Assume the two identical openings and their wall plumes behave symmetrically and do not interact before reaching the interface. If $Q_V^{(1)}$ is the one-way rate through either vent, the global buoyancy balance is
$$
B=2Q_V^{(1)}g'_H.
$$
Each wall plume therefore has buoyancy flux per unit span $\mathcal B=B/(2L)$. The steady volume balance now includes two descending plumes:
$$
C_PB^{1/3}h^{5/3}
=2L\alpha(z_o-h)
\left(\frac{B}{2\alpha L}\right)^{1/3}.
$$
Hence the interface height is determined implicitly by
$$
\boxed{
C_Ph^{5/3}
=2^{2/3}\alpha^{2/3}L^{2/3}(z_o-h),
\qquad
z_o=z_V+\frac{H'}{2\alpha}
}.
$$
For completeness, if each opening retains the same <single-opening exchange flow> coefficient $K_V$, then
$$
Q_V^{(1)}=\left(\frac{K_V^2B}{2}\right)^{1/3},
\qquad
Q_{V,\mathrm{total}}=2Q_V^{(1)}
=2^{2/3}(K_V^2B)^{1/3}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 345 / 3 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-345/3/f/solution