In bioreactor, the growth is autocatalytic in that the more cells you have, the greater the growth rate

Cells $+$ nutrients $$Cell more cells $+$ product

The cell growth rate, $r_g,$ and the rate of nutrient consumption, $r_5,$ are directly proportional to the

concentration of cells for a given set of conditions. A Levenspiel plot of $(\frac{1}{-}{r}_s)$ as a function of nutrient

conversion $x_5=\frac{C_{50}-C_5}{C_{50}}$ is given below in the figure below:

For the nutrient feed rate of $1k\frac{g}{h}r$ with $C_{50}=0\mathrm{.}25\frac{g}{d}{m}^3,$ what chemostat $($CSTR$)$ size is necessary to

achieve

a$)40\%$ conversion of substrate?

b$)80\%$ conversion of substrate?

c$)$ What conversion could you achieve with an $80-$ dm ${?}^{3}CSTR?$ An $80-d{m}^3$ PFR$?$

d$)$ How could you arrange a CSTR and PFR in series to achieve $80\%$ conversion with the minimum

total volume? Repeat for two CSTRs in series.

e$)$ Show the Monod Equation for cell growth

$-r_S=\frac{k{C}_s{C}_c}{{}_M+C_s}$

along with stoichiometric relationship between the cell concentration, $C_c,$ and the substrate

concentration, $C_5,$

$C_c=Y_{\frac{c}{s}}[C_{So}-C_S]+C_{co}=0\mathrm{.}1[C_{So}-C_S]+0\mathrm{.}001$

is consistent with figure above.