GATE 2021 BT – Question 61
Batch E. coli has zero-order Monod growth. At termination dissolved O2 is 10% saturation and $k_La=80\ \mathrm{h^{-1}}$. Saturation is 0.007 kg/m³, maximum growth rate 0.2 h⁻¹ and yield 1.5 kg cells/kg O2. Final biomass (two decimal places kg/m³) is _______.
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Correct answer: 3.76 to 3.80
Explanation
**Step 1: oxygen balance.** At the end of the batch the cells have used oxygen at the rate at which it is transferred. The oxygen transfer rate is
$$OTR=k_La\,(C^*-C_L)=80\times(0.007-0.0007)=80\times0.0063=0.504\ \text{kg O}_2\,\text{m}^{-3}\text{h}^{-1},$$
because the dissolved oxygen is 10% of saturation, so $C_L=0.1C^*=0.0007$ kg/m³.
**Step 2: cell growth rate supported by this oxygen.** The yield is 1.5 kg cells per kg O₂:
$$\mu X=Y_{X/O}\times OTR=1.5\times0.504=0.756\ \text{kg m}^{-3}\text{h}^{-1}.$$
**Step 3: biomass.** The growth is zero-order in substrate, so it grows at the maximum rate $\mu=0.2\ \text{h}^{-1}$:
$$X=\frac{0.756}{0.2}=\mathbf{3.78\ kg/m^3}.$$