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

Exam areas

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

(also written

D_{AB}

D_m

) as the proportionality constant. The minus sign encodes 'downhill': matter flows from high to low concentration.

D

for gases is of order

10^{-5}\,\text{m}^2/\text{s}

, for liquids

10^{-9}\,\text{m}^2/\text{s}

— a four-orders-of-magnitude gap that explains why liquid-phase resistance so often controls.

J_A = -D\,\dfrac{dC_A}{dx}

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 analog of the unsteady heat equation. For a semi-infinite solid initially at

C_0

whose surface is suddenly held at

C_s

, the solution is an error-function profile. The handbook supplies a table of

\operatorname{erf}(z)

; the whole trick is forming the similarity variable

z = x/(2\sqrt{Dt})

correctly and reading the table.

\dfrac{\partial C}{\partial t} = D\,\dfrac{\partial^2 C}{\partial x^2}, \qquad \dfrac{C - C_s}{C_0 - C_s} = \operatorname{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)

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),

N_B = -N_A

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,

(p_B)_{lm}

, which is always less than the total pressure, so stagnant-film flux always exceeds the equimolar flux for the same end pressures.

\underbrace{N_A = \dfrac{D_m}{RT\,\Delta z}\,(p_{A1}-p_{A2})}_{\text{equimolar }(N_B=-N_A)} \qquad \underbrace{N_A = \dfrac{D_m\,P}{RT\,\Delta z\,(p_B)_{lm}}\,(p_{A1}-p_{A2})}_{\text{through stagnant B}}

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

k_G'

and liquid-film resistance in

k_L'

, 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

(

p_A^{*} = H\,C_{AL}

) to convert between a liquid concentration and the gas partial pressure that would be in equilibrium with it.

N_A = k_G'(p_{AG}-p_{Ai}) = k_L'(C_{Ai}-C_{AL}) = K_G'(p_{AG}-p_A^{*})

Adding resistances in series

The overall coefficient combines the film coefficients exactly the way series electrical conductances combine — as reciprocals (resistances) in series. Referenced to the gas,

1/K_G' = 1/k_G' + H/k_L'

; referenced to the liquid,

1/K_L' = 1/(H k_G') + 1/k_L'

. The factor

H

is the lever that decides which film controls: a sparingly soluble gas (large

H

, e.g.,

\mathrm{O_2}

\mathrm{CO_2}

in water) makes the

H/k_L'

term dominate, so the liquid film controls; a very soluble gas (small

H

, e.g.,

\mathrm{NH_3}

\mathrm{HCl}

) makes the gas film control.

\dfrac{1}{K_G'} = \dfrac{1}{k_G'} + \dfrac{H}{k_L'}, \qquad \dfrac{1}{K_L'} = \dfrac{1}{H\,k_G'} + \dfrac{1}{k_L'}

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,

\mathrm{Sh} = 0.023\,\mathrm{Re}^{0.8}\mathrm{Sc}^{1/3}

, 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:

\mathrm{Sc} = \mu/(\rho D_m)

is the mass analog of

\mathrm{Pr}

, and

\mathrm{Sh}

is the analog of

\mathrm{Nu}

j_H = \dfrac{\mathrm{Nu}}{\mathrm{Re}\,\mathrm{Pr}^{1/3}} = j_M = \dfrac{\mathrm{Sh}}{\mathrm{Re}\,\mathrm{Sc}^{1/3}} = \dfrac{f}{8}

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

(p_B)_{lm}

. Carry

R

in consistent units —

8.314\,\text{J/(mol·K)}

pairs with pressures in

\text{Pa}

and

D_m

\text{m}^2/\text{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/(2\sqrt{Dt})

before touching the erf table.

Key equations

Fick's first law

J_A = -D\,\dfrac{dC_A}{dx}

Steady 1-D molar flux ( $\text{mol·m}^{-2}\text{s}^{-1}$ ) of A; $D$ = diffusivity ( $\text{m}^2/\text{s}$ ), $C_A$ = concentration ( $\text{mol/m}^3$ ). Minus sign: flux goes down the gradient.

Fick's second law

\dfrac{\partial C}{\partial t} = D\,\dfrac{\partial^2 C}{\partial x^2}

Unsteady 1-D diffusion (constant $D$ ). The mass-transfer twin of the transient heat-conduction equation.

Semi-infinite solution (erf)

\dfrac{C - C_s}{C_0 - C_s} = \operatorname{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)

Surface held at $C_s$ , bulk initially $C_0$ . Form $z = x/(2\sqrt{Dt})$ , then read $\operatorname{erf}(z)$ from the handbook table.

Equimolar counter-diffusion (gases)

N_A = \dfrac{D_m}{RT\,\Delta z}\,(p_{A1}-p_{A2})

Use when $N_B=-N_A$ (no net molar flow), e.g., binary distillation. $\Delta z$ = path length.

Diffusion through stagnant B

N_A = \dfrac{D_m\,P}{RT\,\Delta z\,(p_B)_{lm}}\,(p_{A1}-p_{A2})

Use when B is inert and motionless (gas absorption). The $(p_B)_{lm}$ term adds the Stefan drift; always gives a larger flux than equimolar.

Log-mean partial pressure of B

(p_B)_{lm} = \dfrac{p_{B2}-p_{B1}}{\ln(p_{B2}/p_{B1})}, \quad p_B = P - p_A

Log mean of the inert's partial pressures at the two film faces.

Two-film flux

N_A = k_G'(p_{AG}-p_{Ai}) = k_L'(C_{Ai}-C_{AL})

Gas-film and liquid-film rate expressions; equal at steady state. $k_G'$ , $k_L'$ are individual coefficients, interface in equilibrium.

Overall coefficients

\dfrac{1}{K_G'} = \dfrac{1}{k_G'} + \dfrac{H}{k_L'}, \quad \dfrac{1}{K_L'} = \dfrac{1}{H k_G'} + \dfrac{1}{k_L'}

Series resistances. Large $H$ (insoluble) -> liquid film controls; small $H$ (soluble) -> gas film controls.

Henry's law (interface)

p_A^{*} = H\,C_{AL}, \qquad C_A^{*} = p_{AG}/H

Links a liquid concentration to the equilibrium gas partial pressure; $H$ has units of pressure per concentration.

Sherwood correlation (turbulent tube)

\mathrm{Sh} = \dfrac{k_m D}{D_m} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}

Here $D$ = tube diameter (NOT the diffusivity $D_m$ ); $\mathrm{Sc}=\mu/(\rho D_m)$ , $\mathrm{Re}=\rho V D/\mu$ . Mass-transfer twin of Dittus-Boelter.

Chilton-Colburn analogy

j_H = \dfrac{\mathrm{Nu}}{\mathrm{Re}\,\mathrm{Pr}^{1/3}} = j_M = \dfrac{\mathrm{Sh}}{\mathrm{Re}\,\mathrm{Sc}^{1/3}} = \dfrac{f}{8}

Estimates $k_m$ 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 $f/2$ ). FE Reference Handbook 10.6 prints it without the name, p. 255 (Use of Friction Factor to Predict Heat-Transfer and Mass-Transfer Coefficients): $j_H = (\mathrm{Nu}/\mathrm{Re\,Pr})\,\mathrm{Pr}^{2/3} = f/8$ and $j_M = (\mathrm{Sh}/\mathrm{Re\,Sc})\,\mathrm{Sc}^{2/3} = f/8$ .

Worked examples

Ammonia diffusing through a stagnant air film

Problem. Ammonia (A) diffuses through a $1.0\,\text{mm}$ stagnant film of air (B) at $298\,\text{K}$ and $101.3\,\text{kPa}$ . The partial pressure of $\mathrm{NH_3}$ is $0.10\,\text{atm}$ at the gas side and $0.02\,\text{atm}$ at the interface, where it is absorbed. With $D_{AB}=0.23\times10^{-4}\,\text{m}^2/\text{s}$ , find the molar flux. How does it compare with the equimolar value?

Solution. Work in

\text{Pa}

p_{A1}=0.10(101{,}325)=10{,}133\,\text{Pa}

p_{A2}=2{,}027\,\text{Pa}

. Inert partial pressures:

p_{B1}=101{,}325-10{,}133=91{,}193\,\text{Pa}

p_{B2}=99{,}299\,\text{Pa}

(p_B)_{lm}=\dfrac{99{,}299-91{,}193}{\ln(99{,}299/91{,}193)}=\dfrac{8{,}106}{0.08516}=95{,}188\,\text{Pa}

N_A=\dfrac{D_{AB}P}{RT\,\Delta z\,(p_B)_{lm}}(p_{A1}-p_{A2})=\dfrac{(0.23\times10^{-4})(101{,}325)}{(8.314)(298)(0.001)(95{,}188)}(8{,}106)

. The coefficient

=\dfrac{(0.23\times10^{-4})(101{,}325)}{(8.314)(298)(10^{-3})(95{,}188)}=9.88\times10^{-6}

, then

\times\,8{,}106

gives

N_A=\mathbf{8.01\times10^{-2}\ \text{mol·m}^{-2}\text{s}^{-1}}

. Equimolar (drop the

P/(p_B)_{lm}

factor):

N_A=7.53\times10^{-2}\,\text{mol·m}^{-2}\text{s}^{-1}

. The stagnant-film flux is

P/(p_B)_{lm}=101{,}325/95{,}188=1.064\times

larger. Sanity check: units

\tfrac{(\text{m}^2/\text{s})(\text{Pa})}{(\text{J/mol·K})(\text{K})(\text{m})(\text{Pa})}\cdot\text{Pa}=\text{mol·m}^{-2}\text{s}^{-1}

(using

\text{J}=\text{Pa·m}^3

). The drift correction is small because A is dilute, exactly as expected.

N_A = \dfrac{(0.23\times10^{-4})(101{,}325)}{(8.314)(298)(10^{-3})(95{,}188)}(8{,}106) = 8.01\times10^{-2}\ \tfrac{\text{mol}}{\text{m}^2\text{s}}

Which film controls? Overall gas coefficient

Problem. For a gas absorbing into water the film coefficients are $k_G'=1.5\times10^{-5}\,\text{mol·m}^{-2}\text{s}^{-1}\text{kPa}^{-1}$ and $k_L'=3.0\times10^{-5}\,\text{m/s}$ , with Henry's constant $H=1.5\,\text{kPa·m}^3/\text{mol}$ . Find $K_G'$ and the fraction of resistance in the gas film. If the bulk gas partial pressure is $5\,\text{kPa}$ and the bulk liquid is essentially solute-free, find the flux.

Solution. Gas-film resistance:

1/k_G'=1/(1.5\times10^{-5})=66{,}667

(units

\text{m}^2\text{s·kPa/mol}

). Liquid-film resistance referred to gas:

H/k_L'=1.5/(3.0\times10^{-5})=50{,}000

. Total:

1/K_G'=66{,}667+50{,}000=116{,}667

, so

K_G'=8.57\times10^{-6}\,\text{mol·m}^{-2}\text{s}^{-1}\text{kPa}^{-1}

. Gas-film fraction

=66{,}667/116{,}667=\mathbf{57.1\%}

— neither film clearly dominates, so both matter. With

C_{AL}\approx0

p_A^{*}=HC_{AL}=0

, so

N_A=K_G'(p_{AG}-p_A^{*})=(8.57\times10^{-6})(5)=\mathbf{4.29\times10^{-5}\ \text{mol·m}^{-2}\text{s}^{-1}}

. Sanity check:

K_G'<k_G'

— the overall coefficient is smaller than the gas-film coefficient

k_G'

it is referenced to, just as adding a series resistance lowers the overall conductance.

\dfrac{1}{K_G'} = \dfrac{1}{1.5\times10^{-5}} + \dfrac{1.5}{3.0\times10^{-5}} = 1.167\times10^{5} \;\Rightarrow\; K_G' = 8.57\times10^{-6}

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 $P/(p_B)_{lm}$ drift factor, which raises the flux.
•Building $(p_B)_{lm}$ from the solute's partial pressures. It is the log mean of the INERT B's partial pressures, $p_B=P-p_A$ , at the two faces — not of $p_A$ .
•Inconsistent gas constant. $R=8.314\,\text{J/(mol·K)}$ goes with pressures in Pa and $D$ in $\text{m}^2/\text{s}$ ; mixing in atm or kPa without converting $R$ corrupts the flux.
•Forgetting the $H$ conversion in the overall coefficient. $1/K_G' = 1/k_G' + H/k_L'$ — the liquid term is multiplied by Henry's constant; dropping $H$ assumes solubility you do not have.
•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 $D\sim10^{-5}\,\text{m}^2/\text{s}$ but liquid $D\sim10^{-9}$ ; this 10000:1 gap is why liquid-phase resistance so often dominates.
•In the unsteady problem, forgetting the factor of 2: the similarity variable is $z=x/(2\sqrt{Dt})$ , not $x/\sqrt{Dt}$ .

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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