Pseudomonas treats phenolic wastewater aerobically in chemostat mode. Influent and effluent phenol are 10000 and 10 ppm; feed is 80 L/h. μmax = 1 h⁻¹, Ks = 100 mg/L and cell death rate is 0.01 h⁻¹. Find working volume in litres, nearest integer.
For dilute wastewater, 10 ppm ≈ 10 mg/L. Monod growth at the exit concentration is $\mu=1(10)/(100+10)=0.090909$ h⁻¹. Steady biomass requires $D=\mu-k_d=0.080909$ h⁻¹. Thus $V=F/D=80/0.080909=988.76$ L, rounding to 989 L. The inlet concentration determines substrate removal, while exit concentration sets the growth rate.
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