Let's solve the problem step-by-step, as requested: --- ## 1.1: Steady-State Effluent Substrate Concentration, \( S \) First, calculate the actual dilution rate: \[ D = \frac{v_}{V} = \frac{20}{100} = .2 \text{ h}^{-1} \] Calculate the effective dilution rate, \( D_{\text{eff}} \): \[ D_{\text{eff}} = D \left( \frac{1 + R - RC}{1 + R} \right) \] Given \( R = .5 \), \( C = 2.5 \): \[ 1 + R - RC = 1 + .5 - (.5 \times 2.5) = 1.5 - 1.25 = .25 \] \[ 1 + R = 1.5 \] \[ D_{\text{eff}} = .2 \left( \frac{.25}{1.5} \right) = .2 \times .1667 = .0333 \text{ h}^{-1} \] From steady-state substrate balance: \[ D(S_ - S) = \frac{\mu X}{Y_{X/S}} \] From Monod kinetics: \[ \mu = \frac{\mu_{\text{max}} S}{K_S + S} \] At steady state for CSTR with recycle, \( \mu = D_{\text{eff}} \), so: \[ \frac{\mu_{\text{max}} S}{K_S + S} = D_{\text{eff}} \] \[ \mu_{\text{max}} S = D_{\text{eff}} (K_S + S) \] \[ \mu_{\text{max}} S = D_{\text{eff}} K_S + D_{\text{eff}} S \] \[ \mu_{\text{max}} S - D_{\text{eff}} S = D_{\text{eff}} K_S \] \[ S (\mu_{\text{max}} - D_{\text{eff}}) = D_{\text{eff}} K_S \] \[ S = \frac{D_{\text{eff}} K_S}{\mu_{\text{max}} - D_{\text{eff}}} \] Plug in the values: \[ S = \frac{.0333 \times 1.}{.5 - .0333} = \frac{.0333}{.4667} \approx .0714 \text{ g/L} \] --- ## 1.2: Steady-State Biomass Concentration, \( X \) Rearrange the substrate balance: \[ D(S_ - S) = \frac{\mu X}{Y_{X/S}} \] \[ X = \frac{D Y_{X/S} (S_ - S)}{\mu} \] Recall that at steady-state, \( \mu = D_{\text{eff}} \): \[ X = \frac{D Y_{X/S} (S_ - S)}{D_{\text{eff}}} \] Plug in the values: \[ D = .2, \quad Y_{X/S} = .4, \quad S_ = 10, \quad S = .0714, \quad D_{\text{eff}} = .0333 \] \[ X = \frac{.2 \times .4 \times (10 - .0714)}{.0333} = \frac{.08 \times 9.9286}{.0333} = \frac{.7943}{.0333} \approx 23.85 \text{ g/L} \] --- ## 1.3: Plot of \( S \) vs. Dilution Rate \( D \) (Sketch Guidance) To sketch the plot, use the Monod equation for \( S \) as a function of \( D_{\text{eff}} \) and thus \( D \): \[ D_{\text{eff}} = D\left(\frac{1+R-RC}{1+R}\right) \] \[ S = \frac{D_{\text{eff}} K_S}{\mu_{\text{max}} - D_{\text{eff}}} \] Plot \( S \) on the y-axis and \( D \) on the x-axis from \( D = \) up to the critical dilution rate (see next part). The curve will increase sharply as \( D \) approaches the critical dilution rate, \( D_{\text{crit}} \). --- ## 1.4: Critical Dilution Rate \( D_{\text{crit}} \) (Washout Condition) Washout occurs when \( D_{\text{eff}} = \mu_{\text{max}} \): \[ D_{\text{eff,crit}} = \mu_{\text{max}} \] Recall: \[ D_{\text{eff}} = D \left(\frac{1+R-RC}{1+R}\right) \] Let \( \alpha = \frac{1+R-RC}{1+R} \): \[ D_{\text{crit}} = \frac{\mu_{\text{max}}}{\alpha} \] Plug in values: \[ \alpha = \frac{.25}{1.5} = .1667 \] \[ D_{\text{crit}} = \frac{.5}{.1667} = 3. \text{ h}^{-1} \] --- # **Summary of Answers** 1. **Steady-state effluent substrate concentration:** \( S = .0714 \) g/L 2. **Steady-state biomass concentration:** \( X = 23.85 \) g/L 3. **Sketch:** Plot \( S \) versus \( D \) using \( S = \frac{D_{\text{eff}} K_S}{\mu_{\text{max}} - D_{\text{eff}}} \) up to \( D_{\text{crit}} = 3. \) h\(^{-1}\). \( S \) will rise sharply near \( D_{\text{crit}} \). 4. **Critical dilution rate (washout):** \( D_{\text{crit}} = 3. \) h\(^{-1}\)