Let's address the given problem step by step: --- ## **Step 1: Understanding the PDEs for the Reactor** The equations given are: \[ \frac{\partial C_A}{\partial t} + u \frac{\partial C_A}{\partial z} = D_{az} \frac{\partial^2 C_A}{\partial z^2} - k_ca(C_A - C_{A_s}) \] \[ a_sp_b \frac{\partial C_{A_s}}{\partial t} = k_ca(C_A - C_{A_s}) \] These describe mass transfer and reaction in a fixed-bed catalytic reactor, including axial dispersion, convection, and inter-phase mass transfer. --- ## **Step 2: Enzyme Reaction with Competitive Inhibition** The enzyme-catalyzed reaction with competitive inhibition has a rate law: \[ r = \frac{V_{max}[S]}{K_m \left(1 + \frac{[I]}{K_I}\right) + [S]} \] where: - \( [S] \): substrate concentration - \( [I] \): inhibitor (product) concentration - \( V_{max} \): maximum rate - \( K_m \): Michaelis constant - \( K_I \): inhibition constant --- ## **Step 3: Answering the Questions** ### **a) Symbolic Expression for Competitive Inhibition Rate** The rate expression is: \[ r = \frac{V_{max} [S]}{K_m \left(1 + \frac{[I]}{K_I}\right) + [S]} \] This equation incorporates competitive inhibition, as the inhibitor competes with the substrate for the enzyme active site. --- ### **b) Numerical Maximum Velocity \( V_{max} \)** The maximum oxygen uptake rate is given as \( .22 \) mol/(m³·s). The working volume is \( 15. \) m³. Thus, the numerical maximum velocity (total for the reactor) is: \[ \text{Total } V_{max} = (.22\ \text{mol}/\text{m}^3\cdot \text{s}) \times 15.\ \text{m}^3 = 3.3\ \text{mol/s} \] If you want the specific \( V_{max} \) per unit volume, it is \( .22 \) mol/(m³·s). --- ### **c) Biomass Yield Coefficient (\( Y_{X/S} \))** The biomass yield coefficient is: \[ Y_{X/S} = \frac{\Delta X}{\Delta S} \] where \( \Delta X \) is the change in biomass, and \( \Delta S \) is the change in substrate. It is usually determined experimentally. If you have the data for biomass formed per substrate consumed, substitute values here. --- ### **d) Stability at the Washout Limit** At the washout limit for a chemostat (continuous reactor), the dilution rate \( D \) equals the maximum specific growth rate \( \mu_{max} \): \[ D = \mu_{max} \] At this point, any further increase in dilution rate leads to loss of biomass from the reactor (washout). To prove stability, analyze the biomass balance: \[ \frac{dX}{dt} = (\mu - D)X \] At washout, \(\mu = D\), so: \[ \frac{dX}{dt} = \] This is a marginally stable equilibrium: any decrease in \( \mu \) (e.g., due to substrate fluctuation) leads to \( \frac{dX}{dt} < \), so \( X \) decreases further, i.e., instability. Thus, the washout point is an unstable equilibrium. --- ## **Final Answers** **a)** \( r = \frac{V_{max} [S]}{K_m \left(1 + \frac{[I]}{K_I}\right) + [S]} \) **b)** \( V_{max} = .22\ \text{mol}/(\text{m}^3\cdot \text{s}) \) (per unit volume) or \( 3.3\ \text{mol/s} \) (total reactor) **c)** \( Y_{X/S} = \frac{\Delta X}{\Delta S} \), substitute experimental values as needed **d)** At the washout limit (\( D = \mu_{max} \)), the system is **unstable** because any perturbation in growth leads to decline in biomass, confirming instability at this threshold.