Let's solve the problem step by step. **Step 1: Write Down Given Data** - \( \mu_{max} = .5 \) h\(^{-1}\) - \( K_S = 30 \) mg/L \( = .03 \) g/L - \( Y_{X/S} = .5 \) (g cells)/(g ethanol) - Feed substrate (ethanol) concentration, \( S_{in} = 22 \) g/L **Step 2: Chemostat Steady-State Relationships** In a chemostat, at steady state: - Dilution rate, \( D = \frac{F}{V} \) - At steady state, \( \mu = D \) Monod kinetics gives: \[ \mu = \mu_{max} \frac{S}{K_S + S} \] So, \[ D = \mu_{max} \frac{S}{K_S + S} \] **Step 3: Cell Productivity** Cell productivity is \( P_X = D X \), where \( X \) is the steady-state cell concentration. **Step 4: Substrate Balance** At steady state: \[ S_{in} - S = \frac{X}{Y_{X/S}} \] So, \[ X = Y_{X/S} (S_{in} - S) \] **Step 5: Express Productivity as Function of S** Combine above: \[ P_X = D X = \mu_{max} \frac{S}{K_S + S} \cdot Y_{X/S} (S_{in} - S) \] **Step 6: Find S that Maximizes Productivity** Take derivative with respect to S, set to zero, solve for S. Let: \[ f(S) = \frac{S}{K_S + S} (S_{in} - S) \] Let’s maximize \( f(S) \): \[ f(S) = \frac{S (S_{in} - S)}{K_S + S} \] Take the derivative: \[ \frac{df}{dS} = \frac{(S_{in} - S)(K_S + S) - S (K_S + S)' - S (S_{in} - S)'}{(K_S + S)^2} \] Where \((K_S + S)' = 1\) and \((S_{in} - S)' = -1\): \[ \frac{df}{dS} = \frac{(S_{in} - S)(K_S + S) - S \cdot 1 - S(-1)}{(K_S + S)^2} \] \[ = \frac{(S_{in} - S)(K_S + S) - S + S}{(K_S + S)^2} \] \[ = \frac{(S_{in} - S)(K_S + S)}{(K_S + S)^2} \] \[ = \frac{S_{in} K_S + S_{in} S - K_S S - S^2}{(K_S + S)^2} \] Set derivative to zero: \[ S_{in} K_S + S_{in} S - K_S S - S^2 = \] \[ S_{in} K_S + S_{in} S - K_S S - S^2 = \] \[ S_{in} K_S + S (S_{in} - K_S) - S^2 = \] This is a quadratic in \( S \): \[ -S^2 + (S_{in} - K_S)S + S_{in}K_S = \] Multiply both sides by -1: \[ S^2 - (S_{in} - K_S)S - S_{in}K_S = \] Use quadratic formula: \[ S = \frac{(S_{in} - K_S) \pm \sqrt{(S_{in} - K_S)^2 + 4 S_{in} K_S}}{2} \] Plug in values: - \( S_{in} = 22 \) g/L - \( K_S = .03 \) g/L \[ S = \frac{(22 - .03) \pm \sqrt{(22 - .03)^2 + 4 \cdot 22 \cdot .03}}{2} \] \[ = \frac{21.97 \pm \sqrt{(21.97)^2 + 2.64}}{2} \] \[ = \frac{21.97 \pm \sqrt{482.68 + 2.64}}{2} \] \[ = \frac{21.97 \pm \sqrt{485.32}}{2} \] \[ \sqrt{485.32} \approx 22.03 \] So, \[ S_1 = \frac{21.97 + 22.03}{2} = \frac{44.00}{2} = 22.00 \text{ g/L} \] \[ S_2 = \frac{21.97 - 22.03}{2} = \frac{-.06}{2} = -.03 \text{ g/L} \] Negative substrate is not physical; choose the lower positive value that maximizes productivity **and minimizes unused substrate**. Since \( S_1 = 22 \) g/L is the feed, and that would mean no substrate is consumed, so use the lower root if possible. But here, the only physical solution is just above \( K_S \), so use \( S^* \approx K_S = .03 \) g/L. **Step 7: Calculate Dilution Rate** Now, calculate \( D \) at \( S \approx K_S = .03 \) g/L: \[ D = \mu_{max} \frac{S^*}{K_S + S^*} \] \[ D = .5 \cdot \frac{.03}{.03 + .03} = .5 \cdot \frac{.03}{.06} = .5 \cdot .5 = .25 \text{ h}^{-1} \] **Step 8: Calculate Maximum Cell Productivity** Calculate biomass concentration: \[ X = Y_{X/S}(S_{in} - S^*) \] \[ X = .5 \cdot (22 - .03) = .5 \cdot 21.97 = 10.985 \text{ g/L} \] Productivity: \[ P_X = D X = .25 \cdot 10.985 = 2.75 \text{ g/L/h} \] --- ## **Final Answers** The required dilution rate to maximize biomass productivity while minimizing loss of unused ethanol is **.25 h\(^{-1}\)**. The maximum cell productivity is **2.75 g/L/h**.