 Solution

Outside, pressure is hydrostatic with slope $-{\rho }_{0}g$. Inside it has the same slope below $h_1$ and the shallower slope $-{\rho }_{1}g$ above $h_1$. Thus $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ is a negative constant below the interface, rises linearly above it with slope ${\rho }_0{g}_{10}^{\prime }$, 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 areaThe ventilation flow isSteady volume and buoyancy balances requireTherefore $h_1=H/2$ precisely when the effective area is chosen asThe individual areas must additionally realize this $A^{\ast }$; their ratio fixes how the total pressure drop is divided between the floor and ceiling vents.