Two chemostats are connected in series, the first having an operational volume of $300L$ and that of the second being $150L.$ The feed to the first fermenter is sterile and the fermenter at $48\frac{L}{h}.$ If the microbial growth can be described by the Monod kinetic model with ${}_{\text{m}\text{a}\text{x}}=0\mathrm{.}2h^{-1}$ and $K_S=230m\frac{g}{L}.$

For the one chemostat, the concentration of the substrate $($S$)$ at its exit is given by:

$S=\frac{D*K_S}{{}_m-D}$

For two chemostats in series, the substrate concentration at the exit of the second reactor $S_2$ is given by the following equation:

$({}_m-D_2)S_2^2+\{\frac{D_1{D}_2{K}_S}{({}_m-D_1)}-D_2{K}_S-{}_m{S}_0\}{S}_2+\frac{D_1{D}_2{K}_S^2}{({}_{m-D_1})}=0$

where $D_1$ and $D_2$ are the dilution rates in each reactor.

Calculate the steady$-$state substrate concentration in the second vessel.