(Solved): The biomass reaction Substrate S+ Cells More cells + Product takes place in a 46-dm 3 CSTR ( ...

The biomass reaction Substrate $\mathrm{S}+$ Cells $??$ More cells + Product takes place in a 46-dm ${}^3$ CSTR (Chemostat) where the entering concentration of substrate is $200\mathrm{g}/{\mathrm{dm}}^3$ and volumetric flowrate of $5{\mathrm{dm}}^3/\mathrm{s}$. The rate law follows the Monod equation with ${?}_{\max }=0.5{\mathrm{s}}^{?1}$ and $\mathrm{Ks}=50\mathrm{g}/{\mathrm{dm}}^3$. Cell mass balance is given by: $V\frac{d{C}_C}{dt}=v_0\left(C_{CO}?C_C\right)+r_g?r_d$ Monod growth rate is given by: ; $r_2={?}_{\max }\frac{C_C{C}_S}{\left(K_S+C_S\right)};r_d=0D=\frac{V_0}{V}$ a)Determine the substrate concentration ${\mathrm{C}}_{\mathrm{S}}\left(\mathrm{g}/{\mathrm{dm}}^3\right)$ to 2 decimal place $C_s=\frac{D{k}_s}{M_{\max }?D}$