10.2 Gas Absorption, Stripping, and Liquid-Liquid Extraction

10.2 Gas Absorption, Stripping, and Liquid-Liquid Extraction

Gas absorption, stripping, and liquid-liquid extraction are key separation processes that involve mass transfer of a solute between two phases. On the FE Chemical exam, questions typically focus on mass balances, equilibrium relationships, operating lines, finding minimum solvent or gas rates, and interpreting ternary extraction diagrams.

Gas Absorption and Stripping Fundamentals

Gas absorption is a process where a gas mixture is contacted with a liquid solvent to selectively dissolve one or more soluble components (solutes) from the gas. Stripping (or desorption) is the physical reverse of absorption, where a liquid mixture is contacted with a gas stream to transfer a solute from the liquid phase into the gas phase.

In a typical industrial column (usually packed or trayed), flow is countercurrent: liquid enters at the top and flows downwards, while gas enters at the bottom and flows upwards.

Equilibrium and Henry's Law

For dilute systems, the equilibrium relationship between the mole fraction of solute in the gas phase ($y^*$) and the liquid phase ($x$) is linear, described by Henry's Law:

$y^* = m x$

Where:

• $y^*$ is the equilibrium mole fraction of solute in the vapor phase.
• $x$ is the mole fraction of solute in the liquid phase.
• $m$ is the equilibrium constant (Henry's law slope).

Molar Material Balances

Consider a countercurrent column where:

• The bottom is designated as Stage 1 and the top as Stage 2.
• Gas enters the bottom at molar flow rate $G$ with solute mole fraction $y_1$.
• Gas leaves the top at molar flow rate $G$ with solute mole fraction $y_2$.
• Liquid enters the top at molar flow rate $L$ with solute mole fraction $x_2$.
• Liquid leaves the bottom at molar flow rate $L$ with solute mole fraction $x_1$.

For dilute systems (typically solute mole fractions $< 5%$), the total gas and liquid flow rates $G$ and $L$ remain approximately constant. A mass balance over the entire column yields:

$L x_2 + G y_1 = L x_1 + G y_2 \implies G(y_1 - y_2) = L(x_1 - x_2)$

A mass balance from the top of the column to an arbitrary cross-section gives the Operating Line equation:

$y = \frac{L}{G} x + \left( y_2 - \frac{L}{G} x_2 \right)$

This operating line represents the actual vapor and liquid compositions passing each other at any point in the column.

Absorption vs. Stripping Operating Lines

• Absorption: Solute is transferred from the gas phase to the liquid. Therefore, at any point in the column, the actual gas phase concentration $y$ must be higher than the equilibrium concentration $y^*$ in contact with liquid concentration $x$. The operating line lies above the equilibrium line ($y > m x$).
• Stripping: Solute is transferred from the liquid phase to the gas. Therefore, the actual liquid phase concentration $x$ must be higher than the equilibrium concentration $x^*$ in contact with gas concentration $y$. The operating line lies below the equilibrium line ($y < m x$).

---

Minimum Liquid-to-Gas Flow Rate $(L/G)_{\min}$ in Absorption

In gas absorption, the liquid solvent rate $L$ represents an operating cost (since the solvent must be regenerated or disposed of). Decreasing $L$ for a fixed gas rate $G$ decreases the slope of the operating line ($L/G$).

If the liquid flow rate is decreased too much, the operating line will rotate downwards until it touches the equilibrium line. The point of contact is called the pinch point. At the pinch point, the operating line and the equilibrium line intersect, meaning the concentration driving force $(y - y^*)$ becomes zero. According to mass transfer principles, a driving force of zero requires an infinite column height (or infinite equilibrium stages) to achieve the desired separation.

For a linear equilibrium curve ($y^* = m x$) and an absorber with a solute-free liquid feed ($x_2 = 0$), the pinch point occurs at the bottom of the column where the liquid leaving ($x_1$) is in equilibrium with the entering gas ($y_1$). Thus, the maximum possible liquid concentration at the outlet is:

$x_1^* = \frac{y_1}{m}$

The minimum liquid-to-gas flow rate ratio is calculated by connecting the inlet liquid composition point $(x_2, y_2)$ to this equilibrium point $(x_1^*, y_1)$:

$\left(\frac{L}{G}\right)_{\min} = \frac{y_1 - y_2}{x_1^* - x_2} = \frac{y_1 - y_2}{(y_1 / m) - x_2}$

Exam Tip: In practice, the operating liquid rate is set to $1.2$ to $1.5$ times the minimum liquid rate. If the liquid flow rate is below $L_{\min}$, the desired separation is physically impossible.

---

Liquid-Liquid Extraction (LLE)

Liquid-liquid extraction separates a solute from a liquid mixture (carrier) by contacting it with another immiscible or partially miscible liquid solvent that has a higher affinity for the solute.

Ternary Systems

LLE involves three components:

1. Solute ($A$): The component being separated.
2. Carrier ($B$): The original solvent containing the solute.
3. Extracting Solvent ($C$): The added liquid solvent.

The separation yields two phases:

• Extract phase: The solvent-rich phase containing the extracted solute.
• Raffinate phase: The residual carrier-rich phase depleted of solute.

Ternary Phase Diagrams

Ternary mixtures are represented on an equilateral triangular diagram, where each vertex represents $100%$ of a pure component ($A$, $B$, or $C$).

• Binodal Curve (Solubility Curve): The envelope dividing the single-phase homogeneous region (outside/above the curve) from the two-phase heterogeneous region (inside the curve).
• Tie Lines: Straight lines within the two-phase region that connect the compositions of the raffinate phase ($R$, on the carrier-rich side of the curve) and the extract phase ($E$, on the solvent-rich side of the curve) that are in equilibrium.
• Plait Point: The point on the binodal curve where the raffinate and extract compositions become identical. At the plait point, the tie line length shrinks to zero.
• Lever-Arm Rule: For a feed mixture $M$ that splits into raffinate $R$ and extract $E$, the ratio of the mass of the phases is given by:

$\frac{\text{Mass of Raffinate (R)}}{\text{Mass of Extract (E)}} = \frac{\overline{ME}}{\overline{MR}}$

Where $\overline{ME}$ is the length of the segment from $M$ to $E$, and $\overline{MR}$ is the length of the segment from $M$ to $R$ along the tie line passing through $M$.

---

Worked Example: Minimum Solvent Rate for an Absorber

Problem: A countercurrent packed column is designed to absorb hydrogen sulfide ($\text{H}_2\text{S}$) from a gas stream using pure water. The entering gas contains $4.0\text{ mol}%\text{ H}_2\text{S}$ ($y_1 = 0.040$), and the exit gas must contain no more than $0.2\text{ mol}%\text{ H}_2\text{S}$ ($y_2 = 0.002$). The total gas flow rate is $150\text{ kmol/h}$. The water entering the column is pure ($x_2 = 0$). The equilibrium relationship is given by $y^* = 1.6 x$. Calculate:

1. The minimum water flow rate ($L_{\min}$) in $ ext{kmol/h}$.
2. The operating water flow rate if the design is set at $1.3$ times the minimum.

Solution:

1. Determine the maximum possible liquid concentration at the bottom of the column ($x_1^*$), which would be in equilibrium with the entering gas ($y_1 = 0.040$): $x_1^* = \frac{y_1}{m} = \frac{0.040}{1.6} = 0.025$

2. Calculate the minimum liquid-to-gas flow rate ratio $(L/G)_{\min}$: $\left(\frac{L}{G}\right)_{\min} = \frac{y_1 - y_2}{x_1^* - x_2} = \frac{0.040 - 0.002}{0.025 - 0} = \frac{0.038}{0.025} = 1.52$

3. Multiply by the gas flow rate $G = 150\text{ kmol/h}$ to find $L_{\min}$: $L_{\min} = 1.52 \times 150\text{ kmol/h} = 228\text{ kmol/h}$

4. Compute the operating liquid flow rate ($L$): $L = 1.3 \times L_{\min} = 1.3 \times 228\text{ kmol/h} = 296.4\text{ kmol/h}$

Thus, the minimum solvent rate is $228\text{ kmol/h}$ and the operating solvent rate is $296.4\text{ kmol/h}$.