GATE 2024 BT Q.55: solution and answer

Official answer: 80Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

1. At steady state, the rate at which oxygen is transferred into the broth equals the rate at which the cells consume it: $k_{L}a(C^{\ast }-C_L)=q_{O_2}X$. Here $k_{L}a$ is the volumetric mass transfer coefficient, $C^{\ast }$ the oxygen solubility (saturation concentration), $C_L$ the dissolved oxygen concentration in the broth, $q_{O_2}$ the specific oxygen uptake rate and $X$ the cell (biomass) concentration.

2. The transfer rate is largest when the driving force $(C^{\ast }-C_L)$ is largest, i.e. when $C_L\to 0$. So the maximum oxygen transfer rate is ${\text{OTR}}_{max}=k_{L}a{C}^{\ast }$, and the largest biomass concentration the aeration can support is $X_{max}=\frac{k_{L}a{C}^{\ast }}{q_{O_2}}$.

3. Put both rates in the same time unit: $k_{L}a=60{\text{min}}^{-1}=\frac{60}{60\text{s}}=1{\text{s}}^{-1}$.

4. Maximum oxygen transfer rate: ${\text{OTR}}_{max}=k_{L}a{C}^{\ast }=1{\text{s}}^{-1}\times 8\times 10^{-3}{\text{kg m}}^{-3}=8\times 10^{-3}{\text{kg O}}_2{\text{m}}^{-3}{\text{s}}^{-1}$.

5. $q_{O_2}=1\times 10^{-4}{\text{g O}}_2{\text{g}}^{-1}{\text{s}}^{-1}$, which is the same as $1\times 10^{-4}{\text{kg O}}_2{\text{kg}}^{-1}{\text{s}}^{-1}$, because a mass-per-mass ratio does not depend on the mass unit.

6. $$X_{max}=\frac{{\text{OTR}}_{max}}{q_{O_2}}=\frac{8\times 10^{-3}{\text{kg O}}_2{\text{m}}^{-3}{\text{s}}^{-1}}{1\times 10^{-4}{\text{kg O}}_2{\text{kg}}^{-1}{\text{s}}^{-1}}=80{\text{kg m}}^{-3}$$

7. The reactor volume (10,000 L) does not enter the answer, because both the transfer rate and the uptake rate are per unit volume of broth.