Q.52 In a chemostat with a dilution rate of 0.8 h−1, the steady state biomass
concentration and the specific product formation rate are 8 mol m−3 and
0.2 (mol product)(mol biomass)−1 h−1, respectively.
The steady state product concentration in mol m−3 is _________.
Chemostats maintain steady-state conditions where the dilution rate equals the specific growth rate, and product formation follows a simple mass balance. For this problem, the steady-state product concentration is calculated as 1.6 mol m-3 using the given values.
Core Calculation
In steady state, the product formation rate equals the dilution rate times product concentration: D × P = qp × X. Rearranging gives P = (qp × X) / D.
Substitute values: P = (0.2 mol product mol-1 biomass h-1 × 8 mol biomass m-3) / 0.8 h-1 = 1.6 mol m-3.
This assumes non-growth-associated product formation (qp constant) and no product in the feed, standard for such problems.
Equation Breakdown
The mass balance for product is: d(VP)/dt = 0 = qp × X × V - D × P × V, simplifying to P = (qp × X)/D.
- qp (specific product formation rate) drives production per biomass.
- X (biomass concentration) scales total biomass.
- D (dilution rate) sets washout rate matching production.
Why This Works
Chemostat steady state requires growth rate μ = D = 0.8 h-1, but product uses qp directly since it’s specified separately from growth-associated kinetics. Biomass X = 8 mol m-3 is given directly, avoiding substrate yield calculations.
Common Mistakes
- Assuming growth-associated product (qP = αμ + β): Here qp is total specific rate, so direct formula applies.
- Units mix-up: mol product mol-1 biomass h-1 × mol m-3 / h-1 = mol m-3.
- Forgetting steady-state balance: Production rate qp X equals output D P.
