Question

A continuous flow stirred tank reactor (CSTR) operates under a steady-state condition with a flow rate of 300 L per hour. The influent concentration of a substrate entering the reactor is 150 mg/L.
To give a treatment efficiency of 76% for the substrate that decays according to half-order kinetics with a rate constant of 0.05 (mg/L)^1/2 per hour, the required volume of the reactor is ______ m³ (answer in integer).

Hint:

First find the effluent concentration from the 76% removal efficiency, then use the steady-state CSTR balance Q(C0-C) = k*sqrt(C)*V to solve for the volume.

Answer 1

Correct Answer: 114

Step 1: Find the effluent substrate concentration from the required efficiency.
The influent concentration is \(C_0 = 150\) mg/L, and the removal efficiency required is 76%, so the effluent concentration is \[ C = C_0 (1 - 0.76) = 150 \times 0.24 = 36\ \text{mg/L} \]

Step 2: Write the steady-state mass balance for the CSTR with half-order decay.
For half-order kinetics, the rate of substrate removal is \(r = k\sqrt{C}\) (mg/L per hour), where \(C\) is the reactor (effluent) concentration, since a CSTR is completely mixed and the reaction proceeds at the effluent concentration everywhere in the tank. The steady-state substrate balance over the reactor volume \(V\) is \[ Q C_0 = Q C + k\sqrt{C}\, V \] which rearranges to \[ V = \frac{Q (C_0 - C)}{k\sqrt{C}} \]

Step 3: Substitute the known values.
\[ \sqrt{C} = \sqrt{36} = 6\ (\text{mg/L})^{1/2} \] \[ V = \frac{300 \times (150 - 36)}{0.05 \times 6} = \frac{300 \times 114}{0.3} = \frac{34200}{0.3} = 114000\ \text{L} \]

Step 4: Convert to cubic metres and state the final answer.
\[ V = \frac{114000\ \text{L}}{1000\ \text{L/m}^3} = 114\ \text{m}^3 \] The required reactor volume is \(114\) m\(^3\).