Let's go through each part step-by-step, explaining the concepts and deriving the equations as required. --- ### 1. Semi-batch Reactor: Second-order Reaction **Given:** - Initial Volume: \( V_ \) (pure liquid B) - Reactant A is fed at rate \( v_ \), concentration \( C_{A} \) - Second-order reaction: \( A + B \rightarrow C \) - Volume at time \( t \): \( V(t) \) **To Find:** *The differential equation for the accumulation of A in the reactor.* **Solution:** Let \( N_A \) be the moles of A in the reactor at time \( t \): \( N_A = C_A V \) Material balance for A: Rate in - Rate out ± Rate of generation = Accumulation For a semi-batch reactor (no outflow): \[ \frac{dN_A}{dt} = \text{in} - \text{consumed} \] Inflow of A: \[ \text{in} = v_ C_{A} \] Consumption by reaction (second-order): \[ \text{consumed} = k C_A C_B V \] But, as only A is being fed, the volume increases: \[ V(t) = V_ + v_ t \] The concentration \( C_B \) can be written in terms of \( N_B \) and \( V \). Initially, only B is present, and it is consumed as A is consumed. So, \[ \frac{d(C_A V)}{dt} = v_ C_{A} - k C_A C_B V \] Expanding the derivative on the left: \[ V \frac{dC_A}{dt} + C_A \frac{dV}{dt} = v_ C_{A} - k C_A C_B V \] Since \( \frac{dV}{dt} = v_ \): \[ V \frac{dC_A}{dt} + C_A v_ = v_ C_{A} - k C_A C_B V \] \[ V \frac{dC_A}{dt} = v_ (C_{A} - C_A) - k C_A C_B V \] So, the final differential equation is: \[ \boxed{V \frac{dC_A}{dt} = v_ (C_{A} - C_A) - k C_A C_B V} \] --- ### 2. Monod Kinetics with Inhibition: Critical Substrate Concentration **Given:** - Monod Kinetics with inhibition - Washout in a chemostat - Dilution rate: \( D \) - Inhibition constant: \( K_I \) **To Find:** *Critical substrate concentration at washout.* **Solution:** Monod with inhibition: \[ \mu = \frac{\mu_{\max} S}{K_S + S + \frac{S^2}{K_I}} \] At washout, microbial growth rate = dilution rate: \[ \mu = D \] So, \[ D = \frac{\mu_{\max} S_{crit}}{K_S + S_{crit} + \frac{S_{crit}^2}{K_I}} \] This is the equation for the critical substrate concentration at washout in terms of \( D, \mu_{\max}, K_S, K_I \): \[ \boxed{D = \frac{\mu_{\max} S_{crit}}{K_S + S_{crit} + \frac{S_{crit}^2}{K_I}}} \] --- ### 3. Shrinking Core Model: Fractional Conversion **Given:** - Non-catalytic reaction, Shrinking Core Model - Fractional conversion: \( X_B \) - Proportional to particle radius \( R \) **To Find:** *Relationship between \( X_B \), time, and \( R \) (for reaction-controlled case).* **Solution:** For reaction-controlled case: \[ 1 - (1 - X_B)^{1/3} = k t \] Where \( k \) is a rate constant dependent on \( R \): \[ k = \frac{k_s C_{A,\infty}}{\rho_B R} \] So, time required to reach conversion \( X_B \): \[ t = \frac{1}{k} [1 - (1 - X_B)^{1/3}] \] Since \( k \propto \frac{1}{R} \): \[ t \propto R [1 - (1 - X_B)^{1/3}] \] --- ### 4. Autothermal CSTR Stability **Given:** - Slope of heat generation and removal curves - Stability determined at steady-state intersection **To Find:** *Critical condition for stability.* **Solution:** Stability requires the slope of the removal line to be greater than the slope of the generation curve at the intersection. Mathematically, \[ \frac{dQ_{rem}}{dT} > \frac{dQ_{gen}}{dT} \] At the critical point (onset of instability): \[ \frac{dQ_{rem}}{dT} = \frac{dQ_{gen}}{dT} \] Given the provided expression: \[ \frac{dQ_{rem}}{dT} = (1 + \beta) > \beta \frac{X(1-X)}{(1+\beta X)^2} \] So, the critical condition is when: \[ \frac{dQ_{rem}}{dT} = \frac{dQ_{gen}}{dT} \] --- ### Summary 1. The differential equation for accumulation of A is: \( V \frac{dC_A}{dt} = v_ (C_{A} - C_A) - k C_A C_B V \) 2. The critical substrate concentration at washout is given by: \( D = \frac{\mu_{\max} S_{crit}}{K_S + S_{crit} + \frac{S_{crit}^2}{K_I}} \) 3. The time for a given conversion in the shrinking core model (reaction-controlled) is: \( t = \frac{1}{k} [1 - (1 - X_B)^{1/3}] \) and \( t \propto R \) 4. The CSTR is stable if: \( \frac{dQ_{rem}}{dT} > \frac{dQ_{gen}}{dT} \) at steady-state. The critical point is when these slopes are equal.