Lecture Notes Chap 2 CHEM3002 2013 2014

7/23/2019 Lecture Notes Chap 2 CHEM3002 2013 2014

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HEM3002 Ahmed Al-Dallal/2013-2014

CHAPTER TWO

Convective Mass Transfer

1

Content


Introduction

Equimolar counter diffusion

A diffusing through stagnant, non-diffusing B

Relationship between mass transfer coefficient

Prediction of mass transfer coefficient

Dimensionless number

Mass Transfer for Flow Inside Pipes

Mass Transfer for Flow Parallel to Flat Plates

Mass Transfer for Flow Past Single Spheres

Mass Transfer to Packed Bed

Flow Perpendicular to Single Cylinder2


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Convective mass transfer occurred when the fluid is flow passing

another immiscible fluid or solid surface. Mass transfer will be

increased.

Previously, for molecular diffusion in stagnant fluid and fluid in

laminar flow

Now, we will considered convective mass transfer.

)(

*

B A A A

AB A

AB A

N N c

cdz

dxcD N

dz

dcA D J

dz

dc D J A

AB A *

3

Unsteady State in Various Geometries1/2

Assuming no convective resistance

Represent by:

NA = k c(c L1-c Li)

Where

k c = mass transfer coefficient, m/s

c L1= bulk fluid concentration, kg mol A/m3

K = c L1/ci where

K = equilibrium distribution coefficient4



Unsteady State in Various Geometries2/2

Charts for unsteady heat transfer

can be used to predict concentration

in unsteady-state mass transfer

by diffusion:

Semi infinite solid (Fig 5.3-3)

Flat plate (Fig 5.3-5/6)

Long cylinder (Fig 5.3-7/8)

Sphere (Fig 5.3-9/10)

Ave. concentration (Fig 5.3-13)

5


For turbulent flow, mass transfer is increase by mass eddy diffusivity, ε

(m/s)

ε is vary with distance, average ε will be used ( ). J*A1 is normally used,

which is flux of A on surface area A1(since the cross sectional area may

vary) relative to the whole bulk phase.

Write using a convective mass transfer coefficient

is mass transfer coefficient ( unit: kgmol/ (s.m2.(kgmol/m3)) or m/s)

dz

dc)D(*J

A

ABA

)(

)(

* 2112

1 A A

M AB

A cc z z

D

J

)(' 21 A Ac A cck N

12

' z z

Dk M AB

c

6



Mass Transfer Coefficient for Equimolar Counter

Diffusion (Table 7.2-1)

GASES:

N A= k’ c (c A1-c A2 )= k’ G (p A1-p A2 )= k’ y(y A1-y A2 )

LIQUID:

N A

= k’ c

(c A1

-c A2

)= k’ L

(c A1

-c A2

)= k’ x

(x A1

-x A2

)

7

Mass Transfer Coefficient for A Diffusing throughStagnant, Non-diffusing B (Table 7.2-1)

GASES:

N A= k c (c A1-c A2 ) = k G (p A1-p A2 )= k y(y A1-y A2 )

LIQUID:

N A= k c (c A1-c A2 ) = k L (c A1-c A2 )= k x(x A1-x A2 )8



Relationship Between Mass Transfer Coefficient

(Table 7.2-1)GASES:

LIQUID:

Unit can be refer to Table 7.2-1

P yk c yk k yk

pk P k RT

pk

RT

P k ck

M BG BM c y BM y

BM GG BM

ccc

'

'''

BM x x L

BM L Lc xk k M

k c xk ck ck '

'''

9

Mass Transfer under High Flux

(case : diffusing through stagnant, nondiffusing B

where diffusion-induced convection is present )

Hence for the film theory,

Combining:

1 2 1 2

'( ) ( )

c

A A A A A

BM

k N c c kc c c

x

' '

1c x

c BM x

k k

k x k

0 0

1

' '

1c x A

c BM x

k k x

k x k

10

ck

Rewriting Eq. (7.2-19) for the flux N A at the surface z = 0 where x A = x A1,



EXAMPLE 7.2-11/3

Vaporizing A and Convective Mass Transfer

A large volume of pure gas B at 2 atm pressure is flowing over a surface from which pure A is vaporizing. The liquid A completely wets the surface, which is a blotting paper. Hence, the partial pressure of A at the surface is the vapor pressure of A at298 K, which is 0.20 atm. The has been estimated to be 6.78 x 10-5 kg mol/s · m2 ·mol frac. Calculate NA, the vaporization rate, and also the value of k y and k G.

Solution:

This is the case of A diffusing through B, where the flux of B normal to the surface iszero, since B is not soluble in liquid A. pA1 = 0.20 atm and pA2 = 0 in the pure gas B.Also, yA1 = pA1/P = 0.20/2.0 = 0.10 and yA2 = 0. We can use Eq. (7.2-12) with molefractions:

Equation 7.2-12

However, we have a value for which is related to k y from Table 7.2-1 byEquation 7.2-16

The term yBM is similar to xBM and is, from Eq. (7.2-11),

11

EXAMPLE 7.2-12/3

Equation 7.2-11

Substituting into Eq. (7.2-11),

Then, from Eq. (7.2-16),

Also, from Table 7.2-1,

Equation 7.2-17

Hence, solving for k G and substituting knowns,

12

3.569 x 10 -5 kg mol/s.m2 . atm



EXAMPLE 7.2-13/3

For the flux, using Eq. (7.2-12),

Also,

Using Eq. (7.2-12) again,

Note that in this case, since the concentrations were dilute, y BM is close to 1.0and k y and differ very little.

13

Dimensionless Number 1/2

1. Reynolds Number

L= D p for sphere; Diameter, D for pipe; Length, L for flat plate

2. Schmidt Number

3. Sherwood Number

N Sc

D AB

L N Re

AB

x

AB

BMc

AB

csh

D

L

c

'k

D

L)yk (

D

L'k N

14



Dimensionless Number 2/2

4. Stanton Number

GM= ρ/MAv=c

5. JD

factor

M

y

M

GcSt

G

'k

G

P'k

v

'k N

J D

k '

c

v N

Sc 23

k 'G P

G M

N Sc

23 N

Sh / N

Re. N

Sc

13

15

Mass Transfer for Flow Inside Pipes1/2

1. Laminar FlowUse Figure 7.3-2

For liquid and gases

flowing inside pipe

N Re < 2100

Use rod-like flow

where

w: mass flow

cA :the exit concentration,

cA0 :inlet concentration,

cAi :concentration at the interface between the wall and the gas 16



Mass Transfer for Flow Inside Pipes2/2

2. Turbulent Flow

N Re > 2100

0.6 <N Sc < 3000

3183.0

Re023.0)('

Sc

AB

BM c

AB

cSh N N

D

D pk

D

Dk N

17

EXAMPLE 7.3-1. Mass Transfer Inside a TubeA tube is coated on the inside with naphthalene and has an inside diameter of 20 mm anda length of 1.10 m. Air at 318 K and an average pressure of 101.3 kPa flows through this

pipe at a velocity of 0.80 m/s. Assuming that the absolute pressure remains essentiallyconstant, calculate the concentration of naphthalene in the exit air. Use the physical

properties given in Example 6.2-4.

Solution: From Example 6.2-4, DAB = 6.92 x 10-6 m2/s and the vapor pressure pAi = 74.0 Paor cAi = pAi/RT = 74.0/(8314.3 x 318) = 2.799 x 10-5 kg mol/m3. For air, from AppendixA.3, μ = 1.932 x 10-5 Pa · s, ρ = 1.114 kg/m3. The Schmidt number is

The Reynolds number is

Hence, the flow is laminar. Then,

Using Fig. 7.3-2 and the rodlike flow line, (cA - cA0)/(cAi - cA0) = 0.55. Also, cA0(inlet) = 0.Then, (cA - 0)/(2.799 x 10-5 - 0) = 0.55. Solving cA (exit concentration) = 1.539 x 10-5 kgmol/m3. 18

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Mass Transfer for Flow Parallel to Flat Plates1/2

For gases or evaporation of liquids in the gas phase

Laminar region; N Re,L= L ν ρ/μ < 15 000

(dimensionless JD factor)

In terms of Sherwood number ( JD= N Sh / (NRe NSc1/3))

5.0

Re,664.0 L D N J

315.0

Re,

'

664.0.

sc L

AB

c sh N N

D

Lk N

19

Mass Transfer for Flow Parallel to Flat Plates2/2

Gases and 15 000 < N Re,L < 300 000

Liquid and 600 < N Re,L < 50 000

2.0

Re,036.0 L D N J

5.0

Re,99.0 L D N J

20



Mass Transfer for Flow Past Single Sphere

1. Gases; 0.6 < N Sc < 2.7; 1 < N Re = D pv/ < 48000

2. Liquid; 2 < N Re = D pv/ < 2000

3. Liquid; 2000 < N Re = D pv/ < 17000

3/153.0

Re552.02 ScSh N N N

3/162.0

Re347.0 ScSh N N N

3/150.0

Re95.02 ScSh N N N

23

EXAMPLE 7.3-3 1/3

Mass Transfer from a SphereCalculate the value of the mass-transfer coefficient and the flux for mass transfer

from a sphere of naphthalene to air at 45°C and 1 atm abs flowing at a velocity of

0.305 m/s. The diameter of the sphere is 25.4 mm. The diffusivity of naphthalene in

air at 45°C is 6.92 x 10-6 m2/s and the vapor pressure of solid naphthalene is 0.555 mm

Hg. Use English and SI units.

Solution: In English units DAB = 6.92 x 10-6(3.875 x 104) = 0.2682 ft2/h. The diameter

D p = 0.0254 m = 0.0254(3.2808) = 0.0833 ft. From Appendix A.3 the physical

properties of air will be used, since the concentration of naphthalene is low:

The Schmidt number is


24






EXAMPLE 7.3-32/3

Equation (7.3-33) for gases will be used:

From Eq. (7.3-3),

Substituting the knowns and solving,

From Table 7.2-1,

Hence, for T = 45 + 273 = 318 K = 318(1.8) = 573°R, 25

EXAMPLE 7.3-33/3

Since the gas is very dilute, yBM 1.0 and .

Substituting into Eq. (7.2-12) for A diffusing through stagnant B and noting that

pA1 = 0.555/760 = 7.303 x 10-4 atm = 74.0 Pa and pA2 = 0 (pure air),

The area of the sphere is

Total amount evaporated = NA A = (1.18 x 10-4)(2.18 x 10-2) = 2.572 x 10-6 lb mol/h =

(1.599 x 10-7)(2.025 x 10-3) = 3.238 x 10-10 kg mol/s

26














Mass Transfer to Packed Bed1/2

1. Gases in packed bed of spheres; 10 < N Re < 10 000;

Where, N Re is

v’ = superficial velocity

4069.0

Re

458.0 N J D

'Re

p D N

27

Mass Transfer to Packed Bed2/2

Liquid in packed bed; 0.0016 < N Re < 55 ; 165 < N Sc < 70 000

Liquid in packed bed; 55 < N Re < 1 500 ; 165 < N Sc < 10 690

Gases or Liquid in Fluidized bed of spheres; 10 < N Re < 4 000

Liquid in Fluidized bed; 1 < N Re < 10

3/2

Re

09.1 N J D

31.0

Re

250.0 N J D

4069.0

Re

458.0 N J D

72.0

Re

1068.1 N J D

28



Mass Transfer Rate for Packed Bed Total external surface area , A

V b = total volume of bed

Mass transfer rate is calculated using log mean driving force at inlet and

outlet of the bed

cAi = concentration at the surface of the solidcA1 = inlet bulk fluid conc.

cA2= outlet bulk fluid concentration

V = Volumetric Flowrate of fluid

A = Total surface area of sphere

b

p

b V D

aV A )1(6

)(

)(

)(

)()(12

2

1

21 A A

A Ai

A Ai

A Ai A Aic A ccV

cc

cc In

cccc Ak A N

29

Flow Perpendicular to Single Cylinder

0.6 < N Sc < 2.6 for Gases

1000 < N Sc < 3000 for Liquid

50 < N Re < 50000

487.0Re600.0 N J D

30



Example 7.3-41/2

Mass Transfer of a Liquid in a Packed Bed Pure water at 26.1°C flows at the rate of 5.514 x 10 -7 m3/s through a packed bed of

benzoic-acid spheres having a diameter of 6.375 mm. The total surface area of thespheres in the bed is 0.01198 m2 and the void fraction is 0.436. The tower diameter is0.0667 m. The solubility of benzoic acid in water is 2.948 x 10-2 kg mol/m3.

1. Predict the mass-transfer coefficient k c. Compare with the experimental value of4.665 x 10-6 m/s by Wilson and Geankoplis.

2. Using the experimental value of k c, predict the outlet concentration of benzoicacid in the water.

Solution: Since the solution is dilute, the physical properties of water will be usedat 26.1°C from Appendix A.2. At 26.1°C, μ = 0.8718 x 10-3 Pa · s, ρ = 996.7 kg/m3.At 25.0°C, μ = 0.8940 x 10-3 Pa · s, and from Table 6.3-1, DAB = 1.21 x 10-9 m2/s.To correct DAB to 26.1°C using Eq. (6.3-9), . Hence,

The tower cross-sectional area = (π/4)(0.0667)2 = 3.494 x 10-3 m2.

Then ν' = (5.514 x 10-7)/(3.494 x 10-3) = 1.578 x 10-4 m/s. Then,

31

Example 7.3-42/2


Using Eq. (7.3-37) and assuming k c = k c’ for dilute solutions,

Then, using Eq. (7.3-5) and solving,

The predicted k c= 4.447 x 10-6 m/s.

This compares with the experimental value of 4.665 x 10-6 m/s.

For part (b), using Eqs. (7.3-42) and (7.3-43),

Equation 7.3-44

The values to substitute into Eq. (7.3-44) are cAi = 2.948 x 10-2, cA1 = 0, A = 0.01198,V = 5.514 x 10-7.

Solving, cA2 = 2.842 x 10-3 kg mol/m3.32















Lecture Notes Chap 2 CHEM3002 2013 2014

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