10.1 Diffusion Fundamentals

10.1 Diffusion Fundamentals

In chemical engineering, mass transfer describes the transport of species from a region of high concentration to a region of low concentration, driven by a chemical potential gradient. The FE Reference Handbook presents mass transfer in two primary regimes: molecular diffusion (governed by Fick's laws) and convective mass transfer (characterized by mass transfer coefficients). Understanding these fundamentals is crucial for solving column sizing, reaction kinetics, and separation process problems on the FE Chemical exam.

Fick's First Law of Molecular Diffusion

Molar diffusion is the thermal motion of molecules in a fluid. For a binary system of components $A$ and $B$, Fick's first law relates the molar diffusion flux of species $A$ to its concentration gradient. The molar flux relative to the molar average velocity of the mixture, denoted as $J_A^*$ (in $\text{kmol}/(\text{m}^2\cdot\text{s})$ or $\text{mol}/(\text{m}^2\cdot\text{s})$), is given by:

$$J_{A}^* = -C D_{AB} \nabla x_A$$

In one dimension ($z$-direction), for a constant total molar concentration $C$ (in $\text{mol}/\text{m}^3$), this simplifies to:

$$J_{Az} = -D_{AB} \frac{dC_A}{dz}$$

Where:

• $J_{Az}$ is the molar flux of $A$ in the $z$-direction relative to the average velocity ($\text{mol}/(\text{m}^2\cdot\text{s})$).
• $D_{AB}$ is the molecular diffusivity of $A$ in $B$ ($\text{m}^2/\text{s}$).
• $C_A$ is the molar concentration of species $A$ ($\text{mol}/\text{m}^3$).
• $z$ is the diffusion distance ($\text{m}$).

General Mass Transfer Equation

In practical engineering systems, bulk fluid motion also contributes to mass transfer. The absolute molar flux of $A$ relative to a stationary coordinate system ($N_A$) is the sum of diffusion flux ($J_A$) and convective flux resulting from bulk fluid flow:

$$N_A = J_A + x_A (N_A + N_B) = -C D_{AB} \frac{dx_A}{dz} + x_A (N_A + N_B)$$

Where:

• $N_A$ is the total molar flux of $A$ relative to a fixed coordinate ($\text{mol}/(\text{m}^2\cdot\text{s})$).
• $N_B$ is the total molar flux of $B$ relative to a fixed coordinate ($\text{mol}/(\text{m}^2\cdot\text{s})$).
• $x_A$ is the mole fraction of component $A$ ($C_A/C$).

Two limiting cases frequently appear on the FE Chemical exam:

1. Equimolar Counterdiffusion (EMD)

In EMD, components $A$ and $B$ diffuse in opposite directions at equal molar rates, typically seen in binary distillation of components with equal latent heats. Under these conditions, $N_A = -N_B$. Consequently, the convective term vanishes:

$$N_A = J_A = -D_{AB} \frac{dC_A}{dz}$$

For gas phases obeying the ideal gas law ($C = P/(RT)$), integrating over a distance $z = z_2 - z_1$ yields:

$$N_A = \frac{D_{AB}}{R T z} (p_{A1} - p_{A2})$$

Where $p_{A1}$ and $p_{A2}$ are the partial pressures of $A$ at positions 1 and 2, $R$ is the universal gas constant, and $T$ is absolute temperature.

2. Unimolar Diffusion (UMD) / Diffusion of A through Stagnant B

Here, component $A$ diffuses while component $B$ remains stationary relative to the boundary ($N_B = 0$). An example is the evaporation of a liquid solvent into a stagnant gas stream. Setting $N_B = 0$ in the general flux equation yields:

$$N_A = -C D_{AB} \frac{dx_A}{dz} + x_A N_A \implies N_A (1 - x_A) = -C D_{AB} \frac{dx_A}{dz}$$

Integrating from $z_1$ to $z_2$ yields:

$$N_A = \frac{D_{AB} C}{z} \ln \left( \frac{1 - x_{A2}}{1 - x_{A1}} \right)$$

For gas phase diffusion, this is expressed using partial pressures and the log-mean partial pressure of stagnant component $B$ ($p_{BM}$):

$$N_A = \frac{D_{AB} P}{R T z p_{BM}} (p_{A1} - p_{A2})$$

Where the log-mean partial pressure of $B$ is defined as:

$$p_{BM} = \frac{p_{B2} - p_{B1}}{\ln(p_{B2}/p_{B1})}$$

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Fick's Second Law of Diffusion

Fick's second law describes unsteady-state (transient) diffusion where concentration changes over time. In one dimension, it is represented as:

$$\frac{\partial C_A}{\partial t} = D_{AB} \frac{\partial^2 C_A}{\partial z^2}$$

For diffusion into a semi-infinite medium ($z \ge 0$) with a constant surface concentration $C_{As}$ and an initial uniform concentration $C_{A0}$, the analytical solution is:

$$\frac{C_A(z,t) - C_{A0}}{C_{As} - C_{A0}} = 1 - \text{erf}\left(\frac{z}{2\sqrt{D_{AB} t}}\right)$$

Where $\text{erf}(x)$ is the error function, which is tabulated in the Mathematics section of the FE Reference Handbook.

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Diffusivity Estimation in Gases and Liquids

Molecular diffusivity is a transport property that depends on temperature, pressure, molecular structure, and solvent characteristics.

Gas Diffusivity ($D_{AB}$)

The FE Handbook details the Fuller-Schettler-Giddings (FSG) correlation for binary gas systems at low pressures:

$$D_{AB} = \frac{0.00143 T^{1.75}}{P M_{AB}^{1/2} [(\Sigma v)_A^{1/3} + (\Sigma v)_B^{1/3}]^2}$$

Where:

• $D_{AB}$ is in $\text{cm}^2/\text{s}$.
• $T$ is in Kelvin ($\text{K}$).
• $P$ is pressure in bar.
• $M_{AB} = \frac{2}{1/M_A + 1/M_B}$ is the effective molecular weight.
• $\Sigma v$ is the diffusion volume of each component (sum of atomic volumes, found in reference tables).

Exam Tip: Note the proportionalities: $D_{AB} \propto T^{1.75}$ and $D_{AB} \propto P^{-1}$. If a gas diffusivity is given at one temperature and pressure, you can quickly estimate it at another using:

$$D_{AB,2} = D_{AB,1} \left(\frac{T_2}{T_1}\right)^{1.75} \left(\frac{P_1}{P_2}\right)$$

Liquid Diffusivity ($D_{AB}$)

For dilute solutes in liquids, the Wilke-Chang correlation is widely used:

$$D_{AB} = \frac{7.4 \times 10^{-8} (\Phi M_B)^{1/2} T}{\eta_B V_A^{0.6}}$$

Where:

• $D_{AB}$ is diffusivity ($\text{cm}^2/\text{s}$).
• $\Phi$ is the association parameter of the solvent $B$ (water = 2.6, methanol = 1.9, ethanol = 1.5, unassociated = 1.0).
• $M_B$ is the molecular weight of solvent $B$.
• $T$ is absolute temperature ($\text{K}$).
• $\eta_B$ is the viscosity of solvent $B$ in centipoise ($\text{cP}$ or $\text{mPa}\cdot\text{s}$).
• $V_A$ is the molar volume of solute $A$ at its normal boiling point ($\text{cm}^3/\text{mol}$).

Exam Tip: In liquids, $D_{AB} \propto T/\eta_B$. Since viscosity decreases exponentially with temperature, liquid diffusivity increases much more rapidly with temperature than gas diffusivity.

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Convective Mass Transfer & Film Theory

When fluid flows past a surface, mass transfer occurs via convective transport. The molar flux is written using a convective mass transfer coefficient, $k_c$ (in $\text{m}/\text{s}$):

$$N_A = k_c (C_{As} - C_{Ab})$$

Where $C_{As}$ is the concentration at the interface (surface) and $C_{Ab}$ is the bulk concentration.

Film Theory

Whitman's Film Theory assumes a thin, stagnant laminar film of thickness $\delta$ adjacent to the phase boundary. All resistance to mass transfer is assumed to lie within this film, where transport is purely by molecular diffusion. Thus, comparing Fick's law and the convective flux equation:

$$k_c = \frac{D_{AB}}{\delta}$$

Under this theory, $k_c$ is directly proportional to the molecular diffusivity ($k_c \propto D_{AB}^1$). Other theories (such as Boundary Layer Theory, where $k_c \propto D_{AB}^{2/3}$, or Penetration/Surface Renewal Theory, where $k_c \propto D_{AB}^{0.5}$) predict different relationships, which is a frequent concept-based exam question.

Interphase Mass Transfer (Two-Film Theory)

For transport across gas-liquid interfaces, resistances in both phases must be summed. Assuming local equilibrium at the interface ($y_i = m x_i$), the overall mass transfer coefficients $K_y$ and $K_x$ relate to individual film coefficients $k_y$ and $k_x$ by:

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

Where $m$ is the slope of the equilibrium curve ($y^* = m x$). If $m \gg 1$ (highly insoluble gas), the liquid film resistance dominates ($\frac{1}{K_x} \approx \frac{1}{k_x}$). If $m \ll 1$ (highly soluble gas), the gas film resistance dominates ($\frac{1}{K_y} \approx \frac{1}{k_y}$).

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Dimensionless Groups in Mass Transfer

Dimensionless groups correlate mass transfer data and establish analogies to heat and momentum transport.

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Worked Example: Diffusion of Water Vapor (UMD)

Problem: Water at $298\text{ K}$ is evaporating from the bottom of a $10\text{ cm}$ deep vertical tube into dry air ($y_{A2} = 0$) flowing across the top. The total pressure is $1.0\text{ bar}$. The vapor pressure of water at $298\text{ K}$ is $0.0317\text{ bar}$. The molecular diffusivity of water in air is $D_{AB} = 0.26 \times 10^{-4}\text{ m}^2/\text{s}$. Compute the steady-state molar flux of water vapor in $\text{mol}/(\text{m}^2\cdot\text{s})$.

Solution: This is a classic case of Unimolar Diffusion (UMD) because air (component $B$) is stagnant inside the tube ($N_B = 0$). Let $A$ represent water and $B$ represent air. At the liquid surface (position 1):

$$p_{A1} = p_{vap} = 0.0317\text{ bar}$$ $$p_{B1} = P - p_{A1} = 1.0 - 0.0317 = 0.9683\text{ bar}$$

At the top of the tube (position 2):

$$p_{A2} = 0\text{ bar}$$ $$p_{B2} = P - p_{A2} = 1.0 - 0 = 1.0\text{ bar}$$

Calculate the log-mean partial pressure of stagnant air ($p_{BM}$):

$$p_{BM} = \frac{p_{B2} - p_{B1}}{\ln(p_{B2}/p_{B1})} = \frac{1.0 - 0.9683}{\ln(1.0/0.9683)} = \frac{0.0317}{\ln(1.03273)} = \frac{0.0317}{0.0322} = 0.9841\text{ bar}$$

Now, apply the UMD flux equation:

$$N_A = \frac{D_{AB} P}{R T z p_{BM}} (p_{A1} - p_{A2})$$

Substitute the known parameters in SI units:

• $D_{AB} = 0.26 \times 10^{-4}\text{ m}^2/\text{s}$
• $P = 1.0 \times 10^5\text{ Pa}$
• $p_{A1} - p_{A2} = 0.0317 \times 10^5\text{ Pa}$
• $p_{BM} = 0.9841 \times 10^5\text{ Pa}$
• $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$
• $T = 298\text{ K}$
• $z = 0.10\text{ m}$

$$N_A = \frac{(0.26 \times 10^{-4}\text{ m}^2/\text{s}) (1.0 \times 10^5\text{ Pa})}{(8.314\text{ J}/(\text{mol}\cdot\text{K})) (298\text{ K}) (0.10\text{ m}) (0.9841 \times 10^5\text{ Pa})} \cdot (0.0317 \times 10^5\text{ Pa})$$

Notice that the pressure terms cancel out:

$$N_A = \frac{(0.26 \times 10^{-4}) (1.0)}{(8.314) (298) (0.10) (0.9841)} \cdot (3170) = \frac{0.08242}{243.83} \approx 3.38 \times 10^{-4}\text{ mol}/(\text{m}^2\cdot\text{s})$$

Thus, the evaporation rate of water is $3.38 \times 10^{-4}\text{ mol}/(\text{m}^2\cdot\text{s})$.