10.2 Convective Mass Transfer and Film Theories

10.2 Convective Mass Transfer and Film Theories

In industrial separation equipment—such as packed absorption columns, tray distillers, spray towers, and bio-aeration basins—fluids are rarely stagnant. Instead, turbulence, eddy mixing, and phase contacting govern transport across phase interfaces. Because solving the coupled Navier-Stokes and species continuity equations in tortuous packed beds or agitated tanks is analytically intractable, chemical engineers utilize convective mass transfer coefficients ($k$) to relate interfacial molar flux to measurable bulk concentration driving forces.

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1. Convective Mass Transfer Coefficients and Rate Equations

Analogous to Newton's law of cooling in heat transfer ($q'' = h \Delta T$), convective mass transfer defines molar flux as the product of a mass transfer coefficient and a concentration driving force:

$N_A = k_c (C_{A,i} - C_{A,b}) = k_y (y_{A,b} - y_{A,i}) = k_G (p_{A,b} - p_{A,i}) = k_x (x_{A,i} - x_{A,b}) = k_L (C_{A,i} - C_{A,b})$

Where:

• Subscript $b$ denotes the bulk fluid core, and subscript $i$ denotes the phase interface.
• $k_c$ = molar concentration-based mass transfer coefficient ($\text{m/s}$ or $\text{ft/hr}$).
• $k_y$ = gas mole-fraction-based coefficient ($\text{kmol/(m}^2\cdot\text{s}\cdot\Delta y)$ or $\text{lbmol/(ft}^2\cdot\text{hr}\cdot\Delta y)$).
• $k_G$ = gas partial-pressure-based coefficient ($\text{kmol/(m}^2\cdot\text{s}\cdot\text{kPa)}$ or $\text{lbmol/(ft}^2\cdot\text{hr}\cdot\text{atm)}$).
• $k_x$ = liquid mole-fraction-based coefficient ($\text{kmol/(m}^2\cdot\text{s}\cdot\Delta x)$ or $\text{lbmol/(ft}^2\cdot\text{hr}\cdot\Delta x)$).
• $k_L$ = liquid volumetric concentration-based coefficient ($\text{m/s}$ or $\text{ft/hr}$).

Interconversion of Mass Transfer Coefficients

For an ideal gas phase where $p_A = y_A P = C_A R T$:

$k_y = k_G P = k_c C = k_c \left( \frac{P}{R T} \right)$

$k_c = k_G R T = k_y \left( \frac{R T}{P} \right)$

For a liquid phase with total molar density $C_t = \rho_L / M_L$:

$k_x = k_L C_t = k_L \left( \frac{\rho_L}{M_L} \right)$

Dimensional and Unit Consistency Reference Table

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2. Classical Mass Transfer Theories & Diffusivity Dependencies

Three foundational physical theories explain how turbulence and molecular diffusion interact near a phase boundary. On the NCEES PE Chemical exam, questions test the theoretical diffusivity exponent ($n$, where $k \propto D_{AB}^n$).

+---------------------------------------------------------------------------------------+
|                                 MASS TRANSFER THEORIES                                |
+-----------------------------+-----------------------------+---------------------------+
|      WHITMAN FILM (1923)    |    HIGBIE PENETRATION (1935)|   DANCKWERTS RENEWAL (1951)|
|                             |                             |                           |
| Stagnant film thickness d   | Eddies reside at interface  | Eddies randomly replaced; |
| Steady-state 1-D diffusion  | for uniform contact time tc | renewal rate distribution s|
|        k_c = D_AB / d       |  k_c = 2 * sqrt(D_AB/(pi*tc))|      k_c = sqrt(D_AB * s)  |
|     k_c ~ (D_AB)^1.0        |      k_c ~ (D_AB)^0.5       |      k_c ~ (D_AB)^0.5     |
+-----------------------------+-----------------------------+---------------------------+

1. Whitman Film Theory (1923)

• Physical Model: A fictitious, stagnant laminar film of thickness $\delta$ exists adjacent to the phase interface. Outside this film, vigorous turbulence maintains a completely uniform bulk concentration. Inside the film, fluid is quiescent, and mass transfer occurs purely by steady-state molecular diffusion.
• Governing Equation:
  $N_A = \frac{D_{AB}}{\delta} (C_{A,i} - C_{A,b}) \implies \mathbf{k_c = \frac{D_{AB}}{\delta}}$
• Diffusivity Exponent: $\mathbf{n = 1.0}$ ($k_c \propto D_{AB}^{1.0}$).
• Engineering Limitations: While mathematically simple and intuitive, predicting the hypothetical film thickness $\delta$ a priori is impossible. Furthermore, experimental mass transfer data in turbulent streams invariably reveals that $k_c$ scales with $D_{AB}^{0.5}$ to $D_{AB}^{0.7}$, proving that the film model overpredicts the influence of molecular diffusivity.

2. Higbie Penetration Theory (1935)

• Physical Model: Turbulent eddies are swept from the bulk core to the gas-liquid interface. Each fluid packet resides at the interface for a fixed, uniform contact time ($t_c$), during which it absorbs or desorbs solute via unsteady-state Fickian diffusion into a semi-infinite slab, before being displaced back into the turbulent bulk.
• Governing Differential Equation:
  $\frac{\partial C_A}{\partial t} = D_{AB} \frac{\partial^2 C_A}{\partial z^2}$
• Instantaneous Flux: $N_A(t) = \sqrt{\frac{D_{AB}}{\pi t}} (C_{A,i} - C_{A,b})$.
• Time-Averaged Flux over Contact Time $t_c$:
  $N_{A,avg} = \frac{1}{t_c} \int_0^{t_c} N_A(t) dt = \frac{C_{A,i} - C_{A,b}}{t_c} \sqrt{\frac{D_{AB}}{\pi}} \int_0^{t_c} t^{-1/2} dt = 2 \sqrt{\frac{D_{AB}}{\pi t_c}} (C_{A,i} - C_{A,b})$
• Mass Transfer Coefficient:
  $\mathbf{k_c = 2 \sqrt{\frac{D_{AB}}{\pi t_c}}}$
• Diffusivity Exponent: $\mathbf{n = 0.5}$ ($k_c \propto D_{AB}^{0.5}$).
• Application: Highly applicable to short-contact systems such as wetted-wall columns, liquid jets, and rapidly coalescing bubble swarms.

3. Danckwerts Surface Renewal Theory (1951)

• Physical Model: Danckwerts recognized that turbulent eddies do not reside at the interface for an identical duration $t_c$. Instead, eddies are randomly swept away and replaced by fresh bulk fluid with a constant probability of replacement per unit time ($s$, the fractional surface renewal rate, $\text{s}^{-1}$).
• Age Distribution Function: $\psi(t) = s \exp(-s t)$.
• Area-Averaged Mass Transfer Flux:
  $N_A = \int_0^\infty N_A(t) \cdot \psi(t) dt = (C_{A,i} - C_{A,b}) \int_0^\infty \sqrt{\frac{D_{AB}}{\pi t}} \cdot s e^{-s t} dt = \sqrt{D_{AB} s} (C_{A,i} - C_{A,b})$
• Mass Transfer Coefficient:
  $\mathbf{k_c = \sqrt{D_{AB} s}}$
• Diffusivity Exponent: $\mathbf{n = 0.5}$ ($k_c \propto D_{AB}^{0.5}$).
• Application: The prevailing modern theoretical model for stirred tank reactors, packed scrubbers, and industrial bubble columns.

4. Boundary Layer Theory (Prandtl / Blasius)

For laminar boundary layer flow along a flat solid surface, solving the convective diffusion equation yields:

$k_c \propto D_{AB}^{2/3} \implies n = 0.67$

For fully turbulent boundary layers, eddy diffusivity models predict $n \approx 0.5$ to $0.75$, bridging the gap between penetration theory ($0.50$) and boundary layer theory ($0.67$).

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3. Two-Film Theory and Overall Mass Transfer Coefficients

When a solute transfers between a gas phase and a liquid phase (e.g., in a gas absorber or stripper), it must cross two distinct fluid boundary layers in series:

1. The gas-phase boundary layer ($k_y$ or $k_G$).
2. The liquid-phase boundary layer ($k_x$ or $k_L$).

              GAS FILM            INTERFACE           LIQUID FILM
        |                    |                 |                    |
  y_A,b |---------------------.                |                    |
(Bulk)  |                      . (y_A,i)       |                    |
        |                       +==============+ (x_A,i)            |
        |                       | (Equilibrium)| .                  |
        |                       | y_i = m*x_i  |  .                 |
        |                       |              |   .                |
        |                       |              |    .---------------| x_A,b (Bulk)
        |<------- delta_G ----->|              |<----- delta_L ---->|

Interfacial Thermodynamic Equilibrium

The two-film theory rests on the fundamental postulate that the phase interface offers zero resistance to mass transfer. Therefore, the fluid at the interface is in instantaneous thermodynamic phase equilibrium:

$y_{A,i} = m x_{A,i} \quad \text{or} \quad p_{A,i} = H C_{A,i}$

Where:

• $m$ = dimensionless slope of the vapor-liquid equilibrium line ($y^* = m x$).
• $H$ = Henry's law constant in pressure-concentration units ($\text{kPa}\cdot\text{m}^3/\text{kmol}$ or $\text{atm}\cdot\text{m}^3/\text{mol}$).

Overall Driving Forces

Because interface compositions ($y_{A,i}, x_{A,i}$) cannot be measured directly in operating equipment, industrial design relies on overall mass transfer coefficients ($K_y, K_x, K_G, K_L$) defined in terms of fictitious equilibrium bulk concentrations:

$N_A = K_y (y_{A,b} - y_A^*) = K_x (x_A^* - x_{A,b})$

Where:

• $y_A^* = m x_{A,b}$ = vapor mole fraction in equilibrium with the actual bulk liquid composition.
• $x_A^* = y_{A,b} / m$ = liquid mole fraction in equilibrium with the actual bulk gas composition.

Rigorous Derivation of Resistances in Series

At steady state, the molar flux crossing the gas boundary layer equals the flux crossing the liquid boundary layer:

$N_A = k_y (y_{A,b} - y_{A,i}) = k_x (x_{A,i} - x_{A,b})$

Expressing the overall gas-phase driving force as the sum of gas-film and liquid-film components:

$(y_{A,b} - y_A^*) = (y_{A,b} - y_{A,i}) + (y_{A,i} - y_A^*)$

Substituting the equilibrium relationships $y_{A,i} = m x_{A,i}$ and $y_A^* = m x_{A,b}$ into the second term:

$(y_{A,i} - y_A^*) = m (x_{A,i} - x_{A,b})$

Therefore:

$(y_{A,b} - y_A^*) = (y_{A,b} - y_{A,i}) + m (x_{A,i} - x_{A,b})$

Dividing through by the molar flux $N_A$:

$\frac{y_{A,b} - y_A^*}{N_A} = \frac{y_{A,b} - y_{A,i}}{N_A} + m \left( \frac{x_{A,i} - x_{A,b}}{N_A} \right)$

Recognizing that $(y_{A,b} - y_A^*)/N_A = 1/K_y$, $(y_{A,b} - y_{A,i})/N_A = 1/k_y$, and $(x_{A,i} - x_{A,b})/N_A = 1/k_x$:

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

Following an identical procedure for the overall liquid-phase driving force $(x_A^* - x_{A,b})$ yields:

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

Relationship Between Overall Coefficients

Multiplying the $1/K_y$ equation by $m$:

$\frac{m}{K_y} = \frac{m}{k_y} + \frac{m^2}{k_x} = m \left( \frac{1}{m k_y} + \frac{1}{k_x} \right) = \frac{m}{K_x} \implies \mathbf{K_x = m K_y}$

In terms of partial pressure and concentration units ($K_G$ and $K_L$):

$\frac{1}{K_G} = \frac{1}{k_G} + \frac{H}{k_L}$

$\frac{1}{K_L} = \frac{1}{H k_G} + \frac{1}{k_L}$

$K_L = H K_G$

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4. Controlling Resistance Regimes: Gas-Film vs. Liquid-Film

On the PE Chemical exam, identifying the controlling resistance dictates how an engineer must troubleshoot or optimize an underperforming separation column.

+---------------------------------------------------------------------------------+
|                           CONTROLLING RESISTANCE CRITERIA                       |
+---------------------------------------+-----------------------------------------+
|       GAS-FILM CONTROLLING            |         LIQUID-FILM CONTROLLING         |
|   (Highly Soluble Solute: m << 1)     |    (Sparingly Soluble Solute: m >> 1)   |
|                                       |                                         |
| Examples: NH3, HCl, SO2 in water      | Examples: O2, CO2, N2, VOCs in water    |
| Total Resistance: 1/Ky approx 1/ky    | Total Resistance: 1/Kx approx 1/kx      |
| Liquid resistance (m/kx) negligible   | Gas resistance (1/(m*ky)) negligible    |
| Debottleneck: Increase gas velocity,  | Debottleneck: Increase liquid flow,     |
|   blower speed, gas turbulence        |   liquid agitation, interfacial area a  |
+---------------------------------------+-----------------------------------------+

1. Gas-Film Controlling Regime ($m \ll 1$)

• Thermodynamic Condition: Solute has extremely high solubility in the liquid solvent (e.g., absorption of $\text{NH}_3$, $\text{HCl}$, or $\text{SO}_3$ into water). Consequently, the equilibrium distribution coefficient is very small ($m \to 0$).
• Mathematical Impact:
  $\frac{m}{k_x} \ll \frac{1}{k_y} \implies \frac{1}{K_y} \approx \frac{1}{k_y} \implies \mathbf{K_y \approx k_y}$
• Physical Interpretation: The liquid absorbs solute so eagerly that virtually the entire concentration gradient resides in the gas boundary layer. The interface concentration $x_{A,i}$ is very close to $x_{A,b}$.
• Operational Strategy: To increase mass transfer, the engineer must increase gas velocity, increase gas Reynolds number, or switch to packing with higher gas turbulence. Increasing liquid pumping rate or liquid agitation produces virtually zero performance improvement!

2. Liquid-Film Controlling Regime ($m \gg 1$)

• Thermodynamic Condition: Solute has very low solubility (sparingly soluble) in the liquid solvent (e.g., stripping of volatile organic compounds like TCE or benzene from water, oxygen aeration in wastewater, or stripping $\text{CO}_2$ from water).
• Mathematical Impact: The Henry's law constant and $m$ are very large ($m \gg 1$): $\frac{1}{m k_y} \ll \frac{1}{k_x} \implies \frac{1}{K_x} \approx \frac{1}{k_x} \implies \mathbf{K_x \approx k_x}$
• Physical Interpretation: Solute easily diffuses across the gas boundary layer, but dissolving into or escaping from the liquid is severely bottlenecked. The gas interface concentration $y_{A,i}$ is nearly identical to bulk $y_{A,b}$.
• Operational Strategy: To increase mass transfer, the engineer must agitate the liquid, increase liquid flow rate, generate smaller bubbles (higher interfacial area $a$), or use high-surface-area packing. Doubling the gas blower rate merely wastes electrical power without boosting mass transfer!

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5. Volumetric Mass Transfer Coefficients & Transfer Units

In industrial packed columns, the actual interfacial contact area per unit volume ($a$, $\text{m}^2/\text{m}^3$ or $\text{ft}^2/\text{ft}^3$) depends complexly on packing geometry, liquid holdup, and surface wetting. Because $k$ and $a$ cannot be isolated independently, engineers design columns using volumetric mass transfer coefficients:

$K_y a \quad (\text{kmol/(m}^3\cdot\text{s}\cdot\Delta y)), \quad K_x a \quad (\text{kmol/(m}^3\cdot\text{s}\cdot\Delta x))$

Height of a Transfer Unit ($HTU$)

For packed column design, the total packed bed height ($Z$) is calculated as:

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

Where:

• $H_{OG} = \frac{G_m}{K_y a}$ = Height of an Overall Gas Transfer Unit ($\text{m}$ or $\text{ft}$).
• $H_{OL} = \frac{L_m}{K_x a}$ = Height of an Overall Liquid Transfer Unit ($\text{m}$ or $\text{ft}$).
• $G_m, L_m$ = superficial molar mass velocities of gas and liquid ($\text{kmol/(m}^2\cdot\text{s)}$).

Connecting individual film transfer units to overall transfer units:

$H_{OG} = H_G + \left( \frac{m G_m}{L_m} \right) H_L = H_G + \lambda H_L$

$H_{OL} = H_L + \left( \frac{L_m}{m G_m} \right) H_G = H_L + \frac{1}{\lambda} H_G$

Where $\lambda = m G_m / L_m$ is the stripping factor.

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6. Summary Comparison Table: Mass Transfer Models

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7. Comprehensive Worked Numerical Example: Absorber Film Resistances for $\text{SO}_2$ and $\text{CO}_2$

Problem Statement

A chemical environmental engineer operates a countercurrent packed column absorber at $T = 298.15\text{ K}$ ($25.0^\circ\text{C}$) and $P = 1.00\text{ atm} = 101.325\text{ kPa}$. The column uses water as the absorbing solvent. Hydrodynamic tests determine the individual film mass transfer coefficients to be:

• Gas-film coefficient: $k_y = 1.25 \times 10^{-3}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta y)$
• Liquid-film coefficient: $k_x = 2.50 \times 10^{-2}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta x)$

The engineer must evaluate two distinct industrial scrubbing operations:

• System 1 (Sulfur Dioxide, $\text{SO}_2$): Moderately soluble in water. Vapor-liquid equilibrium slope: $m = 28.5$ ($y = 28.5 x$).
• System 2 (Carbon Dioxide, $\text{CO}_2$): Sparingly soluble in water. Vapor-liquid equilibrium slope: $m = 1,450.0$ ($y = 1,450.0 x$).

Calculate for each system:

1. The individual gas-film resistance ($1/k_y$) and liquid-film resistance expressed in gas units ($m/k_x$).
2. The overall gas-phase mass transfer coefficient ($K_y$) in $\text{kmol/(m}^2\cdot\text{s}\cdot\Delta y)$.
3. The percentage of total mass transfer resistance residing in the gas film and liquid film.
4. The overall liquid-phase mass transfer coefficient ($K_x$).
5. The engineering recommendation for debottlenecking each absorption system.

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Step 1: Base Resistances

• Individual gas-film resistance:
  $R_g = \frac{1}{k_y} = \frac{1}{1.25 \times 10^{-3}} = \mathbf{800.0\text{ (m}^2\cdot\text{s}\cdot\Delta y)/\text{kmol}}$
• Individual liquid-film base resistance:
  $\frac{1}{k_x} = \frac{1}{2.50 \times 10^{-2}} = \mathbf{40.0\text{ (m}^2\cdot\text{s}\cdot\Delta x)/\text{kmol}}$

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Step 2: System 1 ($\text{SO}_2$ Absorption, $m = 28.5$)

Liquid resistance expressed in gas driving force units:

$R_{l,gas} = \frac{m}{k_x} = 28.5 \times 40.0 = \mathbf{1,140.0\text{ (m}^2\cdot\text{s}\cdot\Delta y)/\text{kmol}}$

Total overall gas-phase resistance:

$\frac{1}{K_y} = \frac{1}{k_y} + \frac{m}{k_x} = 800.0 + 1,140.0 = \mathbf{1,940.0\text{ (m}^2\cdot\text{s}\cdot\Delta y)/\text{kmol}}$

Overall gas mass transfer coefficient:

$K_y = \frac{1}{1,940.0} = \mathbf{5.155 \times 10^{-4}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta y)}$

Resistance breakdown for $\text{SO}_2$:

• Gas film: $\frac{800.0}{1,940.0} \times 100% = \mathbf{41.24%}$
• Liquid film: $\frac{1,140.0}{1,940.0} \times 100% = \mathbf{58.76%}$

Overall liquid mass transfer coefficient:

$K_x = m K_y = 28.5 \times (5.155 \times 10^{-4}) = \mathbf{1.469 \times 10^{-2}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta x)}$

Engineering Recommendation: $\text{SO}_2$ absorption exhibits mixed control (both gas and liquid films offer significant resistance). Debottlenecking requires optimizing both gas flow velocity and liquid distribution.

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Step 3: System 2 ($\text{CO}_2$ Absorption, $m = 1,450.0$)

Liquid resistance expressed in gas driving force units:

$R_{l,gas} = \frac{m}{k_x} = 1,450.0 \times 40.0 = \mathbf{58,000.0\text{ (m}^2\cdot\text{s}\cdot\Delta y)/\text{kmol}}$

Total overall gas-phase resistance:

$\frac{1}{K_y} = \frac{1}{k_y} + \frac{m}{k_x} = 800.0 + 58,000.0 = \mathbf{58,800.0\text{ (m}^2\cdot\text{s}\cdot\Delta y)/\text{kmol}}$

Overall gas mass transfer coefficient:

$K_y = \frac{1}{58,800.0} = \mathbf{1.7007 \times 10^{-5}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta y)}$

Resistance breakdown for $\text{CO}_2$:

• Gas film: $\frac{800.0}{58,800.0} \times 100% = \mathbf{1.36%}$
• Liquid film: $\frac{58,000.0}{58,800.0} \times 100% = \mathbf{98.64%}$

Overall liquid mass transfer coefficient:

$K_x = m K_y = 1,450.0 \times (1.7007 \times 10^{-5}) = \mathbf{2.466 \times 10^{-2}\text{ kmol/(m}^2\cdot\text{s}\cdot\Delta x)}$

Notice that $K_x \approx k_x$ ($2.466 \times 10^{-2} \approx 2.50 \times 10^{-2}$, within $1.4%$).

Engineering Recommendation: $\text{CO}_2$ absorption is heavily liquid-film controlled ($98.6%$ resistance in liquid). To debottleneck the absorber, the plant must increase liquid recirculation rate, improve liquid distributor wetting, or add packing with higher specific surface area ($a$). Increasing gas blower rate will have virtually zero effect on $\text{CO}_2$ removal!

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8. Critical NCEES PE Exam Traps & Pitfalls

Trap 1: Mixing Units of Henry's Law Constant ($H$) and Equilibrium Slope ($m$)
Candidates frequently combine resistances with mismatched units. In the relation $1/K_y = 1/k_y + m/k_x$, $m$ must be dimensionless ($m = y^/x = (p_A^ / P) / (C_A / C_t)$). If Henry's law is given in $\text{atm}\cdot\text{m}^3/\text{kmol}$ or $\text{bar}/(\text{mol fraction})$, you must convert it to $m = H_x / P$ before summing resistances. Summing $1/k_y$ with $H/k_L$ without unit conversion will produce an enormous error.

Trap 2: Assuming Higher Gas Flow Increases Rate for Sparingly Soluble Gases
When asked how to improve stripping or absorption of a sparingly soluble solute (such as $\text{O}_2$, $\text{CO}_2$, $\text{CH}_4$, or chlorinated VOCs), an exam distractor invariably suggests "increasing the gas blower speed." Because sparingly soluble gases are liquid-film controlled ($m \gg 1$), $K_y$ is limited by liquid transport ($K_x \approx k_x$). Changing gas hydrodynamics alters $k_y$, but since gas resistance is $< 2%$ of total resistance, performance remains virtually unchanged.

Trap 3: Applying Whitman's $D_{AB}^{1.0}$ Exponent to Scale Solutes in Industrial Equipment
When predicting the mass transfer coefficient of a new chemical compound from pilot tests on a baseline compound ($k_{c2} = k_{c1} (D_2 / D_1)^n$), remember that modern industrial packings, stirred vessels, and bubble aerators operate under penetration/surface renewal mechanics ($n = 0.50$), NOT film theory ($n = 1.0$). Using $n = 1.0$ severely overstates the impact of molecular diffusivity differences.