Question 4 of 5: Overall vs. Liquid-Film Oxygen Mass-Transfer Coefficient (Two-Film Theory)

Question 4: Overall vs. Liquid-Film Oxygen Mass-Transfer Coefficient (Two-Film Theory) (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.

Approach. Apply Whitman's two-resistance (two-film) theory to derive the overall liquid-phase coefficient K_L in terms of the individual gas- and liquid-film coefficients (k_G, k_L) and Henry's constant, then substitute realistic orders of magnitude for O₂ in water to show the gas-phase resistance term is numerically negligible.

1. Set up the two-film flux balance. At steady state the same molar flux N crosses both films and the interface itself is at local equilibrium (Henry's law, p_i=H_pcC_i): $$N=k_G(p_G-p_i)=k_L(C_i-C_L)$$
2. Eliminate the (unmeasurable) interfacial values. From the gas-film equation, $p_i=p_G-N/k_G$, so $C_i=p_i/H_{pc}=p_G/H_{pc}-N/(H_{pc}k_G)=C_L^{*}-N/(H_{pc}k_G)$, where $C_L^{*}=p_G/H_{pc}$ is the liquid concentration that would be in equilibrium with the bulk gas. Substituting into the liquid-film equation: $$\frac{N}{k_L}=C_i-C_L=C_L^{*}-\frac{N}{H_{pc}k_G}-C_L$$
3. Collect into a single overall liquid-based coefficient. $$N\left[\frac{1}{k_L}+\frac{1}{H_{pc}k_G}\right]=C_L^{*}-C_L \quad\Rightarrow\quad N=K_L(C_L^{*}-C_L),\qquad \boxed{\frac{1}{K_L}=\frac{1}{k_L}+\frac{1}{H_{pc}k_G}}$$ The two resistances (liquid-film, gas-film-as-seen-from-the-liquid-side) add in series, exactly as for two resistors carrying the same current.
4. Substitute realistic O₂-in-water values to show the gas term is negligible. O₂ is a POORLY SOLUBLE gas, so its Henry's constant is large: $H_{pc}\approx0.773\ \text{atm}\cdot\text{m}^3/\text{mol}$ (from the literature value $H\approx4.3\times10^4$ atm/mole-fraction at 25°C, divided by water's molar density $\approx55{,}600$ mol/m³). The film coefficients are usually quoted in m/s, so the gas-film coefficient must be put on a pressure basis before it is combined with H_pc: $k_G=k_G'/RT$, with $RT=8.206\times10^{-5}\times298.15=0.02447$ atm·m³/mol. Equivalently, use the dimensionless Henry constant $H_{cc}=H_{pc}/RT=31.6$ (gas-phase O₂ concentration is about 32× its equilibrium dissolved concentration). Taking typical film coefficients $k_L\sim1\times10^{-4}$ m/s and $k_G'\sim1\times10^{-2}$ m/s (gas-phase diffusivities, and hence film coefficients, are much larger than liquid-phase ones): $$H_{pc}k_G=H_{cc}k_G'=(31.6)(1\times10^{-2})=0.316\ \text{m/s}$$ $$\frac{1}{k_L}=10{,}000\ \text{s/m}\qquad \frac{1}{H_{pc}k_G}=3.16\ \text{s/m}$$ $$K_L=\left(10{,}000+3.16\right)^{-1}=9.997\times10^{-5}\ \text{m/s}\qquad \boxed{\frac{K_L}{k_L}=0.9997\ (99.97\%)}$$ The gas-phase resistance contributes only about 0.03% of the total. Even if k_G' were 100× smaller, the gas film would still add only about 3%. This confirms $K_L\approx k_L$ for O₂ in water, exactly as the question asks to prove.