Mass Transfer and Separation · Study · FE Chemical · FE → PE Prep
Mass Transfer and Separation
8% of exam
Molecular and convective mass transfer with diffusion and mass-transfer coefficients, separation systems (distillation, absorption, extraction, membranes, adsorption), equilibrium-stage methods (McCabe-Thiele, stage efficiency), continuous-contact methods (NTU, HTU, HETP), and humidification, drying, and evaporation.
5 concepts
A. Molecular diffusion
Molecular Diffusion and Convective Mass Transfer
Fick's law, equimolar counter-diffusion versus diffusion through stagnant B, the two-film model and its coefficients, and the momentum-heat-mass analogy.
Every separation in this exam topic — distillation, absorption, extraction, drying — is ultimately rate-limited by how fast one species can move through another. Molecular diffusion is that bottleneck, and convective mass transfer is the engineering shortcut that packages a complicated concentration field into a single coefficient times a driving force. The FE Reference Handbook — Chemical Engineering (Mass Transfer) gives you four or five equations here; the points are lost not because the formulas are hard but because candidates blur two physically different situations (equimolar counter-diffusion versus diffusion through a stagnant gas) and forget that a mass-transfer coefficient is meaningless without the driving force it multiplies. This concept builds the fluency to tell those cases apart and to read a two-film problem at a glance.
Fick's first law: flux follows the gradient
Diffusion is nature's tendency to erase concentration gradients. Fick's first law states that the molar (or mass) flux of a species is proportional to the negative of its concentration gradient, with the diffusion coefficient D
) as the proportionality constant. The minus sign encodes 'downhill': matter flows from high to low concentration.
D
for gases is of order
10−5m2/s
, for liquids
10−9m2/s
— a four-orders-of-magnitude gap that explains why liquid-phase resistance so often controls.
JA=−DdxdCA
Fick's second law: unsteady diffusion
When the concentration field changes in time — a quench, a carburizing step, a transient absorption — a differential mass balance on a slab gives Fick's second law, the diffusion analogue of the unsteady heat equation. For a semi-infinite solid initially at C0 whose surface is suddenly held at Cs, the solution is an error-function profile. The handbook supplies a table of erf(z); the whole trick is forming the similarity variable z=x/(2Dt) correctly and reading the table.
∂t∂C=D∂x2∂2C,C0−CsC−Cs=erf(2Dtx)
Equimolar counter-diffusion versus stagnant B
Two steady gas-phase cases dominate the exam, and they differ by a single physical question: does the other species move? In equimolar counter-diffusion (e.g., the two components of a binary distillation, where one mole of vapor up means one mole of liquid down), NB=−NA and there is no net molar flow, so the flux is simply the gradient form. In diffusion through a stagnant film of B (e.g., ammonia absorbed from air, where the air does not cross the interface), the inert B piles up at the interface and sets up a bulk flow that drags A along — the 'drift' or Stefan flow. That bulk flow is captured by dividing by the log-mean partial pressure of B, (pB)lm, which is always less than the total pressure, so stagnant-film flux always exceeds the equimolar flux for the same end pressures.
The two-film theory and mass-transfer coefficients
Solving the diffusion equation inside every device is hopeless, so we lump the resistance into thin stagnant 'films' on each side of an interface and write the flux as a coefficient times a driving force. In the two-film picture, gas-film resistance lives in kG′ and liquid-film resistance in kL′, and the two phases are assumed to be in equilibrium right at the interface. Because interface compositions are unmeasurable, we fold both films into an overall coefficient referenced to one phase, using Henry's constant H (pA∗=HCAL) to convert between a liquid concentration and the gas partial pressure that would be in equilibrium with it.
The overall coefficient combines the film coefficients exactly the way series electrical conductances combine — as reciprocals (resistances) in series. Referenced to the gas, 1/KG′=1/kG′+H/kL′; referenced to the liquid, 1/KL′=1/(HkG′)+1/kL′. The factor H is the lever that decides which film controls: a sparingly soluble gas (large H, e.g., O2 or CO2 in water) makes the H/kL′ term dominate, so the liquid film controls; a very soluble gas (small H, e.g., NH3 or HCl) makes the gas film control.
KG′1=kG′1+kL′H,KL′1=HkG′1+kL′1
The momentum-heat-mass analogy
Because momentum, heat, and mass all diffuse by the same molecular mechanism, their transfer coefficients are linked. The handbook's Sherwood correlation for turbulent tube flow, Sh=0.023Re0.8Sc1/3, is the mass-transfer twin of the Dittus-Boelter heat correlation. More generally the Chilton-Colburn analogy sets the heat and mass j-factors equal to each other and to f/8 (with f the Darcy friction factor; it is f/2 if the Fanning factor is used), letting you estimate an unknown mass-transfer coefficient from a measured friction factor or heat-transfer coefficient. Watch the dimensionless groups: Sc=μ/(ρDm) is the mass analogue of Pr, and Sh is the analogue of Nu.
jH=RePr1/3Nu=jM=ReSc1/3Sh=8f
Exam strategy
First decide the case: if both species cross (distillation, a pure-A evaporating into pure-B counter-flow) it is equimolar; if one species is inert and trapped (gas absorption into a non-volatile liquid, a film of stagnant air) it is the stagnant-B form and you must build (pB)lm. Carry R in consistent units — 8.314J/(mol⋅K) pairs with pressures in Pa and Dm in m2/s. For two-film problems, write the resistance sum first, identify which term is largest (that film controls), and only then plug numbers. For unsteady problems, compute z=x/(2Dt) before touching the erf table.
Key equations
Fick's first lawJA=−DdxdCA
Steady 1-D molar flux (mol⋅m−2s−1) of A; D = diffusivity (m2/s), CA
Problem. Ammonia (A) diffuses through a 1.0mm stagnant film of air (B) at 298K and 101.3kPa. The partial pressure of NH3 is 0.10atm at the gas side and 0.02atm at the interface, where it is absorbed. With DAB=0.23×10−4m2/s, find the molar flux. How does it compare with the equimolar value?
Solution. Work in Pa: pA1=0.10(101,325)=10,133Pa, pA2=2,027Pa
Problem. For a gas absorbing into water the film coefficients are kG′=1.5×10−5mol⋅m−2s−1kPa−1
Common pitfalls
•Using the equimolar form when one species is stagnant. Gas absorption into a non-volatile liquid is diffusion-through-stagnant-B: you must include the P/(pB)lm drift factor, which raises the flux.
•Building (pB)lm from the solute's partial pressures. It is the log mean of the INERT B's partial pressures, pB=P−pA, at the two faces — not of pA.
•Inconsistent gas constant. R=8.314J/(mol⋅K) goes with pressures in Pa and D in m2/s; mixing in atm or kPa without converting R corrupts the flux.
•Forgetting the H conversion in the overall coefficient. 1/KG′=1/kG′+H/kL′
•Misreading which film controls. Large H (insoluble gas) means the liquid film controls; small H (very soluble gas) means the gas film controls — students routinely flip this.
•Treating D as the same order for gases and liquids. Gas D∼10−5m2/s but liquid D∼10−9
•In the unsteady problem, forgetting the factor of 2: the similarity variable is z=x/(2Dt), not x/Dt
References
NCEES FE Reference Handbook — Chemical Engineering (Mass Transfer)
Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat, and Mass Transfer — Fick's laws, film theory, analogies
Geankoplis, Transport Processes and Separation Process Principles — stagnant-film vs equimolar diffusion
C. Separation systems
Separation Systems: Absorption, Extraction, Membranes and Adsorption
Gas absorption and stripping, liquid-liquid extraction with the distribution coefficient, membrane separations (RO/UF and rejection), and adsorption isotherms.
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D. Equilibrium stage methods
Distillation and the McCabe-Thiele Method
Relative volatility, rectifying and stripping operating lines set by reflux ratio R and feed quality q, stepping off stages, minimum reflux, and stage efficiency.
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E. Continuous contact methods
Continuous Contact: HTU, NTU and HETP
Packed-tower height as HTU times NTU, NTU from the log-mean driving force, the link to HETP, and absorption-factor (Kremser/Colburn) methods.
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F. Humidification, drying, and evaporation
Humidification, Drying and Evaporation
The psychrometric chart, wet-bulb and adiabatic-saturation temperatures, constant- and falling-rate drying periods, and single/multiple-effect evaporation.
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= concentration (
mol/m3
). Minus sign: flux goes down the gradient.
Unsteady 1-D diffusion (constant D). The mass-transfer twin of the transient heat-conduction equation.
Surface held at Cs, bulk initially C0. Form z=x/(2Dt), then read erf(z) from the handbook table.
Use when NB=−NA (no net molar flow), e.g., binary distillation. Δz = path length.
Use when B is inert and motionless (gas absorption). The (pB)lm term adds the Stefan drift; always gives a larger flux than equimolar.
Log mean of the inert's partial pressures at the two film faces.
Gas-film and liquid-film rate expressions; equal at steady state. kG′, kL′ are individual coefficients, interface in equilibrium.
Series resistances. Large H (insoluble) -> liquid film controls; small H (soluble) -> gas film controls.
Links a liquid concentration to the equilibrium gas partial pressure; H has units of pressure per concentration.
Here D = tube diameter (NOT the diffusivity Dm); Sc=μ/(ρDm), Re=ρVD/μ. Mass-transfer twin of Dittus-Boelter.
Estimates km from a friction factor or heat coefficient when a direct correlation is missing. The f/8 form uses the DARCY friction factor (with the Fanning factor it is f/2).
). The drift correction is small because A is dilute, exactly as expected.
and
kL′=3.0×10−5m/s
, with Henry's constant
H=1.5kPa⋅m3/mol
. Find
KG′
and the fraction of resistance in the gas film. If the bulk gas partial pressure is
5kPa
and the bulk liquid is essentially solute-free, find the flux.
Solution. Gas-film resistance: 1/kG′=1/(1.5×10−5)=66,667 (units m2s⋅kPa/mol).
Liquid-film resistance referred to gas: H/kL′=1.5/(3.0×10−5)=50,000.
Total: 1/KG′=66,667+50,000=116,667, so KG′=8.57×10−6mol⋅m−2s−1kPa−1.
Gas-film fraction =66,667/116,667=57.1% — neither film clearly dominates, so both matter.
With CAL≈0, pA∗=HCAL=0, so NA=KG′(pAG−pA∗)=(8.57×10−6)(5)=4.29×10−5mol⋅m−2s−1.
Sanity check: KG′<kG′ — the overall coefficient is smaller than the gas-film coefficient kG′ it is referenced to, just as adding a series resistance lowers the overall conductance.