= Entraining shallow-water current in a triangular channel
{title2=$A=h^2/2$}
For a prismatic channel of width $b(z)=z$, a <Boussinesq> current of depth $h$ occupies area $A=h^2/2$. Ambient fluid at rest enters across the interface at speed $w_e$, supplying volume but no horizontal momentum or excess density. The <volume conservation>, momentum and excess-density balances are
$$
A_t+(Au)_x=hw_e,\qquad
(Au)_t+(Au^2+g'h^3/6)_x=0,\qquad
(Ag')_t+(Aug')_x=0.
$$
Here $g'$ is the <reduced gravity>. With $D=\partial_t+u\partial_x$, these give $Dh+(h/2)u_x=w_e$, $Du+g'h_x+(h/3)g'_x=-2uw_e/h$, and $Dg'=-2g'w_e/h$. The momentum source in primitive form describes acceleration of newly entrained fluid, even when bed drag vanishes.
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