Let's break down the solution step-by-step for each part of the question: --- ## **a) Derive the symbolic expression for the competitive inhibition rate** For a competitive inhibition in enzyme kinetics, the rate equation is: \[ v = \frac{V_{max}[S]}{K_m \left(1 + \frac{[I]}{K_i}\right) + [S]} \] Where: - \( [S] \) = Substrate concentration - \( [I] \) = Inhibitor (product) concentration - \( V_{max} \) = Maximum reaction rate - \( K_m \) = Michaelis-Menten constant - \( K_i \) = Inhibition constant This equation shows how the presence of the inhibitor increases the apparent \( K_m \) (but not \( V_{max} \)), thus reducing the rate. --- ## **b) Determine the numerical maximum velocity \( V_{max} \)** Given Data: - Maximum oxygen uptake rate = \( .22 \) mol/m³·s - Working volume = \( 15. \) m³ The maximum rate in terms of total moles per second: \[ V_{max,\,total} = (\text{max uptake rate}) \times (\text{volume}) = .22 \, \frac{\text{mol}}{\text{m}^3 \cdot \text{s}} \times 15. \, \text{m}^3 = 3.3 \, \frac{\text{mol}}{\text{s}} \] If asked for per volume, it's \( .22 \) mol/m³·s. --- ## **c) Calculate the biomass yield coefficient** The biomass yield coefficient (\( Y_X/S \)) is defined as: \[ Y_{X/S} = \frac{\text{mass of cells formed}}{\text{mass of substrate consumed}} \] This is typically determined experimentally or given. If you have oxygen as the limiting substrate, and assuming you have the substrate yield coefficient on oxygen (\( Y_{X/O_2} \)), you would use that value: \[ Y_{X/O_2} = \frac{\Delta X}{\Delta O_2} \] If not provided, you might need extra data (e.g., grams of biomass formed per mole of oxygen consumed). For example, if for every mole of \( O_2 \) consumed, .5 grams of biomass is produced, \[ Y_{X/O_2} = .5 \, \text{g biomass/mol O}_2 \] --- ## **d) Prove the stability at the washout limit** In a continuous stirred tank reactor (CSTR), washout occurs when the dilution rate \( D \) exceeds the maximum specific growth rate \( \mu_{max} \): \[ D = \frac{F}{V} \] At washout (\( D = \mu_{max} \)), the biomass concentration approaches zero. To analyze stability, consider the balance: \[ \frac{dX}{dt} = (\mu - D)X \] At washout, \( \mu = D \), so: \[ \frac{dX}{dt} = \] If \( D > \mu_{max} \), \( \frac{dX}{dt} < \) and biomass will decrease to zero (stable washout point). If \( D < \mu_{max} \), biomass can be maintained. Thus, the washout point is **stable**—any perturbation that decreases biomass will continue to drive it to zero. --- ### **Summary of Answers** a) The competitive inhibition rate is \( v = \frac{V_{max}[S]}{K_m(1 + \frac{[I]}{K_i}) + [S]} \) b) The numerical maximum velocity \( V_{max} \) is \( .22 \) mol/m³·s, or \( 3.3 \) mol/s for the total reactor c) The biomass yield coefficient \( Y_{X/O_2} \) is calculated as \( \frac{\Delta X}{\Delta O_2} \), with a numerical value requiring more data d) The washout point is stable; for \( D > \mu_{max} \), biomass decays to zero and does not recover If you need the graphing of PDEs or more details on the reactor model, let me know!