Ccb2053 Chap 2 Lecture 1

Mass Transfer Design CCB2053

Dr Sintayehu Mekuria Hailegiorgis

Department of Chemical Engineering

CCB2053 1

Mass Transfer Principles

CCB2053-CHAPTER 2-LECTURE 1

Lesson Outline

Mass transfer principles

Fick’s law of diffusion mass transfer coefficients

Diffusion in gases Equimolar counter diffusion General case of diffusion and convection

Lesson Outcome

At the end of the lesson, the students are able to:

Explain the basics of mass transfer processes and Fick’s law of diffusion.

Apply the principle of diffusion for flux estimation in equimolar counter diffusion processes and general case of diffusion and convection mass transfer in gases.

Separation (recap)

• A physical process in which a mixture is separated into individual

components or group of components.

• Why? Product specification, Recovery, Purification

Mechanical operation

Separation

Homogeneous Heterogeneous

Mass transfer operations

Distillation Extraction Leaching Adsorption

Filtration Centrifugation Settling

Mass transfer principles

• Transport of one component from a region of higher concentration to lower concentration.

• Migration of a substance through another under the influence of concentration gradient.

• Involves the diffusion transport of some component within a single phase or between phases and remain there.

Mass transfer

Dye molecules spread throughout the water in a random fashion

Diffusion • Diffusion (considered in this chapter)

– The movement of a substance from an area of high concentration to an area of low concentration due to concentration difference.

Classification of mass transfer

• Molecular diffusion in stagnate media

• Molecular diffusion in fluids in laminar flow

• Eddy diffusion or turbulent diffusion

• Mass transfer between phases

Fick’s Law of molecular Diffusion

Molecular diffusion: defined as the transfer or movement of individual molecules through a fluid by means of the random, individual movements of the molecules.

This random movement of the molecules is often called a random walk process

Example: If there are a greater number of A molecules near point (1) than at (2), then, since molecules diffuse randomly in both directions, more A molecules will diffuse from point (1) to (2) than from point (2) to (1). The net diffusion of A is from high to low concentration region.


• Mass transfer is characterized by the general form of transport equation.

Resistance

forceDrivingprocesstransferofRate

dzJanddcJ AzAz

1

dz

dxcDJ A

ABAZ

The basic law of diffusion called Fick’s Law by Adolf Eugen Fick in 1885

stated “ the molar flux of a species relative to an observer moving the molar

average velocity is proportional to the concentration gradient of the species”

If A diffuses in a binary mixture of A and B then the flux of A is

• The flux is the amount or concentration of solute carried by a fluid past a plane perpendicular to direction of flow or velocity.

• The unit of flux is given kg mol/m2s (SI system) or g mol/cm2s (cgs system)

Flux

Examples of molecular diffusion flux


• let c is the concentration of A and B in kmol (A+B)/m3, xA the mole fraction of A in a mixture of A and B. Since c is constant, then

AAA dccxdcdx )(

Substituting in to Eq.(2.3)

dz

dcDJ A

ABA

Where D AB is the proportionality constant called the “diffusion coefficient” or the “ diffusivity” of A in a mixture of A and B.

Equation (2.5) is the mathematical representation of Fick’s law of diffusion in a binary mixture.

It is often called mass transfer equation.


AA ccx

Similarities of the three transport equations The three molecular transport process (momentum, heat and mass) are

characterized by the same general type of equation as:

dz

dz

1. Molecular diffusion equation of momentum for constant density, (Newton’s law of viscosity)

2. Fourier’s law of heat conduction for constant and cp

3. Fick’s law of diffusion for constant total concentration, c

dz

dux

zx

dz

TcdAq

p

x

dz

dcDJ A

ABAz


Convective mass transfer and mass transfer coefficient

• When a fluid is flowing outside a solid surface in a forced convection motion, mass transfer occurs under the influence of motion in a fluid medium- convective mass transfer.

)( 1 LiLcA cckN

Where:

NA -Convective molar flux in (kmol/m2s)

kc -Mass-transfer coefficient (m/s)

cl1 -Bulk fluid concentration (kg mol A/m3)

cli -Concentration in the fluid next to the surface of the solid (kg mol A/m3)

(2.10)

Convective mass transfer in liquid-solid system

Flowing liquid Flux

CLi .

i,

CL1

.. …………… …..

Diffusion in Gases

Molecular diffusion in gases can be taken place in different ways as:

1. Equimolar counter-diffusion in gases

2. General case for diffusion of gases A and B plus convection

3. Special case for A diffusing through stagnant, non-diffusing B

4. Diffusion through varying cross-sectional area

1. Equimolar counter-diffusion

– Let molecules A diffuse to the right and molecules B diffuse to the left and P constant throughout the system

21 AA pp

12 BB pp

Consider:

◦ Two gases A and B at constant total pressure (P) in two large chambers connected by a tube and molecular diffusion is occurring at steady state.

◦ Partial pressures: pA1 > pA2 and pB2 > pB1

The moles A diffusing to right is eqaual to Moles of B diffusing to the left since total pressure P is constant,

BzAz JJ

Fick’s law for B for constant total concentration c,

Since total pressure P is constant, then

BA ccc BA dcdc

dz

dcDJ

B

BABz

(2.11)

(2.12)

(1.13)

1. Equimolar counter-diffusion…

Equating Eq.(2.5) and (2.12)

dz

dcDJ

dz

dcDJ B

BABzA

ABAz (2.14)

Substituting (2.13) into (2.14) and canceling

BAAB DD (2.15)

i.e. for binary gas mixture of A and B, the diffusion coefficient (diffusivity) DAB for A diffusing into B is the same as DBA for B diffusing into A

1. Equimolar counter-diffusion…


•Let us consider what happens when the whole fluid is in motion in a bulk or convective flow to the right as shown in the Figure below

AAdA cvJ (m/s)(kgmol A/m3) (kgmol A/sm2)

Diffusion velocity of A

•For diffusion in stationary fluid, the diffusion flux JA passing a fixed point from left (high concn.) to right (low concn.) can be expressed in terms of velocity of diffusion of A,

•The molar average velocity of the whole fluid relative to the stationary point is vM m/s.

vA

(2.16)

vM vAd

2. General case for…

• Component A is still diffusing to the right with its diffusion velocity Ad,

thus for a stationary observer A is moving faster than the bulk velocity M since its diffusion velocity is added to that of the bulk velocity M

The first term represents the total flux relative to the stationary point, NA (kg mol A/s.m2), the second term the diffusion flux, J relative to the moving fluid and the third term is the convective flux of A relative to the stationary point.

MAdA vvv Convective

velocity of the bulk fluid

Velocity of A relative to a

stationary point

Diffusion velocity of A Multiplying by cA,

(2.17)

MAAdAAA vcvcvc

Hence,

MAAA vcJN (2.18)

2. General case for…

If N = total convective flux of the whole stream relative to the stationary point, then

BAM NNcvN c

NNv BA

M

(2.19)

Substituting equation (2.19) and Fick’s law into (2.18),

BAAA

ABA NNc

c

dz

dxcDN

Convection term

Diffusion term

Equation (2.20) is the general form of equation for diffusion plus convection

with relative to a stationary point. It holds for diffusion in gas, liquid, or solid.

(2.20)

2. General case for …

Note: For Equimolar counter-diffusion, Hence,

BABB

BAB NNc

c

dz

dxcDN (2.21)

A similar equation can be written for NB as shown in eq.(2.21)

The convective term in equation becomes zero. Then,

BA NN (2.22)

dz

dxcDN A

ABA (2.23)

2 Fick’s law of diffusion equation for constant total mass,

dz

dcDJ A

ABAz

3. Convective mass transfer and mass transfer coefficient

)( 1 lilcA cckN

Summary 1 Mass transfer is characterized by the general form of transport equation.

Resistance

forceDrivingprocesstransferofRate


4. Equimolar counter-diffusion

BAABBzAz DDandJJ

BAAA

ABA NNc

c

dz

dxcDN

Example-1

1. Diffusion of Methane through Helium.

A gas of CH4 and He is contained in a tube at 101.32 kpa pressure

and 298 K. At one point the partial pressure of methane is pA1 =

60.79 kpa and at a point 0.02m distance away, pA2 = 20.26 kpa. If

the total pressure is constant throughout the tube, calculate the flux

of CH4 (methane) at steady state for equimolar counter-diffusion if

DAB of the CH4-He mixture is 0.675*10-5 m2/s at 101.32 kpa and

298 K.

Solution

Since total pressure P is constant, where the concentration c is as follows for a gas according to ideal gas law

nRTPV (a)

cRT

P

V

n (b)

Where: n is kg mol A plus B, V is volume, m3 , T is temperature, K, R is 8314.3m3 pa/kg mol K, c is in kg mol/m3

For steady state, the flux J and diffusivity DAB for a gas are constant.

dz

dcDJ A

ABAZ (c)

Rearranging and integrating Eq. (c)

2

1

2

1

A

A

c

c

AAB

z

z

AZ dcDdzJ

21

21

zz

ccDJ AAAB

AZ

(d)

Also, from ideal gas law

nRTPV

V

n

RT

pc AA

A 11

(e)

Substituting Eq. (e) into Eq. (d)

21

21

zzRT

ppDJ AAAB

AZ

(f) Final solution

JAZ = 5.52*10-5 kg mol A/s.m2 or 5.52*10-6 g mol A/s.cm2

EXAMPLE-2

2. Equimolar Counterdiffusion of NH3 and N2 at Steady State.

Ammonia gas (A) and nitrogen gas (B) are diffusing in

counterdiffusion through a straight glass tube 0.610 m long with

an inside diameter of 24.4 mm at 298 K and 101.32 kPa. Both

ends of the tube are connected to large mixed chambers at 101.32

kPa. The partial pressure of NH3 is constant at 20.0 kPa in one

chamber and 6.666 kPa in the other. The diffusivity at 298 K and

101.32 kPa is 2.30 × 10−5 m2/s.

Calculate

a) the diffusion of NH3 in kg mol/s. b) the diffusion of N2. c) the partial pressures at a point 0.305 m in the tube and

plot pA, pB, and P versus distance z.

Solution

T=298K P=101.32KPa pA1=20kpa

T=298K P=101.32KPa pA2=6.666kpa

ø=24.4mm

z=0.61m

Ammonia gas (A) Nitrogen gas (B) Equimolar counterdiffusion DAB=2.3x10-5m2/s

Calculate

a) the diffusion of NH3 in kg mol/s. b) the diffusion of N2. c) the partial pressures at a point 0.305 m in the tube

and plot pA, pB, and P versus distance z.

Similar to example-1 above,

Since total pressure P is constant, where the concentration c is as follows for a

gas according to ideal gas law

nRTPV (a)

cRT

P

V

n

(b)

Where: n is kg mol A plus B, V is volume, m3 , T is temperature, K, R is 8314.3m3 pa/kg mol K, c is in kg mol/m3

For steady state, the flux J and diffusivity DAB for a gas are constant.

dz

dcDJ A

ABAZ (c)

Rearranging and integrating Eq. (c)

2

1

2

1

A

A

c

c

AAB

z

z

AZ dcDdzJ

21

21

zz

ccDJ AAAB

AZ

(d)

Also, from ideal gas law

nRTPV

V

n

RT

pc AA

A 11

(e)

Substituting Eq. (e) into Eq. (d)

21

21

zzRT

ppDJ AAAB

AZ

(f)

Substituting DAB = 8314 m2/s, pA1= 2.0x104 pa, pA2= 6.666x103 pa, R=8314.34 m3 pa /kg

mol K, T=298K and z2-z1=0.61m

JAz = 2.03x10-7 kg mol A/s.m2

Rate of diffusion = JAz S,

Where S is surface area = πr2= π(0.0122m)2=4.68x10-4m2

a) Rate of diffusion = JAz πr2 =9.48x10-11 kg mol A/s

b) JB=?

Similarly JB can be given by;

21

21

zzRT

ppDJ BBBA

BZ

(g)

Where: pB1 = P - pA1 = 101.32kPa-20kpa=81.32kpa

pB2 = P - pA2 =101.32kPa-6.666kpa=94.654kpa

Substituting in to Eq.(g) and calculating

JBz = -2.03x10-7 kg mol B/s.m2

Rate of diffusion = JBz S,

Where S is surface area = πr2= π(0.0122m)2=4.68x10-4m2

Rate of diffusion = JBz πr2 =-9.48x10-11 kg mol A/s

The negative value of JBZ means the flux goes from point 2 to point 1

c) pA1 at 0.05m?

2

211 A

AB

AZA p

D

zzRTJp

(h)

Substituting the calculated value of JAz= 2.03X10-7kg mol/s.m2, z1-z2=0.05m and

the above values , then;

pA1= 1.333x104 pa

From equation (f) solving for pA1

Plots P, pA and pB versus distance z

P

pB1

pA1

z

P, p

A, p

B

pB1

pA2

P

pB1

pA1

P, p

A, p

B

Next Lesson

Diffusion of gas A through stagnant, non-diffusing gas B

Diffusion of gases through varying cross-sectional area

Ccb2053 Chap 2 Lecture 1

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