Ch 1 Diffusion

7/27/2019 Ch 1 Diffusion

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MASS TRANSFER 1

CLB 20804

MASS TRANSFER

BASIC PRINCIPLES AND APPLICATIONS

CHAPTER 1


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Topic Outcomes

At the end of Chapter 1, you should:

Define and explain the basic concept of mass transfer.

DeriveFicks Law equation.

Calculate mass transfer rate based on unimolecular

diffusion and equimolar counterdiffusion.

Explain the Inter phase mass transfer.


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Introduction toBasic Principles of

diffusion andapplications

Principles of DiffusionFicks

Law

Application tounimolecular diffusion

and equimolarcounterdiffusion

What are in this chapter?


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Introduction

Mass transfer refers to mass in transit due to a speciesconcentration gradient in a mixture.

When a system contains two or more components whoseconcentration vary from point to point , there is a naturaltendency for mass to be transferred.

Minimizing the concentration differences within the systemand moving it towards equilibrium.


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Principles of Mass transfer

Mass transfer by ordinary molecular diffusion occurs becauseof a concentration difference(driving force for transfer).

The mass transfer rate is proportional to the area normal to

the direction of mass transfer. Not to the volume. So, the rateexpressed as a flux.

Mass transfer stops when the concentration is uniform.Mechanism of mass transfer involves both molecular diffusion

and convection.


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Molecular Diffusion

Net transport of molecules from a region of higher concentration to aregion of lower concentration byrandom molecular motion.

matter wants to "get away" from the other similar matter and go to aplace where there is open space.


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A B

A B

Liquids A and B are separated from each other.

Partition removed.

A goes from high concentration of A to low

concentration of A.

B goes from high concentration of B to low

concentration of B.

Molecules of A and B are uniformly distributed

everywhere in the vessel purely due to the

DIFFUSION.

Molecular Diffusion


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xamp e:Molecular Diffusion

Sprayed air freshener.

drop of liquid dye


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Example of Mass Transfer

At the surface of the lung:

Air Blood

Oxygen

Carbon dioxide

High oxygen concentration

Low carbon dioxide concentrationLow oxygen concentration

High carbon dioxide concentration


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Properties of Mixtures

Mass transfer always involves mixtures.

Consequently, we must account for the variation ofphysical properties which normally exist in a given

system. In order to understand the future discussions, let

us first consider definitions and relations whichare often used to explain the role of components

within a mixture.


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Chemical

Composition

Moles andMolecular

Weight

Mass andMole Fractions

AverageMolecular

Weight

Concentration

Properties of Mixtures


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Definitions

Definitions:iC Molarconcentration of species i. 3kmol/m:i Mass density (kg/m3) of species i.

:iM Molecularweight (kg/kmol) of species i.

i i iC M

*:iJ Molarflux of species i due to diffusion. 2kmol/s m Transport of i relative to molar average velocity (v*) of mixture.

:iN Absolute molar flux of species i. 2kmol/s m Transport of i relative to a fixed reference frame.

:ij Mass flux of species i due to diffusion. 2kg/s m Transport of i relative to mass-average velocity (v) of mixture.

Transport of i relative to a fixed reference frame.

:ix Mole fraction of species i / .i ix C C

:im Mass fraction of species i / .i im

Absolute mass flux of species i. 2

kg/s m:in


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Concentrations

Molar concentration of species A:(liquids,solids) ,(gases)

Where is molecular weight of species A.

Mole fraction:

)(

solids)&(liquids

gases

c

cy

c

cx

A

A

A

A

RT

p

V

n

Mc AA

A

AA

AM


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For a gaseous mixture that obeys the ideal gas law,mole fraction, can be written in terms ofpressures

For gases, P

p

RTP

RTpy AAA

Ay

Daltons

law


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General Ficks Law Equation of diffusion

Ficks Law

dz

dcDJ AABAz

*

The basic empirical relation to estimate the rate of moleculardiffussion:

Molar Flux

Units -

mol/(s.m2)

Diffusivity

(constant)

Units -cm2/s

alternative form

dz

dyCDJ AABA

JAzis the molar flux of A by ordinary molecular

diffusion relative to the molar average velocity of the

mixture in positive zdirection, DAB is the mutual

diffusion coefficient of A in B, and d cA/d z is the

c o n c e n t r a t i o n g r a d i e n t o f A , w h i c h i s

negative in the direction of ordinary molecular diffusion.


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Check your understanding-Theory of diffusion

Think of the last time that you washed the dishes. You placed the firstgreasy plate into the water and the dishwater got a thin film of oil on thetop of it. Assume that there is no oil at the top of the sink yet. Find thediffusion flux, J of oil droplets through the water to the top surface. Thesink is 18 cm deep, and the concentration of oil on the plate is 0.1mol/cm3. Given the diffusivity is 7x10-7 cm2/s.

6.1-1-Geankoplis(413)

Given : D = 7x10-7 cm2/s.

Find : Calculate the flux.


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dC = conc. at the top of the sink conc. of oil on the plate.

The concentration at the top of the sink = 0The concentration of oil on the plate = 0.1 mol/cm3.

dC = 0 0.1 = -0.1 mol/cm3

dx = the depth of the sink = 18 cm

Check your understanding-Theory of diffusion

J = 3.89 x 10-9 mol / (cm2.s)


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1.4 Diffusion Through Moving Bulk Fluid

Movement of A is now due to 2 contributions:

Molecular diffusion fluxes :

Fluxes resulting from bulk flow= CAV (kg-mole/m3. m/s)

Concentration of A at any point inthe mixture = CA (kg-mole/m

3)

(1)

Consider abulk fluid of binary mixture A and B moving in the z-directionas shown, with an average bulk fluid velocity V m/s, as shown in the Figurebelow:


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1.4.1 Diffusion Through MovingBulk Fluid

Molar flux of A :

Similarly for comp. B:

Total molar flux of A and B:

VCJN BBB

NNN BA

VCJN AAA (2)

(3)

(4)

Also, N = C V

Where, C = total molar concentrationV= average molar velocity(m/s)

(5)

Substitute eqn (4) into eqn (5) :

C

NNV

BA

(6)

1 4 1 s on ro g o ng


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1.4.1 us on roug ov ngBulk Fluid

Substitute eqn (1) & (6) into eqn (2) for comp. A: In term mole fraction:

In term concentration

)( BAAA

ABA NNydz

dyCDN

(7)

NA Mass Transfer Relative to a fixed position

J*AzMass Transfer Relative to a mass or molar average velocity

1 5 Stea y State Equimo ar Counter


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1.5 Stea y-State Equimo ar Counter-

Diffusion For binary mixture A and B,

When gas system is closed and are connected by straight tube, and at

constant pressure and temperature.

Molar fluxes A and B areequal, but opposite in direction

Total pressure is constant throughout.

Thus, diffussion flux are also equal but opposite in direction

0 BA NNN

Diffusivity, DAB = DBA

(1)


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Mass transfer in the connecting tube is equimolar counterdiffusion by molecular

1.5 Steady-State Equimolar Counter-

Diffusion


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1.5.1 Steady-State Equimolar Counter-

Diffusion Refer to eqn diffusion through moving bulk fluid:

in term mole fract ion

Integrated at z= z1, cA= cA1 and at z= z2, cA= cA2 to:

after simplify and rearrangement, the equation becomes :

)( BAAA

ABA NNydz

dycDN

0

dzAdyD

ABcJN AA

2

1

Ay

Ay dz

Ady

cD

A

J AB

21)( 12 AC

AC

DA

Jzz

AB


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1.5.2 Steady-State Equimolar Counter-

Diffusion of Ideal Gas Mixture:

Consider 2-component gas mixture (A and B)

Ideal Gas Law:

P = Total pressuren = Total moles of gas

Component-A:

Where;

pA = partial pressure of A

nA = moles of A

nRTPv

RTnVP AA

1 5 2 Steady State Equimolar Counter


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1.5.2 Steady-State Equimolar Counter-Diffusion of Ideal Gas Mixture:

Concentration of A:

Differentiating with respect to distance z

Replacing into the original Fick's Law

Integrated Eqn(3) over a diffusional path from z2 to z1 &pA1 to PA2

RT

p

V

nC AAA

RTdz

dp

dz

dC AA 1

dz

Dp

RT

DJ AABA

(1)

(2)

(3)

(4)

2

1

2

1

.

PA

pA

AAB

z

z

A dpRT

DdzJ


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1.5.2 Steady-State Equimolar Counter-Diffusion of Ideal Gas Mixture:

where :

pA1 = partial pressure of A at point 1

pA2 = partial pressure of A at point 2:

(5) 21)( 12

AAAB

A ppzzRT

DJ

Rearrange equation:

Diffusion flux for components A

Diffusion flux for component-B

)(

)(

12

21

zzRT

ppDJ BBABB

St d St t Eq i l C t


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Steady-State Equimolar Counter-Diffusion

21)( 12

AAAB

A ppzzRT

DJ

Or, In terms of partial pressure

In terms of concentration

21)( 12 AC

AC

DA

Jzz

AB


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Ammonia gas (A) diffusing through a uniform tube 0.10 m longcontaining N2 gas (B) at 1.0132x105 Pa pressure and 298 K. At point 1,

pA1=1.013x104 Pa and point pA2= 0.507x10

4 Pa. The diffusivity DAB =

0.230 x 10-4 m2/s. R = 8314 m3.Pa/kg-mole.K

(a) Calculate the diffusion flux JA at steady-state.

(b) Repeat for calculate the flux JB.

Check your understanding

Given:Total pressure PT = 1.0132 x 10

5 Pa(constant)

Temperature T = 298 KDAB = 0.230 x 10-4 m2/s

R = 8314 m3.Pa/kg-mole.KAt point 1, pA1 = 1.013 x 104 PaAt point 2, pA2 = 0.507 x 104 PaDiffusion path = ( z2 - z1 ) = 0.1 m


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Component-A is diffusing from Point 1 to Point 2, as its partial

pressure is higher at point 1.

Solution:

)()(

12

21

zzRTppDJ AAABA

(a) Calculate the diffusion flux JA at steady-state.


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(b) Calculate the diffussion flux JB.

Use the Dalton's Law of partial pressures to determine the partial pressures

ofcomponent-B at points 1 and 2:

PT = pA+ pBAt point 1:pB1 = PT - pA1PB1 = 1.0132 x 105 - 1.013 x 104 = 91,190 PaAt point 2:

pB2 = PT - pA2PB2 = 1.0132 x 105 - 0.507 x 104 = 96,250 Pa

Component-B is diffusing in the opposite direction to component-A:from Point 2 to Point 1, as the partial pressure for B at point 2 is higher.


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Flux for component-B is the same as the flux for component-A, butwith a negative sign, indicating that it is in the opposite direction.

Calculate the flux of B in A using:

)(

)(

12

21

zzRT

ppDJ BBABB


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1.5.2 Steady-State Equimolar Counter-Diffusion ofIdeal Gas Mixture:

Rate of Diffusion= (molar flux) x (Surface area)

= JAx S

Where ;Rate of Diffusion(unit: kmole/s)

JA= Diffusion flux of component A (kmole/m2 .s)

S= Surface area(unit m2 )


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1.6 Steady-State One componentDiffusions

Opposite of equimolar counterdiffusion.

One component diffuses, while the other remains stagnant(nonmoving).

Fig 1.6, point 0 is saturated with component A.

while a stream of pure B flows past the end of the tube removing any A that

has reached point z.

In this situation, the concentration gradient along the tube is exponential

Fig 1.6


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1.6 Steady-State One componentDiffusions

Example:

Diffusion of Components liquid benzene (A) through stagnant (nonmoving)

component air (B).

Evaporation of a pure (A) at the bottom of a narrowtube.

Where a large amount of nondiffussing air (B) ispassed over the top.

Benzene vapor (A) diffuses through the air (B) in thetube.

Point 1 (Boundary at the liquid surface) ~impermeableto air (since air is insoluble in benzene liquid).

Hence, air (B) cannot diffuse into or away from thesurface.

Point 2~ partial pressure PA2=0(since a large volume ofair is passing by. Component B cannot diffuse,

NB=0


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Rearrange eqn (1) to a ficks law form

dz

dy

y

CDN A

A

ABA

1

:(2)

(1)

1.6 Steady-State One component Diffusions

)( BAAA

ABA NNydz

dyCDN

Refer to equation diffusion through moving bulk fluid

Equation simplifies to

0

)( AAAABA NydzdyCDN

dz

dyCDNyN

AABAAA )(


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1.6 Steady-State One component Diffusions

At quasi steady state condition, rearange eqn becomes in integral form;

Upon integration yields

Log mean (LM) of (1-yA) at the two ends of the stagnants layer :

1

2

12 11

lnA

AAB

yy

zzCD

AN

2

1

2

1 1

yA

yAA

A

A

ABz

z y

dy

N

CDdz

Integration rules

(3)

(4)

)1/(1

112

12

AA

AA

LMAyyIn

yyy

(5)


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Rearrange eqn (5)

Subsitute eqn (6)into eqn (4),

LMA

AA

A

A

y

yy

y

yIn

1)1(

1 21

1

2

LMA

AAABA

y

yy

zz

CDN

)1(

21

12

Finally:the rate of diffusion is given by (in mole fraction terms)


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using the Ideal Gas equation,

RT

pC

RT

PC

AA

(3.1)

General equation Diffusion through moving bulk fluid:

Component B cannot diffuse, NB= 0

)( BAAA

ABA NNC

C

dz

dCDN

(1)

)0(

A

AAABA N

C

C

dz

dCDN (2)

(3.2)

1.6.2 Steady-State One component Diffusions


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Differentiate eqn (3.2) with respect to distance z,

Subsitute eqn (4) and (5) into eqn (2)

AAA

ABA N

P

P

dz

dCDN

Divide eqn (3.2) with eqn (3.1);

re-arrange:

dz

dp

RTdz

dC AA 1

P

p

C

CAA

dz

dp

RT

D

P

pN AABAA 1

(4)

(5)

(6)


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Integrating Eqn (6) from point 1 to point 2: the partial pressure of

A changes from pA1 to pA2 :

Upon integration

2

1

2

1 1

pA

pA A

AAB

z

z

A

Pp

dp

RT

DdzN

Integration rules

(7)


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Log mean (LM) of (P-PA) at the two ends of the stagnants

layer :

Rearange eqn (5)

LMA

AAABA

PP

pP

zzRT

PDN

,

1

12 )(

)(

)(2

LMA

AA

A

A

PPPP

pPPPIn

21

1

2

)(

Subsitute eqn (6)into eqn (4),

Finally:

(9)

(10)

)/( 1221

AA

AA

LMA PPPPIn

PP

PP

Rate of diffusion is given by (inpartial pressure terms)


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An open beaker, 6 cm in height, is filled with liquid benzene at 25C towithin 0.5 cm of the top. A gentle breeze of dry air at 25C and 1 atm is

blown by a fan across the mouth of the beaker so that evaporatedbenzene is carried away by convection after it transfers through astagnant air layer in the beaker. The vapor pressure of benzene at 25Cis 0.131 atm. The mutual diffusion coefficient for benzene in air at 25Cand 1 atm is 0.0905 cm2/s. Determine the initial rate of evaporation ofbenzene as a molar flux in mol/cm2.s

Check your understanding

Evaporation of Benzene from abeaker


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Then

scm

molxNA

.1004.1

933.0

131.0

)06.82)(298(5.0

0905.02

6

LMA

AAABA

PP

pP

zzRT

PDN

,

1

12 )(

)(

)(2

Solution:

Let A = benzene, B = air.

Take Zl = 0.

Then Z2 - Zl= Z = 0.5 cm.

One component equation:

933.0)131.01/(01131.0

)/( 12

21

InPP

PPPPIn

PPPP

LMA

AA

AA

LMA

Log mean (LM) of (P-PA) at the two ends of the stagnants layer :

1.7 Diffusion between phases - Phase


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1.7 Diffusion between phases PhaseEquilibrium

When two immiscible phases are in contact:

Example. gas-liquid, two immiscible liquids, it is possible for

diffusing molecules to pass from one phase to the other across

the interphase boundary.

When diffusion between two phases occurs, there are two

factors to take into account;

1 Equilibrium relationships between the two phases

2 rate at which the diffusion takes place


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1.7 Diffusion between phases - PhaseEquilibrium

Referring to Fig 1.7:

If a component, A is passing between phases 1 & 2, there will ultimately be adynamic equilibrium established where the rate of diffusion of A is the same in

both directions.

Depending on the type of system (liquid vapour, liquid liquid etc.), the

equilibrium concentration may often be expressed by simple relationships.

Fig 1.7


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2 simple relationships are : Distribution Coefficients

1.6

Where: XA = concentration of the component A in Phase 1

YA = concentration of the component A in Phase2

K = constant

YA = KXA


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Henry's Law


Where:

P

A= Partial pressure of A in the gas phase

C

A= Concentration of A in the liquid phase

H = constant - called Henry's law constant

Henry's law is a special case of the distribution coefficient for gas-liquid systems.

In the case of systems where one of the phases is a gas or vapour itis common to express the concentration in the gas phase terms ofpartial pressure.

Thus Henry's Law states that

PA = HCA


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1.8 Rate of diffusion between phases -Mass Transfer

The theory discussed so far enables us to calculate the rates ofdiffusion within a single phase and to calculate the concentrationsof a component in two phases when the system is in a state ofequilibrium.

However, many practical problems concern the rate of diffusion

between two phases when the two phases are not in equilibrium.

The rate of mass transfer between two phases is dependant on anumber of factors including:

1. The diffusivity of the diffusing component in the two

phases

2. How far the system is from equilibrium

3. The resistance to transfer across the interface between the

two phases


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Mass Transfer EquationRate of mass transfer is directly proportional to the driving force for

transfer, and the area available for the transfer process to take place,

that is:

Transfer rate transfer area driving forceThe proportional coefficient in this equation is called the masstransfercoefficient, so that:

Transfer rate = mass-transfer coefficient

transfer area driving force

1.8 Rate of diffusion between phases -Mass Transfer

NA = kACA

Where:

NA = mass transfer rate of A across thephase boundary

K = mass transfer coefficient

C = concentration driving force

1 8 Interphase mass transfer


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1.8 Interphase mass transfer

Two-Film Theory of Mass TransferIn order to illustrate the concept of

interphase mass transfer lets considerthe process of transport of a volatile

chemical across the air/water

interphase.

Gas molecules must diffuse from the

main body of the gas phase to thegas-liquid interface, then cross thisinterface into the liquid side, andfinally diffuses from the interface intothe main body of the liquid.

C t ti d i i f


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The concentration driving force is a measure of how far the systemis from equilibrium and may be developed with reference to the

diagram below. (Fig 1.9)

1.9 Concentration driving force

Consider a component, such as oxygen, diffusing from air to water. Assume the

system is at steady state but not at equilibrium.

PA = partial pressure of oxygen in the air and CA is the concentration ofoxygen in the water.

CA* = concentration of oxygen in water that will be in equilibrium with PA

PA* = partial pressure of oxygen in air that will be in equilibrium with CA

(Fig 1.9)

C t ti d i i f


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1.9 Concentration driving force

The rate of oxygen mass transfer between air and water may be

expressed in two ways. One based on partial pressure of oxygen in air and one based on the

concentration of oxygen in water.

NA = KG (PA P*A)

NA = KL (C*A CA)

Where:

PAP

A

*= partial pressure driving force

CACA

*= concentration driving force

KL = mass transfer coefficient based on the liquid phase.

KG = mass transfer coefficient based on the gas phase.

The approriate equilibrium value PA* or CA* may be calculated using Henry's law

Gas Liquid Equilibrium Partitioning Curve


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Gas-Liquid Equilibrium Partitioning Curve

CA,i

PA,i

PA

CA

P*A

C*A

CA

PA

PA,i = H CA,i

P*A= H CA

PA= H C*A

PA* = HCA

PA= HCA*


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Ch 1 Diffusion