CHE 318 Lecture 23

Mass Transfer In Two-Phase Column: Operating Line

Author

Dr. Tian Tian

Published

March 6, 2026

Note

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 โœ…
  • Required liquid / gas flow rate โœ…
  • Concentration profile โ†’ needs to know \(N_A\)
  • Height of tower needed โ†’ 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:

  1. Know the flow rate \(L'\) โ†’ Determine \(y_2\)
  2. \(y_2\) needs to be at certain value โ†’ 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 Henryโ€™s law:

\[ y_2 = m x_1 \]

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

  1. A is CO\(_2\), \(m = 1.42\times 10^3\)
  2. 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 Henryโ€™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 โ€“ concentration relation
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