CHE 318 Lecture 22

Interface Mass Transfer

Author

Dr. Tian Tian

Published

March 4, 2026

Note

Slides 👉 Open presentation🗒️
PDF version of course note 👉 Open in pdf
Handwritten notes 👉 Open in pdf

Learning outcomes

After this lecture, you will be able to:

Recall the equilibrium conditions that apply at phase interfaces.
Describe equilibrium diagrams for two-phase mass transfer systems.
Apply coupled flux and equilibrium relations to determine interfacial compositions.

What systems have we studies so far?

The most complex case is probably a packed-bed column.

We we have focused on?
- Mass transfer in 1 phase – gas flow over solid spheres
- Solve mass balance equation in flow direction – get outlet concentration
- Solve mass transfer to beds – get concentration profiles using fixed interfacial concentration
What we may miss?
- Real-world applications are mass transfer between 2 phases
- Mass transfer may occur across 2 phase interfaces
- Equilibrium concentration at interfaces are usually not fixed

Examples of 2-phase mass transfer (1)

Absorption tower

Water-soluble gases from industrial reaction mixture is transferred into aqueous solution
Examples:
- ammonia (NH$_3$) from Haber-Bosch process
- CO$_2$ capture (a hot topic!)

Examples of 2-phase mass transfer (2)

Extraction apparatus (liquid-gas)

Volatile chemical compounds originally mixed with water are extracted to vapour phase
Examples:
- Essence oil extraction

Examples of 2-phase mass transfer (3)

Extraction apparatus (liquid-liquid)

Chemical compound in low solubility liquid is transferred to high solubility liquid
Examples:
- Supercritical CO$_2$ extraction of bioactive compounds

Common feature of 2-phase mass transfer

Flow rate and interface transfer are usually orthogonal
Interface concentration has discontinuity

Mass balance equation in 2-phase M.T.

Overall mass balance between liquid and gas

\[\begin{align} \text{In}_{\text{liq}} + \text{In}_{\text{gas}} = \text{Out}_{\text{liq}} + \text{Out}_{\text{gas}} \end{align}\]

In and outlet usually can be described by $\text{[Flow rate]}\times \text{[Concentration]}$
Depends on the direction of flow and control volume!

Mass balance equation in single phase

In each phase, we can use our knowledge from packed bed lecture, e.g.

\[\begin{align} \text{In}_{\text{gas}} - \text{Out}_{\text{gas}} + \text{Gen}_{\text{gas}} &= 0 \\ Q (c_1 - c_2) + A_{\text{eff}} \hat{N}_{\text{eff}} &= 0 \end{align}\]

$\hat{N}_{\text{eff}}$ is the average molar inter-phase flux, and $A_{\text{eff}}$ is the effective contact area
Cannot use packed-bed solution for $\hat{N}_{\text{eff}}$ because interfacial concentration can vary!

How can we describe the interfacial transport?

In Lecture 14 we discussed the interfacial concentration and mass balance. We need to know 1) The equilibrium constant $K$ at the interface 2) The ratio between $k_c'$ in two phases

The equilibrium plot for gas-liquid interface

Most commonly in industry we can use the equilibrium plot between A’s molar fractions in gas $y_A$ (or $p_A$) and liquid $x_A$, respectively.

Simpliest situation is Henry’s law

\[ p_A = H x_A \]

Reading an equilibrium plot (1)

Meaning of points on the equilibrium curve – interfacial concentraion

Reading an equilibrium plot (2)

Points above the equilibrium curve 👉 $N_A$: gas → liquid (vice versa)

Reading an equilibrium plot (3)

Non-equilibrium point + line with slope $-k_x / k_y$ 👉 interfacial concentration

Equilibrium phase: flux balance

The slope + intercept method stems from the flux balance between phases

\[\begin{align} N_A(g) &= N_A(l) \\ k_y (y_{AG} - y_{Ai}) &= k_x (x_{Ai} - x_{AL}) \end{align}\]

We have

\[\begin{align} \text{Slope} &= \frac{y_{AG} - y_{Ai}}{x_{AL} - x_{Ai}} \\ &= - \frac{k_x}{k_y} \end{align}\]

Example 1: finding equilibrium interface concentrations

A solute is being absorbed from a gas mixture of A and B in a wetted-wall tower, with the liquid flowing downwards. At a certain point in the tower, the bulk gas concentration of A is $y_{AG}=0.380$ and the bulk liquid fraction is $x_{AL}=0.100$. The film transfer coefficients for A in gas and liquid phases are: $k_y'=1.465\times 10^{-3}$ kg mol/m$^2$/s and $k_x'=1.967\times 10^{-3}$ kg mol/m$^2$/s. You can assume the $k_x' \approx k_x$ and $k_y' \approx k_y$. The following pairs of equilibrium $(x_Ai, y_Ai)$ data were measured:

$(0.0, 0.0), (0.05, 0.022), (0.10, 0.052) (0.15, 0.087)$

$(0.20, 0.131), (0.25, 0.187), (0.30, 0.265), (0.35, 0.385)$

Find the interface concentrations $y_{Ai}$ and $x_{Ai}$
Calculate the $N_A$ at this point

Solution steps:

Draw the equilbirum plot with interpolation
Draw the current $(x_{AL}, y_{AG})$ point on graph, which direction of mass transfer?
Draw line with slow of $-k_x/k_y$
Read the intersect with equilibrium curve as $(x_{Ai}, y_{Ai})$
Calculate $N_A = k_y (y_{AG} - y_{Ai})$

Example 1: solution plot

Interface concentrations: x_Ai = 0.246589, y_Ai = 0.183180
N_A from gas film:    2.883412e-04 (same units as k_y)
N_A from liquid film: 2.883412e-04 (same units as k_x)

Example 1: answers

Slope of curve $-k_x/k_y = -1.343$
Interfacial concentration $(x_{Ai}, y_{Ai}) = (0.246, 0.180)$
Flux: $N_A =0.29 \times 10^{-3}$ kg mol/m$^2$/s
Direction of flux: gas to liquid

Summary

Real industrial applications involve mass transfer between 2 phases
Equilibrium plots are extremely useful for elucidating the interfacial balance
Describe driving force and interfacial concentrations from the equilibrium plot

--- title: "CHE 318 Lecture 22" subtitle: "Interface Mass Transfer" author: "Dr. Tian Tian" date: "2026-03-04" format: html: {} revealjs: output-file: slides.html pdf: output-file: L22.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L22.pdf) - Handwritten notes 👉 [Open in pdf](./public/L22_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** the equilibrium conditions that apply at phase interfaces. - **Describe** equilibrium diagrams for two-phase mass transfer systems. - **Apply** coupled flux and equilibrium relations to determine interfacial compositions. ## What systems have we studies so far? The most complex case is probably a packed-bed column. - We we have focused on? - Mass transfer in **1 phase** -- gas flow over solid spheres - Solve mass balance equation in flow direction -- get **outlet** concentration - Solve mass transfer to beds -- get concentration profiles using **fixed interfacial concentration** - What we may miss? - Real-world applications are mass transfer between 2 phases - Mass transfer may occur across 2 phase interfaces - Equilibrium concentration at interfaces are usually not fixed ## Examples of 2-phase mass transfer (1) **Absorption tower** - Water-soluble gases from industrial reaction mixture is transferred into aqueous solution - Examples: - ammonia (NH$_3$) from Haber-Bosch process - CO$_2$ capture (a hot topic!) ![](./public/absorption-tower.png) ## Examples of 2-phase mass transfer (2) **Extraction apparatus** (liquid-gas) - Volatile chemical compounds originally mixed with water are extracted to vapour phase - Examples: - Essence oil extraction ![](./public/essense-oil-extraction.png) ## Examples of 2-phase mass transfer (3) **Extraction apparatus** (liquid-liquid) - Chemical compound in low solubility liquid is transferred to high solubility liquid - Examples: - Supercritical CO$_2$ extraction of bioactive compounds ![](./public/sco2-extraction.png) ## Common feature of 2-phase mass transfer - Flow rate and interface transfer are usually orthogonal - Interface concentration has discontinuity ![](./public/common-2phase.svg) ## Mass balance equation in 2-phase M.T. Overall mass balance between liquid and gas ```{=tex} \begin{align} \text{In}_{\text{liq}} + \text{In}_{\text{gas}} = \text{Out}_{\text{liq}} + \text{Out}_{\text{gas}} \end{align} ``` - In and outlet usually can be described by $\text{[Flow rate]}\times \text{[Concentration]}$ - Depends on the direction of flow and control volume! ## Mass balance equation in single phase In each phase, we can use our knowledge from [packed bed lecture](../L21), e.g. ```{=tex} \begin{align} \text{In}_{\text{gas}} - \text{Out}_{\text{gas}} + \text{Gen}_{\text{gas}} &= 0 \\ Q (c_1 - c_2) + A_{\text{eff}} \hat{N}_{\text{eff}} &= 0 \end{align} ``` - $\hat{N}_{\text{eff}}$ is the average molar inter-phase flux, and $A_{\text{eff}}$ is the effective contact area - Cannot use packed-bed solution for $\hat{N}_{\text{eff}}$ because interfacial concentration can vary! ## How can we describe the interfacial transport? In [Lecture 14](../L14) we discussed the interfacial concentration and mass balance. We need to know 1) The equilibrium constant $K$ at the interface 2) The ratio between $k_c'$ in two phases ![](../L14/public/distribution-coeff-profile.png){width="95%"} ## The equilibrium plot for gas-liquid interface Most commonly in industry we can use the equilibrium plot between A's molar fractions in gas $y_A$ (or $p_A$) and liquid $x_A$, respectively. Simpliest situation is Henry's law $$ p_A = H x_A $$ ![](./public/henry-law.png) ## Reading an equilibrium plot (1) Meaning of points on the equilibrium curve -- interfacial concentraion ![](./public/eq-template1.svg) ## Reading an equilibrium plot (2) - Points above the equilibrium curve 👉 $N_A$: gas → liquid (vice versa) ![](./public/eq-template2.svg) ## Reading an equilibrium plot (3) - Non-equilibrium point + line with slope $-k_x / k_y$ 👉 interfacial concentration ![](./public/eq-template3.svg) ## Equilibrium phase: flux balance - The slope + intercept method stems from the flux balance between phases ```{=tex} \begin{align} N_A(g) &= N_A(l) \\ k_y (y_{AG} - y_{Ai}) &= k_x (x_{Ai} - x_{AL}) \end{align} ``` - We have ```{=tex} \begin{align} \text{Slope} &= \frac{y_{AG} - y_{Ai}}{x_{AL} - x_{Ai}} \\ &= - \frac{k_x}{k_y} \end{align} ``` ## Example 1: finding equilibrium interface concentrations A solute is being absorbed from a gas mixture of A and B in a wetted-wall tower, with the liquid flowing downwards. At a certain point in the tower, the bulk gas concentration of A is $y_{AG}=0.380$ and the bulk liquid fraction is $x_{AL}=0.100$. The film transfer coefficients for A in gas and liquid phases are: $k_y'=1.465\times 10^{-3}$ kg mol/m$^2$/s and $k_x'=1.967\times 10^{-3}$ kg mol/m$^2$/s. You can assume the $k_x' \approx k_x$ and $k_y' \approx k_y$. The following pairs of equilibrium $(x_Ai, y_Ai)$ data were measured: $(0.0, 0.0), (0.05, 0.022), (0.10, 0.052) (0.15, 0.087)$ $(0.20, 0.131), (0.25, 0.187), (0.30, 0.265), (0.35, 0.385)$ 1) Find the interface concentrations $y_{Ai}$ and $x_{Ai}$ 2) Calculate the $N_A$ at this point ## Solution steps: 1) Draw the equilbirum plot with interpolation 2) Draw the current $(x_{AL}, y_{AG})$ point on graph, which direction of mass transfer? 3) Draw line with slow of $-k_x/k_y$ 4) Read the intersect with equilibrium curve as $(x_{Ai}, y_{Ai})$ 5) Calculate $N_A = k_y (y_{AG} - y_{Ai})$ ## Example 1: solution plot ```{python} #| echo: false import numpy as np import matplotlib.pyplot as plt # Given bulk compositions y_AG = 0.380 x_AL = 0.100 # Given film coefficients (treat as ky, kx per problem statement) k_y = 1.465e-3 k_x = 1.967e-3 # Equilibrium data (x_Ai, y_Ai) x_eq = np.array([0.0, 0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35]) y_eq = np.array([0.0, 0.022, 0.052, 0.087, 0.131, 0.187, 0.265, 0.385]) # Operating line from two-film theory: # N_A = k_y (y_AG - y_i) = k_x (x_i - x_AL) # => y_i = y_AG - (k_x/k_y) (x_i - x_AL) m = -(k_x / k_y) b = y_AG - m * x_AL # y = m x + b def y_oper(x): return m * x + b # Piecewise-linear interpolation of equilibrium curve and intersection with operating line # Find interval where (y_eq - y_oper(x_eq)) changes sign g = y_eq - y_oper(x_eq) idx = np.where(np.sign(g[:-1]) * np.sign(g[1:]) <= 0)[0] if len(idx) == 0: raise RuntimeError("No intersection found within provided equilibrium data range.") i = idx[0] # Linear interpolation on [x_i, x_{i+1}] for equilibrium: y = y0 + (y1-y0)*(x-x0)/(x1-x0) x0, x1 = x_eq[i], x_eq[i+1] y0, y1 = y_eq[i], y_eq[i+1] s = (y1 - y0) / (x1 - x0) # equilibrium segment slope # Solve y_eq_lin(x) = y_oper(x): # y0 + s (x - x0) = m x + b x_int = (y0 - s * x0 - b) / (m - s) y_int = y_oper(x_int) # Flux at the point N_A_gas = k_y * (y_AG - y_int) N_A_liq = k_x * (x_int - x_AL) print(f"Interface concentrations: x_Ai = {x_int:.6f}, y_Ai = {y_int:.6f}") print(f"N_A from gas film: {N_A_gas:.6e} (same units as k_y)") print(f"N_A from liquid film: {N_A_liq:.6e} (same units as k_x)") # Plot xx = np.linspace(x_eq.min(), x_eq.max(), 400) plt.figure() plt.plot(0.100, 0.380, "s") l, = plt.plot(x_eq, y_eq, "o", label="Equilibrium data") plt.plot(xx, np.interp(xx, x_eq, y_eq), "-", color=l.get_c()) plt.plot(xx, y_oper(xx), "--",) plt.plot([x_int], [y_int], "s") plt.xlabel(r"$x_A$") plt.ylabel(r"$y_A$") plt.title(r"Find $(x_{Ai}, y_{Ai})$ from equilibrium ∩ operating line") plt.xlim(x_eq.min(), x_eq.max()) plt.ylim(min(0, y_oper(x_eq).min(), y_eq.min()), max(y_oper(x_eq).max(), y_eq.max())) plt.grid(True, alpha=0.3) plt.legend() plt.show() ``` ## Example 1: answers - Slope of curve $-k_x/k_y = -1.343$ - Interfacial concentration $(x_{Ai}, y_{Ai}) = (0.246, 0.180)$ - Flux: $N_A =0.29 \times 10^{-3}$ kg mol/m$^2$/s - Direction of flux: gas to liquid ## Summary - Real industrial applications involve mass transfer between 2 phases - Equilibrium plots are extremely useful for elucidating the interfacial balance - Describe driving force and interfacial concentrations from the equilibrium plot