= Solution
Outside, pressure is hydrostatic with slope $-\rho_0g$. Inside it has the same slope below $h_1$ and the shallower slope $-\rho_1g$ above $h_1$. Thus $\Delta p=p_{\rm in}-p_{\rm out}$ is a negative constant below the interface, rises linearly above it with slope $\rho_0g'_{10}$, and is positive at the ceiling. The floor vent admits air and the ceiling vent exhausts it.
Let $C_d$ be a common discharge coefficient and define the effective opening area
$$
A^*=\frac{\sqrt2\,C_dA_fA_c}
{\sqrt{A_f^2+A_c^2}}.
$$
The ventilation flow is
$$
Q_v=A^*\sqrt{g'_{10}(H-h_1)}.
$$
Steady volume and buoyancy balances require
$$
Q_1(h_1)=Q_v,\qquad
g'_{10}=\frac{F_1}{Q_1(h_1)}.
$$
Therefore $h_1=H/2$ precisely when the effective area is chosen as
$$
\boxed{
A^*
=\frac{Q_1(H/2)^{3/2}}
{\sqrt{F_1H/2}}
}.
$$
The individual areas must additionally realize this $A^*$; their ratio fixes how the total pressure drop is divided between the floor and ceiling vents.
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