GATE 2023 BT Q.48: solution and answer

Official answer: 10Source: Official answer key, GATE 2023 BT (IIT Kanpur)

1. At chemostat steady state (sterile feed, negligible cell death) the biomass balance gives $\mu =D=0.2{\text{h}}^{-1}$. Here $\mu$ is the specific growth rate and $D$ is the dilution rate.

2. Volumetric biomass production rate: $r_X=\mu X=DX=0.2\times 10=2{\text{g L}}^{-1}{\text{h}}^{-1}$, where $X=10{\text{g L}}^{-1}$ is the steady-state biomass concentration.

3. The biomass yield on oxygen, $Y_{X/O}=0.2{\text{g biomass per g O}}_2$, links growth to oxygen use: $\text{OUR}=\frac{\mu X}{Y_{X/O}}$, where OUR is the volumetric oxygen uptake rate. The specific uptake rate is $q_{O_2}=\mu /Y_{X/O}=0.2/0.2=1{\text{g O}}_2{\text{g}}^{-1}{\text{h}}^{-1}$.

4. $$\text{OUR}=\frac{2{\text{g L}}^{-1}{\text{h}}^{-1}}{0.2{\text{g g}}^{-1}}=10{\text{g O}}_2{\text{L}}^{-1}{\text{h}}^{-1}$$

5. At steady state dissolved oxygen does not accumulate, and the O${}_2$ carried out dissolved in the outflow is negligible. So the oxygen transfer rate equals the oxygen uptake rate: $\text{OTR}=\text{OUR}=10{\text{g L}}^{-1}{\text{h}}^{-1}$.