Mass transfer is the transport of a chemical species within a material or between phases. In ordinary diffusion, a species moves from higher to lower concentration. Fluid motion can also carry species and increase transfer rates. Distillation, gas absorption, drying and extraction are important applications in chemical engineering.
When sugar spreads through still water, oxygen enters a fermentation broth, or moisture leaves wet granules, a particular component changes location. The engineering question is: how quickly does it move, and what controls that rate?
This guide explains the mechanisms, equations and operations you need to answer that question, with two worked calculations.
What Drives Mass Transfer?
For many introductory problems, the driving force is a concentration difference within a phase or a partial-pressure difference in a gas. More generally, spontaneous species transfer is governed by differences in chemical potential.
A phase boundary provides a location for transfer; it is not itself a driving force. Diffusion can occur entirely within one liquid, gas or solid.
Across two phases, compare the actual composition with the composition that would be in equilibrium with the other phase. Simply subtracting a gas concentration from a liquid concentration does not establish the driving force. The two phases can have different compositions at equilibrium, with no net transfer between them.
Types of Mass Transfer: Diffusion and Convection
Molecular Diffusion
Molecular diffusion results from random molecular motion. In a mixture with a concentration gradient, that motion produces a net movement of species down the gradient under ordinary Fickian conditions.
An example is dissolved dye spreading through still water. Diffusion does not require a pump, an impeller or a phase change.
Convective Mass Transfer
Convective mass transfer combines molecular diffusion with fluid motion. Flow carries material through the fluid and changes the concentration boundary layer near a surface or interface.
Forced convection uses equipment such as pumps, fans or stirrers. Natural convection arises from buoyancy associated with density differences.
For example, airflow carries water vapour away from a wet surface, helping maintain the concentration difference needed for further evaporation. Molecular diffusion still matters near the surface.
Mass Transfer Flux, Rate and Units
Flux is the amount transferred per unit area per unit time. Rate is the total amount transferred per unit time.
| Quantity | Symbol used here | SI unit |
|---|---|---|
| Diffusive molar flux | JA | mol/(m²·s) |
| Total molar flux of species A | NA | mol/(m²·s) |
| Molar transfer rate | ṅA | mol/s |
| Molecular diffusivity | DAB | m²/s |
| Molar concentration | CA | mol/m³ |
| Transfer area | A | m² |
For uniform total flux over area A:
ṅA = NAA
Choose either a mass basis or a molar basis and use it consistently. A flux reported in kg/(m²·s) cannot be inserted unchanged into an equation expecting mol/(m²·s).
Fick’s Laws of Diffusion
Fick’s First Law: Flux and Concentration Gradient
For one-dimensional diffusion in a binary mixture at constant total molar concentration:
JA = −DAB(∂CA/∂z)
Here, JA is the diffusive molar flux relative to the molar-average motion, DAB is the diffusivity of A in B, and z is distance.
The minus sign means the diffusive flux points down the concentration gradient. Fick’s first law relates local flux to the local gradient; it is useful in both steady and unsteady diffusion. It is not restricted to steady state.
Fick’s Second Law: Concentration Changing with Time
For one-dimensional diffusion with constant diffusivity, no reaction and no bulk motion:
∂CA/∂t = DAB(∂²CA/∂z²)
This equation predicts how concentration changes with position and time. Solving it requires initial and boundary conditions. It follows by combining a species balance with Fick’s first law. MIT’s macroscopic diffusion material shows this derivation.
Diffusive Flux Versus Total Flux
In a binary mixture:
NA = JA + xA(NA + NB)
The second term accounts for transport with the mixture’s molar-average motion; xA is the local mole fraction. Under equimolar counterdiffusion, NA + NB = 0, so NA = JA. Diffusion through a stagnant second component generally needs a different total-flux treatment.
Worked Example 1: Calculate Diffusion Flux and Rate
Consider an idealised liquid film with these specified values:
- Film thickness, δ = 1.0 mm = 0.001 m
- Boundary concentrations, CA1 = 100 mol/m³ and CA2 = 20 mol/m³
- Diffusivity, DAB = 2.0 × 10−9 m²/s
- Transfer area, A = 0.50 m²
Assume steady, one-dimensional diffusion, constant diffusivity, no reaction and zero net molar flow. The concentration profile is linear.
Taking positive z from boundary 1 toward boundary 2:
NA = JA = DAB(CA1 − CA2)/δ
NA = (2.0 × 10−9)(100 − 20)/0.001 = 1.6 × 10−4 mol/(m²·s)
The total transfer rate is:
ṅA = (1.6 × 10−4)(0.50) = 8.0 × 10−5 mol/s
The positive result indicates transfer toward boundary 2. If the film thickness halves while the boundary concentrations and diffusivity remain fixed, both flux and rate double. These values are illustrative inputs, not measured plant data.
What Is a Mass Transfer Coefficient?
A mass transfer coefficient relates flux to a specified driving force. For transfer from a surface into a fluid, a concentration-based expression is:
NA = kc(CAs − CAb)
CAs and CAb are the surface and bulk concentrations. With concentrations in mol/m³, kc has units of m/s.
Unlike molecular diffusivity, the coefficient also depends on flow conditions and geometry. Always check its definition: coefficients based on mole fraction or partial pressure have different units from concentration-based coefficients.
Worked Example 2: Calculate an Absorption Rate
Suppose an overall liquid-side model applies:
NA = KL(CA* − CAb)
Take KL = 3.0 × 10−5 m/s, CA* = 2.0 mol/m³, CAb = 0.50 mol/m³ and a uniform effective interfacial area of 10 m².
NA = (3.0 × 10−5)(2.0 − 0.50) = 4.5 × 10−5 mol/(m²·s)
ṅA = (4.5 × 10−5)(10) = 4.5 × 10−4 mol/s
CA* is the liquid concentration in equilibrium with the bulk gas. It is not the bulk gas concentration. This illustrative calculation assumes a uniform coefficient and driving force; a column normally requires allowance for variation along its height.
Two-Film Theory and Overall Resistance
The two-film model represents gas–liquid transfer using a resistance in each phase, with equilibrium at the interface.
For absorption using mole-fraction-based coefficients:
NA = ky(yAb − yAi) = kx(xAi − xAb)
Subscripts b and i denote bulk and interface. For a linear equilibrium relationship y* = mx, the overall gas-side expression is:
NA = Ky(yAb − yA*)
1/Ky = 1/ky + m/kx
Compare 1/ky with m/kx to identify the dominant resistance on this basis. Improving the phase that already contributes little resistance may produce only a small improvement in the overall rate. NPTEL derives the overall-coefficient relationship and its assumptions.
Common Mass Transfer Operations and Applications
| Operation | Material movement | Example |
|---|---|---|
| Distillation | Components redistribute between liquid and vapour according to volatility | Solvent recovery |
| Absorption | Solute moves from gas into liquid | Ammonia absorption into water |
| Stripping | Volatile solute moves from liquid into gas | Removal of dissolved volatile contaminants |
| Liquid–liquid extraction | Solute transfers between two liquid phases | Recovery of a dissolved product using a solvent |
| Leaching | Soluble material leaves a solid and enters a liquid | Sugar extraction from sugar beet |
| Adsorption | Species accumulate on a solid surface | Activated-carbon treatment |
| Drying | Moisture moves through a wet material and into a gas | Drying pharmaceutical granules |
| Humidification | Vapour enters a gas stream | Adding water vapour to air |
| Crystallisation | Dissolved solute joins a growing crystal | Recovery of crystals from solution |
Distillation: Equilibrium and Rate Both Matter
Vapour–liquid equilibrium describes the compositions that can coexist at a given condition. A distillation calculation also needs material balances and operating conditions, such as reflux. An ideal equilibrium stage is a model; actual contacting equipment has finite transfer rates.
Absorption Versus Adsorption
Absorption transfers a component into the bulk of another phase. Adsorption concentrates it on a surface, including internal pore surfaces. In an adsorbent bed, equilibrium capacity and transport resistance are separate considerations: a large capacity does not automatically mean fast uptake.
HTU and NTU in Packed-Column Design
For a suitable packed-column model:
Z = HOGNOG
- Z is the required packing height, in metres.
- HOG is the height of an overall gas-phase transfer unit, in metres.
- NOG is the number of overall gas-phase transfer units, dimensionless.
HTU describes transfer performance under the specified conditions. NTU accounts for the separation requirement and available driving force. For dilute absorption with approximately constant gas molar flow and HOG:
NOG ≈ ∫y,outy,in dy/(y − y*)
The integral runs from the outlet gas mole fraction to the inlet gas mole fraction. At each location, y* is the gas mole fraction in equilibrium with the local bulk liquid.
If HOG = 0.60 m and NOG = 4.0, the calculated packing height is 2.4 m. This is not the total vessel height, which also needs space for distributors, supports and disengagement. Match the phase basis of HTU and NTU.
Sherwood, Reynolds and Schmidt Numbers
Engineers use dimensionless correlations to estimate coefficients for specified geometries and flow regimes.
| Number | Definition | Interpretation |
|---|---|---|
| Reynolds, Re | ρuL/μ | Relative importance of inertia and viscosity |
| Schmidt, Sc | μ/(ρDAB) | Momentum diffusivity divided by molecular mass diffusivity |
| Sherwood, Sh | kcL/DAB | Dimensionless concentration-based mass transfer coefficient |
Here, ρ is density, u is a characteristic velocity, L is a characteristic length and μ is dynamic viscosity.
A correlation may take the form Sh = f(Re, Sc). Its length definition, geometry and validity range are part of the correlation; they cannot be chosen independently.
How Can You Increase the Mass Transfer Rate?
The worked equations suggest three practical approaches:
- Increase the driving force. For example, reduce the receiving fluid’s bulk solute concentration, where the process allows it.
- Increase effective contact area. Packing or dispersed droplets can provide more interfacial area within equipment.
- Reduce the dominant resistance. Improve the relevant phase’s mixing, distribution or contacting conditions.
These choices must also fit the equipment’s hydraulic limits. A packed absorber needs suitable solvent and packing selection, flow rates, column diameter, distributors and transfer height. Increasing flow alone does not establish a workable design. NPTEL outlines the complete packed-absorber design inputs.
Mass Transfer Versus Heat Transfer
Heat transfer concerns energy transport associated with a temperature difference. Mass transfer concerns species transport. Their equations often have analogous forms: Fourier’s law relates heat flux to a temperature gradient, while Fick’s law relates diffusive species flux to a concentration gradient.
Both occur during drying: energy supports evaporation, while moisture moves through the material and into the surrounding gas. For the thermal concepts, read our guide to heat transfer operations. The broader connection is covered in our transport phenomena guide.
Frequently Asked Questions About Mass Transfer
What Is a Simple Example of Mass Transfer?
Dissolved dye spreading through still water is an example of molecular diffusion. Water vapour moving from wet material into flowing air is an example involving convective mass transfer.
Is Diffusion the Same as Mass Transfer?
Diffusion is a mechanism of mass transfer. Species can also be transported by fluid motion, and engineering problems often involve both effects.
What Is the Unit of a Mass Transfer Coefficient?
A coefficient multiplying a molar concentration difference commonly has units of m/s. A coefficient multiplying a partial-pressure or mole-fraction difference has units appropriate to that driving force. Read the defining equation before using a tabulated value.
Does Mass Transfer Require Two Phases?
No. Species can diffuse within one phase. Two-phase contact is important in operations such as absorption, extraction and distillation.
How Should You Start a Mass Transfer Calculation?
Identify the species, phases and direction of transfer. Choose a mass or molar basis, write the species balance, establish the equilibrium relationship where needed, and select a rate equation with matching units and assumptions.
Further Study
Use the source lectures linked throughout this guide to practise diffusion, coefficient and column-design problems. LearnChemE’s separations simulations provide interactive exercises, and our Master Guide to Chemical Engineering places mass transfer within the wider curriculum.
