CHE 318 Lecture 23

Mass Transfer In Two-Phase Column: Operating Line

Dr. Tian Tian

2026-03-06

Learning outcomes

After this lecture, you will be able to:

Recall the overall mass-balance framework for two-phase absorption systems.
Identify the operating line on an equilibrium diagram.
Apply equations to calculate outlet composition and minimum flow rate for an absorption tower.

Recall: reading an equilibrium diagram

Key features:

x-axis & y-axis meaning?
Points on the eq. curve?
Points above and below eq. curve?
How to get interfacial composition?
Slope of line to interfacial points?

Local overall mass transfer coefficients

Knowing \(k_x\) and \(k_y\) (film coefficients) allows us to calculate the interfacial concentration \((x_{Ai}, y_{Ai})\)
However, in many industrial applications exact \(k_x\) and \(k_y\) are hard to find
Use of overall mass transfer coefficients

Relation between overall and exact mass transfer coefficients

Exact coefficient \(k_x\), \(k_y\): driving force \(y_{AG} - y_{Ai}\)

\[ N_A = k_y(y_{AG} - y_{Ai}) \]

Overall coefficient \(K_x\), \(K_y\): driving force \(y_{AG}-y_A^*\)

\[ N_A = K_y (y_{AG} - y_A^*) = K_x (x_A^* - x_{AL}) \]

\(K_x\) and \(K_y\) on a diagram

What are the driving forces associated with \(K_x\) and \(K_y\)?

\((x, y)\) relation in a reactor

Consider an absorption tower with gas inlet at \(y_{A1}\) and outlet at \(y_{A2}\)
The gas solute is continuously absorbed by the flowing water phase
Which is larger, \(y_{A1}\) or \(y_{A2}\)?

Absorption operating line in equilibrium diagram

Tip

Mass balance tells us \(y_{A1} > y_{A2}\)
A series of \((x, y)\) points during the operation forms the operating line

What questions do we want to study?

Given information about the absorption tower, can we answer?

In- and out-let molar fractions ✅
Required liquid / gas flow rate ✅
Concentration profile → needs to know \(N_A\)
Height of tower needed → needs to know \(N_A\)

Mass balance in whole control volume

Mass balance for 2 phases:

\[\begin{align} \text{In}_{\text{liq}} + \text{In}_{\text{gas}} &= \text{Out}_{\text{liq}} + \text{Out}_{\text{gas}} \\ L_2 x_2 + V_1 y_1 &= L_1 x_1 + V_2 y_2 \end{align}\]

\(L_1 = L' + L_{x1}\) (\(L'\): flow rate inert liquid)
\(V_1 = V' + V_{y1}\) (\(V'\): flow rate inert gas)

Explanation for flow rates

Note

Other flow rates: \(Q\) (m\(^3\)/s); \(W\) (kg/s); \(v\) (m/s)

\(L\): molar flow rate (kg mol/s) for liquid phase
- \(L'\): flow rate for inert liquid
- \(L_{x1}\): flow rate for A at molar fraction \(x_1\)
\(V\): molar flow rate (kg mol/s) for gas phase
- \(V'\): flow rate for inert gas
- \(V_{y1}\): flow rate for A at molar fraction \(y_1\)

Mass balance for operating line

The two ends of the operating line \((x_1, y_1)\) and \((x_2, y_2)\) follow:

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

Meaning of the operating line

When \(1 - x_1 \approx 1\) and \(1 - y_1 \approx 1\), we can rewrite the mass balance equation for any \((x, y)\) along the operating line

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

Absorption tower design requirements

In absorption tower, we usually know the following quantities:

Gas inlet fraction \(y_1\) and flow rate \(V_1\)
- \(V'\) can be calculated
Liquid inlet fraction \(x_2\) (usually \(x_2=0\))
The equilibrium curve \(y=f_{\text{eq}}(x)\)

One of the following design goals may be asked:

Know the flow rate \(L'\) → Determine \(y_2\)
\(y_2\) needs to be at certain value → Determine \(L'\)

Minimum operating line for absorption tower

For question 2, we know the requirement for \(y_2\), combine with the equilibrium chart, there is a minimum liquid flow rate \(L'_{\text{min}}\).

Flow rate in absorption tower

From the dilute regime operating line, the slope is determined by \(\left(\dfrac{L'}{V'}\right)\). Although there is a minimal \(\left(\dfrac{L'}{V'}\right)\) requirement, practical operating line has slope that follows

\[ \left(\dfrac{L'}{V'}\right) \approx 1.5 \times \text{[Slope of Eq. Curve]} \]

If \(L'/V'\) is too high, the column usually needs a larger diameter to compensate the pressure drop.
If \(L'/V'\) is too low, a taller absorption tower is needed for sufficient contact area.

Example 1: determine the outlet composition

A mixture of gas A in air kept at total pressure of 1 atm flows through an absorption tower with flowing water at 293 K. The inlet gas flow rate is \(V_1 = 100\) kg mol/h, and inlet \(y_1 = 0.20\). The liquid inlet flow rate is \(L' = 300\) kg mol/h and inlet contains no dissolved gas (\(x_2=0\)). At the outlets the gas-liquid phases reach equilibrium following the Henry’s law:

\[ y_2 = m x_1 \]

Calculate the outlet mole fraction \(y_2\) and \(x_1\) for the following cases:

A is CO\(_2\), \(m = 1.42\times 10^3\)
A is SO\(_2\), \(m \approx 10\)

Example 2: solution process

Write the mass balance equation

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

Obtain \(L'\) and \(V'\)
\(x_1\) and \(y_2\) relation from Henry’s law

Example 2: results

Warning

\(y_1\) and \(y_2\) are supposedly large, so \(1 - y_1 \approx 1\) is not correct!

CO2 system: x_1 = 1.406e-04, y_2= 0.200
SO2 system: x_1 = 1.592e-02, y_2= 0.159

Example 2: quanlitative analysis

Slope of operating line \(L'/V'\)
Slope of equilibrium curve \(m\)
CO\(_2\): \(m \gg L'/V'\), can be treated as if gas phase concentration is fixed!
SO\(_2\): measurable decrease of SO\(_2\) fraction in outlet gas

Summary

Reading equilibrium chart and operating line
Mass balance equation for 2 phases
Solving the flow rate – concentration relation