Q.83 A continuous cell culture being carried out in a stirred tank reactor is described in terms of S. The cell mass concentration of the X substrate in the sterile feed stream is SF = 10 g/L and yield coefficient YX/S = 0.5. (F - flow rate (ml/min) and feed stream and specific growth rate (h-1) = 0.35 in the steady state. The concentration of S is g/L (up to 1 decimal point).

Q.83 A continuous cell culture being carried out in a stirred tank reactor is described in terms of S. The cell mass concentration of the X substrate in the sterile feed stream is SF = 10 g/L and yield coefficient YX/S = 0.5. (F – flow rate (ml/min) and feed stream and specific growth rate (h-1) = 0.35 in the steady state. The concentration of S is g/L (up to 1 decimal point).

Answer: Sf = 10.0 g/L

(reported to one decimal place)


Chemostat Basics

A chemostat maintains a steady-state continuous culture by balancing inflow and outflow
at a defined dilution rate:

D = F / V

Since the feed is sterile (Xi = 0), the steady-state biomass
balance simplifies to:

μX V = D X V

which reduces to:

(μ − D)X = 0

Given μ = 0.3 h−1 and
D = 0.25 h−1, μ ≠ D, so the only feasible steady-state solution is:

X = 0 g/L


Mass Balances Explained

Biomass Balance (steady state)

At steady state (dX/dt = 0):

  • Inflow of biomass = 0
  • Outflow of biomass = D X V
  • Generation of biomass = μ X V

This yields:

(μ − D)X = 0

Since μ > D, biomass cannot accumulate and washout occurs, giving
X = 0 g/L.

Substrate Balance (steady state)

At steady state (dS/dt = 0):

D(S0 − Sf)V = (μ X V) / YX/S

Since X = 0, the consumption term becomes zero:

D(S0 − Sf) = 0

Therefore:

Sf = S0 = 10.0 g/L

The yield coefficient YX/S = 0.5 is unused because no growth occurs.


Why No Growth Occurs

In standard chemostat theory with sterile feed, nonzero biomass at steady state requires
μ = D. When μ is fixed and greater than D (as in this problem),
cells are washed out of the reactor.

Since no Monod parameters (μmax, KS) are provided, μ cannot
adjust to match D. As a result, substrate flows through the system unchanged.


Step-by-Step Solution

  1. Check steady-state biomass condition: μ ≠ D → X = 0 g/L
  2. With no biomass, no substrate is consumed
  3. Exit substrate concentration equals feed concentration
  4. Sf = 10.0 g/L

This problem illustrates key steady-state assumptions commonly tested in
CSIR NET Life Sciences and Biotechnology exams.


Keywords: chemostat steady state, cell mass concentration, substrate yield YX/S,
dilution rate D, continuous culture, Sf calculation

 

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