CHE 318 Lecture 25

Mass Transfer In Two-Phase Column: Concentration Profile

Author

Dr. Tian Tian

Published

March 11, 2026

Note

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Learning outcomes

After this lecture, you will be able to:

Recall the differential mass-balance framework for two-phase columns.
Describe expressions for packed-column height from local transfer relations.
Analyze coupled gas-liquid concentration profiles along the column.

Cheatsheet for packed bed design

Two-phase column: what have we learned so far?

For two-phase mass transfer (solid-state packed bed, liquid-gas absorption tower, etc.), our previous lectures have given:

What we have learned

Equilibrium diagram ✅
Operating line (dilute) ✅
Overall mass balance ✅
Interfacial composition ✅

What are still missing

Concentration profile ❓
Height requirement ❓
Non-linear operating line ❓

In-depth analysis for packed absorption tower

Example goals:

Analyze concentration profiles $x(z)$ and $y(z)$
Design packed-bed height $z$

Solution:

Take a differential element of height $dz$ 👉 differential equation
Effective interfacial area: $A_{\text{eff}} = aSdz$
Gas and liquid compositions both vary with $z$

Mass balance on control volume $z \to z+dz$

Mass balance for control volume still holds:

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

Governing equation for a slab with thickness $dz$:

\[\begin{align} d(Vy)=d(Lx) \end{align}\]

Flux equation in each phase

The relation $d(Vy)=d(Lx)$ can be used to solve for each phase, if we know how to connect with flux equations individually.

Take the gas-phase side, the effective contact area being $A_{\text{eff}}$ (see packed bed example in Lecture 21):

\[\begin{align} d(V y_{AG}) &= d(A_{\text{eff}}N_A) \\ &= \frac{k_y'}{(1-y_A)_{im}}(y_{AG}-y_{Ai})aSdz \end{align}\]

note we need to account for the non-linear concentration profile by $y_{Bm} = (1-y_A)_{im}$ as discussed in last lecture.

Governing differential equation for each phase

The part $d(V y_{AG})$ can be further simplified, because we know $V = V' / (1 - y_{AG})$:

\[\begin{align} d(Vy_{AG}) &= d(V'\frac{y_{AG}}{1-y_{AG}}) \\ &= V' \frac{1}{(1 - y_{AG})^2} d y_{AG} \end{align}\]

The final differential form in gas phase is then:

\[\begin{align} V'\frac{1}{(1-y_{AG})^2}dy_{AG} = \frac{k_y'aS}{(1-y_A)_{im}}(y_{AG}-y_{Ai})dz \end{align}\]

Differential equation for tower height

It is often desired to know the total height $Z$ for the tower given operating line. We can integrate over the differential equation

\[\begin{align} Z &= \int_0^Z dz \\ &= \int_{y_2}^{y_1} \frac{V'}{k_y'aS} \cdot \frac{(1-y_A)_{im}}{(1-y_{AG})^2(y_{AG}-y_{Ai})} \, dy_{AG} \end{align}\]

this is the exact form for the gas-side height profile
we can obtain the same equation using liquid-side conditions
numerical integration is needed

Different forms of height equation

If we drop the subscript $A$, and use $V = V'/(1-y)$ & $L=L'/(1-x)$, the height equation can be expressed in different forms

Gas-side profile

\[ Z = \int_{y_2}^{y_1} \frac{V dy}{\frac{k'_y a S}{(1 - y)_{im}} (1 - y)(y - y_i)} \]

Liquid-side profile

\[ Z = \int_{x_2}^{x_1} \frac{L dx}{\frac{k'_x a S}{(1 - x)_{im}} (1 - x)(x_i - x)} \]

Warning

The order of interface composition differences in liquid $(x_i - x)$ and gas $(y - y_i)$ have opposite signs!

Height equation using overall mass transfer coefficients

It is also possible to express the height equations using $K_y'$ or $K_x'$:

Gas-side profile

\[ Z = \int_{y_2}^{y_1} \frac{V dy}{\frac{K'_y a S}{(1 - y)_{*m}} (1 - y)(y - y^*)} \]

Liquid-side profile

\[ Z = \int_{x_2}^{x_1} \frac{L dy}{\frac{K'_x a S}{(1 - x)_{*m}} (1 - x)(x^* - x)} \]

Warning

The order of pseudo-interface composition differences in liquid $(x^* - x)$ and gas $(y - y^*)$ have opposite signs!

Simplified case: dilute system

The factors $k_x'a$, $k_y'a$, $K_x'a$, $K_y'a$ are usually not constant, making the solution harder to obtain. We will consider a simplified version of absorption of dilute gas.

dilute regime: composition less than 0.1 (10 mol%)
simplification 1: almost constant $V$ and $L$
simplification 2: $(1-x)_{im}/(1-x)$ and $(1-y)_{im}/(1-y)$ are almost independent on the location
simplification 3: change of $V$ (or $L$) over the column is minimal, usually use $V = (V_1 + V_2)/2$

Height equation in dilute systems

Gas-side profile (use $k'_y$)

\[ Z = \left[ \frac{V}{k'_y a S} \frac{(1 - y)_{im}}{1 - y} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} \]

Liquid-side profile (use $k'_x$)

\[ Z = \left[ \frac{V}{k'_x a S} \frac{(1 - x)_{im}}{1 - x} \right] \int_{x_2}^{x_1} \frac{dx}{x_i - x} \]

Replacing $_{im}$ subscripts with $_{*m}$ and using overall coefficients gives another set of equations.

Further simplification: log mean driving force

In practical cases we can even let $\frac{(1 - y)_{im}}{1 - y} \approx 1$ and $\frac{(1 - x)_{im}}{1 - x} \approx 1$. The R.H.S. becomes only dependent on $\int_{y_2}^{y_1} \frac{dy}{y - y_i}$.

\[\begin{align} Z &= \frac{V}{k_y'aS} \int_{y_2}^{y_1} \frac{dy}{y - y_i} \frac{V}{k_y'aS} \ln\left(\frac{y_1-y_i}{y_2-y_i}\right) \end{align}\]

From the packed-bed analysis in Lecture 21 we know it corresponding to a log-mean driving force term $(y - y_i)_{m}$:

\[\begin{align} \frac{V}{S}(y_1-y_2) &= k_y'aZ \frac{(y_1-y_i)-(y_2-y_i)} {\ln\left(\dfrac{y_1-y_i}{y_2-y_i}\right)} \\ &= k_y' aZ (y - y_i)_{m} \end{align}\]

What does the result tell us?

Let’s take a pause and check the meaning for the governing equation

\[\begin{align} \frac{V}{S}(y_1-y_2) = k_y' aZ (y - y_i)_{m} \end{align}\]

L.H.S.: molar flow of A absorbed per column cross-sectional area
R.H.S.: molar flow of A transferred from gas to liquid phase
- average driving force: log-mean of $y-y_i$
- transfer coefficient: $k_y'$

Visualization of log-mean driving force

For dilute gas system, the driving force can be visualized using diagonal lines connecting the operating line and equilibrium line.

Comparison: packed bed with solid spheres

The dilute two-phase system is very close to the packed bed of solid spheres in Lecture 21, see the side-by-side comparison.

Packed bed of solid spheres

Constant $c_{Ai}$
Uses log-mean driving force of $c$

\[ Q(c_2 - c_1) = k_c' a H (c_i - c)_{m} \]

Packed bed of liquid-gas

Varying $y_i$
Uses log-mean driving force of $y$

\[ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} \]

All forms of height equation in dilute system

Gas, use $k_y'$

\[ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} \]

Liquid, use $k_x'$ \[ \frac{V}{S}(x_1 - x_2) = k_x' a Z (x_i - x)_{m} \]
Gas, use $K_y'$

\[ \frac{V}{S}(y_1 - y_2) = K_y' a Z (y - y^*)_{m} \]

Liquid, use $K_x'$ \[ \frac{V}{S}(x_1 - x_2) = K_x' a Z (x^* - x)_{m} \]

Example: calculate height of acetone absorption tower

(Example 10.6-4) Acetone is being absorbed by water in a packed bed column having a cross sectional area of 0.186 m$^2$ at 293 K and 1 atm. The inlet air contains 2.6 mol% acetone and outlet contains 0.5%. The gas flow is 13.65 kg mol inert air per hour. The pure water inlet flow is 45.36 kg mol water per hour. The coefficient $k_y' a$ is estimated to be $3.78 \times 10^{-2}$ kg mol/(s·m$^3$) and $k_x' a$ is $6.16 \times 10^{-2}$ kg mol/(s·m$^3$). The equilibrium line can be approximated by

\[ y = 1.186 x \]

calculate the tower height use $k_y' a$
repeat a) but use $k_x' a$
calculate the tower height use $K_y' a$

Solution steps (1)

Is this system dilute (<10%)?

Yes. Choose linear operating line and height equation
Determine two ends of the operating line

Point 1: $x_1$=?; $y_1=0.026$
Point 2: $x_2=0$; $y_2=0.005$

Calculate $x_1$?

Use mass balance equation

Solution steps (2)

Find the slope of curve

Solutions steps (3)

To use the height equation

\[ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} \]

total flow rate $V=\frac{V_1 + V_2}{2}\approx 3.852 \times{} 10^{-3}$ kg mol/s
log-mean driving force

\[ (y - y_i)_{m} = \frac{(y_1 - y_i) - (y_2 - y_i)}{\ln (\frac{y_1 - y_i}{y_2 - y_i})} = 0.006 \]

Final result: $Z=1.911$ m

Summary

Solving single-phase mass balance in absorption packed tower
Use dilute solution for tower height analysis
Recall the analog between the solid sphere packed bed and liquid-gas tower

\[ (1 - y)_{im} = \frac{(1 - y_{AG}) - (1 - y_{Ai})}{\ln\! \frac{1 - y_{AG}}{1 - y_{Ai}}} \]

\[ (y - y_i)_{m} = \frac{(y_{1} - y_{i1}) - (y_{2} - y_{i2})}{\ln\! \frac{y_{1} - y_{i1}}{y_2 - y_{i2}}} \]

--- title: "CHE 318 Lecture 25" subtitle: "Mass Transfer In Two-Phase Column: Concentration Profile" author: "Dr. Tian Tian" date: "2026-03-11" format: html: {} revealjs: output-file: slides.html pdf: output-file: L25.pdf --- ::: {.content-visible when-format="html" unless-format="revealjs"} ::: {.callout-note} - Slides 👉 [Open presentation🗒️](./slides.html) - PDF version of course note 👉 [Open in pdf](./L25.pdf) - Handwritten notes 👉 [Open in pdf](./public/L25_annotated.pdf) ::: ::: ## Learning outcomes {.center} After this lecture, you will be able to: - **Recall** the differential mass-balance framework for two-phase columns. - **Describe** expressions for packed-column height from local transfer relations. - **Analyze** coupled gas-liquid concentration profiles along the column. ## Cheatsheet for packed bed design ![Physical copies distributed in class](../L25/public/cheatsheet-2phase.svg) ## Two-phase column: what have we learned so far? For two-phase mass transfer (solid-state packed bed, liquid-gas absorption tower, etc.), our previous lectures have given: :::{.columns} :::{.column} What we have learned - Equilibrium diagram ✅ - Operating line (dilute) ✅ - Overall mass balance ✅ - Interfacial composition ✅ ::: :::{.column} What are still missing - Concentration profile ❓ - Height requirement ❓ - Non-linear operating line ❓ ::: ::: ## In-depth analysis for packed absorption tower Example goals: - Analyze concentration profiles $x(z)$ and $y(z)$ - Design packed-bed height $z$ Solution: - Take a differential element of height $dz$ 👉 differential equation - Effective interfacial area: $A_{\text{eff}} = aSdz$ - Gas and liquid compositions both vary with $z$ ## Mass balance on control volume $z \to z+dz$ Mass balance for control volume still holds: ```{=tex} \begin{align} \text{In}_{liq} + \text{In}_{gas} = \text{Out}_{liq} + \text{Out}_{gas} \end{align} ``` Governing equation for a slab with thickness $dz$: ```{=tex} \begin{align} d(Vy)=d(Lx) \end{align} ``` ## Flux equation in each phase The relation $d(Vy)=d(Lx)$ can be used to solve for each phase, if we know how to connect with **flux equations** individually. Take the gas-phase side, the effective contact area being $A_{\text{eff}}$ (see packed bed example in [Lecture 21](../L21)): ```{=tex} \begin{align} d(V y_{AG}) &= d(A_{\text{eff}}N_A) \\ &= \frac{k_y'}{(1-y_A)_{im}}(y_{AG}-y_{Ai})aSdz \end{align} ``` note we need to account for the non-linear concentration profile by $y_{Bm} = (1-y_A)_{im}$ as discussed in last lecture. ## Governing differential equation for each phase The part $d(V y_{AG})$ can be further simplified, because we know $V = V' / (1 - y_{AG})$: ```{=tex} \begin{align} d(Vy_{AG}) &= d(V'\frac{y_{AG}}{1-y_{AG}}) \\ &= V' \frac{1}{(1 - y_{AG})^2} d y_{AG} \end{align} ``` The final differential form in gas phase is then: ```{=tex} \begin{align} V'\frac{1}{(1-y_{AG})^2}dy_{AG} = \frac{k_y'aS}{(1-y_A)_{im}}(y_{AG}-y_{Ai})dz \end{align} ``` ## Differential equation for tower height It is often desired to know the total height $Z$ for the tower given operating line. We can integrate over the differential equation ```{=tex} \begin{align} Z &= \int_0^Z dz \\ &= \int_{y_2}^{y_1} \frac{V'}{k_y'aS} \cdot \frac{(1-y_A)_{im}}{(1-y_{AG})^2(y_{AG}-y_{Ai})} \, dy_{AG} \end{align} ``` - this is the **exact form** for the gas-side height profile - we can obtain the same equation using liquid-side conditions - numerical integration is needed ## Different forms of height equation If we drop the subscript $A$, and use $V = V'/(1-y)$ & $L=L'/(1-x)$, the height equation can be expressed in different forms - Gas-side profile $$ Z = \int_{y_2}^{y_1} \frac{V dy}{\frac{k'_y a S}{(1 - y)_{im}} (1 - y)(y - y_i)} $$ - Liquid-side profile $$ Z = \int_{x_2}^{x_1} \frac{L dx}{\frac{k'_x a S}{(1 - x)_{im}} (1 - x)(x_i - x)} $$ :::{.callout-warning} The order of interface composition differences in liquid $(x_i - x)$ and gas $(y - y_i)$ have opposite signs! ::: ## Height equation using overall mass transfer coefficients It is also possible to express the height equations using $K_y'$ or $K_x'$: - Gas-side profile $$ Z = \int_{y_2}^{y_1} \frac{V dy}{\frac{K'_y a S}{(1 - y)_{*m}} (1 - y)(y - y^*)} $$ - Liquid-side profile $$ Z = \int_{x_2}^{x_1} \frac{L dy}{\frac{K'_x a S}{(1 - x)_{*m}} (1 - x)(x^* - x)} $$ :::{.callout-warning} The order of pseudo-interface composition differences in liquid $(x^* - x)$ and gas $(y - y^*)$ have opposite signs! ::: ## Simplified case: dilute system The factors $k_x'a$, $k_y'a$, $K_x'a$, $K_y'a$ are usually not constant, making the solution harder to obtain. We will consider a simplified version of absorption of dilute gas. - dilute regime: composition less than 0.1 (10 mol%) - simplification 1: almost constant $V$ and $L$ - simplification 2: $(1-x)_{im}/(1-x)$ and $(1-y)_{im}/(1-y)$ are almost independent on the location - simplification 3: change of $V$ (or $L$) over the column is minimal, usually use $V = (V_1 + V_2)/2$ ## Height equation in dilute systems - Gas-side profile (use $k'_y$) $$ Z = \left[ \frac{V}{k'_y a S} \frac{(1 - y)_{im}}{1 - y} \right] \int_{y_2}^{y_1} \frac{dy}{y - y_i} $$ - Liquid-side profile (use $k'_x$) $$ Z = \left[ \frac{V}{k'_x a S} \frac{(1 - x)_{im}}{1 - x} \right] \int_{x_2}^{x_1} \frac{dx}{x_i - x} $$ Replacing $_{im}$ subscripts with $_{*m}$ and using overall coefficients gives another set of equations. ## Further simplification: log mean driving force In practical cases we can even let $\frac{(1 - y)_{im}}{1 - y} \approx 1$ and $\frac{(1 - x)_{im}}{1 - x} \approx 1$. The R.H.S. becomes only dependent on $\int_{y_2}^{y_1} \frac{dy}{y - y_i}$. ```{=tex} \begin{align} Z &= \frac{V}{k_y'aS} \int_{y_2}^{y_1} \frac{dy}{y - y_i} \frac{V}{k_y'aS} \ln\left(\frac{y_1-y_i}{y_2-y_i}\right) \end{align} ``` From the packed-bed analysis in [Lecture 21](../L21) we know it corresponding to a log-mean driving force term $(y - y_i)_{m}$: ```{=tex} \begin{align} \frac{V}{S}(y_1-y_2) &= k_y'aZ \frac{(y_1-y_i)-(y_2-y_i)} {\ln\left(\dfrac{y_1-y_i}{y_2-y_i}\right)} \\ &= k_y' aZ (y - y_i)_{m} \end{align} ``` ## What does the result tell us? Let's take a pause and check the meaning for the governing equation ```{=tex} \begin{align} \frac{V}{S}(y_1-y_2) = k_y' aZ (y - y_i)_{m} \end{align} ``` - L.H.S.: molar flow of A **absorbed** per column cross-sectional area - R.H.S.: molar flow of A **transferred** from gas to liquid phase - _average_ driving force: log-mean of $y-y_i$ - transfer coefficient: $k_y'$ ## Visualization of log-mean driving force For dilute gas system, the driving force can be visualized using diagonal lines connecting the operating line and equilibrium line. ![](./public/dilute-log-mean-chart.svg) ## Comparison: packed bed with solid spheres The dilute two-phase system is very close to the packed bed of solid spheres in [Lecture 21](../L21), see the side-by-side comparison. :::{.columns} :::{.column width="50%} Packed bed of solid spheres - Constant $c_{Ai}$ - **Uses log-mean driving force** of $c$ $$ Q(c_2 - c_1) = k_c' a H (c_i - c)_{m} $$ ::: :::{.column width="50%} Packed bed of liquid-gas - Varying $y_i$ - **Uses log-mean driving force** of $y$ $$ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} $$ ::: ::: ## All forms of height equation in dilute system - Gas, use $k_y'$ $$ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} $$ - Liquid, use $k_x'$ $$ \frac{V}{S}(x_1 - x_2) = k_x' a Z (x_i - x)_{m} $$ - Gas, use $K_y'$ $$ \frac{V}{S}(y_1 - y_2) = K_y' a Z (y - y^*)_{m} $$ - Liquid, use $K_x'$ $$ \frac{V}{S}(x_1 - x_2) = K_x' a Z (x^* - x)_{m} $$ ## Example: calculate height of acetone absorption tower (Example 10.6-4) Acetone is being absorbed by water in a packed bed column having a cross sectional area of 0.186 m$^2$ at 293 K and 1 atm. The inlet air contains 2.6 mol% acetone and outlet contains 0.5%. The gas flow is 13.65 kg mol inert air per hour. The pure water inlet flow is 45.36 kg mol water per hour. The coefficient $k_y' a$ is estimated to be $3.78 \times 10^{-2}$ kg mol/(s·m$^3$) and $k_x' a$ is $6.16 \times 10^{-2}$ kg mol/(s·m$^3$). The equilibrium line can be approximated by $$ y = 1.186 x $$ a) calculate the tower height use $k_y' a$ b) repeat a) but use $k_x' a$ c) calculate the tower height use $K_y' a$ ## Solution steps (1) 1) Is this system dilute (<10%)? Yes. Choose linear operating line and height equation 2) Determine two ends of the operating line - Point 1: $x_1$=?; $y_1=0.026$ - Point 2: $x_2=0$; $y_2=0.005$ 3) Calculate $x_1$? Use mass balance equation ## Solution steps (2) 4) Find the slope of curve ![](./public/ex1-solution.png) ## Solutions steps (3) 5) To use the height equation $$ \frac{V}{S}(y_1 - y_2) = k_y' a Z (y - y_i)_{m} $$ - total flow rate $V=\frac{V_1 + V_2}{2}\approx 3.852 \times{} 10^{-3}$ kg mol/s - log-mean driving force $$ (y - y_i)_{m} = \frac{(y_1 - y_i) - (y_2 - y_i)}{\ln (\frac{y_1 - y_i}{y_2 - y_i})} = 0.006 $$ Final result: $Z=1.911$ m ## Summary - Solving single-phase mass balance in absorption packed tower - Use dilute solution for tower height analysis - Recall the analog between the solid sphere packed bed and liquid-gas tower $$ (1 - y)_{im} = \frac{(1 - y_{AG}) - (1 - y_{Ai})}{\ln\! \frac{1 - y_{AG}}{1 - y_{Ai}}} $$ $$ (y - y_i)_{m} = \frac{(y_{1} - y_{i1}) - (y_{2} - y_{i2})}{\ln\! \frac{y_{1} - y_{i1}}{y_2 - y_{i2}}} $$

Learning outcomes

Cheatsheet for packed bed design

Two-phase column: what have we learned so far?

In-depth analysis for packed absorption tower

Mass balance on control volume \(z \to z+dz\)

Flux equation in each phase

Governing differential equation for each phase

Differential equation for tower height

Different forms of height equation

Height equation using overall mass transfer coefficients

Simplified case: dilute system

Height equation in dilute systems

Further simplification: log mean driving force

What does the result tell us?

Visualization of log-mean driving force

Comparison: packed bed with solid spheres

All forms of height equation in dilute system

Example: calculate height of acetone absorption tower

Solution steps (1)

Solution steps (2)

Solutions steps (3)

Summary