= Cross-section averaged phase momentum balance
{title2=$\partial_t(\alpha\bar\rho\bar v)+\partial_x(\alpha\bar\rho\overline{v^2})=M/A-\partial_x(\alpha\bar p)$}
For a phase occupying area $\alpha A$ of a constant-area pipe, define $\bar\rho=(\alpha A)^{-1}\int\rho\,dA$, $\bar v=\int\rho v\,dA/(\alpha A\bar\rho)$ and $\overline{v^2}=\int\rho v^2\,dA/(\alpha A\bar\rho)$. The <momentum> per length is $A\alpha\bar\rho\bar v$, the advected <momentum flux> is $A\alpha\bar\rho\overline{v^2}$, and end pressures contribute $-A\partial_x(\alpha\bar p)$. Applying <Newton's second law> to a slice proves the displayed balance, with $M$ all other phase forces per length. Interphase pressure terms can cancel the derivative of the area fraction. Equality of the second moment and squared mean requires a separate closure.
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