Mass Transfer Processes S. Majid Hassanizadeh

Department of Earth SciencesFaculty of Geosciences

Utrecht University

Outline:

1. Measures of Concentration2. Volatilization and Dissolution3. Adsorption Processes4. Kinetic Mass Transfer

Measures of Composition

• Mass concentration (of a substance in a fluid):

• Mass fraction (of a substance in a fluid):

• Molar concentration (or molarity):

• Mole fraction:

Measures of Composition

• Mass concentration (of a substance in a fluid):mass of a substance per unit volume of the fluid phase, Ci [ML-3]. For a N-components fluid phase, we have:

• Mass fraction (of a substance in a fluid): mass per unit mass of the fluid phase, ωi [MM-1] or [-]. Units: fractional number, %, ppm, ppb.

Subject to:

1

Ni

i

Cρ=

= ∑

/i iCω ρ=

1

1N

i

i

ω=

=∑

Measures of Composition

• Molar concentration (or molarity): No. of moles of the substance per unit volume of the fluid phase (the solution), ci [molL-3]; g-moles/cm3, g-moles/L, etc.

• Mole fraction: No. of moles of the substance per total number of moles in the fluid phase, [-].

• Molality (or molal concentration): No. of moles of the substance per unit mass of the solvent (not the solution), commonly denoted by mi [mM-1].

/i i iwc C M=

/i ic cχ =

1

i i i ii w w

Ni i w

wi

M MMM

χ χωχ

=

= =

∑

ii

solvw

nmM

=

Measures of Composition for gasses

Ideal Gas Law:

Pg is the gas pressure [MT-2L-1]; (units: Pascal, mbar, etc.)

ng is the number of moles of the gas,

Vg is the gas volume,

T is temperature in degrees Kelvin,

R is the universal gas constant

(in SI system: 8.3144 Joules per Kelvin per mole)

g g gPV n RT=

Measures of Composition for gasses

• Partial pressure: [MT-2L-1]; (units: Pascal, mbar, etc.)is the pressure of a given gas component i as if that component fills the whole gas volume by itself. So, from ideal gas law:

igP

i ig g gP V n RT=

• Dlaton Law: The total gas pressure is equal to the sum of all partial pressures:

ig gn n=∑i

g gP P=∑igi

gnn

χ = =//

ig

g

P V RTP V RT

=ig

g

PP

Mass transfer processes in soil Aqueous

Phase

Solid Phase

Gas

Phase

NAPL

Henry’s Law

Adsorption (K

D )

Vapor Pressure &

Raoult’s Law

Solubility & Raoult’s LawDissolution

Volatilization/Dissolutioon

Volatilization/D

issolutioon

VolatilizationLiquid-Gas Mass Transfer - EquilibriumPure liquid in contact with vacuumIf a pure liquid is in contact with vacuum, it evaporates until the corresponding vapor pressure is reached. Vapor pressure, Pvap, is reported in tables in the literature. It increases with temperature.

Temperature

Vapo

r pre

ssur

e

ExerciseVapor pressures of benzene and PCE are 100 mmHg and 18 mmHg, respectively.

It they are each in a separate closed container in equilibrium with headspace air, what are their mole fractions in the air?

VolatilizationLiquid-Gas Mass Transfer - EquilibriumHow about if a multi-component liquid is in contact with a gas?At equilibrium, both components will be present in the gas phase.

Each component has a partial pressure, which is now smaller than the vapor pressure

How small?That is prescribed by Raoult’s Law.

VolatilizationLiquid-Gas Mass Transfer - EquilibriumRaoult’s Law: i i i

g vap liqP P χ=

i i i ig vap liq g gP P Pχ χ= =

where is the mole fraction of component i in the liquid. iliqχ

ExerciseAn organic liquid is prepared by mixing 100g of benzene and 300g of PCE. They are then put in a closed container. What are the mole fractions of components in the headspace of the container?Molecular weight of Benzene and PCE are 78 g/mole and 165 g/mole, respectively

Henry’s Law

The simple form of Raoult’s Law is not valid for dilute solutions. The complete form is:

/i i i i ig sol v solP a P χ φ=

Equilibrium of dilute multi-component liquid with a gas

If the component is in small concentrations (solute), then we must employ Henry’s Law

Henry’s coefficient Hi depends on temperature and gas pressureVarious forms of Henry’s law are possible, relating different measures of concentration to each other:

i i ig solP Hχ=

i i ig sol pP Hχχ=

i i ig solHχχ χ=

i i ig sol CC C H=

Liquid-liquid mass transfer; Dissolution

• The maximum concentration of a single-component NAPL in water is equal to its solubility, Si [M/L3]

• The maximum concentration of a component from a multi-component NAPL in water is given by a special form of Raoult’s law:

is mole fraction of the component in NAPL

i i imax nC S χ=

inχ

Raoult’s law is applicable only if: 0inχ

Liquid-liquid mass transfer; Dissolution

Maximum aqueous concentration of a component present at small amount in a multi-component NAPL is given by a special form of Henry’s law:

where is the concentration of component i in water and is its concentration in the NAPL.

is Henry’s constant for dissolution.

iwC

inC

i i iw n disC C H=

idisH

Volatilization/Dissolutioon

Distribution of NAPL components in soil Aqueous

Phase

Solid Phase

Gas

Phase

NAPL

Henry’s Law

Adsorption (K

D )

Vapor Pressure &

Raoult’s Law

Solubility & Raoult’s Law

Exercise on Raoult’s Law:

Example: A solvent contains the following components (collectively called BTEX):

The solvent is in equilibrium with air at the pressure of one atmosphere (760 mm Hg). Calculate the mole fraction of BTEX components in the air. If this solvent is in contact with water in a container, what would be the equilibrium concentration of its components in water?

Equilibrium adsorptionOften, it is assumed that adsorption can be modelled as an equilibrium process, i.e. the component in the liquid and the component in the solid are in equilibrium. In fact their chemical potentials are equal. Chemical potential is a function of concentration. This leads to: concentration of adsorbed solute is an algebraic function of the solute mass concentration in the fluid:

which is also called equilibrium adsorption isotherm.

s is mass fraction of adsorbed solute (adsorbed mass per unit mass of solid phase) and C is aqueous phase concentration

If changes in “s” (and “C”) are not large, a linear approximation may be made, in which case we’ll have:

Kd [L3/M] is called distribution coefficient.

( )s f C=

ds K C=

Mass transfer processes; Kinetic• Concentrations change with time.• Initially, a component is present within a phase and starts

to partition into other phase(s), OR there is a change from equilibrium.

• For example, consider kinetic adsorption

Batch experiments on virus adsorption(Data from Reza Sadeghi)

0.00001

0.0001

0.001

0.01

0.1

1

0 10 20 30 40 50 60

Time (h)

C/C 0

L2L10S2S10

Kinetic adsorptionThe change of concentration of a solute in soil is given by the following mass balance equations:

One may assume that adsorption rate depends on C and desorption rate depends on s:

ads desdCn r rdt

= − +

bads des

ds r rdt

ρ = + −

( )adsr f C=

( ) ( )s ads desr r r f C f s= − = −

( )desr f s=

Kinetic adsorption

Commonly, linear approximations are employed such that:

where ka [T-1] and kd [T-1] are adsorption and desorption

rate coefficients, respectively.

This equation may be rewritten as:

bs a dr nk C k sρ= −

( )(1 )

b bas d D

d

nkr k C s K C sn bk

ρ ρ αρ

⎛ ⎞= − = −⎜ ⎟−⎝ ⎠

Mass transfer processes; Linear Kinetic

, ( )i i ikin maxr C Cα αλ= −

A linear kinetic formula, typically has the following form:

where : is a kinetic rate constant [1/T]

is the maximum possible concentration often equal to the equilibrium conc.

is the actual concentraion

imax

i

C

Cα

λ

Mass transfer processes; Linear Kinetic

, (?? )i idis n dis wr Cλ= −

Kinetic dissolution from a multi-component NAPL:

, ( )i i i idis n dis n wr S Cλ χ= −

Kinetic volatilization from water:

, (?? )i ivol w vol gr Cλ= −

, ( )i i ivol w vol C w gr H C Cλ= −

Mass transfer processes; Linear KineticKinetic volatilization from a multi-component NAPL:

, (?? )i ivol n vol gr Cλ= −

i i ig n vapp pχ= = ( ) ,/ / /i i i i i i

mol mol max g moln RT V nW RT VW C RT W= =

, /i i i imax g n vap molC p W RTχ=

, ( / )i i i i ivol n vol n vap mol gr p W RT Cλ χ= −

Consider spread of an adsorbing solute in a pipe

The velocity profile is given by:

2

2 (1 )rv vR

⎛ ⎞= − ⎜ ⎟⎝ ⎠

R

What are sources of kinetic adsorption?

Upscaling is the main cause of kinetic behavior

Transport of an adsorbing solute in a pipe, cylindrical coordinates

2 2

2

12 (1 )

B.C. at the pipe wallr R r R

c r c c cv D rt R z z r r r

c sDr t= =

⎡ ⎤∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞+ − = +⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦∂ ∂

− =∂ ∂

D = diffusion coefficient [L2/T]R = radiusz = longitudinal coordinates = adsorbed solute concentration [mass per unit area]

We assume linear equilibrium adsorption at the wall:

d r Rs k c

==

Example of concentration field, c(z*,r*,t*). Reaction rate is scale-dependent, as is clear from gradient of concentration within the tube. The concentration averaged over the entire radius of the pore, represents the concentration that would be sampled as effluent.

Non-symmetric behavior of breakthrough curve of average concentration resulting from simulation in Single Tube model with an instantaneous concentration source.

Transport of an adsorbing solute in a pipe, equation for average concentration

( )

( )

2

2disp D

D

c c cv D K c st z zs K c st

α

α

∂ ∂ ∂+ = − −

∂ ∂ ∂∂

= −∂

Ddisp= Taylor dispersion coefficient [L2/T]

0.95

9.0(1 4 ) k pe

α =+

KD

Relation between KD and κ

K*D is linearly related to kappa and is independent of peclet number.And the curve obtained by fitting is:

K*D

Presence of stagnant zones and/or low-permeability zones in the porous meidum; Physical Non-equilibrium

( )(1 )b beD k

sC F f K C st x x

φ ρ ρ α∂∂ ∂+ + = − − −

∂ ∂ ∂

Advection-Dispersion- AdsorptionIt is possible to have a combination of equilibrium and kinetic sites

( )(1 )b k bksorption D k

s r f K C sx

ρ ρ α∂= = − − −

∂

e ks s s= + e Ds f K C= −

HOMEWORKA soil sample from a hazardous waste site is analyzed and found to have a total carbon tetrachloride (CTC) mass fraction of 15 ppm (mg/kg). The following soil and NAPL properties are determined: porosity= 0.35, ρb=1600 kg/m3, Sw= 0.52, T = 25˚C, CTC vap. pressure= 15,100 Pa, Mol. Weight of CTC = 153.8 g/mole, HC= 1.24, KD = 0.0022 m3/kg, and ρCTC =1584 kg/m3. Determine whether CTC is present as a separate phase and if so, what is its saturation in the sample.