Question 3 of 6: Batch Culture Growth Kinetics

Question 3: Batch Culture Growth Kinetics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. μ_max, Y_X/S, and the total time to reach stationary phase if S₀ were 2 g/L instead of 10 g/L.

Approach. The doubling time fixes μ_max directly; exponential growth over the (total − lag) interval gives the biomass concentration reached when the substrate is exhausted (no decline phase ⇒ stationary phase begins exactly when S→0), which yields Y_X/S from the overall substrate consumed. The same Y_X/S and μ_max then apply to the S₀=2 g/L case (S≫K_S throughout ⇒ growth stays at μ_max right up to substrate exhaustion).

1. Maximum specific growth rate. Exponential growth doubles X every t_d, so μ_maxt_d=ln 2: $$\mu_{max}=\frac{\ln 2}{t_d}=\frac{0.6931}{2}=\boxed{0.347\ \text{h}^{-1}}$$
2. Biomass at the onset of stationary phase. The exponential phase lasts from t=3 h to t=14 h, i.e. t_exp=11 h: $$X_f=X_0\,e^{\mu_{max}t_{exp}}=0.1\,e^{0.3466(11)}=0.1(45.25)=\boxed{4.53\ \text{g/L}}$$
3. Yield coefficient. With no decline phase, the culture enters stationary phase exactly when the glucose is exhausted (S→0), so all 10 g/L of substrate was converted to the biomass gained: $$Y_{X/S}=\frac{X_f-X_0}{S_0}=\frac{4.53-0.1}{10}=\boxed{0.443\ \text{g/g}}$$
4. Time to stationary phase with S₀=2 g/L. Since S≫K_S holds throughout this fermentation, growth proceeds at μ_max right up until the (now smaller) glucose supply is exhausted, using the SAME Y_X/S and the SAME 3 h lag: $$X_{f,new}=X_0+Y_{X/S}S_{0,new}=0.1+0.443(2)=\boxed{0.985\ \text{g/L}}$$ $$t_{exp,new}=\frac{\ln(X_{f,new}/X_0)}{\mu_{max}}=\frac{\ln(0.985/0.1)}{0.3466}=\frac{2.288}{0.3466}=6.60\ \text{h}$$ $$t_{total,new}=t_{lag}+t_{exp,new}=3+6.60=\boxed{9.60\ \text{h}}$$