mass transfer lecture 09. dispersion vlasis g. mavrantzas ... · bibliography material taken from:...

Mass Transfer

Lecture 09. Dispersion

Vlasis G. Mavrantzas & Sotiris E. Pratsinis

Particle Technology Laboratory, Department of Mechanical and Process Engineering, ETH

Zürich, CH 8092 Zürich, Switzerland

Mass Transfer, Fall 2019 1

Bibliography

Material taken from:

1. E.L. Cussler, Diffusion – Mass transfer in Fluid Systems, Cambridge

Univ. Press, 3rd Ed., 2007 – Chapter 4.

2. Chapter 4: Dispersion

2

1. Introductory Concepts and Examples

Dispersion = to spread widely

• Dispersion is related to diffusion on two very different levels

• First, dispersion is a form of hydrodynamic mixing, and so on a molecular

level it involves diffusion of molecules.

• This molecular dispersion is not understood in detail, but it takes place so

rapidly that it is rarely the most important feature of the process.

• Second, dispersion and diffusion are described with very similar

mathematics.

• Analyses and tools developed for diffusion can often correlate results for

dispersion.


• Paints: pigments are dispersed in water or solvent

• Fog, clouds: water droplets are dispersed in air

• Sooting candle: soot particles in air

• Pollution of rivers, air, ...

• Spreading of diseases, hamsters,...

Mass Transfer, Fall 2019


4


• Dust Clouds from Africa

Global travel of dust carries microbes across oceans and continents.


5


• Pollen Clouds

In 2006, birch pollen from

Denmark traveled across the

North Sea to England.

Water droplets dispersed in air appear as fog/clouds

(and they create rainbows)

• Atmospheric Clouds


6

Suspensions and Emulsions

Paint, for instance, is a dispersion of pigments in a solvent.

Homogeneous dispersion of the pigments in the fluid as well as long-term

stability of the dispersion (shelf-life) are important.


7Mass Transfer, Fall 2019

Soot, smoke

Condensation plumes for 500 ft and 250

ft stacks in Salem (MA), showing the

complex thermal structure in the lower

atmosphere.


8

• Forest fire in southern France (Le Boulou) near the Spanish border, July 2017.

• v0 ≈ 60 km/h.

• Photo: L‘ Indépendant (J. Gallardo)

• Smoke plume of a forest fire follows the dispersion pattern of this pulse decay

9



• On the microscopic level: dispersion

involves diffusion (of droplets, particles,

molecules,...).

• On the macroscopic level: dispersion is

governed by fluid dynamics (laminar,

turbulent flow, eddy formation). The

mechanisms causing mixing and dilution

e.g. of the plumes are hydrodynamic.

• Governing equations are very complex

and an exact solution is impossible,

especially if the variability, e.g. of the

weather is included.


10

A prototype problem for

understanding how plumes

are emitted!

Convection in x

Dispersion in y


2. The dispersion equation

Example: Dispersion from a stack

• A point-source active only at time t

= 0, e.g., emission pulse of smoke

from a stack

• Decay of a pulse in the presence of

convection

• Recall problem addressed in

Lecture 5:


t 1

t 0

t 2

t 3

x=0

x=0

t 0

• It is impossible to calculate the actual concentration at a certain point

for a certain time but we try to calculate the average concentration.

• We try to apply the mathematics of diffusion to dispersion.

Elements of Brownian Motion

t 1

t 0

t 2

t 3

x=0

x=0

t 0

2

1 1

2

, ,

c x t c x tD

t x

Governing equation:

Initial and Boundary conditions:

01 0

1

1

0,

, 0

, 0

Nt c x x

A

x c t

x c t

Solution:

Width of the peak:

2

01 , exp

44

N xc x t

DtA Dt

4L Dt


Extend analysis to the prototype problem

D = 10-5 m2 /s

L = 10 km

v0 = 15 km / h

L 0.3 m

4L Dt

• But measurements show that after about 40 min, a plume has spread

1 km!

• Use the dispersion coefficient E instead of D!

• The dispersion coefficient E must be determined experimentally and

typically has different values in different directions.

13

t = 40 min


2. The dispersion equation

1 11 1

c cj D j

z zE

• Same mathematical form

• D = actual diffusion coefficient

• E = dispersion coefficient (like an effective

diffusion coefficient) caused by wind

• Generalized balance equations keep the

same mathematical form as before


2. Dispersion Coefficients

• Dispersion coefficients: are very different between turbulent and laminar flow

• For turbulent flow: we expect that the dispersion coefficient E should be a

function of the fluid’s velocity v and some characteristic length l:

cons vE t l

• For turbulent dispersion in pipelines of diameter d:1

2E d v

• For laminar flow in pipelines of diameter d:

2 21

192 DE

d vD

• A more general expression (e.g., for packed beds):1 2D d vE


2. Dispersion in one or more dimensions

• Prototype problem of dispersion: The source of the plume is at (x=0, y=0) and

wind is blowing at speed v0 in the x-direction. We want to know the average

pollutant concentration at (x, y):

• Based on the above general equations, a variety of problems can be addressed.

2

01 0 0

0 00 0

; exp4 /4 / yy

N yc y x v t

v EE x vA x

2

01 0, , exp

44

N yc x y t

DtA Dt


• And dispersion in

the x-direction:

2

001 0 0

0 00 0

; exp4 /4 / xx

x xNc x x v t

x E vA E xv

4. The example of the chemical spill

• A container with chemicals breaks, accidentally releasing its content into

a river flowing at v = 0.6 km/h.

• When experts of the environmental protection agency arrive to take water

samples, the maximum concentration of 860 ppm is located 2 km

downstream the release point. 50 m from the maximum, the concentration

is 410 ppm.

• How large is the dispersion coefficient E?

• What will the maximum concentration be 15 km downstream?

t = 0

initial spill

v = 0.6 km/h

t1, s = 2 km: c1,max = 860 ppm



An elementary analysis

• Treat dispersion as (Brownian) diffusion along one dimension only (here both

the x and y directions coincide with the s direction) as the decay of a pulse

• Equation to use:

• Can be also written as:

t = 0

initial spill

v = 0.6 km/h

t1, s0 = 2 km: c1,max= 860 ppm

0

0 00 0

2

0

1 0 0; ex/

p44 /

s sNc s s v t

E s vE s vA


2

0

1 0 0 1,ma

0 0

x 0; exp/4

s sc

Es s c

vv t s

s



• Equation considered:

• So, 50 m away from the maximum:

• Time

t = 0

initial spill

v = 0.6 km/h

t1, s0 = 2 km: c1,max= 860 ppm

0

2

0

50 m410 ppm 860 ppm exp

4 /E s v

0

0

2 km3.33 hr

0.6 km/hr

st

v


2

1 0 0 1,max

0 0

0; exp4 /

sc s s v t c s

E s v



• Solving for E and substituting, we find E = 700 cm2 / s !!!

• Compare with D in liquids: 10-5 cm2 / s! Dispersion is very rapid!

t = 0

initial spill

v = 0.6 km/h

t1, s0 = 2 km: c1,max= 860 ppm

• Also note:

• Thus:

0

0m1, ,max 1 ax

00

1

e

1

distanc/4

Nc

s vAc

E s

1

2 km15 km 860 ppm 314 ppm

15 kmc


The problem

• A 3 km long pipeline with 10 cm diameter is used to transport different

gases from a storage area to a chemical reactor. The gas velocity is 5 m/s.

• After switching from gas A to B, how much will the gases mix?

• First, let us check the condition of flow

• Reynolds number:

• Thus, flow is turbulent!

AB

z

r

v = 5m/s

3

6

kg m1 5 0.1m

m sRe 50,00010 10 Pa s

v d

21

5. Dispersion and Turbulent flow in a gas pipeline

z


Plug flow profile!

Assumptions:

• Well-mixed radially

• Concentration changes only axially!

Mass balance for a point at interface moving

with v = 5 m/s, corresponding to either gas A or

gas B:

Boundary conditions:

2

1 1

2

c cE

t z

1 1,

1 1,0

1 1,

0, all 0 :

0, 0 :

0, :

t z c c

t z c c

t z c c

,1,0 1

1: average radial concentration of the two gases at interface

2cc

AB

z

r

v = 5m/s

22


z



Solution: similar to the general semi-

infinite slab solution!

Dispersion coefficient for turbulent

dispersion in pipelines:

In our case, this gives:

Exercise: compute z so that we have a

significant concentration change:

1 1,0

1, 1,0 4

c c zerf

c c Et

1

2E d v

2 21 m10 cm 5 0.25 m /s = 2500 cm / s

2 sE

1 1,0

1, 1,0

1 1, 1,0 1 1, 1,0 1 1,

1 1 0.844

10.84 1 0.84 0.84 1 0.84 0.92

2

c czerf

c cEt

c c c c c c c c

2m 3000 m4 4 0.25 24 m

m0.5

s

E ts

z

AB

z

r

v = 5m/s

23

z

6. Laminar flow – Taylor dispersion

A typical example of dispersion: the spread of a solute pulse in steady laminar flow

The solution is dilute

The flow is always laminar (no axial change in velocity)

Mass transport is by axial convection and radial diffusion only

Goal: Prediction of the dispersion coefficient E!

We can do an analysis using the following assumptions:

Injection of solute pulse

vzR0



• Mathematical tool: Generalized mass balance in cylindrical coordinates

• Considerations – Assumptions: Cylindrical symmetry, axial flow, radial

diffusion

• Simplified governing equation:

• Need the velocity profile for laminar tube flow:

• Here:

• Need also to apply Fick’s law in the radial direction:

1, 1,11, 1

1 1 z

r

n ncr n r

t r r r z

11

1,

1 z

r

c vcr j

t r r z

2

0

0

2 1z

rv r v

R

a0 m x

1average velocity

2vv

1

1,

, ,r

c r z tj D

r



• Final equation: Generalized mass balance in cylindrical coordinates for

• Boundary conditions:

• To solve it: It helps if we transform to a moving reference frame!

2

1, 1,10

0

2 1r zc cc D r

r vt r r r R z

0 01 2

0

1

10

0, all , : 0

0, 0 : 0

0, : 0

N Nt r z c z z

A R

ct r

r

ct r R

r

1 , , :c r z t



• Assumption: Let us assume that radial variations of c1 are relatively small

compared to axial ones.

• New coordinates:

• New unknowns: At every z, we work with the average radial concentration and the

radially-averaged flux in the flow direction:

• Then, mass balance in the moving reference:

0

0 1 1

01 1 12

00 0

, , 21

, , , 2 2 , ,

r R

r R r

r

r r

c r z t r dr

c z t c r z t r dr r c r z t drA R

0

1 0 12

0 0

12

r R

z

r

j r v v c drR

27

0 01 1

0 0 0

, , , , ,

z v t t vr

r c r z t cR R R

2

0

0

2 1z

rv r v

R

1 1

0

1

c j

v


• Governing equation:

• Substituting:

• Boundary conditions of new equation:

• It reminds us of the Brownian problem (Lecture 5)!

1

1

0 211

0

14

2

r

r

j

vcr r c dr

2

0 01 1

248

v Rc c

D

0

01 2

0

1

1

0, all :

0, : 0

0,

1

0 : 0

Nc

R

c

c

R


Recall Previous Problem (Lecture 5)!

t 1

t 0

t 2

t 3

x=0

x=0

t 0 2

1 1

2

, ,c x t c x tD

t x

Governing equation:

Initial and Boundary conditions:

01 0

1

1

0,

, 0

, 0

Nt c x x

A

x c t

x c t

Solution: 2

01 , exp

44

N x

Dc t

t Dx

tA



• Correspondence with the New Problem:

• Thus, new solution:

•

0 0

1 1

48

, ,

t

x

v RD

D

c x t c

2

01 , exp

44

N xc x t

DtA Dt

2

01

0 000

0

, exp

444848

Nc

v Rv RAR

DD

2

0A R



• We can go back to variables t and z:

• Solution:

• Compare with:

• We find the dispersion coefficient: !!!!

2

001

2 0 0 02 0 0 000

00

0

, exp

444848

R

z v tNc z t

v R tvv R tv RRD RD R

0 0

0 0

, z v t t v

R R

2

001 0 0 2

0 0 00 0//

; exp44

z zNc

E z vE z vz z v t

R

2 2

0 0

48

R v

DE


00v t z


Dispersion Coefficient:

Original derivation: Sir G. Taylor, "Dispersion of soluble matter in solvent

flowing slowly through a tube", Proc. Royal Soc. A 219, 186-203 (1953).

Remarks:

• E is inversely proportional to D!

• Rapid (radial) diffusion leads to small (axial) dispersion!

2 2

0 0

48

R vE

D


7. Laminar flow – Taylor & Aris dispersion

Correction for axial diffusion:

• Governing equation:

• After solving it, we get:

• But now

2

0 11 1 1

2

1v cc c cD r

t z r r r z

Axial diffusion

2 2

0 0

48

R vDE

D

2

001 2

0

, exp44

z zNc

Ez t

tR tE


00v t z

8. Taylor dispersion and Chromatography


• Chromatography is a separation method often used for chemical analysis of

complex mixtures.

• A pulse of mixed solutes is injected into one end of a packed bed of

absorbent (the stationary phase) and washed through the bed with solvent

(the mobile phase). Solutes are absorbed to different degrees, thus they are

washed out of the bed (eluted) at different times.

• The analysis of chromatography is usually empirical, a consequence of the

normally complex geometry of the absorbent.

• One special case: a solute pulse injected into fluid in laminar flow in a

cylindrical tube, just like the solute pulse already discussed. Now, however,

the walls of the tube are coated with a thin film of absorbent.

• The injected solute is retarded by absorption in that thin layer.

• Our goal is to determine the shape of the pulse eluted from this absorbent-

coated tube.


• Mass balance:

• Boundary Conditions:

22

1 1 1 10 2

0

12 1

c r c c cv D r

t R z r r r z

Axial diffusion

included

01

1

0 1 1

1 1

0, all :

0, 0, 0

10, ,

Nt z c z

A

ct r

r

t r R c cH

c cD D

r r



22

1 1 1 10 2

0

12 1

c r c c cv D r

t R z r r r z

• Need a mass balance on the adsorbent (we neglect both convection and

axial diffusion):

• Boundary conditions for the adsorbent:

• Solution: Difficult to solve both problems!

• Need to decouple them!

1 11c cD r

t r r r

1

10

0, all : 0

0, , 0

t r c

ct r R

r

Thickness of

adsorbed layer



• Limiting case: Thin absorbed layer or the Golay equation!

• Parameters:

• Dispersion coefficient:

• Physical meaning:

2

001 0 0 2

0 0 00 0//

; exp44

z zNc

E z vE z vz z v t

R

0

0 0

1 1z

t k kv v

L

0

Hk

R

2 2 2 22

0 0 01 6 111

48 1 3 1

R v vk k kD k

DE

k D k

0 : average residence time of solutet

: capacity factor

equilibrium ratio of solute held in the adsorbent

to that inside the tube itself

k



• 1st term: dispersion caused by axial diffusion

• 2nd term: Taylor dispersion, i.e. coupled radial diffusion and axial convection

• 3rd term: dispersion caused by retardation in the absorbent layer

• Injected solutes will be eluted at different retention times t0 when their

absorption is different. The amount by which the retention times differ is

largely controlled by the difference in the capacity factors k’.

• At the same time, the separation of these solutes can be compromised by

dispersion.

• We need small values of E to keep dispersion at a minimum!


2 2 2 22

0 0 01 6 111

48 1 3 1

R v vk k kE D k

D k D k


Rules to facilitate separation

• Use low velocities → reduce Taylor dispersion and adsorbent-caused

dispersion

• Use small channels (i.e., small R0), though this often means large pressure

drops

• Also, note that as v and R0 become very small, we will always have dispersion

from axial diffusion


Lecture No 9 – Summary

1. Dispersion – Basic Concepts

2. Dispersion and pulse decay

3. Dispersion and turbulent pipe flow

4. Dispersion and laminar pipe flow

5. Dispersion and Chromatography

End of Lecture No 9!

Thank you!!!


mass transfer lecture 09. dispersion vlasis g. mavrantzas ... · bibliography material taken from:...

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