Mass Transfer Dispersion - ETH ZMass Transfer – Dispersion 13-17 Example 1: Chemical spill A...

DispersionLecture 13, 13.12.2017, Dr. K. Wegner

Mass Transfer

Mass Transfer – Dispersion 13-2

to disperse = to spread widely

Dispersion

1. Dispersion is the act or process of dispersing, of distributing one substance (small volume fraction) in another (larger volume fraction).

2. Dispersion is the state of being dispersed.

Examples: 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,...


Dust Clouds from Africa

Examples

Global travel of dust carries microbes across oceans and continents.


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


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.


Soot, smoke

Condensation plumes for 500 ft and 250 ft stacks in Salem (MA), showing the complex thermal structure in the lower atmosphere.


Dispersion on the microscopic level involves diffusion (of droplets, particles, molecules,...). On the macroscopic level dispersion is governed by fluid dynamics (laminar, turbulent flow, eddy formation).

The mechanisms causing the mixing and dilution e.g. of the plumes are hydrodynamic. Moreover, the governing equations are very complex and an exact solution is impossible, especially if the variability, e.g. of the weather is included.

Goal: To understand how e.g. plumes emitted by industrial facilities and ships behave.


The dispersion equation

It is impossible to calculate the actual concentration at a certain point for a certain time but we try to calculate the average concentration.

This has been done before for molecules diffusing in a medium.

We try to apply the mathematics applied to diffusion to dispersion.

Let's start with a rather simple problem: A point source only active at time t = 0, e.g. an emission pulse of smoke from a stack.


The concentrated solute originally located at z = 0 diffuses as the Gaussian profile shown.B.C.:t = 0:

Decay of a pulse (see Chapter 2, Fick's 2nd law)

)z(AMc1 δ=

δ(z): Dirac fct.

t > 0: z = 0:

z → ∞ : c1 = 0

01 =∂∂

zc

z → ∞ : c1 = 0


21

21

zcD

tc

∂∂

=∂∂

Fick's 2nd law:

The concentration profile is (Chap. 2): Dt4z

1

2

eDt4

1AMc

−

π=

where M is the mass of the solute and A the cross-sectional area over which diffusion (dispersion) is occurring.

For particles with concentration N0 (#/m3) released at t = 0 and z = 0:

2

2

p znD

tn

∂∂

=∂∂

and Dt4z

p0

2

etD4

1Nn−

π=

with the particle diffusivity Dp, see "Diffusion coefficients"


However, measurements show that after about 1h a plume has spread 1 km! Use the dispersion coefficient E instead of the diffusion coefficient D! The dispersion coefficient must be determined experimentally and typically has different values in different directions.

As we have seen in Chapter 2, the Fourier number for mass transfer can indicate for an unsteady state problem how far (how long) mass transfer has occurred.

1L

tDFo 2 ≈⋅

=Assume

According to mass transfer by diffusion in gases with D = 10-5 m2/s, the width of the peak roughly should be after 1 hour:

m 2.0s3600sm10tDL

25 ≈⋅=⋅= −


The source of the plume is at (0,0) and wind is blowing at speed v0

in the x-direction. We want to know the average pollutant concentration at (x, y).

Dispersion in one or more dimensions

( )tE4

y

0y

1y

2

ev/xE4

1c ⋅−

⋅⋅π∝

Based on this general equation for one-dimensional dispersion in y direction, a variety of concentration profiles have been derived.


Concentration profiles for free dispersion


Forest fire in southern France (Le Boulou) near the Spanish border in July 2017. Vacation home of the lecturer was x = 3 km away from the point source. v0 ≈ 60 km/h.Photo: L'Indépendant (J. Gallardo)

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


Dispersion coefficients ...

- are rather independent of chemistry (no material property)- are strongly dependent on flow (velocity, lam. / turb.)- depend on position- have different values in different directions- typically must be determined experimentally


In 1905, five muskrats ("Bisamratte") escaped in Bohemia. The animals quickly spread over Europe, as shown in the Figure.

Define a dispersion coefficient similar to the diffusion coefficient(A. Einstein):

Skellam, J.G. (1951) Biometrika 38, 196.

Et2r2 =


Example 1: Chemical spillA container with chemicals breaks, accidently 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.

a) How large is the dispersion coefficient? b) What will the maximum concentration be 15 km downstream?

t=0initial spill

v = 0.6 km/h

t1, s = 2 km: c1max= 860 ppm


Solution:Start with a mass balance for a section of river dz around the maximum concentration, moving with velocity v:

11 j

zdtdc

∂∂

−= withzcEj 1

1 ∂∂

−= 21

21

zcE

tc

∂∂

=∂∂

→

Boundary conditions:

t > 0: z = 0 : z → ∞ : c1 = 001 =∂∂

zc

z → ∞ : c1 = 0)z(AMc :0t 1 δ==


Concentration profile: Decay of a pulse

Et4z

ionconcentrat maximum measured

1

2

e Et4

1AMc

−

π=

a) So, 50 m away from the maximum: ( )

−

−

⋅== hs3600

hkm6.0km2E4

50m

1

2

eppm 608 ppm 410c

scm700E

2

= (compare diffusion coefficients: 10-5 cm2/s)

b) Note: t

1~c max,1 , thus ( ) ppm314km15km2ppm860km15c1 =⋅=


Example 2: Turbulent flow in gas pipelineA 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 check the condition of flow:

000'50s Pa 1010

m 1.0sm 5

mkg 1dvRe 6-

3=

⋅

⋅⋅≈

η⋅⋅ρ

=

Turbulent flow! We'll treat laminar pipe flow afterwards!


Plug flow profile

Assumption: Well mixed radially but concentration changes in axial direction.

ABz

r

v = 5m/s

Mass balance for point at the interface moving with v = 5 m/s:

21

21

zcE

tc

∂∂

=∂∂ B.C.: t = 0, z > 0: c1 = c1∞

t > 0, z = 0: c1 = c10

t > 0, z = ∞: c1 = c1∞

where c10 is the average concentration of the two gases. ∞= 110 c21c


The solution is similar to the general semi-infinite slab solution:

Et4zerf

cccc

101

101 =−−

∞

The dispersion coefficient for turbulent dispersion in pipelines is approximately:

v2dE ⋅≈ ,thus in this case E ≈ 2500 cm2/s

Let us consider a concentration change significant for Et4z =and closer. erf(1) ≈ 0.84, thus c1 = 0.84 c1∞+ 0.16 c10 = 0.92 c1∞

This is the case for m 24

sm5.0

m 3000s

m 25.04z2

≈⋅⋅=


Example 3: Laminar flow – Taylor dispersionA typical example for dispersion is the spread of a solute pulse in steady laminar flow.

The solution is dilute The flow is always laminar, no friction (no axial change in velocity) Mass transport is by axial convection and radial diffusion only

We can do an analysis using the following assumptions:

Injection of solute pulse

vzR0


Goal: Prediction of the dispersion coefficientLet us look at the solute mass balance of a ring-shaped region with dimensions Δr and Δ z, while the radius of the ring is r:

( ) 1z11

r11 r

znn

r1rn

rr1

tc

+∂∂

−θ∂

∂−

∂∂

−=∂∂ θ (1)

This simplifies to:

( )zvcrj

rr1

tc z1

r11

∂∂

−∂∂

−=∂∂

Velocity profile for laminar tube flow:

( )

−=

2

0

0z R

r1v2rv

(2)

(3)average velocity max0 v

21v =


Eq. (2) can be transformed by substituting Fick’s law and eq. (3) into:

zc

Rr1v2

rcr

rrD

tc 1

2

0

011∂∂

−−

∂∂

∂∂

=∂∂ (4)

Boundary conditions:

t = 0, all z: )z(RMc 2

01 δ⋅

π

=

t > 0, r = 0 : 0rc1 =∂∂

r = R0: 0rc1 =∂∂


One solution to this equation is:

−⋅

ζ∂

∂+= ∗∗

==

∗

∗

42

0r

100

0r11 r21rc

DRv

41cc (6)

Let us “sit” on a reference frame moving with velocity v0 to observe the action happening on the z-axis:

0

o

Rtvz −

=ζ0R

rr =∗Substitute and

So eq. (4) becomes:

ζ∂∂

−=

∂∂

∂∂ ∗

∗∗

∗∗12

001 cr

21Rv2

rcr

rrD

(5)


New approach with further simplification:Assuming that the radial variations in c1 are small relative to the axial ones (esp. for fast diffusion), we can work with the radial average:

∫ ∫ ∗∗=⋅⋅ππ

=0R

0

1

0112

01 drcr2dr)z,r(cr2

R1c

Now we think of this concentration as a solute diffusing in the moving reference frame and rewrite the mass balance:

( )0

11Rj

tc

ζ∂∂

−=∂∂

Here 1j (the average flux in flow direction) is equal to:

∫ =∗−−⋅π

π= 0R

0 0r110

z20

1 dr)cc)(vv(r2R1j

(8)

(9)

where tvzR 00 ⋅−=⋅ζ

(7)


Eq. (8) can be written as:

( )( )

⋅⋅

−

ζ∂∂

−=ζ∂

∂−=

τ∂∂

=∂

∂ ∗∗∗∫ drcr21r4vjc

Rtvc

1

1

0

011

00

1 2

Inserting eq. (6) and using eq. (7) we have thus:

21

20

01 c

D48Rvc

ζ∂∂

=

τ∂∂

The boundary conditions for the new equation are similar to the ones for the "decaying pulse" problem, namely:

(10)

( )ζδ⋅π

=ζ∀=τ

=ζ∂

∂=ζ>τ=±∞→ζ>τ

30

1

11

RMc : ,0

0c :0 ,0 ;0c : ,0


The solution to this problem is a Gaussian curve:

tE4)tvz(

1

20

11

20

etE4)R(Mc

−−

ππ

= (11)

The dispersion coefficient is given by (Taylor, 1953):

=

D48RvRvE 0

0

00

1 (12)

Note that E depends inversely on the diffusion coefficient! Rapid (radial) diffusion leads to small (axial) dispersion.

Taylor, Sir G. (1953), "Dispersion of soluble matter in solvent flowing slowly through a tube", Proc. Royal Soc. A 219, 186-203


Rapid (radial) diffusion leads to small (axial) dispersion.

=

D48RvRvE 0

0

00

1

radial diffusion


Correction for axial diffusion: Taylor-Aris Dispersion

∂∂

+

∂∂

∂∂

=∂

∂+

∂∂

21

21

011

zc

rcr

rr1D

zvc

tc

axial diffusion

tE4)tvz(

1

20

11

20

etE4)R(Mc

−−

ππ

=also yields but now

+=

D48RvDE 0

0

1

Aris, R (1958), "On the dispersion of a solute in a fluid flowing through a tube", Proc. Royal Soc. A 235, 67-77.


UV Absorbance by particlesin the sample zone as fct of time

MovingsampleZone

PressureGradient

Particle-freecarrier solution

Belongia B.M. and Baygents, J.C. (1997): "Measurements on the Diffusion Coefficient of Colloidal Particles by Taylor-Aris Dispersion"

Belongia B.M., Baygents, J.C. (1997), J. Colloid Interface Sci. 195, 19-31.

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Mass Transfer Dispersion - ETH ZMass Transfer – Dispersion 13-17 Example 1: Chemical spill A...

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