Chapter 2: Problem 8
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\(_{\mathrm{B}}.\)
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