <density> varies within the <phase>; it becomes the ordinary intrinsic mean for an incompressible <phase> of constant <density>.
Integrate over the <cross section>. The transverse <divergence> contributes no boundary mass <flux> at the impermeable pipe wall, and the material interface has no mass leakage. The section area is constant, so
$$
\boxed{\partial_t(\rho_1\alpha)+\partial_x(\rho_1\alpha u_1)=0.}
$$
For <phase> 2 use $\chi_2=1-\chi_1$ and fraction $1-\alpha$, with the analogous intrinsic <density> and <density>-weighted <velocity>:
$$
\boxed{\partial_t[\rho_2(1-\alpha)]+\partial_x[\rho_2(1-\alpha)u_2]=0.}
$$
The intrinsic quantities are defined where their <phase> fraction is nonzero. These are the <intrinsic phase averages in a two-fluid pipe model>.
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