GATE 2022 BT Q.58: solution and answer

Official answer: 150Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

1. Symbols: $X_0=50\text{g}$ is the total biomass in the reactor when feeding starts, $F=1{\text{L h}}^{-1}$ is the volumetric feed rate, $S_0=50{\text{g L}}^{-1}$ is the substrate concentration in the feed, $Y_{XS}^M=0.4{\text{g biomass (g substrate)}}^{-1}$ is the maximum biomass yield on substrate, and $X_t$ is the total biomass (in grams, not a concentration) at time $t$.

2. Quasi-steady state in fed-batch: the cells consume substrate as fast as it is fed, so the substrate concentration in the reactor stays close to zero ($S\approx 0$). Every gram of substrate fed is therefore converted into biomass at the yield $Y_{XS}^M$.

3. The total biomass then grows at a constant rate: $\frac{d{X}_t}{dt}=Y_{XS}^{M}F{S}_0$. Integrating from $X_t=X_0$ at $t=0$ gives $X_t=X_0+Y_{XS}^{M}F{S}_{0}t$, so total biomass rises linearly with time. The growing culture volume does not enter, because the question asks for the total amount of biomass, not its concentration.

4. Substrate fed in 5 h: $F{S}_{0}t=1{\text{L h}}^{-1}\times 50{\text{g L}}^{-1}\times 5\text{h}=250\text{g}$.

5. Biomass formed from this substrate: $Y_{XS}^{M}F{S}_{0}t=0.4\times 250=100\text{g}$.

6. $$X_t=X_0+Y_{XS}^{M}F{S}_{0}t=50+100=150\text{g}$$ This is the maximum possible amount because the maximum yield was used, which assumes no substrate is diverted to maintenance or product formation.