AEROSOL NANOPARTICLE PLUG FLOW REACTORS ...elements/528/probsets/aerosol.pdf† Fogler, H. S., Elements of Chemical Reaction Engineering, University of Michigan, Ann Arbor MI. 2 528F03AerosolReactor.doc

H. Scott FoglerChapter 4 10/13/03!!!!

1528F03AerosolReactor

AEROSOL NANOPARTICLE PLUG FLOW REACTORS†

Synthesis of Particles in an APFRThere is considerable interest in the synthesis and use of nanosized particles for a

variety of applications including superalloys and thick film conductors for theelectronics industry (Kear, 1986). Furthermore, other areas of interest includemeasurements of magnetic susceptibility, far-infrared transmission and nuclearmagnetic resonance (Granqvist and Buhrman, 1976). For these systems, it is necessary toproduce fine particles of controlled size. Particle sizes can typically be in the range from10 to 500nm.

Owing to their size, shape, and high specific surface area these particles can alsobe used in pigments in cosmetics, membranes, photo catalytic reactors, catalysts andceramic and catalytic reactors. Examples of uses of nanoparticles include SnO2 forcarbon monoxide gas sensors, TiO2 for fiber optics, SiO2 for fumed silica and opticalfibers, C for carbon black fillers in tires Fe for recording materials, Ni for batteries andto a lesser extent Pd, Mg, Bi and others; all these materials have been synthesized inaerosol reactors. In the bioarea, nanoparticles are used to prevent and treat woundinfections in artificial bone implants, and for use in imaging the brain.

We will use the production of aluminum particles as an example of an APFRoperation, however, the analysis to other systems is analogous. A stream of argon gassaturated with Al vapor is cooled in a Aerosol Plug Flow Reactor (APFR) [Panda andPratsinis, 1995], with a diameter of 18 mm and a length of 0.5 m, from 1600°C at a rateof 1000°C/sec. As the gas stream flows through the reactor, the nucleation and growthof Al particles take place. Flow rate of the carrier gas is 2 dm3(STP)/min and thepressure inside the PFR is 1 atm (1.013 Pa). Moving with the gas velocity U, the coolingrate inside the reactor is 1000 K/s and hence the temperature profile down the reactor isgiven by

dTdx

= -1000

UMonomers Nuclei Particles

V

T

x

CarrierGas

Crucible

Al

Figure 1. Aerosol Reactor and Temperature Profile.

† Fogler, H. S., Elements of Chemical Reaction Engineering, University of Michigan, Ann Arbor MI.

2528F03AerosolReactor.doc

As we move down the reactor the gas is cooled and becomes supersaturated. Thussupersaturation leads to the nucleation of particles. This nucleation is a result ofmolecules colliding, escaping (evaporating) and agglomerating until a critical nucleussize is reached and a particle is formed. As these particles move down thesupersaturated gas molecules condense on the particles causing them to grow in size.

200 nm [scale for all frames]3 mmmm

200 nm

Figure 2. Nucleation and evolution of particle growth in a APFR. Photos courtesy of Prof. Sotiris

E. Pratsinis, Swiss Federal Institute of Technology (ETH Zuerich), Institute of ProcessEngineering (IVUK), CH-8092 Zurich, Switzerland.


The growth of titania particles generated by oxidation of titanium isopropoxide in apremixed flame aerosol reactor collected at different positions (3-200 mm) down thereactor.

Frame (3mm) in Figure 2 shows the smaller particles near the reactor entranceand subsequent frames show the increase in particle size as they move down thereactor. This growth is also shown schematically in Figure 1. Particles obtained in theexit stream in a reactor at the University of Michigan are shown in Figure 3.

!!!!!!!!!!Mullite Mullite

Strontium AluminosilicateFigure 3. Nanoparticles formed in an aerosol reactor. Photos courtesy of Prof. Richard M. Laine,

Department of Materials Science and Engineering, University of Michigan, Ann Arbor,Michigan 48109.

In the development that follows we will model the formation and growth ofaluminum nanoparticles in an AFPR. However, the algorithm applies to other reactingsystems such as flame spray pyrolosis, (FPS) where the monomers are formed, thennucleate to form particles which simultaneously grow and flocculate.


1. MOLE BALANCE ON MONOMERS (ALUMINUM MOLECULES)A mole balance on the moles of monomer, m, (i.e., aluminum in this case) in the

reactor is

†

dFmdV1

= rm (1)

Multiply both sides by Avogadro’s number

†

d NAvoFm( )dV1

= rmNAvo (2)

†

dfmdV1

= ˜ r m (3)

where fm is the molecular flow rate of monomer, (molecules/s), and

†

˜ r m is the rate offormation of monomer (aluminum) (molecules/s/dm3).

Base everything on per kg of gas

†

fm = ˙ m n m (4)

where ̇ m is the mass flow rate of gas, kg/s, and nm is the molecular concentration ofaluminum molecules per kg of gas.

d ˙ m nm( )dV1

=˙ m dnmdV1

= rgUAcdnmdV1

(5)

If one were to generalize, one could think of the aluminum molecules as monomermolecules, in which nm is the concentration of monomer in terms of monomermolecules per kg of gas. Writing the reactor volume, V1, in terms of cross sectional areaAc and distance x down the reactor we have

V1 = Acx (6)Combining Eqns (5) and (6) we obtain

†

r gUdn mdx

= ˜ r m (7)

If we move with the gas, U = dx

dt, then

†

r gdxdt

dn mdx

= ˜ r m

†

r gdn m

dt= ˜ r m

Dividing by rg, (kg of gas/m3)

†

dn mdt

=˜ r mr g

= ˜ ¢ r m


†

˜ ¢ r m =˜ r mr g

=

moleculess•dm 3

Ê

Ë Á

ˆ

¯ ˜

kg dm 3=

moleculeskg•s

dnm

dt= ˜ ¢ r A (8)

2. RATE LAWSWe now focus on the rate of disappearance of monomer molecules. We lose the

monomer molecules by two mechanisms: Nucleation to form particles and bycondensation of monomer molecules on the newly formed particles.

†

- ¢ ˜ r A =Loss of monomermolecules during

Nucleation

È

Î Í Í

˘

˚ ˙ ˙ +

Loss of monomer moleculesduring condensation(i.e. particle growth)

È

Î Í Í

˘

˚ ˙ ˙

-rA = Term 1( ) + Term 1( )

2A. Nucleation KineticsLoss of monomer by nucleation.Term 1. Rate of formation of nuclei.

Here, the monomer molecules aglomerate to form a nuclei containing n*monomer molecules.

n*

The following equation is proposed for the nucleation kinetics (Girshick andChiu, 1990). The rate of formation of stable nuclei per kg of gas is

†

¢ r N =v1rg

2spm1

È

Î Í Í

˘

˚ ˙ ˙

12

nms2 S exp q - 4q

3

27 lnS( )2È

Î

Í Í Í

˘

˚

˙ ˙ ˙

! # of nuclei( )/kg(gas)/s Derive (9)

wherev1 and m1 are the molecular volume and mass of aluminum, rg is the gasdensity, nms is the saturation concentration of aluminum molecules,

†

n ms =PVRT

,

PV the vapor pressure, S is the supersaturation ratio, q is the

dimensionless surface tension,

†

q =ss1k BT

Ê

Ë Á

ˆ

¯ ˜ and s1 is the surface area of the


aluminum molecules. The saturation ratio is just the partial pressure ofmonomer, Pm, divided by the vapor pressure, PV

†

S = Pm PV( ) = n m n msr g( )( )[ ]The units and orders of magnitude of each of these symbols is given at the end inTable!A. For the conditions given for Al, the rate of formation of stable nuclei falls in therange.

†

0 < ¢ r N < 1.45 ¥1015 nuclei kg s[ ] (10)

We now need to find the critical nucleus size, n*, above which the particles are stableand can grow and below which the g-mer, i.e. particle, is unstable. A stable particlecannot revert back to the individual molecules where an unstable particle can. The rateof disappearance of monomer (Al atom) by nucleation is just,

†

- ¢ r N n *( ) where n* is thecritical nucleus size, i.e., number of monomer molecules per nuclei.

Finding the critical nucleus size n*

A schematic of a g-mer of diameter dg made up of g monomers (aluminum molecules)each with diameter d1 is shown below.

dg

d1g–mer monomer (Al atom)

The volume of the g mer is

†

Vg = g v1

p d g( )3

6= g pd1

3

6

dg = g( )1 3 d1

The surface area of the g-mer is

sg = p dg2 = p g1 3d1( )

2= g2 3pd1

2

sg = g2 3s1

Figure 4. A g–mer cluster made up g monomers

We now determine the change is free energy, DG, to form a cluster of size g from gindividual monomer molecules. It consists of two terms, one to bring the dispersedmonomer molecules close together, DG1, and the other to coalesce the liquid monomerinto a drop, DG2. 1

†

DG = DG1 + DG2

1 [See p57 Basic Principles of Colloid Science by D. H. Everett Royal Society of Chemistry ISBN 0-

85186-443-0 1988 for the details.]


†

DG1 =

Change in Free Energy toTransfer g moles (molecules)

from the vapor phase atactivity P1 concentration nm( )

to the liquid phaseactivity Pv concentration nms( )

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

= -gk BTlnP1Pv

= -gk BTlnr gn mn ms

Ê

Ë Á

ˆ

¯ ˜ = -gkTlnS!!!!! Derive

DeriveCalculate DG, to bring g molecules from vapor phase at P1 to liquid phase at PV.

By definitionDG = VdP – SdT

At constant temperature TdG = Vdp

We want to calculate the change in free energy as g molecules in the gas phase asthe move to form a nuclei containing g molecules. The volume of gas containing gmolecules at pressure P and temperature T is

†

V = g k BTP

†

dG = g k BTP

dP

The partial pressure of the molecules far from the nuclei is P1 (state 1) from which themolecules are brought to the nuclei at which the partial pressure is the vapor pressure,Pv (state 2). Integrating between state 1 and state 2

†

DG = G2 - G1 = gk BTdPPP1

PvÚ = gRT dPPP1PvÚ d ln P = gk BTln

PvP1

Rearranging

†

= -gk BTlnP1Pv

†

DG = -gk BTlnS

where

†

S = PmPv

Ê

Ë Á

ˆ

¯ ˜ =

n mn ms r g

Ê

Ë Á Á

ˆ

¯ ˜ ˜ =

n mr gn ms

Ê

Ë Á

ˆ

¯ ˜ =

n1n1

e where n1 is the concentration of monomer

molecules (molecules/dm3) and n1e is the concentration of monomer molecules at

saturation equilibrium (molecules/dm3).For DG2

†

DG2 =Change in Free Energy toform a drop surface area

s g and with surface tension s

È

Î

Í Í

˘

˚

˙ ˙

= ss g = spd g2 = sg2 3s1

Substituting for DG1 and DG2

†

DG = sg2 3s1 - gkTlnS


Dividing by kBT and replacing

†

ss1 k BT( ) by the dimensionless surface tension q

†

DGk BT

= g2 3q - g lnS, Derive (11)

where

†

q =ss1k BT

We now use Eqn. (11) to plot

†

DGk BT

Ê

Ë Á

ˆ

¯ ˜ as a function of particle (i.e. cluster) size, g, as

shown below for a fixed value of the supersaturation, S. One observes that a maximumexists in the Gibbs free energy.

†

DGkBT

Ê

Ë Á

ˆ

¯ ˜

g* @ n*

gAt the maximum, g*!≡!n*,

†

d DGk BT

Ê

Ë Á

ˆ

¯ ˜

dg

ˆ

¯

˜ ˜ ˜ ˜

g*

= 0 = 23

g*-1 3 q - lnS,

Solving for g*

†

g* ≡ n* = 23

q

lnSÊ

Ë Á

ˆ

¯ ˜

3(12)

Particles of size greater than the critical cluster (i.e. particle) size g* (n*) are “stable.”Generally, particles larger than n* will grow while particles small than n* are not stableand can break up, i.e., dissolve.

†

DGkBT

Ê

Ë Á

ˆ

¯ ˜

n1*

n2*

High Supersaturation

Low Supersaturation

n2* > n1

*

S1 > S2

g

Figure 5. Dimensionless free energy as a function of cluster size g showing the variationof critical nucleus size n* with degree of supersaturation, (see Eqn. 12).

We note that the higher the supersaturation, S, the smaller the critical nucleussize n*


Summary for nucleation.

†

The rate of loss of monomermolecules by nucleation

È

Î Í

˘

˚ ˙ = - ¢ r N n *

†

¢ r N =v1r g

2spm1

È

Î Í

˘

˚ ˙

1 2

n ms2 S exp q - 4q

3

27 lnS( )2È

Î

Í Í

˘

˚

˙ ˙ ! # of nuclei( )/kg(gas)/s (9)

†

Term 1= ¢ r N =v1r g

2spm1

È

Î Í

˘

˚ ˙

1 2

n ms2 S exp q - 4q

3

27 lnS( )2È

Î

Í Í

˘

˚

˙ ˙ n* (13)

2B. Growth Kinetics (Condensation Kinetics)Term 2

The newly formed nuclei can grow in size as a result of monomers hittingthe particles and sticking, causing the particles to increase in size.

From the kinetic theory of gases we know the net rate of addition of the gasmolecules to the particles is

†

¢ ¢ r C =[n mr g Um4Rate of

moleculesstriking the

surface

1 2 4 3 4 - n ms

Um4

Rate ofmolecules

evaporatingfrom thesurface

1 2 4 3 4 ]f(Kn)

Factoring out nms

†

¢ ¢ r C =Um

4n ms S-1( )f(Kn) (14)

†

molecules[ ]m 2 •s

=ms

Ê

Ë Á

ˆ

¯ ˜

moleculesm 3

Ê

Ë Á

ˆ

¯ ˜

The 4 in the denominator of Um in Eqn. (14) is a result of averaging themolecular velocities hitting the surface from all directions. [See (Moore, W.Physical Chemistry, 2nd Ed. p167-168) to obtain the rate of molecules striking thesurface per unit area.]

Where

¢ ¢ r C !=!Flux of monomer molecules to particle surface (Net number ofmolecules striking and sticking on the particle)/(m2 of particle)/s

†

Um =8k BTpm1

= gas phase velocity of monomer molecules


Because the gas molecules are in transition range between having small meanfree paths (continuum) and large mean free paths [Knudsen Flow], we need acorrection factor.

f(Kn)!=!a correction for the transition from the free molecule to thecontinuum regime.

f(Kn) =

1.333Kn + 1.333Kn2( )1+ 1.71Kn +1.333Kn2( )

provided [0.59 < f(Kn) < 1.0] (15)

[Warren and Seinfeld (1984)]Kn is the Knudsen number Kn = 2l dp ,l is the mean free path of the gas and dp is the particle diameter

The mean free path is given by

†

l =1

g 2k BT

P, m (Phy. Chem, Atkins page 653, eq. 26.2.6)

For the condition in this example, [1.8¥10-7 < l < 3.9¥10-7 m]

where g = pdcg2 collision cross-section of the gas, m2

dcg is the molecular diameter of the carrier gas (m)

dcg = 3.84¥10-10 m for argenP is the total pressure, Pa

kB is Boltzmann’s constant = 1.38x10–23 kg/s2/molecule/K

The total volume of gas molecules that have formed a liquid, V, is the volume ofa molecule, v1, times the number that have been depleted from the gas phase sinceentering the reactor

V=v1 (nmo – no)We could also obtain the rate at which the volume is growing by multiplying the netrate of loss of monomer molecules by the volume of a monomer molecule

†

dVdt

= v1˜ ¢ r m

The volume of particle is the total volume of liquid V divided by the number ofproducts

†

vP =VN

The diameter of a particle is then

†

pd P3

6= vP =

VN

Solving for the particle diameter


†

d P =6VpN

Ê

Ë Á

ˆ

¯ ˜

1 3

The rate of condensation per unit surface area is written as

†

¢ ¢ r C = n ms S-1( )k BT

2pm1f(Kn) , [(# of molecules)/m2 of particle/s] (14)

where [0 < ¢ ¢ r c < 1.27¥1024 molecules/m2 of particle/s for Al and the conditions given]

Net loss by condensationEquation (14) gives the rate of transport of molecules per unit surface area. The

surface area of a particle is

†

pd P2 . So the product (

†

¢ ¢ r C •pd P2 ) is the loss or monomermolecules per particle (molecules)/particle/s. We now need to multiply by the numberof particles per kg of gas.

†

Term 2= moleculesm 2 •s

Ê

Ë Á

ˆ

¯ ˜ •

surface areaparticle

Ê

Ë Á

ˆ

¯ ˜ •

number of particleskg gas

Ê

Ë Á

ˆ

¯ ˜

= ¢ ¢ r C • pd P2 • N (16)

Where N is the number of particles per kg gas and dp is the particle diameter, the netrate of loss of monomer molecules is the sum of terms

Summary

Term 1

†

Loss of monomerfrom nucleation

È Î Í

˘ ˚ ˙

= ¢ r N n* = -

v1r g

2spm1

È

Î Í

˘

˚ ˙

1 2

n ms2 S exp q - 4q

3

27 lnS( )2È

Î

Í Í

˘

˚

˙ ˙ n* (9)

Term 2

†

Loss of monomermolecules by condensation

È

Î Í

˘

˚ ˙ = - ¢ ¢ r C ⋅ N ⋅ pdp

2

The total rate of formation of monomer molecules is

†

¢ ˜ r m = - ¢ r N n * - ¢ ¢ r C ⋅ N ⋅ pd p2 (17)

combining Equations (8) and (17)

†

dn mdt

= - ¢ r N n * - ¢ ¢ r CNpd p2 (18)π

3. FLOCCULATION (COLLISION) KINETICSThe particle can also increase is size as a result of collision and subsequent

coagulation with other particles. The rate of agglomeration of particles is

+


A balance on the number of particles N. Where N is the number of particles per kg ofgas, is

†

Rate ofAccumulation

Ê Ë Á ˆ

¯ ˜ = Rate of Formation (Gain)by Nucleation

Ê Ë Á ˆ

¯ ˜ - Rate of Lossby Flocculation

Ê Ë Á ˆ

¯ ˜

dNdt

= ¢ r N -12

rF (19)

The 1/2 is included so we don’t count collisions twice.The rate of flocculation (collision) of the particles is proportional to the square of

the concentration of particles is

†

¢ r F =bN2r g

Ws # of collisions/kg(gas)/s

where [0 < r’F < 5.9¥1010] [Seinfeld, 1986] (19)

Where b is the collision coefficient and Ws a correction factor that can eitherincrease or decrease the flocculation rate depending on the gas and particle properties.

The Flocculation ParametersAgain the parameter values for the APFR properties and operating conditions are givenjust above Figure 1.b is the collision coefficient

†

b = 8pDd pd p

d p + g1 2+

De 4 2cd p

È

Î Í Í

˘

˚ ˙ ˙

-1

m3/s

where [5¥10-16 < b < 7.2¥10-15 m3/s]dp is the particle diameter, m

D is the particle diffusion coefficient

†

D = k BT3pm d p

5 + 4Kn + 6Kn 2 +18Kn 3

5 - Kn + (8 + p)Kn 2È

Î Í

˘

˚ ˙ (m2/s)

where [1.5¥10-10 < D < 8.5¥10-4 m2/s]c is the particle velocity

c = 8kBT

prpvp (m/s), where [0.0146 < c < 1408 m/s]

rp is the density of the solid (molten liquid) particle = 2700 kg/m3

vp is the particle volume vp =

pdp3

6, m3 where [1.2¥10-29 < vp < 5.2¥10-20]

g1 is the transition parameter

†

g1 =1

3d plad p + la( )

3- d p

2 + la2( )

3 2È Î Í

˘ ˚ ˙

- d p , (m)

where [1.35¥10-8 < g1 < 1.53¥10-6 m]la is the mean free path for the particle

la =8Dpc

where m [2.6¥10-8 < la


m is the viscosity of the career gas

†

m =26.7 ¥10-7 Mw T

pd cg2 , kg/m/s where dcg is in

Angstroms, Å in this equation for viscosity see BSLwhere [3.4¥10-5 < m < 4.9¥10-5 kg/m/s]

Ws is the stability factor that can either retard or accelerate flocculation.1 < Ws < 10

6

We note Ws can be smaller 1 for bipolarity charged aerosols.

4. BALANCE EQUATIONS – SUMMARY

4A. Disappearance of Al molecules

dnmdx

= ¢ ˜ r A = Term 1(nucleation) + Term 2(condensation)

Nucleation

Term 1 = - ¢ r Nn* =

no. of nucleikg of gas ⋅ s

•no. of molecules

nuclei=

moleculeskg of gas ⋅s

Condensation

†

Term 2 = - ¢ ¢ r C ⋅ N ⋅ pd p2 =

moleculesm 2of particle ⋅ s

=no. of particles

kg of gas¥

m 2of surfaceparticle

=moleculeskg of gas ⋅ s

Substituting for terms 1 and terms 2 the balance on the Al monomers becomes

dnmdt

= - ¢ r Nn * - ¢ ¢ r CNpdp2 (16)

with

r' N =

v1rg

2spm1

È

Î Í Í

˘

˚ ˙ ˙

12

nms2 Sexp q - 4q

3

27 ln S( )2È

Î Í Í

˘

˚ ˙ ˙ !

†

number of nuclei( )/kg(gas)/s[ ] (9)

¢ ¢ r C = nms S -1( )

kBT2pm1

f(Kn) ,

†

number of molecules( )/m2 (particle)/s[ ][0 < ¢ ¢ r c < 1.27¥10

24] (14)where

n* is the number of molecules in a nucleus

n* = 2q

3 lnS( )È

Î Í Í

˘

˚ ˙ ˙

3

!![0 < n* < 2.4¥106] (17)


U is the carrier gas velocity (no pressure drop, P = P0) for an entering volumetricflow rate

†

U = 4QAC

TT0

Ê

Ë Á

ˆ

¯ ˜ (m/s) (Q ≡ vo in the text) (18)

†

AC = pd R2

4B. Balance on number of particles, N (# of particle/kg(gas))

†

dNdt

= ¢ r N -12

¢ r F (19)

†

¢ r F =bN2r g

Ws

OTHER PARAMETERS TO BE CALCULATED1. Volume density of Particles, V, (volume of particles/kg(gas))

V = nmo - nm( )v1 m3/kg (gas)

[0 < V < 3.4¥10-7 m3/kg] (20)

where nmo is the initial number density of Al molecules

†

nmo = 2.6 ¥ 1022 molecules

kgÈ

Î Í Í

˘

˚ ˙ ˙

2. Particle Diameter, dp,

†

dp =6VNp

È

Î Í

˘

˚ ˙ 1 3

m

[2.86¥10-10 < dp < 4.65¥10-7 m] (21)We will solve Equation (9), (14) through (21) [the double boxed equations]simultaneously using an ODE solver. The polymath problem can be loaded directlyfrom the CDROM.

Equations for PolymathBalance Equation

(1)

†

dTdt

= -1000 K s( )

(2)

†

dn mdt

= - ¢ r N n* - ¢ ¢ r C pd P

2 N

(3)

†

dNdt

= -rN -12

rF

(4)

†

dVdt

= rN n* + ¢ ¢ r C pd P

2 N


(5)

†

d P =6VpN

Ê

Ë Á

ˆ

¯ ˜

1 3

(6)

†

dNdt

= rN - rF 2Nucleation

(7)

†

¢ r N =v1r g

2spm1

Ê

Ë Á

ˆ

¯ ˜

1 2

n ms2 S exp q - 4q

3

27 lnS( )2È

Î

Í Í

˘

˚

˙ ˙

Parameters in Equation (7)nms = PS/RT

†

q =ss1kBT

†

S = n m r g n msFor Aluminum

†

s = 948 - 0.2 TK

Ê

Ë Á

ˆ

¯ ˜

gs2

=948 - 0.2 T K( )

1000, kg

s2Ê

Ë Á

ˆ

¯ ˜ (sigma)

†

s1 = 2.575 ¥10-19, m 2 molecule( )

†

v1 =1.23¥10-29, m 3 molecule( )

†

m1 = 4.48 ¥10-26, kg molecule( )

†

r g =MW( )P

RT 1kg

molÊ

Ë Á

ˆ

¯ ˜

molm 3

Ê

Ë Á

ˆ

¯ ˜ (rhog)

†

MW = 39 1000( ) kg mol

†

P =1.013¥105 Pa

†

p = 3.14

†

PS = P exp 13.07 - 36373TK

È

Î Í ˘

˚ ˙

Condensation

†

- ¢ ¢ r C =k BT

2pm1= n ms S-1[ ]f K n( ) (9)

Parameters for Equation (9)Carrier gas

†

f K n( ) =1.33 K n + K n2( ) 1+ 1.7K n +1.33K A2( )( )

†

f K n( )

†

K n = 2 l d P


†

l =1g

k BTP

(Lamda)

†

k B =1.38 ¥10-23 kg s2 molecule K

†

g = p d cg2 (gamma) Collision cross section of carrier gas

†

d cg = 3.84 ¥10-10 m

Flocculation

†

rF = bN2rg

Parameters

b = 8pDedp

dpdp + g1 2

+De 4 2

cdp

È

Î

Í Í

˘

˚

˙ ˙

-1

, (m3/s) (Beta)

De =

k BT3pm dp

5 + 4Kn + 6Kn2 + 18Kn3

5 - Kn + (8 + p)Kn2È

Î Í Í

˘

˚ ˙ ˙

, (m2/s)

c = 8kBT

prpvp, (m/s),

†

m p = rsvp

†

vp =pd p

3

6

†

g1 =1

3d plad p + la( )

3- d p

2 + la2( )

3 2È Î Í

˘ ˚ ˙

- d p , (m)

†

la = 8Dpc

, (m)

(MW)

†

(kg m s)

†

m =2.67 MWT

d cg ¥1010( )

2 , kg

m •sÊ

Ë Á

ˆ

¯ ˜ mu( )

Reactor

†

d =18mm ¥ 1m1000mm

=18

1000m

†

OrT =2dm 3

min¥

1 m 3

1000 dm 3¥

1min60s

=2

1000( ) 60( ) (Note : QrT = v0 in text)

†

Q = QrTT

TrT


Preliminary Resultsa. Effect of flow rate: No effect observed on the particle diameter.b. Effect of Cooling Rate: Particle size decreases with an increase in the cooling rate.c. Effect of Pressure: Particle size increases with an increase in pressure.

Figure 6. Temperature Profile

Figure 7. Super Saturation Profile

Figure 8. Monomer Concentration Profile


Figure 9. Particle Concentration Profile

Figure 10. Particle Size Profile


Table A. Definitions and orders of magnitude

Nucleation Kinetics

nm = aluminum concentration per unit mass of gas (molecules/kg of gas)[6.9¥1012 < nm < 2.634¥1022]

nms = aluminum saturation concentration (molecules/m3)

[3.2¥1012 < nms < 6.8¥1021]

S = super saturation Ratio =

nmnms rg

=nmrgnms

!!!![0.999 < s < 2.74]

N = number of Al particles per kg of gas [0 < N < 6.1¥1012]

̃ ¢ r A = net rate of formation (loss) of Al molecules resulting from particle nucleationand from particle growth (condensation) (molecules/kg gas•s)

¢ r N = rate of nucleation of particles (no. of nuclei formed/kg of gas/s)[0 < ¢ r N < 1.45¥10

15]dp = diameter of particle (m)

V1 = volume reactor (m3)

Ac = cross sectional area of reactor (m2)

rg =

PM wRT

= carrier gas density (kg/m3)!!!![0.26 < rg < 0.56]

q is the dimensionless surface tension. q =

s s1kBT

[5.7 < q < 16.5]

s is the surface tension (Rhee, 1970), s = 948 - 0.202T kg/s2[0.57 < s < 0.77]v1, s1, m1 are the volume, surface area and mass of an Al molecule respectively.

v1 = 1.23 ¥10-29m3 , s1 = 2.575 ¥ 10

-19 m2 and m1 = 4.48 ¥ 10-26 kg

ps is the vapor pressure and P is the total pressure (Pa) and temperature T.

where ps = exp 13.07 -

36373T

Ê

Ë Á

ˆ

¯ ˜ P (22)

nms =

psk BT

(molecules/m3(gas)) [3.2¥1012 < nms < 6.8¥1021]

¢ ¢ r C != rate of condensation on particles (i.e. rate of molecules hitting particle surfaceand sticking) = (no. of molecules/m2 of particles/s)

[0 < ¢ ¢ r c < 1.27¥1024]

¢ r F = rate of flocculation of particles = (no. of particle collisions/kg of gas/s)[0 < ¢ r F < 5.9¥10

10]n* = no. of molecules per nuclei (Note in development of nucleation rate g* was used

but g*=n*)


5. FLAME PYROLYSIS REACTORSOne can also extend this analysis to flame pyrolysis by including a reaction step.

Here we mix two streams A and B in a flame that reacts to form species C which is themonomer.

A + B æ Æ æ CSpecies then flows the same pathway as the Al molecules of nucleation, growth andflocculation.

Figure 11. Flame pyrolysis reaction

Examples of materials formed by flame pyrolysis include TiO2 and Mullite (Figure 3).

REFERENCESKear, B. H., Scientific American, p. 159 (Oct. 1986).Granqvist, C. G. and R. A. Buhrman, J. Appl. Phys., 47, p. 2200 (1976).Panda, S. and S. E. Pratsinis, Nanostructured Materials, 5, p. 755 (Dec. 1995).Girshick, S. L. and C. P. Chiu, J. Chem. Phys., 93, p. 1273 (1990).S. K. Rhee, J. Amer. Ceram. Soc., 53, p. 386 (1970).Atkins, P. W., Physical Chemistry, 3rd Edition, p. 653, eq. 26.2.6.D. R. Warren and J. H. Seinfeld, Aerosol Sci. Tech., 3, p. 135 (1984).J. H. Seinfeld, Atmospheric Chemistry and Physics of Air Pollution, 14 (1986).

Exercises1. Review the description of the aerosol reactor on page 1. Assume the inlet gaseous

stream is always saturated with Al vapor. The entering concentration of Almolecules is 2.64 ¥ 1022 molecules/kg of gas.

a) For a cooling rate of 500 K/s and an initial temperature of 1600°C, plot particlediameter, (dp) nucleation rate, ( ¢ r N ) , and ( ¢ r Nn*) , condensation rate, ( ¢ ¢ r C ) and


( ¢ ¢ r C N pdP2 ) , molecule concentrations (nm) and (nms), saturation ratio, S,

molecules per nuclei, (n*), gas velocity, (U), gas density, (rg), number ofparticles, (N), and flocculation rate ( ¢ r F) as a function of time (moving with thefluid) and of distance down the reactor.

b) Determine the number of particles, Al concentration and particle diameter in theexit stream as a function of cooling rate between 50°C/s and 3000°C/s for aninitial temperature of 1600°C.

c) Determine the number of particles, Al concentration and particle diameter in theexit stream as a function of pressure between 0.1 atm and 20 atm for an initialtemperature of 1600°C and cooling rate 1000°C/s.

d) Determine the number of particles, Al concentration and particle diameter in theexit stream as a function of initial temperature between 1500°C and 2500°C for 1atm pressure cooling rate 1000 K/s.

e) For a cooling rate of 1000 K/s find the relationship between inlet temperatureand pressure.

f) Explore the problem to see the effect of the various operational variables. on dp, ( ¢ ¢ r C N pdP

2 ) , ( ¢ r Nn*) , ¢ r F , etc. Write a paragraph describing operation of an APFRand how to vary the particle sizes.

e.g.!!!!To

Pofor which dp = 500 nm.The Polymath program is shown in the appendix.

2) It is proposed to replace the carrier gas by helium.

a) What changes do you need to make in the polymath code given in the CD and inthe appendix? Before carrying out any computer runs, discuss qualitatively whatdifferences using Helium rather than Argon.

b) Compare your plots (He versus Ar) of the number of Al particles as a function oftime. Explain the shape of the plots. Print out your Polymath code and plot.

c) How does the final value of dp compare with that when the carrier gas wasArgon? Explain.

d) Compare the time at which the rate of nucleation reaches a peak in the two cases[carrier gas = Ar and He]. Explain the comparison.

For parts (b) and (c), write down or sketch the results as appropriate.


Data for a He molecule:Mass = 6.64!¥!10–27 kgVolume = 1.33!¥!10–29 m3

Surface area = 2.72!¥!10–19 m2

Bulk density = 0.164 kg/ m3, at normal temperature (25°C) and pressure (1 atm)

3) What additional equations will be needed in the flame pyrolysis model? Write downas many as you can. Explain qualitatively any differences you expect.


Appendix


AEROSOL REACTORSClick Back from Page 5

ProcedureStep 1. Find the rate of formation of g-mers by summing at g-mer formation from g!=!2

to a very large number G to obtain

†

J = 1dgb Sg ng

e Sg0•

ÚStep 2. Define H(g) = ln

†

b Sg nge Sg( ) and expand H(g) in a Taylor series about g*

Step 3. Change variables to let

†

y = 12

g- g *( )d2H g( )

dy2ˆ

¯

˜ ˜

g*and approximate the lower limit of integration, –a, by –• and use the definitionof the probability integral to obtain

†

J = eH g*( )d2H g( )

dg2Ê

Ë

Á Á

ˆ

¯

˜ ˜

g*

12p

Step 4. Find the critical nucleus size

†

g* = 23

q

lnSÊ

Ë Á

ˆ

¯ ˜

3

Step 5. Evaluate terms in H(g*) and

†

d2Hdg2

ˆ

¯ ˜ ˜

g* then neglect

†

g-2( ) w.r.t.

†

qg-4 3( ) to obtain

†

J = b S1 N13

q

p

Ê

Ë Á

ˆ

¯ ˜ 1 2

exp A[ ]Step 6. Evaluate A at g* to obtain

†

J = N1 b S13

q

p

Ê

Ë Á

ˆ

¯ ˜ 1 2

exp - 4q3

27 lnS( )2È

Î

Í Í Í

˘

˚

˙ ˙ ˙

Step 7. Simplify the pre-exponential to obtain

†

J = n12spm

Ê

Ë Á

ˆ

¯ ˜

1 2S n1

e N1 exp -4q3

27 ln S( )2È

Î

Í Í Í

˘

˚

˙ ˙ ˙

Step 8. Evaluate N1 and recall

†

¢ r N =J

rgto obtain

†

¢ r N =n1rg

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2nms

2 S exp q - 4q3

27 lnS( )2È

Î

Í Í Í

˘

˚

˙ ˙ ˙


End result.Nucleation Kinetics

A1 + Ag-1

æ Æ æ ¨ æ æ Ag

We will now derive the following equation for the nucleation rate ¢ r N

¢ r N = n1

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2nms

2 S exp q - 4q3

27 lnS( )2È

Î Í Í

˘

˚ ˙ ˙

number of nucleikg gas •s

È

Î Í Í

˘

˚ ˙ ˙

[See Girshick and Chiu (1990)]

1. Calculate the rate of formation of g-mers. JgFrom kinetic collision theory we have

†

Net Rate ofFormation of

g – mers

Ï Ì Ô

Ó Ô

¸ ˝ Ô

˛ Ô =

Rate of Collisionof A1 g = 1( ) with

g - 1( ) mers with surfacearea sg-1

Ï

Ì Ô Ô

Ó Ô Ô

¸

˝ Ô Ô

˛ Ô Ô

-

Rate ofEvaporationof monomersfrom g - mers

Ï

Ì Ô Ô

Ó Ô Ô

¸

˝ Ô Ô

˛ Ô Ô

no. nucleidm3 •s

Ê

Ë Á Á

ˆ

¯ ˜ ˜ =

(Collision Velocity)¥ (cross sectionalarea)¥ (concentration of A1 )

¥ (concentration of Ag-1( ) - mers)

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

-(Evaporation Rate

per g – mer)¥ (no. of g - mer)

È

Î

Í Í Í

˘

˚

˙ ˙ ˙

Jg =kT

2pm1

Ê

Ë Á Á

ˆ

¯ ˜ ˜

1 2sg-1 n1ng-1

Rate of Collision of monomer(g=1) and (g–1)mer to form

g– mer

1 2 4 4 4 4 3 4 4 4 4

- Eg ngRate of loss

of g– mers byEvaporation

1 2 3 (1)

Recall that the collision velocity is (Um/4)

†

U4

=14

8kT2pm1

=kT

2pm1

Ê

Ë Á Á

ˆ

¯ ˜ ˜

1 2

A1 + Ag-1

æ Æ æ ¨ æ æ Ag

g–1sg–1

where n1 is the monomer concentration (molecules/dm3), ng–1 is the concentration of

(g–1) mers (number/dm3), sg–1 is the surface area (dm2) of the (g–1)–mer and Eg theevaporation rate constant for the g–mers, (seconds–1).

At equilibrium Jg = 0 and n1 = n1e


Eg = sg-1

kT2pm1

Ê

Ë Á Á

ˆ

¯ ˜ ˜

1 2 n1eng-1

e

nge

n1 is the concentration of monomers, n1e is the equilibrium concentration of monomers

(molecules/dm3), and nge is the equilibrium concentration of g–mers (number/dm3)

Let b =

kT2pm1

Ê

Ë Á Á

ˆ

¯ ˜ ˜

1 2n1

Jg = b sg-1 ng-1

e ng-1ng-1

e -1S

ngng

e

È

Î

Í Í

˘

˚

˙ ˙ (2)

At steady state the rates of formation of g–mers are the same J1 = J2 = … Jg, rearranging

Jb sg-1 ng-1

e =ng-1ng-1

e -1S

ngng

e (3)

We will now evaluate Equation (3) for different values of g, starting with g=2

g=2 ,

Jbs1n1

e =n1n1

e -1S

ngn2

e = S 1 -1

S2n2n2

e

Ê

Ë Á Á

ˆ

¯ ˜ ˜

Dividing by S

Jbs1n1

eS= 1- 1

S2n2n2

e (4)

g=3 ,

Jbs1n2

e =n2n2e

-1S

n3n3

e = S2 1

S2n2n2

e -1

S3n3n3

e

Ê

Ë Á Á

ˆ

¯ ˜ ˜ (5)

We continue writing the equationsanalogous to equation (4) and (5) up tosome arbitrary value of g, g = G

.

.

.

.

†

Jbsg-1ng-1

e SG-1=

1SG-1

ng-1ng-1

e -ng

SGnge (6)

When we sum all these equations (4) through (6), we see all terms on the right handside cancel out (e.g. circled terms in Equations (4) and (5) when added) except

1 -

ngSGng

e

Ê

Ë

Á Á

ˆ

¯

˜ ˜ of the rhs

J

g=1

G-1Â 1

b sgnge Sg

È

Î

Í Í

˘

˚

˙ ˙ = 1-

ngSGng

e

Ê

Ë

Á Á

ˆ

¯

˜ ˜ (7)

For large G and S!>!1, we can neglect second term on right hand side so that r.h.s.becomes unity.


Next we take the reciprocal and sum over all g–mers and then for large G we canreplace the summation Â by an integral Ú.

J =

g=1

G-1Â 1

b sg nge Sg

Ê

Ë

Á Á

ˆ

¯

˜ ˜

-1

=0•

Údg

b sg nge Sg

È

Î

Í Í

˘

˚

˙ ˙

-1

(8)

J = 1

0•

Ú e-H g( )dg(9)

where

H g( ) = ln b sg nge Sg( ) (10)

It will be shown [see Eqn. (29)] that

†

H g( ) = lnbs1N1 - qg2 3 - ln g2 3 - glnS( ) (11)We observe that this function will go through a minimum and consequently the firstderivative will be zero

dHdg

= 0 @ g = g* ≡ n *

g* ≡ n *

g

H(g)

0

This minimum corresponds to the critical g-mer size. Particles larger than the criticalsize n* are stable while those smaller than n* are not.

Returning to the equation for the rate of formation of a g-mer,

J = 1

0•

Ú e-H g( )dg(9)

We focus on evaluating the denominator, D,

D =

0•

Ú e-H g( )dg (11A)

2. First we use a Taylor series to expand H(g) about the critical size g*

H g( ) = H g*( ) + g - g*( ) dH g( )dg +

12

g - g *( )2 d2H g( )dg2

ˆ

¯ ˜ ˜

g*= H g *( ) + 12 g - g

*( )d2H g( )dg2

ˆ

¯ ˜ ˜

g*(12)

Using Eqn. (12) to substitute for H(g) in the denominator D.0


D =

0•

Ú e-H g( )dg = e

-H g*( )0•

Ú exp-12

g - g*( )2 d2H g( )dg2

È

Î Í Í

˘

˚ ˙ ˙ dg (13)

We now want to evaluate the integral in D.

Let y = 1

2g - g*( ) d

2H g( )dg2

ˆ

¯ ˜ ˜

g* , then dy = 1

2

d2H g( )dg2

ˆ

¯ ˜ ˜

g*dg

3. We now consider the lower limit of the integration

when g = 0 then y = -g* 1

2 d2H g( )

dg2ˆ

¯ ˜ ˜

g*= -a , a constant

D =0•

Ú e-H g( )dg = e

-H g*( )

12

d2H g( )dg2

ˆ

¯ ˜ ˜

g*

-a•

Ú e-y2dy (14)

Now let’s look at a typical value of the lower limit of integration a.

a = -g* 12

d2H g( )

dg2ˆ

¯ ˜ ˜

g*= -g* q

9 g*( )4 3 ( ¢ ¢ H from Eqn. 27)

Even choosing conservative values for q (q=18) and g* (g*=1000) we see –a is a largenegative number the order of (–10)

a @ g*( )1 3 q9 =

g1 3

3q =

103

⋅ 3 = 10

This value of the lower limit (a!=!10, –a!=!–10) is very large for the function we have, e-y

2, so therefore we can essentially replace it with minus infinity. Actually any value

greater than a=3 is essentially equivalent to setting the lower limit at –•. Therefore seta!ª!• and we obtain the probability integral

2 0 •Ú e-y 2 dy

ProbabilityIntegral6 7 4 8 4 = 2 p2 = p

(15)

†

D =0

•

Ú e-H g( )dy = e-H g*( ) 2p

¢ ¢ H g*( )

J = 1

0•

Ú e-H g*( )dy=

1

e-H g*( )

2p¢ ¢ H g*( )


J = e

H g*( ) ¢ ¢ H g*( )

2p (16)

where

¢ ¢ H ≡ d

2Hdg2

ˆ

¯ ˜ ˜

g*

Before we can evaluate H″(g*) we need to find the critical radius.

4. Finding the critical nucleus size g*Now we will assume the equilibrium concentration of g-mers will follow a Boltzmanndistribution,

ng

e = N1 exp -DGkT

Ê

Ë Á

ˆ

¯ ˜ (17)

where N1 is a normalizing factor to be determined later.Recalling Eqn. (11) [Click back to Eqn. 11]

DGkT

= g 2 3q - gln S, Derive (18)

where q =

ss1kT

We previously showed that DG goes through a maximum at g*

g* = 2

3q

ln SÊ

Ë Á

ˆ

¯ ˜

3(19)

As previously noted, particles of size greater than the critical size g* are stable. Particleslarger than g* will grow while particles small than g* are not stable.

Expanding H(g)

H g( ) = lnb + lnsg + ln nge + gln S (20)

sg = s1 g2 3 (21)

Applying Eqn. (21) at stable equilibrium S=1, for the kinetic theory derivation equation(7) becomes

DG g( )

kT= g 2 3q (22)

The number of g–mers per unit volume at equilibrium is given by a Boltzmanndistribution

nge = N1 e

-DG g( )

kT = N1 e-g2 3q (23)

where DG(g) is the Gibbs free energy of g-mer formation in the equilibrium state. N1 is anormalization factor


for g = 1Equation (21) becomes

n1e = N1e

-q (24)or

N1 = n1eeq (25)

5. Evaluating Terms in H(g)

Evaluating H(g*) and d2H(g*)/dg2

Substituting Eqns. (21, 22, and 23) into Eqn. (20)

H g( ) = lnb + lns1 +

23

ln g + ln N1 - g2 3q + g ln S (26)

J = eH g*( )

¢ ¢ H g *( )2p

(16)

Differentiating Equation (26) twice wrt g

dHdg

= 0 + 0 + 23

g-1 + 0 - 23

qg-1 3 + ln S

d2Hdg2

= -23

g-2 + 29

qg-4 3

Since g > 10 we neglect g–2 wrt qg-4 3

d2Hdg2

ª29

g-4 3q (27)

J = eH g*( ) 1

2pd2Hdg2

Ê

Ë Á Á

ˆ

¯ ˜ ˜

g*

1 2

= e-H g*( ) 12p

29

g *( )-4 3qÈ

Î Í

˘

˚ ˙

1 2

J = 1

3q

p

Ê

Ë Á

ˆ

¯ ˜

1 2 eH g*( )

g *2 3(28)

Evaluating exp [H(g*)]Replacing H(g*) in the exponential using Eqn. (10).

eH g*( ) = e

lnb+ lns1+23

lng*+ lnN1 -qg*2 3 +g *lnSÈ

Î ˘ ˚ (29)

†

= bs1g*2 3 N1 exp -qg*

2 3 +g lnS[ ] (30)Combining Equations (3_) and (3_)


J = eH g*( ) 1

2pd2Hdg2

Ê

Ë Á Á

ˆ

¯ ˜ ˜

g*

1 2

=13

q

p

Ê

Ë Á

ˆ

¯ ˜

1 2 1g *2 3

bs1g *2 3 N1 exp -qg *

2 3 +gln S[ ][ ]

†

J = bs1N13

q

p

Ê

Ë Á

ˆ

¯ ˜

1 2exp A [ ] (31)

where A = -qg *

2 3 -g * ln S6. Evaluate A at g*

Recall g*!=!

23

q

ln SÊ

Ë Á

ˆ

¯ ˜

3

A = -q 4

9q2

ln S( )2È

Î Í Í

˘

˚ ˙ ˙ +

23

Ê

Ë Á

ˆ

¯ ˜

3 q3

ln S( )3ln S

A = - 12

27q3

ln S( )2+

827

q3

ln S( )2

A = - 4

27q3

ln S( )2(32)

Substituting Eqn. (32) back into Eqn. (31)

J = N1bs1

3q

p

Ê

Ë Á

ˆ

¯ ˜

1 2exp - 4q

3

27 ln S( )2È

Î Í Í

˘

˚ ˙ ˙ (33)

7. Simplify the Pre-exponentialWe now will simplify the pre-exponential term.

s13

bq

p

Ê

Ë Á

ˆ

¯ ˜

1 2=

pd123

kT2pm

Ê

Ë Á

ˆ

¯ ˜

1 2n1

spd12pkT

Ê

Ë Á Á

ˆ

¯ ˜ ˜

1 2

=pd13

3•

22

s

2pmÊ

Ë Á

ˆ

¯ ˜

1 2n1 = n1

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2n1

J = n12spm

Ê

Ë Á

ˆ

¯ ˜

1 2 n1n1eS

{

n1e N1 exp -4q3

27 ln S( )2È

Î Í Í

˘

˚ ˙ ˙

(34)

J = n1

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2S n1

e N1 exp -4q3

27 ln S( )2È

Î Í Í

˘

˚ ˙ ˙ (35)

8. We now need to eliminate N1

Recall N1,

N1 = n1e eq = nms e

q


where n1e ≡ nms( ) , the concentration of monomers at saturation. Therefore the

rate of formation of nuclei per unit volume (number of nuclei/s/dm3) is

J = n1

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2nms

2 S exp q - 4q3

27 ln S( )2È

Î Í Í

˘

˚ ˙ ˙ (36)

Rate of loss of monomers to form nuclei per unit mass of gas is

¢ r N =

Jrg

¢ r N =

n1rg

2spm

Ê

Ë Á

ˆ

¯ ˜

1 2nms

2 S exp q - 4q3

27 lnS( )2È

Î Í Í

˘

˚ ˙ ˙ (37)

This is the result we have been looking for


Insert 1 APFR Web

Derive

Consider a cluster consisting of ng molecules with ns molecules attached to thesurface.

ng

ns

The Gibbs free energy, G(kJ) of a cluster of ng molecules plus ns surface molecules is thesum of the Gibbs free energy of a cluster (Gc) (kJ/molecule) plus the Gibbs free energyof the surface moles (Gs) (kJ/molecule)

†

G g( ) = ngGc + nsGs (1)Adding and subtracting nsGc to the r.h.s. of Equation (1)

†

G g( ) = ng + ns( )Gc + ns Gs -Gc( ) (2)Neglecting the number of monomers on the surface wrt to the number monomers in thecluster (ns


where mi is the chemical potential of the monomer molecule. For dilute solutions, thechemical potential of the monomer in the dispersed phase is

†

mi =mo + kT lnX (8)where X is the mole fraction of monomer in the gas. The chemical potential of the bulksolid corresponds to saturation corresponds to saturated conditions, Xsat

†

mg = mo + kT ln Xsat (9)

Combining Equations (7), (8), and (9) and setting ng ≡ n.

†

DG = -nkT ln XXsat

+ ss1n2 3 (10)

but S = X/Xsat

†

DG = -nkT lnS + ss1n2 3 (11)

where again n is the number of monomers in the cluster

n

DG

†

DGkT

=ss1

q}

kTn2 3 - n lnS = qn2 3 - n lnS

†

DGkT

= sn2 3 - nS (12)

In terms of g-mers (n ≡ g)

†

DGkT

= qg2 3 - n lnSBack to Text.


Insert 2 APFR

Derive

†

dG = VdP - SdTAt constant T

†

dG = VdP

V = g kTP

V = gNAw

NAvokTR6 7 4 8 4 È

Î

Í Í Í

˘

˚

˙ ˙ ˙

dG = gkT dPP

Bring a molecule from the dispersed state P1 to saturation Pv.

†

DG = gkT dP1

PvÚ lnP

DG = gkT ln PvP1

= -gkT ln P1Pv

=

†

DG = -gkT ln P1Pv

Back to Text.

Documents

AEROSOL NANOPARTICLE PLUG FLOW REACTORS ...elements/528/probsets/aerosol.pdf† Fogler, H. S., Elements of Chemical Reaction Engineering, University of Michigan, Ann Arbor MI. 2 528F03AerosolReactor.doc