GATE 2026 BT Q.56: solution and answer

Official answer: 100Source: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Symbols: $V$ = reactor volume, $F$ = volumetric flow rate, $X$ = steady-state biomass concentration inside the reactor, $X_e$ = biomass concentration in the effluent, ${\theta }_c$ = mean cell retention time (also called solids retention time).

2. Definition: ${\theta }_c=\frac{\text{biomass held in the reactor}}{\text{rate at which biomass leaves the system}}$. The question gives no separate wastage stream, so all the biomass that leaves goes out in the effluent, at flow $F$ and concentration $X_e$.

3. Biomass held in the reactor: $VX=100{\text{m}}^3\times 200{\text{mg L}}^{-1}$.

4. Rate at which biomass leaves: $F{X}_e=10{\text{m}}^3{\text{day}}^{-1}\times 20{\text{mg L}}^{-1}$. Both concentrations are in ${\text{mg L}}^{-1}$ and both volumes in ${\text{m}}^3$, so these units cancel and only $\text{day}$ is left.

5. $${\theta }_c=\frac{VX}{F{X}_e}=\frac{100\times 200}{10\times 20}=\frac{20000}{200}=100\text{days}$$

6. Check: the hydraulic retention time is $V/F=100/10=10$ days. Recycle keeps cells in the reactor $X/X_e=200/20=10$ times longer than the liquid, so ${\theta }_c=10\times 10=100$ days.