An organism grows on glucose in a Monod-model chemostat, with specific growth-rate parameter 0.45 h⁻¹ and Ks = 0.5 g/L. Yield Yxs = 0.4 g/g. Volume is 0.9 L and feed is 1 L/h containing 20 g/L glucose. Find steady-state biomass concentration in g/L.
The dilution rate is $D=F/V=1/0.9=1.111\ \mathrm{h}^{-1}$. A nonzero steady biomass requires growth to balance removal, μ = D. The supplied growth-rate limit 0.45 h⁻¹ is below D, so that balance is impossible and the culture washes out. The physical steady-state biomass is 0 g/L; inserting D into an algebraic nonwashout formula would yield an unphysical result.
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