In bioreactors. the growth is autocatalytic in that the more cells you have, the greater the growth rate
$$\text { Cells + nutrients cells more cells + product }$$
The cell growth rate, $r_{e}$ and the rate of nutrient consumption. $r_{s}$, are directly proportional to the concentration of cells for a given set of conditions. A Levenspiel plot of $\left(1 /-r_{s_{s}}\right)$ a function of nutrient conversion $X_{s}=\left(C_{S 0}-\right.$ $\left.C_{S}\right) / C_{S O}$ is given below in Figure 2-8.
For a nutrient feed rate of $1 \mathrm{kg} / \mathrm{hr}$ with $C_{S 0}=0.25 \mathrm{g} / \mathrm{dm}^{3},$ what chemostat (CSTR) size is necessary to achieve.
(a) $40 \%$ conversion of the substrate.
(b) $80 \%$ conversion of the substrate.
(c) What conversion could you achieve with an $80-\mathrm{dm}^{3}$ CSTR? An $80-\mathrm{dm}^{3}$ 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 that Monod Equation for cell growth
$$-r_{s}=\frac{\mathrm{k} C_{s} C_{C}}{K_{M}+C_{S}}$$
along with the stoichiometric relationship between the cell concentration. $C_{C}$ and the substrate concentration, $C_{n}$
$$C_{C}=Y_{C S}\left[C_{S 0}-C_{S}\right]+C_{\kappa}=0.1\left[C_{S 0}-C_{S}\right]+0.001$$
is consistent with Figure P2-8$_{\mathrm{B}}.$