Q.53 The following schematic diagram shows a chemostat with cell recycle
Where F0 and Fr are the volumetric flow rates (in L.h−1) of feed and recycle streams, respectively. X1, X0 and X are the cell concentrations (in g.L−1) in the reactor, recycle-stream and product-stream, respectively. If (X0 / X1) = 1.5, (Fr / F0) = 0.7 and X1 is 7.3 g.L−1, the value of X (in g.L−1, rounded off to one decimal place) is ____________.
Problem Summary
The following data is provided for a chemostat system operating with cell recycle:
- X1 = 7.3 g L-1 (reactor concentration)
- X0 / X1 = 1.5 (recycle concentration ratio)
- Fr / F0 = 0.7 (recycle to feed ratio)
Our goal is to calculate X, the cell concentration in the product stream, rounded to one decimal place.
Step-by-Step Solution
1. Mass Balance on Cell Separator
Total cell mass entering = total cell mass leaving.
Input:
(F0 + Fr) X1
Output:
Fr X0 + F0 X
Mass balance equation:
(F0 + Fr) X1 = Fr X0 + F0 X
2. Substitute Given Values
X0 = 1.5 X1
Fr = 0.7 F0
Equation becomes:
1.7F0 X1 = 1.05F0 X1 + F0 X
Divide by F0:
1.7X1 = 1.05X1 + X
3. Solve for X
X = (1.7 − 1.05) X1
X = 0.65 × 7.3 = 4.745
Final Answer
X = 4.7 g L-1
Explanation of Typical Options
| Option | Value (g L-1) | Explanation |
|---|---|---|
| A | 3.0 | Too low; inconsistent with given ratios. |
| B | 4.7 | Correct—matches mass balance calculation. |
| C | 7.3 | Reactor concentration, not product. |
| D | 10.9 | Too high; unrealistic for given recycle. |
Conclusion
By applying a steady-state mass balance to the chemostat cell separator, we calculated the product outlet concentration to be:
X = 4.7 g L-1
This completes the chemostat cell recycle concentration calculation.
