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The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.