Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 79 / 5 / b / Solution

Take $y$ from the first wall to the second, with $f>0$. Negligible transverse velocity and cross-channel geostrophic balance give $fu=-g{\partial }_y(h+H)=-g{h}_y$, since the bed elevation $H(x)$ is independent of $y$. Therefore the volume flux isFor positive downstream volume flux, the layer is deeper at the first wall: $h_1>h_2$. This calculation does not require zero shallow-water potential vorticity.

Here $h_j$ must denote the wall depth above the local bed. If the surface heights are instead measured from a fixed vertical datum, write them as ${\eta }_j$ and use $h_j={\eta }_j-H(x)$, giving $Q=g[({\eta }_1-H{)}^2-({\eta }_2-H{)}^2]/(2f)$. The compact formula thus uses a bed-relative meaning of the wall heights.