22. A 5 liter chemostat is fed fresh medium at 0.2 liters/minute having a substrate concentration of 25
grams/liter. At steady state, the outgoing stream has substrate concentration of 2.5 grams/liter. The
rate of consumption (grams/liter/minute) of the substrate in the reactor is ___________.
Chemostat Basics
A chemostat maintains steady-state microbial growth by balancing inflow and outflow of medium. The dilution rate D = F/V equals the specific growth rate μ at steady state, where F is the volumetric flow rate and V is the reactor volume.
For this setup, V = 5 liters and F = 0.2 liters/minute, so D = 0.04 min-1.
Mass Balance Derivation
At steady state, the substrate mass balance equation is F Sin = F Sout + rs V, where rs is the volumetric consumption rate (g/L/min), Sin = 25 g/L, and Sout = 2.5 g/L.
Rearranging gives rs = D (Sin − Sout). Substituting values: rs = 0.04 × (25 − 2.5) = 0.04 × 22.5 = 0.9 g/L/min.
Step-by-Step Calculation
- Compute dilution rate:
D = 0.2/5 = 0.04min-1. - Determine substrate difference:
25 − 2.5 = 22.5g/L. - Multiply:
rs = 0.04 × 22.5 = 0.9g/L/min. This matches exam solutions directly.
Common Misconceptions
No growth parameters like yield or Monod constants are needed, as the balance uses measured concentrations only. Total consumption rate is F (Sin − Sout) = 4.5 g/min, but per liter is requested.
Options implying other values (e.g., 0.045 or 22.5) confuse total rate, volume, or units.
