CHE 318 Lecture 25

Mass Transfer In Two-Phase Column: Concentration Profile

Dr. Tian Tian

2026-03-11

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

Physical copies distributed in class

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

\[ y = 1.186 x \]

  1. calculate the tower height use \(k_y' a\)
  2. repeat a) but use \(k_x' a\)
  3. 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: \(x_2=0\); \(y_2=0.005\)
  1. Calculate \(x_1\)?

Use mass balance equation

Solution steps (2)

  1. Find the slope of curve

Solutions steps (3)

  1. 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

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

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