Question

A CSTR with a volume of 100 m\(^3\) is operated in cell recycle mode. At a volumetric flow rate of 10 m\(^3\) day\(^{-1}\) and effluent biomass of 20 mg L\(^{-1}\), the steady state biomass concentration is 200 mg L\(^{-1}\). The mean cell retention time in the reactor is ____ days. (answer in integer)

Hint:

Mean cell retention time \(\theta_c = VX/(FX_e)\), where \(X\) is the reactor biomass and \(X_e\) is the effluent biomass concentration.

Answer 1

Correct Answer: 100

Step 1: Recall the definition of mean cell retention time.
In a cell-recycle CSTR (also called a chemostat with recycle), the mean cell retention time, or sludge age, \(\theta_c\), is the average time a cell spends inside the reactor before leaving in the effluent stream. It is defined as the total biomass held in the reactor divided by the rate at which biomass leaves the system:
\[ \theta_c = \frac{V \, X}{F \, X_e} \]
where \(V\) is the reactor volume, \(X\) is the biomass concentration inside the reactor, \(F\) is the volumetric flow rate of the effluent stream, and \(X_e\) is the biomass concentration in that effluent.

Step 2: Identify the given values.
\[ V = 100\ \text{m}^3, \quad F = 10\ \text{m}^3\,\text{day}^{-1}, \quad X = 200\ \text{mg L}^{-1}, \quad X_e = 20\ \text{mg L}^{-1} \]
Since \(X\) and \(X_e\) both appear as a ratio, their units (mg L\(^{-1}\)) cancel out and there is no need to convert them to match \(V\) and \(F\)'s units.

Step 3: Substitute into the formula.
\[ \theta_c = \frac{V\,X}{F\,X_e} = \frac{100 \times 200}{10 \times 20} \]
\[ = \frac{20000}{200} = 100 \]

Final Answer:
The mean cell retention time in the reactor is
\[ \boxed{\theta_c = 100\ \text{days}} \]