CHEMOSTAT Design Problem:

You are designing a new $10$L pilot CHEMOSTAT for a cell line. The specific growth rate constant can be modeled as ${}_g={}_{\text{m}\text{a}\text{x}}(1-e^{-S})$ Where $S$ is your outlet substrate concentration, ${}_g$ is your growth rate and ${}_{\text{m}\text{a}\text{x}}$ is the mass growth rate.

a$.)$ Derive an expression $($i$.$e$.$ solve for $S)$ for outlet substrate concentration $($S$)$ in terms of ${}_{\text{m}\text{a}\text{x}}$ and $D.$

b$.)$ The initial substrate concentration at the inlet is $10\frac{g}{L}.$ The process requires the outlet substrate concentration to be $1\frac{g}{L}.$ What is the outlet cell concentration and the flow rate needed, in this process.$($Assume standard operation conditions: steady$-$state, no cell death, and no cell feeding in the

inlet$).$ The other constants for the cells are ${}_{\text{m}\text{a}\text{x}}=\frac{0\mathrm{.}5}{h}r$ and $Y_{\frac{x}{s}}=0\mathrm{.}4.$ If you don't know how to do part a$,$ you can calculate these by assuming the Monod model for partial credit.

Useful relationships: $x=Y_{\frac{x}{s}}(S_o-S)$ ; $D=\frac{F}{N}$