GATE 2026 BT Q.61: solution and answer

Official answer: 2.4 to 2.5Source: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Symbols: ${\mu }_{max}=0.5{\text{h}}^{-1}$ is the maximum specific growth rate, $K_s=0.1{\text{mg L}}^{-1}$ is the Monod constant, $S_0=10{\text{g L}}^{-1}$ is the substrate concentration in the feed, $S$ is the substrate concentration in the reactor, $V=5\text{L}$ is the working volume, $F$ is the feed flow rate and $D=F/V$ is the dilution rate.

2. Put the concentrations in the same units: $K_s=0.1{\text{mg L}}^{-1}=1\times 10^{-4}{\text{g L}}^{-1}$.

3. Steady-state biomass balance with sterile feed: $\mu =D$. With Monod kinetics, $\mu =\frac{{\mu }_{max}S}{K_s+S}$, so the substrate level in the reactor is $$S=\frac{K_{s}D}{{\mu }_{max}-D}$$

4. The largest value $S$ can reach is $S_0$, the feed value (no biomass is left to use any substrate). Washout begins when the $S$ needed to sustain $\mu =D$ reaches $S_0$. Setting $S=S_0$ gives the critical dilution rate $$D_{crit}=\frac{{\mu }_{max}{S}_0}{K_s+S_0}$$ Any $D\ge D_{crit}$ washes the culture out, so $D_{crit}$ is the minimum dilution rate for washout.

5. Substitute: $$D_{crit}=\frac{0.5\times 10}{10+0.0001}=0.499995{\text{h}}^{-1}$$

6. Minimum feed flow rate for washout: $$F_{crit}=D_{crit}V=0.499995\times 5=2.499975{\text{L h}}^{-1}\approx 2.5{\text{L h}}^{-1}$$

7. Check: $K_s$ is 100,000 times smaller than $S_0$, so $D_{crit}\approx {\mu }_{max}$ and $F_{crit}\approx {\mu }_{max}V=0.5\times 5=2.5{\text{L h}}^{-1}$.