1

363 Class Notes

Keith P. Johnston

Chemical Engineering

2

ChE 363 Mass Transport

E.L. Cussler, Diffusion, Chpt. 1.

As shown in Cussler in section 1.1, there are two basic models for treating diffusion as shown in Figure 1.2-1. flux = mol/cm2/s (amount/area/time)

Model 1 flux = k (concentration difference)More applied. Easier to use. Less rigorous.

Model 2 flux = D (concentration difference/length) More fundamental, sometimes more difficult to apply

Look at interesting examples in 1.3 to make this physically meaningful to you.

Ex. 1.3-1 Ammonia scrubbing- easier to use k since the diffusion length is not well defined unless detailed flows are well known.

Ex. 1.3-3 Corrosion of marble by acid rain- D allows one to predict conc. vs. position in marble. k- only flux at interface

3

Ex. 1.3-4 Protein size in solution- diameter is proportional to diffusion coefficient.

Ex. 1.3-5 Antibiotic production- oxygen uptake for aerobic fermentation. k- changes in bubble size, non-Newtonian flow, foam caused by biological surfactants

Ex. 1.3-6 Facilitated transport across membranes-

Diff. and reversible chem. Rxn. couple in non-linear fashion. Need D to understand system.

4

Diffusion and Mass Transfer

• Flux relative to stationary plane:

• Flux relative to solvent flow at uo:

JA = cA uA – cA uo = cA (uA – uo)

e.g. canoe relative to river• Fick’s Law

• Ideal gas cA + cB = = P/RT

At const T, P, is const. dcA + dcB = 0

M totalmolar density

M o A A AN u For species A, N c u

3 2

g mol cm g mol =

cm s cm s

2

A AB A 3 2

cm g mol g molJ D dc / db

s cm cm cm s

M

M

5

JA + JB = 0 (to maintain const and uniform M)

- DAB dcA/db – DBA dcB / db = 0

• DAB = DBA

• Total flux species A = Convective flux A + Diffusive flux A

NA = cA uo + cA (uA – uo)

= yA (NA + NB) + JA (M const)

Suppose JA is 0. Then the convective flux of A is yAN, where N is the total flux.

6

Cussler Chapter 2

2.2 Steady state diffusion across a thin film

Fig. 2.2-1 Fig. 2.2-2

7

Accumulation= diff. into the layer at z – diff. out of layer at z + z.

0 = A (j1z - j1z+z)

Divide by Az and rearrange

0 = -(j1z+z - j1z)/ [(z + z) – z] = -dj1/dz

From Fick’s law, -j1 = D dc1/dz, thus 0 = D d2c1/dz2

Boundary conditions: z = 0 c1 = c10, z = l c1 = c1l

Integrate twice: c1 = a + bz

Flux expression:

c1 = c10 + (c1l – c10) z/l

j1 = (D/l) (c10 – c1l). At steady state, the flux is constant.

8

Thin film diffusion or equimolar One-way diffusioncounter diffusion NA = const

= Diff. flux + Conv. fluxNo convection dyA/dz = const. Diffusive flux const.

yA large conv. dominant

dyA/dz small

yA small

dyA/dzLarge

bulk fluid

9

Equimolar counter-diffusion (e.g. binary distillation)

NA → ← NB NA = -NB

NA + NB = N = 0

NA = yA N + JA where N = 0. Thus NA = JA See Fig 17.1 (6th, p. 517)

Same as thin film w/ no convection

10

ChE 363 Summary of Mass Transfer

Total flux = Convective flux + Diffusive flux

NA = cAuA = cAuo + cA(uA-uo)

= yA(NA + NB) + JA if M is constant

where JA = -cDv dyA/dz (Fick’s Law)

Steady state

a. Thin film with no convection- flux is constant. NA = JA

Or Equimolar diffusion which eliminates convection

JA = (Dv/l ) (cAi - cA) = (Dvm/l)(yAi – yA) const.

cAint = cAi + (cA – cAi) z/l (int = intermediate) linear

b. One-way diffusion of A only with convection (NB = 0)

JA, yAN = yANA = f(z)

11

NA = (DvM/l) ln [(1-yA)/( 1-yAi)] constant

= (DvM/l) (yAi – yA)/(yB)lm

where (yB)lm = (yB – yBi)/ln (yB/yBi)

yBint/yBi = (yB/yBi)z/l [= f(z)]

Diff. flux JA is largest at z where yA is smallest. Here the conc. profile is most curved. As JA decreases yAN increases. The sum NA is constant.

c. Steady diffusion into falling film with short contact time

Identical to penetration theory below where time becomes a residence time,

= x/vmax. Here vmax is a constant.

Unsteady Diffusion in a Semi-infinite slab (Penetration theory)

cA/t = - jA/z = Dv 2cA/ z2 Diffusion eqn. with no convection

(cAint – cAi)/(cA - cAi) = erf , erf = (2/) 0

exp (-s2) ds = z /(4Dt)1/2 JA = -Dv cA/ z = (Dv/t)1/2 exp(-z2/4Dt) (cAi – cA)

JAz=0 = (Dv/t)1/2 (cAi – cA) JAz=0time avg = 2(Dv/tT)1/2 (cAi – cA)

12

BSL One-Way Diffusion/Convection Thru Gas Film (const T, P)

Transport Eqn (not mass balance)

diffusion convection

NA = -cDdxA/dz + xA (NA + NB) (a)

Assume B insoluble in A.

NB << NA NB ~ O

From eq (a)

Steady State Mass balance:

S NA |z – S NA |z + z = 0

A AA

cDN dx / dz (17.2-1)

1 x

13

Divide by z -dNA/dz = 0 NA f(z) and is constant

For ideal gas, const T, P, c is const

BC1, 2 z = z1, xA = xA1 = PAsat / P z = z2 xA = xA2

Solve for c1, c2 (fair amount algebra)

A1 A 1 2

A

dx1 c ln(1 x ) c z c

1 x dz

1 1

2 1 2 1

z - z z - z

z - z z - zA A2 B B2

A1 A1 B1 B1

1-x 1-x x x = or =

1-x 1-x x x

2 A 2

1 A1

z x

A A A2

A A1z x

(N ) dx 1 xFrom 17.2-1 const dz ln

cD 1 x 1 x

14

B2A

2 1 B1

xcDN n (17.2 14)

z z x

1 2 1Let (z z ) /(z z ) d dz

2

1

1

z B2 B1BAvg 0B B1

1B1 z

0

(x / x ) dxMean value (x / x )

dz theorem x dz

d

u u a du a / ln aB Avg B2 B1 B2 B1

B1 B2 B1 B2 B1

x 1(x / x ) x / x 1

x ln(x / x ) 0 ln(x / x )

B2 B1B,Avg B lm

B2 B1

x xx (x ) (2)

ln(x / x )

1

Subst (2) → 17.2-14 for ln (xB2/xB1)

A1 A2A B2 B1 A1 A2

2 1 B lm

x xcDN BSL Note : x x x x

z z (x )

15

1 2

A1 Ai A2 A

z 0 zOur notation

x x x x

Ai AA

B lm

(x x )cDN

(x )

A A Ai A B m

A A Ai A

N (one way) N (y y ) /(y )

N (equimolal) J y y

B m(1/ y ) p.661

One-way denoted by ‘

c c y y B mk '/ k k '/ k 1/(y )

16

2.3 Unsteady Diffusion in a Semi-infinite Reservoir (Penetration theory)Solute accum. = diff. in – diff. out no convection, A

perpendicular toplane of

paper

/t (Azc1) = A(j1z – j1z+z)

Divide by Az and use definition of derivative

c1/t = - j1/z = D 2c1/ z2 Diffusion eqn. Fig 2.3-2 Cussler

t = 0 all z c1 = c1 (bulk)

t > 0 z = 0 c1 = c10

z = c1 = c1

For small z or large t (small ), erf linear in .

Linear concentration profile.

17

Define a new variable to generate ordinary differential equation (Boltzman, 1894):

Method of combination of variables

= z (4Dt)-1/2 Thus: /t = -(1/2) /t /z = /z

Use the chain rule for introduction of new variable:

c1/ /t = D 2c1/ 2 (/z)2

Appendix 1: (Subst for two partial derivatives from above)

2 2 2 21 1c / ( 1/ 2) / t D c / ( / z )

22 2

1 1

1 zc / c /

2 Dt

2 2Subst z (4Dt)

2 21 12 c / c / 2.3 11

18

One integration of the eqn 2c1/ 2 +2 c1/ = 0 (2.3-11) gives

c1/ = a exp(-2) Subst into above eqn. to see it is correct.

(c1 – c10)/(c1 - c10) = erf (see Appendix 2) (2.3-15)

where erf = (2/) 0 exp (-s2) ds

erf (0) = 0; z = 0 c1 = c10

erf (∞ ) = 1; z = ∞ c1 = c1∞

The flux may be obtained by taking the derivative of this integral (Appendix 2)

j1 = -D c1/ z = (D/t)1/2 exp(-z2/4Dt) (c10 – c1)

Note: Don’t forget to multiply integrand by deriv. of limit, /z.

At interface where z = 0 t incr., j1 decr., abs. value slope decr.

j1z=0 = (D/t)1/2 (c10 – c1) a useful result!

19

Appendix 2

1 1 10j D(c c ) derf / dz

1/ 2 2derf2 exp( ) d / dz

dz

upper limit of integration

1/ 2Recall d / dz / z (4Dt)

1 1 10

Subst :

j D(c c ) 2 1/ 2 exp( )( 4 1/ 2Dt)

Insert this result into previous page.

1

10

c2

1

c 0

proportional to erf

dc a exp( )d For ,integral / 2

1 10Insertion of bound. cond. yields : a (c c ) /( / 2)

Second Fundamental Theorem of Calc. to obtain j1 = - D dc1/dz

20

2.5 Convection and Dilute Diffusion

2.5.2 Steady Diffusion into a falling film- forced convection mass transfer(BSL p. 537; Cussler p. 45) Semi-infinite film: limited penetration into film

Assumptions: 1. Dilute soln. 2. Diffusion in z direction; convect. in x direction.3. Pure gas. 4. Contact between gas and liquid is short

Constant solvent velocity, vx , result of assumptions 1 and 4.

Accum. = diffusion in z direction + convection in x dir.

0 = (j1Wx)z - (j1Wx)z+z + (c1vxWz)x - (c1vxWz)x+x

W = width perpendicular to plane of paperDivide by volume Wxz

0 = -j1/z - (c1vx)/x

21

Let vx be a constant the maximum velocity at the gas surface.

c1/(x/vmax) = D 2c1/ z2 (Analogous to diffusion eqn) = x/vmax

B.C. x = 0, all z, c1 = 0

x > 0, z = 0 c1 = c1sat z = l c1 = 0 (short contact)

c1/c1sat = 1 – erf [z/(4Dx/vmax)1/2]

j1z=0 = (Dvmax/x)1/2 c1sat

Final perspective:1. Approximation for USS diffusionl = (Dt)1/2

2. j1 = (D/l) c (thin film) (steady state)

j1z=0 = (D/t)1/2 c (thick slab)

22

Diffusivities

Gases

M

21/2

D 1/ 3 u u avg molec vol

1/ RT /P mean free path

~1/P , ~T ( goes down in both cases)

1 u ~ T mu related to kT, k is Boltzmann const

2

Av

1.5

., k=R/N .

D ~ T /P for ideal gas

23

For real gas

See eq 17.28 p. 519

Lennard-Jones potential

See empirical Wilke – Chang (17.31)

F = Faraday constant,

Table 17.1 cations less hydrated as size incr.

p. 522

12 6u 4 ( / r) ( / r)

0Liquids: Stokes Einstein Friction 6 r

o

thermal activation energy for hopskTD

6 r friction

0 0 2Electrolytes D 2 RT /(1/ 1/ )F 17.33

ATurbulent Diffusion J (D )dc / db

eddy diff

= f(u, roughness, position)

0 limiting conduc tance for anion

24

Mass Transfer Coefficients

c A Ai A y A Ai Ak J /(c c ) k J /(y y )

3

2 2

cm g mol cm g mol

s cm s g mol cm s

y c M A A M

A A2 3

k k since c y

g mol cm g mol y =c /c

cm s s cm

MFor an ideal gas P (n / V)RT or P /RT

y c x c av gk k (P /RT) For a liq, k k /M

y

A M T Ai A

k

J D /B y y

Diffusion to wall in laminar layer (membrane or electrode)

Fast mixing in core turbulent zone.

Resistance in stagnant film.

25

Dimensional Analysis- Route to Correlations (Cussler, Diffusion, p. 235)

Show NSh = f(NRe NSc)

kc = f(u,,,Dv, D) All variables that affect mass transfer.

= (const.) (u , , , Dv , D)

L/t = (Lt-1) (ML-3) (ML-1t-1) (L2t-1) L

Exponents must match on both sides:

For L 1 = - 3 - + 2 + (1)For M 0 = + (2)For t -1 = - - - (3)

Solve in terms of and (arbitrary choice) = , =

= 1 from eq. 3 (4) (5)

26

-1 (from 4, 5 substituted into 1) (6)

kc1 = (const) u - 1 1 - Dv

D - 1

= (const) -1D-1 (uD/) (Dv/)

kc D/ = (const) NRe NSc

Multiply both sides by NSc = Dv

kcD/Dv = NSh = (const) NReNSc

(1-) Fit const, to data.

27

Dynamic light scattering (1 nm to 1 m, polymers, proteins, colloidsIncident light particle diffusesLaser (He-Ne visible) by Brownian motion

in and out ofscattering volume

Scattered light

For two particles near each other constructive and destructive interferencet = 0 particles certain distance apartt + t relative distance changesAs t → 0 initial positions correlated As t →∞ no correlation between pairsAuto correlation fxn < A(0) A(t) > exp - (q2Dt) Higher D, faster decayMeasure q2 D → Know q, Get D

<A(0) A(t) >

o4

q sin 90 usually2

28

Mass Transfer Heat Transfer

Mass transfer to pipe walls (turbulent)

Convenience:

Sh c Nu

m.t.velocity convectionN k D /D N hD /k

diffus.vel. conduction

Sc Pr p

viscous viscousN / D N c /k

diffusion diffusivity energy

0.8 1/3 0.14sh Re Sc w

Re c

N 0.023 N N ( / )

As N k

Shm 1/3

Re Sc

Nj

N N

1/3 2 /3

c ck D D k

D uD u D

29

For pipes

Flow past single solid spheres p. 537

0.2m Rej 0.023 N

1/3 cSc Sh

k DAs D , N (a little), N (a little)

D

ReN 0 Sh 2 analytical

pp / 2 Sh

T

DDeffective film thickness D 2 N ~

B D

1/ 2Re Sh ReN 1000 N f(N ) boundary layer theory

ReN turbulent exponent 1/2 greater flow rate effect

kc incr.

30

Mass transfer to drops + bubbles:

Penetration theory for time avg kc

T1/ 2 t1/ 2 1/ 2

c c

0

Dk (D / t) k t dt / dt

1/ 2T2(D / t )

T p 0For = D /u (flow past entire drop)

1/ 2c 0 pk 2(D u / D )

1/ 2

p 0Sh c p

1/ 2 1/ 2Re Sc

D u2N k D /D

D

1.13 N N

Re Sc p 0 PeN N D u /D N (Peclet) inertial forces / diffusion

Cussler: Small liquid drops

NSh = 1.13 (NRe NSc)0.8 behave like rigid sphere drops

Circulation currents raise exponents depending upon viscosities of the phases.

31

Controlled Release Cussler, 2nd Ed (Review of mass transfer) Diffusion

Eq. 18.1-1 (thin film) – dissolution of solute

Accum Flux .A

Matrix system Diffusion thru

Eq. 18.1-2 (thick slab) at small times, lim t 0

s1 1 c 1 1

s satdVc AN Ak (c c )

dt 1 drug

1

t1

cs01 1 0

c dc(Ak / V)dt

c c

ss1 1 1

c 1 1 cs s s1 1

c c c Mln Ak t / V 1-c / c exp( k At / V)

c 0 c M

ck At / Vsamt. released : M M (1 e ) first order release kinetics

time avg 1/ 21 v T 1i 1 1

time avg 1/2 1/21 T v T 1i 1 1i

J 2(D / t ) (c c ), c is external to slab

M J At 2(D t / ) A(c -c ) t kinetics M =c AL

2 1/ 2/ 2( / ) :v TM M D t L Saltzman dual slabis twice this value

32

Controlled Release by Solvent Diffusion

18.21.1 Drug dispersed in polymer (Matrix release –hydrogels)

Control cross-links

diffusion through swollen polymer leads to release that has t ½

Glassy polymers – zero order (tortuous unfolding of chains)

33

Chapter 20 Equil Stage Operations

Total mat bal (mb) on top to tray n

La + Vn+1 = Ln + Va

Component mat bal on more vol. comp.

La xa + Vn+1 yn+1 = Lnxn + Va ya (20.2)

Total mb (entire column)

La + Vb = Lb + Va

Solve 20.2 → operating line

a a a an nn 1

n 1 n 1

V y L xL xy 20.7

V V

b N 1y y b N

b N

L L

x x

a a a an n

n 1n 1 n 1

V y L xL xy 20.7

V V

VN+1

34

Rectification at Absorption of gas Desorptiontop distil. column into liq. or stripping

from liq. to gasOperating to equilibrium line:

y ↑ x ↓ y ↓ x ↑ y ↑ x ↓(opposite of absorpt.)

Enriched at top Rich gas- bottom Rich liquid at top Lean gas at top Lean liquid at after absorption bottom after desorption

35

Gas Absorption (Chpt. 18) (No accum.-interface)

Rate of Adsorption / vol. (a = interfacial area/vol)

Likewise

2

y i3 2 3

g mol g mol cm r = k a (y y )

cm s cm s cm

x i

y y i i

x

k a(x x)

K a(y y*) K a[(y y ) (y y*)]

K a(x * x)

i i

y

y y y y *1

K a r r

i i

y i x i

y y y y *

k a(y y ) k a(x x)

y x

1 m

k a k a

x x y

1 1 1

K a k a mk a

m = slope eq curve

36

Note: xi > yi even though mass transfer is to the right due to phase equilibria (solubility of gas in liquid)

At interface

For low solubility gases (O2 in water), y = mx, m is large and x is small.

Major resistance is in liq phase where concentrations are low

For high solubility gases (NH3 in water), m is small and x is large. Major resistance is likely to be in the gas phase where concentrations are low

v LA A

37

Transfer unit concept:

Material balance on gas

bubbles in shell (define upwards for in):

Both sides are positive quantities

dV << V (dilute systems)

dy is + in Z direction downwards

+ Vdy = Kya (y-y*) SdZ

= Hoy . Noy

For straight operating + equil curves (e.g. dilute systems), analytical integration

In Out Absorp. to liquid

(V dV)(y dy) Vy rSdZ

TZ b by

Ty0 a a

K aS dy V / S dy dz or Z

V y y * K a y y *

oy b a m b b a aN (y y ) / y Note : lm for y y * and y y * driving forces

S = cross sect. area of column

As driving forces incr., smaller Noy needed to achieve overall change in y from b to a

38

See p. 564

molal mass velocity = molar flux

two-film theory

Note: LM = L/S : molar flux

Note: mGM/LM =ratio of slope of eq. line to op. line, Eq. 18.28

Notice the H values take ratio of convection to kya

y y x x oy oy ox ox yy

V / SH ,N ; H ,N ; H ,N ; H ,N H

k a

M MG u V / S

MMM M

y y x M

L1 1 mG G G

K a k a k a L

Moy y x

M

mGH H H

L

y y x

1 1 m

K a k a k a

39

Absorption: Comparison of plates (discrete) versus packed bed (differential) (optional minor point)

Fig 18.13 (b) p. 563

NTP = 1 for a column w/plates

Since lm avg driving force is < yb – ya

More transfer units are required to achieve yb – ya since driving forces are smaller. This distinction is simply a result of the two slopes shown- not general.

The discrete plate system achieves a huge driving force from op line to equil line at state yb → yb*.

b bb alm

a alm

y -y *y yNTU 1 y for

y -y *y

40

Mass transfer correlations

To focus on liquid resistance control: Desorption of O2 from H2O

m (slope eq curve) > > L/V

very low xi so most resistance is in liquid

i i i iyP x H

i i i i i iy / x H /P H large, since x small

mxox x y

M

LH H H H

mG

41

Choice of determination of Hox from absorption versus desorption

Why is desorption preferable to absorption here?

For H0x, focus is on liquid phase

For absorption xb* - xb gives little accuracy (loss of significant figs)

In contrast, for desorption xb* = 0

More significant figures are obtained in xb – 0 vs. xb* - xb

Goal of engineering:

Run absorption pilot unit (e.g. 2 inch diameter). Measure concentrations and flow rates. Develop correlation for Hox. Use mass transfer model to predict other conditions and systems, and to design large columns

ox T

dxH Z /

x * x

42

Fig 18.21 p. 715 Correlation for Hx in terms of Gx = u = mass flow rate/S = GMM

KL is a Kc [cm/s]

As Gx ↑, convection ↑, NRe ↑, therefore, KLa ↑. However numer. dominates and Hox ↑

To convert from O2 in water to other systems:

is an empirical constant

Gas film resistance (highly soluble gases)

Absorption of NH3 in water

xi is large, m is small → most resistance is in vapor phase where concent. is small

Fig 18.22 – focus is on vapor phase.

Absorption from rich gases not covered

v

n

0.5xx

G1H ( / D )

x x xox x L m

x L

m x

G /M G /H ; K a K a

K a K a

M

i i i iyP Hx

43

LaVa

ya

Lx Vy

xa = 0.0002Xa = 0.0002

Lb

xb = 0.05

Vb

yb = 0.14 359 ft3 / lb mol (STP)500 ft3/m (STP)

18.1

Basis: 1 hour

Acetone in: 0.14 x 83.57 = 11.70Acetone out: 0.05 x 11.70 = 0.585 5% loss

Acetone absorbed = 11.115 95%

Moles air in: 83.57 (1-0.14) = 71.87 V’ (const)

ya = 0.585/(71.87 + 0.585) = 0.008074

Lb = La + 11.115 Total Liquid Gain0.05 Lb – 0.0002 La = 11.115 Acetone GainLa = 212.03 mol / h Lb = 223.145

b

500V x 60 83.57 lb mol /h

349

vapor

44

(a) Water to tower:212.03(1-0.0002) x 18.02 = 3820 lb / h L’ (const)

(b) Acetone balance Lx + Va ya = La xa + Vy‘

Solute free basis: (V’, L’ constant)V’ = V (1-y) L’ = L(1-x)

a a

a a

y xy xV ' L '

1 y 1 y 1 x 1 x

Solute-free basis:

E.g. Va = V’ / (1-ya)

From this: y = F(x) / (1+F(x)) where

const L’/ V’

a a

a a

y xy L ' xF(x)

1 y 1 y V ' 1 x 1 x

ya = 0.008074 xa =0.0002L’ = 3820 / 18.02 = 211.98 V’ = 71.87

45

Equilibrium relationship: Acetone

P = 1 atm

For values of x between 0.0002 and 0.08, calculate y and y* and solveEq. (18.21) by numerical integration over dy. The computer solution is Noy = 9.34

21.95(1 x)A Ay* P' x 0.33e x

b

a

y

oy

y

dyN

y y *

(c) Calculate the operating mass velocities from the flooding velocities basedon flow rates at the bottom of the column (where both rates are greatest).The mass flow rates,

x y

x

y

m and m are

lb/h MW water + acetone

Liquid : m 3820 (11.115x58.1) 4465 lb /h

Gas : m (71.87x29) (11.7x58.1) 2764 lb /h

air

MW

acetone inUse Fig 18.6, p. 554

(18.21)

46

For 7% acetone in air, y = 0.0791. Assume x = 60.5

yx

y x

G 4465 0.0791 0.0584

G 2764 60.42

(ideal mixing) m

3

lb

ft

p. 549From Table 18.1, for 1-inch Raschig rings, Fp = 155. Since this isgreater than 60, the pressure drop at flooding, as recommended on p. 553,is 2.0 in. H2O / ft. By extrapolation of the data in Fig. 18.6 to this P,

2 0.1y p x

c y x y

G F ( )0.115

g ( )

c y x yy 0.1

0.115 g (G

F

For water at 80°F, ux = 0.862 cP (Appendix 6)

8

y 0.1

2

0.115x4.17x10 x0.0791x(60.5 0.0791)G

155x0.862

= 1225 lb / ft - h

Actual rates: 8

2c

2m

2 2f

g 32.174(3600) 4.17x10

lb ft s

lb s h

47

½ flooding: Gas: Gy = 1225/2 = 613 lb / ft2 – h

Liquid: Gx = 613 x 4465 / 2764 = 990 lb / ft2 – h

Calculate HTU’s, using Eq. (18.44) for Hx. Find Dvx from Eq. (17.31).At 80°F (26.7°C), x = 0.862 cP or 2.086 lb/ft – h (App 9)(assuming is the same as for water). Also

T = 299.7 K MB = 58.1A at 20°C: 0.792 g / cm3 or 0.792 / 58.1= 0.0136 g mol / cm3

Estimated x at normal boiling point (56.1°C),is 0.0128 g mol/cm3.

Hence VA = 1/0.0128 = 78.1 cm3 / g mol(see Perry, 5th ed., p.3-230).Substitution in Eq. (17.31) gives, since Wilke – ChangB = 2.6, 8 1/ 2

5 2vx 0.6

7.4x10 x(2.6x18) x299.7D 1.29x10 cm / s

0.862x78.1

or 5.0 x 10-5 ft2 / h (Appendix 1)1/ 2

8 B Bv 0.6

A

( M ) TD 7.4x10

V

Solute normal b.p.

assoc parameter

flux ratio

48

p. 581From Fig. 18.21, Gx = 990 (above) , 1” rings

For acetone in water,Sc = 2.086 / (60.5x5.0x10-5) = 689 = / D

For O2, Sc = 381 (see p. 582)

for water at 25°C: 0.90 cP (Appendix 6)

From Eq. (18.44), using n = 0.3

2x,OH 0.68 ft

1/ 2n

xx

v

G1H

D

Penetration theory

18.44

Hx = 0.68 (0.9/0.862)0.3 (689/381)0.5= 0.93 ft

To find Hy, use Fig. 18.22. For Gy = 613 and Gx = 990 lb / h-ft2, byinterpolation:

3y,NHH 2.0 ft

49

Use Sc for acetone-air for 0°C (Appendix 18). In view of the uncertaintiesin the correlations, correction to 80°F is not worthwhile. Then Sc = 1.6.from Eq. (18.47)

Hy = 2.0 x 1.6 / 0.66)0.5 = 3.11 ftFind the average slop of the equilibrium curve for use in Eq (18.28). From theequation given in part (b),

at x = 0.05, y* = 0.09589 x = 0.0002, y* = 0.000464

3

1/ 2

y y,NH

ScH H

0.6

0.09589 0.000464m 1.916

0.05 0.0002

yM [(0.008074 0.14) / 2]x58.1 0.926x29 31.15

xM (0.0251x58.1) (0.9749x18.02) 19.02 2

MG 613 / 31.15 19.68 lb mol /h ft

ML 990 /19.02 52.05

moy y x

m

GH H m H (18.28)

L

50

From Eq. (18.28)

Hoy = 3.11 + 1.916 (19.68 / 52.05) x 0.93 = 3.78 ft

Column diameter D = (4my / Gy)1/2

= ((4 x 2764) / (x 613))1/2 = 2.4 ft

NOTES: Since the equilibrium and operating lines are nearly parallelthe number of transfer units about equals the number of theoreticalstages. The column height is 3.78 x 9.34 = 35.3 ft.

2

y

m D

4 G

51

Phase Equilibria ReviewL Vi i L V sati i i i i i if f x P y P

sati i i i iK y / x P /P

sat satij i j i i j j K /K P / P

At azeotrope i i iy x K 1

ij i jK /K 1/1

sat sati j j iIn / ln P /P

52

i i i

i i

y 1 K x

P,x T,y

i i i

i i

x 1 y /K

P,y T,x

53

Flash calculation (after thermo done to determine yD, xB)

Thermo notation

1 1 1z = y V + x (1-V)

MSH notation

F D Bx = fy + (1 - f) x

FD B

x(1 - f)y = x

f f

54

B F

D F

s. . f = 0 x = x

s.v. f = 1 y = x

ABf

AC

xB

x, y

55

56

Total F = D + B

Component:

FxF = DxD + BxB

Elimin. B

D/F = (xF – xB)/(xD - xB)

Elimin. D

B/F = (xD – xF)/(xD - xB)

Condenser D = Va – La

Va

La D

B

F

Binary Distillation

57

Rectification (Enriching) Section (Above feed line):D = Vn+1 – Ln

DxD = Vn+1 yn+1 – Ln xn = Va ya – La xa

a a a an n n Dn 1 n

n 1 n 1 n n

V y L xL L x Dxy x = + (20.12,14)

V V L + D L + D

Figure: Seader and Henley, Sep. Proc. Principles

Va

58

Constant molar overlow: simplifying assumption to obtain linear operating line- leads to simple graphs (pedagogical and conceptual value)

Recyle ratio

n n Dn+1

n n

L x Dxy = + (20.14)

L + D L + D

DR L /D 1L, 1H

1L, 1H

vap

D Dn 1 n

D D

m Bm+1 m

m m

H per mol ~ const

L,V const

R xy x

R 1 R 1

Likewise in stripping section

L Bxy = x

L -B L B

59

60

61

Feed tray energy balance to determine changes in L and V

For sat’d liq.

For sat’d vapor

q > 1 subcooled liq TF < Tbub pt

q = 1 sat’d liq TF = Tbub pt

q = 0 sat’d vap TF = Tdew pt

q < 0 superheated vapor TF > Tdew pt

Energy balance on feed tray for sub cooled liq.

L L F V V

L L V V F

L L qF (21.26) V V (1 q) F (21.27)

Let q mol. liq added to L by F

vapPL b F

F

F

(q 1) H C (T T )

bubble T ( T )

As T q L

62

Determination of feed line from material balances and feed tray energy balance

Let m = n at feed tray where the two sections meet:

Subst (21.26) , (21.27)

(same as 1 stage distillation)

n+1 n D

m+1 m B

Enriching section Vy =Lx + Dx

Stripping section Vy =Lx - Bx

D B(V V) y (L L) x Dx Bx

F(1 q) Fy qF x Fx

Fxqy x

1 q 1 q

f 1 q

V L

Reboiler energy balance

At 20 psia,

Condenser

Rise in cooling water T

vap vaps sm H V H

vapSH 522 cal / g, 939 Btu / lb

vapw p 2 1m C (T -T ) V H

63

Feed tray location (optimization)

Given RD, q, xF, F, xD, xB → calculate D, B

Both operating lines and feed line are fixed

Fig. 21.15 p. 660

Switch may be delayed past optimum, yet operating + feed lines are fixed

Pinch pt limits: a and b. Here concentrat. driving forces go to zero.

If feed tray changed at const q

Tf or ratio of s.l. and s.v. in feed must be changed.

e.g. (q-1)Hvap = CPL (Tb – Tf)

↑ ↑ ↑

fixed changes must be changed

64

Plate EfficiencyMurphree

ηM = (xn – xn-1) / (xn*– xn-1)

η

Local effic. (‘ means local

location on tray)

η ‘

n n 1M

n n 1

y y

y * y

n n 1

n n 1

y ' y '

y * y '

mVlarge

L

n n 1 Mx ,y '

can be 1

mV/L small

65

DE : effective D, Z: flow path, tL : res time

NPe = NRe NSc

Using same concepts as in absorption an d definition of ’

2

E L E

uZ Z

D t D

convec visc

visc diff

M

'

Pe M

Pe

Low N fast diffus. ' 1, '

High N slow diffus. ' small

Pe M Pe

Fig 18.32

As Z , N , ' / ' f( ', N )

M Pe/ ' vs N

M

ys

s

-K aZ1-η' = exp where V is thesuperficial velocity

V ρ

66

Analysis of McCabe - Thiele

Variables F, xF, xD, xB, D, B,

(10)

Nenrich, Ntotal or Nstripping

RD, q

Equations (4) 2 total mat. Bal. column, 1 op line, 1 equil curve

(Enrich + q + xB + L → Stripping Line)

1. Design column

Variables Calculate

F, xF, xD, xB D, B, Nenrich, Nstrip

RD, q

2. Existing column (variable feed tray)

F, xf, q, Ntotal, RD, xD → xB, D, B, Nenrich

Vary feed tray until Nenrich + Nstrip = Ntotal

3. Existing column (fixed feed tray)

F, xF, q, Nenrich, Nstrip, RD → xD, xB, D, B

67

Energy Balances

DHD + B HB – F HF = qr + qc

As RD = L/D ↑

V = L + D ↑

E.B. Around enriching section

Vn+1 Hy, n+1 = Ln Hx,n + VaHy,a – RHD

rV V (1 q) F q

n n Dn 1

n n

op line curvedL x Dxy +

L + D L + D see fig 18.25

yn+1 = f (Ln)Ln = f (Hy, n+1) where Hy,n+1 = f (yn+1)

Simultaneous eqns

Skip books tedious method – use professional software

68

Multicomponent Distillation

Dew Bubble

Flash p.717 or thermo book

Keys – choose more concentrated components

Sharp separation – only keys distribute

Ki < KHK by ~10%, B only

Ki > KLK by ~10%, D only

satsat i i i

i i i i ii

y PVLE y P = x P K = =

x P

sat sati i i i i

satij satjj j j j

y /x P P = Py /x P

ix 1 iy 1

ij ij ij

3ij D F B

Fenske Min plates 21.42

69

Underwood (min reflux) – not covered

Gilliland p. 732 Fig 22.5 (empirical)

Do problem 22.4 (Fenske part only)

minN Nlog

N 1

min

D

towards N

R

D Dmin

D

R Rlog

R 1

70

Tray by Tray Calculation

C components, N trays

On each tray, { xi } { yi } L,V,T unknowns (2C+3)Material balance for each C C

Binary: Given xD, xB, xF → Solve D, B

Multicomp: Given xDH, xBL

cannot solve for D, BGreater number of degrees of freedom

Invariant zones may be found at other locations than feed tray

L Vi i

i

i

ˆ ˆf f C

1 energy balance 1

x 1 1

y 1 1

Per tray 2C + 3

71

Methods

1. Desired { xDi } {xBi } and RD → Determine Nenrich, Nstrip

2. Given Nenrich, Nstrip

Calc {xDi }, {xBi } – PREFERRED

M. Doherty Conceptual Design of Dist. McGraw Hill, 2001

72

Leaching and Extraction

1. Methylene chloride from poly (glycolic – co – lactic acid) biodegradable microcapsules, soybean oils from seeds

aromatics from aliphatics, natural substances (caffeine from coffee, taxol from yew bark, antibiotics from fermentation broth, PEG-dextran protein extraction, metals from ore

Leaching (optional)

Substrate entrains liquid. If , of solution ↑ with solute concentration, amount entrained increases. Ln = f(xn)

If , constant amount entrained relatively constant. Constant molar under flow. Ln constant

Equilibrium y* = x (solids – free basis) entrained solution has the same composition as overflow. The solution fills pores. No adsorption on surface. Often concentration for solids-free basis

Ln-1Ln

73

Extraction - Bioseparations

Biomolecules often non-distillable thermally sensitive or nonvolatile

Belter, Cussler et al. BioseparationsK = CL / CH L = light (organic phase)

Antibiotics, etc. AH ↔ A- + H+

low pH high pHKa = [ A-] [H+] / AH

pH < pKa favors protonation (nonionic)pH > pKa favors dissociation andgreater water soln

For COOH pKa ~ 4.5See Table 5.1-1Fig 1.2-4

higher water solubility

74

Pure solvent

Raffinate

One stage dilute extraction (mat bal stage 1) (L,V constant)

Assume y2 = 0 (Pure solvent) (Stages 2-n are not present)

xF L + y2 V = x1 L + y1 V Subst. y1= Kx1

x1/xF = 1/(1 + E)

Fractional recovery

As E ↑ x1 (less in raffinate) ↓, y1 ↑, fractional recovery ↑

11 F

F

x KV E y V / x L

x L 1+E

F F1

where E KV/L x L x

x Recall absorptionL KV 1 E

factor A mV/L

n 1yn+1y1,V

xF,L

2V yn y2

x1xn-1

extract

xn

75

Multistage extraction

Left stage (n): xn-1 = (1 + E) xn (1)

Next stage (n-1): ynV + xn-2L = xn-1(L + KV) (2)

Subst yn = Kxn and (1) into (2) to yield xn-2 = (1 + E + E2) xn

After n sequential material balances, we can show

xF = (1 + E + E2 +…En) xn

Geom series

(like 20.17)

This result is simpler than A method since the solvent was not pure in absorption.

n 1

F n

E 1x x

E 1

n 1 F

n

xE (E 1) 1

x

n 1 ln[RHS] / ln E

n 1yn+1y1,V

xF,L

2V yn y2

x1xn-1

extract

xn

76

Belter, Cussler, H u Bioseparations

Example 5.3-1 Actinomycin D

At pH 3.5, protonated non-ionic, K = 57

99% recovery. Calculate n

E = KV/L

R = Recovery

By total mat bal

xFL = xnL + y1V or 1 = xn/xF + R (definition of R)

xn = (1-R) xF (non-recovered material is in raffinate)

FL 450 /h x 260 mg/ feed

V 37 /h

1 F y V / x L

77

Problem V/L E=KV/L % R n y1/xF= RL/V type (input) (input) (input) concentrat. factor

Base 37/450 4.69 99 2.8 12case

Lower 37/450 4.69 90* 1.3 11recovery

Greater 100/450* 12.7 99 1.8 4.5Solvent

Less 10/450* 1.26 99 13.3 45Solvent

As V/L incr., L/V decr., n decr.

78

Adsorption/Fixed Bed Separation

Langmuir isotherm

For a given time interval from 0 to t*

Ideal breakthrough Kca/u0 → ∞Square wave – vertical profile (Pre-calc)Accum / Area IN – OUT

Non ideal breakthrough

max

W kc

W 1 kcs

W

g ads

g

b sat 0 0 0

adsorp. capacityt* L (W W )/u c

inlet flux

b sat 0 0 0L (W W ) u c t * 0 s3

s

gcm g

cm g

oC C

c

79

(Area above curve)

b vs. pdensity : external voids

q = b W (g/cm3) p: interval voids only

b = p (1-) → 0 b → p

→ 1 b → 0

b gives mass of solid in a given volume for adsorption

bt

b b 0 0 00L (W W ) u (c c)dt breakthrough

0 0 0

IN OUT

u c (1 c / c )dt

80

Summary of Adsorption

1. Ideal (fast mass transfer)

2. Irrev Adsorption W = Wsat for C > 0

bt

uo0

ct 1 dt

c

amt adsorbed

81

Choose c/c0,

b

sat

W LUBAs N , 1, ideal 1

W L

u bz t LV tN

L t * L

3. Linear adsorption Large N, Small HTU sharp

Small N Large HTU diffuse

pink blue

82

Ex 25.2

t*: time equiv of total capacity

Definition of Lb: used bed length (fully saturated)

b sat o

o o

o

0

L (W W )t*

u cin notes

(1 c / c )dt

bt

uo0

ct (1 )dt

c

ub T

t L L where

t *

usable capacity time equivalent

LT : actual physical length LUB = LT – Lb

83

uT T

b

tL LUB L

t *L

uT

LUBt / t* 1 25.4

L

u b TAs t t* LUB 0 L L (HTU 0)

8.5

o0

cp. 825 For 8-cm bed t*= (1- )dt = 4.79h

c

bt

u bo0

ct (1 )dt 2.37h t 2.4h

c

b

LUB2.4 / 4.79 1 LUB 4.04 cm

8

b T b satL /L W / W

b u bt / t* ~ t / t * o(small c / c )

if sliver on rt. negligible

compared to area of rec tangle

84

For 16 cm bed

Notice LUB did not change much (expt’l uncert). All we did was added a larger saturated zone.

Summary

LT

tb

tu/t* =

LUB

ut 7.07 LUB

0.74 1 t * 9.59 16

LUB = 4.2 cm

b

sat

W

W

8 16 32

2.4 7.1 16.7

0.5 0.74 0.87

~const�������������� �

85

Material Balance Including Mass Transfer Model

voids particles

Mass Transfer model:c*: eq conc. at surfacea: external area/volume

Dp = particle diameter De = effective diffusivity in pores (De: porosity, tortuosity, pore diameter)

p o o

cdL (1 )dL W / dt u c u (c dc)

t

3

cm g

cm s

p cdq/ dt (1 )dW / dt K a(c c*)

s3

s

g g

cm g s

2

3 3 3

cm cm g g

s cm cm cm s

cc,ext c,int c,ext e

D1 1 11/K

k k k 10 D

p 0

c W(1 ) u c / L (25.6)

t t

86

Water vapor, HC vapors – both resistances are important

Solutes in aq soln – slow diff in pores; often kc,int limiting

Surface area 100 – 1000 m2/g

87

Irrev Adsorption W = Wsat for all c

Accum. Fluid << Accum in solid, since isotherm is so strong (irrev)

c* = 0 for any W for irrev adsorption

L is end of profile and not Ltotal

(25.18)

No time dependence since wave has constant exponential shape as it moves thru.

*0 c u c / L K a(c c )

0 sat

c Lc

0c L

K adc / c dL

u

c0 sat

0

K aln c / c (L L )

u

88

From t1 to t:

Amount adsorb. = u0c0 (t-t1) = cross sect’l areaVelocity of mass transfer profile (wave)

= inlet flow/ads capacityFrom prev page,

Define MHS chooses

L = L(relative to Lsat),time is only variable

= N (1) – 1 see Figure 25.11

satsat pW (1 )L

sat 0 0z 1

p sat

u cv L /(t t )

(1 )W

c 0 0 cN K aL /u L /HTU HTU u /K a

co sat z 1

0

K aln(c / c ) (L L ) ( N/L)[L v (t t )]

u

zv t /L t / t *

0 1

1(see below)

N N Lln(c / c ) N (L ) t

L L t

89

Ways to think of t1 (time to establish sat. zone in first part of bed

For ideal adsorption (vertical wave): t*

Approach of MSH to describe t1

zt* L / v t / or t / t *

z1 z 1

v t Lt t * /N Optional: v t HTU

N N t / N

sat p sat p o

1c 0 z o o c

W (1 ) W (1 )L uequil capacity t Lt

K a(c 0) initial rate N v N u c K aL

1 ct as capacity , K a

90

Linear isotherm W = k1C

Solve 25.6 (material balance)

p. 827 25.7 (mass transfer rate)

Bessel fxns (p 833)

N = L/HTU = KcaL/u0

Kc = f (Kc,cxt, Kc,int) as before

As N ↑, HTU ↓, steeper

breakthrough curve

z b b u b b

satb

v t L t W t t~

L L t * W t * t *

91

For c/c0 =0.05 at tb:

Greater fraction of bed is used due to stronger isotherm.

bN

10 0.35Linear isotherm

20 0.5As N incr, better mass tran. - sharper profile

100 0.8Note: N( -1) = 1.6

IRREV :

10 0.84

92

Adsorption Chapter 15 p.783

Cylindrical pores (dp = pore diameter)

Sg Ξ specific surface area =

p ~ 0.5 p ~ 1 g/cm3 dp = 2 nm

Sg = 1000 m2/g

BET eqn. (Brunauer, Emmett, Teller)

Liq N2 -195.8°C (15-6)

Measure P, → Regress m

m = volume of monomolecular layer of adsorbed gas

p

p2p

d LS / V 4 / d

d L / 4

area

mass

pg

p p

1

pp

S particlevolume pore volume 4S

V mass particle volume d

pd

93

Sg = m NA/V V = 22, 400 cm3/mol

Molecular model for (2-D closest packed)

= 1.091 (MW(N2) / NA L)2/3

2 3 2

3

m cm (STP) molec mol m

molec g mol cm (STP) g

94

Ion Exchange

Henley & Seader Fig 15.5, 15.6McCabe C+ + Li+ R- = C+ R- + Li+

K’Li+ 1Na+ 1.98K+ 2.90 - less hydrated, greater charge/volume, stronger ion pairing

Regeneration – Add salt or change pHUltrapure water (low conductivity) anion + cation exchange resinsE.g.

Ca+2 → 2H+

Cl- → OH-

iC R Li

C Li R

(C )(C ) neglecting ' sK '

(C )(C ) if dilute

4NH4

Li

K ' 2.55Let C NH K '

K ' 1

95

MembranesMcCabe, Smith, Harriott

Porous membranes: Knudson diffusion through small pores

DA = 9,700r(T/MA)0.5 [cm2/s]where r is pore radius in cm.

DeA = DAwhere is the tortuosity.

JA = DeA(cA/z) = DeA([pA/RT]/z)

Dense non-porous uniform membrane layers where gases do not interact with each other,

JA = -DA(dcA/dz) = DA (cA1 – cA2)/z

Equilibrium (ideal solution) HAcA = pA or cA = pASA

SA = solubility coefficient

cA

pA

Gas-liq equilibria

Slope = SA = 1/HA

96

JA = DASA (pA1 – pA2)/z. Let qA = DASA (permeability coeff.)

qA units: 1 Barrer = 10-10 cm3 (STP) cm-2s-1 / (cmHg cm-1)

Flux / (pressure/ length)

JA = qA (pA1 – pA2)/z = QA (pA1 – pA2) where QA is permeability.

QA is used when z is unknown and qa cannot be determined.

For a given pA1 – pA2

(QA/QB) = (DASA/DBSB) See Fig. 26.2.

If S values are similar as in case of O2, N2 then DA/DB can play a large role as shown in Table 26.1. See Fig. 26.4.

97

Product purity and yield

Countercurrent Flow Membrane

Assume asymmetric membrane and uniform distribution of feed with negligible boundary layer resistance. Assume no axial P in shell side and tube side.

x, y refer to component A. At a particular axial position

y is average composition of gas (all permeate to this position)

y’ is the local composition of gas (see Fig. 26.6)

y = y’ at the closed end of the tube with only new permeate

R = P2/P1 = QA/QB y’ = JA/(JA + JB)

JA = QA (P1x – P2y) = QAP1 (x – Ry)

JB = QB (P1[(1-x) – P2[1-y]) = QB P1[1-x – R(1-y)]

98

Subst. above eqns to get local permeate comp. y’= JA/(JA + JB)

y’ = QAP1 (x – Ry) /{ QAP1 (x – Ry) + QBP1 [1-x – R(1-y)] }

= f (x,y, = QA/QB, R) since P1 cancels out.

As x increases, JA incr. and JB decr., thus y’ incr.

Upper limit on y’ so that driving force in partial pressure is positive :

P1x > P2y’ = P1Ry’ or y’max = x/R (see Fig 26.7)

At R = 1, y’ = x and there is no permeation and no separation since there is no driving force in expression for JA.

Membrane area may be obtained from flux for more permeable gas

A = Voutyout/QA(P1x – P2y)average

At closed end, y = y’; therefore, y’ = f(, x, R) (26.17)

99

Reverse Osmosis

equilibrium

cw = 1/vw

Desalinization, Removal of heavy metal ions, sulfites in pulp effluent, recovery of sugars, etc. in wastewater, municipal water (salts, bacteria) amino acid concentration

ww

PJ ( P )

z

1 1 2 2(c ) (c )

wIf P J 0

w w ww

c D vP

RT

s s s sJ D S c / z ww

w

c (membrane)S

c (soln)

wc sol' y in membrane

ss

s

c (membrane)S ~ 0.035

c (soln)dense polymer films

c(right) > c(left)

100

Osmotic pressure at equilibrium at constant T

wo(P) = w(P+)

= wo(P+) + RT ln xw

Superscript o means pure water.

In limit xw approaches 1, approaches 0.

wo(P+) - w

o(P) = -RT ln xw

Vw = -RT ln (1 - xs)= RT xs

s ss

w

n nRT RT c RT

nV V

Pure water

Water and salt, cs

Hydration of salt s pulls in water

101

18.2.2 Osmotic Pump

Fig 18.2.1 – constant flux for osmosis: incoming water pushes drug out hole with zero order release

The brine solution remains saturated as water comes in due to excess salt.

P = permeability

0.5 solubility coeff

Since Jw ≠ f(t)

drug release rate is zero order.

See Ex 18.2-1

p. 473 Cussler, Diffusion

w s

PJ A A c

w

sw

Note c RTV

102

Reservoir system polymer or liposome

Permeation through

membrane

Avoid burst + rupture

Do Ex 18.1-1 p. 471 Cussler

1

PM N At c (sat)At zero order releaselinear in time!1 l

M/ t const P D S

solubility coeff membrane

1s1

cS

c

103

Crystallization, Interfacial Phenomena, Nanotechnology

Surface Tension A liquid draws into a drop to minimize surface area (surface is in tension) Therefore, surface has higher energy than bulk. Molecules are drawn inwards for approximately three molecular diameters.

Move barrier by x TOP VIEW

dW = FdxdG = Ldx (rev work)Fdx = Ldx or = F/L Langmuir trough (top view) units: mN/m = dyn/cm = erg/cm2

Surface tension or line tensionSurface free energy

free energy

dG dAA

L F

dx

dx

104

Surface tension values at 20 oC

Water 72.94 Hg 486.5

Ethanol 22.39 Ag 878 (1100°C)

Octane 21.62 Pt 1800 (mp)

Perfluoropentane 9.89

Pendant drop tensiometer (shown here)

Spinning drop tensiometer (centrifugal forces)

(UT- Wade and Schechter)

surface tension favors a sphere

Gravity produces Elongation

105

Laplace Pressure P = Po – Pw > 0

Consider ↑ in r of a sphere at const T:

A = 4 r2 dA = 8 r dr

V = 4/3 r3 dV = 4r2 dr dA/dV = 2/r

dW = - PodVo - PwdVw + ow dA = 0 at mechanical equilibrium

Since dVo = -dVw

(Po – Pw) dVo= ow dA

P = Po – Pw = 2 ow/r

Laplace P

Since > 0 dG = dA surface tension raises G. Molecules pull towards the inside.

P > 0: curving interface raises P on inside; raises due to surface tension →P↑

2P

r

wP

r

oP

dr

2P

r

106

Laplace P becomes important for micron sized drops

r(nm) (mN/m) P (MPa) P(bar)

1,000 50 0.1 1

100 50 1 10

10 50 10 100

10 0.5 0.1 1

9 6

2

2 mN 10 nm 10 N = = MPa

r m nm m m

107

Kelvin eqn Solid-Liquid (SL) or Liquid-vapor (LV) equilibria

Solid cluster formed in case 1. Liquid cluster formed in case 2.

Equilibrium – Objective: solubility in soln.: y = f (r) because of Laplace P

curved droplets Bulk or “flat” particles

r , y > ys (shown later) r →∞ , ys

P = 2/r P → 0

Higher s since P ↑

State function- change in around the thermo. square is zero:

SM

SM

s ssd V dPV P(pure) (pure)

L LL

s

dy(curved)RTlny (bulk)

RTdlny(in solution) (in solution)

= =

108

sM

RTlnP 2 / r

V Kelvin eqn

sMln 4V / RTL (McCabe) L 2r

In that case 1 ion /molecule

s

s s

=fractional supersaturation

y yy 1 1 S

y y

Implications of Kelvin eqn.

y = f(r): As r ↓ y ↑ small crystals are more soluble since large fraction of atoms at surface raises μs (Laplace P)

= f(r): A higher supersaturation = y/ys is needed to keep smaller crystals from dissolving

μs (small crystals) > μs(large crystals)Ostwald ripening – mass transfer from small crystals to large crystals to lower μ.

Nanocrystals will grow to reduce surface area unless stabilized with ligands.

Supersaturation in temperature or concentration on phase diagram (Fig 27.5)Tc = Ts – Tc = k (y – ys) where k is slope of T vs y line

109

Classical Nucleation TheoryAdamson, Gast. P Chem Surfaces pp. 328-332

Cluster of nmolecules: size rNucleation: V → L or L → S

If = GV – GL (+); transfer from V → Lper molecular favorable (-)(or L → S for solid cluster)

Nucleation of cluster of radius n:

G(nucleation) = G(bulk transfer) + G (surface)

G

0

24 r

3n r

V

L

Favorable bulk driving force

Surf. energy penalty

r or n1/3

110

For a spherical nucleus or cluster,

3AV

4ˆ ˆn r (3) N /M number density

3n = number molecules in a cluster

g molec. mol molec.

vol mol g volume

(L) (S)

satV LcondensationG n n(G G ) P P

2

( )

n 4 r (1)

favorable bulk change penalty to create interface

satsnkT ln P /P or nkT ln y / y (2)

1 1

111

Max G

As ↓, or that is ↑rc decr. and it is easier to pass rc

in 4r2 term.

large surface area/volume : interfacial penalty dominant for small nucleiiBarrier to form critical radius = Gmax

Once r is greater than rc, it is easy for the nucleus to grow(downhill).tiny amt of catalyst can lower barrier(cloud seeding)– heterogeneous nucleation orders of magnitude faster than homogeneous nucleation

2 ˆd G/ dr 4 r 8 r 0 (4)

From(4)

c

2r (5) Subst(5) (3)

ˆ

3

c 2 3

32n

ˆ3

G

barrier

cr1/3r or n

Bulk term eventually wins due to r3 or small surface area/volume

112

Rate of nucleation (Adamson & Gast) text (optional)

Frequency factor from kinetic theory of gases

Barrier ↓ as ↓, ↑faster nucleation

1/ 2

0 max max

c

G GZB exp Becker 1935

n 3 KT KT

2max c cG n 4 r (classical theory)

23

2 3

32 2 4

ˆ ˆ3

3

2 2

3216

ˆ3

A A MˆN R /k N / V 23

m2 2 2 2

A

V16

3 N k T (ln )

3 2

0 M2 2 2

16 V B Cexp where C is freq. factor

3 R T (ln ) kT

3 2M A

3 2

16 V NCexp 27.9

3 (RT) (ln )

0, : B

113

Crystal Growth – releases heat at surface

Diffusion to surface sp: surf. area cryst.

Surface rxn (definition of ks) layer bylayer incl. dislocations

NA = ks (y’ – ys)Sum of resistances

For invariant crystals (same growth in all directions)

vp = aL3 moles: m = aL3 M

sp = 6vp/L = 6aL3/L

= dm/dt = 3 a L2 M dL/dt where G = dL/dt

G = 2K (y-ys) / M

A yp

mN k (y y ')

s

p s y s

m 1 K

s (y y ) 1/k 1/k

diff rxn

m

bulksy y ' y

2M

2s

3aL G K

6aL (y y )

2p p

3p p p

s D 6

v D / 6 D

114

Continuous Stirred Tank Crystallizer (ChE 372 Reactor)

Mixed Susp Mixed Prod Removal Cont Stirred Tank Crystallizer

Steady State, uniform supersat., L = Gt, G ≠ f (L)

Start: p934 Table 27.2 See Fig. 27.14

N/V = xn cumulative N/V

d(N/ V)n population density

dL

L

0

N/ V ndLN/ V ndL, between sieves Land L dL

nN/V

Sum of each bar in n yields N/V.

115

For crystals from L to L+dL

d(N/V) = ndL = differential number/volume in this size interval

(between sieves)

Withdraw a fraction of the crystals in dL size interval: dndL

(Here dL is const) Q is volumetric flow rate leaving crystallizer

The crystals grow during dt: dL = Gdt

N~

V

c

dndL Qdt Volume removed

ndL V total volume

c c

dn Gdt Qdt -dn Qn or

ndL V dL GV

0

n L o

0n

dn 1 n LdL ln z

n G n G

cLet V / Q 0 zn n e

116

lim d(N/ V) dL d(N/ V)

L o dt dt dL

j Cumulative Name Differential

0 0 number dist xn=N/V d0/dz

1 size dist xL

2 area dist xa

3 3 mass or volume dist xm d3/dz

Fig 27.14 dxm/dz = 0 where distribution changes the fastest

Kineticcoefficients(const volume basis)Thus B0 = Gn0 B0 G n0

Nucleation rate size where nucleii are formedThus, no=Bo/G where both Bo and G are known. Larger no for large Bo and

small G. High nucleation rate and slow growth.

z zj j z

0 0j

j j z

0 0

nz dz z e dz

nz dz z e dz

Normalized jth moment (cumulative 0 to z)

Integrate slope of cumul. dist to determine number, size, mass or volume cumulative values

117

Total number of crystals/mass Recall: z = L/GdL = Gdz

greatest change in

cumulative dist for mass Fig 27.14

0 z 0c

10 0

1n N/ V ndL n G e dz n G 1 1

4 0 3 z 3c c c

60 0

m mndL a (G ) n z e dz where m =aL

c

3c c

n 1

m 6a (G )

predomL 3G

0c c

c c

m mproduction rate #nucleii mass B

V vol. mother liq. n vol time no.cyrstals

c3

c c pr

n 9

m 2a L

oc c c3

c c c pr c

ratio is fxn of G above (27.44)n m m (9)B

m V 2a L V

Here we assume one nucleus produces one final crystal

0 zn n e

00 0 o

pr c L

BGiven G, L , Q, m / V B , n ; from n pred. size distrib. (ex.27.6)

G