= Solution
The final state is <displacement ventilation>: fresh air of density $\rho_0$ forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height $h$. Steady <volume conservation> requires the plume volume flux there to equal the imposed ventilation flux,
$$
V(h)=A.
$$
Below the interface the ambient is uniform, so $B_0=F_0/\rho_0$ is constant. Put $J=M/\rho_0$. The triangular-profile <wall line plume> equations reduce to
$$
\frac{dV}{dz}=\alpha W=\frac{3\alpha J}{2V},
\qquad
\frac{dJ}{dz}=\frac{B_0V}{J}.
$$
The <pure plume> conditions $V,J\to0$ at the radiator select the <similarity solution>
$$
V(z)=C_LB_0^{1/3}z,
\qquad
C_L=\left(\frac{3\alpha}{2}\right)^{2/3}.
$$
Consequently the <displacement-ventilation interface height> is
$$
\boxed{h=\frac{A}{C_L(F_0/\rho_0)^{1/3}}
=\left(\frac{2}{3\alpha}\right)^{2/3}
\frac{A}{(F_0/\rho_0)^{1/3}}.}
$$
The upper-layer <reduced gravity> follows from its steady buoyancy balance as $g'_u=B_0/A=F_0/(\rho_0A)$. The two-layer solution applies when the calculated interface satisfies $0<h<H$.
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