Unit Operations (CHE 425 Course Notes)

Unit Operations

(CHE 425 Course Notes)

T.K. Nguyen

Chemical and Materials Engineering Cal Poly Pomona

(Fall 2016)

Contents CHE 425

Chapter 1: Fundamentals of Mass Transfer

1.1 Molecular Mass Transfer 1-1

Example 1.1-1 Flux of oxygen at steady state 1-1

Example 1.1-2 Rate of evaporation of water 1-2

1.2 Flue Gas Desulfurization (FGD) Process 1-3

The enhancement factor of the liquid film coefficient 1-7 Example 1.2-1 Evaporation of a water droplet 1-9 Example 1.2-2 Absorption of SO

2

into a lime slurry droplet 1-10

1.3 Diffusivity Measurement 1-14

Example 1.3-1 Diffusion in a capillary tube 1-17

Example 1.3-2 Diffusion time in a capillary tube 1-18

1.4 Diffusion with Chemical Reaction 1-19

Example 1.4-1 Reaction time with carbon particles 1-19 Example 1.4-2 Consumption rate with carbon pellets 1-21

Example 1.4-3 Biofilms 1-23

1.5 Unsteady State Diffusion: Differential Mass Balance 1-27 1.6 Unsteady State Diffusion: Approximate Solutions 1-32

Example 1.6-1 Unsteady diffusion in a slab 1-37

Example 1.6-2 Unsteady diffusion in white pine 1-38

Example 1.6-3 Diffusion in spray tower 1-42

Example 1.6-4 Diffusion in a finite cylinder 1-43 Example 1.6-5 Semi-infinite solution approximation 1-45

Chapter 2: Convective Mass Transfer

2.1 Introduction 2-1

Example 2.1-1 Mass transfer from a flat container 2-2

Example 2.1-2 Rate of absorption of H

2

S. 2-3

Example 2.1-3 Mass transfer rate of napthalene. 2-5 Example 2.1-4 Mass transfer coefficient from experiment. 2-6 Example 2.1-5 Mass transfer in bubble column. 2-7

2.2 Packed Column 2-11

Example 2.2-1 Height of an absorption tower 2-17

Chapter 3: Phase Equilibrium Calculation

3.1 Introduction 3-1

3.2 Vapor-Liquid Equilibrium calculation 3-1

Example 3.2-1: Bubble point temperature calculation 3-4

Example 3.2-2: Bubble point pressure calculation 3-5

Example 3.2-3: Dew point temperature calculation 3-6

Example 3.2-4: Dew point pressure calculation 3-8

Example 3.2-5: Dew point pressure with Van Laar equations 3-9

3.3 Isothermal Flash Calculation Using K-values 3-11

Case 1: Fraction of the feed, f, vaporized is specified 3-12 Case 2: Fraction of the feed, f, vaporized is determined 3-12 Example 3.3-1: Compositions of streams leaving a flash drum 3-14 Example 3.3-2: Fraction of the feed vaporized in a flash drum 3-21 3.4 Distribution of a Solute between Liquid Phases 3-25

3.4a Solubility of a Solid in a Liquid Phase 3-25

Example 3.4-1: Solubility of a Drug 3-27

3.4b Distribution of a Solute between Liquid Phases 3-30 Example 3.4-2: Drug Extraction from the Aqueous Phase 3-33 Example 3.4-3: Purification of an Antibiotic 3-34

3.4c Single-Stage Equilibrium Extraction 3-35

Example 3.4-4: Drug Extraction from the Aqueous Stream 3-36

3.5 Non-isothermal Flash 3-37

Example 3.5-1: Adiabatic flash drum 3-37

Example 3.5-2: Non-adiabatic flash drum 3-39

Chapter 4: Distillation

4.1 Vapor Liquid Equilibrium Relations 4-1

Example 4.1-1: Txy diagram for a benzene-toluene mixture 4-2 4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System 4-8 Example 4.2-1: Single equilibrium stage calculation 4-8

4.3 Simple Batch or Differential Distillation 4-12

Example 4.3-1: A semi-batch distillation unit 4-12

4.4 Distillation with Reflux 4-15

4.4a Introduction 4-15

4.4b McCabe-Thiele Method 4-16

Example 4.4-1: Rectifying section calculation 4-18 Example 4.4-2: Stripping section calculation 4-24

Example 4.4-3: Number of equilibrium stages 4-33

Example 4.4-4: Total and minimum flux ratios 4-45

4.5 Tray Efficiency 4-49

Example 4.5-1: Overall column efficiency 4-51

Example 4.5-2: Plate efficiency 4-52

4.6 Enthalpy-Concentration Diagram 4-53

Example 4.6-1: Enthalpy-concentration plot for benzene-toluene 4-54

4.7 Distillation Using Enthalpy-Balance 4-59

Example 4.7-1: Enthalpy-balance for equilibrium stages 4-60 Example 4.7-2: Enthalpy diagram for methanol-water system 4-67 Example 4.7-3: Malab codes for Enthalpy diagram 4-71

Chapter 5: Approximate Methods for Multicomponent Distillation

5.1 Introduction 5-1

5.2 Fenske-Underwood-Gilliland Method 5-3

Example 5.2-1: Butane-pentane distillation 5-6

5.3 Underwood Equation for Minimum Reflux 5-15

Example 5.3-1: Underwood’s method for a depropanizer 5-19

Chapter 6: Rigorous Methods for Distillation

6.1 Theoretical Model for an Equilibrium Stage 6-1

6.2 Rigorous Computational Methods 6-5

Example 6.2-1: Tridiagonal matrix equations 6-8

Example 6.2-2: Stage by stage calculation 6-11

Algorithm for Wang-Henke BP methods for distillation 6-21

Example 6.2-3: BP method 6-22

6.3 Rate-Based Model for Distillation 6-29

Example 6.3-1: Rate-Based Ternary Distillation Calculations 6-34 6.4 Equilibrium Distillation Calculations using ChemSep program 6-36 Example 6.4-1: Distillation Calculations using ChemSep 6-36

Chapter 7: Heat Exchangers

7.1 Introduction 7-1

7.2 Heat Exchanger types 7-4

7.3 Analysis of Heat Exchangers 7-9

Example 7.3-1: Different exchanger configurations 7-17 Example 7.3-2: Length of tube in heat exchanger 7-19 Example 7.3-3: Heat transfer rate for a counter flow exchanger 7-20

7.4 The Effectiveness-NTU Method 7-22

Example 7.4-1: Effectiveness-NTU relation for exchanger 7-26 Example 7.4-2: Counter-flow heat exchange in whales 7-29 Example 7.4-3: LMTD and Effectiveness-NTU method 7-31 7.5 Special Operations for Liquid/Gas Heat Exchangers 7-33

Chapter 8: Absorption and Stripping

8.1 Introduction 8-1

8.2 Single-Component Absorption 8-3

Example 8.2-1: Equilibrium and operating lines 8-4 Example 8.2-2: Number of trays in an absorber 8-9 8.3 Multicomponent Absorption Using Shortcut Methods 8-11

Example 8.3-1: Composition of the exit gas stream 8-16

Appendix

A. Distillation Calculation using ChemSep Software A-1 B. Previous Exams (2009)

Quiz 1 B-1

Quiz 2 B-4

Quiz 3 B-7

Quiz 4 B-11

Quiz 5 B-14

Answers to 2009 quizzes B-17

C. Previous Exams (2010)

Quiz 1 C-1

Quiz 2 C-4

Quiz 3 C-7

Quiz 4 C-11

Quiz 5 C-15

Answers to 2010 quizzes C-18

D. Previous Exams (2012)

Quiz 1 C-1

Quiz 2 C-4

Quiz 3 C-7

Quiz 4 C-10

Quiz 5 C-13

Chapter 1

Fundamentals of Mass Transfer

When a single phase system contains two or more species whose concentrations are not uniform, mass is transferred to minimize the concentration differences within the system. In a multi-phase system mass is transferred due to the chemical potential differences between the species. In a single phase system where temperature and pressure are uniform, the difference in chemical potential is due to the variation in concentration of each species. Mass transfer is the basis for many chemical and biological processes. We will discuss the application of mass transfer in the removal of sulfur dioxide from the flue gas, a chemical process, and the design of an artificial kidney, a biological process.

1.1 Molecular Mass Transfer

For a binary mixture of A and B, the molar flux, N

A,z

, of species A relative to the z axis is N

A,z

=  cD

AB^dyA

+ y

A

(N

A,z

+ N

B,z

) (1.1)

dz

In this equation, c is the total concentration, D

AB

is the diffusivity of A in B, y

A

is the mole fraction of A, and N

B,z

is the molar flux of B. For a binary mixture D

AB

= D

BA

.The term  cD

AB^dyA

is the molar flux, J

A

, resulting from the concentration gradient and the term

dz

y

A

(N

A,z

+ N

B,z

) is the molar flux resulting from the bulk flow of the fluid.

Example 1.1-1 ---

A mixture of oxygen and nitrogen gas is contained in a pipe at 298 K and 1 atm total pressure which is constant throughout. At one end of the pipe the partial pressure p

_A1

of oxygen is 0.70 atm and at the other end 0.8 m, p

A2

= 0.2 atm. Calculate the flux of oxygen at steady state if D

AB

of the mixture is 0.206 cm

²

/s.

Solution --- Since the temperature and pressure is constant throughout the pipe, the total concentration is a constant.

c =

^P

= = 4.0910

^-5

mol/cm

³ RT

1 82.057 298 

Note: R = 82.057 cm

³

atm/molK

The total concentration is a constant therefore N

A,z

=  N

B,z

. This condition is known as

equimolar counterdiffusion. The molar flux of A is

N

A,z

=  cD

ABdy^A

+ y

A

(N

A,z

+ N

B,z

) =  cD

AB

dz

dyA

dz

For steady state and constant area of mass transfer N

A,z

= constant. Separating the variable and integrating

N

A,z

=  cD

AB 0

L

dz

 

y^y_A^A₁²

dy

^A

N

A,z

=

^cDAB

⁽

^{yA1 yA2}

⁾ =

L

 (4.09 10 )(0.206)(0.7 0.2)

⁵

80







N

A,z

= 5.2710

^-8

mol/cm

²

s

Example 1.1-2 --- Water in the bottom of a narrow metal tube is held at a constant

temperature of 298 K. The dry ambient air outside the tube is at 1 atm (101.3 kPa) and 298 K. Water evaporates and diffuses through the air in the tube, and the diffusion path z

2

 z

1

is 50 cm long.

Calculate the rate of evaporation at steady state in mol/scm

²

. The diffusivity of water vapor (A) in air (B) at 1 atm and 298 K is 0.250 cm

²

/s. Assume that air is insoluble in water.

Solution --- The molar flux of A (water vapor) is

N

A,z

=  cD

AB^dyA

+ y

A

(N

A,z

+ N

B,z

)

dz

Since air is insoluble in water, it is stagnant (or nondiffusing) and N

B,z

= 0. Solving for N

A,z

give

N

A,z

(1  y

A

) =  cD

ABdy^A

 N

A,z

dz =  dy

A

dz

1

AB A

cD



y

N

A,z

=  cD

AB

 N

A,z

L = cD

AB

ln

0 L

dz



y^y_A^A₁²

₁

^A A

dy



y



²

1

1 1

A A

y y





The ambient air is dry so y

A2

= 0. Vapor pressure of water at 298 K is 3.17 kPa, therefore y

A1

= 3.17/101.3 = 0.0313.

N

A,z

=

^cDAB

ln = ln = 6.510

^-9

mol/cm

²

s

L

2 1

1 1

A A

y y





0.250 82.057 298 50  

1 0 1 0.313





Water (A) Air (B)

1 2

N

A

1.2 Flue Gas Desulfurization (FGD) Process

The removal of sulfur dioxide from the flue gas is known as the Flue Gas Desulfurization (FGD) Process. As industry emissions of SO

x

led to a significant increase in acid rain, the government has mandated that industry restrict SO

_x

emissions to various levels, depending on locality. These mandates are enforced through organizations at the federal, state, and local levels. In response to these mandates, government, industry, and scholastic institutes have endeavored to model the SO

x

absorption / reaction process within a Ca(OH)

2

slurry droplet in an FGD reactor.

A typical FGD system shown in Figure 1.2-1 is used to contact SO

2

laden flue gas with spray droplets of a lime-slurry prepared by a rotary atomizer. The amount of lime added depends on the stoichiometric ratio to be used and the inlet flue gas SO

2

concentration

¹

. The amount of water added to the system is controlled by inlet flue-gas temperature and humidity, and the desired approach to saturation at the FGD outlet. The approach to saturation is defined as the difference between the outlet gas temperature and the wet-bulb temperature for the process, with only pure water in the spray. As lime slurry droplets are sprayed into relatively dry hot flue gas, the water will evaporate and the temperature of the gas and the unevaporated droplets both decrease until the gas is saturated with water vapor. The temperature of the flue gas at this limiting condition is called the adiabatic saturation temperature, which is essentially the wet-bulb temperature for the process. In a FGD operation, the outlet gas temperature is always higher than the adiabatic saturation temperature by an amount called the approach to saturation.

Lime slurry is prepared in a slaker to obtain a slurry of fine-grained lime particles. The lime particle size is about

¹

4-6  m. The lime slurry is then diluted further with water and atomized into 20-150  m droplets

²

. There is a limit to the mass fraction of solid in the droplet to prevent clogging and abrasion

³

. The mass fraction of lime in the form of calcium hydroxide Ca(OH)

2

for a typical slurry is about 0.20.

The required fraction of lime in a slurry is a constraint to the maximum amount of lime that can be used to remove SO

2

from the flue gas. If lime is added to the slurry at a rate higher than the limit, the corresponding water rate required will push the approach to saturation closer to zero and not all the water will evaporate from the solid leaving the FGD unit. If the dew point of the outlet flue gas is approached too closely, condensation may also occur inside the downstream equipment, causing problems. However SO

2

removal efficiency improves with FGD operation nearer to flue-gas saturation condition.

A 100  m droplet dries within a few seconds inside a FGD unit, and the typical residence time is about 10 seconds

²

. In the FGD unit, two processes occur simultaneously: water evaporates from the droplet; and SO

2

is absorbed in, and reacts with Ca

⁺⁺

ions from the Ca(OH)

2

. Most of the Ca(OH)

2

is present as solid particles when the drop is formed since the solubility of Ca(OH)

2

is low.

1 Damle, A. S., “Modeling of SO2 removal in Spray-Dryer Flue-Gas Desulfurization System”, EPA-600/7-85- 038, December (1985)

2 Hariot, P. and M. Kinzey, “Modeling the Gas and Liquid Phase Resistance in the Dry Scrubbing Process for CO2 Removal,” Proceedings: Third Pittsburgh Coal Conference, 220 (1986)

3 Engineering Institute at the California State Polytechnic University, Pomona, 2004-2005 BP Team, 125 (2005)

Lime Slurry

Dilution Water

Slaker Temperature = 180 F

^o

Flue Gas from the Waste Heat Boiler

380 F

^o

Flue Gas to Baghouse

Ash

223 F

^o

Figure 1.2-1 A typical FGD reactor.

The operation of a FGD unit is essentially a spray drying process for SO

₂

control. The evaporation from a slurry droplet proceeds in two main stages: a first stage, called the constant rate drying period, and a second stage, called the falling rate period. In the constant rate period, the evaporation rate is constant and water is the continuous phase. The rate of droplet drying is determined by the simultaneous heat transfer from the gas phase to the droplet and water vapor transfer from the droplet to the gas phase. Figure 1.2-2 shows an idealized slurry droplet in the initial stage of the drying process.

To summarize the Constant-Rate period, the processes to be considered are:

 SOx flux from the flue gas into the droplet.

 Heat Flux from the flue gas into the droplet.

 Water flux from the slurry droplet into the flue gas.

 Shrinking diameter (reduction in surface area) of the slurry droplet associated with

the flux of water.

Calcium hydroxide particle Water

Figure 1.2-2: A slurry droplet in the Constant-Rate Period.

In the falling rate period, the particles contact one another as a continuous phase. The moisture content of the droplet at this point is called the critical moisture content. The drop diameter does not change significantly during this period, while evaporation continues until the droplet moisture content reaches equilibrium with the surrounding flue gas. As water recedes, a coating of calcium sulfites and sulfates may be deposited on the outer surface of the solid matrix. This coating inhibits further reaction by limiting the accessibility of the un- reacted particle interior. After the moisture content in the droplet falls to the equilibrium moisture content, the droplet may be considered a dry-porous solid to determine further SO

2

removal. After the spray dryer, the flue gases pass through a bag-house where SO

2

removal continues in the filter cake as the unused lime is collected on the bag.

To summarize the Falling-Rate period, the processes to be considered are:

 SOx flux from the flue gas into the solid mass of particles.

 Heat Flux from the flue gas into the solid mass of particles.

Figure 1.2-3: A Slurry Droplet in the Falling-Rate Period

Although droplet evaporation and SO

2

absorption occur simultaneously, the droplet drying process is relatively independent of the SO

2

absorption process. However, the SO

2

absorption reaction process is strongly related to the drying process and the droplet moisture content.

The overall process is a function of inlet gas temperature, relative humidity, approach to saturation temperature, equilibrium moisture content of the solid, particle and slurry droplet diameter, calcium hydroxide concentration, SO

_x

concentration, and residence time through the reactor. SO

x

removal begins with a constant rate phase and transitions to a falling rate once the critical moisture content is reached.

Water Flux out of Droplet

Heat Flux into Droplet SOx Flux into Droplet

Heat Flux SO₂Flux

Ca(OH)

₂

Particles

Heat Flux into the Solid Particle SOx Flux into the SolidParticle

Figure 1.2-4 summarizes the main process in the FGD process. Flue gas with high concentration of SO

₂

is in contact with a spray of lime slurry. SO

₂

is absorbed into the lime slurry. As the calcium hydroxide dissolves, the sulfur and calcium species are then able to react, producing a calcium sulfite solution that precipitates onto the lime particle. Heat is transferred from the hot flue gas to evaporate the water in the slurry. At the outlet, the slurry becomes solid waste while the lower temperature flue gas gains moisture and contains less SO

2

.

Lime Slurry

Dilution Water

Slaker Temperature = 180 F^o

Flue Gas from the Waste Heat Boiler

380 F^o

Flue Gas to Baghouse

Ash

223 F^o

Figure 1.2-4 Schematic of the FGD process.

We will now derive many formulas that can be used to model the FGD process. The evaporation of a slurry droplet can be model by a quiescent droplet-gas system as shown in Figure 1.2-5.

Droplet surface

Bulk gas phase y_d,w

y_g,w T_d

T_g

Figure 1.2-5 Schematic of the FGD process.

We assume a pure water droplet with radius R. y

d,w

is the water mole fraction at the surface in the gas phase. y

_g,w

is the mole fraction of water in the bulk gas phase. The molar flux of water vapor (A) into the gas (B) phase is

N

A,r

=  cD

AB^dyA

+ y

A

(N

A,r

+ N

B,r

)

dr

In this equation, r is the radial distance from the center of the drop. For quiescent droplet-gas system, the gas is stagnant and N

B,r

= 0. Therefore

N

A,r

=  1

AB A

cD



y ^A dy

dr

The system is at steady state, however the molar flux is not a constant since the area of mass transfer 4r

²

is not a constant. For steady state, the mass (mole) transfer rate, 4r

²

N

A,r

, is a constant

W

A

= 4r

²

N

A,r

=  4r

²

= constant 1

AB A

cD



y dyA

dr

Separating the variables and integrating gives

W

A ₂

=  4 cD

AB R

dr r



 ,^,

1

g w d w

y A

y A

dy



y



= 4 cD

AB

ln  W

A

= 4RcD

AB

ln

WA

R

, ,

1 1

g w d w

y y





, ,

1 1

g w d w

y y





This equation provides the moles of water evaporated from the droplet when the radius of the drop is R.

The enhancement factor of the liquid film coefficient

Figure 1.2-6 shows the concentration profile for SO

2

in the liquid film without and with

chemical reaction. If there is no chemical reaction in the liquid, SO

2

must travel the entire

length δ of the liquid film before it can react with lime. However dissolved into water in the

form of calcium ion will react with SO

2

at a location within the liquid film. The reaction

plane is a distance λ from the droplet surface. Since λ < δ, the absorption rate of SO

2

is higher

with chemical reaction within the liquid film.

C*_SO2 C^*_SO2

Film thickness Film thickness

Reaction plane

C*_lime Lime Lime

Water

Figure 1.2-6 Absorption of SO

2

without and with chemical reaction in the liquid phase.

The mass transfer coefficient without reaction is given by k

_L

=

^DSO2



For the very fast reaction of dissolved SO

2

with dissolved lime in the bulk liquid, the mass transfer coefficient is then

k*

L

=

^DSO2



From the film theory model, the film thickness is divided in two parts, of width  and (    ), such that the stoichiometric balance at the reaction plane is fulfilled:

= D

lime SO2

D ²

*

CSO



*

Clime

  

Rearranging the equation to solve for  we have (    ) = D

lime



SO2

D 2

*

CSO C_lime^*

 = 

^{2 2}

2 2

*

SO SO

* *

SO SO lime lime

D C

D C  D C

Therefore the mass transfer coefficient when chemical reaction occurs in the liquid film is given as

k*

L

=

^SO2

= = k

L

= k

L

D



²

DSO



2 2

* lime lime

*

SO SO

1 D C

D C

 

  

 

 

^{2 2}

* lime lime

*

SO SO

1 D C

D C

 

  

 

  

In this equation =  = the enhancement factor of the liquid film coefficient k

L

2 2

* lime lime

*

SO SO

1 D C

 D C

due to the very fast reaction of dissolved SO

2

with dissolved lime in the bulk liquid.

Chapter 1

Example 1.2-1 ---

A water droplet having a diameter of 0.10 mm is suspended in still air at 50

^o

C, 1.0132×10

⁵

Pa (1 atm), and 20% relative humidity. The droplet temperature can be assumed to be at 50

^o

C and its vapor pressure at 50

^o

C is 7.38 kPa.

1) Calculate the initial rate of evaporation of water if D

AB

of water vapor in air is 0.288 cm

²

/s.

2) Determine the time for the water droplet to evaporate completely.

Solution ---

Droplet surface

Bulk gas phase yd,w

y_g,w

T_d Tg

The molar flux of water vapor (A) into the air (B) is N

A,r

=  cD

AB^dyA

+ y

A

(N

A,r

+ N

B,r

)

dr

In this equation, r is the radial distance from the center of the drop. For still air N

B,r

= 0, we have

N

A,r

=  1

AB A

cD



y dyA

dr

The system is not at steady state, the molar flux is not independent of r since the area of mass transfer 4r

²

is not a constant. Using quasi steady state assumption, the mass (mole) transfer rate, 4r

²

N

A,r

, is assumed to be independent of r at any instant of time.

W

A

= 4r

²

N

A,r

=  4r

²

= constant 1

AB A

cD



y dyA

dr

Separating the variables and integrating gives

W

A ₂

=  4 cD

AB R

dr r



 y^y_{d w}^{g w}_,^,

₁

^A A

dy



y



= 4 cD

AB

ln  W

A

= 4RcD

AB

ln

WA

R

, ,

1 1

g w d w

y y





, ,

1 1

g w d w

y y





This equation provides the moles of water evaporated from the droplet when the radius of the drop is R.

1) Calculate the initial rate of evaporation of water if D

AB

of water vapor in air is 0.288 cm

²

/s.

The initial rate of evaporation occurs when the diameter of the drop is 0.10 mm.

R g 82.057 cm³ atm mol K

 R .005 cm P 1 atm T (50 273) K

D AB .288cm²

s Mw 18 g

c P mol

R g T c 3.773 105 mol cm³



P A 7.38

101.32atm y dw P A

P y dw 0.073

For 20% relative humidity

_{y gw .2}^{P A}

P

 y gw 0.015

W A 4  R c D ABln 1 y gw 1 y dw

 W A 4.161 10 8 s1 mol

The initial mass evaporation rate is

4.161 10 ⁸18 7.49 107 g s

2) Determine the time for the water droplet to evaporate completely.

Making a material balance around the water droplet gives = − 4MwRcD

AB

ln

d dt

4

3

3  

R ^A

 

 

 

, ,

1 1

g w d w

y y





4ρ

A

R

^2dR

= − 4MwRcD

_AB

ln

dt

, ,

1 1

g w d w

y y



 Simplifying and rearranging we have

RdR = − Mwc

^AB

ln dt = − Kdt

A

D



, ,

1 1

g w d w

y y





In this equation, K = Mwc

^AB

ln

A

D



, ,

1 1

g w d w

y y





Integrating the equation from the initial radius to zero gives the time for complete evaporation of the droplet.

= − K => t =

0

Ri

 RdR

0 t

dt

 ₂

^R_Kⁱ²

Substituting the numerical values we have

y dw .0728 y gw .0146 D AB .288cm² s

D AB ln 1 y gw 1 y dw

 0.0175 cm²

 s

Density of liquid water at 50

^o

C is 0.988 g/cm

³

K =

^AB

ln (Mw)c = (0.0175)(18)(3.773×10

^-5

)/(0.988) = 1.207×10

^-5

cm

²

/s

A

D



, ,

1 1

g w d w

y y





t = = 0.005

²

/(2×1.207×10

^-5

) = 1.04 s

2

2

Ri

K

Example 1.2-2 ---

A lime (CaO) slurry droplet having a diameter D

d

of 0.10 mm is suspended in a gas containing SO

2

at 50

^o

C, 1.0132×10

⁵

Pa (1 atm). The mole fraction of SO

2

in the gas phase is 10

^-4

. Rate of diffusion of SO

2

from the gas to the droplet surface is given by

N

s

= k

d

(πD

d2

) c (y

g,s

− y

d,s

)

In this equation c is the total gas concentration and k

d

is the mass transfer coefficient for SO

2

in the gas phase given by k

d

= ²

^{SO2 gas}

d

D D



Rate of diffusion of SO

2

from the droplet surface into the interior droplet is given by N

s

= k

L

 (πD

d2

) C

^*SO2

In this equation C

^*SO2

is the equilibrium SO

2

concentration at the droplet surface k

L

 is the

mass transfer coefficient for SO

2

in the liquid phase given by

k

L

 = k

L

2 2

* lime lime

*

SO SO

1 D C

D C

 

  

 

 

where

k

L

=

^DSO2

and the liquid film thickness δ at the surface is estimated to be 1 μm.



= = the enhancement factor of the liquid film coefficient k

L

due



2 2

* lime lime

*

SO SO

1 D C

D C

 

  

 

 

to the very fast reaction of dissolved SO

2

with dissolved lime in the bulk liquid.

The gas- and liquid-phase surface concentrations are related by Henry’law as follows:

c y

d,s

= HC

^*SO2

The Henry’law constant for the solubility of SO

2

in water at 50

^o

C is given by H = 0.054 mol/cm in gas phase

₃³

mol/cm in liquid phase

Other data at 50

^o

C:

Equilibrium concentration of lime, C

^*lime

= 1.35×10

^-5

mol/cm

³

Diffusivity of SO

2

and lime in water: D

lime

= D

SO2

= 2×10

^-5

cm

²

/s Diffusivity of SO

2

in gas: D

SO2-gas

= 0.20 cm

²

/s

Estimate the rate of diffusion of SO

2

, N

s

.

Solution ---

We will estimate the enhancement factor =  by using the equilibrium

2 2

* lime lime

*

SO SO

1 D C

 D C concentration C

^*SO2

from the equation

c y

g,s

= HC

^*SO2

Note that in this equation y

g,s

, a known value, is used instead of y

d,s

.

R g 82.057 cm³ atm mol K

 R .005 cm P 1 atm T (50 273) K

c P

R g T c 3.773 10 5 mol cm³



C

^*SO2

= c y

g,s

/H = = 7.0×10

^-8

mol/cm

³

5 4

(3.773 10 )(10 ) .054

 



Since D

lime

= D

SO2

, =  = = 1 + = 194

2 2

* lime lime

*

SO SO

1 D C

 D C

2

* lime

* SO

1 C

 C ^{1.35 10}

₈⁵

7.0 10









The diffusion equation for SO

2

in the gas phase is rewritten as

= cy

g,s

− cy

d,s

(E-1)

(

2

)

s

d d

N k



D

The diffusion equation for SO

2

in the liquid phase is rewritten as = C

^*SO2

=

(

2

)

s

L d

N k

 

D

,

cyd s

H

= cy

d,s

(E-2)

(

2

)

s

L d

HN k

 

D

Adding equation (E-1) to equation (E-2) gives

= cy

g,s 2

s d

N



D

1

d L

H k k 

 

  

 

Therefore

N

s

= (πD

d2

)

^,

1

g s

d L

cy H k k 

 

  

 

k

d

= ²

^{SO2 gas}

= = 40 cm/s

d

D D



4

(2)(0.20) 100 10 

^

k

L

=

^DSO2

= = 0.2 cm/s



5 6

2 10 10







= + = .025 + .00139 = .0264 s/cm 1

d L

H k

^

k



1 40

.054 (.2)(194)

The rate of diffusion of SO

2

is then

N

s

= π×(100×10

^-4

)

²

= 4.49×10

^-11

mol/s

5 4

(3.773 10 )(10 ) .0264

 



---

1.3 Diffusivity Measurement

For liquids of high volatility, A. Stefan (1879) devised a convenient means of measuring the diffusivity D

AB

of their vapor A through a stagnant gas B. If the volatile substance A (e.g.

ethyl ether or ethanol) is placed in the lower part of a vertical capillary, then liquid A will evaporate and, by the mechanism of diffusion, travel to the end of the capillary. Maintaining the mouth of the capillary at a given composition automatically establishes the concentration gradient in the capillary, and the falling rate of the meniscus in the capillary provides the rate of transport. The capillary is placed in an envelope through which air is passed. At the meniscus the gaseous phase composition is specified by the vapor pressure of liquid A, the diffusing constituent. At the mouth of the capillary, the gaseous phase is essentially air. The gradient in the capillary is thus obtained by circulating sufficient air to reduce the substance A concentration at the mouth to a negligible quantity. The air rate should be low, constant, and not turbulent. The falling rate of the meniscus can be observed remotely with a cathetometer.

z = 0

z = z1 at t = 0 z = zt at t

pure liquid A PA1

PA2 gas B

(air)

Figure 1.3-1 Diffusion cell with moving liquid surface.

If the length of the diffusion path changes a small amount over a long period of time, a pseudo-steady state assumption may be used. This condition can be quantified as follow:

Let t

d

be the average time it takes for molecule A to travel the diffusion path z

1

. t

d

can be estimated by Einstein ‘s formula

t

d

=

2 1

2

_AB

z D

Let ∆z be the change in diffusion path (or the drop in liquid level) during the time t

d

. The

pseudo-steady state assumption may be used if

∆z/ z

₁

< 0.01

The molar flux of A is given by

N

A,z

=  cD

AB^dyA

+ y

A

(N

A,z

+ N

B,z

)

dz

We will assume that air is insoluble in liquid A, it is stagnant (or nondiffusing) and N

B,z

= 0.

Solving for N

A,z

give

N

_A,z

(1  y

_A

) =  cD

_AB^dyA

 N

_A,z

dz =  dy

_A

dz

1

AB A

cD



y

N

A,z

=  cD

AB

 N

A,z

z = cD

AB

ln

0 z

dz



y^y_A^A₁²

₁

^A A

dy



y



²

1

1 1

A A

y y





Using the definition of log mean average, y

B,lm

=

^{2 1}

=

2 1

ln( / )

B B

B B

y y

y y



_{2 1}

2 1

(1 ) (1 )

ln[(1 ) /(1 )]

A A

A A

y y

y y

  

 

we have

ln

²

=

1

1 1

A A

y y





^A1_{B lm,} ^A2 y y

y



The molar flux of A is then written as N

A,z

=

^{1 2}

,

( )

AB A A

B lm

cD y y zy



The molar flux is related to the amount of A leaving the liquid by the material balance N

A,z

=

^{1 2}

=

,

( )

AB A A

B lm

cD y y zy



_{A L,}

MA



_dz dt

Separating the variables and integrating the equation from t = 0 to t gives

0

=

t

dt



^{A L,}

MA



_,

1 2

( )

B lm

AB A A

y

cD y



y ¹

zt

z

zdz



We have

t =

^{A L,} MA



_,

1 2

( )

B lm

AB A A

y

cD y



y





 



 

2

2 1

2

z

z

_t

Replacing c by P/RT yields

t =

^{A L,}

=

MA



_,

1 2

( )

B lm

AB A A

y RT

D P y



y





 



 

2

2 1

2

z

z

_{t A L,}

MA



_,

1 2

( )

B lm

AB A A

y RT

D P



P





 



 

2

2 1

2

z

z

_t

Since

y

B,lm

=

^{1 2}

= =

2 1

ln[(1 ) /(1 )]

A A

A A

y y

y y



 



_{1 2}



2 1

ln[ (1 ) / (1 )]

A A

A A

P y y

P P y P y



 

,

PB lm

P

,

=

1 2

B lm

A A

y P



P

,

1 2

( )

B lm

A A

P P P



P

Therefore

t =

^{A L,} MA



_,

1 2

( )

B lm

AB A A

P RT

D P P



P

 

 



 

2

2 1

2

z

z

_t

Solving for the diffusivity we have

D

AB

=

)

(

_{1 2}

, ,

A A A

lm B L A

P P tPM

RT P



  

 



 

2

2 1

2

z

z

_t

In this equation

P

_B,lm

= [(P - P

_A1

) - (P - P

_A2

)]/ln [(P - P

_A1

)/(P - P

_A2

)]

P

_A1

= vapor pressure of liquid A at temperature T.

P

_A2

= partial pressure of vapor A at the mouth of the capillary.

R = gas law constant.

t = time during which the meniscus fall from z

₁

to z

_t

.

z

₁

= distance from the mouth of the capillary to the meniscus at t = 0.

z

_t

= distance from the mouth of the capillary to the meniscus at t.

P = ambient atmospheric pressure



A,L

= density of liquid A at T.

z = 0

z = L y_AL

y_A0

Chapter 1

Example 1.3-1 ---

A long glass capillary tube, of diameter 0.01 cm, is in contact with water at one end and dry air at the other. Water vapor evaporates at the wet end within the capillary, and the vapor diffuses through the capillary toward the dry end. How long is required for one gram of water to evaporate through this system? The vapor pressure of water is 17.5 mmHg at 20

^o

C, the temperature at which the entire system is maintained. Take the diffusivity of water vapor in air at 20

^o

C to be 0.20 cm

²

/s, and assume that the dry air is at a pressure of 760 mmHg.

Assume that the distance from the wet interface within the capillary to the dry end is always 10 cm. (

Ref. An Introduction to Mass and Heat transfer by Stanley Middleman.

)

Solution --- The molar flux of A (water vapor) is given by

N

_A,z

=  cD

_AB^dyA

+ y

_A

(N

_A,z

+ N

_B,z

)

dz

Since air is insoluble in water, it is stagnant (or nondiffusing) and N

B,z

= 0.

Solving for N

A,z

give

N

A,z

(1  y

A

) =  cD

ABdy^A

 N

A,z

dz =  dy

A

dz

1

AB A

cD



y

N

A,z

=  cD

AB

 N

A,z

L = cD

AB

ln

0 L

dz



y^y_A^AL₀

₁

^A A

dy



y



0

1 1

AL A

y y





The ambient air is dry so y

AL

= 0. Vapor pressure of water at 20

^o

C is 17.5 mmHg, therefore y

A0

= 17.5/760 = 0.023. Note: R = 82.057 cm

³

atm/mol

^o

K

N

A,z

=

^cDAB

ln = ln = 1.93510

^-8

mol/cm

²

s

L ₀

1 1 

y_A

0.20

82.057 293.15 10  

1 1 0.023 

Time for 1 g of water to evaporate through this system is then

t = = = 9.14×10

⁹

s

,

1/18

( )

NA z Area ⁸

4(1/18)

(1.935 10 )( )(0.01^ 2) 

^



The average time it takes for molecule A to travel 10 cm can be estimated by t

d

= = = 250 s.

2

2

_AB

L D

10

2

2(.2)

Example 1.3-2 --- Water evaporates at the surface of a column confined to a long

narrow capillary tube, closed at the bottom. The system is at 1 atm and 25

^o

C as shown. The inside diameter of the tube is 1 mm. Dry air is blow over the top of the tube. For the case that L

o

= 10 cm and L

w

(t = 0) = 10 cm, how long will it take for the column of water to vanish. Diffusivity of water vapor in air is 0.22 cm

²

/s. Vapor pressure of water at 25

^o

C is 3.17 kPa. (

Ref

.

An Introduction to Mass and Heat transfer by Stanley Middleman

.)

Solution --- The molar flux of A (water vapor) is

N

A,z

=  cD

AB^dyA

+ y

A

(N

A,z

+ N

B,z

)

dz

Since air is insoluble in water, it is stagnant (or nondiffusing) and N

B,z

= 0.

Solving for N

A,z

give

N

_A,z

(1  y

_A

) =  cD

_AB^dyA

 N

_A,z

dz =  dy

_A

dz

1

AB A

cD



y

N

A,z

=  cD

AB

 N

A,z

L = cD

AB

ln

0 L

dz



y^y_A^AL₀

₁

^A A

dy



y



0

1 1

AL A

y y





The dry air is blow over the top y

AL

= 0, y

A0

= 3.17/101.3 = 0.0313 N

A,z

=

^cDAB

ln

L ₀

1 1 

y_A

The molar flux is related to the amount of A leaving the liquid by the material balance A

_c ^{A L,}

=  A

_c

N

_A,z

=  A

_c

ln

MA



_dLw dt

cDAB

L ₀

1 1 

y_A

In this equation, A

_c

is the cross sectional area of the tube and L

_w

is the length of the liquid column. Since

^dLw

=  , we have

dt

dL dt

= ln

, A L

MA



_dL dt

cDAB

L ₀

1 1 

y_A

Separating the variables and integrating the equation from t = 0 to t gives

Pure water L

w

(t)

L

o