CSTR Reactor Volume Calculation

This problem requires calculating the volume of a Continuous Stirred Tank Reactor (CSTR) needed to achieve a specific conversion for an enzyme-catalyzed reaction.

Reactor Volume Calculation

The design equation for a CSTR is given by:

$ V = \frac{F_0 (C_{A0} - C_A)}{-r_A} $

Where:

• $V$ is the reactor volume ($L$).
• $F_0$ is the volumetric flow rate ($L.min^{-1}$).
• $C_{A0}$ is the inlet substrate concentration ($mol.L^{-1}$).
• $C_A$ is the outlet substrate concentration ($mol.L^{-1}$).
• $-r_A$ is the reaction rate ($mol.L^{-1}.min^{-1}$).

Step 1: Calculate Outlet Substrate Concentration ($C_A$)

Given the feed substrate concentration $C_{A0} = 2 \ mol.L^{-1}$ and desired conversion $X_A = 0.95$:

$ C_A = C_{A0} (1 - X_A) $

$ C_A = 2 \ mol.L^{-1} \times (1 - 0.95) = 2 \ mol.L^{-1} \times 0.05 = 0.1 \ mol.L^{-1} $

Step 2: Calculate Reaction Rate ($-r_A$) at Outlet Concentration

The given rate equation is $-r_A = \frac{0.1C_A}{1+0.5C_A}$. Substitute $C_A = 0.1 \ mol.L^{-1}$:

$ -r_A = \frac{0.1 \times 0.1}{1 + (0.5 \times 0.1)} = \frac{0.01}{1 + 0.05} = \frac{0.01}{1.05} \ mol.L^{-1}.min^{-1} $

Step 3: Calculate Reactor Volume ($V$)

Using the CSTR design equation with $F_0 = 25 \ L.min^{-1}$, $C_{A0} = 2 \ mol.L^{-1}$, $C_A = 0.1 \ mol.L^{-1}$, and $-r_A = \frac{0.01}{1.05} \ mol.L^{-1}.min^{-1}$:

$ V = \frac{25 \ L.min^{-1} \times (2 - 0.1) \ mol.L^{-1}}{\frac{0.01}{1.05} \ mol.L^{-1}.min^{-1}} $

$ V = \frac{25 \times 1.9}{\frac{0.01}{1.05}} \ L = \frac{47.5 \times 1.05}{0.01} \ L $

$ V = \frac{49.875}{0.01} \ L = 4987.5 \ L $

The calculated reactor volume is $4987.5 \ L$, which falls within the provided range.