1 363 Class Notes Keith P. Johnston Chemical Engineering
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.
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Cussler Chapter 2
2.2 Steady state diffusion across a thin film
Fig. 2.2-1 Fig. 2.2-2
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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
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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
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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)
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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)
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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
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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 )
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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 )
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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.
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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
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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!
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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
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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
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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)
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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
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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
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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.
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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)
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-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.
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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
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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
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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.
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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.
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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
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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)
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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
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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
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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
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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