12.1 Gas Absorption, Stripping, and the Kremser Equation

12.1 Gas Absorption, Stripping, and the Kremser Equation

In chemical manufacturing, environmental remediation, and refining, gas absorption (scrubbing) transfers one or more soluble components from a gas stream into a relatively non-volatile liquid solvent. Conversely, stripping (desorption) transfers volatile solutes from a liquid phase into an inert or condensable stripping gas (such as steam, air, or nitrogen). Both operations rely on the same fundamental thermodynamic phase equilibria and interfacial mass transfer mechanisms governed by the NCEES PE Chemical Reference Handbook.

On the PE Chemical exam, problems frequently evaluate whether a column is operating above or below its minimum solvent rate ($(L/G){min}$), require analytical calculation of theoretical stages via the Kremser-Brown-Souders equation, or demand packed column height sizing using the **Height of a Transfer Unit ($H{OG}$)** and Number of Transfer Units ($N_{OG}$).

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1. Thermodynamic Phase Equilibria & Henry's Law

In dilute gas-liquid systems, the equilibrium distribution of a sparingly soluble gaseous solute $A$ between the vapor phase (mole fraction $y$) and the liquid phase (mole fraction $x$) follows Henry's Law:

$p_A = y_A P = H_A' \cdot x_A$

Where:

• $p_A$ = partial pressure of solute $A$ in the gas phase ($\text{atm}$, $\text{bar}$, or $\text{kPa}$).
• $P$ = total system operating pressure ($\text{atm}$, $\text{bar}$, or $\text{kPa}$).
• $H_A'$ = Henry's law constant in pressure units ($\text{atm/mole fraction}$, $\text{bar}$, or $\text{kPa}$).
• $x_A$ = mole fraction of solute $A$ in the liquid phase.
• $y_A$ = mole fraction of solute $A$ in the gas phase.

Dividing by total pressure $P$ yields the linear equilibrium line in mole fraction coordinates:

$y_A^* = m \cdot x_A$

Where the equilibrium distribution slope $m$ is:

$m = \frac{H_A'}{P}$

[!NOTE] Henry's Constant Sensitivity:
Henry's constant increases significantly with rising temperature (dissolution of gases in liquids is typically exothermic, $\Delta H_{soln} < 0$). Consequently, absorption is favored at low temperatures and high pressures (which decreases $m$, flattening the equilibrium line and maximizing liquid solubility). Conversely, stripping is favored at high temperatures and low pressures (which increases $m$, steepening the equilibrium line and driving the solute out of the liquid phase).

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2. Operating Lines & Material Balances: Absorption vs. Stripping

Consider a continuous, steady-state countercurrent column. The gas phase flows upward at molar rate $G$ ($\text{mol/s}$ or $\text{kmol/h}$), entering at the bottom with solute mole fraction $y_{in}$ and exiting at the top at $y_{out}$. The liquid solvent flows downward at molar rate $L$, entering at the top with solute mole fraction $x_{in}$ and exiting at the bottom at $x_{out}$.

               Top of Column (Stage 1)
             Liquid In (L, x_in)  Gas Out (G, y_out)
                    |                 ^
                    v                 |
             +-------------------------------+
             |                               |
             |      COUNTERCURRENT           |
             |      MASS TRANSFER            |
             |                               |
             +-------------------------------+
                    |                 ^
                    v                 |
             Liquid Out (L, x_out)  Gas In (G, y_in)
               Bottom of Column (Stage N)

Dilute System Operating Line

When the solute concentration is low ($y < 0.05$ to $0.10$ and $x < 0.05$), the total molar flow rates of gas ($G$) and liquid ($L$) remain essentially constant throughout the column. A steady-state solute material balance from the top of the column down to an arbitrary internal cross-section yields:

$G(y - y_{out}) = L(x - x_{in})$

Rearranging into the linear operating equation ($y = f(x)$):

$y = \frac{L}{G} x + \left( y_{out} - \frac{L}{G} x_{in} \right)$

The slope of the operating line on an arithmetic $y-x$ plot is strictly the liquid-to-gas molar flow ratio, $L/G$.

Absorption vs. Stripping Geometry

The geometric orientation of the operating line relative to the equilibrium line ($y^* = m x$) determines the direction of mass transfer:

   y (Gas Mole Fraction)                     y (Gas Mole Fraction)
     ^
     |      / Operating Line (Slope L/G)       |          / Equilibrium (Slope m)
     |     /                                   |         /
     |    /                                    |        /     / Operating Line (Slope L/G)
     |   /      / Equilibrium (Slope m)        |       /     /
     |  /      /                               |      /     /
     | /      /                                |     /     /
     +------------------------> x              +------------------------> x
          ABSORPTION (y > y*)                       STRIPPING (y < y*)

Minimum Solvent Rate $(L/G)_{min}$ in Absorption

To achieve a specified separation (reducing gas concentration from $y_{in}$ to $y_{out}$ using a solvent entering at $x_{in}$), decreasing the liquid rate $L$ flattens the operating line slope $L/G$. The theoretical minimum solvent rate, $(L/G){min}$, occurs when the operating line touches the equilibrium line at the bottom of the column, creating an equilibrium pinch point where driving force $\Delta y = y{in} - y_{in}^* \to 0$, requiring an infinite number of stages ($N \to \infty$):

$\left( \frac{L}{G} \right)_{min} = \frac{y_{in} - y_{out}}{x_{out}^* - x_{in}} = \frac{y_{in} - y_{out}}{\left(\frac{y_{in}}{m}\right) - x_{in}}$

If the entering solvent is pure solute-free liquid ($x_{in} = 0$):

$\left( \frac{L}{G} \right)_{min} = m \left( \frac{y_{in} - y_{out}}{y_{in}} \right) = m \cdot (\text{Fractional Recovery})$

In industrial process design, economic optimization balances capital equipment cost (column diameter and height) against recurring solvent circulation and regeneration utility costs. Standard practice specifies an actual operating solvent rate of:

$\left( \frac{L}{G} \right)_{actual} = 1.20 \text{ to } 1.50 \times \left( \frac{L}{G} \right)_{min}$

Minimum Stripping Gas Rate $(G/L)_{min}$ in Stripping

In a stripping column, decreasing the stripping gas rate $G$ steepens the operating line ($L/G$ increases). The minimum stripping gas rate $(G/L){min}$ occurs when the operating line touches the equilibrium line at the top of the column ($x{in}, y_{out}^* = m x_{in}$):

$\left( \frac{G}{L} \right)_{min} = \frac{x_{in} - x_{out}}{y_{out}^* - y_{in}} = \frac{x_{in} - x_{out}}{m x_{in} - y_{in}}$

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3. The Absorption Factor ($A$) and Stripping Factor ($S$)

The performance of equilibrium-stage and packed mass transfer columns is governed by the dimensionless ratio of the operating line slope to the equilibrium line slope.

The Absorption Factor ($A$)

$A \equiv \frac{L}{m \cdot G}$

Where:

• $L$ = molar liquid flow rate ($\text{kmol/h}$).
• $G$ = molar gas flow rate ($\text{kmol/h}$).
• $m$ = slope of the linear equilibrium line ($y^* = m x$).

Physical Significance of $A$:

• $A > 1.0$: The operating line is steeper than the equilibrium line ($L/G > m$). The two lines diverge toward the bottom of the column. A high fractional solute recovery ($> 99%$) is theoretically and practically achievable with a finite, reasonable number of stages.
• $A = 1.0$: The operating line and equilibrium line are strictly parallel ($L/G = m$). The driving force remains constant across all stages. Recovery increases linearly with stage count.
• $A < 1.0$: The operating line is flatter than the equilibrium line ($L/G < m$). The lines pinch at the bottom of the column. Complete solute recovery is mathematically and physically impossible, even with an infinitely tall column ($N \to \infty$). The maximum possible recovery is asymptotically limited to $A$ (i.e., $\text{Fractional Recovery}_{max} = A$).
• Economic Optimum: In plant design, the optimal absorption factor typically ranges between $1.2 \le A \le 2.0$, with $A = 1.4$ commonly adopted as the starting rule of thumb.

The Stripping Factor ($S$)

The stripping factor is the exact reciprocal of the absorption factor:

$S \equiv \frac{1}{A} = \frac{m \cdot G}{L}$

For stripping columns, $S > 1.0$ is mandatory to achieve deep solute removal without encountering an equilibrium pinch at the top of the column. The standard economic design range is $1.2 \le S \le 2.0$ (optimal $S \approx 1.4$).

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4. The Kremser-Brown-Souders Equations

When the equilibrium line ($y = m x$) and operating line are linear, stage-by-stage mass balances can be solved analytically. The resulting closed-form expressions are known as the Kremser-Brown-Souders (or simply Kremser) equations, heavily featured on the NCEES PE Chemical exam.

Theoretical Stages for Absorption ($A \neq 1$)

To calculate the number of ideal (theoretical) equilibrium stages $N$ required to reduce gas composition from $y_{in}$ to $y_{out}$ using solvent with inlet composition $x_{in}$:

$N = \frac{\ln\left[ \left( \frac{y_{in} - m x_{in}}{y_{out} - m x_{in}} \right) \left( 1 - \frac{1}{A} \right) + \frac{1}{A} \right]}{\ln(A)}$

When pure, solute-free solvent is fed to the top of the column ($x_{in} = 0$):

$N = \frac{\ln\left[ \left( \frac{y_{in}}{y_{out}} \right) \left( 1 - \frac{1}{A} \right) + \frac{1}{A} \right]}{\ln(A)}$

Fractional Unabsorbed Solute ($\phi_A$)

The fraction of solute entering in the gas stream that escapes unabsorbed in the exiting overhead gas is:

$\phi_A = \frac{y_{out} - m x_{in}}{y_{in} - m x_{in}} = \frac{A - 1}{A^{N+1} - 1}$

For an infinite number of stages ($N \to \infty$):

• If $A > 1$: $\phi_A \to 0$ (100% absorption is achievable).
• If $A < 1$: $\phi_A \to 1 - A$ (maximum achievable absorption is $1 - \phi_A = A$).

Theoretical Stages for Stripping ($S \neq 1$)

To strip a liquid from inlet concentration $x_{in}$ down to outlet concentration $x_{out}$ using stripping gas with solute content $y_{in}$:

$N = \frac{\ln\left[ \left( \frac{x_{in} - y_{in}/m}{x_{out} - y_{in}/m} \right) \left( 1 - \frac{1}{S} \right) + \frac{1}{S} \right]}{\ln(S)}$

When clean stripping gas is utilized ($y_{in} = 0$):

$N = \frac{\ln\left[ \left( \frac{x_{in}}{x_{out}} \right) \left( 1 - \frac{1}{S} \right) + \frac{1}{S} \right]}{\ln(S)}$

Special Case: Balanced Capacity ($A = 1$ or $S = 1$)

When $L/G = m$, the operating and equilibrium lines are parallel, creating a $0/0$ indeterminacy in the logarithmic Kremser relation. Applying L'Hôpital's rule reveals that stage count collapses to a simple ratio of overall concentration change to driving force:

$N = \frac{y_{in} - y_{out}}{y_{out} - m x_{in}} \quad (\text{for } A = 1)$

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5. Rate-Controlled Continuous Packed Columns: HTU and NTU

While trayed towers contact vapor and liquid in discrete equilibrium stages, packed columns provide continuous, differential contact over structured or random packing (e.g., Raschig rings, Pall rings, Mellapak). Column height is evaluated using the concept of Transfer Units.

   Packed Bed Column Height: Z = H_OG * N_OG

   |==================| <--- Gas Out (y_out), Liquid In (x_in)
   |  Liquid Spray    |
   |  Distributor     |
   |------------------|
   |                  |
   |  PACKED BED      |  Z = Total Bed Height
   |  (Structured or  |  H_OG = Height of an Overall Gas Transfer Unit
   |   Random Rings)  |  N_OG = Number of Overall Gas Transfer Units
   |                  |
   |------------------|
   |  Packing Support |
   |==================| <--- Gas In (y_in), Liquid Out (x_out)

Height of the Packed Bed ($Z$)

The required packed height ($Z$) is the product of the Height of an Overall Transfer Unit ($H_{OG}$ or $H_{OL}$) and the Number of Overall Transfer Units ($N_{OG}$ or $N_{OL}$):

$Z = H_{OG} \cdot N_{OG} = H_{OL} \cdot N_{OL}$

Number of Transfer Units ($N_{OG}$)

$N_{OG}$ represents the overall difficulty of the separation, defined by integrating the differential driving force across the gas phase:

$N_{OG} = \int_{y_{out}}^{y_{in}} \frac{dy}{y - y^*}$

Where $(y - y^*)$ is the local vertical distance between the operating line and the equilibrium line. For dilute systems with linear operating and equilibrium lines, this integration yields the exact logarithmic mean driving force formula:

$N_{OG} = \frac{y_{in} - y_{out}}{\Delta y_{lm}}$

Where the log-mean concentration difference $\Delta y_{lm}$ is:

$\Delta y_{lm} = \frac{(y_{in} - y_{in}^*) - (y_{out} - y_{out}^*)}{\ln\left[ \frac{y_{in} - y_{in}^*}{y_{out} - y_{out}^*} \right]}$

Here $y_{in}^* = m x_{out}$ and $y_{out}^* = m x_{in}$.

Relationship Between $N_{OG}$ and Kremser Stages $N$

For linear systems with constant absorption factor $A$, the analytical relationship connecting continuous transfer units ($N_{OG}$) to discrete theoretical stages ($N$) is:

$N_{OG} = N \cdot \frac{\ln(A)}{1 - \frac{1}{A}} = N \cdot \frac{A \ln(A)}{A - 1}$

• When $A > 1$: $N_{OG} > N$ (transfer units exceed theoretical stages).
• When $A = 1$: $N_{OG} = N$.
• When $A < 1$: $N_{OG} < N$.

Height of a Transfer Unit ($H_{OG}$) and Two-Film Theory

$H_{OG}$ characterizes the mass transfer efficiency of the packing hardware. A smaller $H_{OG}$ indicates a more efficient packing requiring less height for a given separation:

$H_{OG} = \frac{G_m}{K_y a}$

Where:

• $G_m$ = superficial molar gas mass velocity ($\text{kmol/(m}^2\cdot\text{s)}$ or $\text{lbmol/(ft}^2\cdot\text{hr)}$).
• $K_y$ = overall gas-phase mass transfer coefficient ($\text{kmol/(m}^2\cdot\text{s}\cdot\Delta y)$).
• $a$ = specific interfacial surface area of packing per unit volume ($\text{m}^2\text{/m}^3$).

According to Whitman's Two-Film Theory, the overall mass transfer resistance is the sum of the individual gas-film resistance ($1/k_y$) and liquid-film resistance ($1/k_x$):

$\frac{1}{K_y} = \frac{1}{k_y} + \frac{m}{k_x}$ $\frac{1}{K_x} = \frac{1}{m k_y} + \frac{1}{k_x}$

Multiplying through by flow rates expresses total transfer unit height as a function of individual film heights ($H_G$ and $H_L$):

$H_{OG} = H_G + \left( \frac{m G}{L} \right) H_L = H_G + S \cdot H_L = H_G + \frac{H_L}{A}$ $H_{OL} = H_L + \left( \frac{L}{m G} \right) H_G = H_L + A \cdot H_G$

[!IMPORTANT] Controlling Film Resistance:

• Gas-Film Controlling ($m \ll 1$, highly soluble gases like $\text{NH}_3$ or $\text{HCl}$ in water): The equilibrium slope $m$ is small, so $m/k_x \to 0$. Thus $1/K_y \approx 1/k_y$, and $H_{OG} \approx H_G$. Turbulence in the gas phase controls the rate of absorption.
• Liquid-Film Controlling ($m \gg 1$, sparingly soluble gases like $\text{O}_2$, $\text{CO}_2$, or hydrocarbons in water): The equilibrium slope $m$ is large, so the liquid-side resistance dominates: $1/K_y \approx m/k_x$, and $H_{OG} \approx (m G / L) H_L$. Liquid-phase agitation controls the separation.

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6. Concentrated Gas Absorption (Solute-Free Coordinates)

When gas mixtures contain high concentrations of soluble gas ($> 5-10\text{ mol}%$), significant solute transfer causes total gas flow $G$ and liquid flow $L$ to decrease markedly along the column height. As a result, the operating line plotted in mole fractions ($y$ vs. $x$) becomes curved, invalidating simple linear analysis.

To restore a strictly linear operating line, balances are written in terms of solute-free mole ratios:

$Y \equiv \frac{y}{1 - y} = \frac{\text{moles solute}}{\text{moles inert carrier gas}}$ $X \equiv \frac{x}{1 - x} = \frac{\text{moles solute}}{\text{moles non-volatile solvent}}$

Defining $G_s$ as the constant molar flow rate of inert carrier gas ($G_s = G(1 - y)$) and $L_s$ as the constant molar flow rate of solute-free solvent ($L_s = L(1 - x)$), the steady-state solute balance yields:

$G_s (Y_{in} - Y_{out}) = L_s (X_{out} - X_{in})$

Rearranging gives the rigorous, linear operating line in ratio coordinates:

$Y = \frac{L_s}{G_s} (X - X_{in}) + Y_{out}$

The slope on a $Y-X$ diagram is strictly constant at $L_s / G_s$.

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7. Summary Comparison Table: Absorption & Stripping Column Design

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8. Comprehensive Worked Numerical Example: Acetone Scrubbing Column Design

Problem Statement

A chemical processing facility must scrub $100.0\text{ kmol/h}$ ($0.02778\text{ kmol/s}$) of an air-acetone mixture containing $4.0\text{ mol}%$ acetone ($y_{in} = 0.040$) to achieve $95.0%$ removal ($y_{out} = 0.0020$) using pure water ($x_{in} = 0$) at $25.0^\circ\text{C}$ and $1.0\text{ atm}$.

System Data:

• Gas flow rate: $G = 100.0\text{ kmol/h}$
• Inlet gas mole fraction: $y_{in} = 0.040$
• Outlet gas mole fraction: $y_{out} = 0.040 \times (1 - 0.950) = 0.0020$
• Entering liquid mole fraction: $x_{in} = 0.000$
• Henry's law constant: $H' = 1.20\text{ atm/mole fraction} \implies m = H'/P = 1.20 / 1.00 = 1.20$
• Overall gas-phase transfer unit height: $H_{OG} = 0.750\text{ m}$

Calculate:

1. The minimum solvent rate $(L/G){min}$ and $L{min}$ in $\text{kmol/h}$.
2. The actual operating solvent rate $L$ using a design factor of $1.40 \times L_{min}$, and the resulting liquid outlet mole fraction $x_{out}$.
3. The absorption factor $A$.
4. The number of theoretical equilibrium stages $N$ required using the Kremser equation.
5. The log-mean driving force $\Delta y_{lm}$ and the number of transfer units $N_{OG}$.
6. The total required packed bed height $Z$ in meters.

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Step 1: Minimum Solvent Rate $(L/G)_{min}$

At the column bottom, the maximum possible liquid outlet concentration is in equilibrium with the entering gas ($y_{in}$):

$x_{out}^* = \frac{y_{in}}{m} = \frac{0.040}{1.20} = 0.03333$

Applying the minimum solvent rate formula:

$\left( \frac{L}{G} \right)_{min} = \frac{y_{in} - y_{out}}{x_{out}^* - x_{in}} = \frac{0.040 - 0.0020}{0.03333 - 0} = \frac{0.0380}{0.03333} = \mathbf{1.140}$

$L_{min} = 1.140 \times G = 1.140 \times 100.0\text{ kmol/h} = \mathbf{114.0\text{ kmol/h}}$

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Step 2: Actual Liquid Rate and Exit Concentration

Using the design margin of $1.40 \times L_{min}$:

$L = 1.40 \times 114.0\text{ kmol/h} = \mathbf{159.6\text{ kmol/h}}$ $\frac{L}{G} = \frac{159.6}{100.0} = \mathbf{1.596}$

From the overall solute balance:

$x_{out} = x_{in} + \frac{G}{L} (y_{in} - y_{out}) = 0 + \frac{1}{1.596} (0.040 - 0.0020) = \frac{0.0380}{1.596} = \mathbf{0.02381} \quad (2.381\text{ mol}\%)$

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Step 3: Absorption Factor ($A$)

$A = \frac{L}{m \cdot G} = \frac{1.596}{1.20} = \mathbf{1.330}$

Because $A = 1.330 > 1.0$, the operating line is steeper than the equilibrium line, ensuring robust mass transfer without an equilibrium pinch.

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Step 4: Theoretical Stages via Kremser Equation

For $x_{in} = 0$:

$\frac{y_{in}}{y_{out}} = \frac{0.040}{0.0020} = 20.0$ $1 - \frac{1}{A} = 1 - \frac{1}{1.330} = 1 - 0.75188 = 0.24812$

Evaluating the logarithmic numerator argument:

$\text{Arg} = \left( \frac{y_{in}}{y_{out}} \right) \left( 1 - \frac{1}{A} \right) + \frac{1}{A} = (20.0 \times 0.24812) + 0.75188 = 4.9624 + 0.75188 = 5.7143$ $\ln(\text{Arg}) = \ln(5.7143) = 1.74297$ $\ln(A) = \ln(1.330) = 0.28518$

$N = \frac{1.74297}{0.28518} = \mathbf{6.11\text{ theoretical stages}}$

Specifying hardware requires rounding to $7$ ideal stages (or $\approx 10-12$ actual trays assuming $60%$ tray efficiency).

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Step 5: Log-Mean Driving Force and Transfer Units ($N_{OG}$)

Calculate driving forces at column terminals:

• Bottom of column: $\Delta y_1 = y_{in} - y_{in}^* = y_{in} - m x_{out} = 0.040 - (1.20 \times 0.02381) = 0.040 - 0.02857 = 0.01143$
• Top of column: $\Delta y_2 = y_{out} - y_{out}^* = y_{out} - m x_{in} = 0.0020 - 0 = 0.0020$

Log-mean driving force:

$\Delta y_{lm} = \frac{\Delta y_1 - \Delta y_2}{\ln(\Delta y_1 / \Delta y_2)} = \frac{0.01143 - 0.0020}{\ln(0.01143 / 0.0020)} = \frac{0.00943}{\ln(5.715)} = \frac{0.00943}{1.7431} = \mathbf{0.005410}$

Number of overall gas transfer units:

$N_{OG} = \frac{y_{in} - y_{out}}{\Delta y_{lm}} = \frac{0.040 - 0.0020}{0.005410} = \frac{0.0380}{0.005410} = \mathbf{7.024\text{ transfer units}}$

(Cross-check via stage conversion: $N_{OG} = N \frac{A \ln(A)}{A - 1} = 6.112 \times \frac{1.330 \times 0.28518}{0.330} = 6.112 \times 1.1494 = 7.025$. Matches perfectly).

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Step 6: Total Packed Bed Height ($Z$)

$Z = H_{OG} \cdot N_{OG} = 0.750\text{ m} \times 7.024 = \mathbf{5.27\text{ m}} \quad (17.3\text{ ft})$

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9. Critical PE Exam Traps & Pitfalls

Trap 1: Confusing Absorption Factor $A$ with Stripping Factor $S$
Remember that $A = L / (m G)$, whereas $S = m G / L = 1/A$. In an absorption problem, you must ensure $A > 1$ for efficient operation. If an exam question asks about stripping a contaminated groundwater stream with air, you must verify that the stripping factor $S > 1.0$. Using the absorption equation for a stripping problem inverts the logarithmic arguments, producing mathematical errors or negative stage counts.

Trap 2: Ignoring Solute-Free Coordinates in Concentrated Feeds
When feed gas contains $15\text{ mol}%$ solute, mole fractions do not yield a straight operating line because $G$ decreases significantly as solute dissolves. Applying the standard Kremser equation directly in mole fractions introduces errors exceeding $20-40%$. You must convert to solute-free ratios: $Y = y/(1-y)$, $X = x/(1-x)$, $G_s = G(1-y)$, and $L_s = L(1-x)$.

Trap 3: Inverting the Film Resistances in $H_{OG}$
When combining transfer unit heights, the equilibrium slope $m$ multiplies the liquid film: $H_{OG} = H_G + (m G / L) H_L = H_G + H_L / A$. A common exam error is writing $H_{OG} = H_G + A H_L$. Remember that if liquid solubility is very low ($m$ is huge), liquid resistance dominates the gas-phase overall height, requiring a large term multiplying $H_L$.

Trap 4: Misidentifying the Equilibrium Pinch Location
For absorption with $A < 1$, the operating line slope is flatter than equilibrium, pinching at the bottom of the column ($x_{out}, y_{in}$). For stripping with $S < 1$, the operating line is steeper than equilibrium, pinching at the top of the column ($x_{in}, y_{out}$). Never place the pinch at the wrong end of the tower.