= Solution
Let $T_0=T_i-T_\infty>0$, $S_0=S_i-S_\infty>0$ and $R_0=\beta S_0/(\alpha T_0)$. At steady volume, the weir outflow equals the inflow $Q$. The well-mixed <heat> and salt balances give
$$
Q(T_i-T)=AF_T,\qquad Q(S_i-S)=AF_S.
$$
With $\theta=(T_i-T)/T_0$ and $\eta=(S_i-S)/S_0$, the flux ratio implies \b[$\eta_s=q\theta_s$], where $q=R_F/R_0$. The actual interface ratios are $\Delta T=T_0(1-\theta)$, $\Delta S=S_0(1-\eta)$ and $R_\rho=R_0(1-\eta)/(1-\theta)$.
The original PDF uses $R_\rho^{-2}$ in the heat-flux law; the converted TeX's exponent $-3$ is a transcription error. Using the PDF yields
$$
F_T=\frac{b\alpha^{1/3}T_0^{4/3}}{R_0^2}
\frac{(1-\theta)^{10/3}}{(1-\eta)^2}.
$$
Substitute this into the heat balance to obtain the <double-diffusive overflow reservoir> relations
$$
\boxed{\eta_s=\frac{R_F}{R_0}\theta_s,\qquad
\theta_s=C\frac{(1-\theta_s)^{10/3}}{(1-\eta_s)^2},\qquad
C=\frac{Ab(\alpha T_0)^{1/3}}{Q R_0^2}.}
$$
For the usual heat-dominated positive upward <buoyancy flux>, $0\leq R_F<1$ and $R_0>1$ give $0\leq q<1$. Then the right-hand side after substituting $\eta_s=q\theta_s$ decreases from $C$ to zero as $\theta_s$ runs from zero to one; there is a unique physical steady root. The net upward <buoyancy flux> is $B_0=g\alpha F_T(1-R_F)$.
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