prof. r. shanthini 21 feb 2013 1 course content of mass transfer section lta diffusion theory of...

Prof. R. Shanthini 21 Feb 2013

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Course content of Mass transfer section

L T A

Diffusion Theory of interface mass transfer Mass transfer coefficients, overall coefficients and transfer units Application in absorption, extraction and adsorption

08 02 05

Concept of continuous contacting equipment Simultaneous heat and mass transfer in gas-liquid contacting, and solids drying CSK

06 01 05

CP302 Separation Process PrinciplesMass Transfer - Set 1


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CP302 Separation Process PrinciplesReference books used for ppts

1. C.J. Geankoplis Transport Processes and Separation Process Principles4th edition, Prentice-Hall India

2. J.D. Seader and E.J. HenleySeparation Process Principles2nd edition, John Wiley & Sons, Inc.

3. J.M. Coulson and J.F. RichardsonChemical Engineering, Volume 15th edition, Butterworth-Heinemann

4. Chapter 10 of R.P. Singh and D.R. Heldman Introduction to Food Engineering 3rd edition, Academic Press


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Mass transfer could occur by the following three ways:

Diffusion is the net transport of substances in a stationary solid or fluid under a concentration gradient.

Advection is the net transport of substances by the moving fluid. It cannot therefore happen in solids. It does not include transport of substances by simple diffusion.

Convection is the net transport of substances caused by both advective transport and diffusive transport in fluids.

Modes of mass transfer


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Stirring the water with a spoon creates forcedforced convectionconvection.

That helps the sugar molecules to transfer to the bulk water much faster.

DiffusionDiffusion(slower)(slower)

ConvectionConvection(faster)(faster)

Modes of mass transfer


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Objectives of the slides that follow:

Understanding diffusion in one dimension


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Mass transfer by diffusion occurs when a component in a stationary solid or fluid goes from one point to another driven by a concentration gradient of the component.

Solute ASolvent B

concentration of Ais high

concentration of Ais low

Diffusion


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concentration of A is the same everywhere

Diffusion

Diffusion is the macroscopic result of random thermal motion on a micoscopic scale (Brownian motion). It occurs even when there is no concentration gradient (but there will be no net flux).

Solute ASolvent B


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A B

A B

Liquids A and B are separated from each other.

Separation 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 DIFFUSION. Equilibrium is reachedEquilibrium is reached


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Examples of Diffusion

A diffusion problem that occurs in the field of microelectronics is the oxidation of silicon according to the reaction Si + O2 SiO2.

When a slab of the material is exposed to gaseous oxygen, the oxygen undergoes a first-order reaction to produce a layer of the oxide. The task is to predict the thickness d of the very slowly-growing oxide layer as a function of time t.

Silicon

Silicon oxide layer

OxygenCA = CA0 at Z = 0

CA = CAδ at Z = δ


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Examples of Diffusion

If an insect flight muscle contains tracheal tubules which allow air to diffuse into all parts of the muscle, and the tracheal tubules make up 20% of the volume of the muscle, how large can the muscle be?


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Mathematical modelling of steady-state one dimensional diffusive mass transfer


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CA

CA + dCA

dz

Mixture of A & B

For mass transfer occurring only in z-direction

(1)

Fick’s First Law of Diffusion

JA

What is JA?

JA = - DAB dzdCA


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diffusion coefficient (or diffusivity) of A in B

diffusion flux of A in relation to the bulk motion in z-directionUnit: mass (or moles) per area per time

concentration gradient of A in z-directionUnit: mass (or moles) per volume per distance

What is the unit of diffusivity?

JA = - DAB dzdCA

Fick’s First Law of Diffusion


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For dissolved matter in water: D ≈ 10-5 cm2/s

For gases in air at 1 atm and at room temperature: D ≈ 0.1 to 0.01 cm2/s

Diffusivity depends on the type of solute, type of solvent, temperature, pressure, solution phase (gas,

liquid or solid) and other characteristics.

Unit and Scale of Diffusivity


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Molecular diffusion of Helium in Nitrogen: A mixture of He and N2 gas is contained in a pipe (0.2 m long) at 298 K and 1 atm total pressure which is constant throughout. The partial pressure of He is 0.60 atm at one end of the pipe, and it is 0.20 atm at the other end. Calculate the flux of He at steady state if DAB of He-N2 mixture is 0.687 x 10-4 m2/s.

Example 6.1.1 from Ref. 1

Solution:

Use Fick’s law of diffusion given by equation (1) as

JA = - DAB dzdCA

Rearranging the Fick’s law and integrating gives the following:


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JA = - DAB dz dCA

DAB is given as 0.687 x 10-4 m2/s

(z2 – z1) is given as 0.2 m

(CA2 – CA1) = ?

⌠⌡z1

z2

⌠⌡ CA1

CA2

At steady state, diffusion flux is constant. Diffusivity is taken as constant.

(2)

Therefore, equation (2) gives

JA = - DAB(z2 – z1) (CA2 – CA1) (3)


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Even though CA is not given at points 1 and 2, partial pressures are given. We could relate partial pressure to concentration as follows:

CA =nA

V

Number of moles of A

Total volume

pA V = nA RT

Partial pressure of A

Absolute temperature

Gas constant

Combining the above we get CA =pA

RT


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Equation (3) can therefore be written as

JA = - DAB(z2 – z1)

(pA2 – pA1)

RT

which gives the flux as

JA = - DAB

(pA2 – pA1)

RT(z2 – z1)

JA = 5.63 x 10-6 kmol/m2.s

JA = - (0.687x10-4 m2/s)(0.6 – 0.2) x 1.01325 x 105 Pa

(8314 J/kmol.K) x (298 K) x (0.20–0) m


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Summary: Modelling diffusion in z-direction

= - DAB

(pA2 – pA1)

RT(z2 – z1)JA = - DAB

(CA2 – CA1)

(z2 – z1)

zz1 z2

CA2CA1

JA

Longitudinal flow: Flow area perpendicular to the flow direction is a constant.

CA1 and pA1 at z1 and CA2 and pA2 at z2 remain unchanged with time (steady state).

DAB is constant


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Derivation of DAB = DBA under certain conditions

DAB: diffusivity of A in B

DBA: diffusivity of B in A


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JA = - DAB

dCA

dz

Consider steady-state diffusion in an ideal mixture of 2 ideal gases A & B at constant total pressure and temperature.

JA and JB : molar diffusive flux of A and B, respectively (moles/area.time)

CA and CB : concentration of A and B, respectively (moles/volume)

DAB and DBA : diffusivities of A in B and of B in A, respectively

z : distance in the direction of transfer

Molar diffusive flux of A in B:

Molar diffusive flux of B in A:

(4)

(5)dCB

dz JB = - DBA

Equimolar counter-diffusion in gases


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For an ideal gas mixture at constant pressure,

CA + CB = pA/RT + pB/RT = P/RT = constant

Therefore, dCA + dCB = 0

JA + JB = 0

Since the total pressure remains constant, there is no net mass transfer. That is, (6)

(7)

Substituting (6) and (7) in (4) and (5), we get

DAB = DBA = D (say)

Therefore, (4) & (5) giveJA = - D

dCA

dz

JB = - DdCB

dz

(8a)

(8b)


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This describes the mass transfer arising solely from the random motion of the molecules (i.e., only diffusion)

It is applicable to stationary mediumstationary medium or a fluid in fluid in streamline flowstreamline flow.

This is known as equimolar counter diffusion.

JA = - DdCA

dz

JB = - DdCB

dz

(8a)

(8b)


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Mathematical modelling of steady-state one dimensional convective mass transfer


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Diffusion is the net transport of substances in a stationary solid or fluid under a concentration gradient.

Advection is the net transport of substances by the moving fluid, and so cannot happen in solids. It does not include transport of substances by simple diffusion.

Convection is the net transport of substances caused by both advective transport and diffusive transport in fluids.

Diffusion of gases A & B plus convection

JA is the diffusive flux described by Fick’s law, and we have already studied about it.

Let us use NA to denote the total flux by convection (which is diffusion plus advection.


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JA = - DAB

dCA

dzMolar diffusive flux of A in B: (4)

The velocity of the above diffusive flux of A in B can be given by

vA,diffusion (m/s) = JA (mol/m2.s)

CA (mol/m3)

The velocity of the net flux of A in B can be given by

vA,convection (m/s) = NA (mol/m2.s)

CA (mol/m3)

The velocity of the bulk motion can be given by

vbulk (m/s) = (NA + NB) (mol/m2.s)

(CT) (mol/m3)

(9)

(10)

(11)

Total concentration


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vA,diffusion + vA,convection = vbulk

(NA + NB)

CT

CA vA,diffusion + CA vA,convection = CA vbulk

JA + NA = CA (12)

Substituting JA from equation (4) in (12), we get

(NA + NB)

CT

NA = + CA(13)dCA

dz-DAB

Multiplying the above by CA, we get

Using equations (9) to (11) in the above, we get


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Using (14a) and (14b), equation (13) can be written as

(NA + NB)NA = - + (15)dpA

dz

DAB

Let us introduce partial pressure pA into (13) as follows:

CA =nA

V(14a)=

pA

RT

CT =nT

V(14b)=

P

RT

Total number of molesTotal pressure

RT

pA

P


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Using (16), equation (13) can be written as

(NA + NB)NA = + xA(17)dxA

dz-CT DAB

Let us introduce molar fractions xA into (13) as follows:

xA = NA

(NA + NB) =

CA

CT

(16)


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(NA + NB)NA = (17)dxA

dz-CT DAB

Diffusion of gases A & B plus convection:Summary equations for (one dimensional) flow in z direction

(NA + NB)NA = - + (15)dpA

dz

DAB

RT

pA

P

NA = + (13)dCA

dz-DAB

In terms of concentration of A:

In terms of partial pressures (using pA = CART and P = CTRT):

In terms of molar fraction of A (using xA = CA /CT):

xA

(NA + NB)CT

CA

+


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Evaporation of a pure liquid (A) is at the bottom of a narrow tube.

Large amount of inert or non-diffusing air (B) is passed over the top.

Vapour A diffuses through B in the tube.

The boundary at the liquid surface (at point 1) is impermeable to B, since B is insoluble in liquid A.

Hence, B cannot diffuse into or away from the surface.

Therefore, NB = 0

A diffusing through stagnant, non-diffusing B

Liquid Benzene (A)

Air (B)

1

2

z2 – z1


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Substituting NB = 0 in equation (15), we get

(NA + 0)NA = - + dpA

dz

DAB

RT

pA

P

Rearranging and integrating

NA (1 - pA/P) = - dpA

dz

DAB

RT

NA dz = - dpA

(1 - pA/P)

DAB

RT⌠⌡

z1

z2

⌠⌡ pA1

pA2

NA = ln P - pA2 DAB P

RT(z2 – z1) P – pA1

(18)


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Introduce the log mean value of inert B as follows:

NA = (pA1 - pA2 )

DAB P

RT(z2 – z1) pB,LM

(19)

pB,LM = = (pB2 – pB1 )

ln(pB2 /pB1 )

(P – pA2 ) – (P – pA1 )

ln[(P - pA2 )/ (P - pA1 )]

Equation (18) is therefore written as follows:

Equation (19) is the most used form.

(pA1 – pA2 )

ln[(P - pA2 )/ (P - pA1 )]

=


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NA = - (xA1 - xA2 )

DAB CT

(z2 – z1) xB,LM

(21)

NA = ln 1 - xA2 DAB CT

(z2 – z1) 1 – xA1

(20)

Using xA = CA /CT, pA = CART and P = CTRT,

equation (18) can be converted to the following:

Introduce the log mean value of inert B as follows:

xB,LM = = (xB2 – xB1 )

ln(xB2 /xB1 )

(1 – xA2 ) – (1 – xA1 )

ln[(1 - xA2 )/ (1 - xA1 )]

(xA1 – xA2 )

ln[(1 - xA2 )/ (1 - xA1 )]

=

Therefore, equation (20) becomes the following:


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Diffusion of water through stagnant, non-diffusing air: Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed to be dry) is 1 atm and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path is 0.1524 m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapour at 1 atm and 293 K is 0.250 x 10-4 m2/s. Assume that the vapour pressure of water at 293 K is 0.0231 atm.

Answer: 1.595 x 10-7 kmol/m2.s

Example 6.2.2 from Ref. 1

prof. r. shanthini 21 feb 2013 1 course content of mass transfer section lta diffusion theory of...

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