Maximum Oxygen-Transfer Flux in a Stirred-Tank Bioreactor: 23-Chem-B4 Biochemical Engineering, Undated paper, Q1

Question 1: Maximum Oxygen-Transfer Flux in a Stirred-Tank Bioreactor (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
Working volume	V_L	10 m³
Impeller diameter	D	0.936 m
Rotational speed	N	1 rev/s
Broth density	ρ	1000 kg/m³
Broth viscosity	μ	10⁻³ Pa·s
Power number	N_p	6
Gassed/total power ratio	P_g/P	0.6
O₂ mole fraction in air	y_O2	0.21
Total pressure	P	760 Torr (1 atm)
Temperature	t	20 °C

Find. The maximum volumetric oxygen-transfer rate OTR_max (g O₂ per m³ per h) the impeller/sparger system can supply to the broth.

Fig. 1 — single Rushton-turbine stirred tank with sparged air; OTR_max is set by the achievable k_La at this power input and the saturation driving force from the sparged air.

Approach. Compute the impeller power draw from the given power number, apply the gassed/ungassed ratio to get P_g, use the given empirical correlation to convert volumetric power input into k_La, and combine with the maximum achievable driving force (saturation dissolved oxygen, taking the bulk liquid as oxygen-free) to get the maximum transfer flux.

Impeller (ungassed) power draw. Rearranging the given power-number definition, $$P=N_p\,\rho\,N^3D^5=6\times1000\times(1)^3\times(0.936)^5$$ $$P=6000\times0.71840\ \text{W}=\boxed{4310.5\ \text{W}}$$ (Reynolds number $N_{Re}=\rho N D^2/\mu=1000\times1\times0.936^2/10^{-3}=8.76\times10^5$, fully turbulent — consistent with using a constant power number.)
Gassed power and volumetric power input. With P_g/P = 0.6, $$P_g=0.6\times4310.5=2586.3\ \text{W}=2.5863\ \text{kW}$$ $$\frac{P_g}{V_L}=\frac{2.5863\ \text{kW}}{10\ \text{m}^3}=0.25863\ \text{kW/m}^3$$
Volumetric mass-transfer coefficient. Substituting into the given correlation (P_g/V_L in kW/m³, k_La in s⁻¹), $$k_La=9.09\times10^{-4}\left(\frac{P_g}{V_L}\right)^{0.7}=9.09\times10^{-4}\times(0.25863)^{0.7}$$ $$k_La=9.09\times10^{-4}\times0.3881=\boxed{3.528\times10^{-4}\ \text{s}^{-1}}\ (=1.270\ \text{h}^{-1})$$
Saturation dissolved-oxygen concentration. The question labels P and p as the total and the (oxygen) partial pressure, so for air at 1 atm $P=760$ Torr and $p=y_{O2}P=0.21\times760=159.6$ Torr. At $t=20^\circ$C, which is inside the stated 0–30 °C range, $$DO^*=\frac{(P-p)\times0.678}{35+t}=\frac{(760-159.6)\times0.678}{35+20}=\frac{600.4\times0.678}{55}$$ $$\boxed{DO^*=7.401\ \text{ppm}=7.401\ \text{g/m}^3}$$ (1 ppm ≈ 1 mg/L = 1 g/m³ in dilute aqueous solution.)
Assumption (exam Note 1): p is used exactly as the question labels it, the oxygen partial pressure. The equation is the classical formula for air-saturated water, in which p is normally the saturated water-vapour pressure at t. On that reading p = 17.54 Torr at 20 °C, giving DO* = 9.15 ppm (close to the tabulated 9.1 mg/L) and OTR_max = 11.6 g O₂/(m³·h). The method is identical; only the driving force changes.
Maximum volumetric oxygen flux. The flux is largest when the bulk liquid holds no dissolved oxygen ($C_L=0$), so the full saturation value is the driving force. Converting k_La to per hour for the requested units, $$OTR_{max}=k_La\,(DO^*-0)=3.528\times10^{-4}\ \text{s}^{-1}\times3600\ \text{s/h}\times7.401\ \text{g/m}^3$$ $$\boxed{OTR_{max}=9.40\ \text{g O}_2/(\text{m}^3\cdot\text{h})}$$

Quantity	Value
Ungassed impeller power, P	4310.5 W
Gassed power, P_g	2586.3 W (0.2586 kW/m³)
Volumetric mass-transfer coefficient, k_La	3.528×10⁻⁴ s⁻¹ (1.270 h⁻¹)
Saturation DO, DO*	7.401 g/m³
Maximum O₂ flux, OTR_max	9.40 g O₂/(m³·h)

Topic: Stirred-tank aeration — power draw, empirical k_La correlations, and maximum oxygen-transfer rate.

Key relations: $P=N_p\rho N^3D^5$; $k_La=9.09\times10^{-4}(P_g/V_L)^{0.7}$ (s$^{-1}$); $DO^*=(P-p)(0.678)/(35+t)$; $OTR=k_La(DO^*-C_L)$, maximized at $C_L=0$.

Why this works: The Rushton-turbine power number is essentially constant in the fully turbulent regime ($N_{Re}\gtrsim10^4$), so P can be found algebraically from N_p, ρ, N and D alone; the gassed-power fraction then converts this to the actual aerated power draw, which the empirical correlation maps to a volumetric mass-transfer coefficient. OTR is driving-force × k_La, and is largest when the bulk dissolved-oxygen concentration is smallest — hence "maximum flux" corresponds to C_L=0.

Common pitfall: Forgetting to convert P_g/V_L to kW/m³ before applying the correlation, and forgetting that the correlation returns k_La per second, so it must be multiplied by 3600 to give a flux per hour. Missing the first changes k_La by a factor of 1000^0.7 ≈ 125; missing the second puts the flux out by a factor of 3600.

Source: Shuler & Kargi, Bioprocess Engineering, 2nd ed., Ch. 8 (oxygen transfer, stirred-tank aeration correlations); Perry's Chemical Engineers' Handbook, 9th ed. (impeller power number correlations).

Question 1 of 5: Maximum Oxygen-Transfer Flux in a Stirred-Tank Bioreactor

Question 1: Maximum Oxygen-Transfer Flux in a Stirred-Tank Bioreactor (20 marks)