GATE 2020 BT Q.49: solution and answer

Official answer: 0Source: Official answer key, GATE 2020 BT (IIT Delhi)

1. Symbols: the printed specific growth rate is the Monod maximum, ${\mu }_{max}=0.45{\text{h}}^{-1}$; saturation constant $K_s=0.5{\text{g L}}^{-1}$; biomass yield on glucose $Y_{X/S}=0.4{\text{g g}}^{-1}$; feed glucose $S_0=20{\text{g L}}^{-1}$; feed flow $F=1{\text{L h}}^{-1}$; volume $V=0.9\text{L}$. The feed contains only medium, so it is sterile ($X_0=0$).

2. Dilution rate: $D=\frac{F}{V}=\frac{1{\text{L h}}^{-1}}{0.9\text{L}}\approx 1.11{\text{h}}^{-1}$.

3. Biomass balance on a chemostat with sterile feed: $\frac{dX}{dt}=(\mu -D)X$. A steady state with $X>0$ needs $\mu =D$. But Monod gives $\mu =\frac{{\mu }_{max}S}{K_s+S}<{\mu }_{max}$ for any finite $S$, so $\mu$ can never reach $1.11{\text{h}}^{-1}$ when ${\mu }_{max}=0.45{\text{h}}^{-1}$.

4. Critical (washout) dilution rate, reached when the reactor substrate equals the feed: $D_{crit}=\frac{{\mu }_{max}{S}_0}{K_s+S_0}=\frac{0.45\times 20}{0.5+20}\approx 0.439{\text{h}}^{-1}$.

5. Since $D\approx 1.11{\text{h}}^{-1}>D_{crit}\approx 0.439{\text{h}}^{-1}$, cells are removed faster than they can grow: $\frac{dX}{dt}=(\mu -D)X<0$ for every $X>0$, so the culture washes out.

6. The only steady state is the washout one: $S=S_0=20{\text{g L}}^{-1}$ and $X=0{\text{g L}}^{-1}$.

7. Check: blindly using $S=\frac{K_{s}D}{{\mu }_{max}-D}=\frac{0.5\times 1.11}{0.45-1.11}\approx -0.84{\text{g L}}^{-1}$ gives a negative concentration, which confirms that no non-trivial steady state exists. The biomass concentration at steady state is $0{\text{g L}}^{-1}$.