Monod Growth Rate Calculation: 23-Chem-B4 Biochemical Engineering, May 2013, Q6

Question 6: Monod Growth Rate Calculation (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.

Quantity	Symbol	Value
Maximum specific growth rate	μ_max	0.5 h^-1
Monod (saturation) constant	K_S	25 mg/L
Glucose concentration	S	40 g/L = 40,000 mg/L
Cell (biomass) concentration	X	5 g/L (dry weight)

Find. The cell growth rate: specific growth rate μ and volumetric growth rate r_x.

Approach. Convert S and K_S to the same units, apply the Monod equation for the specific growth rate μ, then multiply by the biomass concentration X to get the volumetric growth rate r_x=μX.

Fig. 7 — Monod curve μ vs. S; the operating point (S=40,000 mg/L) sits far to the right of K_S=25 mg/L.

Unit conversion. Express glucose concentration in the same units as K_S: $$S=40\ \text{g/L}\times1000\ \text{mg/g}=40{,}000\ \text{mg/L}$$
Specific growth rate (Monod equation). $$\mu=\frac{\mu_{max}S}{K_S+S}=\frac{0.5(40{,}000)}{25+40{,}000}=\frac{0.5(40{,}000)}{40{,}025}=0.5\times0.99938=\boxed{0.4997\ \text{h}^{-1}}$$ Because S/K_S=1600≫1, the culture is deep in the substrate-saturated (zero-order) region of the Monod curve, so μ≈μ_max to within 0.06%.
Volumetric growth rate. The rate of new biomass formation per unit reactor volume is r_x=μX: $$r_X=\mu X=0.4997\ \text{h}^{-1}\times5\ \text{g/L}=\boxed{2.50\ \text{g biomass}/(\text{L}\cdot\text{h})}$$

Quantity	Value
Specific growth rate, μ	0.4997 h^-1 (≈ μ_max)
Volumetric growth rate, r_X	2.50 g/(L·h)

Topic: Direct application of the Monod equation away from the chemostat context — computing an instantaneous growth rate from a single (S, X) snapshot.

Key relations: μ=μ_maxS/(K_S+S); volumetric rate r_X=μX; at S≫K_S, μ→μ_max (zero-order in substrate); at S=K_S, μ=μ_max/2 by definition of the saturation constant.

Why this works: The Monod equation is a saturation (Michaelis–Menten-type) form: at low S growth is substrate-limited (μ roughly proportional to S), while at high S (S≫K_S, as here) the growth machinery is saturated and μ plateaus at μ_max regardless of further increases in S.

Common pitfall: Mixing units (S in g/L, K_S in mg/L) without converting first — here that error would make S look 1000× smaller relative to K_S and give a wildly wrong, substrate-limited μ. Also confusing the specific growth rate μ (units h^-1) with the volumetric growth rate r_X=μX (units g/(L·h)) — "growth rate" is ambiguous between the two, so both are reported here.

Source: Shuler & Kargi, Bioprocess Engineering, 2nd ed., Ch. 6 (Monod kinetics).

Question 6 of 6: Monod Growth Rate Calculation

Question 6: Monod Growth Rate Calculation (20 marks)