CHE 318 Lecture 24

Mass Transfer In Two-Phase Column: Realistic Situations

Author

Dr. Tian Tian

Published

March 9, 2026

Note

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Learning outcomes

After this lecture, you will be able to:

Recall the stagnant-film flux relation for two-phase mass transfer.
Apply film and equilibrium relations to determine interfacial compositions and fluxes.
Describe how internal column mass balances shape local driving forces.

Recall of last week: two-phase mass transfer

Useful tool: equilibrium diagram

Key features:

x-axis & y-axis meaning?
Points on and below curve?
Equilibrium line and operating line?
Points above and below eq. line?
Meaning of operating line with flow rates?

Recap: key equation 1 – flux relation

For very diluted (EMCD-like) system, the flux equation in each phase follows:

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

This leads to the slope to the eq. line as:

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

Recap: key equation 2 – mass balance (2-phase)

If we only care what possible $(x, y)$ points are in the tower, can use 2-phase mass balance:

\[\begin{align} L'\left(\frac{x_2}{1 - x_2}\right) + V'\left(\frac{y_1}{1 - y_1}\right) = L'\left(\frac{x_1}{1 - x_1}\right) + V'\left(\frac{y_2}{1 - y_2}\right) \end{align}\]

In very diluted (EMCD-like) system, we have operating line with a slope of $L'/V'$.

Given inlet $y_1$ and target $x_2, y_2$ 👉 minimal operating $L'$ (see Assignment 6!)
Given actual $L'$ 👉 predict outlet $x_1$

What can we improve from last week’s picture?

Applicability

Instead of focusing on very diluted ($1-x\approx 1-y \approx 1$), derive equations for general, diffusion through stagnant film case
Solving interfacial composition for general case

Case study: height requirement in packed absorption tower

How tall should the tower / column be, given the mass transfer coefficients?

More accurate slope for interfacial connection

For non-dilute systems, $1-x_{AL}$ and $1-y_{AG}$ may not be close to

The relation between $(x_{AL}, y_{AG})$ and $(x_{Ai}, y_{Ai})$ is no longer linear
Need correction using log-mean composition terms
Usual description: diffusion of $A$ through non-diffusing $B$

Practical relation 1: flux relation for stagnant-film case

Still require the same flux through gas and liquid films:

\[\begin{align} N_A = \frac{k_y'}{(1-y)_{im}} (y_{AG}-y_{Ai}) = \frac{k_x'}{(1-x)_{im}} (x_{Ai}-x_{AL}) \end{align}\]

So the line connecting bulk point to interface point has slope:

\[\begin{align} \text{Slope} &= \frac{y_{AG}-y_{Ai}}{x_{AL}-x_{Ai}} \\ &= -\frac{k_x'/(1-x)_{im}}{k_y'/(1-y)_{im}} \end{align}\]

uses: log mean correction $(1-x){im}$ and $(1-y){im}$
depend on the actual location

Note: log mean correction to interfacial flux

The notations $(1 - x)_{im}$ and $(1 - y)_{im}$ are log mean values for inert composition between bulk and interface
Just $x_{Bm}$ in the steady-state diffusion problems
Will be frequently used in this week’s lecture!

\[\begin{align} (1-x)_{im} &= x_{Bm} \\ &= \frac{(1 - x_{Ai}) - (1-x_{AL})}{ \ln\!\left( \frac{(1 - x_{Ai})}{(1-x_{AL})} \right) } \end{align}\]

Example 1: finding the interfacial composition (stagnant B)

Solution steps for solving interfacial composition (general case)

Manual trial-and-error steps:

Start from bulk point $P$
Guess slope using current correction terms (initially $(1-x)_{im} = (1-y)_{im} = 1$)
Connect to equilibrium curve to get $(x_{Ai}, y_{Ai})$
Update $(1-x){im}$ and $(1-y){im}$, calculate new slope
Is new slope in 4) converged?
- No 👉 go back to step 2
- Yes 👉 continue to step 6
Get final $(x_{Ai}, y_{Ai})$ and $N_A$

Example 1: step 1 (follow textbook)

Example 1: step 2 (follow textbook)

Practical relation 2: link overall coefficient to film coefficients

We learned last week that writing flux equations using $K_x'$ and $K_y'$ are usually easier than $k_x'$ and $k_y'$
What is the relation between them?
Take gas-phase, diffusion through stagnant film

\[\begin{align} N_A &= \frac{k_y'}{(1-y)_{im}}(y_{AG}-y_{Ai}) \\ &= \frac{K_y'}{(1-y)_{*m}}(y_{AG}-y_A^*) \\ &= K_y (y_{AG}-y_A^*) \end{align}\]

$(1-y)_{*m}$: log mean between $(1-y_{AG})$ and $(1-y_A^*)$

Geometric interpretation of overall $K$

Use gas-phase example, from geometry of the equilibrium diagram:

\[\begin{align} \frac{1}{K_y}=\frac{1}{k_y}+\frac{m'}{k_x} \end{align}\]

Transport resistance and overall mass transfer coefficients

The overall mass transfer coefficient $K$ basically tells which transport resistance is dominant ($k$ inversely proportional to resistance)
The transport equation becomes a “resistance-in-series” analog

\[\begin{align} \text{[Total resistance in gas]} &= \sum \text{resistance in each phase} \\ \frac{1}{K_y} &= \sum f_i \frac{1}{k_i} \end{align}\]

Case 1: overall $K_y$ in gas phase for highly liquid-soluble A

Local slope $m'$ is small
Transfer resistance is mainly in gas!
Design rule: tune $k_y$ –> more efficient mass transfer

\[\begin{align} \frac{1}{K_y} &= \frac{1}{k_y}+\frac{m'}{k_x}\\ &\approx \frac{1}{k_y} \end{align}\]

Case 2: overall $K_x$ in liquid phase for low solubility A

Local slope $m''$ is large
Transfer resistance is mainly in liquid!
Design rule: tune $k_x$ –> more efficient mass transfer

\[\begin{align} \frac{1}{K_x} &= \frac{1}{m'' k_y}+\frac{1}{k_x}\\ &\approx \frac{1}{k_x} \end{align}\]

Example 2: estimate overall mass transfer coefficients

Use the conditions from example 1: bulk phase point $P=(0.10, 0.380)$, $k_x'=1.967\times 10^{-3}$ kg mol/m$^2$/s and $k_y' = 1.465\times 10^{-3}$ kg mol/m$^2$/s, estimate $K_y'$?

Example 2: solutions

Summary

In-depth analysis of diffusion through stagnant interfacial equilibrium
Geometric interpretation of equilibrium diagram
Case studies for interfacial composition & mass transfer coefficient

So far we have built almost all prerequisite for solving the concentration profile in the absorption tower! We will discuss that in upcoming Lecture 25

--- title: "CHE 318 Lecture 24" subtitle: "Mass Transfer In Two-Phase Column: Realistic Situations" author: "Dr. Tian Tian" date: "2026-03-09" format: html: {} revealjs: output-file: slides.html pdf: output-file: L24.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L24.pdf) - Handwritten notes 👉 [Open in pdf](./public/L24_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** the stagnant-film flux relation for two-phase mass transfer. - **Apply** film and equilibrium relations to determine interfacial compositions and fluxes. - **Describe** how internal column mass balances shape local driving forces. ## Recall of last week: two-phase mass transfer Useful tool: equilibrium diagram :::{.columns} :::{.column width="50%"} Key features: - x-axis & y-axis meaning? - Points on and below curve? - Equilibrium line and operating line? - Points above and below eq. line? - Meaning of operating line with flow rates? ::: :::{.column width="50%"} ![](./public/blank-eq-diag.png){width="95%"} ::: ::: ## Recap: key equation 1 -- flux relation For **very diluted** (EMCD-like) system, the flux equation in each phase follows: ```{=tex} \begin{align} N_A = k_y' (y_{AG} - y_{Ai}) = k_x' (x_{Ai} - x_{AL}) \end{align} ``` This leads to the slope to the eq. line as: ```{=tex} \begin{align} \text{Slope} &= - \frac{y_{AG} - y_{Ai}}{x_{Ai} - x_{AL}} \\ &= -\frac{k_x'}{k_y'} \end{align} ``` ## Recap: key equation 2 -- mass balance (2-phase) If we only care what possible $(x, y)$ points are in the tower, can use 2-phase mass balance: ```{=tex} \begin{align} L'\left(\frac{x_2}{1 - x_2}\right) + V'\left(\frac{y_1}{1 - y_1}\right) = L'\left(\frac{x_1}{1 - x_1}\right) + V'\left(\frac{y_2}{1 - y_2}\right) \end{align} ``` In **very diluted** (EMCD-like) system, we have operating line with a slope of $L'/V'$. - Given inlet $y_1$ and target $x_2, y_2$ 👉 minimal operating $L'$ (see Assignment 6!) - Given actual $L'$ 👉 predict outlet $x_1$ ## What can we improve from last week's picture? **Applicability** - Instead of focusing on **very diluted** ($1-x\approx 1-y \approx 1$), derive equations for general, diffusion through stagnant film case - Solving interfacial composition for general case **Case study**: height requirement in packed absorption tower - How tall should the tower / column be, given the mass transfer coefficients? ## More accurate slope for interfacial connection For non-dilute systems, $1-x_{AL}$ and $1-y_{AG}$ may not be close to - The relation between $(x_{AL}, y_{AG})$ and $(x_{Ai}, y_{Ai})$ is no longer linear - Need correction using log-mean composition terms - Usual description: diffusion of $A$ through non-diffusing $B$ ## Practical relation 1: flux relation for stagnant-film case Still require the same flux through gas and liquid films: ```{=tex} \begin{align} N_A = \frac{k_y'}{(1-y)_{im}} (y_{AG}-y_{Ai}) = \frac{k_x'}{(1-x)_{im}} (x_{Ai}-x_{AL}) \end{align} ``` So the line connecting bulk point to interface point has slope: ```{=tex} \begin{align} \text{Slope} &= \frac{y_{AG}-y_{Ai}}{x_{AL}-x_{Ai}} \\ &= -\frac{k_x'/(1-x)_{im}}{k_y'/(1-y)_{im}} \end{align} ``` - uses: log mean correction $(1-x){im}$ and $(1-y){im}$ - depend on the actual location ## Note: log mean correction to interfacial flux - The notations $(1 - x)_{im}$ and $(1 - y)_{im}$ are log mean values for inert composition between bulk and interface - Just $x_{Bm}$ in the steady-state diffusion problems - Will be frequently used in this week's lecture! ```{=tex} \begin{align} (1-x)_{im} &= x_{Bm} \\ &= \frac{(1 - x_{Ai}) - (1-x_{AL})}{ \ln\!\left( \frac{(1 - x_{Ai})}{(1-x_{AL})} \right) } \end{align} ``` ## Example 1: finding the interfacial composition (stagnant B) ![](./public/question-10-4-1.png) ## Solution steps for solving interfacial composition (general case) Manual trial-and-error steps: 1. Start from bulk point $P$ 2. Guess slope using current correction terms (initially $(1-x)_{im} = (1-y)_{im} = 1$) 3. Connect to equilibrium curve to get $(x_{Ai}, y_{Ai})$ 4. Update $(1-x){im}$ and $(1-y){im}$, calculate new slope 5. Is new slope in 4) converged? - No 👉 go back to step 2 - Yes 👉 continue to step 6 6. Get final $(x_{Ai}, y_{Ai})$ and $N_A$ ## Example 1: step 1 (follow textbook) ![](./public/ex1-step1.png) ## Example 1: step 2 (follow textbook) ![](./public/ex1-step2.png) ## Practical relation 2: link overall coefficient to film coefficients - We learned last week that writing flux equations using $K_x'$ and $K_y'$ are usually easier than $k_x'$ and $k_y'$ - What is the relation between them? - Take gas-phase, diffusion through stagnant film ```{=tex} \begin{align} N_A &= \frac{k_y'}{(1-y)_{im}}(y_{AG}-y_{Ai}) \\ &= \frac{K_y'}{(1-y)_{*m}}(y_{AG}-y_A^*) \\ &= K_y (y_{AG}-y_A^*) \end{align} ``` $(1-y)_{*m}$: log mean between $(1-y_{AG})$ and $(1-y_A^*)$ ## Geometric interpretation of overall $K$ Use gas-phase example, from geometry of the equilibrium diagram: ```{=tex} \begin{align} \frac{1}{K_y}=\frac{1}{k_y}+\frac{m'}{k_x} \end{align} ``` ![](./public/local-coeff-stagB.png) ## Transport resistance and overall mass transfer coefficients - The overall mass transfer coefficient $K$ basically tells which transport resistance is dominant ($k$ inversely proportional to resistance) - The transport equation becomes a "resistance-in-series" analog ```{=tex} \begin{align} \text{[Total resistance in gas]} &= \sum \text{resistance in each phase} \\ \frac{1}{K_y} &= \sum f_i \frac{1}{k_i} \end{align} ``` ## Case 1: overall $K_y$ in gas phase for highly liquid-soluble A - Local slope $m'$ is small - Transfer resistance is mainly in gas! - Design rule: tune $k_y$ --> more efficient mass transfer ```{=tex} \begin{align} \frac{1}{K_y} &= \frac{1}{k_y}+\frac{m'}{k_x}\\ &\approx \frac{1}{k_y} \end{align} ``` ![](./public/high-solu.png) ## Case 2: overall $K_x$ in liquid phase for low solubility A - Local slope $m''$ is large - Transfer resistance is mainly in liquid! - Design rule: tune $k_x$ --> more efficient mass transfer ```{=tex} \begin{align} \frac{1}{K_x} &= \frac{1}{m'' k_y}+\frac{1}{k_x}\\ &\approx \frac{1}{k_x} \end{align} ``` ![](./public/low-solu.png) ## Example 2: estimate overall mass transfer coefficients Use the conditions from example 1: bulk phase point $P=(0.10, 0.380)$, $k_x'=1.967\times 10^{-3}$ kg mol/m$^2$/s and $k_y' = 1.465\times 10^{-3}$ kg mol/m$^2$/s, estimate $K_y'$? ![](./public/ex1-plot.png) ## Example 2: solutions ![](./public/solution-ex2.png) ## Summary - In-depth analysis of diffusion through stagnant interfacial equilibrium - Geometric interpretation of equilibrium diagram - Case studies for interfacial composition & mass transfer coefficient So far we have built _almost all_ prerequisite for solving the concentration profile in the absorption tower! We will discuss that in upcoming [Lecture 25](../L25)