= Solution
Define the indoor-to-outdoor <reduced gravity>
$$
g'_V=\frac{g(\rho_1-\rho_H)}{\rho_0}>0.
$$
Because the whole opening lies above the interface, its indoor side contains upper-layer fluid of density $\rho_H$. Under <hydrostatic pressure>, the pressure difference varies linearly about a <neutral pressure level>. Equal opening geometry and equal <discharge coefficients> for inflow and outflow put that level at the vent midpoint. At vertical distance $y$ from it, the ideal <Bernoulli equation> gives speed $\sqrt{2g'_V|y|}$.
Writing $Q_V$ for either one-way <volumetric flow rate>, integration over one half of the opening gives
$$
Q_V
=C_dL\int_0^{H'/2}\sqrt{2g'_Vy}\,dy
=\frac{C_d}{3}LH'^{3/2}\sqrt{g'_V}.
$$
The total unsigned exchange is $2Q_V$. The ideal sharp-edged inviscid model has $C_d=1$; an empirical $C_d<1$ represents contraction and losses.
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