chapter 3. d concentrated solutions · 2019. 10. 15. · mass transfer –diffusion in concentrated...

Mass Transfer – Diffusion in Concentrated Solutions 3-1

Chapter 3. DIFFUSION IN CONCENTRATED SOLUTIONS

Diffusion causes convection in fluids

Convective flow occurs because of pressure gradients (most

common) or temperature differences (buoyancy or free or natural

convection). However even in isothermal and isobaric systems,

convection can occur due to diffusion.

Maxwell (1860) said: “Mass transfer is due partly to the motion of

translation and partly that of agitation.”

Diffusion and convection always occur together in fluids.

3.1 Theory

Mass Transfer – Diffusion in Concentrated Solutions

Example: Evaporation of Benzene:

3-2

At 6°C the benzene vapor is dilute and

evaporation is limited by diffusion.

At 80.1°C benzene boils (p = 1atm).

Evaporation is controlled by convection.

At 60°C an intermediate case occurs in which

both diffusion and convection are important.


Separating Convection from Diffusion

3-3

Assume that these two effects are additive:

convection by

dtransporte mass

diffusion by

dtransporte mass

dtransporte

mass total

If the total mass flux is n1, the mass transported per area per time

relative to fixed coordinates:

111 vcn

where v1 is the total (= effective) solute velocity (velocity due to

convection and superimposed diffusion).


The total average solute velocity can be split into one part due to

diffusion and one due to convection, called reference velocity va:

convection

a

1

fluxdiffusive

a

1

a

1

a

111 vcjvcvvcn

The art is to select va in such a way that the convection term is

simplified (or, ideally, even va=0).

For example, in a flowing solution, va is the velocity of the solvent

because the solvent is usually in excess so its transfer is minimal (in

other words the difference in solvent concentration is too small

across the solution). That way we eliminate convection and deal with

a SIMPLER problem.


Two-bulb apparatus (Diaphragm-

cell) for understanding different

definitions of reference velocities.

Volume average velocity = 0

Molar average velocity = 0

Mass average velocity ≠ 0

Volume average velocity = 0

Molar average velocity ≠ 0

Mass average velocity ≈ 0


For gases (e.g. H2 and N2) at equal T and p the number of

moles is always the same in both sides because the volume is

the same in both sides. As a result, the

v0 = 0 volume average velocity

v* = 0 molar average velocity

v 0 mass average velocity,

because the masses of N2 and H2 are different.

As a result, as time goes by the center-of-mass in the two-

bulb apparatus moves away from the bulb containing N2

initially. Thus the mass average velocity v is not zero.


For liquids: The volume is nearly always constant.

v0 = 0 volume average velocity

v = 0 mass average velocity. This is usually correct as liquid

densities differ little.

e.g. H2O=1 g/cm3

Glycerol=1.1 g/cm3

However, the molar concentration is usually quite different following

large differences in molecular weight.

e.g. MWH2O=18 g/mol and

MWGlycerol=92 g/mol

So v*0 molar average velocity for liquids.

In conclusion: For gases use as reference velocity va the volume-

average v0 or molar-average v*, while for liquids use the volume-

average v0 or the mass-average v.


3-9

ω i : mass fraction of species i y i: mole fraction of species

iic V : volume fraction of species i, : partial specific volume

Precisely: Partial specific volume:jmTpi

im

VV

,,

Partial molar volume:jnTpi

in

VV

,,

The partial specific or molar volume expresses how much a volume changes

upon addition of a certain mass or number of moles of a given component.

iV

Partial molar volume for ideal gases (pV = nRT) :

cp

TR

n

p

RTnn

n

VV

nTp

nTp

1

2

2

,,

1

21

,,11


3.2 Examples for Parallel Diffusion and Convection

3-10

Example 3.2.1: Fast diffusion through a stagnant film

Goal: Calculate the flux and

the concentration profile

Now both diffusion and

convection are important!

Remember that at intermediate

temperatures both diffusion

and convection affect the

evaporation of benzene (or any

other solute).


1. Step: Mass balance

zzatout

dtransportesolute

zatin

dtransportesolute

zAvolumein

daccumulatesolute

zz1z11 |An|AnczAt

Divide by Az and as volume 0

z

nc

t1

1

At steady state:

11

n0 n const.

z


Now the flux is affected by both diffusion and convection. For

simplicity we choose va = v0 (volume-average velocity)

)vVcvVc(cdz

dcDvcjn 2221111

10111

2. Step: Choose and simplify mass transport equation

Also, n1 = c1v1 and n2 = c2v2

The total average flux of the solvent (air) is zero (it seems to be

stagnant), since it cannot penetrate into the liquid phase and does

not accumulate.

Therefore n2=0 and v2=0.


So 1111

1 nVcdz

dcDn

If the vapor is an ideal gas, then 1

11 1 1

c1c V = c = = y

c c

dz

dcD)Vc1(n 1

111

dz

dyDc)y1(n 1

11 (1)

3. Step: Boundary conditions z=0: y1=y10

z=l: y1=y1l

(2)

(3)

or

and


Solve eqn. (1) subject to BC’s to determine n1

10

11

y1

y1ln

Dcn

(4)

Note that doubling the concentration difference DOES NOT double

the flux, as in dilute systems.

Integrating eq. (1) also for z=0 to z and y1=y10 to y1 and

considering that n1 does not change with height z as the cross-

sectional area does not change gives:

z

10

1

10

1

y1

y1

y1

y1

(5)Concentration profile


10

1

z

10

110

y1

y1ln

y1

y1y1Dc

dz1

dyDc

1j

(6)

Diffusive Flux of benzene

Now does this result (eqn. 5,6) reduce to that for dilute solutions?

Expansion into series for small y (dilute system small conc. y):

32a

y!3

2a1aay

!2

1aaya1y1

(7)

(8)

ya1y!3

2a1aay

!2

1aaya1y1 32a

Here a=1: y1y1

1

0 0


(9) y3

y

2

yyy1ln

32

Let´s apply eqn. (8) to eqn. (5)

z

101

z

101

10

1 yy1y1y1y1

y1

z

101101 yy1)y1(y1

1101010110

yyz

y1yz

yz

1y1

(10)

with approx. (7) becomes


If we rearrange and multiply both sides of eqn. (10) with c

)cc(z

cc 101101

(11)

Likewise for the flux from eqn. (4)

)cc(D

yyDc

)y1ln()y1ln(Dc

n

1101019Eq

1011

Eqn. (11) and (12) are identical to the dilute limit ones!

(12)


Example 3.2.2: Calculate the error associated with the neglect of

diffusion-driven convection when estimating the evaporation rate of

benzene @ 6°C and @ 60°C.

a) At 6°C the saturation vapor pressure is p1(sat) = 37 mmHg

Mole fraction 049.0760

37

p

)sat(p

c

cy 11

10

Total flux at steady-state for concentrated solution:

cD

05.0049.01

01ln

cD

y1

y1ln

cDn

10

11

Total flux for dilute solution:

cD

049.00049.0cD

yycD

jn 11011 Only 2% error!


b) At 60°C the saturation vapor pressure is p1(sat) = 395 mmHg

760

395y10 Mole fraction

Concentrated solution:

cD

73.0760/3951

01ln

cD

y1

y1ln

cDn

10

11

Dilute solution:

cD

52.00760

395cDyy

cDjn 11011

There is 40% error!!!


10

11

y1

y1ln

Dcn

z

10

1

10

1

y1

y1

y1

y1

10

1

z

10

110

y1

y1ln

y1

y1y1Dc

dz1

dyDc

1j

(6)

Physical picture

(4)

(5)

y1


Example 3.2.3: Hydrogen production by catalytic cracking of CH4

Methane gas is cracked at the surface of a solid catalyst forming

hydrogen and a solid carbon deposit.

Goal: Total methane (molar) flux per unit area at steady state?

n1 Catalyst surface

n2

z 0

CH4

2H2 Carbon deposit

CH4(g) → C(s) + 2 H2(g)

Mass Transfer – Diffusion in Concentrated Solutions 3-223-22


zzatout

dtransportesolute

zatin

dtransportesolute

zAvolumein

daccumulatesolute

zz1z11 |An|AnczAt

Divide by Az and as volume 0

z

nc

t1

1

At steady state:

11

n0 n const.

z


Note: For processes with chemical reactions, it is best to use the

molar flux and the molar average velocity!

Thus, from Table 3.2.1:* *1

1 1 1 1

dcn j c v D c v

dz (1)

with * 1 2 1 21 1 2 2 1 2

c c n nv y v y v v v

c c c

(2)

Now 1 mole of CH4 gives 2 moles of H2, flowing in the opposite

direction. Therefore,

12 2 nn in eq. (2):c

nv 1* and 1 1

1 1

dc nn D c

dz c



Using that:1

1

cy

c

11 1 1

dyn D c y n

dz 1

1 11dy

n y D cdz

→ (3)

B.C.: z = 0: y1 = 0 (due to decomposition)

z = L: y1 = y1,L (some measured conc. at L)

Integration of (3) subject to B.C.s yields: 1 1ln 1 L

Dcn y

L

or, the general form if y1,0 ≠ 0: 1,

1

1,0

1ln

1

LyDcn

L y


Example 3.2.4: Fast Diffusion into Semi-Infinite Slab

A volatile liquid solute evaporates into a long capillary

Initially the capillary contains no

solute. As the solute evaporates

the interface between the vapor

and the liquid solute drops.

Goal: Calculate the solute

evaporation rate accounting for

diffusion-induced convection and

the effect of moving interface.


outtransport

solute

intransport

solute

zAin

ionaccummulatsolute

zz1z11 nAnAzAct

Divide by A z and as z 0

z

n

t

c 11

(1)

There is no solvent (air) flow across the capillary, blowing the solute

away. As a result, the solute accumulates in the capillary.



n1 = j1+c1 v02211222111

0 VnVnvVcvVcv with

In the unsteady case, the solvent flux varies with position and time

but the solvent gas does not dissolve in the liquid, thus at the

interface (z=0): n2 = 0.

012

12

1vc

zz

cD

t

c

In (1):

0z1

0z111z

cDnVc1

(2)

(3)

z

c

Vc1

z

cVD

z

cD

t

c 1

0z

11

11

2

12

1

(4)



t = 0 z > 0 c1 = 0

t > 0 z = 0 c1 = c1(sat)

z = c1 = 0

Boundary conditions:

Define combined variable:

(as in the dilute case) tD

z

4

0d

dc2

d

cd 1

2

12

(5)

with B.C. = 0 c1 = c1(sat)

= c1 = 0


where

0

11

11

Vc1

cV

2

1

(6)

In eq. (5) is a dimensionless velocity characterizing the

convection by diffusion and the movement of the interface. Note

that if = 0 the problem reduces to that of diffusion in dilute

concentrations !!

Eqn. (5) is integrated to give:

21)(expttancons

c


2nd integration and insertion of B.C.:

erf

erf

satc

c

1

1

)(1

1

(7)

eqn. (6) (7):

1

211

experf1

11)sat(cV

Calculate now also the interfacial flux (see eq. 3)

0z

11

1

4.eqn

0z11Vc1

z

c

DnN

erf1

exp

)sat(cV1

1)sat(ct/D

2

11itlimdilute

1

(see next

figure)

chapter 3. d concentrated solutions · 2019. 10. 15. · mass transfer –diffusion in concentrated...

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