Consider the flowsheet in the figure for manufacturing C via the reaction \( A + B \rightarrow C \) in an isothermal CSTR. The split in the separator is perfect so that the recycle stream is free of C and the product stream is pure C. Let \( x_i \) denote the mole fraction of species \( i \) (where \( i = A, B, C \)) in the CSTR, which is operated in excess B with \( x_B/x_A = 4 \). The reaction is first-order in A with the reaction rate \( (-r_A) = k x_A \), where \( k_x = 5.0 \, {kmol}/({m}^3 \cdot {h}) \).
The reactor volume \( V \) in \( {m}^3 \) is to be optimized to minimize the cost objective \( J = V + 0.25 R \), where \( R \) is the recycle rate in \( {kmol/h} \). For a product rate \( P = 100 \, {kmol/h} \), the optimum value of \( V \) is ____ \( {m}^3 \) (rounded off to the nearest integer).
Given:
\[
\frac{d}{dz} \left( \frac{z}{(1-z)^2} \right) = \frac{1}{(1 - 2z)^2}
\]
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Solution and Explanation
Step 1: Write the rate equation for the CSTR.
We know the rate of reaction in the CSTR is given by: \[ r_A = k_x x_A \] The molar flow rate of \( A \) in the reactor inlet stream is: \[ F_A = F \cdot x_A \] The outlet flow rate of \( A \) is \( F_A - r_A \cdot V \), where \( F_A \) is the inlet flow rate of A, and \( r_A \) is the rate of consumption of A.
Step 2: Derive the optimization objective.
The cost objective \( J \) is given as: \[ J = V + 0.25 R \] Where \( R \) is the recycle rate. To minimize \( J \), we must express \( R \) in terms of \( V \). The relationship between \( R \) and \( V \) is derived from the material balance and reaction kinetics. By using the provided differential equation, we can calculate the optimal reactor volume.
Step 3: Solve for the optimum volume.
After solving the equations and optimizing the objective, we find the optimum value of \( V \) to be: \[ \boxed{150 \, {m}^3} \] Final Answer: The optimum value of \( V \) is \( \boxed{150} \, {m}^3 \).
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