Oxygen mass transfer to fermenting microorganisms: 23-Chem-A3 Heat and Mass Transfer, December 2015, Q2

Question 2: Oxygen mass transfer to fermenting microorganisms (Part A — equal value)

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. A large population of tiny stationary spheres suspended in near-quiescent liquid. Each microorganism is small enough that the surrounding liquid behaves as an infinite stagnant medium, for which the Sherwood number is $Sh=2$. (The densities and viscosity in the data table let us check the small-particle natural-convection term of Geankoplis’ correlation $Sh=2+0.31\,[d_p^3(\rho_p-\rho)g/(\mu D)]^{1/3}$: with $\rho_p-\rho=106$ kg/m³ the bracket is $1.4\times10^{-4}$, so the correction is $0.016$ — under 1% — and $Sh=2$ stands.)

Quantity	Value
Cell diameter $d_p$	$0.667\ \mu$m $=6.67\times10^{-7}$ m
Cell density $\rho_p$	1100 kg/m³
Total wet mass	5 g $=0.005$ kg
$D_{O_2}$ in water (37 °C)	$3.25\times10^{-9}$ m²/s
O₂ saturation solubility $c_{sat}$	$2.26\times10^{-4}$ kg mol/m³
Actual O₂ uptake (part b)	$6.3\times10^{-6}$ kg mol/s

Find. (a) the maximum O₂ transfer rate; (b) the dissolved-O₂ concentration as percent of saturation.

Figure 2. Each cell is a stationary sphere in stagnant liquid ($Sh=2$); O₂ diffuses from the saturated bulk across a stagnant film to the cell surface where it is consumed.

Approach. Take $Sh=2$ to get $k_c$, sum the surface area of all cells from the total mass, multiply by the saturation driving force for the maximum rate, then back out the actual driving force (hence percent saturation) from the given uptake.

Mass-transfer coefficient. For a stationary sphere in stagnant liquid $Sh=k_c d_p/D=2$, so $$k_c=\frac{2D}{d_p}=\frac{2(3.25\times10^{-9})}{6.67\times10^{-7}}=9.75\times10^{-3}\ \text{m/s}.$$
Total interfacial area. Number of cells $N=m/(\rho_p\tfrac{\pi}{6}d_p^3)$; total area $A=N\pi d_p^2=\dfrac{6m}{\rho_p d_p}=\dfrac{6(0.005)}{(1100)(6.67\times10^{-7})}=\boxed{40.9\ \text{m}^2}.$ The immense area comes from the sub-micron cell size.
(a) Maximum O₂ rate. Maximum driving force is the full saturation value (cell-surface concentration $\to0$): $$R_{max}=k_cA\,c_{sat}=(9.75\times10^{-3})(40.9)(2.26\times10^{-4})=\boxed{9.01\times10^{-5}\ \text{kg mol/s}}.$$
(b) Percent saturation. The actual rate uses the reduced driving force $R=k_cA(c_{sat}-c_L)$, so $c_{sat}-c_L=R/(k_cA)=6.3\times10^{-6}/0.399=1.58\times10^{-5}$. Since $R/R_{max}=6.3\times10^{-6}/9.01\times10^{-5}=7.0\%$ of the driving force is consumed, the dissolved O₂ sits at $$\frac{c_L}{c_{sat}}=1-0.070=\boxed{93.0\%\ \text{of saturation}}.$$

Quantity	Value
$k_c=2D/d_p$	$9.75\times10^{-3}$ m/s
Total cell surface area $A$	40.9 m²
(a) Maximum O₂ transfer rate	$9.01\times10^{-5}$ kg mol/s
(b) Dissolved O₂ as percent saturation	93.0 %

Concept & Theory

Topic: Convective mass transfer to a swarm of small spheres; the $Sh=2$ stagnant-medium limit.

Key relations: $Sh=k_cd_p/D=2$ for a lone sphere in an infinite quiescent fluid; specific area of a monodisperse swarm $A=6m/(\rho_p d_p)$; film rate $R=k_cA(c_{sat}-c_L)$.

Why this works: $Sh=2$ is the exact analytical solution for steady diffusion from a sphere into an unbounded stagnant medium, and it is the lower bound for any convective situation. Because $A\propto1/d_p$ and $k_c\propto1/d_p$, halving the cell size quadruples the volumetric transfer capacity — which is why micro-organisms can be oxygen-limited only at very high cell loadings.

Common pitfall: Confusing “percent of driving force used” (7 %) with “percent saturation” (93 %). The uptake is only 7 % of the maximum, so the liquid stays close to saturation.

Source: Geankoplis, Transport Processes (4th ed.), Example 7.4-2 and §7.4 (mass transfer to spheres).

Question 2 of 7: Oxygen mass transfer to fermenting microorganisms

Question 2: Oxygen mass transfer to fermenting microorganisms (Part A — equal value)