unit operationstknguyen/che425/lecture/che-425.pdfin chemical potential is due to the variation in...

Unit Operations(CHE 425 Course Notes)

T.K. Nguyen

Chemical and Materials EngineeringCal Poly Pomona

(Fall 2016)

i

Contents CHE 425

Chapter 1: Fundamentals of Mass Transfer

1.1 Molecular Mass Transfer 1-1Example 1.1-1 Flux of oxygen at steady state 1-1Example 1.1-2 Rate of evaporation of water 1-2

1.2 Flue Gas Desulfurization (FGD) Process 1-3The enhancement factor of the liquid film coefficient 1-7Example 1.2-1 Evaporation of a water droplet 1-9Example 1.2-2 Absorption of SO2 into a lime slurry droplet 1-10

1.3 Diffusivity Measurement 1-14Example 1.3-1 Diffusion in a capillary tube 1-17Example 1.3-2 Diffusion time in a capillary tube 1-18

1.4 Diffusion with Chemical Reaction 1-19Example 1.4-1 Reaction time with carbon particles 1-19Example 1.4-2 Consumption rate with carbon pellets 1-21Example 1.4-3 Biofilms 1-23

1.5 Unsteady State Diffusion: Differential Mass Balance 1-271.6 Unsteady State Diffusion: Approximate Solutions 1-32

Example 1.6-1 Unsteady diffusion in a slab 1-37Example 1.6-2 Unsteady diffusion in white pine 1-38Example 1.6-3 Diffusion in spray tower 1-42Example 1.6-4 Diffusion in a finite cylinder 1-43Example 1.6-5 Semi-infinite solution approximation 1-45

Chapter 2: Convective Mass Transfer

2.1 Introduction 2-1Example 2.1-1 Mass transfer from a flat container 2-2Example 2.1-2 Rate of absorption of H2S. 2-3Example 2.1-3 Mass transfer rate of napthalene. 2-5Example 2.1-4 Mass transfer coefficient from experiment. 2-6Example 2.1-5 Mass transfer in bubble column. 2-7

2.2 Packed Column 2-11Example 2.2-1 Height of an absorption tower 2-17

Chapter 3: Phase Equilibrium Calculation

3.1 Introduction 3-13.2 Vapor-Liquid Equilibrium calculation 3-1

Example 3.2-1: Bubble point temperature calculation 3-4Example 3.2-2: Bubble point pressure calculation 3-5Example 3.2-3: Dew point temperature calculation 3-6Example 3.2-4: Dew point pressure calculation 3-8Example 3.2-5: Dew point pressure with Van Laar equations 3-9

3.3 Isothermal Flash Calculation Using K-values 3-11

ii

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

3.4 Distribution of a Solute between Liquid Phases 3-253.4a Solubility of a Solid in a Liquid Phase 3-25

Example 3.4-1: Solubility of a Drug 3-273.4b Distribution of a Solute between Liquid Phases 3-30

Example 3.4-2: Drug Extraction from the Aqueous Phase 3-33Example 3.4-3: Purification of an Antibiotic 3-34

3.4c Single-Stage Equilibrium Extraction 3-35Example 3.4-4: Drug Extraction from the Aqueous Stream 3-36

3.5 Non-isothermal Flash 3-37Example 3.5-1: Adiabatic flash drum 3-37Example 3.5-2: Non-adiabatic flash drum 3-39

Chapter 4: Distillation4.1 Vapor Liquid Equilibrium Relations 4-1

Example 4.1-1: Txy diagram for a benzene-toluene mixture 4-24.2 Single-Stage Equilibrium Contact for Vapor-Liquid System 4-8

Example 4.2-1: Single equilibrium stage calculation 4-84.3 Simple Batch or Differential Distillation 4-12

Example 4.3-1: A semi-batch distillation unit 4-124.4 Distillation with Reflux 4-15

4.4a Introduction 4-154.4b McCabe-Thiele Method 4-16Example 4.4-1: Rectifying section calculation 4-18Example 4.4-2: Stripping section calculation 4-24Example 4.4-3: Number of equilibrium stages 4-33Example 4.4-4: Total and minimum flux ratios 4-45

4.5 Tray Efficiency 4-49Example 4.5-1: Overall column efficiency 4-51Example 4.5-2: Plate efficiency 4-52

4.6 Enthalpy-Concentration Diagram 4-53Example 4.6-1: Enthalpy-concentration plot for benzene-toluene 4-54

4.7 Distillation Using Enthalpy-Balance 4-59Example 4.7-1: Enthalpy-balance for equilibrium stages 4-60Example 4.7-2: Enthalpy diagram for methanol-water system 4-67Example 4.7-3: Malab codes for Enthalpy diagram 4-71

Chapter 5: Approximate Methods for Multicomponent Distillation5.1 Introduction 5-15.2 Fenske-Underwood-Gilliland Method 5-3

Example 5.2-1: Butane-pentane distillation 5-65.3 Underwood Equation for Minimum Reflux 5-15

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

iii

Chapter 6: Rigorous Methods for Distillation6.1 Theoretical Model for an Equilibrium Stage 6-16.2 Rigorous Computational Methods 6-5

Example 6.2-1: Tridiagonal matrix equations 6-8Example 6.2-2: Stage by stage calculation 6-11Algorithm for Wang-Henke BP methods for distillation 6-21Example 6.2-3: BP method 6-22

6.3 Rate-Based Model for Distillation 6-29Example 6.3-1: Rate-Based Ternary Distillation Calculations 6-34

6.4 Equilibrium Distillation Calculations using ChemSep program 6-36Example 6.4-1: Distillation Calculations using ChemSep 6-36

Chapter 7: Heat Exchangers7.1 Introduction 7-17.2 Heat Exchanger types 7-47.3 Analysis of Heat Exchangers 7-9

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

7.4 The Effectiveness-NTU Method 7-22Example 7.4-1: Effectiveness-NTU relation for exchanger 7-26Example 7.4-2: Counter-flow heat exchange in whales 7-29Example 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 Stripping8.1 Introduction 8-18.2 Single-Component Absorption 8-3

Example 8.2-1: Equilibrium and operating lines 8-4Example 8.2-2: Number of trays in an absorber 8-9

8.3 Multicomponent Absorption Using Shortcut Methods 8-11Example 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-1Quiz 2 B-4Quiz 3 B-7Quiz 4 B-11Quiz 5 B-14

Answers to 2009 quizzes B-17

iv

C. Previous Exams (2010)Quiz 1 C-1Quiz 2 C-4Quiz 3 C-7Quiz 4 C-11Quiz 5 C-15

Answers to 2010 quizzes C-18

D. Previous Exams (2012)Quiz 1 C-1Quiz 2 C-4Quiz 3 C-7Quiz 4 C-10Quiz 5 C-13

1-1

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, NA,z, of species A relative to the z axis is

NA,z = cDAB + yA(NA,z + NB,z) (1.1)Adydz

In this equation, c is the total concentration, DAB is the diffusivity of A in B, yA is the mole fraction of A, and NB,z is the molar flux of B. For a binary mixture DAB = DBA.The term

cDAB is the molar flux, JA, resulting from the concentration gradient and the term Adydz

yA(NA,z + NB,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 pA1 of oxygen is 0.70 atm and at the other end 0.8 m, pA2 = 0.2 atm. Calculate the flux of oxygen at steady state if DAB of the mixture is 0.206 cm2/s.

Solution ----------------------------------------------------------------------------------------------

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

c = = = 4.0910-5 mol/cm3PRT

182.057 298

Note: R = 82.057 cm3atm/molK

The total concentration is a constant therefore NA,z = NB,z. This condition is known as equimolar counterdiffusion. The molar flux of A is

1-2

NA,z = cDAB + yA(NA,z + NB,z) = cDABAdy

dzAdy

dz

For steady state and constant area of mass transfer NA,z = constant. Separating the variable and integrating

NA,z = cDAB0

Ldz

2

1

A

A

y

Aydy

NA,z = = 1 2( )AB A AcD y yL

5(4.09 10 )(0.206)(0.7 0.2)80

NA,z = 5.2710-8 mol/cm2s

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 z2 z1 is 50 cm long. Calculate the rate of evaporation at steady state in mol/scm2. The diffusivity of water vapor (A) in air (B) at 1 atm and 298 K is 0.250 cm2/s. Assume that air is insoluble in water.

Solution ----------------------------------------------------------------------------------------------

The molar flux of A (water vapor) is

NA,z = cDAB + yA(NA,z + NB,z)Adydz

Since air is insoluble in water, it is stagnant (or nondiffusing) and NB,z = 0. Solving for NA,z give

NA,z(1 yA) = cDAB NA,zdz = dyAAdy

dz 1AB

A

cDy

NA,z = cDAB NA,zL = cDABln0

Ldz

2

1 1A

A

yA

yA

dyy 2

1

11

A

A

yy

The ambient air is dry so yA2 = 0. Vapor pressure of water at 298 K is 3.17 kPa, therefore yA1 = 3.17/101.3 = 0.0313.

NA,z = ln = ln = 6.510-9 mol/cm2sABcDL

2

1

11

A

A

yy

0.25082.057 298 50

1 01 0.313

Water (A)

Air (B)

1

2

NA

1-3

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 SOx led to a significant increase in acid rain, the government has mandated that industry restrict SOx 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 SOx 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 SO2 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 SO2 concentration1. 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 about1 4-6 m. The lime slurry is then diluted further with water and atomized into 20-150 m droplets2. There is a limit to the mass fraction of solid in the droplet to prevent clogging and abrasion3. 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 SO2 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 SO2 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 seconds2. In the FGD unit, two processes occur simultaneously: water evaporates from the droplet; and SO2 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)

1-4

LimeSlurry

DilutionWater

Slaker Temperature = 180 Fo

Flue Gas fromthe Waste Heat Boiler

380 Fo

Flue Gasto Baghouse

Ash

223 Fo

Figure 1.2-1 A typical FGD reactor.

The operation of a FGD unit is essentially a spray drying process for SO2 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.

1-5

Calcium hydroxide particleWater

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 SO2 removal. After the spray dryer, the flue gases pass through a bag-house where SO2 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 SO2 absorption occur simultaneously, the droplet drying process is relatively independent of the SO2 absorption process. However, the SO2 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, SOx concentration, and residence time through the reactor. SOx 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

SO2 Flux

Ca(OH)2 Particles

Heat Flux into the Solid Particle

SOx Flux into the Solid Particle

1-6

Figure 1.2-4 summarizes the main process in the FGD process. Flue gas with high concentration of SO2 is in contact with a spray of lime slurry. SO2 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 SO2.

LimeSlurry

DilutionWater

Slaker Temperature = 180 Fo

Flue Gas fromthe Waste Heat Boiler

380 Fo

Flue Gasto Baghouse

Ash

223 Fo

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

yd,w

yg,wTd

Tg

Figure 1.2-5 Schematic of the FGD process.

1-7

We assume a pure water droplet with radius R. yd,w is the water mole fraction at the surface in the gas phase. yg,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

NA,r = cDAB + yA(NA,r + NB,r)Adydr

In this equation, r is the radial distance from the center of the drop. For quiescent droplet-gas system, the gas is stagnant and NB,r = 0. Therefore

NA,r = 1

AB

A

cDy

Adydr

The system is at steady state, however the molar flux is not a constant since the area of mass transfer 4r2 is not a constant. For steady state, the mass (mole) transfer rate, 4r2NA,r, is a constant

WA = 4r2NA,r = 4r2 = constant1

AB

A

cDy

Adydr

Separating the variables and integrating gives

WA = 4 cDAB2R

drr

,

, 1g w

d w

yA

yA

dyy

= 4 cDAB ln WA = 4RcDAB lnAWR

,

,

11

g w

d w

yy

,

,

11

g w

d w

yy

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 SO2 in the liquid film without and with chemical reaction. If there is no chemical reaction in the liquid, SO2 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 SO2 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 SO2 is higher with chemical reaction within the liquid film.

1-8

C*SO2 C*SO2

Film thickness Film thickness

Reactionplane

C*lime

LimeLime

Water

Figure 1.2-6 Absorption of SO2 without and with chemical reaction in the liquid phase.

The mass transfer coefficient without reaction is given by

kL = 2SOD

For the very fast reaction of dissolved SO2 with dissolved lime in the bulk liquid, the mass transfer coefficient is then

k*L = 2SOD

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:

= Dlime2SOD 2

*SOC

*limeC

Rearranging the equation to solve for we have

( ) = Dlime 2SOD

2

*SOC *

limeC

= 2 2

2 2

*SO SO

* *SO SO lime lime

D CD C D C

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

k*L = = = kL = kL2SOD

2SOD

2 2

*lime lime

*SO SO

1 D CD C

2 2

*lime lime

*SO SO

1 D CD C

In this equation = = the enhancement factor of the liquid film coefficient kL 2 2

*lime lime

*SO SO

1 D CD C

due to the very fast reaction of dissolved SO2 with dissolved lime in the bulk liquid.

1-9

Chapter 1 Example 1.2-1 ------------------------------------------------------------------------------A water droplet having a diameter of 0.10 mm is suspended in still air at 50oC, 1.0132×105 Pa (1 atm), and 20% relative humidity. The droplet temperature can be assumed to be at 50oC and its vapor pressure at 50oC is 7.38 kPa. 1) Calculate the initial rate of evaporation of water if DAB of water vapor in air is 0.288 cm2/s.2) Determine the time for the water droplet to evaporate completely.Solution ----------------------------------------------------------------------------------------------

Droplet surface

Bulk gas phase

yd,w

yg,w

Td Tg

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


In this equation, r is the radial distance from the center of the drop. For still air NB,r = 0, we have

NA,r = 1

AB

A

cDy

Adydr

The system is not at steady state, the molar flux is not independent of r since the area of mass transfer 4r2 is not a constant. Using quasi steady state assumption, the mass (mole) transfer rate, 4r2NA,r, is assumed to be independent of r at any instant of time.

WA = 4r2NA,r = 4r2 = constant1

AB

A

cDy

Adydr

Separating the variables and integrating gives

WA = 4 cDAB2R

drr

,

, 1g w

d w

yA

yA

dyy

1-10

= 4 cDAB ln WA = 4RcDAB lnAWR

,

,

11

g w

d w

yy

,

,

11

g w

d w

yy

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 DAB of water vapor in air is 0.288 cm2/s.

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

R g 82.057 cm3 atmmol K

R .005 cm P 1 atm T 50 273( ) K

D AB .288cm2

sMw 18

gmolc

PR g T

c 3.773 10 5 mol

cm3

P A7.38

101.32atm y dw

P AP

y dw 0.073

For 20% relative humidity y gw .2P AP

y gw 0.015

W A 4 R c D AB ln1 y gw1 y dw

W A 4.161 10 8 s 1 mol

The initial mass evaporation rate is 4.161 10 8 18 7.49 10 7gs

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

Making a material balance around the water droplet gives

= − 4MwRcDAB lnddt

343 AR

,

,

11

g w

d w

yy

4ρAR2 = − 4MwRcDAB lndRdt

,

,

11

g w

d w

yy

Simplifying and rearranging we have

RdR = − Mwc ln dt = − KdtAB

A

D

,

,

11

g w

d w

yy

1-11

In this equation, K = Mwc lnAB

A

D

,

,

11

g w

d w

yy

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

= − K => t = 0

iR

RdR 0

tdt

2

2iRK

Substituting the numerical values we have

y dw .0728 y gw .0146 D AB .288cm2

s

D AB ln1 y gw1 y dw

0.0175cm2

s

Density of liquid water at 50oC is 0.988 g/cm3

K = ln (Mw)c = (0.0175)(18)(3.773×10-5)/(0.988) = 1.207×10-5 cm2/sAB

A

D

,

,

11

g w

d w

yy

t = = 0.0052/(2×1.207×10-5) = 1.04 s2

2iRK

Example 1.2-2 ------------------------------------------------------------------------------

A lime (CaO) slurry droplet having a diameter Dd of 0.10 mm is suspended in a gas containing SO2 at 50oC, 1.0132×105 Pa (1 atm). The mole fraction of SO2 in the gas phase is 10-4. Rate of diffusion of SO2 from the gas to the droplet surface is given by

Ns = kd(πDd2) c (yg,s − yd,s)

In this equation c is the total gas concentration and kd is the mass transfer coefficient for SO2 in the gas phase given by

kd = 22 SO gas

d

DD

Rate of diffusion of SO2 from the droplet surface into the interior droplet is given by

Ns = kL (πDd2) C*

SO2

In this equation C*SO2 is the equilibrium SO2 concentration at the droplet surface kL is the

mass transfer coefficient for SO2 in the liquid phase given by

1-12

kL = kL2 2

*lime lime

*SO SO

1 D CD C

where

kL = and the liquid film thickness δ at the surface is estimated to be 1 μm. 2SOD

= = the enhancement factor of the liquid film coefficient kL due 2 2

*lime lime

*SO SO

1 D CD C

to the very fast reaction of dissolved SO2 with dissolved lime in the bulk liquid.

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

c yd,s = HC*SO2

The Henry’law constant for the solubility of SO2 in water at 50oC is given by

H = 0.054 3

3

mol/cm in gas phasemol/cm in liquid phase

Other data at 50oC:

Equilibrium concentration of lime, C*lime = 1.35×10-5 mol/cm3

Diffusivity of SO2 and lime in water: Dlime = DSO2 = 2×10-5 cm2/sDiffusivity of SO2 in gas: DSO2-gas = 0.20 cm2/s

Estimate the rate of diffusion of SO2, Ns.

Solution ----------------------------------------------------------------------------------------------

We will estimate the enhancement factor = by using the equilibrium 2 2

*lime lime

*SO SO

1 D CD C

concentration C*SO2 from the equation

c yg,s = HC*SO2

Note that in this equation yg,s, a known value, is used instead of yd,s.

R g 82.057 cm3 atmmol K

R .005 cm P 1 atm T 50 273( ) K

cP

R g Tc 3.773 10 5 mol

cm3

1-13

C*SO2 = c yg,s/H = = 7.0×10-8 mol/cm3

5 4(3.773 10 )(10 ).054

Since Dlime = DSO2, = = = 1 + = 1942 2

*lime lime

*SO SO

1 D CD C

2

*lime*SO

1 CC

5

8

1.35 107.0 10

The diffusion equation for SO2 in the gas phase is rewritten as

= cyg,s − cyd,s (E-1)2( )s

d d

Nk D

The diffusion equation for SO2 in the liquid phase is rewritten as

= C*SO2 = 2( )

s

L d

Nk D

,d scyH

= cyd,s (E-2)2( )s

L d

HNk D

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

= cyg,s2s

d

ND

1

d L

Hk k

Therefore

Ns = (πDd2) ,

1g s

d L

cyH

k k

kd = = = 40 cm/s22 SO gas

d

DD

4

(2)(0.20)100 10

kL = = = 0.2 cm/s2SOD

5

6

2 1010

= + = .025 + .00139 = .0264 s/cm1

d L

Hk k

140

.054(.2)(194)

The rate of diffusion of SO2 is then

Ns = π×(100×10-4)2 = 4.49×10-11 mol/s5 4(3.773 10 )(10 )

.0264

1-14

----------------------------------------------------------

1.3 Diffusivity Measurement

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

PA2gas 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 td be the average time it takes for molecule A to travel the diffusion path z1. td can be estimated by Einstein ‘s formula

td = 21

2 AB

zD

Let ∆z be the change in diffusion path (or the drop in liquid level) during the time td. The pseudo-steady state assumption may be used if

1-15

∆z/ z1 < 0.01

The molar flux of A is given by


We will assume that air is insoluble in liquid A, it is stagnant (or nondiffusing) and NB,z = 0. Solving for NA,z give


dz 1AB

A

cDy

NA,z = cDAB NA,zz = cDABln0

zdz

2

1 1A

A

yA

yA

dyy 2

1

11

A

A

yy

Using the definition of log mean average,

yB,lm = = 2 1

2 1ln( / )B B

B B

y yy y

2 1

2 1

(1 ) (1 )ln[(1 ) /(1 )]

A A

A A

y yy y

we have

ln = 2

1

11

A

A

yy

1 2

,

A A

B lm

y yy

The molar flux of A is then written as

NA,z = 1 2

,

( )AB A A

B lm

cD y yzy

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

NA,z = = 1 2

,

( )AB A A

B lm

cD y yzy

,A L

AM dz

dt

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

= 0

tdt ,A L

AM ,

1 2( )B lm

AB A A

ycD y y 1

tz

zzdz

We have

1-16

t = ,A L

AM ,

1 2( )B lm

AB A A

ycD y y

2

21

2 zzt

Replacing c by P/RT yields

t = = ,A L

AM ,

1 2( )B lm

AB A A

y RTD P y y

2

21

2 zzt ,A L

AM ,

1 2( )B lm

AB A A

y RTD P P

2

21

2 zzt

Since

yB,lm = = = 1 2

2 1ln[(1 ) /(1 )]A A

A A

y yy y

1 2

2 1ln[ (1 ) / (1 )]A A

A A

P y yP P y P y

,B lmPP

= ,

1 2

B lm

A A

yP P

,

1 2( )B lm

A A

PP P P

Therefore

t = ,A L

AM ,

1 2( )B lm

AB A A

P RTD P P P

2

21

2 zzt

Solving for the diffusivity we have

DAB = )( 21

,,

AAA

lmBLA

PPtPMRTP

2

21

2 zzt

In this equation

PB,lm = [(P - PA1) - (P - PA2)]/ln [(P - PA1)/(P - PA2)]PA1 = vapor pressure of liquid A at temperature T.PA2 = partial pressure of vapor A at the mouth of the capillary.R = gas law constant.t = time during which the meniscus fall from z1 to zt.z1 = distance from the mouth of the capillary to the meniscus at t = 0.zt = distance from the mouth of the capillary to the meniscus at t.P = ambient atmospheric pressureA,L = density of liquid A at T.

1-17

z = 0

z = L yAL

yA0

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 20oC, the temperature at which the entire system is maintained. Take the diffusivity of water vapor in air at 20oC to be 0.20 cm2/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




dz 1AB

A

cDy


Ldz

0 1AL

A

yA

yA

dyy

0

11

AL

A

yy

The ambient air is dry so yAL = 0. Vapor pressure of water at 20oC is 17.5 mmHg, therefore yA0 = 17.5/760 = 0.023. Note: R = 82.057 cm3atm/moloK

NA,z = ln = ln = 1.93510-8 mol/cm2sABcDL 0

11 Ay

0.2082.057 293.15 10

11 0.023

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

t = = = 9.14×109 s,

1/18( )A zN 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

td = = = 250 s. 2

2 AB

LD

2102(.2)

1-18

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 25oC 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 Lo = 10 cm and Lw(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 cm2/s. Vapor pressure of water at 25oC is 3.17 kPa. (Ref. An Introduction to Mass and Heat transfer by Stanley Middleman.)

Solution ----------------------------------------------------------------------------------------------

The molar flux of A (water vapor) is




dz 1AB

A

cDy


Ldz

0 1AL

A

yA

yA

dyy

0

11

AL

A

yy

The dry air is blow over the top yAL = 0, yA0 = 3.17/101.3 = 0.0313

NA,z = lnABcDL 0

11 Ay

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

Ac = AcNA,z = Ac ln ,A L

AM wdL

dtABcD

L 0

11 Ay

In this equation, Ac is the cross sectional area of the tube and Lw is the length of the liquid

column. Since = , we havewdLdt

dLdt

= ln,A L

AM dL

dtABcD

L 0

11 Ay

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

Pure water Lw(t)

Lo

1-19

K = where 0

tdt

20

10LdL

K = ln = ln = 5.149×10-6 cm2/s,

AB A

A L

cD M 0

11 Ay

(0.22)(18)(82.06)(298)

11 0.0313

t = (202 − 102) = 2.91×107 s = 8093 hr12K

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

td = = = 227 s. 2

2 AB

LD

2102(.22)

At the end of this time the diffusion path can be determine from

227 = (L2 − 102) => L = (100 + 227×5.149×10-6)0.5 = 10.0001 cm12K

The pseudo-steady state assumption is valid since the diffusion path changes very little during the average time the water molecule travel from the water interface to the top of the tube.

1.4 Diffusion with Chemical Reaction

Example 1.4-1 ------------------------------------------------------------------------------

A fluidized coal reactor operates at 1145oK and 1 atm. The process will be limited by the diffusion of oxygen countercurrent to the carbon dioxide, CO2, formed at the particle surface. Assume that the coal is pure solid carbon with a density of 1280 kg/m3 and that the particle is spherical with an initial diameter of 1.5×10-4 m. Air (21% O2 and 79% N2) exists several diameters away from the sphere. The diffusivity of oxygen in the gas mixture at 1145oK is 1.3×10-4 m2/s. If a quasi-steady state process is assumed, calculate the time necessary to reduce the diameter of the carbon particle to 5.0×10-5 m. (Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001, pg. 496.)Solution ----------------------------------------------------------------------------------------------

The reaction at the carbon surface is

C(s) + O2(g) → CO2(g)

We have diffusion of oxygen (A) toward the surface and diffusion of carbon dioxide (B) away from the surface. The molar flux of oxygen is given by

1-20

NA,r = cDAmix + yA(NA,r + NB,r)Adydr

In this equation, r is the radial distance from the center of the carbon particle. Since NA,r = − NB,r, we have

NA,r = cDAmixAdy

dr


WA = 4r2NA,r = 4r2cDAmix = constantAdydr

Rr

yA,R yA,inf

At the surface of the coal particle, the reaction rate is much faster than the diffusion rate to the surface so that the oxygen concentration can be considered to be zero: yA,R = 0. Separating the variables and integrating gives

WA = 4 cDAmix 2R

drr

,

0

Ay

Ady

WA = 4 cDAmixyA,∞ => WA = 4cDAmixyA,∞R1Rr

Since one mole of carbon will disappear for each mole of oxygen consumed at the surface

WC = WA = 4cDAmixyA,∞R

Making a carbon balance gives

= 4πR2 = 4cDAmixyA,∞RC

CM 34

3d Rdt

C

CM dR

dt

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

= 0

tdt

Amix ,

C

C AM cD y

f

i

R

RRdR

1-21

t = (Ri2 Rf

2)Amix ,2C

C AM cD y

The total gas concentration can be obtained from the ideal gas law

c = = = 0.0106 kmol/m3PRT

1(0.08206)(1145)

Note: R = 0.08206 m3atm/kmoloK

The time necessary to reduce the diameter of the carbon particle from 1.5×10-4 m to 5.0×10-5 m is then

t = = 0.92 s 2 25 5

4

(1280) 7.5 10 2.5 10

2(12)(0.0106)(1.3 10 )(0.21)

Example 1.4-2 ------------------------------------------------------------------------------

Pulverized coal pellets, which may be approximated as carbon spheres of radius R = 1 mm, are burned in a pure oxygen atmosphere at 1450 K and 1 atm. Oxygen is transferred to the particle surface by diffusion, where it is consumed in the reaction C(s) + O2(g) → CO2(g). The reaction rate is first order and of the form = k1”CO2|R where k1” = 0.1 m/s. This is "Rthe reaction rate per unit surface area of the carbon pellets. Neglecting change in R, determine the steady-state O2 molar consumption rate in kmol/s. At 1450 K, the binary diffusion coefficient for O2 and CO2 is 1.71×10-4 m2/s.(Ref. Fundamentals of Heat Transfer by Incropera and DeWitt.)Solution ----------------------------------------------------------------------------------------------

We have diffusion of oxygen (A) toward the surface and diffusion of carbon dioxide (B) away from the surface. The molar flux of oxygen is given by


In this equation, r is the radial distance from the center of the carbon particle. Since NA,r = − NB,r, we have

NA,r = cDABAdy

dr

1-22


WA = 4r2NA,r = 4r2cDAB = constantAdydr

Rr

yA,R yA,inf

The oxygen concentration at the surface of the coal particle, yA,R, will be determined from the reaction at the surface. The mole fraction of oxygen at a location far from the pellet is 1. Separating the variables and integrating gives

WA = 4 cDAB 2R

drr

,

,

A

A R

y

Aydy

WA = 4 cDAB(yA,∞ yA,R) => WA = 4cDAB(1 yA,R)R1Rr

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction

WA = 4R2 = 4R2k1”CO2|R = 4R2k1” c yA,R"R

4cDAB(1 yA,R)R = 4R2k1” c yA,R

DAB(1 yA,R) = Rk1”yA,R => yA,R = 1 "

AB

AD

DD Rk

yA,R = = 0.6314

4 3

1.71 101.71 10 10 .1

The total gas concentration can be obtained from the ideal gas law. (Note: R = 0.08206 m3atm/kmolK)

c = = = 0.008405 kmol/m3PRT

1(0.08206)(1450)

The steady-state O2 molar consumption rate is

WA = 4cDAB(1 yA,R)R = 4(0.008405)(1 0.631)(10-3) = 6.66×10-9 kmol/s

1-23

Example 1.4-3 ------------------------------------------------------------------------------

A biofilm consists of living cells immobilized in a gelatinous matrix. A toxic organic solute (species A) diffuses into the biofilm and is degraded to harmless products by the cells within the biofilm. We want to treat 0.1 m3 per hour of wastewater containing 0.1 mole/m3 of the toxic substance phenol using a system consisting of biofilms on rotating disk as shown below.

Waste water feed stream

Treatedwaste water

biofilm

Well-mixed contactorbiofilm

z=0

C (z)A

CA0

CA0

Inertsolidsurface

Determine the required surface area of the biofilm with 2 mm thickness to reduce the phenol concentration in the outlet stream to 0.02 mole/m3. The rate of disappearance of phenol (species A) within the biofilm is described by the following equation

rA = − k1cA where k1 = 0.019 s-1

The diffusivity of phenol in the biofilm at the process temperature of 25oC is 2.0×10-10 m2/s. Phenol is equally soluble in both water and the biofilm.(Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001, pg. 496.)Solution ----------------------------------------------------------------------------------------------

The rate of phenol processed by the biofilms, WA, is determined from the material balance on the process unit

WA = 0.1 m3/h(0.1 − 0.02) mol/m3 = 8.0×10-3 mol/h

WA is then equal to the rate of phenol diffused into the biofilms and can be calculated from

WA = S∙NA,z = S∙0

AAB

z

dcDdz

In this equation, S is the required surface area of the biofilm and NA,z is the molar flux of phenol at the surface of the biofilm. The molar flux of A (phenol) is given by


1-24

Since the biofilm is stagnant (or nondiffusing), NB,z = 0. Solving for NA,z give

NA,z(1 yA) = cDABAdy

dz

The mole fraction of phenol in the biofilm, yA, is much less than one so that c can be considered to be constant. Therefore

NA,z = cDAB = DABAdy

dzAdc

dz

z

Solid surface

Biofilm

NA,z

Making a mole balance around the control volume S∙∆z gives

S∙NA,z|z − S∙NA,z|z+∆z + S∙∆z∙rA = 0

Dividing the equation by S∙∆z and letting ∆z → 0 yields

= rA = k1cA (E-1),A zdNdz

Substituting NA,z = DAB into equation (E-1) we obtainAdcdz

DAB = k1cA => = cA (E-2)2

2Ad c

dz

2

2Ad c

dz1

AB

kD

The solution to the homogeneous equation (E-2) has two forms

1) cA = C1exp + C2exp1

AB

k zD

1

AB

k zD

2) cA = B1sinh + B2cosh1

AB

k zD

1

AB

k zD

1-25

The first exponential form (1) is more convenient if the domain of z is infinite: 0 z while the second form using hyperbolic functions (2) is more convenient if the domain of z is finite: 0 z δ. The constants of integration C1, C2, B1, and B2 are to be determined from the two boundary conditions. We use the hyperbolic functions as the solution to Eq. (E-2).

cA = B1sinh + B2cosh (E-3)1

AB

k zD

1

AB

k zD

At z = 0, cA = cAs = cA0 = B2

At z = δ, = 0 = B1 cosh + B2 sinhAdcdz

1

AB

kD

1

AB

kD

1

AB

kD

1

AB

kD

Therefore

B1 = B2 = cA0

1

1

sinh

cosh

AB

AB

kD

kD

1

1

sinh

cosh

AB

AB

kD

kD

Equation (E-3) becomes

cA = cA0 sinh + cA0cosh

1

1

sinh

cosh

AB

AB

kD

kD

1

AB

k zD

1

AB

k zD

cA = cA0

1 1 1 1

1

cosh cosh sinh sinh

cosh

AB AB AB AB

AB

k k k kz zD D D D

kD

Using the identity cosh(A – B) = cosh(A)cosh(B) – sinh(A)sinh(B) we have

cA = cA0

1

1

cosh ( )

cosh

AB

AB

k zD

kD

1-26

= – cA0 = – cA00

A

z

dcdz

1 1

1

0

sinh ( )

cosh

AB AB

AB z

k k zD D

kD

1 1tanhAB AB

k kD D

The molar flux of phenol at the biofilm surface is given by

NA,z = = δ 0

AAB

z

dcDdz

0A ABc D

1 1tanhAB AB

k kD D

The dimensionless parameter δ represents the ratio of reaction rate to diffusion rate. 1

AB

kD

For this problem we have

δ = 0.002 m = 19.491

AB

kD 2

10

10.019sm2 10s

This value indicates that the rate of reaction is very rapid relative to the rate of diffusion. The flux of phenol into the biofilm is then

NA,z = (19.49) tanh(19.49) = 3.9×10-8 mol/(m2∙s)10(0.02)(2 10 )

0.002

The required surface area of the biofilm is finally

S = = = 57.0 m2

,

A

A z

WN

3

8

8.0 10(3.9 10 )(3600)

1-27

Chapter 1

1.5 Unsteady State Molecular Diffusion: Differential Mass Balance

When the internal concentration gradient is not negligible or Bi << 1, the microscopic or differential mass balance will yield a partial differential equation that describes the concentration as a function of time and position. For a binary system with no chemical reaction, the unsteady state molecular diffusion is given by

= (DABcA) (1.5-1)Act

For one-dimensional mass transfer in a slab with constant DAB and convective conditions of hm and cA,, equation (1.5-1) is simplified to

= DAB (1.5-2)Act

2

2Ac

x

x=0 L-L

h , cm A,infh , cm A,inf

Figure 1.5-1 One-dimensional unsteady mass transfer in a slab.

Equation (1.5-2) can be solved with the following initial and boundary conditions

I. C. t = 0, cA(x, 0) = cAi

B. C. x = 0, = 0; x = L, DAB = hm(cAf cA,)0

A

x

cx

A

x L

cx

In general, the concentration within the slab depends on many parameters besides time t and position x.

cA = cA(x, t, cA,i, cA,, L, DAB, hm)

The differential equation and its boundary conditions are usually changed to the dimensionless forms to simplify the solutions. We define the following dimensionless variables

1-28

Dimensionless concentration: * = cA =K’cA, + *(cA,i K’cA,),

, ,

''

A A

A i A

c K cc K c

Dimensionless distance: x* = x = L x*Lx

Dimensionless time or Fourier number: t* = Fo = t = Fo2ABD t

L

2

AB

LD

K’ is the equilibrium distribution coefficient. Substituting T, x, and t in terms of the dimensionless quantities into equation (1.5-2) yields

(cA,i cA,) = (cA,i cA,)1

ABD 2ABD

L Fo *

21L 2

*2

*x

= (1.5-3)Fo

*2

*2

*x

Similarly, the initial and boundary conditions can be transformed into dimensionless forms

*(x*, 0) = 1

= 0; = Bim**(1, t*), where Bim = 0

*

*

*

xx

1*

*

*

xx

'm

AB

h LK D

Therefore * = f(x*, Fo, Bim)

The dimensionless concentration depends * only on x*, Fo, and Bim. The mass transfer Biot number, Bim, denotes ratio of the internal resistance to mass transfer by diffusion to the external resistance to mass transfer by convection. Equation (1.5-3) can be solved by the method of separation of variables to obtain

* = exp( Fo) cos(nx*) (1.5-4)

1nnC 2

n

where the coefficients Cn are given by

Cn = )2sin(2

sin4

nn

n

and n are the roots of the equation: n tan(n) = Bim.

1-29

Table 1.5-1 lists the Matlab program that evaluates the first ten roots of equation n tan(n) = Bim and the dimensionless concentrations given in equation (1.5-4). The program use Newton’s method to find the roots (see Review).

Table 1.5-1 Matlab program to evaluate and plot * = exp( Fo) cos(nx*)

1nnC 2

n

% Plot the dimensionless concentration within a slab%% The guess for the first root of equation z*tan(z)=Bi depends on the Biot number%Biot=[0 .01 .1 .2 .5 1 2 5 10 inf]';alfa=[0 .0998 .3111 .4328 .6533 .8603 1.0769 1.3138 1.4289 1.5707];zeta=zeros(1,10);cn=zeta;Bi=1;fprintf('Bi = %g, New ',Bi)Bin=input('Bi = '); if length(Bin)>0;Bi=Bin;end% Obtain the guess for the first root if Bi>10 z=alfa(10); else z=interp1(Biot,alfa,Bi); end% Newton method to solve for the first 10 roots for i=1:10 for k=1:20 ta=tan(z);ez=(z*ta-Bi)/(ta+z*(1+ta*ta)); z=z-ez; if abs(ez)<.00001, break, end end % Save the root and calculate the coefficients zeta(i)=z; cn(i)=4*sin(z)/(2*z+sin(2*z)); fprintf('Root # %g =%8.4f, Cn = %9.4e\n',i,z,cn(i))% Obtain the guess for the next root step=2.9+i/20; if step>pi; step=pi;end z=z+step; end%% Evaluate and plot the concentrationshold onFop=[.1 .5 1 2 10];xs=-1:.05:1;cosm=cos(cn'*xs);for i=1:5 Fo=Fop(i);

1-30

theta=cn.*exp(-Fo*zeta.^2)*cosm; plot(xs,theta)endgridxlabel('x*');ylabel('Theta*')

Bi = .5Root # 1 = 0.6533, Cn = 1.0701e+000Root # 2 = 3.2923, Cn = -8.7276e-002Root # 3 = 6.3616, Cn = 2.4335e-002Root # 4 = 9.4775, Cn = -1.1056e-002Root # 5 = 12.6060, Cn = 6.2682e-003Root # 6 = 15.7397, Cn = -4.0264e-003Root # 7 = 18.8760, Cn = 2.8017e-003Root # 8 = 22.0139, Cn = -2.0609e-003Root # 9 = 25.1526, Cn = 1.5791e-003Root # 10 = 28.2920, Cn = -1.2483e-003

Figure 1.5-2 shows a plot of dimensionless concentration * versus dimensionless distance x* at various Fourier number for a Biot number of 0.5.

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x*

Thet

a*

Temperature distribution in a slab for Bi = 0.5

Fo=1

Fo=2

Fo=10

Fo=0.1

Fo=0.5

Figure 1.5-2 Dimensionless concentration distribution at various Fourier number.

For the roots of equation n tan(n) = Bim, let

f = tan() Bim

Then f’ = tan() +(1 + tan()2);

1-31

The differential conduction equation for mass transfer in the radial direction of an infinite cylinder with radius R is

= DAB (1.5-5)Act

r

1 Acrr r

The differential conduction equation for mass transfer in the radial direction of a sphere with radius R is

= DAB (1.5-6)Act

2

1r

2 Acrr r

Equations (1.5-5) and (1.5-6) can be solved with the following initial and boundary conditions

I. C. t = 0, cA(r, 0) = cA i

B. C. r = 0, = 0; r = R, DAB = hm(cAf cA,)0

A

r

cr

A

r R

cr

The solution of equation (1.5-5) for the infinite cylinder is given as

* = exp( Fo) J0(nx*) (1.5-7)

1nnC 2

n

where J0(nx*) is Bessel function of the first kind, order zero. The coefficient Cn are not the same as those in a slab. The solution of equation (1.5-6) for a sphere is given as

* = exp( Fo) (1.5-8)

1nnC 2

n**)sin(

rr

n

n

Since = = 1, it should be noted that at r* = 00*

limr *

*)sin(r

rn

n

0*limr n

nn r

*)cos(

* = exp( Fo)

1nnC 2

n

For one-dimensional mass transfer in a semi-infinite solid as shown in Figure 1.5-3, the differential equation is the same as that in one-dimensional mass transfer in a slab

= DAB Act

2

2Ac

x

1-32

x

Semi-Infinite Solid

Figure 1.5-3 One-dimensional mass transfer in a semi-infinite solid.

We consider three cases with the following initial and boundary conditions

Case 1: I. C.: cA(x, 0) = cAiB. C.: cA(0, t) = cAs, cA(x , t) = cAi

Case 2: I. C.: cA(x, 0) = cAi

B. C.: DAB = NA0, cA(x , t) = cAi0

A

x

cx

Case 3: I. C.: cA(x, 0) = cAi

B. C.: DAB = hm(cAf cA,), cA(x , t) = cAi0

A

x

cx

All three cases have the same initial condition cA(x, 0) = cAi and the boundary condition at infinity cA(x , t) = cAi. However the boundary condition at x = 0 is different for each case, therefore the solution will be different and will be summarized in a table later.

1.6 Unsteady State Molecular Diffusion: Approximate Solutions

The summation in the series solution for transient diffusion such as equation (1.5-4) can be terminated after the first term for Fo > 0.2. The full series solution is

* = exp( Fo) cos(nx*) (1.5-4)

1nnC 2

n

The first term approximation is

= C1exp(- Fo) cos(1x*) (1.6-1)* 21

where C1 and 1 can be obtained from Table 1.6-1 for various value of Biot number. Table 1.6-2 lists the first term approximation for a slab, an infinite cylinder, and a sphere. Table 1.6-3 lists the solution for one-dimensional heat transfer in a semi-infinite medium for three different boundary conditions at the surface x = 0. Table 1.6-4 shows the combination of one-dimensional solutions to obtain the multi-dimensional results.

1-33

Table 1.6-1 Coefficients used in the one-term approximation to the seriessolutions for transient one-dimensional conduction or diffusion

PLANE WALL INFINITE CYLINDER SPHERE

Bim 1(rad) C1 1(rad) C1 1(rad) C10.010.020.030.040.050.060.070.080.090.10.150.20.250.30.40.50.60.70.80.91.02.03.04.05.06.07.08.09.010.020.030.040.050.0100.0500.01000.0∞

0.09980.14100.17320.19870.22170.24250.26150.27910.29560.31110.37790.43280.48010.52180.59320.65330.70510.75060.79100.82740.86031.07691.19251.26461.31381.34961.37661.39781.41491.42891.49611.52021.53251.54001.55521.56771.56921.5708

1.00171.00331.00491.00661.00821.00981.01141.01301.01451.01601.02371.03111.03821.04501.05801.07011.08141.09191.10161.11071.11911.17951.21021.22871.24021.24791.25321.25701.25981.26201.26991.27171.27231.27271.27311.27321.27321.2732

0.14120.19950.24390.28140.31420.34380.37080.39600.41950.44170.53760.61700.68560.74650.85160.94081.01851.08731.14901.20481.25581.59951.78871.90811.98982.04902.09372.12862.15662.17952.28812.32612.34552.35722.38092.40002.40242.4048

1.00251.00501.00751.00991.01241.01481.01731.01971.02221.02461.03651.04831.05981.07121.09321.11431.13461.15391.17251.19021.20711.33841.41911.46981.50291.52531.54111.55261.56111.56771.59191.59731.59931.60021.60151.60201.60201.6020

0.17300.24450.29890.34500.38520.42170.45500.48600.51500.54230.66080.75930.84480.92081.05281.16561.26441.35251.43201.50441.57082.02882.28892.45562.57042.65372.71652.76542.80442.83632.98573.03723.06323.07883.11023.13533.13853.1416

1.00301.00601.00901.01201.01491.01791.02091.02391.02681.02981.04451.05921.07371.08801.11641.14411.17131.19781.22361.24881.27321.47931.62271.72011.78701.83381.86741.89211.91061.92491.97811.98981.99421.99621.99902.00002.00002.0000

1-34

Table 1.6-2 Approximate solutions for diffusion and conduction (valid for Fo>0.2)

= = , = , = C1exp(- Fo)Fo 2ABD t

L 20

ABD tr

* ,

, ,

''

A A

A i A

c K cc K c

*0 2

1

Diffusion in a slabL is defined as the distance from the center of the slab to the surface. If one surface is insulated, L is defined as the total thickness of the slab.

= cos(1x*) ; = 1 − * *0 tM

M 1

1)sin(

*0

Diffusion in an infinite cylinder

= J0(1r*) ; = 1 − J1(1)* *0 tM

M 1

*02

Diffusion in a sphere

= sin(1r*) ; = 1 − [sin(1) − 1cos(1)]* *1

1r

*0 tM

M 3

1

*03

If the concentration at the surface cA,s is known K’cA, will be replaced by cA,s

1 and C1 will be obtained from table at Bim = ∞

Notation: cA = concentration of species A in the solid at any location at any timecA,s = concentration of species A in the solid at the surface for t > 0cA,i = concentration of species A in the solid at any location and at t = 0cA, = bulk concentration of species A in the fluid surrounding the solidK’cA, = cA

* = concentration of species A in the solid that is in equilibrium with cA, Mt = amount of A transferred into the solid at any given timeM∞ = amount of A transferred into the solid as t → ∞ (maximum amount transferred)

Bim = = ratio of internal resistance to mass transfer by diffusion to external mass 'm

AB

h LK D

transfer by convectionhm = kc = mass transfer coefficientL = L for a slab with thickness 2L or a slab with thickness L and an impermeable surfaceL = ro for radial mass transfer in a cylinder or sphere with radius ro

K’ = equilibrium distribution coefficientDAB = diffusivity of A in the solid

1-35

Table 1.6-3 Semi-infinite mediumConstant Surface Concentration: cA(0, t) = cA,s

= erf ; NA0 = − DAB = ,

, ,

A A s

A i A s

c cc c

2 AB

xD t

0

A

x

cx

,( , )AB A s A i

AB

D c cD t

Constant Surface Flux: NA(x=0) = NA0

cA(x, t) − cA,i = 2NA0 − AB

tD

2

exp4 AB

xD t

0A

AB

N xD 2 AB

xerfcD t

The complementary error function, erfc(w), is defined as erfc(w) = 1 – erf(w)

Surface Convection: DAB = hm(cAf cA,)0

A

x

cx

= ,

, ,'A A i

A A i

c cK c c

2 AB

xerfcD t

2

exp' 'm m

AB AB

h x h tK D K D

'2

m

ABAB

x h terfcK DD t

Notation: cA = concentration of species A in the solid at any location at any timecA,s = concentration of species A in the solid at the surface for t > 0cA,i = concentration of species A in the solid at any location and at t = 0cAf = concentration of species A in the liquid at the solid-liquid interface at any timecA, = bulk concentration of species A in the fluid surrounding the solidK’cA, = cA

* = concentration of species A in the solid that is in equilibrium with cA, hm = kc = mass transfer coefficientK’ = equilibrium distribution coefficientDAB = diffusivity of A in the solid

1-36

Table 1.6-4 Multidimensional Effects

L

L

L

L

x(r,x)

r

ro ro

c (r,x,t)A

The concentration profiles for a finite cylinder and a parallelpiped can be obtained from the concentration profiles of infinite cylinder and slabs.[ finite cylinder ] = [ infinite cylinder ] × [ slab 2L ][ parallelpiped ] = [ slab 2L1 ] × [ slab 2L2 ] × [ slab 2L3 ]

S(x, t) ,

Semi-infinite, ,solid

( , ) ''

A A

A i A

c x t K cc K c

P(x, t) ,

Plane, ,wall

( , ) ''

A A

A i A

c x t K cc K c

C(r, t) ,

Infinite, ,cylinder

( , ) ''

A A

A i A

c r t K cc K c

1-37

Chapter 1 Example 1.6-1 ------------------------------------------------------------------------------

A solid slab of 5.15 wt% agar gel at 278oK is 10.16 mm thick and contains a uniform concentration of urea of 0.1 kmol/m3. Diffusion is only in the direction through two parallel flat surfaces 10.16 mm apart. The slab is suddenly immersed in pure turbulent water so that the surface resistance can be assumed to be negligible; i.e, the convective coefficient hm is very large. The diffusivity of urea in the agar is 4.72×10-10 m2/s.

1) Calculate the concentration at the midpoint of the slab (5.08 mm from the surface) and 2.54 mm from the surface after 10 hr.2) If the thickness of the slab is halved, what would be the midpoint concentration in 10 hr?(Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)Solution ----------------------------------------------------------------------------------------------

1) Calculate the concentration at the midpoint of the slab (5.08 mm from the surface) and 2.54 mm from the surface after 10 hr.

The concentration at any point is given by the approximate solution

= cos(1x*) where = C1exp(- Fo) and Fo = * *0 *

0 21 2

ABD tL

2L = 10.16 mm => L = 5.08 mm = 5.08×10-3 m

Fo = = = 0.6584 > 0.2 2ABD t

L

10

3 2

(4.72 10 )(36000)(5.08 10 )

At the midpoint of the slab x = 0 => x* = = 0Lx

Bim = = ∞ => 1 = 1.5708 and C1 = 1.2732'm

AB

h LK D

= cos(1x*) = C1exp(- Fo) = 1.2732 exp(−1.57082×0.6584)* *0 2

1

= = 0.251 * ,

, ,

''

A A

A i A

c K cc K c

Since cA,∞ = 0 => = 0.251 => cA = (0.251)(0.1) = 0.0251 kmol/m3.,

A

A i

cc

At x = 2.54 mm => x* = = 0.5Lx

1-38

= C1exp(- Fo)cos(1x*) = 1.2732 exp(−1.57082×0.6584)cos(1.5708×0.5)* 21

= = (0.251)cos(π/4) = 0.177*,

A

A i

cc

cA = (0.177)(0.1) = 0.0177 kmol/m3.

2) If the thickness of the slab is halved, what would be the midpoint concentration in 10 hr?

Fo = = = 2.63362ABD t

L

10

3 2

(4.72 10 )(36000)(2.54 10 )

= cos(1x*) = C1exp(- Fo) = 1.2732 exp(−1.57082×2.6336)* *0 2

1

= = 0.00192*,

A

A i

cc

cA = (0.00192)(0.1) = 0.000192 kmol/m3.

Example 1.6-2 ------------------------------------------------------------------------------A slab of white pine, 5 cm thick, has a moisture content of 20 wt% at the start of the drying process. The equilibrium moisture content is 5 wt% for the humidity conditions in the drying air. The ends and edges are covered with moisture-resistance coating to prevent evaporation. The diffusivity of water through pine is 1.0×10-9 m2/s.1) If the two large surfaces are exposed to the drying air and the resistance to mass transfer outside the pine is negligible, estimate the time required to reduce the moisture content of the slab at the center to 10 wt% and the time to reduce the (average) water content to 10 wt%.2) Repeat part (1) if the external mass transfer resistance is equal to the internal mass transfer resistance.3) Repeat part (1) if only one large surface is exposed to the drying air while the other surface is also covered with moisture resistance coating. (Ref. Fundamentals of Momentum, Heat, and Mass Transfer by Welty, Wicks, and Wilson, 4th Edition, 2001)Solution ----------------------------------------------------------------------------------------------1) If the two large surfaces are exposed to the drying air and the resistance to mass transfer outside the pine is negligible, estimate the time required to reduce the moisture content of the slab at the center to 10 wt%.

The dimensionless concentration can be expressed in terms of mass fraction

= = ,

**

A A

A i A

c cc c

,

**

A A

A i A

,

/ * // * /

A B A B

A i B A B

1-39

ωA = => = A

A B

A

B

1

A

A

= = 0.25 = = 0.0526,A i

B

0.21 0.2

*A

B

0.051 0.05

= = 0.1111A

B

0.11 0.1

= = = 0.2964,

**

A A

A i A

c cc c

,

/ * // * /

A B A B

A i B A B

0.1111 .05260.2500 0.0526

Bim = = ∞ => 1 = 1.5708 and C1 = 1.2732'm

AB

h LK D

0.2964 = cos(1x*) = C1exp(- Fo) = 1.2732 exp(−1.57082 Fo)*0 2

1

Solving for Fo gives

Fo = = 0.591 2ABD t

L

Since 2L = 5 cm => L = 0.025 m

t = = 3.69×105 sec.2

9

(0.591)(0.025)10

Estimate the time to reduce the (average) water content to 10 wt%.

= 1 − = = = 0.7036tMM 1

1)sin(

*0

,

, *A i A

A i A

VV

0.2500 0.11110.2500 0.0526

Solving for gives*0

= = (1 − 0.7036) = 1.2732 exp(−1.57082 Fo)*0 1 tM

M

1

1sin( )

1.5708

sin(1.5708)


Fo = = 0.408 2ABD t

L

1-40

t = = 2.55×105 sec.2

9

(0.408)(0.025)10

2) Repeat part (1) if the external mass transfer resistance is equal to the internal mass transfer resistance.

Bim = = 1 => 1 = 0.8603 and C1 = 1.1191'm

AB

h LK D

0.2964 = cos(1x*) = C1exp(- Fo) = 1.1191 exp(−0.86032 Fo)*0 2

1


Fo = = 1.795 2ABD t

L

t = = 1.12×106 sec.2

9

(1.795)(0.025)10


= 1 − = = = 0.7036tMM 1

1)sin(

*0

,

, *A i A

A i A

VV

0.2500 0.11110.2500 0.0526

Solving for gives*0

= = (1 − 0.7036) = 1.1191 exp(−0.86032 Fo)*0 1 tM

M

1

1sin( )

0.8603

sin(0.8603)


Fo = = 1.624 2ABD t

L

t = = 1.015×106 sec.2

9

(1.624)(0.025)10

1-41

3) Repeat part (1) if only one large surface is exposed to the drying air while the other surface is also covered with moisture resistance coating.

Bim = = ∞ => 1 = 1.5708 and C1 = 1.2732'm

AB

h LK D

Since mass transfer occurs only on one side we have L = 0.050 m and x is zero at the coating surface. At the center of the slab x = 0.025 m and x* = 0.5

0.2964 = cos(1x*) = C1exp(- Fo) cos(1x*) *0 2

1

0.2964 = 1.2732 exp(−1.57082 Fo) cos(1.5708×0.5)


Fo = = 0.45 2ABD t

L

t = = 1.126×105 sec.2

9

(0.45)(0.05)10


= 1 − = = = 0.7036tMM 1

1)sin(

*0

,

, *A i A

A i A

VV

0.2500 0.11110.2500 0.0526

Solving for gives*0

= = (1 − 0.7036) = 1.2732 exp(−1.57082 Fo)*0 1 tM

M

1

1sin( )

1.5708

sin(1.5708)


Fo = = 0.408 2ABD t

L

t = = 1.02×105 sec.2

9

(0.408)(0.05)10

1-42

Example 1.6-3 ------------------------------------------------------------------------------A pure water stream is to be chlorinated by contacting it with pure chlorine gas in a counter current spray tower. The spray tower is operated at 15.5oC and 1.2 atm. The spray droplets are spherical with a diameter of 2.0×10-4 m and the tower height is 1.5 m. The concentration of chlorine on the surface of the droplet can be expressed in terms of the chlorine pressure as cA,s (kmol/m3) = 0.05 PA (atm). The diffusion coefficient of chlorine in water at this condition is 4.1×10-9 m2/s. The residence time of the water droplets in the column can be determined from h = 0.5gt2, where h is the tower height and g is the acceleration of gravity, 9.81 m2/s.1) Determine the chlorine concentration at the center of the droplet at the column exit.2) Determine the moles of chlorine transferred to a droplet at the column exit. (Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)Solution ----------------------------------------------------------------------------------------------1) Determine the chlorine concentration at the center of the droplet at the column exit.

The approximate concentration for a spherical system is given by

= = sin(1r*)*,

**

A A

A i A

c cc c

*

1

1r

*0

At r* = 0, = = C1exp(- Fo)* *0 2

1

There is no external resistance to mass transfer:

Bim = = ∞ => 1 = π and C1 = 2.0'm o

AB

h rK D

The contact or diffusion time for chlorine within the water droplet is determined from

t = = = 0.553 sec0.5

hg

1.50.5 9.81

Since pure water enters the tower, cA,i = 0. The equilibrium relation is cA,s (kmol/m3) = 0.05 PA (atm). Therefore

cA* = (0.05)(1.2) = 0.06 kmol/m3

Fo = = = 0.226732AB

o

D tr

9

4 2

(4.1 10 )(0.553)(10 )

= = C1exp(- Fo) = 2.0 exp(-π2×0.22673) = 0.2134,

**

A A

A i A

c cc c

*0 2

1

cA = cA* 0.2134 cA* = (0.06)(1 0.2134) = 0.0472 kmol/m3

1-43

2) Determine the moles of chlorine transferred to a droplet at the column exit.

= 1 − [sin(1) − 1cos(1)] = 1 − [sin(π) − π cos (π)]tMM

31

*03

3

3 0.2134

= 1 − = 0.935tMM

2

3 0.2134

The maximum moles that can be transferred to a water drop is

M∞ = πro3 cA* = π (10-4)3(0.06) = 2.513×10-13 kmol4

343

Moles of chlorine transferred to a droplet at the column exit is then

Mt = 0.935 M∞ = 2.35×10-13 kmol

Example 1.6-4 ------------------------------------------------------------------------------A wet cylinder of agar gel at 278oK containing a uniform concentration of urea of 0.1 kmol/m3 has a diameter of 30.48 mm and 38.1 mm long with flat parallel ends. The diffusivity is 4.72×10-10 m2/s. Calculate the concentration at the midpoint of the cylinder after 100 hr for the following cases if the cylinder is suddenly immersed in turbulent pure water.1) For radial diffusion only.2) Diffusion occurs radially and axially. (Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)Solution ----------------------------------------------------------------------------------------------1) Calculate the concentration at the midpoint of the cylinder after 100 hr for radial diffusion only.

For radial diffusion only, the solution of an infinite cylinder is used

= = J0(1r*) *,

**

A A

A i A

c cc c

*0

At the center of the cylinder, r* = 0 => J0(0) = 1

= = C1exp(- Fo)* *0 2

1

The cylinder is immersed in turbulent pure water so that cA* = 0.

Bim = = ∞ => 1 = 2.4048 and C1 = 1.6020'm o

AB

h rK D

1-44

Fo = = = 0.7322AB

o

D tr

10

2

(4.72 10 )(360,000)(0.01524)

= = C1exp(- Fo) = 1.6020 exp(-2.40482×0.732) = 0.0233,

00

A

A i

cc

*infinite0cylinder

21

cA = 0.0233 cA,i = (0.0233)(0.1) = 0.00233 kmol/m3

2) Calculate the concentration at the midpoint of the cylinder after 100 hr if diffusion occurs radialy and axially.

For a slab with thickness 2L = 38.1 mm, L = 0.01905 m

P(x, t) = cos(1x*)*

slab *

0

At the center, x* = 0, = = C1exp(- Fo) *

slab *

0 21

Bim = = ∞ => 1 = 1.5708 and C1 = 1.2732'm

AB

h LK D

Fo = = = 0.4682ABD t

L

10

2

(4.72 10 )(360,000)(0.01905)

= C1exp(- Fo) = 1.2732 exp(−1.57082×0.468) = 0.401*

slab 2

1

= = = (0.0233)(0.401) = 0.00934*finite0cylinder

,

**

A A

A i A

c cc c

*infinite0cylinder

*

slab

= 0.00934,

00

A

A i

cc

The concentration at the midpoint of the finite cylinder is then

cA = 0.00934 cA,i = (0.00934)(0.1) = 9.34×10-4 kmol/m3

1-45

Semi-infinite solution approximation

Consider the semi-infinite solution for the case of constant surface concentration given by

= erf,

, ,

A A s

A i A s

c cc c

2 AB

xD t

The solutions are shown in Figure 1.6-1 as a parametric family of “penetration” curve ,

, each plotted for a different value of time-diffusivity product = (4Dt)1/2.,

, ,

A A s

A i A s

c cc c

Figure 1.6-1 Normalized concentration field versus distance

The penetration achieved in the material is negligible for distances beyond about 2. For a large slab with finite thickness 2L and mass transfer through both surface, if the time for

mass transfer is less than the “penetration” time given by = 2 or t = , then the 2 AB

LD t

2

16 AB

LD

semi-infinite solution can be applied to this slab.

Example 1.6-5 ------------------------------------------------------------------------------A slab of white pine, 5 cm thick, has a moisture content of 20 wt% at the start of the drying process. The equilibrium moisture content is 5 wt% for the humidity conditions in the drying air. The ends and edges are covered with moisture-resistance coating to prevent evaporation. The diffusivity of water through pine is 1.0×10-9 m2/s. If the two large surfaces are exposed

1-46

to the drying air and the resistance to mass transfer outside the pine is negligible, determine the moisture concentration in the slab at 0.5 cm from the surface after 5 h.Solution ----------------------------------------------------------------------------------------------

For a slab with thickness 2L = 5 cm, L = 0.025 m

Bim = ∞ => 1 = 1.5708 and C1 = 1.2732 and cA,s = K’cA,∞

Fo = = = 0.0288 < 0.22ABD t

L

9

2

(10 )(5 3,600)(0.025)

We need more than one term of the series solution:

* = = = exp( Fo) cos(nx*),

, ,

''

A A

A i A

c K cc K c

,

, ,

A A s

A i A s

c cc c

1nnC 2

n

The first six terms of the series solution are given in the following table for x* = 20/25 = 0.8

n n Cn Cn exp( Fo) cos(nx*)2n

1 1.5708 1.2732 0.366462 4.7124 0.42441 0.181133 7.8540 0.25465 0.043094 10.9956 0.18189 0.004525 14.1372 0.14147 0.000146 17.2788 0.11575 0.00001

The series solution with the first six terms is

= 0.36646 + 0.18113 + 0.04309 + 0.00452 + 0.00014 0.00001 = 0.5953,

, ,

A A s

A i A s

c cc c

Since 5 h is less than the “penetration” time t = = = 39,063 s = 10.85 h, the 2

16 AB

LD

2

9

0.02516 10

semi-infinite solution can be used:

= erf,

, ,

A A s

A i A s

c cc c

2 AB

xD t

= erf = erf(5.8926) = 0.5953,

, ,

A A s

A i A s

c cc c

9

0.0052 10 5 3600

2-1

Chapter 2

Convective Mass Transfer

2.1 Introduction

The mass transfer coefficient for the transport of species A between two locations within a fluid may be defined from the following relations:

(Gases): NA = kc(cA1 − cA2) = kG(pA1 − pA2) = ky(yA1 − yA2)

(Liquids): NA = kc(cA1 − cA2) = kL(cA1 − cA2) = kx(xA1 − xA2)

In these equations, NA is the molar flux of species A and the mass transfer coefficient k has different subscript and different units depending on the units of the driving force used in the expression. Since many mass operations involve the transfer of material between two contacting phases, different subscript for the mass transfer coefficient is also used to distinguish between the phases.

The mass transfer coefficients might be obtained from the correlations given in chapter 8 where the Prandtl number (Pr = ν/α) is replaced by Schmidt number (Sc = ν/DAB) and the Nusselt number (Nu = hL/k) is replaced by Sherwood number (Sh = kcL/DAB). Table 2.1-1 lists some correlations to determine the mass transfer coefficient for external forced convection flow. The expressions for the flat plate are obtained from the solutions of the boundary layer equations. The other formulas are experimental correlations.

--------- Table 2.1-1 Mass transfer coefficients for external forced convection flow. -------Correlations Geometry Conditions

Shx = 0.332Rex1/2 Sc1/3 Flat plate Laminar, local, Tf, 0.6 Sc 50

= 0.664Rex1/2 Sc1/3

xSh Flat plate Laminar, average, Tf, 0.6 Sc 50

Shx = 0.0296Rex4/5 Sc1/3 Flat plate Turbulent, local, Tf, Rex 108

0.6 Sc 50 = (0.037Rex

4/5 871) Sc1/3xSh Flat plate Mixed, average, Tf, Rex,c = 5105

Rex 108, 0.6 Sc 50 = 0.3 + [0.62 ReD

1/2Sc1/3DSh

[1 + (0.4/Sc)2/3]-1/4] [1 + (ReD/282,000)5/8]4/5

Cylinder Average, Tf, ReDSc > 0.2Cross flow

= 2 + (0.4ReD1/2 + 0.06ReD

2/3)Sc0.4DSh

(/s)1/4

Sphere Average, T, 3.5 < ReD < 7.6104

0.71 < Sc < 380, 1.0 < (/s) < 3.2

These correlations are valid for equimolar mass transfer or low mass transfer rate where the mole fraction of species A is less than about 0.05. For higher mass transfer rate the coefficients might be corrected by the log mean concentration difference. Instead of using kc one should use kc/(1 − yA)lm where

2-2

(1 − yA)lm = 1 2

1

2

(1 ) (1 )1ln1

A A

A

A

y yyy

Example 2.1-1 ------------------------------------------------------------------------------Air at 32oC is humidified by flowing over a 1.2-m-long container filled with water. The interfacial temperature is 20oC. If the initial humidity of the air is 25% and its velocity is 0.15 m/s, calculate (a) the convective mass transfer coefficient, and (b) the amount of water evaporated per unit width of the container. (Ref. Fundamentals of Heat Transfer by Incropera and DeWitt, Wiley, 5th Edition, 2002)Solution ---------------------------------------------------------------------------------------------- The film temperature is Tf = (32 + 20)/2 = 26oC

Air at 26oC: ν = 1.51×10-5 m2/s, DAB = 2.77×10-5 m2/s.

Water vapor pressure: pAsat(20oC) = 0.02308 atm, pA

sat(32oC) = 0.04696 atm.

Partial pressure of water vapor at the air-water interface is pAs = pAsat(20oC) = 0.02308 atm.

Mole fraction of water vapor at this location is yAs = 0.02308.

Partial pressure of water vapor in the ambient air is pA∞ = 0.25×pAsat(32oC) = 0.25×0.04696

atm = 0.01174 atm. Mole fraction of water vapor at this location is yA∞ = 0.01174.

Since both yAs and yA∞ are less than 0.05, we could use kc without the correction factor (1 − yA)lm.

For parallel flow to a flat plate, laminar flow exists with Re < 300,000.

Re = = = 1.192×104U L

5

(0.15)(1.2)1.51 10

The average Sherwood number over the container may be obtained from the following correlation:

= 0.664Re1/2Sc1/3 = 0.664(1.192×104)1/2 = 59.22 = LSh1/ 31.51

2.77

c

AB

k LD

The convective mass transfer coefficient is then

kc = 59.22 = 59.22 = 1.367×10-3 m/sABDL

52.77 101.2

The molar flux of water is given by

NA = kc(cAs − cA∞)

2-3

The molar concentrations can be evaluated from the ideal gas law with gas constant R = 0.08205 atm∙m3/(kmol∙K):

cAs = = = 9.6×10-4 kmol/m3.As

s

pRT

0.02308(0.08205)(273 20)

cA∞ = = = 4.691×10-4 kmol/m3.ApRT

0.01174(0.08205)(273 32)

NA = kc(cAs − cA∞) = 1.367×10-3(9.6×10-4 − 4.691×10-4) = 6.711×10-7 kmol/m2∙s

The amount of water evaporated per m width of the container is

WA = (1)(1.2)(18) NA = 1.449×10-5 kg/s

Example 2.1-2 ------------------------------------------------------------------------------In a wetted-wall tower, an air-H2S mixture is flowing by a film of water which is flowing as a thin film down a vertical plate. The H2S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and 30oC. The value of kc of 9.567×10-4 m/s has been predicted for the gas-phase mass-transfer coefficient. At a given point the mole fraction of H2S in the liquid at the liquid-gas interface is 2.0×10-5 and pA of H2S in the gas is 0.05 atm. The Henry’s law equilibrium relation is pA(atm) = 609xA (mole fraction in liquid). Calculate the rate of absorption of H2S. (Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)Solution ----------------------------------------------------------------------------------------------

Vapor Liquid

y

yi

xi

x y

yi xi

x

Mass transfer from the liquid to the gas phase

Mass transfer from the gas to the liquid phase

Vapor LiquidAi Ai

The rate of absorption of H2S per unit area of the thin film is given by

NA = (cA − cAi) = (yA − yAi) = (yA − yAi)(1 )c

A lm

ky (1 )

y

A lm

ky (1 )

c

A lm

k cy

NA = (yA − yAi)(1 )c

A lm

k Py RT

2-4

The mole fraction of H2S in the gas phase is given by

yA = = = 0.0333App

0.051.5

The partial pressure of H2S in the gas phase at the interface is determined from Henry’s law and the mole fraction of H2S in the liquid at the liquid-gas interface.

pAi = 609xAi = 609×2.0×10-5 = 1.218×10-2 atm

The mole fraction of H2S in the gas phase at the interface is then

yAi = = = 0.00812Aipp

0.012181.5

(1 − yA)lm = ≈ = 0.979(1 ) (1 )1ln1

A Ai

A

Ai

y yyy

(1 ) (1 )2

A Aiy y

NA = (yA − yAi) = NA = (pA − pAi)(1 )c

A lm

k Py RT (1 )

c

A lm

ky RT

NA = = 1.486×10-3 kmol/m2∙s49.576 10

0.979

3

0.05 .01218(82.06 10 )(273 30)

2-5

Chapter 2 Example 2.1-3 ------------------------------------------------------------------------------Calculate the mass transfer from a sphere of naphthalene to air at 45oC 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 45oC is 6.92×10-6 m2/s and the vapor pressure of solid naphthalene is 0.555 mmHg. The mass transfer coefficient may be obtained from the following correlation:

= 2 + 0.552Re0.53Sc1/3DSh

(Ref: Transport Processes and Separation Process by C.J. Geankoplis, Prentice Hall, 4th Edition, 2003)Solution ----------------------------------------------------------------------------------------------

Let A denote naphthalene and B denote air. Since the mole fraction of naphthalene is very small, the physical properties of air at 45oC and 1 atm will be used for the gas mixture.

μ = μB = 1.93×10-5 Pa.s, ρ = ρB = 1.113 kg/m3

We now evaluate the dimensionless numbers:

Sc = = = 2.506ABD

5

6

1.93 10(1.113)(6.92 10 )

Re = = = 446.8VD 5

(1.113)(0.0254)(0.305)1.93 10

= 2 + 0.552Re0.53Sc1/3 = 21 = DSh c

AB

k DD

kc = = 5.72×10-3 m/s6(21)(6.92 10 )

0.0254

NA = (cAi − cA) = (pAi − pA)(1 )

c

A lm

ky (1 )

c

A lm

ky RT

(1 − yA)lm = ≈ 1(1 ) (1 )1ln1

A Ai

A

Ai

y yyy

NA = = 1.60×10-7 kmol/m2∙s35.72 10

(8314)(318)

50.555 1.013 10 0760

The mass transfer rate from the sphere is then

WA = πD2NA = π(0.0254)2(1.60×10-7) = 3.24×10-10 kmol/s

2-6

Example 2.1-4 ------------------------------------------------------------------------------An experiment can be performed to determine the mass transfer coefficient by flowing pure water through a tube constructed of solid benzoic acid. The saturated concentration of benzoic acid is 2.0×10-2 g/cm3. The water velocity is 10 cm/s and the mass of the tube is reduced by 0.62 g after 3 hr. Determine the mass transfer coefficient for the dissolution of benzoic acid in water if the tube diameter is 1.0 cm and the tube length is 20 cm.Solution ----------------------------------------------------------------------------------------------

x

V V

Making a material balance of benzoic acid over the control volume πD2∆x/4 we have

CVπ – CVπ + kc(C* – C)πD∆x = 0 (E-1)2

4 x

D 2

4 x x

D

In this expression, C is the bulk concentration of benzoic acid in water within the tube and C* is the saturated benzoic acid concentration in water at the solid and liquid interface. kc is the mass transfer coefficient for the dissolution of benzoic acid in water. Dividing equation (E-1) by πD2∆x/4 gives

– V + kc(C* – C) = 0 (E-2)| |x x xC Cx

4D

Taking the limit as ∆x → 0, we have

– V + kc(C* – C) = 0 dCdx

4D

Separating the variables and integrating to obtain

= 0 *

LC dCC C

4 ckDV 0

Ldx

– ln = => CL = C* **

LC CC

4 ck LDV

41 exp ck LDV

The mass transfer rate to the water is then

W = Vπ ( CL – 0) = Vπ CL

2

4D 2

4D

2-7

The mass change of the tube during the time t is given by

∆m = Wt = Vπ tC* 2

4D 41 exp ck L

DV

For x << 1, exp(– x) ≈ 1 – x, therefore

∆m = Vπ tC* = πDL tC*kc

2

4D 4 ck L

DV

The mass transfer coefficient is evaluated

kc = = = 4.57×10-5 cm/s*

mDLtC 0.62

(1)(20)(3 3600)(0.02)

Now we need to check the condition that x = << 14 ck LDV

= = 3.65×10-4 << 14 ck LDV

5(4 4.57 10 )(20)(1)(10)

Example 2.1-5 ------------------------------------------------------------------------------Liquid water at 25oC is to be aerated in a bubble column where finely air bubbles with diameter dB of 0.5 mm are injected cocurrently with the liquid. The interfacial contact area, a, between air and water can be calculated from the expression a = 6ε/dB, where ε is the volume fraction of the injected air. The bubble column is 1.8 m high with a superficial liquid velocity of 0.2 m/s. The oxygen concentration of the inlet water is 0.12×10-4 kmol/m3. The saturated oxygen concentration is 2.67×10-4 kmol/m3. Determine the oxygen concentration of the outlet water if the mass transfer coefficient for the transfer of oxygen from the liquid interface to the bulk water is 5.8×10-6 m/s. The diffusivity of oxygen in water is 2.42×10-9 m2/s. The volume fraction of the injected air is 0.2.Solution ----------------------------------------------------------------------------------------------

Making a steady state material balance for dissolved oxygen in water we have

CVA|x – CVA|x+∆x + (1 – ε)A – (1 – ε)A + kc(C* – C)aA∆x = 0A xN A x x

N

In this expression, C is the bulk concentration of oxygen in water, C* is the saturated oxygen concentration, A is the cross sectional area of the bubble column, V is the superficial liquid velocity, NA is the molar flux of oxygen in the x direction, and kc is the mass transfer coefficient. Dividing this equation by A∆x and taking the limit as ∆x → 0 we obtain

– (1 – ε) – V + kc(C* – C)a = 0 (E-1)AdNdx

dCdx

2-8

x

x

V|x

V|x+ x

Substituting NA = – DAB into equation (E-1) we havedCdx

(1 – ε)DAB – V + kc(C* – C)a = 0 (E-2)2

2

d Cdx

dCdx

The mass transfer contribution by axial diffusion, DAB , is usually much smaller than the 2

2

d Cdx

other terms. If we neglect the axial diffusion, equation (E-2) becomes

– V + kc(C* – C)a = 0dCdx

Separating the variables and integrating to obtain

= 0 *LC

C

dCC C cak

V 0

Ldx

– ln = => CL = C* – (C*– C0) 0

**

LC CC C

cak LV

exp cak LV

We have L = = = 0.1253cakV

6 c

B

k Ld V 6

3

(6)(0.2)(5.8 10 )(1.8)(0.5 10 )(0.2)

CL = 2.67×10-4 – (2.67 – 0.12)×10-4exp(– 0.1253) = 0.42×10-4 kmol/m3

2-9

2.2 Packed Column

Packed towers can be used for continuous countercurrent contacting of gas and liquid in absorption and for vapor-liquid contacting in distillation. In a packed column used for gas-liquid contact, the liquid flows downward over the surface of the packing and the gas flows upward in the void space of the packing material. A low pressure drop and, hence, low energy consumption is very important in the performance of packed towers. The packing material provides a very large surface area for mass transfer, but it also results in a pressure drop because of friction. The performance of packed towers depends upon the hydraulic operating characteristics of wet and dry packing. In dry packing, there is only the flow of a single fluid phase through a column of stationary solid particles. Such flow occurs in fixed-bed catalytic reactor and sorption operations (including adsorption, ion exchange, ion exclusion, etc.) In wet packing, two-phase flow is encountered. The phases will be a gas and a liquid in distillation, absorption, or stripping. When the liquid flows over the packing it occupies some of the void volume in the packing normally filled by the gas, therefore the performance of wet packing is different from that of dry packing.

For dry packing, the pressure drop may be correlated by Ergun equation

hP

= 150 + 1.75 (2.2-1)

sf

cp

vgD

1

3

Re

1N

whereP = pressure drop through the packed bedh = bed heightDp = particle diameterf = fluid densityvs = superficial velocity at a density averaged between inlet and outlet conditions = bed porosity

NRe = average Reynolds number based upon superficial velocity When the

fspvD

packing has a shape different from spherical, an effective particle diameter is defined

Dp = = (2.2-2)p

p

AV6

sA)1(6

whereAs = interfacial area of packing per unit of packing volume, ft2/ft3 or m2/m3

The effective particle diameter Dp in Eq. (2.2-1) can be replaced by sDp where Dp now represents the particle size of a sphere having the same volume as the particle and s the shape factor. The bed porosity, , which is the fraction of total volume that is void is defined as

2-10

bedentireofvolume

voidsvolume

bedentireofvolume

particlesofvolumebedentireofvolume

= (2.2-3)hR

densityparticleparticlesallofweighthR

2

2

where R = inside radius of column, As and are characteristics of the packing. Experimental values of can easily be determined from Eq. (2.2-3) but As for non-spherical particles is usually more difficult to obtain. Values of As and can be found in various references6,7 for the common commercial packing. As for spheres can be computed from the volume and surface area of a sphere.

For wet packing, the pressure drop correlation is given by Leva8

= (2.2-4)

hP LL /10

v

vG

2

where P is the pressure drop (psf), h is the packing height (ft), L is the liquid mass flow rate per unit area (lb/hr-ft2), Gv is the gas mass flow rate per unit area (lb/hr-ft2), L is the liquid density (lb/ft3), V is the gas density (lb/ft3), and and are packing parameters9.The initial procedure for designing a packed column is similar to that for a plate column. However we will need to follow different procedure in the calculation of the column diameter and height.

Vapor Liquid

y

yi

xi

x y

yi xi

x



Vapor LiquidAi Ai

Figure 2.2-1 Mass transfer across the interface.

6 Mc Cabe W. L. et al , Unit Operations of Chemical Engineering, McGraw-Hill, 1993, pg. 6897 Perry, J. H., Chemical Engineers’ Handbook, McGraw-Hill, 1984, pg. 18-238 Leva M., Chem. Eng. Prog. Symp. Ser. 50(10): 51 (1954)9 Wankat, P. C., Equilibrium Staged Separations, Elsevier, 1988, pg.420

2-11

Chapter 2 The mass transfer rate, , of species A across the interfacial area for mass transfer Ai is Amgiven by

= (Area for mass transfer)(mass transfer coefficient)(driving force)Am

The driving force for mass transfer can be expressed in many different ways. It could be based on the mole or mass fraction in the gas, or liquid phase, or both. The mass transfer rate for mass transfer from the liquid to the gas phase can be written as

= Aiky(yAi yA) = Aikx(xA xAi) (2.2-5)Am

In this expression, ky and kx are the individual mass transfer coefficients based on the gas and the liquid phase, respectively. The mole fractions yAi, yA, xA, and xAi are defined in Figure 2.2-1. The mass transfer rate for mass transfer from the gas to the liquid phase can be written as

= Aiky(yA yAi) = Aikx(xAi xA) (2.2-6)Am

If we do not know the direction of mass transfer, we could use either equation (2.2-5) or equation (2.2-6). If the mass transfer rate calculated to be positive, our assumption of the mass transfer direction is correct. For example, if we use equation (2.2-5) and is positive, Amthen species A is being transferred from the liquid to the gas phase.

Let a be the interfacial area per unit volume of packing (m2/m3). Multiplying both sides of equation (2.2-5) by a, we obtain

a = Aiaky(yAi yA) = aky(yAi yA) (2.2-7)AmaA

m

i

A

/

Ai/a = (Interfacial area)/(Interfacial area/volume of packing) = Vpack = Volume of packing.

Equation (2.2-7) can be written as

= aky(yAi yA) (2.2-8)A

pack

mV

Since, the interfacial concentrations are difficult to measure, the mass transfer rate is usually written in terms of the overall mass transfer coefficient that is based on the overall driving force for mass transfer (yA yA

*) or (xA* xA) as shown in Figure 2.2-2.

2-12

Equilibriumdata

x

y

yA

yAi

y *A

xA xAi x *A

m2

m1

Figure 2.2-2 Concentration driving forces in interphase mass transfer.

For mass transfer from the gas to the liquid phase (as shown in Figure 2.2-2)

= aKy(yA yA*) = aKx(xA

* xA) (2.2-9)A

pack

mV

For mass transfer from the liquid to the gas phase

= aKy(yA* yA) = aKx(xA xA

*) (2.2-10)A

pack

mV

The term is the total resistance to mass transfer based on the gas phase, and the yaK

1

xaK1

total resistance to mass transfer based on the liquid phase. We have neglected the resistance to mass transfer at the interface. In Figure 2.2-2, m1 is the average slope of the equilibrium curve between two points (xA, yA

*) and (xAi, yAi), m2 the average slope of the equilibrium curve between two points (xAi, yAi) and (xA

*, yA).

2-13

We now want to determine the height of a packed bed required to change the concentration of the inlet gas from yA,in to yA,out in a distillation column.

z

yA,out

yA,in

V|z+dz

V|z

L|z+dz

L|z

dz

Figure 2.2-3 Material balance over Acdz

We will assume constant molar overflow so that the vapor molar flow rate, V, and the liquid molar flow rate, L, are constant over the height of the packed column. Let Ac be the cross-sectional area of the column, the material balance over the differential volume Acdz gives

xAL|z+dz + yAV|z = xAL|z + yAV|z+dz

Rearranging the liquid and vapor flow rates gives

xAL|z+dz xAL|z = yAV|z+dz yAV|z

Dividing the equation by dz and letting the control volume Acdz approach zero, we have

d(LxA) = d(VyA)

For constant L and V

Ld(xA) = Vd(yA)

The molar flux of A across the interface is

NA = Ky(yA yA*)

Multiplying the expression by aAcdz gives

NAaAcdz = Ky(yA yA*)aAcdz

Since NAAiadz = volume = rate of A transfer from volume Acdz, we ))(( timearea

transferMassvolume

area

have

2-14

Ky(yA yA*)aAcdz = Vd(yA)

Solving for dz gives

dz = AAAy

dyyyaK

V*)(

Integrating over the height of the packed bed we have

h = = (2.2-11)hdz

0 outA

inA

y

y AAAcy

dyyyaAK

V,

,*)(

The height of an overall gas transfer unit, HOG, is defined as

HOG = cyaAK

V

If HOG is a constant, equation (2.2-11) becomes

h = (2.2-12)cyaAK

V

outA

inA

y

yAA

A

yydy

,

,*)(

For distillation column where species A is transferred from the liquid to the gas phase, h is given by

h = = HOGnOG (2.2-13)cyaAK

V

outA

inA

y

yAA

A

yydy

,

,)*(

In this expression, nOG is the number of overall gas transfer unit. If the driving force is based on the driving force in the liquid phase

NAaAcdz = Kx(xA xA*)aAcdz = Ld(xA) (2.2-14)

The height of the packed bed is then given by

h = = HOLnOL (2.2-15)cxaAK

L

outA

inA

xx

AA

A

xxdx

,

,*)(

HOL is the height of an overall liquid transfer unit and nOL the number of overall liquid

transfer unit. We now need to evaluate the integral or . outA

inA

y

yAA

A

yydy

,

,)*(

outA

inA

xx

AA

A

xxdx

,

,*)(

2-15

z

yA,out

yA,in

V L

yA xA

xA,in

xA,out xA

yA

y *- yA A

Equilibrium curve

Operating line

yA,out

yA,in

xA

Figure 2.2-4 Material balance over the lower section of the tower.

Assuming L and V are constant and making an A balance over the lower section of the tower as shown in Figure 2.2-4 we have

xA,outL + yAV = xAL + yA,inV

Solving for yA we obtain an equation called the operating line

yA = xA + yA,in xA,out (2.2-16)VL

VL

At any location in the packed bed the bulk concentration in the vapor and liquid phase are yA and xA respectively. The point (xA, yA) is on the operating line shown in Figure 2.2-4. The

number of transfer unit nOG or the integral can be evaluated where (yA*

outA

inA

y

yAA

A

yydy

,

,)*(

yA) is the difference between the concentration yA* that is in equilibrium with the liquid at xA

to the vapor concentration at the point (xA, yA). If the operating line and equilibrium curves are straight, nOG can be evaluated analytically. This condition might occur in gas absorption with dilute solution where the equilibrium curve is straight and L and G are constant. We will drop the subscript A with the understanding that y and x are the mole fractions of the diffusing species in the gas and liquid phase, respectively. For gas absorption, we use G for the gas flow rate instead of V for the vapor flow used in distillation.

Making a material balance for the diffusing species over the top section of the column as shown in Figure 2.2-5 we have

xinL + yG = xL + youtG

Solving for y we obtain a straight operating line since L/G = constant

y = x + yout xin (2.2-17)GL

GL

2-16

z

yout

yin

G L

y x

xin

xout x

y

y - y *A A

Equilibrium curve

Operating lineyout

yin

xFigure 2.2-5 Material balance over the upper section of the tower.

For straight equilibrium line

y* = mx

The number of overall gas transfer unit can be written as

nOG = = ( * )

out

in

Ay

y

dyy y

in

out

y

y yydyA

*)(

Combining the operating line and the equilibrium line, we have

y y* = y mx = y m = y + yout mxin

inout xy

LGy

LG

LmG1

LmG

nOG =

in

out

y

yinout

A

mxyL

mGyL

mGdy

1

Performing the integration gives

nOG = ln

LmG

1

1

inoutout

inoutin

mxyL

mGyL

mG

mxyL

mGyL

mG

1

1

nOG = ln (2.2-18)

LmG

1

1

inout

inoutin

mxy

mxyL

mGyL

mG1

2-17

Since y*out = mxin, we have

yin + yout mxin = yin + yout y*out

LmG1

LmG

LmG1

LmG

yin + yout mxin = yin + yout y*out + y*out y*out

LmG1

LmG

LmG1

LmG

LmG

LmG

yin + yout mxin = yin y*out + (yout y*out)

LmG1

LmG

LmG1

LmG1

LmG

Substituting the above expression into equation (2.2-18), the number of overall gas transfer unit is finally

nOG = ln (2.2-19)

LmG

1

1

LmG

yyyy

LmG

outout

outin

**1

We can follow a similar procedure to obtain the number of overall liquid transfer unit

nOL = out

in

xx

A

xxdx

*)(

nOL = ln (2.2-20)

mGL

1

1

mGL

xxxx

mGL

outout

outin

**1

In the above expression x*out = yin/m.

Example 2.2-110. ----------------------------------------------------------------------------------Acetone in air is being absorbed by water in a packed tower having a cross-sectional area of 0.186 m2 at 293 K and 1 atm. At these conditions, the equilibrium relation is given by ye = mxe. The inlet air contains 2.6 mole % acetone and the outlet 0.5 %. The air flow is 13.65 kmol/hr and the pure water inlet flow is 43.56 kmol/hr. Film coefficients for the given flow in the tower are kya = 3.8×10-2 kmol/s∙m3 and kxa = 6.2×10-2 kmol/s∙m3. Determine the tower height.

Solution -----------------------------------------------------------------------------------------The tower height can be obtained from the following relation

h = = HOGnOGy c

GK aA

outA

inA

y

yAA

A

yydy

,

,)*(

10 C. J. Geankoplis, Transport Processes and Separation Process Principle, Prentice Hall, 2003, pg. 672

2-18

For straight operating and equilibrium lines, the number of overall gas transfer unit is given by

nOG = = ln outA

inA

y

yAA

A

yydy

,

,)*(

LmG

1

1

LmG

yyyy

LmG

outout

outin

**1

We have m = 1.186, G =13.65 kmol/hr, L = 43.56 kmol/hr, yin = 0.026, yout = 0.005, y*out = mxin = 1.186(0) = 0.

= 0.3569mGL

nOG = ln = ln

LmG

1

1 1 in

out

mG y mGL y L

11 0.3569

0.0261 0.3569 0.35690.005

nOG = 2.0348

The overall mass transfer coefficient is determined from the film coefficients.

= + => Kya = 1

yK a1

yk a x

mk a

11y x

mk a k a

Kya = = 0.022 kmol/s∙m311 1.186

0.038 .062

Therefore

HOG = = = 0.9264 my c

GK aA

(13.65/ 3600)(0.022)(0.186)

The tower height is then

h = HOGnOG = (0.9264)(2.0348) = 1.89 m

3-1

Chapter 3 Phase Equilibrium Calculation

3.1 Introduction

Consider a mixture of different components in a gas, liquid, or solid phase. One way to remove one or more of the components from its original mixture, we could let this mixture to be in intimate contact with another phase so that a solute or solutes can diffuse from one to the other. The two phases will eventually reach equilibrium and the components of the original mixture redistribute themselves between the two phases. The compositions in each phase can be determined from the equilibrium requirements of equal values for the temperature, pressure, and fugacity of each component in both phases. By choosing the proper conditions and phases, one phase is enriched while the other is depleted in one or more components. Table 3.1 lists some of the common separation processes.

Table 3.1 Common separation processes.ProcessAbsorption Transfer of solute(s) from gas to liquid phase.Desorption or Stripping Transfer of solute(s) from liquid to gas phase.Distillation Transfer of the more volatile component(s) to the vapor phase and

the less volatile component(s) to the liquid phase.Extraction Transfer of solute(s) from a liquid to another liquid phase.Leaching Transfer of solute(s) from solid to liquid phase.

3.2 Vapor-Liquid Equilibrium Calculation

The most common problems requiring contact between phases chemical engineers encounter in the chemical, petroleum, and related industries involve vapor-liquid equilibrium. At equilibrium, the fugacity of species i in the vapor phase is equal to that in the liquid phase

= (3.2-1)vif

lif

The fugacity of species i in the vapor phase can be expressed in terms of the mole fraction, yi, fugacity coefficient, , and total pressure, P, asv

i

= yi P (3.2-2)vif

vi

The fugacity of species i in the liquid phase can be expressed in terms of the mole fraction, xi, activity coefficient, i, and fugacity of pure component, fi, as

= xii fi (3.2-3)lif

Thereforeyi P = xii fi (3.2-4)v

i

3-2

If the vapor phase obeys ideal gas law, = 1, and the liquid solution is ideal, i = 1, Eq. vi

(3.2-4) becomes

yiP = xifi (3.2-5)

At low pressure, fi = Pisat, Eq. (3.2-5) becomes Raoult’s law

yiP = xiPisat (3.2-6)

The equilibrium ratio or K-value is defined as

Ki = (3.2-7)i

i

xy

When Raoult’s law applies we have

Ki = (3.2-8)P

P sati

In general, the K-values depend on temperature, pressure, and the composition in both phases. For light hydrocarbon system (methane to decane), the K-values have been determined semi-empirically and can be evaluated from the equations given in Table 3.2-11. In general, K is a function of temperature, pressure, and composition.

Table 3.2-1 Equilibrium K values for light hydrocarbon systems=============================================================

(1) ln K = A/T2 + B C ln(P) + D/P2

(2) ln K = A/T2 + B C ln(P) + D/P(3) ln K = A/T + B C ln(P) , where P is in psia, T is in oR

Compound A B C D Form=============================================================Methane 292860 8.2445 .8951 59.8465 (1)Ethylene 600076.9 7.90595 .84677 42.94594 (1)Ethane 687248.2 7.90694 .866 49.02654 (1)Propylene 923484.7 7.71725 .87871 47.67624 (1)Propane 970688.6 7.15059 .76984 6.90224 (2)i-Butane 1166846 7.72668 .92213 0 (1)n-Butane 1280557 7.94986 .96455 0 (1)i-Pentane 1481583 7.58071 .93159 0 (1)n-Pentane 1524891 7.33129 .89143 0 (1)n-Hexane 1778901 6.96783 .84634 0 (1)n-Heptane 2013803 6.52914 .79543 0 (1)n-Octane 7646.816 12.48457 .73152 (3)n-Nonane 2551040 5.69313 .67818 0 (1)n-Decane 9760.457 13.80354 .7147 (3)=============================================================

1 Wankat, P. C., Equilibrium Staged Separations, Elsevier, 1988

3-3

In flash distillation, a liquid mixture is partially vaporized and the vapor is allowed to come to equilibrium with the liquid. The process flow diagram is shown in Figure 3.2-1. The vapor and liquid phases are then separated.

F, xiF

V, yi

L, xi

Q

Figure 3.2-1 Flash distillation.

Besides flash calculation, which will be discussed later, there are four types of vapor-liquid equilibrium calculations: (1) Bubble point temperature calculation, (2) Bubble point pressure calculation, (3) Dew point temperature calculation, and (4) Dew point pressure calculation.

Bubble point temperature calculation

In a bubble point temperature calculation, the pressure and liquid phase composition are specified. We will solve for the temperature and the vapor composition. The solution provides the composition of the first bubble of vapor that forms when heat is supplied to a saturated liquid. Since the vapor mole fractions are unknown, we start with the equation

= 1 (3.2-9)

m

iiy

1

Using the K-values: Ki = , Eq. (3.2-9) becomesi

i

xy

= 1 (3.2-10)

m

iii xK

1

If the system contains more than two components, we might want to solve the log form of equation (3.2-10) for better convergence

ln = 0 (3.2-11)

m

iii xK

1

3-4

Example 3.2-1 ----------------------------------------------------------------------------------Determine the temperature and composition of the first bubble created from a saturated liquid mixture of benzene and toluene containing 45 mole percent benzene at 200 kPa. Benzene and toluene mixtures may be considered as ideal.

Data: Vapor pressure, Psat, data: ln Psat = A B/(T + C), where Psat is in kPa and T is in oK.

Compound A B CBenzene (1) 14.1603 2948.78 44.5633Toluene (2) 14.2515 3242.38 47.1806

Solution ------------------------------------------------------------------------------------------

We start with the equation

y1 + y2 = 1 (E-1)

Substituting yi = xiPisat/P into equation (E-1) yields

x1P1sat + x2P2

sat = P (E-2)

With the numerical values for mole fractions and pressure, equation (E-2) becomes

0.45exp(14.1603 2948.78/(T 44.5633)) + 0.55exp(14.2515 3242.38/(T 47.1806)) = 200 (E-3)

The bubble point temperature should be between the boiling points of benzene and toluene given by

T1boil = + 44.5633 = 377.31oK

)200log(1603.1478.2948

T2boil = + 47.1806 = 409.33oK

)200log(2515.1438.3242

The solution of the nonlinear algebraic equation (E-3) can be determined using Matlab function fsolve with inline function as follows:

fun=inline('0.45*exp(14.1603 - 2948.78/(T - 44.5633))+ 0.55*exp(14.2515 - 3242.38/(T - 47.1806)) - 200');>> T=fsolve(fun,400,optimset('Display','off'))T = 391.7925

The bubble point temperature of the benzene-toluene mixture is 391.8oK. At this temperature, the vapor pressure of benzene is

P1sat = exp(14.1603 2948.78/(391.7925 44.5633)) = 289.45 kPa

3-5

The mole fraction of benzene in the vapor phase is then

y1 = = = 0.6513PPx sat

11

200)45.289)(45.0(

The mole fraction of toluene in the vapor phase is

y2 = 1 y1 = 0.3487

--------------------------------------------------------------------------------------------------------

Bubble point pressure calculation

In a bubble point pressure calculation, the temperature and liquid phase composition are specified. We will solve for the pressure and the vapor composition. Since the vapor mole fractions are unknown, we start with the equation

= 1 (3.2-12)

m

iiy

1


i

xy

1 = 0 (3.2-13)

m

iii xK

1

Example 3.2-2 ----------------------------------------------------------------------------------Determine the pressure and composition of the first bubble created from a saturated liquid mixture of benzene and toluene containing 45 mole percent benzene at 400oK. Benzene and toluene mixtures may be considered as ideal.



Solution ------------------------------------------------------------------------------------------


y1 + y2 = 1 (E-1)


x1P1sat + x2P2

sat = P (E-2)

3-6

With the numerical values for mole fractions and temperature, the bubble point pressure is

P = 0.45exp(14.1603 2948.78/(400 44.5633)) + 0.55exp(14.2515 3242.38/(400 47.1806)) = 245.284 kPa

At 400oK, the vapor pressure of benzene is

P1sat = exp(14.1603 2948.78/(400 44.5633)) = 352.160 kPa

The mole fraction of benzene in the vapor phase is then

y1 = = = 0.6461PPx sat

11

284.245)160.352)(45.0(

The mole fraction of toluene in the vapor phase is

y2 = 1 y1 = 0.3539---------------------------------------------------------------------------------------------------

Dew point temperature calculation

In a dew point temperature calculation, the pressure and vapor phase composition are specified. We will solve for the temperature and the liquid composition. The solution provides the composition of the first drop of dew that forms from a saturated vapor. Since the liquid mole fractions are unknown, we start with the equation

= 1 (3.2-14)

m

iix

1


i

xy

= 1 (3.2-15)

m

i i

i

Ky

1

Example 3.2-3 ----------------------------------------------------------------------------------Determine the temperature and composition of the first dew created from a saturated vapor mixture of benzene and toluene containing 45 mole percent benzene at 200 kPa. Benzene and toluene mixtures may be considered as ideal.



Solution ------------------------------------------------------------------------------------------

3-7


x1 + x2 = 1 (E-1)

Substituting xi = yiP/Pisat into equation (E-1) yields

+ = 1 (E-2)satPPy

1

1satPPy

2

2


90/exp(14.1603 2948.78/(T 44.5633)) + 110/exp(14.2515 3242.38/(T 47.1806)) = 1 (E-3)

The dew point temperature should be between the boiling points of benzene and toluene given by

T1boil = + 44.5633 = 377.31oK

)200log(1603.1478.2948

T2boil = + 47.1806 = 409.33oK

)200log(2515.1438.3242


>> fun=inline('90/exp(14.1603 - 2948.78/(T - 44.5633))+ 110/exp(14.2515 - 3242.38/(T - 47.1806)) - 1');>> T=fsolve(fun,400,optimset('Display','off'))T = 398.0874

The dew point temperature of the benzene-toluene mixture is 398.1oK. At this temperature, the vapor pressure of benzene is

P1sat = exp(14.1603 2948.78/(398.0874 44.5633)) = 336.70 kPa

The mole fraction of benzene in the liquid phase is then

x1 = = = 0.2673satPPy

1

1

7.336)200)(45.0(

The mole fraction of toluene in the liquid phase is

x2 = 1 x1 = 0.7327

3-8

--------------------------------------------------------------------------------------------------------

Dew point pressure calculation

In a dew point pressure calculation, the temperature and vapor phase composition are specified. We will solve for the pressure and the liquid composition. Since the liquid mole fractions are unknown, we start with the equation

= 1 (3.2-16)

m

iix

1


i

xy

= 1 (3.2-17)

m

i i

i

Ky

1

If the system contains more than two components, we might want to solve the log form of equation (3.2-17) for better convergence

ln = 0 (3.2-18)

m

i i

i

Ky

1

Example 3.2-4 ----------------------------------------------------------------------------------Determine the temperature and composition of the first dew created from a saturated vapor mixture of benzene and toluene containing 45 mole percent benzene at 400oK. Benzene and toluene mixtures may be considered as ideal.



Solution ------------------------------------------------------------------------------------------


x1 + x2 = 1 (E-1)


+ = 1 P = (E-2)satPPy

1

1satPPy

2

2

satsat Py

Py

2

2

1

1

1

3-9

With the numerical values for mole fractions and temperature, equation (E-2) becomes

P =

))1806.47400/(38.32422515.14exp(55.0

))5633.44400/(78.29481603.14exp(45.0

1

P = 209.98 kPa

At 400oK, the vapor pressure of benzene is

P1sat = exp(14.1603 2948.78/(400 44.5633)) = 352.160 kPa

The mole fraction of benzene in the liquid phase is then

x1 = = = 0.2683satPPy

1

1

16.352)98.209)(45.0(

Example 3.2-5 ----------------------------------------------------------------------------------Find the bubble point pressure and vapor composition for a liquid mixture of 41.2 mol% ethanol (1) and n-hexane (2) at 331oK.2Data: Activity from Van Laar equations:

ln 1 = ln 2 = 2

21 )/(1 BxAxA

212 )/(1 AxBx

B

A = 2.409, and B = 1.970

Vapor pressure: (Pisat in kPa and T in oK)

ln P1sat = 16.1952 ln P2

sat = 14.0568 7152.5553.3423

T 7089.4242.2825

T

Solution ------------------------------------------------------------------------------------------

At 331oK

P1sat = exp[16.1952 3423.53/(331 55.7152)] = 42.9 kPa

P2sat = exp[14.0568 2825.42/(331 42.7089)] = 70.54 kPa

For the liquid solution with x1 = 0.412 and x2 = 0.588, we have

2 Kyle, B.G., Chemical and Process Thermodynamics, Prentice Hall, 1999, pg. 279

3-10

ln 1 = = 0.699 1 = 2.011 2

21 )970.1/409.2(1409.2

xx

ln 2 = = 0.4195 2 = 1.521 2

12 )409.2/790.1(1790.1

xx

The partial pressure of ethanol is evaluated from

P1 = x11P1sat = (0.412)(2.011)(42.9) = 35.55 kPa

Similarly, the partial pressure of n-hexane is given by

P2 = x22P2sat = (0.588)(1.521)(70.54) = 63.09 kPa

The bubble point pressure is then

P = P1 + P2 = 35.55 + 63.09 = 98.64 kPa

Mole fraction of ethanol in the vapor phase is calculated from

y1 = P1/P = 35.55/98.64 = 0.360

The actual data for this system is Pexp = 101.3 kPa, and y1,exp = 0.350

3-11

Chapter 3

3.3 Isothermal Flash Calculation Using K-values

In flash distillation, a liquid mixture is partially vaporized and the vapor is allowed to come to equilibrium with the liquid. The process flow diagram is shown in Figure 3.3-1. The vapor and liquid phases are then separated.

F, xiF

V, yi

L, xi

Q


Making a component i balance gives

FxiF = Vyi + Lxi = Vyi + (F V)xi (3.3-1)

Defining f = V/F, Eq. (3.3-1) becomes

xiF = fyi + (1 f)xi (3.3-2)

The above equation can be solved for yi,

yi = Kixi = xi + (3.3-3)f

f 1f

xiF

or for xi,

xi = (3.3-4)1)1( i

iF

Kfx

We will discuss the solution for two cases of isothermal flash calculation. In the first case, the fraction of the feed vaporized, f, is specified and in the second case f is determined.

3-12

Case 1: Fraction of the feed, f, vaporized is specified

The feed composition xiF and the fraction f of the feed vaporized are given at a specified separator pressure P, the temperature T and compositions xi and yi can be calculated by solving the equation:

= = 1

m

iix

1 1)1),(( PTKf

xi

iF

The procedure for the calculation is as follows:

1) Determine the bubble, Tb, and dew, Td, point temperatures at the feed composition.

2) Assume a temperature T = fTd + (1 f)Tb

3) Evaluate Ki = Ki (T, P)

4) Evaluate f(T) = 1 1)1),(( PTKfx

i

iF

5) Evaluate Ki = Ki (T + dT, P) and f(T + dT)

6) Since the slope of the curve f(T) versus T is approximated by

Slope = = dT

TfdTTf )()(

calTTTf

)(

The calculated Tcal is given by

Tcal = T )()(

)(TfdTTf

dTTf

Steps (3-6) are repeated until |Tcal T| error tolerance

Case 2: Fraction of the feed, f, vaporized is determined

If the feed composition xiF, temperature T and pressure P of separator are given, then the fraction of the feed vaporized V/F and compositions xi and yi can be calculated. Eqs. (3.3-3) and (3.3-4) can be arranged so that f = V/F is the only unknown.

= 0 (3.3-5) iy ix

= 0 (3.3-6) 1)1( i

iFi

KfxK 1)1( i

iF

Kfx

3-13

F = = 0 (3.3-7)

1)1()1(

i

iFi

KfxK

Equation (3.3-7), which is known as the Rachford-Rice equation, has excellent convergent properties and can be solved by Newton’s method. Take the derivative of the function F with respect to V/F (or f),

= (3.3-8)dfdF

2

2

1)1()1(

i

iFi

KfxK

The following procedure can be used to solve for V/F:

1) Evaluate Ki = Ki (T, P)

2) Check to see if T is between Tb and Td.

If all K-values are greater than 1, the feed is a superheated vapor above the dew point1. If all K-values are less than 1, the feed is a subcooled liquid below the bubble point. If one or more K-values are greater than 1 and one or more K-values are less than 1 we need to evaluate Eq. (3.3-7) at f = 0 and at f = 1.

2a) If < 0, the feed is below its bubble point. iFi xK )1(

2b) > 0 the feed is above its dew point.

i

iFi

KxK )1(

3) Assume f = 0.5

4) Evaluate F =

1)1()1(

i

iFi

KfxK

5) Evaluate = dfdF

2

2

1)1()1(

i

iFi

KfxK

6) Let ER = F/ , and f = f ER dfdF

7) If abs(ER) > .001 go to step 4, otherwise

xi = and yi = Kixi1)1( i

iF

Kfx

1 Seader J. D., and Henley E. J., Separation Process Principles, Wiley, 1998, pg. 180.

3-14

Example 3.3-1 ----------------------------------------------------------------------------------The following mixture is fed to a flash drum at 200 psia.

Ethylene C2H4: 20% Ethane C2H6: 20% Propane C3H8: 40% n-Buane C4H10: 20%

Determine the temperature and the composition of the vapor and liquid streams leaving the flash drum if 40% of the feed is vaporized.

Data: Equilibrium K values for light hydrocarbon systems

=============================================================ln K = A/T2 + B C ln(P) + D/P + E/P2, where P is in psia, T is in oR

Compound A B C D E=============================================================Ethylene 600076.9 7.90595 .84677 0 42.94594Ethane 687248.2 7.90694 .866 0 49.02654Propane 970688.6 7.15059 .76984 6.90224 0n-Butane 1280557 7.94986 .96455 0 0=============================================================

Solution

The vapor fraction f is specified, therefore the temperature of the flash drum or the temperature of the vapor and liquid streams leaving the flash drum can be calculated from the equation:

= = 1

m

iix

1 1)1),(( PTKf

xi

iF

This is a nonlinear equation in term of T. We will solve this equation using Newton’s method with numerical derivative approximation for a slope of the function

f(T) = 1 1)1),(( PTKfx

i

iF

Slope = dT

TfdTTf )()(

The initial guess of the flash temperature can be obtained from the bubble point and dew point of the feed mixture.

Dew point calculation:

We will first determine the dew point of the mixture from the equation:

= 1 (3.2-15)

m

i i

i

Ky

1

3-15


f(T) = 1 = 0

m

i i

i

Ky

1

Slope = dT

TfdTTf )()(

The initial guess of the dew point temperature can be obtained from the boiling points of the pure species at the flash pressure of 200 psia where K = 1. Since this is just a guess value for the dew point temperature, the boiling points are estimated from the first three terms of the equation for K

ln K = A/T2 + B C ln(P) = 0 for K = 1

T = (A./(B-C*ln(P)))0.5

The estimated boiling points for the hydrocarbons at 200 psia are given in the following table:

Species A B C Tboil(K) at 200 psiaEthyleneEthanePropanen-Butane

600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

418.91455.07 562.14 671.57

The initial guess for the dew point temperature is

Tguess = 0.2×418.91 + 0.2×455.07 + 0.4×562.14 + 0.2×671.57 = 533.97 K

The equilibrium ratios are then evaluated at 533.97 K

Kequi = exp[A/T2 + B C ln(P) + D/P + E/P2]

Species A B C D E KequiEthyleneEthanePropanen-Butane

600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.7283 2.4830 0.74214 0.19168

f(T) = f(533.97) = 1 = + + + 1 = 0.71660

m

i i

i

Ky

1

0.23.7283

0.22.4830

0.40.74214

0.20.19168

Using dT = 1 K, T + dT = 534.97 K, the equilibrium ratios at this temperature are

3-16


600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.7577 2.5054 0.75164 0.19492

f(534.97) = 1 = + + + 1 = 0.69129

m

i i

i

Ky

1

0.23.7577

0.22.5054

0.40.75164

0.20.19492

The slope at 533.97 K is then

Slope = (0.69129 0.71660)/1 = 0.02531

The new temperature from Newton’s method is

T = Tguess f(533.97)/Slope = 533.97 + 0.71660/0.02531 = 533.97 + 28.313 = 562.28 K

The procedure is repeated until the magnitude difference between the new and the previous temperature is less than 0.1 K and the final dew point temperature is 574.80 K.

Bubble point calculation:

The bubble point of the mixture can be calculated from the equation:

ln = 0 (3.2-11)

m

iii xK

1


f(T) = ln = 0

m

iii xK

1

Slope = dT

TfdTTf )()(

The initial guess of the bubble point temperature can be obtained from the boiling points of the pure species at the flash pressure of 200 psia where K = 1. Since this is just a guess value for the bubble point temperature, the boiling points are estimated from the first three terms of the equation for K

ln K = A/T2 + B C ln(P) = 0 for K = 1

T = (A./(B-C*ln(P)))0.5

3-17

The estimated boiling points for the hydrocarbons at 200 psia are given in the following table:

Species A B C Tboil(K) at 200 psiaEthyleneEthanePropanen-Butane

600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

418.91455.07 562.14 671.57

The initial guess for the bubble point temperature is

Tguess = 0.2×418.91 + 0.2×455.07 + 0.4×562.14 + 0.2×671.57 = 533.97 K




600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.7283 2.4830 0.74214 0.19168

f(T) = f(533.97) = ln

m

iii xK

1

f(533.97) = 0.2×3.7283 + 0.2×2.4830 + 0.4×0.74214 + 0.2×0.19168 = 0.45580



600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.7577 2.5054 0.75164 0.19492

f(534.97) = ln

m

iii xK

1

f(534.97) = 0.2×3.7577 + 0.2×2.5054 + 0.4×0.75164 + 0.2×0.19492 = 0.46516


Slope = (0.46516 0.45580)/1 = 0.00936

The new temperature from Newton’s method is

3-18

T = Tguess f(533.97)/Slope = 533.97 0.45580/0.00936 = 533.97 48.734 = 485.23 K

The procedure is repeated until the magnitude difference between the new and the previous temperature is less than 0.1 K and the final bubble point temperature is 490.69 K.

The initial guessed temperature for the flash drum is based on the vapor fraction f:

T = fTdew + (1 f)Tbubble = 0.4×574.80 + 0.6×490.69 = 524.34 K




600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.4484 2.2707 0.65414 0.16227

f(T) = f(524.34) = 1 1)1),(( PTKfx

i

iF

f(524.34) = + + 0.20.4(3.4484 1) 1

0.20.4(2.2707 1) 1

0.40.4(0.65414 1) 1

+ 1 = 0.00134150.20.4(0.16227 1) 1

f(533.97) = 0.2×3.7283 + 0.2×2.4830 + 0.4×0.74214 + 0.2×0.19168 = 0.45580



600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.4771 2.2924 0.66298 0.16517

f(525.34) = = 1 1)1),(( PTKfx

i

iF

f(525.34) = + + 0.20.4(3.4771 1) 1

0.20.4(2.2924 1) 1

0.40.4(0.66298 1) 1

+ 1 = 0.00510550.20.4(0.16517 1) 1

3-19


Slope = = ( 0.0051055 + 0.0013415)/1 = 0.003764dT

TfdTTf )()(

The new flash temperature from Newton’s method is

T = Tguess f(524.34)/Slope = 524.34 ( 0.0013415)/( 0.003764)

T = 524.34 0.35639 = 523.98 K

The procedure is repeated until the magnitude difference between the new and the previous temperature is less than 0.1 K and the final flash temperature is 523.98 K.

The equilibrium ratios at 523.98 K are


600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.4382 2.2630 0.65101 0.16125

The mole fraction of each species in the liquid stream and vapor stream is given by


iF

Kfx

xEthylene = = 0.10125 yEthylene = 3.4382×0.10125 = 0.348120.20.4(3.4382 1) 1

xEthane = = 0.13287 yEthane = 2.2630×0.13287 = 0.300690.20.4(2.2630 1) 1

xPropane = = 0.46490 yPropane = 0.65101×0.46490 = 0.302650.40.4(0.65101 1) 1

xn-Butane = = 0.30098 yn-Butane = 0.16125×0.30098 = 0.0485330.20.4(0.16125 1) 1

Matlab codes ------------------------------------------------------------------------------------------

% Example 3.3-1, Isothermal Flash with f=V/F specified% Dew point temperature calculation%P=200; % psiaf=0.4;

3-20

A=[600076.9 687248.2 970688.6 1280577]; B=[7.90595 7.90694 7.15059 7.94986]; C=[0.84677 0.866 0.76984 0.96455];D=[0 0 6.90224 0];E=[42.94594 49.02654 0 0];% Estimate the saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)));zi=[.2 .2 .4 .2];yi=zi;Te=yi*Tsat'; % Estimate pressure from vapor pressure and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P+E/(P*P))';dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=sum(yi./Ki)-1; Tsav=T;T=T+dT; Ki=eval(fK);fdT=sum(yi./Ki)-1; eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;fprintf('Dew point temperature, T(R) = %8.2f\n',T)disp(' ')% Bubble point temperature calculationxi=zi;eT=1;T=Te;while abs(eT)>.1 Ki=eval(fK);fT=log(xi*Ki'); Tsav=T;T=T+dT; Ki=eval(fK);fdT=log(xi*Ki'); eT=fT*dT/(fdT-fT); T=Tsav-eT;endTb=T;fprintf('Bubble point temperature, T(R) = %8.2f\n',T)disp(' ')T=f*Td+(1-f)*Tb;eT=1;for i=1:20 Ki=eval(fK);fT=sum(zi./(f*(Ki-1)+1))-1; Tsav=T;T=T+dT; Ki=eval(fK);fdT=sum(zi./(f*(Ki-1)+1))-1; eT=fT*dT/(fdT-fT); T=Tsav-eT; if abs(eT)<.1, break, endendKi=eval(fK)x=zi./(f*(Ki-1)+1);y=x.*Ki;fprintf('Flash temperature, T(R) = %8.2f, V/F = %8.4f\n',T,f)disp('Species: Ethylene Ethane Propane n-Butane')fprintf('Feed z =');disp(zi)

3-21

fprintf('x = ');disp(x)fprintf('y = ');disp(y)

>> e1d3d1Dew point temperature, T(R) = 574.80 Bubble point temperature, T(R) = 490.69 Flash temperature, T(R) = 523.98, V/F = 0.4000Species: Ethylene Ethane Propane n-ButaneFeed z = 0.2000 0.2000 0.4000 0.2000x = 0.1013 0.1329 0.4649 0.3010y = 0.3481 0.3007 0.3027 0.0485Ki = 3.4382 2.2630 0.6510 0.1613-----------------------------------------------------------------------------------------------------------

Example 3.3-2 ----------------------------------------------------------------------------------The following mixture is fed to a flash drum at 200 psia.

Ethylene C2H4: 20% Ethane C2H6: 20% Propane C3H8: 40% n-Buane C4H10: 20%

Determine the fraction of the feed vaporized and the composition of the vapor and liquid streams leaving the flash drum if the drum is at 400oR, 600oR, and 525oR.

Data: Equilibrium K values for light hydrocarbon systems

=============================================================ln K = A/T2 + B C ln(P) + D/P + E/P2, where P is in psia, T is in oR

Compound A B C D E=============================================================Ethylene 600076.9 7.90595 .84677 0 42.94594Ethane 687248.2 7.90694 .866 0 49.02654Propane 970688.6 7.15059 .76984 6.90224 0n-Butane 1280557 7.94986 .96455 0 0=============================================================

Solution

We first need to check if the feed mixture is in the two phase region. At 400oR and 200 psia, the K values are given in the following table:


600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

0.71899 0.377010.051790 0.0057174

3-22

Since all the K values are less than 1, < 0, the feed is below its bubble point. iFi xK )1(The feed mixture is a subcooled liquid.

At 600oR and 200 psia, the K values are given in the following table:


600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

5.7759 4.0992 1.5066 0.48783

If > 0, the feed is above its dew point.

i

iFi

KxK )1(

= + + +

i

iFi

KxK )1( (5.7759 1) 0.2

5.7759 (4.0992 1) 0.2

4.0992 (1.5066 1) 0.2

1.5066 (0.48783 1) 0.2

0.48783

= 0.24111 >0, the feed is above its dew point. The feed is a superheated vapor.

i

iFi

KxK )1(

At 525oR and 200 psia, the K values are given in the following table:


600076.9687248.2970688.61280557

7.905957.906947.150597.94986

.84677

.866

.76984

.96455

00 6.90224 0

42.9459449.0265400

3.4674 2.2851 0.659990.16419

Check if the feed is below its bubble point:

= (3.4674 1)×0.2 + (2.2851 1)×0.2 + (0.65999 1)×0.4 + (0.16419 1)×0.2 iFi xK )1(

= 0.44734 > 0, the feed is above its bubble point. iFi xK )1(

Check if the feed is above its dew point:

= + + +

i

iFi

KxK )1( (3.4674 1) 0.2

3.4674 (2.2851 1) 0.2

2.2851 (0.65999 1) 0.4

0.65999 (0.16419 1) 0.2

0.16419

= 0.96935 < 0, the feed is below its dew point. The feed is a mixture of

i

iFi

KxK )1(

saturated liquid and vapor. We can now solve the following equation for the vapor fraction f:

F = = 0 (3.3-7)

1)1()1(

i

iFi

KfxK

3-23

F = + + + = 0(3.4674 1) 0.2(3.4674 1) 1f

(2.2851 1) 0.2(2.2851 1) 1f

(0.65999 1) 0.4(0.65999 1) 1f

(0.16419 1) 0.2(0.16419 1) 1f

We can solve the above nonlinear equation using Newton’s method with the slope of the function given by

= (3.3-8)dfdF

2

2

1)1()1(

i

iFi

KfxK

Starting with an initial guessed value for f = 0.5, we have

F = + + +(3.4674 1) 0.20.5(3.4674 1) 1

(2.2851 1) 0.20.5(2.2851 1) 1

(0.65999 1) 0.40.5(0.65999 1) 1

(0.16419 1) 0.20.5(0.16419 1) 1

F = 0.073630

= dfdF

2

2(3.4674 1) 0.20.5(3.4674 1) 1

2

2(2.2851 1) 0.20.5(2.2851 1) 1

2

2(0.65999 1) 0.40.5(0.65999 1) 1

= 0.84593

2

2(0.16419 1) 0.20.5(0.16419 1) 1

The new value for the vapor fraction is then

f = 0.5 F/ = 0.5 ( 0.073630)/( 0.84593) = 0.5 0.08704 = 0.41296dfdF

The procedure is repeated until the magnitude difference between the new and the previous vapor fraction is less than 0.0001 and the final vapor fraction is f = 0.4116.

The mole fraction of each species in the liquid stream and vapor stream is given by


iF

Kfx

xEthylene = = 0.099230 yEthylene = 3.4674×0.09923 = 0.344070.20.4(3.4674 1) 1

xEthane = = 0.13081 yEthane = 2.2851×0.13081 = 0.298920.20.4(2.2851 1) 1

xPropane = = 0.46508 yPropane = 0.65999×0.46508 = 0.306950.40.4(0.65999 1) 1

3-24

xn-Butane = = 0.30488 yn-Butane = 0.16419×0.30488 = 0.0500580.20.4(0.16419 1) 1

Matlab codes ------------------------------------------------------------------------------------------% Example 3.3-2, Isothermal Flash with T specified% Dew point temperature calculation%P=200; % psiaT=input('Drum temperature T(R) = ');A=[600076.9 687248.2 970688.6 1280577]; B=[7.90595 7.90694 7.15059 7.94986]; C=[0.84677 0.866 0.76984 0.96455];D=[0 0 6.90224 0];E=[42.94594 49.02654 0 0];zi=[.2 .2 .4 .2];fK='exp(-A/(T*T)+B-C*log(P)+D/P+E/(P*P))';Ki=eval(fK);Km=Ki-1;f0=Km*zi';f1=sum(Km.*zi./Ki);if f0<0 disp('Feed is subcooled liquid')elseif f1>0 disp('Feed is superheated vapor')else f = 0.5; for i=1:20 Km=Ki-1; F=sum((Km.*zi)./(f*Km+1)); dF=-sum((Km.^2).*zi./(f*Km+1).^2); eR=F/dF;f=f-eR; if abs(eR)<.0001, break, end end x=zi./(f*Km+1);y=x.*Ki; fprintf('Flash temperature, T(R) = %8.2f, V/F = %8.4f\n',T,f) disp('Species: Ethylene Ethane Propane n-Butane') fprintf('Feed z =');disp(zi) fprintf('x = ');disp(x) fprintf('y = ');disp(y) fprintf('Ki = ');disp(Ki)end

3-25

Chapter 3 3.4 Distribution of a solute between two liquid phases

3.4a Solubility of a Solid in a Liquid Phase The solubility of a solid in a liquid solvent and the distribution of a solute between two liquid phases will be considered in this section. When a solute is transported from one phase to another, the solute must cross the interface between phases as shown in Figue 3.4-1. We assume that the solute at the interface is in phase equilibrium. If the mole fraction yi is known at a given temperature, xi can be determined from the equilibrium relation and vice-versa.

Vapor Liquid

y

yi

xi

x y

yi xi

x



Vapor LiquidAi Ai

Figure 3.4-1 Solute transport across the interface.

3.4a Solubility of a solid in a liquid solvent Consider a binary system with solute (2) in equilibrium with solvent (1) as shown in Figure 3.4-2. We assume that the solvent is not soluble in the solid so that the solid solute will exist as a pure phase. At equilibrium, the fugacity of the solute in the solid phase, , is equal to the fugacity of the solute in solution, 2

sf. The fugacity is related to the fugacity of pure liquid solute, , at the equilibrium 2

solf 2solf 2

Lftemperature and pressure of the solution by

= 2x2 = (3.4-1)2solf 2

Lf 2sf

solute (2)

Figure 3.4-2 Solute (2) in solvent (1).

In equation (3.4-1) x2 is the solubility or the equilibrium mole fraction of solute in solution and 2 is the activity coefficient of the solute. Solving for the solubility gives

x2 = (3.4-2)2

1

2

2

s

Lff

3-26

The ratio / may be estimated by212sf 2

Lf

= exp (3.4-3)2

2

s

Lff

1 1m

m

HR T T

In this expression, Hm is the enthalpy of fusion at the normal melting temperature Tm. For an ideal solution the activity 2 is equal to 1. For non-ideal solutions, an appropriate activity coefficient model must be used to calculate the solubility. For non-polar solutes and solvents, the activity coefficient might be obtained from the Scatchard-Hildebrand equation:

2 = exp (3.4-4)2 2

2 1 2 1( )LVRT

In this equation is the molar volume of the solute as a subcooled liquid at the temperature 2LV

of the solution. However, is usually assumed to be the same as the molar volume of the 2LV

solute as a liquid at the melting point. The ’s are the solubility parameters for the solute and the solvent, and 1 is the volume fraction of the solvent defined by the following equation.

1 = (3.4-5)1 1

1 1 2 2

L

L Lx V

x V x V

The solubility parameter is given by

i = (3.4-6)1/ 2vap

iL

i

H RTV

The heat of evaporation can be obtained from the heat of sublimation and the heat of fusion.

= (3.4-7)vapiH sub

iH miH

If the vapor pressure of the solid is known as a function of temperature, the heat of sublimation can be estimated.

ln Psat = A (3.4-8)BT

If the solid is in equilibrium with the vapor we have

dGV = SVdT + VVdPsat = SSdT + VSdPsat = dGS (3.4-9)

Rearranging this equation gives

21 Fournier, R. L., “Basic Transport Phenomena in Biomedical Engineering”, Taylor & Francis, 2007, p. 57

3-27

= = (3.4-10)satdP

dT

V S

V SS SV V

SV

SVSV

Since SSV = HSV/T, and we can neglect the volar volume of the solid in comparision to the volume of the vapor, equation (3.4-10) becomes

= (3.4-11)satdP

dT 2

SV satH PRT

We assume ideal gas law in equation (3.4-11), which may be written as follows:

= (3.4-12)ln1

satd P

dT

SVHR

Taking derivative of equation (3.4-8), ln Psat = A , we obtainBT

= B = HSV = = RB (3.4-13)ln1

satd P

dT

SVHR

subiH

Example 3.4-1.22 ----------------------------------------------------------------------------------A drug has a molecular weight of 230 and a melting temperature of 155oC. Estimate the solubility of this drug in benzene and in n-hexane at 25oC assuming

a) Ideal solutionb) Nonideal solution using the Scatchard-Hildebrand equation

Data:Heat of fusion of the drug 4300 cal/molDensity of the drug 1.04 g/cm3 at 25oC

Vapor pressure of the solid drug ln Psat(mm Hg) = 27.3 8926(K)T

Molar volume of benzene 89.4 cm3/molSolubility parameter for benzene 9.2 (cal/cm3)1/2

Molar volume of n-hexane 131.6 cm3/molSolubility parameter for n-hexane 7.3 (cal/cm3)1/2

Solution ------------------------------------------------------------------------------------------

a) The ideal solubility of the drug is given by

x2 = = exp where 2 = 12

1

2

2

s

Lff

1 1m

m

HR T T


3-28

x2 = exp = 0.110-14300 cal/mol 1 1 K1.987 cal/(mol K) 273.15+155 298.15

The solubility is the same whether the solvent is benzene or n-hexane.

b) Nonideal solution using the Scatchard-Hildebrand equation

ln Psat(mm Hg) = 27.3 8926(K)T

= 8926= HSV = = 8926Rln1

satd P

dT

SVHR

subiH

= (8926 K)(1.987 cal/molK) = 17736.9 cal/mol2subH

The heat of evaporation of the drug is then estimated:

= = 17,736.9 4,300 = 13,436.9vapiH sub

iH miH

The solubility parameter of the drug is given by

i = 1/ 2vap

iL

i

H RTV

i = = 7.62 (cal/cm3)1/2

1/ 2

3

13,436.9 cal/mol (1.987 cal/mol K)(298 K)1 cm / 230 g/mol

1.04g

The drug solubility is calculated from

x2 = = exp2

1

2

2

s

Lff 2

1

1 1m

m

HR T T

Substituting 2 = exp and 1 = into the above equation we 2 2

2 1 2 1( )LVRT

1 1

1 1 2 2

L

L Lx V

x V x Vhave

3-29

x2 = 2

2 1 2 2 1

2 1 2 2

1 1exp

( ) (1 )exp(1 )

m

m

L L

L L

HR T T

V x VRT x V x V

The above equation is implicit in the solubility x2 and can be written as

f(x2) = x2 = 02

2 1 2 2 1

2 1 2 2

1 1exp

( ) (1 )exp(1 )

m

m

L L

L L

HR T T

V x VRT x V x V

The nonlinear equation f(x2) = 0 can be solved using Matlab function fzero with the ideal solution as the initial guess. The solubility of drug in benzene is evaluated using the fzero('solubility',0.11) with the function solubility representing f(x2) = 0.

>> x2=fzero('solubility',0.11)

x2 =

5.3999e-002

function y=solubility(x)dHm=4300; % cal/molR=1.987; % cal/(mol*K)TK=273.15; % KTm=155+TK;dHsub=R*8926; % cal/moldHvap=dHsub-dHm;VL2=230/1.04; % molar volume of drug (cm3/mol)T=25+TK; del2=((dHvap-R*T)/VL2)^0.5; % solubility parameter for the drug (cal/cm3)^0.5VL1=89.4; % molar volume of solvent, benzene (cm3/mol)del1=9.2; % solubility parameter for benzene (cal/cm3)^0.5tem1=VL2*(del1-del2)^2/(R*T);tem2=(1-x)*VL1/((1-x)*VL1+x*VL2);gamma2=exp(tem1*tem2^2);y=x-exp(dHm*(1/Tm-1/T)/R)/gamma2;

The solubility of drug in n-hexane is evaluated using the fzero('solhexane',0.11) with the function solhexane representing f(x2) = 0.

>> x2=fzero('solhexane',0.11)

x2 = 1.0748e-001

3-30

function y=solhexane(x)dHm=4300; % cal/molR=1.987; % cal/(mol*K)TK=273.15; % KTm=155+TK;dHsub=R*8926; % cal/moldHvap=dHsub-dHm;VL2=230/1.04; % molar volume of drug (cm3/mol)T=25+TK; del2=((dHvap-R*T)/VL2)^0.5; % solubility parameter for the drug (cal/cm3)^0.5VL1=131.6; % molar volume of solvent, n-hexane (cm3/mol)del1=7.3; % solubility parameter for n-hexane (cal/cm3)^0.5tem1=VL2*(del1-del2)^2/(R*T);tem2=(1-x)*VL1/((1-x)*VL1+x*VL2);gamma2=exp(tem1*tem2^2);y=x-exp(dHm*(1/Tm-1/T)/R)/gamma2;

The drug and n-hexane form an ideal solution since the solubility calculated using non-ideal model (x2 = 0.1075) is very close to the ideal model (x2,ideal = 0.11).

3.4b Distribution of a solute between liquid phases One or more of the components in a liquid mixture might be separated by contacting the mixture with another liquid in the process of liquid extraction. The separation is due to the unequal distribution of a solute between two partially miscible liquid phases. Through the process of liquid extraction, a product such as penicillin produced in fermentation mixtures can be extracted into a suitable solvent and purified from the fermentation broth. Choice of solvent extraction would depend on toxicity, cost, degree of miscibility with the fermentation broth, and selectivity for the solute.

We normally have three components, A, B, and C and two phases in equilibrium in a liquid-liquid system. From the phase rule, the degree of freedom F is given by

F = C + 2 P = 3 + 2 2 = 3 (3.4-14)

The variables are temperature, pressure, and four concentrations. Four concentrations occur since the mole fraction of the components in a phase must be equal to one:

xA + xB + xC = 1

If pressure and temperature are fixed, which is the usual case, then, at equilibrium, setting one concentration in either phase defines the system. Consider an equilibrium system from the mixing of N1 moles of solutes, N2 moles of solvent 2, and N3 moles of solvent 3. At equilibrium, the fugacity of component i in phase I is equal to its fugacity in phase II.

(T, P) = (T, P) (3.4-15)Ii I

ix Iif

IIi II

ix IIif

3-31

In this equation (T, P) = (T, P) = fi(T, P) = fugacity of pure component i at the same Iif

IIif

temperature and pressure of the system. Equation (3.4-15) becomes

= (3.4-16)Ii I

ix IIi II

ix

The distribution coefficient or the equilibrium constant Ki is defined as the ratio of the mole fraction of component i in the two phases.

Ki = = (3.4-17)IiIIi

xx

IIiIi

In liquid-liquid equilibrium system, the two partially miscible liquid phases usually form a non-ideal solution. The activity coefficients should be determined from multi-component activity models that can describe liquid-liquid equilibrium system.

Equilateral triangular coordinates can be used to represent the equilibrium data for a three-component system as shown in Figure 3.4-3. Each of the three corners represents a pure component, A, B, or C. The point M represents a mixture with xA = 0.4, xB = 0.2, and xC = 0.4.The perpendicular distance from the point M to the base AB represents the mole fraction xC of C in the mixture at M, the distance to the base CB represents the mole fraction xA of A, and the distance to the base AC represents the mole fraction xB of B.

C1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

B1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

A 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

M

xB

xA

xC

Figure 3.4-3 Equilateral triangular coordinates.

3-32

A common phase diagram where components A and B are partially miscible is shown in Figure 3.4-4. In this system, liquid C dissolves completely in A or B. Liquid A is only slightly soluble in B and B slightly soluble in A. The phase diagram is separated into two regions by a curved or phase envelope. The region outside the curved envelope is the one phase region and the region inside the curved envelope is the two-phase region. Any original mixture with composition in the two-phase region will separate out into two phases with the equilibrium compositions connected by the tie line. For example, a mixture with 5 moles of A, 3 moles of B, and 2 moles of C will not exist at equilibrium as a solution at point M. Instead this mixture will separate into liquid phase I and liquid phase II with compositions given by point a and b respectively. Liquid phase I has the compositions xA,I = 0.79, xB,I = 0.03, and xC,I = 0.18. Liquid phase II has the compositions xA,II = 0.08, xB,II = 0.68, and xC,II = 0.24. The moles of liquid in each phase can be determined from the materials balance.

xA,ILI + xA,IILII = 5 moles

C1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

B1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

A 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

P

M

ab

one-phase region

equilibrium tie line

two-phase region

Figure 3.4-4 Liquid-liquid phase diagram where componentsA and B are partially miscible.

Since LI + LII = 10 moles, we have

0.79LI + (0.08)(10 LI) = 5 LI = = 5.916 moles5 0.80.79 0.08

LII = 4.084 moles

3-33

The liquid-liquid equilibrium system will be simplified if the two solvents are immiscible. Usually solvents are selected to minimize their mutual solubility so immiscible system is a reasonable assumption. In many biological applications the solute concentration is so low that the activity coefficients approach their infinite dilution values and the distribution coefficient is a constant. Consider an equilibrium system from the mixing of N1 moles of solutes, N2 moles of solvent 2, and N3 moles of solvent 3. If the solute concentration is low then N1 << N2 and N1 << N3. For a system with two immiscible liquid phases at equilibrium, a mole balance of the solutes is given by

N1 = x1,ILI + x1,IILII (3.4-18)

In this equation LI is the moles of solvent 2 and some solute in liquid phase I and LII is the moles of solvent 3 and some solute in liquid phase II. Since the solute is present at such small quantity the value of LI and LII are assumed to be constant and equal to N2 and N3 respectively. Equation (3.4-18) becomes

N1 = x1,I N2 + x1,II N3 (3.4-18)

If the distribution coefficient is known, x1,I = K1 x1,II, then the mole fraction of the solute in phase II can be solved:

N1 = K1 x1,II N2 + x1,II N3 x1,II = (3.4-19)1

1 2 3

NK N N

Example 3.4-2.23 ----------------------------------------------------------------------------------We have 0.01 moles (N1) of drug dissolved in 100 moles of water (N3). We then add to this phase 100 moles (N2) of octanol. The octanol-water partition coefficient for the drug is 89.

Ki = = 89IiIIi

xx

For this system, octanol (phase I) and water (phase II) are immiscible. Estimate the mole fractions of the drug in the two phases once equilibrium has been attained and the % extraction of the drug from the aqueous phase.

Solution ------------------------------------------------------------------------------------------

Making a mole balance for the drug gives

N1 = x1,I N2 + x1,II N3

Substituting x1,I = K1 x1,II into the above equation we obtain

N1 = K1 x1,II N2 + x1,II N3

The mole fraction of the drug in the water phase is


3-34

x1,II = = = 1.1110-61

1 2 3

NK N N

0.0189 100 100

The mole fraction of the drug in the octanol phase is

x1,I = K1 x1,II = (89)( 1.1110-6) = 9.88910-5

The % extraction of the drug from the aqueous phase is given by

% extraction = 100 = 100 = 98.89%1, 2

1

Ix NN

59.889 10 1000.01

Example 3.4-3.24 ----------------------------------------------------------------------------------Purification of an Antibiotic

Benzylpenicillin is an older antibiotic effective against pneumococcal and meningoccal infections, anthrax, and Lyme disease. As part of a purification process, 200 mg of benzylpenicilin is mixed with 25 ml of n-octanol and 25 mL of water. After equilibrium is established, there is a water-rich phase that contains essentially no n-octanol and an octanol-rich phase that contains 74 mol % n-octanol and 26 mol % water. Determine the concentrations of benzylpenicillin in each of these phases.

Data: The molecular weight of benzylpenicillin is 334.5, that of n-octanol is 130.23, the liquid density of n-octanol is 0.826 g/cm3, and

KOW,B = = = 65.5OBWB

CC

mg B/(ml n-octanol)mg B/(ml water)

Solution ------------------------------------------------------------------------------------------

Since n-octanol is insoluble in water, the number of moles of n-octanol in the octanol-rich phase is

= 0.1586 mol25 mL 0.826 g/mL130.23 g/mol

The amount of water in the octanol-rich phase is

0.1586 mol octanol = 0.0557 mol water0.26 mol water0.74 mol octanol

Assuming no change in volume upon mixing, the volume of the octanol-rich phase is

VO = + = 26.0028 mL0.1586 mol 130.23 g/mol0.826 g/mL

0.0557 mol 18 g/mol1 g/mL

24 Sandler, S. I., “Chemical, Biochemical, and Engineering Thermodynamics”, Wiley, 2006, p. 643

3-35

The volume of the water-rich phase is

VW = 50 mL 26.0028 mL = 23.997 mL

The total number of moles of benzylpenicillin is

= 5.98110-4 mol0.2 g334.5 g/mol

Making a mole balance on of benzylpenicillin gives

5.98110-4 mol = VW + VO = VW + KOW,B VOWBC O

BC WBC W

BC

5.98110-4 mol = (23.997 + 65.526.003) = 1727.52 mLWBC W

BC

= 3.46210-7 mol/mL = 1.15810-4 g/mL = 0.1158 mg/mLWBC

= KOW,B = 65.53.46210-7 = 2.26810-5 mol/mLOBC W

BC

= 7.58510-3 g/mL = 7.585 mg/mLOBC

------------------------------------------------------------------------------------------

3.4c Single-Stage Equilibrium Extraction

Aqueous Stream L , xII

II,in

L , xII,out

L , xIIII,out

L I Extractor

Figure 3.4-5 Single stage equilibrium liquid-liquid extraction.

Figure 3.4-5 shows a single stage liquid extractor where a pure flowing solvent stream at molar flow rate LI is contacted with an aqueous stream flowing at LII with a solute of mole fraction xII,in. We assume that the solvent and water are immiscible and there is no change in flowrates of LI or LII. The streams leaving the liquid extractor are at equilibrium so that xI,out = KxII,out. Making a mole balance on the solute gives

LII xII,in = LI xI,out + LII xII,out = LIKxII,out + LII xII,out

3-36

LII xII,in = (LIK + LII) xII,out (3.4-20)

Solving for xII,out / xI,out we obtain

= = = (3.4-21),

,

II out

II in

xx

II

I IIL

KL L1

1I

IIKLL

11 E

In equation (3.4-21), E is defined as the extraction factor, . The amount of solute I

IIKLL

entering the extractor is LII xII,in. The amount of solute extracted from the aqueous phase II is LII xII,in LII xII,out. The % extraction of the solute from phase II is then given by

% extraction = 100 = 100, ,

,

II IIII in II out

IIII in

L x L xL x

,

,

1 II out

II in

xx

Example 3.4-4.25 ----------------------------------------------------------------------------------A drug is in an aqueous (phase II) stream flowing at 100 moles/min at a drug mole fraction of 0.01. The aqueous stream is then contacted with an extractor with a pure solvent (phase I) flowing at 200 moles/min. The distribution coefficient for this particular drug is given by

Ki = = 6IiIIi

xx

Determine the equilibrium mole fraction of the drug in the streams exiting the extractor and the % extraction of the drug from the aqueous stream.

Solution ------------------------------------------------------------------------------------------Aqueous Stream L = 100 moles/min, x = 0.01II

II,in

L , xII,out

L , xIIII,out

L = 200 moles/min I

Extractor

Making a mole balance on the solute gives

LII xII,in = LI xI,out + LII xII,out = LIKxII,out + LII xII,out

LII xII,in = (LIK + LII) xII,out


3-37

Solving for xII,out / xI,out we obtain

= = = = = 0.0769,

,

II out

II in

xx

II

I IIL

KL L1

1I

IIKLL

11 E

1(6)(200)1

100

xII,out = (0.01)( 0.0769) = 7.6910-4, xI,out = (6)( 7.6910-4) = 4.6210-3

% extraction = 100 = 100 = (1 0.0769) 100 = 92.3 %, ,

,

II IIII in II out

IIII in

L x L xL x

,

,

1 II out

II in

xx

3.5 Non-isothermal Flash

In flash distillation, a liquid mixture is partially vaporized and the vapor is allowed to come to equilibrium with the liquid. The process flow diagram is shown in Figure 3.5-1. The vapor and liquid phases are then separated. For non-isothermal flash calculation, energy balance is also required in addition to mass balance and equilibrium relations. It is easy to do non-isothermal flash calculations on an Hxy diagram for binary system.

F, xiF

V, yi

L, xi

Q


Example 3.5-1.22 ----------------------------------------------------------------------------------10,000 lb/h of saturated liquid at 1 atmosphere is fed adiabatically into a flash drum operating at 0.1 atmosphere. The feed stream is a binary mixture of 29 weight percent ethanol and 71 weight percent water. Determine the temperature in the flash drum, the compositions of the vapor and liquid phases, and the vapor and liquid flow rates leaving the flash drum.

22 Luyben, W. L. and Wenzel, L. A., “Chemical Process Analysis”, Prentice-Hall, 1988, p. 406

3-38

F

V

L

Solution ------------------------------------------------------------------------------------------The feed point F is located on the saturated liquid curve at 1 atm and 29 wt% ethanol. The vapor V and liquid L streams are located on the saturated vapor and saturated liquid curves at 0.1 atm respectively. The points V and L are connected by a tie line through the F point. We now prove that the three points V, L, and F are indeed connected by a straight line. Let V, L, and F are the molar flow rates of the vapor, liquid, and feed streams respectively. We have

Total mass balance: F = L + V

3-38

Chapter 3

Component balance: FxF = Vy + Lx

Energy balance: FhF = VH+ Lh

To prove the three points V, L, and F are on the same line we can show that the slope of line LF is the same as the slop of line FV. Combining the total and component balances gives

(L + V)xF = Vy + Lx

(xF y)V = (x xF)L = LV

F

F

x yx x

Combining the total mass and energy balances gives

(L + V)hF = VH+ Lh

(hF H)V = (h hF)L = LV

F

F

h Hh h

Therefore = = = LV

F

F

x yx x

F

F

h Hh h

F

F

H hy x

F

F

h hx x

The three points V, L, and F have coordinates (y, H), (x, h), and (xF, hF) respectively. The

slope of line FV is and the slope of line LF is . Therefore the three points V, F

F

H hy x

F

F

h hx x

L, and F are on the same straight line.

The temperature of the feed (before the flash) can be read off the VLE tie-line to be 185oF. Note that the solid tie-lines are for the system at 1 atm and the dash tie-lines are for the system at 0.1 atm. The VLE tie-line through F is drawn to the saturated liquid and saturated vapor lines at 0.1 atmosphere. The temperature of the flash drum is the temperature on the tie line at 0.1 atmosphere which is 97oF. The compositions are x = 22 weight percent and y = 68 weight percent ethanol.

The flow rates of liquid and vapor are calculated from total and component balances as follows.

F = L + V = 10,000 lb/hr


(0.29)(10,000) = (0.68)V + (0.22)(10,000 V)

3-39

V = 1,522 lb/hr and L = 8478 lb/hr

About 15 percent of the liquid feed becomes vapor. Therefore the temperature drops from 185oF to 97oF during the flash because of the conversion of sensible to latent heat.

Example 3.5-2.22 ----------------------------------------------------------------------------------10,000 lb/h of saturated liquid at 1 atmosphere is fed into a flash drum operating at 0.1 atmosphere. 3.05×106 Btu/hr of heat is transferred to the flash drum. The feed stream is a binary mixture of 29 weight percent ethanol and 71 weight percent water. Determine the temperature in the flash drum, the compositions of the vapor and liquid phases, and the vapor and liquid flow rates leaving the flash drum.Solution ------------------------------------------------------------------------------------------

If heat is transferred into the flash drum at a rate of Q, the energy equation becomes

Q + FhF = VH + Lh (E-1)

This equation can be written as:

F(Q/F + hF) = VH+ Lh

In the previous section, we have shown that for a system where the mass and energy balances applies:


Energy balance: FhF = VH+ Lh

The three points V, L, and F with coordinates (y, H), (x, h), and (xF, hF) respectively are on the same straight line. When the energy balance becomes F(Q/F + hF) = VH+ Lh, the three points V, L, and F’ with coordinates (y, H), (x, h), and (xF, Q/F + hF) will be on the same straight line. If Eq. (E-1) is written as

FhF = V(H Q/V) + Lh

Then the three points V’, L, and F with coordinates (y, H Q/V), (x, h), and (xF, hF) will be on the same straight line.

The point F is on the saturated liquid line at 1 atm and 29 wt% ethanol with enthalpy hF = 140 Btu/lb. The pseudo enthalpy point hF’ must be calculated:

hF’ = Q/F + hF = 3.05×106/10,000 + 140 = 445 Btu/lb

We thus locate the point F’(xF, Q/F + hF) at 29 wt% ethanol with an enthalpy of 445 Btu/lb. The VLE tie-line for 0.1 atmosphere pressure that goes through this point has a temperature of 105oF. This is the temperature of the flash tank.

22 Luyben, W. L. and Wenzel, L. A., “Chemical Process Analysis”, Prentice-Hall, 1988, p. 408

3-40

F'

V

L

F

The x and y values at the two ends of this tie-line are x = 7 wt. % ethanol and y = 48 wt. % ethanol. Overall and ethanol component balances are solved to get the flow rates:

(0.29)(10,000) = (0.07)(10,000 V) + (0.48)(V)

Therefore V = 5,366 lb/hr and L = 4,634 lb/hr. The addition of heat to the flash tank has results in a higher temperature (105oF) than when the process was run adiabatically (97oF). Also more vapor is formed (5,366 lb/hr versus 1,522 lb/hr).

3-41

Tie line interpolation

Consider the situation where the two tie lines AB and CD are given and a tie line EF with x = 0.15 is interpolated using these two tie lines. In this case, the line through point F (x = 0.15) is drawn so that it intersects the saturated vapor curve a point E where the ratio

= = = FBDB

EACA

0.15 0.100.30 0.10

14

In other words, the five points A (y = 0.217), B (x = 0.1), C (y = 0.517), D (x = 0.3), and F (x = 0.15) are given, the point E is located on the saturated vapor curve with coordinate yE obtained from linear interpolation:

= E A

F B

y yx x

C A

D B

y yx x

yE = yA + (yC yA) = 0.217 + (0.517 0.217) F B

D B

x xx x

0.15 0.100.30 0.10

yE = 0.217 + (0.517 0.217) = 0.29214

4-1

Chapter 4 Distillation

4.1 Vapor Liquid Equilibrium Relations

The equilibrium in vapor-liquid equilibrium is restricted by the phase rule. For a binary mixture in two phases, the degree of freedom F is given by

F = m + 2 = 2 2 + 2 = 2 (4.1-1)

As an example we will use the benzene-toluene, vapor-liquid system. The four variables are temperature, pressure, the mole fraction of the more volatile component in the vapor and liquid phases. Benzene is the more volatile component in the benzene-toluene system. If the pressure is fixed at 200 kPa, only one more variable can be set. If we set the temperature, the mole fraction of benzene (component 1) can be determined from the following relations:

= and = (4.1-2a, b)1

1

yx

1satPP

1

1

11

yx

2satPP

In these expressions, P = 200 kPa, P1sat and P2

sat are the vapor pressures of benzene and toluene at the specified temperature, respectively. For ideal solution such as benzene-toluene system, the bubble and dew point temperatures of the mixture will be between the saturation temperatures of benzene and toluene. At 200 kPa, the saturation temperatures of benzene and toluene are 377.31 K and 409.33 K, respectively. If we set the mixture temperature at 400 K, P1

sat = 352.160 kPa and P2sat = 157.8406 kPa. Equations (4.1-2a, b) become

= and = (4.1-3a, b)1

1

yx

352.16200

1

1

11

yx

157.841200

Solving these two equations, we obtain x1 = 0.2170 and y1 = 0.3820. The vapor-liquid equilibrium relations for benzene (1)-toluene (2) at a total pressure of 200 kPa are given as a boiling-point Txy diagram shown in Figure 4.1-1. The upper curve is the saturated vapor curve (the dew-point curve) and the lower curve is the saturated liquid curve (the bubble-point curve). The region between these two curves is the two-phase region.

In Figure 4.1-1, if we start with a cold liquid mixture of x1 = 0.217 and heat the mixture, it will start to boil at 400 K, and the composition of the first vapor in equilibrium is y1 = 0.382. As we continue boiling, the composition x1 in the remaining liquid mixture will decrease since y1 is richer in benzene. If we start with a hot gas mixture of y1 = 0.382 and cool the mixture, it will start to condense at 400 K, and the composition of the first liquid drop in equilibrium is x1 = 0.217. As we continue condensing, the composition y1 in the remaining gas mixture will increase since x1 is richer in toluene. The Txy diagram for an ideal mixture can be calculated from the pure vapor-pressure data as illustrated in Example 4.1-1.

4-2

Figure 4.1-1 Calculated Txy diagram of benzene and toluene at 200 kPa.

Example 4.1-1 ----------------------------------------------------------------------------------Construct a Txy diagram for a mixture of benzene and toluene at 200 kPa. Benzene and toluene mixtures may be considered as ideal.

Data: Vapor pressure, Psat, data: ln Psat = A B/(T + C), where Psat is in kPa and T is in K.


Solution ------------------------------------------------------------------------------------------

The temperature for the Txy diagram should be between the boiling points of benzene and toluene given by

T1boil = + 44.5633 = 377.31 K

)200log(1603.1478.2948

T2boil = + 47.1806 = 409.33 K

)200log(2515.1438.3242

4-3

The simplest procedure is to choose a temperature T between 377.31 K and 409.33 K, evaluate the vapor pressures, and solve for x and y from the following equations:

= (E-1)i

i

xy

PP sat

i

At 400 K, the vapor pressure of benzene and toluene are given by

P1sat = exp(14.1603 2948.78/(400 44.5633)) = 352.160 kPa

P2sat = exp(14.2515 3242.38/(400 47.1806)) = 157.8406 kPa

Therefore

= y1 = K1 x1 = 1.7608x1 (E-1)1

1

xy 352.16

200

= 1 y1 = K2 (1 x1) = 0.7892 (1 x1) (E-1)1

1

11

xy

157.841

200

Substituting Eq. (E-2) into (E-1) yields

1 K1 x1 = K2 (1 x1)Therefore

x1 = = = 0.217021

21KK

K

7892.07608.1

7892.01

y1 = K1 x1 = 1.7608x1 = 0.3820

The Matlab program listed in Table E4.1-1 plots the Txy diagram shown in Figure 4.1-1.

Table E4.1-1--------------------------------------------------------------------------% Example 4.1-1, Txy diagram for benzene-toluene mixture at 200 kPa%P=200; % kPaA=[14.1603 14.2515]; B=[2948.78 3242.38]; C=[-44.5633 -47.1806];% Boling point at 200 kPaTb=B./(A-log(P))-C;fprintf('Boiling point of Benzene at P = %g, Tb = %6.2f C\n',P,Tb(1))fprintf('Boiling point of Toluene at P = %g, Tb = %6.2f C\n',P,Tb(2))T=linspace(Tb(1),Tb(2),50);K1=exp(A(1)-B(1)./(T+C(1)))/P;K2=exp(A(2)-B(2)./(T+C(2)))/P;x=(1-K2)./(K1-K2); y = K1.*x;ymin=round(Tb(1)-1);ymax=round(Tb(2)+1);

4-4

% Solve for x and y at T = 400 KT1=400;K1=exp(A(1)-B(1)./(T1+C(1)))/P;K2=exp(A(2)-B(2)./(T1+C(2)))/P;x1=(1-K2)/(K1-K2); y1 = K1*x1;plot(x,T,y,T,':',[0 y1],[T1 T1],'--',[x1 x1],[ymin T1],'--',[y1 y1],[ymin T1],'--')axis([0 1 ymin ymax]);grid onxlabel('x,y');ylabel('T(K)')legend('Saturated liquid','Saturated vapor')-----------------------------------------------------------------------------------------

When deviations from Raoult’s law are large enough, the Tx and Ty curves can go through a maximum or a minimum. The extreme point (either minimum or maximum) is called azeotrope where the liquid mole fraction is equal to the vapor mole fraction for each species:

xi = yi (4.1-1)

A system that exhibits a maximum in pressure (positive deviations from Raoult’s law) will exhibits a minimum in temperature called minimum boiling azeotrope as shown the top part of Figure 4.1-1 for a mixture of chloroform and hexane. The Pxy diagram is plotted at 318 K and the Txy diagram is plotted at 1 atm. This is the case when the like interaction is stronger than the unlike interaction between the molecules. The mixture will require less energy to go to the vapor phase and hence will boil at a lower temperature that that of the pure components.

If the unlike interaction is stronger than the like interaction we have negative deviations from Raoult’s law and the system will exhibit a minimum in pressure or a maximum in temperature called maximum boiling azeotrope. A mixture of acetone and chloroform shows this behavior in the bottom part of Figure 4.1-1. The Pxy diagram is plotted at 328 K and the Txy diagram is plotted at 1 atm. The data for vapor pressure and Wilson model are from the Thermosolver program by Koretsky. This program can also plot Pxy and Txy diagrams for different mixtures. Table 4.1-1 lists the Matlab program used to produce Figure 4.1-2.

4-5

0 0.2 0.4 0.6 0.8 1

334

336

338

340

342

x,y

T(K

)

Chloroform-Hexane

xy

0 0.2 0.4 0.6 0.8 10.4

0.45

0.5

0.55

x,y

P(a

tm)

Chloroform-Hexane

xy

0 0.2 0.4 0.6 0.8 1

330

332

334

336

338

340

x,y

T(K

)

Acetone-Chloroform

xy

0 0.2 0.4 0.6 0.8 10.65

0.7

0.75

0.8

0.85

0.9

0.95

1

x,y

P(a

tm)

Acetone-Chloroform

xy

Figure 4.1-2 Top: minimum boiling azeotrope for chloroform and n-Hexane system.Bottom: maximum boiling azeotrope for acetone and chloroform system

Table 4.1-2 -----------------------------------------------------------------------------------% Figure 4.1-2: Construct a Txy diagram for chloroform (1) and hexane (2) mixture % at a total pressure (atm) ofP = 1;% Use the Wilson equation with parametersG12 = 1.2042; G21 = 0.39799;% Vapor pressure data: P(atm), T(K)p1sat = 'exp(9.33984-2696.79/(T-46.14))';p2sat = 'exp(9.20324-2697.55/(T-48.78))';% Estimate boiling pointsTb1 = 2696.79/(9.33984-log(P))+46.14;Tb2 = 2697.55/(9.20324-log(P))+48.78;x1=[0 .002 .004 .006 .008 .01 .015];x2=linspace(.02,.92,46);x3=[.93 .94 .95 .96 .97 .98 .985 .990 .995 1];xp=[x1 x2 x3];np=length(xp);yp=xp;Tp=xp;Tp(1)=Tb2;Tp(np)=Tb1;dT = .01;T=Tb2;

4-6

for i=2:np x1=xp(i);x2=1-x1;% Evaluate activity coefficients tem1 = x1 + x2*G12; tem2 = x2 + x1*G21; gam1 = exp(-log(tem1)+x2*(G12/tem1-G21/tem2)); gam2 = exp(-log(tem2)+x1*(G21/tem2-G12/tem1)); for k=1:20 fT=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; T=T+dT; fT2=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; dfT=(fT2-fT)/dT; eT=fT/dfT; T=T-dT-eT; if abs(eT)<.001, break ,endendTp(i)=T;yp(i)=x1*gam1*eval(p1sat)/P;fprintf('T(K) = %8.2f , x=%8.4f;y = %8.4f, iteration = %g\n',T,x1,yp(i),k)endsubplot(2,2,2); plot(xp,Tp,yp,Tp,':')axis([0 1 333 343])xlabel('x,y');ylabel('T(K)');title('Chloroform-Hexane')grid onlegend('x','y')%% Construct a Pxy diagramfor chloroform (1) and hexane (2) mixture % at a temperature (K) ofT=318;p1vap=eval(p1sat);p2vap=eval(p2sat);xp2=1-xp;tem1 = xp + xp2*G12; tem2 = xp2 + xp*G21; gam1 = exp(-log(tem1)+xp2.*(G12./tem1-G21./tem2)); gam2 = exp(-log(tem2)+xp.*(G21./tem2-G12./tem1)); p1=xp.*gam1*p1vap;p2=xp2.*gam2*p2vap; Pp=p1+p2;yp=p1./Pp; subplot(2,2,1); plot(xp,Pp,yp,Pp,':')axis([0 1 0.4 0.6])xlabel('x,y');ylabel('P(atm)');title('Chloroform-Hexane')grid onlegend('x','y',2)%% Figure 4.1-2: Construct a Txy diagramfor acetone (1) and chloroform (2) mixture % at a total pressure (atm) ofP = 1;% Use the Wilson equation with parametersG12 = 1.324; G21 = 1.7314;% Vapor pressure data: P(atm), T(K)p1sat = 'exp(10.0179-2940.46/(T-35.93))';p2sat = 'exp(9.33984-2696.79/(T-46.14))';% Estimate boiling points

4-7

Tb1 = 2940.46/(10.0179-log(P))+35.93;Tb2 = 2696.79/(9.33984-log(P))+46.14;x1=[0 .002 .004 .006 .008 .01 .015];x2=linspace(.02,.92,46);x3=[.93 .94 .95 .96 .97 .98 .985 .990 .995 1];xp=[x1 x2 x3];np=length(xp);yp=xp;Tp=xp;Tp(1)=Tb2;Tp(np)=Tb1;dT = .01;T=Tb2;for i=2:np x1=xp(i);x2=1-x1;% Evaluate activity coefficients tem1 = x1 + x2*G12; tem2 = x2 + x1*G21; gam1 = exp(-log(tem1)+x2*(G12/tem1-G21/tem2)); gam2 = exp(-log(tem2)+x1*(G21/tem2-G12/tem1)); for k=1:20 fT=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; T=T+dT; fT2=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; dfT=(fT2-fT)/dT; eT=fT/dfT; T=T-dT-eT; if abs(eT)<.001, break ,endendTp(i)=T;yp(i)=x1*gam1*eval(p1sat)/P;fprintf('T(K) = %8.2f , x=%8.4f;y = %8.4f, iteration = %g\n',T,x1,yp(i),k)endsubplot(2,2,4); plot(xp,Tp,yp,Tp,':')axis([0 1 329 340])xlabel('x,y');ylabel('T(K)');title('Acetone-Chloroform')grid onlegend('x','y')%% Construct a Pxy diagramfor acetone (1) and chloroform (2) mixture % at a temperature (K) ofT=328;p1vap=eval(p1sat);p2vap=eval(p2sat);xp2=1-xp;tem1 = xp + xp2*G12; tem2 = xp2 + xp*G21; gam1 = exp(-log(tem1)+xp2.*(G12./tem1-G21./tem2)); gam2 = exp(-log(tem2)+xp.*(G21./tem2-G12./tem1)); p1=xp.*gam1*p1vap;p2=xp2.*gam2*p2vap; Pp=p1+p2;yp=p1./Pp; subplot(2,2,3); plot(xp,Pp,yp,Pp,':')axis([0 1 0.65 1.0])xlabel('x,y');ylabel('P(atm)');title('Acetone-Chloroform')grid onlegend('x','y',2)

4-8

4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System

Consider an equilibrium stage where a vapor stream V2 is in contact with a liquid stream L0. The two streams V1 and L1 leaving the equilibrium stage is in equilibrium. The compositions in streams V1 and L1 must be solved from the material balance, the energy balance, and the equilibrium relations. For a binary mixture of A and B, if sensible heat effects are small and the latent heats of both compounds are the same, then when 1 mole of A condenses, 1 mole of B must vaporize. Hence, the total molar flow V1 will equal the total molar flow V2. Similarly, the total molar flow L1 will equal the total molar flow L0. This situation is called constant molal overflow (CMO). When CMO is valid, the compositions in streams V1 and L1 can be solved from only the material balance and the equilibrium relations. The energy balance is not required since it is satisfied when the material balance is satisfied.

Equilibrium stage

V2

V1L0

L1

Figure 4.1-1 Schematic of an equilibrium stage.

Example 4.2-1 ----------------------------------------------------------------------------------A vapor at the dew point and 200 kPa containing a mole fraction of 0.40 benzene (1) and 0.60 toluene (2) and 100 kmol total is brought into contact with 110 kmol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molar overflow, calculate the amounts and compositions of the exit streams.



Solution ------------------------------------------------------------------------------------------

For CMO, L1 = L0 = 110 kmol, V2 = V1 = 100 kmol. Making a balance on benzene gives

L0x0 + V2y2 = L1x1 + V1y1

110(0.30) + 100(0.40) = 110x1 + 100y1

11x1 + 10y1 = 7.3 y1 = 0.73 1.1x1 (E-1)

Since the two streams V1 and L1 are in equilibrium, we have

4-9

= 200y1 = x1exp(14.1603 2948.78/(T 44.5633)) (E-2)1

1

xy 1

200

satP

= 200(1 y1) = (1 x1)exp(14.2515 3242.38/(T 47.1806)) (E-3)1

1

11

xy

2

200

satP

The three equations (E-1,2,3) can be solved for T, x1, and y1 either by graphical or numerical method.

A) Graphical methodWe can choose a range of temperatures between the boiling point of pure benzene (377.31 K) and pure toluene (409.33 K) and solve equations (E-1,2) for x1, and y1 then plot y1 versus x1 to obtain the equilibrium curve. We then plot the material balance equation (E-1) to obtain the operating line. The intersection of the equilibrium curve and the operating line provides the composition x1 and y1 of the streams exiting the equilibrium stage. The following is a procedure to obtain a point on the equilibrium curve:

At 400 K, the vapor pressure of benzene and toluene are given by

P1sat = exp(14.1603 2948.78/(400 44.5633)) = 352.160 kPa

P2sat = exp(14.2515 3242.38/(400 47.1806)) = 157.8406 kPa

Therefore

= y1 = K1 x1 = 1.7608x1 (E-2)1

1

xy 352.16

200

= 1 y1 = K2 (1 x1) = 0.7892 (1 x1) (E-3)1

1

11

xy

157.841

200

Substituting Eq. (E-3) into (E-2) yields

1 K1 x1 = K2 (1 x1)Therefore

x1 = = = 0.217021

21KK

K

7892.07608.1

7892.01

y1 = K1 x1 = 1.7608x1 = 0.3820

Figures E4.2-1a, and b show the intersection of the equilibrium curve and the operating line. The intersection point can be zoom in as shown in Figure E4.2-1b with x1 = 0.2615 and y1 = 0.4425. The Matlab program is listed in Table E4.2-1.

4-10

Figure E4.2-1a Equilibrium diagram for benzene-toluene mixture at 200 kPa

Figure E4.2-1b Equilibrium diagram for benzene-toluene mixture at 200 kPa

Table E4.2-1 Matlab program for Figure E4.2-1a,b-----------------------------------------% Example 4.2-1, Composition of benzene-toluene mixture at 200 kPa leaving an equilibrium stage%P=200; % kPaA=[14.1603 14.2515]; B=[2948.78 3242.38]; C=[-44.5633 -47.1806];% Boling point at 200 kPaTb=B./(A-log(P))-C;

4-11

fprintf('Boiling point of Benzene at P = %g, Tb = %6.2f C\n',P,Tb(1))fprintf('Boiling point of Toluene at P = %g, Tb = %6.2f C\n',P,Tb(2))% Solve for equilibrium curveT=linspace(Tb(1),Tb(2),50);K1=exp(A(1)-B(1)./(T+C(1)))/P;K2=exp(A(2)-B(2)./(T+C(2)))/P;x=(1-K2)./(K1-K2); y = K1.*x;% Solve for operating linexp=[0.2 0.6];yp = 0.73 - 1.1*xp;plot(x,y,':',xp,yp)axis([0 1 0 1]);grid onxlabel('x, mole fraction of benzene in liquid phase');ylabel('y')legend('Equilibrium curve','Operating line',2)------------------------------------------------------------------

B) Numerical method. We need to solve the following equations:

y1 = 0.73 1.1x1 (E-1)

200y1 = x1exp(14.1603 2948.78/(T 44.5633)) (E-2)

200(1 y1) = (1 x1)exp(14.2515 3242.38/(T 47.1806)) (E-3)

Substituting (E-1) into (E-2) and (E-3) we obtain

146 220x1 x1exp(14.1603 2948.78/(T 44.5633)) = 0 (E-2a)

54 + 220x1 (1 x1)exp(14.2515 3242.38/(T 47.1806)) = 0 (E-3a)

Equations (E-2a, and E-3a) can be solved by the following Matlab statement

>> p=fminsearch('estage',[.5 400])

p = 0.2614 398.3056

The function estage is written as follow

function y=estage(p)x=p(1);T=p(2);f1=146-220*x-x*exp(14.1603-2948.78/(T-44.5633));f2=54+220*x-(1-x)*exp(14.2515-3242.38/(T-47.1806));y=f1*f1+f2*f2;

Therefore x1 = 0.2614 and y1 = 0.73 1.1x1 = 0.73 1.1(0.2614) = 0.4425

4-12

4.3 Simple Batch or Differential Distillation

In a simple batch (or, more precisely, semi-batch) distillation unit, liquid is first charged to a heated kettle. The liquid charge is boiled slowly and the vapors are withdrawn as rapidly as they form to a condenser, where the condensed vapor (distillate) is collected. The first portion of vapor condensed will be richest in the more volatile component A. As vaporization proceeds, the vaporized product becomes leaner in A. Initially, a charge of Li moles of a binary mixture with a mole fraction xi of the more volatile component A is placed in the still. At any given time, L is the moles of liquid left in the still with composition x. V is the instant vapor flow rate leaving the still with the mole fraction y of the more volatile component A. The vapor leaving the still may be considered to be in equilibrium with the remaining liquid.

Example 4.3-1. ---------------------------------------------------------------------------------- A semi-batch distillation unit is charged with 100 mol of a 60 mol% benzene-40 mol% toluene mixture. At any given instant, the benzene mole fraction in the vapor flow rate, y, and the benzene mole fraction in the remaining liquid, x, are related by the equilibrium relation

y = x

x6.11

6.2

Derive an equation relating the amount of liquid remaining in the still to the mole fraction of benzene in this liquid.

Solution ------------------------------------------------------------------------------------------Step #1: Define the system.

System: Moles L of liquid left in the still with composition x.

Let = moles vapor rate to condenserVy = mole fraction of the more volatile component in

the vaporL = moles of liquid in the stillx = mole fraction of the more volatile component in

the liquid

The composition in the still pot changes with time with x as the mole fraction of benzene in the liquid phase.

L

4-13

Step #2: Find equation that relates L to x.

At any given instant, the mole of benzene in the still is Lx. A benzene balance will give an equation relating the amount of liquid remaining in the still to the mole fraction of benzene in this liquid.

Step #3: Apply the benzene balance.

= y (E-1)dtLxd )( V

Total balance is required since there are more than one unknown in Eq. (E-1)

= dtdL V

The product rule of derivative can be applied to the LHS of Eq. (E-1)

L + x = y (E-2)dtdx

dtdL V

and y can be eliminated from Eq.(E-2) by using the total balance and the V

equilibrium relation.

L + x = dtdx

dtdL

xx6.11

6.2 dt

dL

L = x = dtdx

xx6.11

6.2 dt

dLdtdL

x

xx6.11

6.2dtdL

= = = dxL

dL

xx

xdx

6.11

6.2x

xxxdx

6.116.16.2 2

)1(6.1

6.11xx

x

The expression can be expanded by partial fraction)1(6.1

6.11xx

x

= + 1 + 1.6x = a(1 x) + b(1.6x))1(6.1

6.11xx

x

x

a6.1 x

b1

x = 0 => 1 = a

x = 1 => 2.6 = 1.6b => b = 2.6/1.6

4-14

= + )1(6.1

6.11xx

x

x6.1

1)1(6.1

6.2x

= dx + dxL

dLx6.1

1)1(6.1

6.2x

Step #4: Specify the boundary conditions for the differential equation.

At L = 100, x = 0.6

Step #5: Solve the resulting equation and verify the solution.

= + L

LdL

100 x

xdx

6.0 6.1

x

xdx

6.0 )1(6.16.2

ln = 100

L x

xx6.0

)1ln(6.16.2ln

6.11

ln = 0.625 ln 1.625 ln = ln100

L6.x

4.1 x

625.1625.0

4.1

6.xx

L = 100

625.1625.0

4.1

6.xx

A plot of L versus (1 x) shows that the liquid remaining in the still becomes progressively leaner in benzene; when L 0, (1 x) 1 (pure toluene).

4-15

Chapter 4

4.4 Distillation with Reflux

4.4a Introduction: Distillation with reflux can be considered to be a process in which equilibrium stages are arranged in a series in such a manner that the vapor and liquid product from each stage flow counter-current to each other. Each stage receives the liquid flow from the stage above and the vapor flow from the stage below. The liquid and vapor are mixed in each stage and they are assumed to be in equilibrium when they leave a stage. Figure 4.4-1 shows the process flow diagram for a distillation tower containing sieve trays.

Feed

Reboiler

Condenser

RefluxOverheadproduct

Bottomliquidproduct

Liquid

Vapor

Figure 4.4-1 Process flow of a distillation tower containing sieve trays.

The feed is normally introduced into the distillation at a location between the top and the bottom trays. For a total condenser, the vapor from the top tray is condensed and part of the condensate is returned to the top of the column to provide liquid flow above the feed point. Part of the liquid from the bottom tray is vaporized in the reboiler and returned to the bottom of the column to provide the vapor flow.

4-16

4.4b McCabe-Thiele Method

F, xF

v1 QC

L0 D, xD

B, xBQR

feed plate

plate n

plate m

Ln-1

Ln

Vn

Vn+1

Lm-1 Vm

Lm Vm +1

plate n-1

plate n

plate n+1

plate m-1

plate m

plate m+1

Figure 4.4-2 Distillation column notations.

The McCabe Thiele method is a graphical procedure to solve alternately the equilibrium relation and the material balance for binary distillation. Consider the distillation column shown in Figure 4.4-2 with the tray above and below the feed tray using the index n and m, respectively. The section of the column above the feed tray is the rectifying section and the section below the feed tray is the stripping section. The two streams (Ln and Vn or Lm and Vm) leaving a theoretical (or equilibrium) tray are in equilibrium. The two passing streams (for example Ln-1 and Vn) are related by the material balance or operating line. A total material balance for tray n gives

Vn+1 + Ln-1 = Vn + Ln (4.4-1)

A component balance on the more volatile component, A, gives

yn+1Vn+1 + xn-1Ln-1 = ynVn + xnLn (4.4-2)

In this expression, yn+1 is mole fraction of component A in Vn+1 and xn-1 is the mole fraction of component A in Ln-1 and so on. The McCabe-Thiele method is valid when we have constant molal overflow in both the rectifying and stripping sections of the distillation column. Hence

V1 = V2 = V3 = … = Vn = Vn+1 = constant

L0 = L1 = L2 = … = Ln-1 = Ln = constant

Vm-1 = Vm = Vm+1 = constant

4-17

Lm-1 = Lm = Lm+1 = constant

Even for CMO, Vn Vm and Ln = Lm. The constant molal overflow is justified if the heat of mixing is negligible for the two components in the mixture (ideal solution) and the molar heats of vaporization for the two components are almost the same.

Equations for rectifying section

Figure 4.4-3 shows the rectifying or enriching section which is the distillation tower section above the feed. The vapor from the top tray with a composition y1 is condensed in the total condenser so that the resulting liquid is at the boiling point. Part of the liquid is taken out as the overhead product D and the remaining liquid is returned to the first tray with flow rate L0.

v1 QC

L0 D, xD

x0 y1

x1 y2

x2 y3

L , xn n V , yn+1 n+1

Figure 4.4-3 Schematic of the rectifying section with a total condenser.

Making a balance over the top part of the tower (the dashed-line section) on component A gives

yn+1Vn+1 = xnLn + xDD (4.4-3)

Since CMO is assumed, Vn+1 = V = constant and Ln = L = constant. Eq. (4.4-3) can be written as

yn+1 = xnL/V + xDD/V (4.4-4)

From the total balance V = L + D D/V = 1 L/V. In terms of the reflux ratio R = L/D

= = = LV

LL D

// 1L D

L D 1R

R

4-18

= = = DV

DL D

1/ 1L D

11R

Eq. (4.4-4) becomes

yn+1 = xn + xD (4.4-5)1

RR

11R

This equation which is the material balance for component A is called the operating line. Given xD, xF, R, and the equilibrium data, the equilibrium stages can be determined by successively solving the equilibrium relation and the material balance at each stage. Starting at xD, y1 is evaluated, for a total condenser y1 = xD. The liquid composition x1 is then determined form the equilibrium curve at y = y1. This is a dew point calculation with known y1. Next y2 is determined from the mass balance or operating line

y2 = x1 + xD1R

R 1

1R

x2 is again determined from the equilibrium curve and y3 is determined from

y3 = x2 + xD1R

R 1

1R

This process continues until xn < xF. The number of equilibrium stages in the rectifying section is n including the feed tray, n is the number of equilibrium calculations during the process.

Example 4.4-1 ----------------------------------------------------------------------------------A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed flow rate is 2,500 lbmol/hr and feed temperature is 30oC. The column is at 1 atm. A distillate that is 97 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. The external reflux ratio is L0/D = 3. Bottoms from the partial reboiler is 98 mole % n-hexane. Determine the number of equilibrium trays and their compositions for the rectifying section.


Compound A B Cn-pentane (1) 13.9778 2554.6 36.2529n-hexane (2) 14.0568 2825.42 42.7089

Heat of evaporation for n-pentane, C5 = 11,369 Btu/lbmol, CpL,C5 = 39.7 Btu/lbmoloFHeat of evaporation for n-hexane, C6 = 13,572 Btu/lbmol, CpL,C6 = 51.7 Btu/lbmoloF

Solution ------------------------------------------------------------------------------------------

Overall material balance over the entire tower gives

4-19

D + B = 2,500

Material balance for n-pentane over the entire tower gives

0.97D + 0.02B = (0.4)(2,500) = 1,000 0.97(2,500 B) + 0.02B = 1,000

Solving for B and D from the above equations we have B = 1,500 lbmoles/hr and D = 1000 lbmoles/hr.

F, xF

v1 QC

L0 D, x = 0.97D

B, x = 0.02B QR

feed plate

F = 2,500 lbmol/hrT = 30 CP = 1 atm

o

L/D = 30

Since the column has a total condenser, y1 = xD = 0.97. The liquid composition x1 is then determined form the equilibrium curve at y = y1 = 0.97. The equilibrium curve y versus x for n-pentane and n-hexane system is plotted in Figure E-1a where x and y are the mole fraction of n-pentane in the liquid and vapor phase, respectively. The equilibrium curve is expanded near the region x = 1 in Figure E-1b. At y = y1 = 0.97, x = x1 = 0.9107. We can also perform a dew point temperature calculation with y = y1 = 0.97.


x1 + x2 = 1 (E-1)


+ = 1 (E-2)satPPy

1

1satPPy

2

2


4-20

.97×101.325/exp(13.9778 2554.6/(T 36.2529)) + 0.03×101.325/exp(14.0568 2825.42/(T 42.7089)) = 1 (E-3)

Figure E-1a: Equilibrium curve for n-pentane and n-hexane system.

Figure E-1b: Equilibrium curve for n-pentane and n-hexane system near x = 1.

4-21

The dew point temperature should be between the boiling points of n-pentane and n-hexane given by

T1boil = + 36.2529 = 309.20 K2554.6

13.9778 ln(101.325)

T2boil = + 42.7089 = 342.06 K2825.42

14.0568 ln(101.325)


>>fun=inline('.97*101.325/exp(13.9778-2554.6/(T-36.2529))+ 0.03*101.325/exp(14.0568 -2825.42/(T-42.7089))-1');>> T=fsolve(fun,309.5,optimset('Display','off'))

T = 311.0479

The dew point temperature of the n-pentane and n-hexane mixture is 311.0479 K. At this temperature, the vapor pressure of n-pentane is

P1sat = exp(13.9778 - 2554.6/( 311.0479 - 36.2529)) = 107.9228 kPa

The mole fraction of n-pentane in the liquid stream leaving state 1 is then

x1 = = = 0.9107satPPy

1

1 (0.97)(101.325)107.9228

Next y2 is determined from the mass balance or operating line

y2 = x1 + xD = x1 + xD = 0.75x1 + 0.25xD 1

RR

11R

33 1

13 1

y2 = (0.75)(0.9107) + (0.25)(0.97) = 0.9255

x2 is again determined from the equilibrium curve and y3 is determined from

y3 = x2 + xD1R

R 1

1R

This process continues until xn < xF. The following Matlab program alternately solves the equilibrium relation and material balance for the mole fraction of n-pentane leaving each equilibrium stage:

4-22

------------------------ Table 4.4-1: Matlab program e2d4d1b ------------------------------% Example 4.4-1%A=[13.9778 14.0568];B =[2554.6 2825.42];C=[-36.2529 -42.7089];xd=.97;xb=.02;xf=.4;R=3;am=R/(R+1);bm=xd/(R+1);P=101.325;pl=log(P);Tb=B./(A-pl)-C;dT=.01;yi=xd;for i=1:20; % Assume a temperature for the dew point calculationT=yi*Tb(1)+(1-yi)*Tb(2);% Solve for the dew point temperature using Newton's method for n=1:20; f=P*(yi/exp(A(1)-B(1)/(T+C(1)))+(1-yi)/exp(A(2)-B(2)/(T+C(2))))-1; T1=T+dT;f1=P*(yi/exp(A(1)-B(1)/(T1+C(1)))+(1-yi)/exp(A(2)-B(2)/(T1+C(2))))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end % Solve for the mole fraction in the liquid phase using equilibrium relation for ideal system xi=yi*P/exp(A(1)-B(1)/(T+C(1))); fprintf('Stage #%g x = %8.5f , y = %8.5f\n',i,xi,yi) % Solve for the mole fraction in the vapor stream leaving a stage yi=am*xi+bm; if xi<xf, break, endend ---------------------------------------------------------------------

>> e2d4d1bStage #1 x = 0.91070 , y = 0.97000Stage #2 x = 0.79889 , y = 0.92552Stage #3 x = 0.63454 , y = 0.84167Stage #4 x = 0.46085 , y = 0.71840Stage #5 x = 0.32841 , y = 0.58814

4-23

Chapter 4

Equations for stripping section

The distillation tower section below the feed is shown schematically in Figure 4.4-3 with a partial reboiler.

B, x = 0.02B QR

m

m+1

L , xm m V , ym +1 m+1

NV , yB BLN

Figure 4.4-3 Schematic of the stripping section with a partial reboiler.

Making a mole balance on component A gives

ym+1V m+1 + xBB = xmL m (4.4-6)

Again, since CMO is assumed, L m = = constant and V m+1 = = constant. The stripping-L Vsection operating line is a straight line when plotted as y versus x in Figure 4.4-4, with a slope of / ,L V

ym+1 = xm xB (4.4-7)LV

BV

Given xB, xF, / , B/ , and the equilibrium data, the equilibrium stages can be determined L V Vby successively solving the equilibrium relation and the material balance. Starting at xB, yB is evaluated from the equilibrium relation. The liquid composition leaving tray N, xN, is then solved from the operating line,

yB = xN xBLV

BV

We then determine yN from the equilibrium curve and xN-1 is determined from

yN = xN-1 xBLV

BV

4-24

Operating linewith slope = /L V

xB

yB

yN

xN

Equilibrium curve

xN-1

Figure 4.4-4 Operating line and equilibrium data (curve).

The process continues until xm > xF. The number of equilibrium stages in the stripping section is equal to the number of equilibrium calculations during the process.

Example 4.4-2 ----------------------------------------------------------------------------------A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed is saturated liquid at 2,500 lbmol/hr. The column is at 1 atm. A distillate that is 97 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. The external reflux ratio is L0/D = 3. Bottoms from the partial reboiler is 98 mole % n-hexane. Determine the number of equilibrium trays and their compositions for the stripping section.



Solution ------------------------------------------------------------------------------------------


D + B = 2,500


0.97D + 0.02B = (0.4)(2,500) = 1,000 0.97(2,500 B) + 0.02B = 1,000

4-25


F, xF

v1 QC

L0 D, x= 0.999D

B, x= 0.02B QR

feed plate

F = 2,500 lbmol/hrT = 30CP = 1 atm

o

L/D = 30

Since the column has a partial reboiler, the vapor composition yB is determine from the equilibrium curve at xB = 0.02. The equilibrium curve y versus x for n-pentane and n-hexane system is plotted in Figure E-2 where x and y are the mole fraction of n-pentane in the liquid and vapor phase, respectively. At x = xB = 0.02, y = yB = 0.053. We can also perform a bubble point temperature calculation with x = xB = 0.02.


y1 + y2 = 1 (E-1)


x1P1sat + x2P2

sat = P (E-2)


.02exp(13.9778 2554.6/(T 36.2529)) + 0.98exp(14.0568 2825.42/(T 42.7089)) = 101.325 (E-3)

4-26

Figure E-2: Equilibrium curve for n-pentane and n-hexane system.

The buble point temperature should be between the boiling points of n-pentane and n-hexane given by

T1boil = + 36.2529 = 309.20 K2554.6

13.9778 ln(101.325)

T2boil = + 42.7089 = 342.06 K2825.42

14.0568 ln(101.325)


>>fun=inline('.02*exp(13.9778-2554.6/(T-36.2529))+ 0.98*exp(14.0568 -2825.42/(T-42.7089))- 101.325');>> T=fsolve(fun,340,optimset('Display','off'))

T = 340.9753

The bubble point temperature of the n-pentane and n-hexane mixture is 340.9753 K. At this temperature, the vapor pressure of n-pentane is

P1sat = exp(13.9778 - 2554.6/( 340.9753 - 36.2529)) = 268.9249 kPa

4-27

The mole fraction of n-pentane in the vapor stream leaving the reboiler is then

yB = = = 0.05311sat

Bx PP

(0.02)(268.9249)101.325

The liquid composition leaving tray N, xN, is then solved from the operating line,

yB = xN xBLV

BV

We need to determine the liquid and vapor flow rates and , respectively, for the L Voperating line. Since the external reflux ratio R is 3, we have

L = 3D = (3)(1000) = 3000 lbmoles/hr

The vapor flow rate in the rectifying section is then

V = L + D = 3000 + 1000 = 4000 lbmoles/hr

Feed plateSaturatedliquid feed

L V

L VF

Making materials balance over the feed plate gives

= V = 4000 lbmoles/hrV

= L + F = 3000 + 2500 = 5500 lbmoles/hrL

The mole fraction of n-pentane in the liquid stream leaving plate N is then

yB = xN xBLV

BV

xN = yB + xB = ×0.0531 + ×0.02 = 0.0441VL

BL

40005500

15005500

Next yN is determined from the equilibrium curve and xN-1 is determine from the operating line

4-28

yN = xN-1 xBLV

BV

This process continues until xm > xF. The following Matlab program alternately solves the equilibrium relation and material balance for the mole fraction of n-pentane leaving each equilibrium stage in the stripping section:

------------------------ Table 4.4-2: Matlab program e2d4d2 ------------------------------% Example 4.4-2%A=[13.9778 14.0568];B =[2554.6 2825.42];C=[-36.2529 -42.7089];xd=.97;xb=.02;xf=.4;R=3;am=R/(R+1);bm=xd/(R+1);P=101.325;pl=log(P);Tb=B./(A-pl)-C;F=2500;DB=[1 1;xd xb]\[F;xf*F];D=DB(1);Bot=DB(2);L=R*D;Vbar=(R+1)*D;Lbar=L+F;dT=.01;xi=xb;for i=0:20; % Assume a temperature for the dew point calculationT=xi*Tb(1)+(1-xi)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=xi*exp(A(1)-B(1)/(T+C(1)))+(1-xi)*exp(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=xi*exp(A(1)-B(1)/(T1+C(1)))+(1-xi)*exp(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end % Solve for the mole fraction in the vapor phase using equilibrium relation for ideal system yi=xi*exp(A(1)-B(1)/(T+C(1)))/P; fprintf('Stage #N-%g x = %8.5f , y = %8.5f\n',i,xi,yi) % Solve for the mole fraction in the liquid stream leaving a stage xi=(Vbar*yi+Bot*xb)/Lbar; if xi>xf, break, endend ---------------------------------------------------------------------

>> e2d4d2Stage #N-0 x = 0.02000 , y = 0.05308Stage #N-1 x = 0.04406 , y = 0.11291Stage #N-2 x = 0.08757 , y = 0.21110Stage #N-3 x = 0.15898 , y = 0.34848Stage #N-4 x = 0.25890 , y = 0.50188Stage #N-5 x = 0.37046 , y = 0.63380

4-29

Effect of feed conditions

Feed plate

L V

L V

FHF

qF(1-q)F

hL HV

hL HVLiquid Vapor

Figure 4.4-5 Relationships between flows above and below the feed entrance.

The condition of the feed stream F determines the relation between the flow rates in the stripping and enriching sections. Let q be the ratio of the moles of saturated liquid in the feed to the total amount of feed then

F = qF + (1 q)F (4.4-8)

= L + qF L = qF (4.4-9)L L

V = + (1 q)F V = (q 1)F (4.4-10)V V

The magnitude of q is related to the thermal condition of the feed. Making enthalpy and material balance around the feed plate in Figure 4.4-5, we obtain

HFF + + hLL = + HVV (4.4-11)VH V Lh L

F + + L = + V (4.4-12)V L

Assuming = HV and = hL, Eq. (4.4-11) can be rearranged toVH Lh

HFF + HV( V) = hL( L) (4.4-13)V L

Substituting V = (q 1)F and L = qF into Eq. (4.4-13) we obtainV L

HFF + HV(q 1)F = hL qF (4.4-14)

Solving for q gives

q = , where (4.4-15)V F

V L

H HH h

HV = specific enthalpy of the feed at the dew point (saturated vapor)HF = specific enthalpy of the feed at its entrance conditions

4-30

hL = specific enthalpy of the feed at the bubble point (saturated liquid)HV HF = heat needed to vaporize 1 mole of feed at its entrance conditionsHV hL = molar latent heat of vaporization of feed

From equation (4.4-15) q is the ratio of the heat needed to vaporize 1 mole of feed at its entrance conditions to the molar latent heat of vaporization of feed. From this definition, q can be negative when the feed is superheated vapor or it can be greater than 1 when the feed is sub-cooled liquid.

The feed-line or q-line equation, which is the locus of the intersection of the enriching and stripping operating-line equations, can be derived as follow:

yn+1V = xnL + xDD (4.4-16)

ym+1 = xm xBB (4.4-17)V L

At the intersection of the two operating lines y = yn+1 = ym+1 and x = xn = xm.

yV = xL + xDD (4.4-18)

y = x xBB (4.4-19)V L

Subtracting Eq. (4.4-18) from (4.4-19) we obtain

y( V) = x( L) (xBB + xDD) (4.4-20)V L

Substituting V = (q 1)F, L = qF, and (xBB + xDD) = xFF into Eq. (4.4-20) and V Lrearranging we obtain

y = x (4.4-21)1

qq 1

Fxq

There are five possible feed conditions listed below and illustrated in Figure 4.4-6

1) Feed is a saturated liquid: HF = hL q = 1, q-line is vertical2) Feed is a saturated vapor: HF = HV q = 0, q-line is horizontal3) Feed is a mixture of saturated vapor and liquid: hL < HF < HV 0 < q < 14) Feed is a superheated vapor: HF > HV q < 05) Feed is a sub-cooled liquid: HF < hL q > 1

4-31

y

Equilibrium curve

x

q = 1q > 1

0 < q < 1

q = 0

q < 0

Figure 4.4-6 Slope of the q-line for various feed conditions.

Usual Specified Variables for Binary Distillation

In design problems, the desired separation is set and a column is designed to achieve this separation. The following five conditions are usually specified:

1) Column pressure2) Feed flow rate3) Feed composition4) Feed temperature or enthalpy or quality5) Reflux temperature or enthalpy (usually saturated liquid)

In addition to the above conditions, four more variables must be set. Two cases are listed in Table 4.4-3 with the calculation results.

Table 4.4-3 Specified Variables Designer Calculates

Case A1. xD2. xB3. External reflux ratio L0/D4. Use optimum feed plate

D, B: distillate and bottoms flow ratesQR, QC: heating and cooling loadsN: number of stages, Nfeed: optimum feed plateDC: column diameter

Case B1. (FRA)dist2. (FRB)bot3. External reflux ratio L0/D4. Use optimum feed plate

xD, xB, D, B: distillate and bottoms flow ratesQR, QC: heating and cooling loadsN: number of stages, Nfeed: optimum feed plateDC: column diameter

xD = mole fraction of more volatile component A in distillatexB = mole fraction of more volatile component A in bottoms(FRA)dist = fractional recovery of component A in distillate(FRB)bot = fractional recovery of component B in bottoms

4-32

F, hF, xF

V1, H1 QC

L0 D, hD, xD

B, hB, xB QR

Figure 4.4-7 Specified variables for binary distillation column.

To determine the distillation process completely, the “description rule” procedure provided by Hanson et al1 can be applied. This rule states that “to determine a separation process completely, the number of independent variables that must be set (by the designer) will equal the number that are set in the construction of the column or that can be controlled by external means in operation.” Therefore the designer should consider the column in operation and list the number of variables fixed by the column construction as well as the variables fixed by the process and those that have to be controlled for the column to operate steadily and produce product within specification. Figure 4.4-7 shows a conventional distillation column with a total condenser and a partial reboiler. The designer will fix the number of stages above and below the feed point (two variables). The total enthalpy and the feed composition will be fixed by the processes upstream [{1 + (C 1)} variables, where C is the number of components]. Four more variables will be controlled: feed rate, column pressure, cooling water flow for condenser, and steam flow for reboiler. The total number of variables need to be specified is then [2 + 1 + (C 1) + 4 = C + 6]. This is also F, the degree of freedom for adistillation column.

With binary distillation, the degree of freedom is then F = 2 + 6 = 8. For a design problem we can specify feed rate F, feed composition xF, feed quality q, distillate composition xD, distillate temperature (saturated liquid at the specified pressure), bottoms composition xB, external reflux ratio L0/D, and optimum feed plate. With these variables defined, D and B can be obtained from the external mass balances and QR and QC can be solved from the energy balances. Then the number of equilibrium stages can be solved from the stage-by-stage calculations or internal balances. A stage-by-stage calculation involves the successive solutions of the equilibrium relations and the material balance. The procedure is straight forward with no iteration.

1 Hanson, D. N. et al Computation of Multistage Separation Processes , Reihnold, 1962

4-33

Chapter 4 Example 4.4-3 ----------------------------------------------------------------------------------A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed flow rate is 2,500 lbmol/hr and feed temperature is 30oC. The column is at 1 atm. A distillate that is 97 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. The external reflux ratio is L0/D = 3. Bottoms from the partial reboiler is 98 mole % n-hexane. Find D, B, QR, QC, and the number of equilibrium stages.




Solution ------------------------------------------------------------------------------------------

(a) Distillate and bottoms flow rates


D + B = 2,500


0.97D + 0.02B = (0.4)(2,500) = 1,000 0.97(2,500 B) + 0.02B = 1,000


(b) Heating and cooling loads

QC = V1(H1 hD) V1Hevap = V1(0.97C5 + 0.03C6)

V1 = L0 + D = (L0/D + 1)D = 4D = (4)(1000) = 4000 lbmoles/hr

QC = (4,000)(0.97×11,369 + 0.03×13,572) = 4.574×107 Btu/hr

The temperatures of the reflux stream and the reboiler must be known to solve for the heat load of the reboiler. Since the distillate is almost pure pentane and the bottoms product is almost pure hexane, the boiling temperatures of pure pentane and hexane at 1 atm are used for the temperatures of the reflux stream and the reboiler, respectively. Hence

TD 309 K and TB 342 K

4-34

F, xF

v1 QC

L0 D, x = 0.97D

B, x = 0.02B QR

feed plate

F = 2,500 lbmol/hrT = 30 CP = 1 atm

o

L/D = 30

Figure E-1. Distillation column with total condenser and partial reboiler.

The equilibrium data for n-pentane and n-hexane at 1 atm are listed in Table 2.4-4. The data were generated with the Matlab codes listed in Table 4.4-5 assuming ideal solution.

Table 4.4-4 Equilibrium data for n-pentane and n-hexane system at 1 atm.x = mole fraction of n-pentane in the liquidy = mole fraction of n-pentane in the vapor

x = 0.00000 , y = 0.00000, T(K) = 342.06x = 0.05000 , y = 0.12705, T(K) = 339.40x = 0.10000 , y = 0.23699, T(K) = 336.91x = 0.15000 , y = 0.33263, T(K) = 334.58x = 0.20000 , y = 0.41626, T(K) = 332.39x = 0.25000 , y = 0.48975, T(K) = 330.32x = 0.30000 , y = 0.55462, T(K) = 328.38x = 0.35000 , y = 0.61214, T(K) = 326.53x = 0.40000 , y = 0.66335, T(K) = 324.79x = 0.45000 , y = 0.70911, T(K) = 323.14x = 0.50000 , y = 0.75016, T(K) = 321.56x = 0.55000 , y = 0.78711, T(K) = 320.07x = 0.60000 , y = 0.82048, T(K) = 318.64x = 0.65000 , y = 0.85070, T(K) = 317.28x = 0.70000 , y = 0.87816, T(K) = 315.97x = 0.75000 , y = 0.90317, T(K) = 314.72x = 0.80000 , y = 0.92601, T(K) = 313.53x = 0.85000 , y = 0.94692, T(K) = 312.38x = 0.90000 , y = 0.96610, T(K) = 311.28x = 0.95000 , y = 0.98374, T(K) = 310.22x = 1.00000 , y = 1.00000, T(K) = 309.20

4-35

-------Table 4.4-5 Matlab codes for n-pentane and n-hexane system -------%A=[13.9778 14.0568];B =[2554.6 2825.42];C=[-36.2529 -42.7089];dT=.01;Tb=B./(A-pl)-C;x=0:0.05:1;for i=1:21; xi=x(i); % Assume a temperature for the buble point calculationT=xi*Tb(1)+(1-xi)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=xi*exp(A(1)-B(1)/(T+C(1)))+(1-xi)*exp(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=xi*exp(A(1)-B(1)/(T1+C(1)))+(1-xi)*exp(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end % Solve for the mole fraction in the vapor phase using equilibrium relation for ideal system yi=xi*exp(A(1)-B(1)/(T+C(1)))/P; fprintf('x = %8.5f , y = %8.5f, T(K) = %8.2f\n',xi,yi,T)end

-------------------------------------------------------------

Since the boiling point of a 40 mole % n-pentane is 324.79 K, the feed enters the column at 30oC (or 303.15 K) is a subcooled liquid. Making an energy balance over the column, we obtain

FhF + QR = DhD + QC + BhB

Let the reference state be liquid at 30oC, then hF = 0.

hD = CpL,C5(TD TF) = (39.7)(309 303)(1.8) = 428.8 Btu/lbmol

hB = CpL,C6(TB TF) = (51.7)(342 303)(1.8) = 3,629.3 Btu/lbmol

QR = DhD + QC + BhB = (1,500)( 3,629.3) + 4.574×107 + (1,000)(428.8) = 5.1613×107 Btu/hr

(c) Number of equilibrium stages

Making a balance over the top part of the tower on n-pentane gives the operating line

yn+1Vn+1 = xnLn + xDD

Since CMO is assumed, Vn+1 = V = constant and Ln = L = constant, we have

yn+1 = xnL/V + xDD/V

From the total balance V = L + D D/V = 1 L/V. In terms of the reflux ratio R = L/D

4-36

= = = LV

LL D

// 1L D

L D 1R

R

= = = DV

DL D

1/ 1L D

11R

The operating line becomes

yn+1 = xn + xD1R

R 1

1R

Since R = 3 and xD = (0.97) we have

yn+1 = xn + (0.97) = 0.75 xn + 0.242533 1

13 1

The q-line is determined next. Since feed is subcooled liquid we have

q = = = 1 + V F

V L

H HH h

V L L F

V L

H h h HH h

( )pL boil F

V L

C T TH h

In this equation

HV hL = latent heat = 0.4C5 + 0.6C6 = (0.4)(11,369) + (0.6)( 13,572)

HV hL = 12,691 Btu/lbmol

CpL = 0.4CpL,C5 + 0.6CpL,C6 = (0.4)(39.7) + (0.6)(51.7) = 46.9 Btu/lbmoloF

q = 1 + = 1.08(46.9)(324.79 303.15)12,691

The q-line (4.4-21) derived in earlier section is

y = x = x = 13.5x 51

qq 1

Fxq

1.081.08 1

0.41.08 1

At the intersection of the q-line and the enriching operating line y = yn+1 or

13.5x 5 = 0.75x + 0.2425 x = 0.41118, y = 0.55088

The stripping operating line is

yN = xN-1 xBLV

BV

4-37

At xN-1 = xB, yN = xB = xB = xBL B

V V

V

The stripping operating line can be plotted by connecting two points (x = xB, y = xB) and (x= xq = 0.41118, y = yq = 0.55088). Starting from the bottoms, point (x = xB = 0.02, y = xB), the following graphical procedure, illustrated in Figure E-2, will determine the number of equilibrium stages

1) Draw a vertical line to meet the equilibrium curve. This determines the mole fraction of n-pentane in the vapor flow leaving the equilibrium stage.

2) Draw a horizontal line to meet the (stripping) operating line. This determines the mole fraction of n-pentane in the liquid flow entering the equilibrium stage.

3) Repeat step 1-2 until the mole fraction of n-pentane in the vapor flow exceeds yq (the vapor mole fraction of n-pentane at the intersection of the q-line and the operating lines. The horizontal line now will meet the rectifying operating line .

4) Repeat step1-2 with the stripping operating line replaced by the rectifying operating line until the mole fraction of n-pentane in the liquid flow entering the equilibrium stage exceeds xB.

The number of equilibrium stages is number of times the vertical line intersect the equilibrium curve. From Figure E-2, there are 10 equilibrium stages consisting of 1 partial reboiler and 9 equilibrium trays. This graphical method is known as the McCabe-Thiele method to determine the number of equilibrium stages.

Figure E-2 Graphical method to determine the number of equilibrium stages.

4-38

Table 4.4-6 lists the Matlab codes for the number of equilibrium stages. Starting from the bottoms, point (x = xB = 0.02, y = xB), a bubble point calculation with x = 0.02 is performed to determine the vapor mole fraction of n-pentane leaving the first equilibrium stage (the partial reboiler), then the stripping operating line is used to calculate liquid the mole fraction of pentane entering the equilibrium stage. A linear interpolation is used between two points (x = xB, y = xB) and (x = xq = 0.41118, y = yq = 0.55088).

= B

B

x xy x

q B

q B

x xy y

Let c1 = then x = xB + c1(y xB) and q B

q B

x xy y

When the bubble point calculation using the liquid mole fraction from the stripping operating line produces the vapor mole fraction of n-pentane exceeding yq, the liquid mole fraction of pentane entering an equilibrium stage is determined from the rectifying operating line.

yn+1 = xn + xD1R

R 1

1R

Let slope = and yint = xD then x =

1R

R 1

1R inty y

slope

The vapor mole fraction of n-pentane is then determined from the bubble point calculation. The process is repeated until the mole fraction of n-pentane in the liquid flow entering the equilibrium stage exceeds xB.

The specification of optimum feed plate requires that the stripping operating line is used for the materials balance as soon as the vapor mole fraction of n-pentane exceeding yq. We could choose to continue with the stripping operating line. However this action will require a higher number of equilibrium stages to reach the same separation since the distance from the equilibrium curve is closer to the stripping than to the rectifying operating line. A condition further away from equilibrium indicates a higher driving force for mass transfer and hence requires less equilibrium stages.

-------Table 4.4-6 Matlab codes for number of equilibrium stages -------% Example 4.4-3%A=[13.9778 14.0568];B =[2554.6 2825.42];C=[-36.2529 -42.7089];P=101.325;pl=log(P);dT=.01;Tb=B./(A-pl)-C;xf=0.4;xq=0.41118;yq= 0.55088;xb=.02;c1=(xq-xb)/(yq-xb);xd=.97;R=3;yint=xd/(R+1);slop=R/(R+1);x=0:0.05:1;y=x;for i=1:21; xi=x(i);

4-39

% Assume a temperature for the bubble point calculationT=xi*Tb(1)+(1-xi)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=xi*exp(A(1)-B(1)/(T+C(1)))+(1-xi)*exp(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=xi*exp(A(1)-B(1)/(T1+C(1)))+(1-xi)*exp(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end % Solve for the mole fraction in the vapor phase using equilibrium relation for ideal system yi=xi*exp(A(1)-B(1)/(T+C(1)))/P; y(i)=yi; end % Plot the equilibrium curve, the rectifying and the stripping operating lines % plot(x,y,[0 1],[0 1],[xq xd],[yq xd],[xb xq],[xb yq]) axis([0 1 0 1]);xlabel('x');ylabel('y') hold on % Plot the q line line([xf xq],[xf yq]) xi=xb;yo=xb; for i=1:20; T=xi*Tb(1)+(1-xi)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=xi*exp(A(1)-B(1)/(T+C(1)))+(1-xi)*exp(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=xi*exp(A(1)-B(1)/(T1+C(1)))+(1-xi)*exp(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end % Solve for the mole fraction in the vapor flow leaving an equilibrium stage % using equilibrium relation for ideal system % yi=xi*exp(A(1)-B(1)/(T+C(1)))/P; % Draw a vertical line from the operating line to meet the equilibrium curve % line([xi xi],[yo yi]) if yi<yq % Solve for the mole fraction in the liquid flow entering an equilibrium stage % using the stripping operating line xn=xb+c1*(yi-xb); else % Solve for the mole fraction in the liquid flow entering an equilibrium stage % using the rectifying operating line xn=(yi-yint)/slope; end yo=yi; % Draw a horizontal line from the equilibrium curve to meet the operating line

4-40

% line([xi xn],[yi yi]) xi=xn; if xi>xd,break, end end fprintf('Number of equilibrium stages = %g\n',i)

>> e2d4d3Number of equilibrium stages = 10

-------------------------------------------------------------

The number of equilibrium stages can also be determined from the program “bdist4”. The equilibrium data for n-pentane and n-hexane system must first be generated. The Peng-Robinson equation of state is used by “bdist4” for equilibrium calculation. This program also calculates the minimum reflux ratio and the minimum number of equilibrium stages which will be discussed in the next section.

4-41

Chapter 4 Total Reflux and Minimum Reflux Ratio

a. Total Reflux. In design problems, the desired separation is specified and a column is designed to achieve this separation. In addition to the column pressure, feed conditions, and reflux temperature, four additional variables must be specified.

Specified Variables Designer CalculatesCase A

1. xD2. xB3. External reflux ratio L0/D4. Use optimum feed plate

D, B: distillate and bottoms flow ratesQR, QC: heating and cooling loadsN: number of stages, Nfeed: optimum feed plateDC: column diameter

xD, xB = mole fraction of more volatile component A in distillate and bottoms, respectively

The number of theoretical stages depends on the reflux ratio R = L0/D. As R increases, the products from the column will reduce. There will be fewer equilibrium stages needed since the operating line will be further away from the equilibrium curve. The upper limit of the reflux ratio is total reflux, or R = ∞. The rectifying operating line is given as

yn+1 = xn + xD1R

R 1

1R

When R = ∞, the slop of this line becomes 1 and the operating lines of both sections of the column coincide with the 45o line. In practice the total reflux condition can be achieved by reducing the flow rates of all the feed and the products to zero. The number of trays required for the specified separation is the minimum which can be obtained by stepping off the trays from the distillate to the bottoms.

Figure 4.4-8 Minimum numbers of trays at total reflux.

4-42

y , VN-1 N-1 x , LN N

1

2y x1 2

N

N-1

yN xN+1

x1y0

Totalreboiler

Totalcondenser

Figure 4.4-9 Distillation column operation at total reflux.

The minimum number of equilibrium trays can also be approximated by Fenske equation,

Nm = (4.4-22)

1log1log

D B

D B

ave

x xx x

In this equation ave = (1B)1/2 where 1 is the relative volatility of the overhead vapor and B is the relative volatility of the bottoms liquid. We can derive Fenske equation using the notation shown in Figure 4.4-9 where stages are numbered from the bottom up. The vapor leaving stage N is condensed and returned to stage N as reflux. The liquid leaving stage 1 is vaporized and returned to stage 1 as vapor flow. For steady state operation with no heat loss, heat input to the reboiler is equal to the heat output from the condenser. From material balance, vapor and liquid streams passing between any pair of stages have equal flow rates and compositions, for example, VN-1 = LN and yN-1 = xN. In general, molar vapor and liquid flow rates will change from stage to stage unless the assumption of constant molar overflow is valid. At stage 1, the equilibrium relation is written as

y1 = K1x1 (4.4-23)

4-43

From the material balance

y1 = x2 (4.4-24)

Combine Eqs. (4.4-23) and (4.4-24)

x2 = K1x1 (4.4-25)

Similarly for stage 2

y2 = K2x2 (4.4-26)

Combine Eqs. (4.4-25) and (4.4-26)

y2 = K2K1x1 (4.4-27)

The procedure can be repeated to stage N where

yN = KNKN-1…K2K1x1 (4.4-28)

Similarly for the less volatile component i

1 yN = Ki,NKi,N-1…K i,2K i,1 (1 x1) (4.4-29)

Dividing Eq. (4.4-8) by Eq. (4.4-9), we have

= NN-1…21 (4.4-30)1

N

N

yy

1

11x

x

In this equation k = = relative volatility between the two components on stage k. ,

k

i k

KK

Rearranging Eq. (4.4-30) we obtain

= or = (4.4-31)1

Nyx 1

11

Nyx

min

1

N

kk

1

1

Nxx

1

1

11

Nxx

min

1

N

kk

Since xN+1 = xD, x1 = xB, and assuming constant relative volatility, Eq. (4.4-31) becomes

= (4.4-32)minN1

D

D

xx

1 B

B

xx

Solving for the minimum number of equilibrium trays gives

Nmin = (4.4-33)

1log1log

D B

D B

x xx x

4-44

Eq. (4.4-33) is the Fenske equation (4.4-22) where = ave = (1B)1/2

Nm = (4.4-22)

1log1log

D B

D B

ave

x xx x

b. Minimum reflux ratio. As the reflux ratio is reduced, the distance between the operating line and the equilibrium curve becomes smaller. The minimum reflux ratio Rm is the limiting reflux where the operating line either touches the equilibrium curve or intersects the equilibrium curve at the q-line. The minimum reflux ratio will require an infinite number of trays to attain the specified separation of xD and xB. Figure 4.4-10 shows an equilibrium plate n with streams Ln-1 and Vn+1 entering and streams Ln and Vn leave the plate. If the two steams Ln-1 and Vn+1 are at equilibrium there will be no net mass transfer between the liquid and vapor streams. The equilibrium curve will touch or intersect the operating line at this point.

n

n-1

n+1

Ln-1 Vn

LnVn+1

Figure 4.4-10 Equilibrium plate n with vapor and liquid streams.

Given q, xD and xF, the feed line is fixed and the upper operating line depends on the reflux ratio R. At total reflux, the operating line coincides with the 45o line. As R is decreased, the slope of the enriching operating line R/(R + 1) is decreased. The operating line will rotate clockwise around the point (x = xD, y = xD ) until it is tangent to the equilibrium curve or it intersects the q-line at the equilibrium point whichever comes first. The location where the operating line touches or intersects the equilibrium curve is called the pinch point. The enriching operating line at minimum reflux is then defined.

yn+1 = xn + xD1m

m

RR

11mR

The minimum reflux Rm can be obtained from either the intercept of the slope of the enriching operating line. The operating flux ratio is between the minimum Rm and total reflux. Usual value is between 1.2Rm to 1.5Rm. Figure 4.4-11 shows the pinch points for case 1 where the operating line intersects the equilibrium curve and case 2 where the operating line touches the equilibrium curve.

4-45

Figure 4.4-11 The pinch points for minimum reflux.

Example 4.4-4 ----------------------------------------------------------------------------------A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane. Feed is saturated liquid with a flow rate of 2,500 lbmol/hr. The column is at 1 atm. A distillate of 90 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated liquid. Bottoms from the reboiler is 98 mole % n-hexane. Determine the minimum number of equilibrium trays and the minimum reflux ratio.




Solution ------------------------------------------------------------------------------------------

(a) Minimum number of equilibrium trays

The equilibrium data for n-pentane and n-hexane at 1 atm are listed in Table 4.4-4. The data were generated with the Matlab codes listed in Table 4.4-5 assuming ideal solution.

4-46

Table 4.4-4 Equilibrium data for n-pentane and n-hexane system at 1 atm.x = mole fraction of n-pentane in the liquidy = mole fraction of n-pentane in the vapor

x = 0.00000 , y = 0.00000, T(K) = 342.06x = 0.05000 , y = 0.12705, T(K) = 339.40x = 0.10000 , y = 0.23699, T(K) = 336.91x = 0.15000 , y = 0.33263, T(K) = 334.58x = 0.20000 , y = 0.41626, T(K) = 332.39x = 0.25000 , y = 0.48975, T(K) = 330.32x = 0.30000 , y = 0.55462, T(K) = 328.38x = 0.35000 , y = 0.61214, T(K) = 326.53x = 0.40000 , y = 0.66335, T(K) = 324.79x = 0.45000 , y = 0.70911, T(K) = 323.14x = 0.50000 , y = 0.75016, T(K) = 321.56x = 0.55000 , y = 0.78711, T(K) = 320.07x = 0.60000 , y = 0.82048, T(K) = 318.64x = 0.65000 , y = 0.85070, T(K) = 317.28x = 0.70000 , y = 0.87816, T(K) = 315.97x = 0.75000 , y = 0.90317, T(K) = 314.72x = 0.80000 , y = 0.92601, T(K) = 313.53x = 0.85000 , y = 0.94692, T(K) = 312.38x = 0.90000 , y = 0.96610, T(K) = 311.28x = 0.95000 , y = 0.98374, T(K) = 310.22x = 1.00000 , y = 1.00000, T(K) = 309.20

The minimum number of trays required is obtained by stepping off between the 45o line and the equilibrium curve from xB = 0.1 to xD = 0.9. The answer is 4.1 equilibrium trays.

Figure E-1 Minimum numbers of trays at total reflux.

4-47

The minimum number of trays can also be estimated by Fenske equation,

Nm = =

=

1log1log

D B

D B

ave

x xx x

0.9 1 0.1log1 0.9 0.1log ave

log 81

log ave

In this equation ave = (DB)1/2 where D is the relative volatility of the overhead vapor and B is the relative volatility of the bottoms liquid.

y , VN-1 N-1 x , LN N

1

2y x1 2

N

N-1

yN xN+1

x1y0

Totalreboiler

Totalcondenser

Figure E-2 Distillation column operation at total reflux.

From the notation in Figure E-2, at the top yN = xD = 0.9 0.9032, xN = 0.75000. At the bottoms x1 = xB = 0.1, y1 = 0.237 (equilibrium data from Table 4.4-4).

D = = = 3.1102

/1 / 1

N N

N N

y xy x

0.9032 / 0.75001 0.9032 / 1 0.7500

B = = = 2.7955

1 1

1 1

/1 / 1

y xy x

0.237 / 0.1001 0.237 / 1 0.100

4-48

ave = (DB)1/2 = (3.1102×2.7955)1/2 = 2.9487

Nm = = = 4.1

log 81log ave

log 81log 2.9487

(b) Minimum reflux ratio.

Figure E-3 The pinch point for minimum reflux.

Since feed is saturated liquid, the feed-line is vertical and intersects the equilibrium curve at the point x = 0.4, y = 0.6633. The enriching operating line for minimum reflux passes through this point (x = 0.4, y = 0.6633) and the point (x = xD = 0.9, y = 0.9). The slope of the rectifying operating line is given by

= = 0.47341

m

m

RR

0.9000 0.66330.9000 0.4000

The minimum reflux ratio is then

Rm = = 0.900.47341.0000 0.4734

4-49

Chapter 4 4.5 Tray Efficiency

For an equilibrium tray, the vapor and liquid leaving the tray are in thermodynamic equilibrium. In a real tray, equilibrium will rarely be attained. The concept of tray efficiency is used to link the performance of a real tray to an equilibrium tray. Tray efficiencies can be defined in several ways:

The overall efficiency, Eo, is given by

Eo = Nequil/Nactual

The overall efficiency is extremely easy to measure and use; thus, it is the most commonly used efficiency value in the plant. However, the overall efficiency is not representative of column operation because the different compositions on the various trays result in different tray efficiencies. The overall column efficiency6 can be estimated from

Eo = 0.52782 – 0.27511log10() + 0.044923[log10()]2

where = relative volatility of key components, and = viscosity of feed in cP. These properties are determined at the average temperature and pressure of the column.

Murphree plate efficiency

The Murphree vapor efficiency is defined as

EMV = yn - yn+1

yn* - yn+1= actual change in vapor

change in vapor at equilibrium

while the Murphree liquid efficiency is defined as

EML = xn - xn-1xn* - xn-1

= actual change in liquidchange in liquid at equilibrium

yn

yn+1 xn

n-1x

Tray n Murphree efficiency model

Point efficiency

6 Wankat, Equilibrium Staged Separation, Elsevier, 1988, pg. 383

4-50

If the vapor and liquid compositions are taken at a point on the plate, the above equation gives the local or point efficiency

Estimating Murphree tray efficiencies

The prediction methods available in the open literature are limited either to binary systems or to the efficiency of separation between the key components in a multi-component system. In a binary system, the efficiency obtained for each component must be the same. For a multi-component system; the heavier components will usually exhibit lower efficiencies than the lighter components. The following guide rule can be used to estimate the efficiencies for a multi-component system from binary data:

- If the components are similar, the multi-component efficiencies will be similar to the binary efficiency.

- If the predicted efficiencies for the binary pairs are high, the multi-component efficiency will be high.

- If the resistance to mass transfer is mainly in the liquid phase, the difference between the binary and multi-component efficiencies will be small.

- If the resistance is mainly in the vapor phase, as it normally will be, the difference between the binary and multi-component efficiencies can be substantial.

For mixtures with dissimilar compounds, the efficiency can be very different from the predicted for each binary pair, and laboratory or pilot-plant studies should be made to confirm any predictions.

Van Winkle's correlation can be used to predict plate efficiencies for binary systems. The data used to derive the correlation covered both bubble-cap and sieve plates.

EMV = 0.07 Dg0.14 Sc0.25 Re0.08

where Dg = surface tension number = (L/Luv),uv = superficial vapor velocity,L = liquid surface tension,L = liquid viscosity,Sc = liquid Schmidt number = (L/LDLK),L = liquid density,DLK = liquid diffusivity, light key component,Re = Reynolds number = (hwuvv/L(FA)),hw = weir height,v = vapor density,(FA) = fractional area = (area of holes or risers)/(total column cross-sectional area)

To evaluate the terms for the plate efficiency (if data are not available)- Calculate vapor density from the ideal gas equation- Calculate the density of the liquid mixture as the mole fraction average molar

volume- Calculate liquid mixture viscosity by using the expression

4-51

L,mix = 3

1

3/1

n

iiix

- Calculate the diffusivity of the liquid light key component by the dilute solution of Wilke and Chang.

Example 4.5-1. ----------------------------------------------------------------------------------Determine the overall efficiency of a distillation column used to separate acetone from water5. The average temperature and pressure of the column are 83oC and 1 atm, respectively. The feed is saturated liquid with mole fraction of acetone equal to 0.033 corresponding to equilibrium vapor mole fraction of 0.421.

Solution -----------------------------------------------------------------------------------------

The overall column efficiency can be estimated from

Eo = 0.52782 – 0.27511log10() + 0.044923[log10()]2

= = = 21.3)1/()1(

/xy

xy

0.421/ 0.033(1 9.421) /(1 0.033)

Viscosities of acetone and water can be obtained from T.K. Prop4 program at 83oC and 1 atm.

acetone = 0.190 cp, water = 0.350 cp

The mixture viscosity is then

L,mix = = 3

1

3/1

n

iiix

31/3 1/30.033 0.190 0.0967 0.350

mix = 0.343 cp

= 7.31 log10() = 0.864

The overall column efficiency is then

Eo = 0.52782 – 0.27511log10() + 0.044923[log10()]2 = 0.324

5 Sinnot, R.K. Coulson & Richardson’s, Chemical Engineering. Boston: Butterworth and Heinemann, 1999.

4-52

Example 4.5-2. ----------------------------------------------------------------------------------Determine the plate efficiency of a distillation column used to separate acetone from water5. The plate is at 79oC and 1.14 bar. The cross-sectional area of the column is 0.50 m2. The vapor volumetric flow rate is 0.81 m3/s. The following data are provided

L (kg/m3) v (kg/m3) L(kg/ms) v(kg/ms) hw (m) Ah (m2)925 1.35 0.34×10-3 10.0×10-6 50×10-3 0.038

Atomic volume of C H Om3/kmol 0.0148 0.0037 0.0074

Solution -----------------------------------------------------------------------------------------

We will first calculate the diffusivity of acetone in water at 79oC using Wilke Chang equation:

DL = 0.516

0.6

1.173 10

m

Mw TV

In this equation, = 2.26 for water solvent, Mw = 58, T = 273 + 79 = 352 K, and for acetone (C3H6O)

Vm = 0.0148×3 + 0.0037×6 + 0.0074 = 0.074 m3/kmol

DL = = 6.63×10-9 m2/s

0.516

0.63

1.173 10 2.26 58 3520.34 10 0.074

The plate efficiency can be estimated from

EMV = 0.07 Dg0.14 Sc0.25 Re0.08

Dg = surface tension number = (L/Luv),

uv = superficial vapor velocity = = 1.62 m/s0.810.50

L = liquid surface tension = 60×10-3 N/m

Dg = = 109 3

3

60 100.34 10 1.62

Sc = (L/LDLK) = = 55.4

3

9

0.34 10925 6.63 10

5 Sinnot, R.K. Coulson & Richardson’s, Chemical Engineering. Boston: Butterworth and Heinemann, 1999.

4-53

Re = Reynolds number = (hwuvv/L(FA)),

FA = = = 0.076h

c

AA

0.0380.50

Re = = 4232

3

3

50 10 1.62 1.35

0.34 10 0.076

EMV = 0.07 Dg0.14 Sc0.25 Re0.08 = 0.07 (109)0.14 (55.4)0.25 (4232)0.08 = 0.72

The efficiency of the plate is 72%.

4.6 Enthalpy-Concentration Diagram

The calculations for the number of stages in the case of constant molal overflow (CMO) only involve the equilibrium relations and the material balances. CMO requires the molar latent heat to be the same for each component with no heat of mixing. For non-ideal system, where the CMO is no longer valid, the calculations involve the equilibrium relations, the material balances, and the energy balances. For non-ideal binary system, the number of stages can be obtained graphically on an enthalpy-composition chart known as the Ponchon and Savarit method. A typical enthalpy-composition diagram or H-xy chart is shown in Figure 4.6-1, with the dew point and bubble (boiling) curves.

Figure 4.6-1 Enthalpy-composition diagram, showing enthalpiesof saturated liquid and vapor.

4-54

The saturated liquid enthalpy for a binary mixture is given by

h = x + (1 x) = xhA + (1 x)hB + hmix (4.6-1)Ah Bh

where

x = liquid mole fraction of the more volatile component A= partial molar enthalpy of component A in the saturated liquid mixtureAh

hA = molar enthalpy of pure saturated liquid Ahmix = enthalpy change on mixing

The saturated vapor enthalpy for a binary mixture is given by

H = y + (1 y) = yHA + (1 y)HB + Hmix (4.6-2)AH BH

where

y = vapor mole fraction of the more volatile component A= partial molar enthalpy of component A in the saturated vapor mixtureAH

HA = molar enthalpy of pure saturated vapor AHmix = enthalpy change on mixing

Example 4.6-1. ----------------------------------------------------------------------------------Prepare an enthalpy-concentration plot for benzene-toluene at 1 atm pressure assuming ideal solution. Physical property data are given in the following Table1.Component Tb(oC) Cp(kJ/kmolK) Cpy(kJ/kmolK) (kJ/kmol)Benzene (A)Toluene (B)

80.1110.6

138.2167.5

96.3138.2

30,82033,330

Vapor pressure: Psat (mmHg), T(oC)

log10 PAsat = 6.90565 1211.033/(T + 220.79)

log10 PBsat = 6.95334 1343.943/(T + 219.377)

Solution -----------------------------------------------------------------------------------------Let reference temperature Tref = 80.1 = boiling point of benzene (A)

hA = CpA(T Tref) = 138.2(T 80.1)

hB = CpB(T Tref) = 167.5(T 80.1)

HA = A + CpyA(T Tref) = 30,820 + 96.3(T 80.1)

HB = B,Tref + CpyB(T Tref)

1 Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, p. 732

4-55

The latent heat, B, of vaporization for toluene is 33,330 kJ/kmol at 110.6oC, the latent heat, B,Tref, with respect to the reference temperature must be evaluated.

Liquid at TbB Vapor at TbB

Liquid at Tref Vapor at Tref

TbB = 110.6 Co

Tref = 80.1 Co

B

B,ref

From the above diagram

B,Tref = CpB(TbB Tref) + B + CpyB(Tref TbB)

B,Tref = 167.5(110.6 80.1) + 33,330 + 138.2(80.1 110.6) = 34,224 kJ/mol

Therefore HB = 34,224 + 138.2 (T 80.1)

For an ideal binary mixture, the liquid and vapor enthalpies can be determined from

h = xhA + (1 x)hB and H = yHA + (1 y)HB

To prepare an enthalpy-concentration plot we need to know the equilibrium compositions at 1 atm as a function of temperature. At a given mole fraction of benzene in the liquid phase, we can perform a bubble point calculation to determine the equilibrium temperature and compositions in the vapor phase. The saturated liquid enthalpy h and the saturated vapor enthalpy H can then be determined from the temperature and equilibrium compositions. The equilibrium temperature is determined from the following equation

P = xPAsat + (1 x)PB

sat

760 = x×10^[6.90565 1211.033/(T + 220.79)] + (1 x) ×10^[6.95334 1343.943/(T + 219.377)]

At x = 0.1, the above equation can be solved with an initial guess of (0.1)(80.1) + (0.9)(110.6) = 107.6oC. The calculated bubble point is 106.0oC with a vapor mole fraction of 0.2085 for benzene. The results for the calculations at various value of x from 0 to 1 are given in Table 4.6-1.

At T = 106.0oC, PAsat = 10^[6.90565 1211.033/(T + 220.79)] = 1584.2

y = = = 0.2085sat

AxPP

(0.1)(1,584.2)760

4-56

Table 4.6-1 Enthalpy-concentration data for benzene-tolueneMixture at 1 atm total pressure

x T(oC) h(J/mol) y H(J/mol) 0

0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 0.5500 0.6000 0.6500 0.7000 0.7500 0.8000 0.8500 0.9000 0.9500 1.0000

110.4633 108.1749 106.0083 103.9538 102.0025 100.1466 98.3788 96.6926 95.0821 93.5418 92.0670 90.6530 89.2960 87.9921 86.7379 85.5304 84.3666 83.2441 82.1602 81.1129 80.1000

5.0859e+003 4.6614e+003 4.2637e+003 3.8907e+003 3.5403e+003 3.2110e+003 2.9010e+003 2.6091e+003 2.3339e+003 2.0743e+003 1.8292e+003 1.5976e+003 1.3787e+003 1.1716e+003 9.7571e+002 7.9026e+002 6.1465e+002 4.4833e+002 2.9076e+002 1.4146e+002

0

0 0.1103 0.2085 0.2963 0.3751 0.4460 0.5101 0.5681 0.6208 0.6688 0.7127 0.7529 0.7898 0.8237 0.8550 0.8839 0.9107 0.9355 0.9585 0.9800 1.0000

3.8420e+004 3.7599e+004 3.6868e+004 3.6216e+004 3.5630e+004 3.5102e+004 3.4623e+004 3.4188e+004 3.3792e+004 3.3428e+004 3.3094e+004 3.2787e+004 3.2502e+004 3.2238e+004 3.1993e+004 3.1764e+004 3.1551e+004 3.1351e+004 3.1163e+004 3.0986e+004 3.0820e+004

The equilibrium data in Table 4.6-1 are plotted in Figure 4.6-1 to obtain the boiling point curve and the dew point curve. Table 4.6-2 lists the Matlab program to produce Figure 4.6-1

----------------------------------------Table 4.6-2---------------------------------------------% Enthalpy-composition diagram%A=[6.90565 6.95534];B =[1211.033 1343.943];C=[220.79 219.377];CpA=138.2;CpB=167.5;CpyA=96.3;CpyB=138.2;dHA=30820;dHB=34224;P=760;pl=log10(P);dT=.01;Tb=B./(A-pl)-C;xf=0.4;x=0:0.05:1;y=x;Tv=x;for i=1:21; xi=x(i); % Assume a temperature for the bubble point calculationT=xi*Tb(1)+(1-xi)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=xi*10^(A(1)-B(1)/(T+C(1)))+(1-xi)*10^(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=xi*10^(A(1)-B(1)/(T1+C(1)))+(1-xi)*10^(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end Tv(i)=T;

4-57

% Solve for the mole fraction in the vapor phase using equilibrium relation for ideal system yi=xi*10^(A(1)-B(1)/(T+C(1)))/P; y(i)=yi; end Tref=Tb(1); hA=CpA*(Tv-Tref);hB=CpB*(Tv-Tref); HA=dHA+CpyA*(Tv-Tref);HB=dHB+CpyB*(Tv-Tref); h=x.*hA+(1-x).*hB; H=y.*HA+(1-y).*HB;plot(y,H,':',x,h)xlabel('x, y');ylabel('Enthalpy(J/mol)')legend('H(J/mol)','h(J/mol)')

---------------------------------------------------------------------------------

The use of the enthalpy-concentration chart to solve distillation problems depends upon the property that any three streams which are connected by the material and energy balances will be on a straight line when their enthalpies are plotted versus the compositions. Consider the following material and energy balances:

xFF = yV + xL and HFF = HV + hL

The three points F(xF, HF), V(y, H), and L(x, h) must be on a straight line. Consider the following material and energy balances:

xFF = yV + xL and (HF + Q/F)F = HV + hL

Then the three points F(xF, HF + Q/F), V(y, H), and L(x, h) will be on a straight line. This property could be used to solve a binary distillation problem when both material and energy balances are required. Consider a distillation column shown in Figure 4.6-2 with a feed stream F, distillate stream D, and bottoms stream W. Making overall material and energy balances on the entire column we have

FxF = DxD + WxW (4.6-3)

FhF + QR = DhD + QC + WhW (4.6-4)

Rearranging,

FhF = D(hD + QC/D)+ W(hW QR/W) (4.6-5)

From equations (4.6-3) and (4.6-5), three points F(xF, hF), D’(xD, hD+QC/D), and W’(xW, hW QR/W) must be on the same straight line with F in between D’ and W’.

4-58

F

D'

D12345678

H or h

W

W'

xW xF xD

QR

QC

F, hF, xF

D, hD, xD

W, hW, xW

V1,H1

Lo

Figure 4.6-2 Analysis of a complete distillation tower.

The graphical construction for the number of theoretical stages can start at the tower top or bottom and is shown in Fig. 4.6-2. The feed in this case is a saturated vapor with mole fraction xF. When a tie line crosses the line , which goes through the feed point F, the W'FD'other operating point is used for the rest of the tower. This is the optimum feed plate location. In Fig. 4.6-2 the feed is on the fifth plate from the top. The tower has a total of eight theoretical stages or seven theoretical trays plus a reboiler.

The reflux ratio R = L0/D can be determined by making material and energy balances around the condenser.

V1 = L0 + D (4.6-6)

V1H1 = (L0 + D)H1 = L0hD + DhD + QC (4.6-7)

Rearranging,

L0(H1 hD) = D(hD + QC/D - H1) (4.6-7)

or

R = = (4.6-8)0LD

1

1

/D C

D

h Q D HH h

4-59

Chapter 4

4.7 Distillation Using Enthalpy-Balance

We will consider the distillation calculations where the equilibrium relations, the material and energy balances must be solved simultaneously.

v1 QC

L0 D, xD

x0 y1

x1 y2

x2 y3

L , x , hn n n V , y , Hn+1 n+1 n+1

n

n+1

Figure 4.7-1 Schematic of the rectifying section with a total condenser.

Making an overall and a component balance over the enriching section as shown in Figure 4.7-1, we obtain

Vn+1 = Ln + D (4.7-1)

yn+1Vn+1 = xnLn + xDD (4.7-2)

Equation (4.7-2) can be rearranged to solve for yn+1,

yn+1 = xn + xD (4.7-3)1

n

n

LV 1n

DV

Making an enthalpy balance over the enriching section

Hn+1Vn+1 = hnLn + hDD + Qc (4.7-4)

In this equation, Qc is the condenser duty in kW. An enthalpy balance can be made just around the condenser

4-60

H1V1 = hDL0 + hDD + Qc (4.7-5)

Subtracting equation (4.7-5) from equation (4.7-4) to eliminate Qc we have

Hn+1Vn+1 H1V1 = hnLn hDL0 (4.7-6)

Hn+1Vn+1 = hn(Vn+1 D) + H1V1 hDL0

Vn+1 = (H1V1 hDL0 hnD)/(Hn+1 hn) (4.7-7)

Example 4.7-1. ----------------------------------------------------------------------------------A liquid mixture of benzene-toluene is to be distilled in a fractionating tower at 760 mmHg. The feed is liquid with a flow rate of 100 kmol/h, containing 45 mol % benzene and 55 mol % toluene, and enters at 327.6 K. We want to obtain a distillate containing 95 mol % benzene and 5 mol % toluene and a bottoms containing 10 mol % benzene and 90 mol % toluene. The reflux ratio is 1.755. The average heat capacity of the feed is 159 kJ/kmol and the average latent heat 32,099 kJ/kmol. Use enthalpy balances to calculate the number of equilibrium stages in the rectifying section.

Physical property data are given in the following Table 11.Component Tb(oC) Cp(kJ/kmolK) Cpy(kJ/kmolK) (kJ/kmol)Benzene (A)Toluene (B)

80.1110.6

138.2167.5

96.3138.2

30,82033,330

Vapor pressure: Psat (mmHg), T(oC)

log10 PAsat = 6.90565 1211.033/(T + 220.79)

log10 PBsat = 6.95334 1343.943/(T + 219.377)

Solution -----------------------------------------------------------------------------------------

Making an overall and a component balance over the distillation column

D + B = 100

0.95D + 0.10B = 45 D = 41.18 kmol/h, B = 58.82 kmol/h

L0/D = 1.755 L0 = (1.755)(41.18) = 72.27 kmol/h

V1 = L0 + D = 72.27 + 41.18 = 113.45 kmol/h

The boiling point of the distillate D (and stream L0) is obtained from Table 4.6-1 and is 81.11oC.

----------------------------------------------------


4-61

The boiling point of the distillate with x = 0.95 can be obtained from the following Matlab codes:

clearA=[6.90565 6.95534];B =[1211.033 1343.943];C=[220.79 219.377];P=760;pl=log10(P);dT=.01;Tb=B./(A-pl)-C;x=0.95; % Mole fraction of benzene % Assume a temperature for the bubble point calculationT=x*Tb(1)+(1-x)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=x*10^(A(1)-B(1)/(T+C(1)))+(1-x)*10^(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=x*10^(A(1)-B(1)/(T1+C(1)))+(1-x)*10^(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endfprintf(' Bubble point T(C) = %8.2f\n',T)

>> e4d7d1a Bubble point T(C) = 81.11

----------------------------------------------------

For a total condenser, the mole fraction of benzene in the vapor stream V1, y1 = xD = 0.95. We can calculate the number of equilibrium stages in the rectifying section by the following procedure:

Step 1: The temperature of stage 1 can then be determined from the dew point calculation by the following Matlab codes.

clearA=[6.90565 6.95534];B =[1211.033 1343.943];C=[220.79 219.377];P=760;pl=log10(P);dT=.01;Tb=B./(A-pl)-C;x=0.95; % Mole fraction of benzene % Assume a temperature for the dew point calculationT=x*Tb(1)+(1-x)*Tb(2);% Solve for the dew point temperature using Newton's method for n=1:20; f=x*P/10^(A(1)-B(1)/(T+C(1)))+(1-x)*P/10^(A(2)-B(2)/(T+C(2)))-1; T1=T+dT;f1=x*P/10^(A(1)-B(1)/(T1+C(1)))+(1-x)*P/10^(A(2)-B(2)/(T1+C(2)))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endfprintf(' Dew point T(C) = %8.2f\n',T)

>> e4d7d1b Dew point T(C) = 82.57

The dew point temperature of the benzene and toluene mixture is 82.57oC, which is also the temperature of stage 1.

4-62

Step 2: At this temperature, the vapor pressure of benzene is

P1sat = 10^(6.90565 1211.033/(T + 220.79)) = 819.49 mmHg

The mole fraction of benzene in the liquid stream leaving stage 1 is then

x1 = = = 0.88104satPPy

1

1 (0.95)(760)819.49

Step 3: Next y2 is determined from the component mass balance

y2 = x1 + xD1

2

LV 2

DV

Since L1 and V2 are not known, y2 is estimated from

y2 = x1 + xD = 0.88104 + 0.95 = 0.906070

1

LV 1

DV

72.27113.45

41.18113.45

Step 4: V2 can be evaluated from the energy balance (4.7-7)

V2 = (H1V1 hDL0 h1D)/(H2 h1)

The enthalpies hD, h1, and H1 are obtained from the following relation

hA = CpA(T Tref) = 138.2(T 80.1)

hB = CpB(T Tref) = 167.5(T 80.1)

HA = A + CpyA(T Tref) = 30,820 + 96.3(T 80.1)

HB = B,Tref + CpyB(T Tref) = 34,224 + 138.2(T 80.1)

h = xhA + (1 x)hB

H = yHA + (1 y)HB

For hD, T = 81.11oC, x = 0.95, therefore hD = 141.46 kJ/kmol

For h1, T = 82.57oC, x = 0.88104, therefore h1 = 349.52 kJ/kmol

For H1, T = 82.57oC, y = 0.95, therefore H1 = 31232.93 kJ/kmol

For H2 temperature at stage 2 is required. This temperature can be estimated from a dew point calculation with y2 = 0.90607. The result is T2,estimated = 84.57oC. H2 = 31587.84 kJ/kmol. Substituting the numerical values,

4-63

V2 = (H1V1 hDL0 h1D)/(H2 h1)

V2 = (31232.93×113.45 141.46×72.27 349.52×41.18)/( 31587.84 141.46)

V2 = 112.64 kmol/h

Step 5: L1 is then evaluated from the overall mass balance

L1 = V2 D = 112.64 41.18 = 71.46 kmol/h

Step 6: Next y2 is determined from the component mass balance

y2 = x1 + xD = 0.88104 + 0.95 = 0.906251

2

LV 2

DV

71.46112.64

41.18112.64

Since the new value of y2 (0.90625) is closed to the estimated y2 (0.90607), this value is accepted as the mole fraction of benzene in the vapor stream leaving stage 2.

We can now go back to step 1 and determine the temperature at stage 2. A new dew point calculation with y2 = 0.90625 is performed with the result T2 = 84.56oC. Step 1-6 can be repeated until the mole fraction of benzene in the liquid stream leaving a stage is less than xf = 0.45. The following Matlab codes perform the required calculations:

---------------------------------------------% Example 4.7-1: Calculate the number of equilibrium stages in the rectifying section. clearA=[6.90565 6.95534];B =[1211.033 1343.943];C=[220.79 219.377];P=760;pl=log10(P);xf=.45;%CpA=138.2;CpB=167.5;CpyA=96.3;CpyB=138.2;dHA=30820;dHB=34224;%dT=.01;Tb=B./(A-pl)-C;x=0.95; % Mole fraction of benzene % Assume a temperature for the bubble point calculationT=x*Tb(1)+(1-x)*Tb(2);% Solve for the bubble point temperature using Newton's method for n=1:20; f=x*10^(A(1)-B(1)/(T+C(1)))+(1-x)*10^(A(2)-B(2)/(T+C(2)))-P; T1=T+dT;f1=x*10^(A(1)-B(1)/(T1+C(1)))+(1-x)*10^(A(2)-B(2)/(T1+C(2)))-P; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endfprintf(' Bubble point T(C) = %8.2f\n',T)%% Enthalpy of the liquid distillateTref=Tb(1);

4-64

hA=CpA*(T-Tref);hB=CpB*(T-Tref); hD=x*hA+(1-x)*hB; fprintf('hD(kJ/kmol) = %8.2f\n',hD) y1=x; % Mole fraction of benzene (A) in the vapor leaving the first tray % D = 41.18 ; Bot = 58.82 ; Reflux=1.755;xD=x; L0=Reflux*D;V1=L0+D; fprintf(' V1 = %8.2f, L0 = %8.2f\n',V1,L0) % % Evaluate the temperature at the first (top) equilibrium tray % % Assume a temperature for the dew point calculationT=y1*Tb(1)+(1-y1)*Tb(2);% Solve for the dew point temperature using Newton's method for n=1:20; f=y1*P/10^(A(1)-B(1)/(T+C(1)))+(1-y1)*P/10^(A(2)-B(2)/(T+C(2)))-1; T1=T+dT;f1=y1*P/10^(A(1)-B(1)/(T1+C(1)))+(1-y1)*P/10^(A(2)-B(2)/(T1+C(2)))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endfprintf(' First stage T(C) = %8.2f\n',T)PAsat=10^(6.90565 -1211.033/(T + 220.79));x1=y1*P/PAsat;y2=L0*x1/V1+D*xD/V1;fprintf('Stage 1, x1 = %8.5f, y2(estimated) = %8.5f\n',x1,y2)hA=CpA*(T-Tref);hB=CpB*(T-Tref);HA=dHA+CpyA*(T-Tref);HB=dHB+CpyB*(T-Tref); h1=x1*hA+(1-x1)*hB; H1=y1*HA+(1-y1)*HB;fprintf('h1(kJ/kmol) = %8.2f, H1(kJ/kmol) = %8.2f\n',h1,H1)% Evaluate the temperature at the second equilibrium tray % % Assume a temperature for the dew point calculation % for i=1:10 % Iteration for the material and energy balances at a stageT=y2*Tb(1)+(1-y2)*Tb(2);% Solve for the dew point temperature using Newton's method for n=1:20; f=y2*P/10^(A(1)-B(1)/(T+C(1)))+(1-y2)*P/10^(A(2)-B(2)/(T+C(2)))-1; T1=T+dT;f1=y2*P/10^(A(1)-B(1)/(T1+C(1)))+(1-y2)*P/10^(A(2)-B(2)/(T1+C(2)))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endHA=dHA+CpyA*(T-Tref);HB=dHB+CpyB*(T-Tref); H2=y2*HA+(1-y2)*HB; V2=(H1*V1-hD*L0-h1*D)/(H2-h1);

4-65

L1=V2-D; fprintf(' V2 = %8.2f, L1 = %8.2f\n',V2,L1) y2save=y2; y2=L1*x1/V2+D*xD/V2; if abs((y2-y2save)/y2)<.001, break, endend % End material and energy balance at a stagefprintf(' Estimated second stageT(C) = %8.2f, estimated H2(kJ/kmol) = %8.2f\n',T,H2)% fprintf('Stage 1, x1 = %8.5f, y2 = %8.5f\n\n',x1,y2)%% Continue iteration from the second stage%fprintf('Continue iteration from the second stage\n');yn=y2;Lnm1=L1;Vn=V2;for nstage=2:100% Estimate the temperature at stage nT=yn*Tb(1)+(1-yn)*Tb(2);% Solve for the temperature at stage n using Newton's method% This is a dew point calculation for n=1:20; f=yn*P/10^(A(1)-B(1)/(T+C(1)))+(1-yn)*P/10^(A(2)-B(2)/(T+C(2)))-1; T1=T+dT;f1=yn*P/10^(A(1)-B(1)/(T1+C(1)))+(1-yn)*P/10^(A(2)-B(2)/(T1+C(2)))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end endfprintf(' Stage n = %g, T(C) = %8.2f\n',nstage,T)PAsat=10^(6.90565 -1211.033/(T + 220.79));xn=yn*P/PAsat;if xn<xf, break, endynp1=Lnm1*xn/Vn+D*xD/Vn;fprintf('Stage n= %g, x(n) = %8.5f, estimated y(n+1) = %8.5f\n',nstage,xn,ynp1)hA=CpA*(T-Tref);hB=CpB*(T-Tref); hn=xn*hA+(1-xn)*hB; fprintf('hn(kJ/kmol) = %8.2f\n',hn)% Evaluate the temperature at the next equilibrium tray % for i=1:10 % Iteration for the material and energy balances at a stage % Assume a temperature for the dew point calculationT=ynp1*Tb(1)+(1-ynp1)*Tb(2);% Solve for the dew point temperature using Newton's method for n=1:20; f=ynp1*P/10^(A(1)-B(1)/(T+C(1)))+(1-ynp1)*P/10^(A(2)-B(2)/(T+C(2)))-1; T1=T+dT;f1=ynp1*P/10^(A(1)-B(1)/(T1+C(1)))+(1-ynp1)*P/10^(A(2)-B(2)/(T1+C(2)))-1; fp=(f1-f)/dT;eT=f/fp;T=T-eT; if abs(eT)<0.001,break, end end

4-66

HA=dHA+CpyA*(T-Tref);HB=dHB+CpyB*(T-Tref); Hnp1=ynp1*HA+(1-ynp1)*HB; Vnp1=(H1*V1-hD*L0-h1*D)/(Hnp1-h1); Ln=Vnp1-D; fprintf(' V(n+1) = %8.2f, L(n) = %8.2f\n',Vnp1,Ln) ynp1save=ynp1; ynp1=Ln*xn/Vnp1+D*xD/Vnp1; if abs((ynp1-ynp1save)/ynp1)<.001, break, end%end % End material and energy balances for a stage% fprintf('Stage = %g, estimated T(C) = %8.2f, estimated H(n+1)[kJ/kmol] = %8.2f\n',nstage,T,Hnp1)fprintf('Stage = %g, x(n) = %8.5f, y(n+1) = %8.5f\n\n',nstage,xn,ynp1)yn=ynp1;Lnm1=Ln;Vn=Vnp1;%end % End the calculation for the number of stage%fprintf('Number of equilibrium stages in enriching section = %g\n', nstage)

>> e4d7d1 Bubble point T(C) = 81.11hD(kJ/kmol) = 141.46 V1 = 113.45, L0 = 72.27 First stage T(C) = 82.57Stage 1, x1 = 0.88104, y2(estimated) = 0.90607h1(kJ/kmol) = 349.52, H1(kJ/kmol) = 31232.93 V2 = 112.64, L1 = 71.46 Estimated second stageT(C) = 84.57, estimated H2(kJ/kmol) = 31587.84Stage 1, x1 = 0.88104, y2 = 0.90625

Continue iteration from the second stage Stage n = 2, T(C) = 84.56Stage n= 2, x(n) = 0.79146, estimated y(n+1) = 0.84942hn(kJ/kmol) = 643.99 V(n+1) = 111.05, L(n) = 69.87Stage = 2, estimated T(C) = 86.97, estimated H(n+1)[kJ/kmol] = 32037.12Stage = 2, x(n) = 0.79146, y(n+1) = 0.85025

Stage n = 3, T(C) = 86.93Stage n= 3, x(n) = 0.69212, estimated y(n+1) = 0.78775hn(kJ/kmol) = 1005.86 V(n+1) = 109.39, L(n) = 68.21 V(n+1) = 109.42, L(n) = 68.24Stage = 3, estimated T(C) = 89.32, estimated H(n+1)[kJ/kmol] = 32506.67Stage = 3, x(n) = 0.69212, y(n+1) = 0.78917

Stage n = 4, T(C) = 89.32

4-67

Stage n= 4, x(n) = 0.59912, estimated y(n+1) = 0.73117hn(kJ/kmol) = 1382.41 V(n+1) = 107.93, L(n) = 66.75 V(n+1) = 107.97, L(n) = 66.79Stage = 4, estimated T(C) = 91.36, estimated H(n+1)[kJ/kmol] = 32939.32Stage = 4, x(n) = 0.59912, y(n+1) = 0.73294



Stage n = 7, T(C) = 93.91Number of equilibrium stages in enriching section = 7>>

Example 4.7-2. ----------------------------------------------------------------------------------Using the enthalpy-composition diagram for the methanol-water system (Figure 4.7-2), calculate the number of stages required and the feed plate location to produce 98 wt% methanol liquid product and 5 wt% methanol bottoms product from a feed of 45 wt% methanol entering the column at 50oC. The column has a total condenser and a partial reboiler, and operates at 1.5 times minimum reflux. What is the reboiler duty in Btu/hr if the feed rate is 5,000 lb/hr? What is the condenser duty?

Solution -----------------------------------------------------------------------------------------

The feed point F is located on the liquid curve at 50oC and x = 45 wt% methanol. The reflux ratio R = L0/D can be determined by making material and energy balances around the condenser.

V1 = L0 + D V1H1 = (L0 + D)H1 = L0hD + DhD + QC

Rearranging,

L0(H1 hD) = D(hD + QC/D - H1)

4-68

or R = = 0LD

1

1

/D C

D

h Q D HH h

Figure 4.7-2 Hxy diagram for methanol-water system at 1 atm.

Draw a tie line through the feed point F. This tie line intersects the vertical line at xD = 0.98 corresponding to the point H’min = (hD + QC/D)min = 540 cal/g.

Rmin = = = 0.96 R = 1.5Rmin = 1.44 1min

1

/D C

D

h Q D HH h

540 300300 50

H’ = (hD + QC/D) = H1 + 1.44(H1 hD) = 300 + 1.44(300 50) = 660 cal/g

QC/D = 660 hD = 660 50 = 610 cal/g

4-69

Chapter 4

F

H'min

H'

h'WFigure E-1Enthalpy-concentration diagram for methanol-water system

The line H’F intersects the vertical line at xB = 0.05 mass fraction corresponding to the lower operating point h’W.

From the chart h’W = hW + QR/W = 450 cal/g QR/W = hW h’W = 90 + 450 = 540 cal/g

4-70

Making overall and methanol balances we obtain

D + W = 5,000

0.98D + 0.05W = (0.45)(5,000) D = 2,150 lb/hr and W = 2,850 lb/hr

The condenser duty is then

QC = 610D = (610)(2,150)(454)(3.986×10-3 Btu/cal) = 2.37×106 Btu/hr

The reboiler duty is

QR = 540W= (540)(2,850)(454)(3.986×10-3 Btu/cal) = 2.77×106 Btu/hr

To find the number of equilibrium stages we start at the saturated vapor curve at xD and draw a tie line to intersect the saturated liquid curve at point 1 (See Figure E-2 for illustration). Connect point 1 to the upper operating point D’ (or H’ from Figure E-1) then draw a tie line to intersect the saturated liquid curve at point 2. Continue this process until you reach point 5. At this location, the tie line crosses the feed line D’W’ (or H’h’W from Figure E-1). Connect point 5 with the lower operating point W’ (or h’W from Figure E-1) to intersect the saturated vapor curve. From this point, draw a tie line to intersect the saturated liquid curve at point 6. Continue this process until the liquid composition becomes less than or equal to the composition of the liquid leaving at the bottoms, xW.

If you apply this process to the enthalpy concentration diagram of the methanol and water (Figure E-1) you will obtain 9 equilibrium stages or 8 equilibrium trays and a reboiler.

F

D'

D12345678

H or h

W

W'

xW xF xD

QR

QC

F, hF, xF

D, hD, xD

W, hW, xW

V1,H1

Lo

Figure E-2 A typical enthalpy-concentration diagram

4-71

Example 4.7-3. ----------------------------------------------------------------------------------For the methanol-water system with given data at 1 atm, calculate the number of stages required and the feed plate location to produce 95 mol% methanol liquid product and 4 mol% methanol bottoms product from a feed of 45 mol% methanol entering the column as saturated liquid. The column has a total condenser and a partial reboiler, and operates at 1.6 times minimum reflux. Data:

Mol % x or y Saturated Hv Saturated hl Mol % x Mol % yBtu/lbmol Btu/lbmol 0 0

0 20,720 3,240 2.0 13.45 20,520 3,070 4.0 23.010 20,340 2,950 6.0 30.415 20,160 2,850 8.0 36.520 20,000 2,760 10.0 41.830 19,640 2,620 15.0 51.740 19,310 2,540 20.0 57.950 18,970 2,470 30.0 66.560 18,650 2,410 40.0 72.970 18,310 2,370 50.0 77.980 17,980 2,330 60.0 82.590 17,680 2,290 70.0 87.0100 17,390 2,250 80.0 91.5

90.0 95.895.0 97.9100.0 100.0

Solution -----------------------------------------------------------------------------------------

This problem can be solved graphically similar to problem 4.7-2. The graph can be obtained from the following Matlab codes:

% Methanol-water enthalpy concentration diagramxy=[0 5 10 15 20 30 40 50 60 70 80 90 100];% Saturated vapor enthalpy at xy valuesHv=[20720 20520 20340 20160 20000 19640 19310 18970 18650 18310 17980 17680 17390];% Saturated liquid enthalpy at xy valueshl=[3240 3070 2950 2850 2760 2620 2540 2470 2410 2370 2330 2290 2250];% Equilibrium datax=[0 2 4 6 8 10 15 20 30 40 50 60 70 80 90 95 100];y=[0 13.4 23 30.4 36.5 41.8 51.7 57.9 66.5 72.9 77.9 82.5 87 91.5 95.8 97.9 100];% Spline fitting to determine tie linextoy=spline(x,y);ytox=spline(y,x);% Fit the saturated vapor curve by a second order polynomialVapor=polyfit(xy,Hv,2);v1=Vapor(1);v2=Vapor(2);v3=Vapor(3);% Fit the saturated liquid curve by a second order polynomialLiquid=polyfit(xy,hl,2);a1=Liquid(1);a2=Liquid(2);a3=Liquid(3);

4-72

% Problem specifications with saturated liquid feed at xf, Reflux ratio = 1.6RminxB=4;RRatio=1.6;xf=45;xD=95;%% Determine the enthalpy of the saturated liquid feedhf=a1*xf^2+a2*xf+a3;% Determine the mole fraction of the vapor in equilibrium with the liquid feedyf=ppval(xtoy,xf);Hvf=v1*yf^2+v2*yf+v3;% Determine the enthalpy Hmin of the top operating point at minimum reflux% This is the intersection of the tie line through the feed point and the vertical line at xDht=Hvf+(Hvf-hf)*(xD-yf)/(yf-xf);Hmin=ht;dy=(ymax-ymin)/20;% Evaluate the enthalpies of saturated liquid hD and saturated vapor H1 at mole percent xDhD=a1*xD^2+a2*xD+a3;H1=v1*xD^2+v2*xD+v3;% Determine minimum reflux ratio RminRmin=(Hmin-H1)/(H1-hD);Reflux=RRatio*Rmin;% Determine enthalpy of the top operating point Htop at operating reflux ratioHtop=H1+Reflux*(H1-hD);% Determine enthalpy of the lower operating point hbot at operating reflux ratiohbot=hf+(hf-Htop)*(xB-xf)/(xf-xD);ytop=ymax-dy;ybot=ymin+dy;xmin=0;xmax=100;ymin=-30000;ymax=50000;% Plot the saturated vapor and liquid enthalpy, the tie line through the feed point,% and the line connecting the top and lower operating pointsplot(xy,Hv,xy,hl,[xf xD],[hf Hmin],'--',[xB xD],[hbot Htop],'--')axis([xmin xmax ymin ymax])xlabel('x or y (mol%)');ylabel('Enthalpy (Btu/lbmol)')grid on% Plot the vertical lines at xD and xBline([xD xD],[ybot ytop]);line([xB xB],[ybot ytop]);yi=xD;Hvi=H1;% Determine the number of equilibrium stages in the enriching section% starting from the vapor leaving the first equilibrium stagefor i=1:50; % Determine the composition of the liquid stream leaving the equilibrium stage xi=ppval(ytox,yi); % Determine the enthalpy of the liquid stream leaving the equilibrium stage hLi=a1*xi^2+a2*xi+a3; % Plot the tie line line([xi yi],[hLi Hvi]); % Determine the slope (a) and intercept (b) of the top operating line a=(Htop-hLi)/(xD-xi); b=hLi-a*xi;

4-73

% Solve for the intersection of the top operating line and the saturated vapor curve for k=1:20 fy=v1*yi^2+v2*yi+v3-(a*yi+b); ff=2*v1*yi+v2-a; ey=fy/ff; yi=yi-ey; if abs(ey)<0.0001,break,end end % Determine enthalpy of vapor stream Hvi=v1*yi^2+v2*yi+v3; % If xi<xf need to exit the loop to use the lower operating line if xi<xf, break, end % Plot the top operating line line([xi xD],[hLi Htop]);end % for n=i:100; % Determine the slope (a) and intercept (b) of the lower operating line a=(hLi-hbot)/(xi-xB); b=hbot-a*xB; % Solve for the intersection of the lower operating line and the saturated vapor curve for k=1:20 fy=v1*yi^2+v2*yi+v3-(a*yi+b); ff=2*v1*yi+v2-a; ey=fy/ff; yi=yi-ey; if abs(ey)<0.0001,break,end end Hvi=v1*yi^2+v2*yi+v3; % Plot the lower operating line line([xB yi],[hbot Hvi]); xsave=xi; % Determine the composition of the liquid stream leaving the equilibrium stage xi=ppval(ytox,yi); hLi=a1*xi^2+a2*xi+a3; % Plot the tie line line([xi yi],[hLi Hvi]); % If xi<xB need to exit the loop if xi<xB, break, end end % Determine fraction of equilibrium stage fstage=(xsave-xB)/(xsave-xi); nstage=n+fstage; fprintf('# of stage = %8.3f\n',nstage)

4-74

Figure E-1 Enthalpy-concentration diagram for methanol-water system

>> methanolw# of stage = 8.328

5-1

Chapter 5 Approximate Methods for Multi-component Distillation

5.1 Introduction

Software is available to solve multi-component distillation problems by rigorous methods. However approximate methods continue to be used in practice for various results, including preliminary design, parametric studies to establish optimal design conditions, and for obtaining an initial approximation for a rigorous method. The general principles of design for multi-component distillation column are very similar to those described for binary systems. Material, enthalpy, and equilibrium relations are used to solve the problem. The concepts of minimum reflux and total reflux as limiting cases are also used.

For binary distillation we can start out by solving the external balances first and then do the internal stage-by-stage calculations. Multi-component distillation is more complex than binary distillation since the external mass and energy balances might be couple with the internal stage-by-stage calculations.

F, hF, xF

V1, H1 QC

L0 D, hD, xD

B, hB, xB QR

For a distillation column, degrees of freedom = C + 6Binary distillation C = 2 8 degrees of freedomIn a design problem, you can specify F (feed rate) xF (feed composition) q (feed quality) xD (distillate composition) Distillate temperature (saturated liquid) xB (bottoms composition) L0/D (external reflux ratio) Optimum feed plate

With these eight variables chosen, the problem is completely defined, and we can solve for the number of equilibrium stages.

Figure 5.1-1 Specified variables for binary distillation column.

Consider a conventional distillation column with a total condenser and a partial reboiler. If the column is adiabatic with constant molal overflow and constant pressure, it will have C + 6 degrees of freedom, where C is the number of components. For binary distillation this is 8 degrees of freedom. Figure 5.1-1 shows the usual specified variables for a design problem. With these variables defined, D and B can be solved from the external mass balance and QC and QR can be solved from the energy balance. Then the number of equilibrium stages can be solved from the stage-by-stage calculations (internal balances). The procedure is straight forward with no iteration.

5-2

For ternary distillation the degrees of freedom is 9. Figure 5.1-2 and Table 5.1-1 show the usual specified variables for the design of a ternary distillation column. Figure 5.1-2 is similar to that of a binary distillation except that the extra degree of freedom is used to completely specify the feed. Note that the compositions of the product streams D and B are not completely specified.

F, hF, xF

V1, H1 QC

L0 D, hD, xD

B, hB, xBQR

x x1F 2F,

x1D

x2B

Figure 5.1-2 Specified variables for ternary distillation column.(Note: x1F and x2F instead of xF, x1D instead of xD, and x2B instead of xB)

Table 5.1-1 Specified variables for ternary distillation column.For a distillation column, degrees of freedom = C + 6Ternary distillation C = 3 9 degrees of freedomIn a design problem, you can specify F (feed rate) x1F and x2F (feed composition) q (feed quality) x1D (distillate composition or one fractional recovery) Distillate temperature (saturated liquid) x2B (bottoms composition or one fractional recovery) L0/D (external reflux ratio) Optimum feed plate

With these nine variables chosen, the problem is completely defined, and we can solve for the number of equilibrium stages.

The external material and energy balances for the ternary distillation column are

F = D + B (5.1-1)

5-3

x1FF = x1DD + x1BB (5.1-2)

x2FF = x2DD + x2BB (5.1-3)

hFF + QR = hDD + hBB + QC (5.1-4)

There are four unknowns D, B, x2D, and x1BB and there are three independent mass balance equations. The additional energy equation adds additional variables QR and QC. So we cannot start the calculation with the external mass and energy balances. This is the first major different between binary and ternary (or multi-component) distillation.

For a mixture of A, B, and C, a separation in one column can only be made between A and B or B and C. The components separated are called the light key, which is the more volatile (identified by subscript LK), and the heavy key (HK). The components more volatile than the light key are call light components or light non-keys and the components less volatile than the heavy key are called the heavy components or heavy non-keys. The light and heavy components are the non-key components.

For ternary distillation, the external mass and energy balances can be solved if one additional composition is assumed. The internal stage-by-stage calculation can then be started from one end of the column to the other end. The results from the internal balance can then be compared with that from the external balance and the assumed composition adjusted. Thus the calculations for multi-component problems are trial-and-error. This is the second major difference between binary and multi-component distillations.

Fortunately, it is simple to make an excellent first guess in many cases depending on the separation of the key components. These are the components that do have their distillate and bottoms compositions or fractional recoveries specified (such as component 1 in the distillate and component 2 in the bottoms in Figure 5.1-2. If a sharp separation of the keys is required, then almost all of the heavy components will appear in the bottoms, and almost all of the light components will appear only in the distillate. For these cases we can complete the external mass balances by assuming that all light components appear only in the distillate and all heavy components appear only in the bottoms. Thus

xL,bot = 0 and xH,dist = 0.

5.2 Fenske-Underwood-Gilliland Method

The Fenske-Underwood-Gilliland or FUG method is commonly used for preliminary design and optimization of simple distillation1. The steps in the FUG method are

1) Specify feed and splits of the two key components;2) Estimate splits of the non-key components;3) Determine column pressure and type of condenser using bubble and dew point

calculations;4) Flash the feed adiabatically at column pressure;5) Calculate the minimum number of equilibrium stages using Fenske equation;

1 J. D. Seader and E. J. Henley, Separation Process Principles, , Wiley, 2006, pg. 344

5-4

6) Calculate splits of non-key components and repeat steps 3-6 if estimated and calculated splits of non-key components differ considerably;

7) Calculate minimum reflux ratio using Underwood equations;8) Calculate actual equilibrium stages for specified reflux ratio using Gilliland

correlation;9) Calculate feed stage location using Kirkbride equation;10) Calculate condenser and reboiler duties using energy balances.

For each single-phase stream containing C components, a complete specification of intensive variables consists of C mole fraction plus temperature and pressure, or C + 2 variables. Even thought the summation of the mole fraction is equal to one, it is preferable to include all the mole fractions in the list of stream variable and then to include in the list of equations the mole fraction constraint. The total flow rate is needed to completely specify the stream. Thus associated with each stream are C + 3 variables.

Equilibrium stageQ

Ln-1

Ln

Vn

Vn+1

Figure 5.2-1 Equilibrium stage with heat addition.

For a single equilibrium stage with possible heat transfer Q, two entering streams, and two exit streams, as shown in Figure 5.2-1, the variables are those associated with the four streams plus the heat transfer rate. Thus the number of variables NV is

NV = 4(C + 3) + 1 = 4C + 13

For an equilibrium stage, the two exit streams are in equilibrium, so there are equilibrium restrictions as well as material balances for C components, an energy balance, and mole fraction constraints. Thus the equations, NE, relating NV variables are

Equations Number of Eq.Pressure equality: PVn = PLn 1Temperature equality: TVn = TLn 1Phase equilibrium relationships: (yi)n = Ki(xi)n CComponent material balances: Ln-1(xi)n-1 + Vn+1(yi)n+1 = Ln(xi)n + Vn(yi)n CEnergy balance: Q + Ln-1hn-1 + Vn+1Hn+1 = Lnhn + VnHn 1

Mole fraction constraints in streams: ; 1

1C

ii

x

1

1C

ii

y

4

Total equations NE = 2C + 7

The degree of freedom, F, for an equilibrium stage is then

5-5

F = NV NE = (4C + 13) (2C + 7) = 2C + 6

Table 5.2-1 lists the degree of freedom for some common operation units. For a distillation column with a total condenser, a partial reboiler, one inlet stream, and two product (exit) streams, the number of degrees of freedom is 2N + C + 9. Table 5.2-2 lists the common specifications for a distillation column.

Table 5.2-1 Degree of freedom for separation unitsUnit NV NE F

Total condenser 2C + 7 C + 3 C + 4Partial reboiler 3C + 10 2C + 6 C + 4Adiabatic equilibrium stage

4C + 12 2C + 7 2C + 5

Equilibrium stage with feed

5C + 15 2C + 8 3C + 7

N-connected adiabatic equilibrium stages

6N + 2NC + 2C + 7 4N + 2NC + 2 2N + 2C + 5

Stream divider 3C + 9 2C + 4 C + 5

For the N-connected adiabatic equilibrium stages, the number of variables is

NV = N(4C + 12) 2(N 1)(C + 3) + 1 = 6N + 2NC + 2C + 7

The first term {N(4C + 12)} accounts for the variables in each equilibrium stages, the second term {2(N 1)(C + 3)} is the redundant variables since N equilibrium stages are connected by 2(N 1) streams, and the last term {1} is the number of stages. The number of independent relationships for N stages needs to subtract the redundant mole-fraction constraints.

NE = N(2C + 7) 2(N 1)

Table 5.2-2 Common specifications for distillation columnVariables or constraints Number of specifications

Feed flow rateFeed mole fractionsFeed temperatureFeed pressureAdiabatic stages (excluding reboiler)Stage pressures (including reboiler)Split of light key componentSplit of heavy key componentFeed-stage locationReflux ratioReflux temperatureAdiabatic reflux dividerPressure of total condenserPressure at reflux divider

1C 1

11

N 1N11111111

Total 2N + C + 9

5-6

Example 5.2-1. ----------------------------------------------------------------------------------We wish to distill 2000 kmol/hr of a saturated liquid feed. The feed is 0.056 mole fraction propane, 0.321 n-butane, 0.482 n-pentane, and the remainder n-hexane. The column operates at 101.3 kPa with a total condenser and a partial reboiler. Reflux ratio is L0/D = 3.5, and reflux is a saturated liquid. A fractional recovery of 99.4% n-butane is desired for the distillate and 99.7% n-pentane recovery for the bottoms. Calculates the followings:

(a) Distillate and bottoms compositions and flow rates;(b) Temperature (dew point) of distillate and boiling point of bottoms;(c) Minimum stages for total reflux and distribution of other components in the distillate

and bottoms;(d) Calculate minimum reflux ratio using Underwood equations;(e) Calculate actual equilibrium stages for specified reflux ratio using Gilliland

correlation;(f) Calculate feed stage location using Kirkbride equation;(g) Calculate condenser and reboiler duties using energy balances.

Solution -----------------------------------------------------------------------------------------

(a) Distillate and bottoms compositions and flow rates

There are 10 unknowns: D, B, xi,dist(4), and xi,bot(4) and there are 10 equations to solve for these unknowns.

Assume xL,bot = 0 xC3,bot = 0 (E-1)

Assume xH,dis = 0 xC6,dist = 0 (E-2)

Overall balance 2,000 = B + D (E-3)

C3 balance (0.056)(2,000) = xC3,distD (E-4)

C4 balance, top (0.994)(0.321)(2,000) = xC4,distD (E-5)

C4 balance, bottoms (0.006)(0.321)(2,000) = xC4,botB (E-6)

C5 balance, top (0.003)(0.482)(2,000) = xC5,distD (E-7)

C5 balance, bottoms (0.997)(0.482)(2,000) = xC5,botB (E-8)

D = (E-9),i distDx

B = (E-10),i botBx

From Eq. (E-9) we have

D = xC3,distD + xC4,distD + xC5,distD + xC6,distD

D = (0.056)(2,000) + (0.994)(0.321)(2,000) + (0.003)(0.482)(2,000) + 0

5-7

D = 112 + 638.15 + 2.892 = 753.04 kmol/hr

xi,dist = ,i distDxD

xC3,dist = 112/753.04 = 0.14873

xC4,dist = 638.15/753.04 = 0.84743

xC5,dist = 2.892/753.04 = 0.00384

xC6,dist = 0/753.04 = 0.0

B = 2,000 D = 1247 kmol/hr

xi,bot = ,i botBxB

xC3,bot = 0/1247 = 0.0

xC4,bot = (0.006)(0.321)(2,000)/1247 = 0.00309

xC5,bot = (0.997)(0.482)(2,000)/1247 = 0.77076

xC6,bot = (0.141)(2,000)/1247 = 0.22615

(b) Temperature (dew point) of distillate and boiling point of bottoms;

Dew point of distillate can be obtained from the following Matlab codes:

---------------------------------------% Dew point temperature calculationP=14.69; % psiaA=[970688.6 1280577 1524891 1778901]; B=[7.15059 7.94986 7.33129 6.96783]; C=[0.76984 0.96455 0.89142 0.84634];D=[6.90224 0 0 0];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)+D/P));zi=[0.14873 0.84743 0.00384 0];yi=zi;Te=yi*Tsat'; % Estimate temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P)';dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=sum(yi./Ki)-1; Tsav=T;T=T+dT;

5-8

Ki=eval(fK);fdT=sum(yi./Ki)-1; eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;TK=T/1.8;fprintf('Dew point temperature, T(R) = %8.2f, T(K) = %8.2f\n',T,TK)Ki=eval(fK);alfai=Ki/Ki(3);fprintf('C3 KC3 = %8.5f, alfa_C3 = %8.4f\n',Ki(1),alfai(1))fprintf('C4 KC4 = %8.5f, alfa_C4 = %8.4f\n',Ki(2),alfai(2))fprintf('C5 KC5 = %8.5f, alfa_C5 = %8.4f\n',Ki(3),alfai(3))fprintf('C6 KC3 = %8.5f, alfa_C6 = %8.4f\n',Ki(4),alfai(4))

>> e5d2d1_dewDew point temperature, T(R) = 483.98, T(K) = 268.88C3 KC3 = 4.08664, alfa_C3 = 19.7243C4 KC4 = 0.89668, alfa_C4 = 4.3279C5 KC5 = 0.20719, alfa_C5 = 1.0000C6 KC3 = 0.05498, alfa_C6 = 0.2653---------------------------------------

Boiling point of bottoms can be obtained from the following Matlab codes:

---------------------------------------% Bubble point temperature calculationP=14.69; % psiaA=[970688.6 1280577 1524891 1778901]; B=[7.15059 7.94986 7.33129 6.96783]; C=[0.76984 0.96455 0.89142 0.84634];D=[6.90224 0 0 0];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)+D/P));zi=[0 0.00309 0.77076 0.22615];xi=zi;Te=xi*Tsat'; % Estimate temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P)';dT=1;eT=1;dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=log(xi*Ki'); Tsav=T;T=T+dT; Ki=eval(fK);fdT=log(xi*Ki'); eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;TK=T/1.8;fprintf('Bubble point temperature, T(R) = %8.2f, T(K) = %8.2f\n',T,TK)Ki=eval(fK);alfai=Ki/Ki(3);fprintf('C3 KC3 = %8.5f, alfa_C3 = %8.4f\n',Ki(1),alfai(1))fprintf('C4 KC4 = %8.5f, alfa_C4 = %8.4f\n',Ki(2),alfai(2))

5-9

fprintf('C5 KC5 = %8.5f, alfa_C5 = %8.4f\n',Ki(3),alfai(3))fprintf('C6 KC3 = %8.5f, alfa_C6 = %8.4f\n',Ki(4),alfai(4))

>> e5d2d1_bubbleBubble point temperature, T(R) = 564.45, T(K) = 313.59C3 KC3 = 12.24498, alfa_C3 = 10.5414C4 KC4 = 3.81400, alfa_C4 = 3.2834C5 KC5 = 1.16161, alfa_C5 = 1.0000C6 KC3 = 0.41075, alfa_C6 = 0.3536---------------------------------------

(c) Minimum stages for total reflux and distribution of other components in the distillate and bottoms;

Minimum stages for total reflux is given by

Nmin = ,

FR of LK FR of HKlog1 - FR of HK 1 - FR of LK

log L ave

In this equation, FR of LK = fractional recovery of light key and FR of HK = fractional recovery of heavy key.

L,ave = (L,distL,bot)0.5 = (4.3279×3.2834)0.5 = 3.7696

Nmin = = 8.23

0.994 0.997log1 - 0.997 1 - 0.994

log 3.7696

The estimation of other components in the distillate and bottoms at total reflux is given by the following formula

= (i,ave)Nmin (E-5)id

ib

x Dx B

hd

hb

x Dx B

In this equation, xhd is the mole fraction of heavy key in the distillate, xhb is the mole fraction of heavy key in the bottoms

For nC6

= (nC6,ave)Nmin6,

6,

nC d

nC b

x Dx B

hd

hb

x Dx B

nC6,ave = = (0.2653×0.3536)0.5 = 0.30636 6nC d nC b

5-10

= (0.3063)8.23 = 1.776×10-76,

6,

nC d

nC b

x Dx B

2.892(0.997)(0.482)(2,000)

xnC6,bB = 5.631×106 xnC6,dD

xnC6,dD + xnC6,bB = (1 0.056 0.321 0.482)(2,000) = 282

xnC6,dD = 282/5.631×106 = 5.008×10-5

xnC6,d = 5.008×10-5/753.04 = 6.65×10-8

For nC3

= (nC3,ave)Nmin3,

3,

nC d

nC b

x Dx B

hd

hb

x Dx B

nC3,ave = = (19.7243×10.5414)0.5 = 14.423 3nC d nC b

= (14.42)8.23 = 1.0392×1073,

3,

nC d

nC b

x Dx B

2.892(0.997)(0.482)(2,000)

xnC3,dD = 1.0392×107 xnC3,bB

xnC3,dD + xnC3,bB = (0.056)(2,000) = 112

xnC3,bB = 112/1.0392×107 = 1.0777×10-5

xnC3,b = 1.0777×10-5/1247 = 8.64×10-9

We can now update the distribution of other components in the distillate and bottoms.

Table E-1 Product distributionDistillate xd Bottoms xb

C3nC4, light key nC5, heavy keynC6

112638.152.892

5.008×10-5

0.148730.84743 0.00384

6.65×10-8

1.0777×10-5

3.852961.11

282

8.64×10-9

0.00309 0.77076 0.22615

753.04 1.0000 1247 1.0000

5-11

Chapter 5

(d) Calculate minimum reflux ratio using Underwood equations

The two equations to be solved in order to determine the minimum reflux ratio are

1 q = (d-1)i iF

i

x

Rmin + 1 = (d-2)i iD

i

x

The values of xi,dist for each component in the distillate in Eq. (d-1) are supposed to be the values at the minimum reflux. However as an approximation, the values obtained from part (a) are used. Since I is a function of temperature, the average value i = (i,disti,bot)0.5 will be used for the calculation.

xi,F i,dist i,bot i = (i,disti,bot)0.5

0.056 19.7243 10.5414 14.41950.321 4.3279 3.2834 3.76960.482 1.0 1.0 1.00.141 0.3536 0.2653 0.3063

To solve for Rmin, the value of in Eq. (d-1) is first obtained by trial and error. Since feed is saturated liquid, q = 1and Equation (d-1) becomes:

= 0 (d-3)i iF

i

x

Equation (d-3) can be solved by the following Matlab codes:

>> fun=inline('14.4195*.056/(14.4195-t)+3.7696*.321/(3.7696-t)+.482/(1-t)+0.3063*.141/(0.3063-t)');

>> theta=fsolve(fun,2.5,optimset('Display','off'))

theta = 1.7583


Rmin = 1 i iD

i

x

Rmin = + + 1 = 0.7526(14.4195)(0.14873)14.4195 1.7583

(3.7696)(0.84743)3.7696 1.7583

0.003841 1.7583

5-12

(e) Calculate actual equilibrium stages for specified reflux ratio using Gilliland correlation

Gilliland correlation can be obtained from the equation developed by Molokanov et al [2].

Y = = 1 exp , where X = min

1N N

N

0.5

1 54.4 111 117.2

X XX X

min

1R R

R

X = = 0.61053.5 0.75263.5 1

Y = 1 exp = 0.1866 0.5

1 54.4 (0.6105) (0.6105) 111 (117.2)(0.6105) (0.6105)

Number of equilibrium stages at reflux ratio of 3.5 is then

N = = = 10.35 = 11min

1N Y

Y

8.23 0.1866

1 0.1866

(f) Calculate feed stage location using Kirkbride equation;

The feed stage location can be estimated from the following correlation

= e

s

NN

0.2062

, ,

, ,

H F L B

L F H D

x xBx D x

In this equation, Ne is the number of equilibrium stages above the feed plate and Ns is the number of equilibrium stages below the feed plate.

= = 1.1031e

s

NN

0.2062.482 1247 0.00309.321 753 0.00384

Ne + Ns = 11 1.1031Ns + Ns = 11 Ns = 5

Ne = 6

The feed tray is 6 trays from the top.

(e) Calculate condenser and reboiler duties using energy balances.

5-12

The duty of a total condenser might be estimated from the heat of condensation at the temperature of the distillate which is about 269 K or 4oC. Table e-1 lists the heat of condensation at 4oC. These values are obtained from Prop4 program by T.K.

Table e-1 Heat of condensation at 4oC.Propane n-butane n-pentane n-hexane

hevap(kJ/kmol) 1.665×104 2.261×104 2.820×104 3.308×104

hevap,mixture = 0.14873×1.665×104 + 0.84743×2.261×104 + 0.00384×2.820×104

hevap,mixture = 2.1745×104 kJ/kmol

Qc = (L0 + D)hevap,mixture = (R + 1)Dhevap,mixture = (4.5)(753)( 2.1745×104)

Qc = 7.37×107 kJ/hr

The reboiler duty can be obtained from the enthalpy balance

hFF + QR = hDD + hBB + QC QR = hDD + hBB + QC hFF

Feed is saturated liquid so its temperature can be obtained from a bubble point calculation. The following Matlab codes determine the boiling point of the feed.

% Bubble point temperature calculationP=14.69; % psiaA=[970688.6 1280577 1524891 1778901]; B=[7.15059 7.94986 7.33129 6.96783]; C=[0.76984 0.96455 0.89142 0.84634];D=[6.90224 0 0 0];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)+D/P));zi=[0.056 0.321 0.482 0.141];xi=zi;Te=xi*Tsat'; % Estimate temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P)';dT=1;eT=1;dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=log(xi*Ki'); Tsav=T;T=T+dT; Ki=eval(fK);fdT=log(xi*Ki'); eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;TK=T/1.8;fprintf('Bubble point temperature, T(R) = %8.2f, T(K) = %8.2f\n',T,TK)Ki=eval(fK);alfai=Ki/Ki(3);

5-13

fprintf('C3 KC3 = %8.5f, alfa_C3 = %8.4f\n',Ki(1),alfai(1))fprintf('C4 KC4 = %8.5f, alfa_C4 = %8.4f\n',Ki(2),alfai(2))fprintf('C5 KC5 = %8.5f, alfa_C5 = %8.4f\n',Ki(3),alfai(3))fprintf('C6 KC3 = %8.5f, alfa_C6 = %8.4f\n',Ki(4),alfai(4))

>> e5d2d1_feedBubble point temperature, T(R) = 507.46, T(K) = 281.92C3 KC3 = 5.94395, alfa_C3 = 15.9259C4 KC4 = 1.46992, alfa_C4 = 3.9384C5 KC5 = 0.37322, alfa_C5 = 1.0000C6 KC3 = 0.10923, alfa_C6 = 0.2927

Table e-2 lists the enthalpies of the feed, distillate, and bottoms streams. Temperature at the distillate (-4oC) is the reference point. The enthalpies are obtained from Prop4 program by T.K with the assumption of ideal mixture. The inputs require temperature and mole fractions of the chemical species.

Table e-2 Enthalpies of feed, distillate, and bottoms streams.Feed at 9oC Distillate at - 4oC Bottom at 40oC

xC3 0.056 0.14873 0xC4 0.321 0.84743 0.00309xC5 0.482 0.00384 0.77076xC6 0.141 0 0.22615h(kJ/kmol) 1,999 0 7,576Rate (kmol/hr) 2,000 753 1,247

The reboiler duty is then

QR = hDD + hBB + QC hFF

QR = (0)(753) + (7,576)(1,247) + 7.37×107 (1,999)(2,000)

QR = 7.91×107 kJ/hr

5-15

Chapter 5

5.3 Underwood Equation for Minimum Reflux

For binary distillation at minimum reflux, most of the stages are crowded into a constant-composition zone that bridges the feed stage. In this zone, all liquid and vapor streams have compositions essentially identical to those of flashed feed. This zone constitutes a single pinch point, as shown in Figure 5.3-1 for case 1. If non-ideal phase conditions cause a point of tangency between equilibrium curve and the operating line in the rectifying section, as shown in Figure 5.3-1 for case 2, the pinch point will occur in the rectifying section. The single pinch point can also occur in the stripping section.

Figure 5.3-1 The pinch points for minimum reflux.

Consider the case where the pinch point is in the rectifying section at or away from the feed stage as shown in Figure 5.3-2. The mass balance for component i around the top portion of the rectifying section as illustrated in Figure 5.3-2 is

yi,n+1Vmin = x i,nLmin + x i,DD (5.3-1)

At the pinch point where the compositions do not change,

x i,n-1 = x i,n = x i,n+1 and y i,n-1 = y i,n = y i,n+1 (5.3-2)

The equilibrium expression can be written as

5-16

v1 QC

L0 D, xi,D

x0 y1

x1 y2

x2 y3

L , xmin i,n V , ymin i,n+1

Pinchpointzone

Figure 5.3-2 Rectifying-section pinch-point zone

y i,n+1 = Ki x i,n-1 (5.3-3)

Substituting x i,n = x i,n+1 = y i,n+1/ Ki into Eq. (5.3-1) gives

yi,n+1Vmin = y i,n+1 + x i,DD (5.3-4)min

i

LK

Defining the relative volatility i = , Eq. (5.3-4) becomesi

HK

KK

yi,n+1Vmin = y i,n+1 + x i,DD (5.3-5)min

i HK

LK

Rearranging Eq. (5.3-5) yields

yi,n+1Vmin = x i,DD (5.3-6)min

min

1i HK

LV K

Solving for yi,n+1Vmin gives

yi,n+1Vmin = (5.3-7),

min

min

i i D

iHK

DxL

V K

5-17

The total vapor flow in the enriching section at minimum flux is obtained by summing over all components:

Vmin = = (5.3-8), 1 mini ni

y V ,

min

min

i i D

ii

HK

DxL

V K

Similarly, in the stripping section we have

Vst,min = (5.3-9), ,

,min,

,min ,

i st i B

stii st

st HK st

BxL

V K

Defining = and st = , Eqs. (5.3-8) and (5.3-9) becomemin

min HK

LV K

,min

,min ,

st

st HK st

LV K

Vmin = and Vst,min = (5.3-10),i i D

i i

Dx , ,

,

i st i B

i i st st

Bx

Assuming constant molar overflow and constant relative volatilities and adding both equations in (5.3-10) we have

Vfeed = Vmin Vst,min = + (5.3-11),i i D

i i

Dx ,i i B

i i

Bx

From the overall column mass balance ziF = x i,DD + x i,BB, Eq. (5.3-11) becomes

Vfeed = (5.3-12)i i

i i

Fz

If the ratio of saturated liquid over the total feed, q, is known we have

Vfeed = F(1 q) = (5.3-13)i i

i i

Fz

or

1 q = (5.3-14)i i

i i

Fz

Eq. (5.3-14) is known as the first Underwood equation, which can be used to calculate . Equation (5.3-10) is known as the second Underwood equation and is used to calculate Vmin.

5-18

Equation (5.3-14) can be solved for one value of between the relative volatilities of the two keys, HK < < LK. This value of is then substituted into equation (5.3-10) to calculate Vmin. The minimum reflux ratio is then

Rmin = = = 1 = 1minL

DminV D

D minV

Di iD

i

x

or

Rmin + 1 = (5.3-15)i iD

i

x

The above procedure is valid for the following cases:

A) Asumes that none of the non-keys distribute. In this case the amounts of non-keys in the distillate are

DxHNK,D = 0 and DxLNK,D = FzLNK

The balances for the keys are

DxLK,D = FRLK,DFzLK

DxHK,D = [1 FRHK,B]FzHK

B) Assume that the distributions for the non-keys determined from the Fenske equation at total reflux are also valid at minimum reflux. In this case the DxHNK,D values are obtained from the Fenske equation.

Equation (5.3-14) can be solved for all values of lying between the relative volatility of all components. This gives C 1 valid roots. Equation (5.3-10) is then written C 1 times, once for each value of .

Vmin = (5.3-10),i i D

i i

Dx

There are C 2 values of the non-keys DxNK. We have (C 1) equations in (C 1) unknowns (Vmin and DxNK). These equations can be solved simultaneously to obtain the minimum reflux ratio.

5-19

Example 5.3-1.3 ----------------------------------------------------------------------------------The feed to a depropanizer is 66% vaporized at the column inlet. The feed composition and average relative volatilities are given in table below. It is required that 98% of the propane in the feed is recovered in the distillate, and 99% of the pentane is to be recovered in the bottoms product. Calculate the minimum reflux ratio for this case using Underwood’s methodComponent Mole fraction in feed Relative volatilityMethane (C1)Ethane (C2)Propane (C3)Butane (C4)Pentane (C5)Hexane (C6)

0.260.090.250.170.110.12

39.4710.004.082.111.000.50

Solution -----------------------------------------------------------------------------------------

For this problem, propane and pentane are the light key and heavy key, respectively. Methane and ethane are light non-keys (LNK), hexane is a heavy non-key (HNK), while butane is a “sandwich component,” which is a component with a volatility intermediate between the keys. For this case, the distribution of the non-keys at minimum reflux must be determined simultaneously with the minimum reflux ratio. The following relation can be used to determine whether or not a component is distributed at minimum reflux4

Di,R = FRLK,D + FRHK,D (E-1)11

i

LK

1

LK i

LK

In this equation, the relative volatilities are based on a reference value of 1.0 for the heavy key component. Equation (E-1), developed by Shiras et al., applies at minimum reflux as follows:

Di,R > 1 Component is nondistributed; contained entirely in distillate.0 < Di,R < 1 Component is distributed; appears in both distillate and bottoms.Di,R < 0 Component is nondistributed; contained entirely in bottoms.

For methane,

DC1,R = ×0.98 + ×0.01 = 12.1339.47 14.08 1

4.08 39.474.08 1

DC2,R = ×0.98 + ×0.01 = 2.8410.00 14.08 1

4.08 10.004.08 1

DC4,R = ×0.98 + ×0.01 = 0.362.11 14.08 1

4.08 2.114.08 1

3 Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, pg. 3774 Shiras, R. N., et al., Ind. Eng. Chem., 42, 871 (1950)

5-20

DC6,R = ×0.98 + ×0.01 = 0.150.50 14.08 1

4.08 0.504.08 1

According to Shiras criterion, methane and ethane appear only in the distillate, hexane appears only in the bottoms, and butane is the only distributed non-key. The two values of are determined from the first Underwood equation

1 q = i i

i i

z

0.66 = + + + + + (39.49)(0.26)39.49

(10)(0.09)10

(4.08)(0.25)4.08

(2.11)(0.17)2.11

0.111

(0.5)(0.12)0.5

+ + + + + 0.66 = 0(39.49)(0.26)39.49

(10)(0.09)10

(4.08)(0.25)4.08

(2.11)(0.17)2.11

0.111

(0.5)(0.12)0.5

These two values of are between the two intervals 1.0 < < 2.11 and 2.11 < < 4.08. The Underwood equation can be solved by the following Matlab codes:

>> fun=inline('39.49*.26/(39.49-t)+10*.09/(10-t)+4.08*.25/(4.08-t)+2.11*.17/(2.11-t)+.11/(1-t)+.5*.12/(.5-t)-.66');>> theta1=fsolve(fun,1.5,optimset('Display','off'))

theta1 = 1.2629

>> theta2=fsolve(fun,3,optimset('Display','off'))

theta2 = 2.8460

For 100 mol of feed we have

DxC3,D = (100)(0.25)(0.98) = 24.5 mol propane

DxC3,D = (100)(0.11)(0.01) = 0.11 mol pentane

DxC1,D = (100)(0.26) = 26 mol methane

DxC2,D = (100)(0.09) = 9 mol ethane

DxC4,D = unknown (butane)

DxC6,D = (100)(0.12)(0) = 0 mol hexane

We now apply the second Underwood equation

Vmin = =,i i D

i i

Dx

5-21

For = 1.263

Vmin = + + + + (39.49)(26)39.49 1.263

(10)(9)10 1.263

(4.08)(24.5)4.08 1.263

4,2.112.11 1.263

C DDx

0.111 1.263

For = 2.846

Vmin = + + + + (39.49)(26)39.49 2.846

(10)(9)10 2.846

(4.08)(24.5)4.08 2.846

4,2.112.11 2.846

C DDx

0.111 2.846

Simplifying the above equations yields

Vmin = 72.243 + 2.494(DxC4,D)

Vmin = 121.614 2.863(DxC4,D)

Hence

DxC4,D = (121.614 72.243)/(2.494 + 2.863) = 9.22 moles

Vmin = 72.243 + (2.494)(9.22) = 95.23

D = = 26 + 9 + 24.5 + 9.22 + 0.11 = 68.83 mol,1

C

i Di

Dx

Lmin = Vmin D = 95.23 68.83 = 26.40 mol


Rmin = Lmin /D = 26.40/68.83 = 0.384

6-1

Chapter 6 Rigorous Methods for Distillation

6.1 Theoretical Model for an Equilibrium Stage

Consider a general, continuous, steady-state vapor-liquid separator consisting of a number of equilibrium stages arranged in a countercurrent cascade. There is no chemical reaction and entrainment of liquid drops in vapor or vapor bubbles in liquid are negligible. A general schematic representation of an equilibrium stage n is shown in Figure 6.1-1 for a vapor-liquid separator, where the stages are number down from the top.

Stage n

ValveFFeed

Fn

zhTP

i,n

F,n

Fn

Fn yi,n+1

n+1

n+1

n+1

HTP

Vn+1

ValveV

Vapor fromstage below

yi,n

n

n

n

HTP

Vaporside streamWn

Vn

HeadHs

xhTP

i,n

n

n

n

xhTP

i,n-1

n-1

n-1

n-1

Liquid fromstage above

Ln-1

Ln

Liquidside stream

Un

Heat transfer Q

(+) if from stage

n

(-) if to stage

Figure 6.1-1 General equilibrium stage1

Entering stage n can be one single- or two-phase feed of molar flow rate Fn, with overall composition zi,n of component i, temperature TFn, pressure PFn, and corresponding overall molar enthalpy hFn. Feed pressure is assumed equal to or greater than stage pressure Pn. Any excess pressure (PFn Pn) is reduced to zero adiabatically across valve F.

Liquid molar flow rate Ln-1 from stage n1 enters stage n with composition in mole fraction xi,n-1, enthalpy hn-1, temperature Tn-1 and pressure Pn-1, which is


6-2

equal to or less than the pressure of stage n. Pressure of liquid from stage n1 is increased adiabatically by hydrostatic head change across head Hs.

Stage n

ValveFFeed

Fn

zhTP

i,n

F,n

F n

F n yi,n+1

n+1

n+1

n+1

HTP

Vn+1

ValveV


yi,n

n

n

n

HTP

Vaporside streamWn

Vn

HeadHs

xhTP

i,n

n

n

n

xhTP

i,n-1

n-1

n-1

n-1


Ln-1

Ln

Liquidside stream

Un

Heat transfer Q

(+) if from stage

n

(-) if to stage

Figure 6.1-1 General equilibrium stage

Vapor molar flow rate Vn+1 from stage n+1 enters stage n with composition in mole fraction yi,n+1, enthalpy Hn+1, temperature Tn+1 and pressure Pn+1, which is greater or equal the pressure of stage n. Any excess pressure (Pn+1 Pn) is reduced to zero adiabatically across valve V.

Leaving stage n is vapor of mole fraction yi,n, enthalpy Hn, temperature Tn and pressure Pn. This stream can be divided into a vapor side stream of molar flow rate Wn and an inter-stage stream of molar flow rate Vn to stage n-1. If n = 1, the vapor stream leaves the separator as a product. Also leaving stage n is liquid with mole fraction xi,n, enthalpy hn, temperature Tn and pressure Pn. This liquid stream can also be divided into a liquid side stream of molar flow rate Un and an inter-stage stream of molar flow rate Ln to stage n+1. If n = N = number of stages in the separator, the liquid stream leaves the separator as a product.

Heat can be transferred at a rate Qn from (+) or to (-) stage n to simulate stage intercoolers, interheaters, intercondenser, interreboilers, condensers, or reboilers.

6-3

The theoretical model for an equilibrium stage involves the MESH equations which are all the equations used to describe the steady state operation of a distillation column. MESH stands for:

1) Material or flow rate balance equations, both component and total.2) Equilibrium equations including the bubble-point and dew-point

equations.3) Summation or stoichiometric equations or composition constraints.4) Heat or enthalpy or energy balance equations.

1) M equations- Material balance for each component (C equations for each stage).

Ln-1xi,n-1 + Vn+1yi,n+1 + Fnzi,n (Ln + Un)xi,n (Vn + Wn)yi,n = 0 (6.1-1)

2) E equations- phase equilibrium relation for each component (C equations for each stage).

yi,n Ki,n xi,n = 0 (6.1-2)

3) S equations-mole fraction summations (one for each stage)

(6.1-3),1

1 0C

i ni

y

(6.1-4),1

1 0C

i ni

x

4) H equation- energy balance (one for each stage)

Ln-1hn-1 + Vn+1Hn+1 + FnhFn (Ln + Un)hn (Vn + Wn)Hn Qn = 0 (6.1-5)

In general Ki,n = Ki,n{Tn, Pn, xi,n, yi,n}, Hn = Hn{Tn, Pn, yi,n}, and hn = hn{Tn, Pn, xi,n}. If these relations are not counted as equations and the three properties are considered known, each equilibrium stage is defined by the 2C + 3 MESH equations. Hence, there are N(2C + 3) such equations in a counter-current separator with N equilibrium stages. For each equilibrium stage the variables are listed in Table 6.1-1.

Table 6.1-1 Variables for each equilibrium stagezi,n xi,n yi,n Tn Pn TFn PFn Vn Wn Ln Un Fn QnC C C 1 1 1 1 1 1 1 1 1 1

6-4

There are 3C + 10 variables for each equilibrium stage. For a separator with N equilibrium stages there are N(3C + 10) +1 variables. (Note: The extra 1 is for the number of equilibrium stages). If N and all zi,n, Fn, TFn, PFn, Pn, Wn, Un, and Qn are specified, the model is represented by N(2C + 3) simultaneous algebraic equations in N(2C + 3) variables. Since the M, E, and H equations are nonlinear they must be solved by iterative techniques.

Stage 1F1 Q1

V2W2U1

L1

V1

Stage 2F2 Q2

V3W3U2

L2

Stage nFn Qn

Vn+1

Wn+1

UnLn

VnWnUn-1

Ln-1

W1

Figure 6.1-2 General countercurrent cascade of n stages.

A total material balance equation can be used instead of (6.1-3) or (6.1-4). This equation is obtained by making a material balance over the first n stages as shown in Figure 6.1-2.

Vn+1 + = V1 + Ln + 1

n

mm

F

1

n

m mm

U W

Ln = Vn+1 + V1 (6.1-6) 1

n

m m mm

F U W

6-5

6.2 Rigorous Computational Methods

Many methods for solving the MESH equations utilize the tridiagonal matrix algorithm that results from a modified form of the M equations (6.1-1) when Tn and Vn are selected as the iteration (tear) variables. The modified M equations are linear in the unknown liquid mole fractions. These equations are obtained by substituting (6.1-2) into (6.1-1) to eliminate yi and by substituting (6.1-6) into (6.1-1) to eliminate Ln.

xi,n-1 + Vn+1 Ki,n+1 xi,n+1 + Fnzi,n 1

11

n

n m m mm

V F U W V

xi,n + Unxi,n (Vn + Wn)Ki,n xi,n = 0 1 11

n

n m m mm

V F U W V

The above equation can be rearranged to

xi,n-1 xi,n 1

11

n

n m m mm

V F U W V

1 1 ,

1

n

n m m m n n i nm

V F U W V V W K

+ Vn+1 Ki,n+1 xi,n+1 = Fnzi,n (6.1-7)

Let

An = , 2 n N 1

11

n

n m m mm

V F U W V

Bn = , 1 n N 1 1 ,1

n

n m m m n n i nm

V F U W V V W K

Cn = Vn+1 Ki,n+1, 1 n N 1

Dn = Fnzi,n, 1 n N

Equation (6.1-7) becomes

An xi,n-1 + Bn xi,n + Cn xi,n+1 = Dn (6.1-8)

In these expressions xi,0 = 0, VN+1 = 0, W1 = 0, and UN = 0. The modified M equations will be written for each component over the entire countercurrent cascade of N stage. The resulting set of equations can be arranged in matrix form, equation (6.1-9)2. The coefficient matrix is a tridiagonal matrix that is solved by a highly efficient Thomas algorithm.


6-6

= (6.1-9)

1 1

2 2 2

3 3 3

1 2 2

1 1 1

0 0 0 ... 00 0 ... 0

0 0 ... 0... ...... ...... ...... ...0 ... ... 00 ... ... 00 ... ... 0 0

N N N

N N N

N N

B CA B C

A B C

A B CA B C

A B

,1

,2

,3

, 2

, 1

,

...

...

...

...

i

i

i

i N

i N

i N

xxx

xxx

1

2

3

2

1

...

...

...

...

N

N

N

DDD

DDD

If K-values are independent of compositions, the constant Bn and Cn depend only on tear variables Tn and Vn. If K-values are dependent on compositions they can be estimated from the compositions in the previous iteration.

The Thomas algorithm consists of two steps. In the first forward elimination step, xi,1 from the first equation is solved in terms of xi,2 and is used in the second equation to solve for xi,2 in terms of xi,3. The process continues toward stage N to obtain a solution for xi,N. In the second backward substitution step, the value of xi,N is substituted into the N-1 equation to solve for of xi,N-1. Other values of xi,n are then obtained by a similar procedure. For five stages, the matrix equation at the beginning of the procedure is

1 1

2 2 2

3 3 3

4 4 4

5 5

0 0 00 0

0 00 00 0 0

B CA B C

A B CA B C

A B

1 1

2 2

3 3

4 4

5 5

x Dx Dx Dx Dx D

At the end of the forward elimination step, the matrix equation is in the following form

1 11

2 22

3 33

4 44

5 5

1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0 1

x qpx qpx qpx qpx q

At the end of the backward substitution step, the matrix equation is in the form

6-7

1 1

2 2

3 3

4 4

5 5

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

x rx rx rx rx r

For stage 1, equation (6.1-9) is B1 xi,1 + C1 xi,2 = D1, which can be solved for xi,1 in terms of unknown xi,2 to give

xi,1 = = q1 p1xi,2, where q1 = and p1 = (6.1-10)1 1 ,2

1

iD C xB

1

1

DB

1

1

CB

The coefficients in the first row of the matrix become B1 1, C1 p1, and D1 q1, where means “is replaced by.” Only values for p1, and q1 are needed for the backward substitution step.

For stage 2, equation (6.1-9) is A2 xi,1 + B2 xi,2 + C2 xi,3 = D2, which can be combined with (6.1-10) and solve for xi,2 to give

A2(q1 p1xi,2) + B2 xi,2 + C2 xi,3 = D2

xi,2 = xi,32 2 1

2 2 1

D A qB A p

2

2 2 1

CB A p

Let q2 = and p2 = then xi,2 = q2 p2xi,32 2 1

2 2 1

D A qB A p

2

2 2 1

CB A p

The coefficients in the second row of the matrix become A1 0, B2 1, C2 p2, and D2 q2. Only values for p2 and q2 are needed. The elimination procedure can be repeated with

the values for pn and qn given by

pn = and qn = 1

n

n n n

CB A p

1

1

n n n

n n n

D A qB A p

Then xi,n = qn pnxi,n+1 (6.1-11)

Thus the coefficients in the matrix become An 0, Bn 1, Cn pn, and Dn qn. Only values for pn and qn are needed. Starting with stage 1, values of pn and qn are computed recursively in the order p1, q1, p2, q2, …, pN-1, qN-1, qN. For stage N, xi,N is calculated as

xi,N = qN

Successive values of xi,n are computed recursively by backward substitution from (6.1-11) in the form

6-8

xi,n-1 = qn-1 pn-1xi,n = rn-1 (6.1-12)

Example 6.2-1 ------------------------------------------------------------------------------Solve the tridiagonal matrix equations where the coefficients are given by

An = 1, n = 1, , 9

Bn = [2 + 0.22(1 ], n = 1, , 9; zn = 1.2:.2:2.8;)5nz

Cn = 1, n = 1, , 9

D1 = 0.22(1.2) 2; Dn = 0.22zn, n = 2, , 8; D9 = 0.22(2.8) + 1

Solution ----------------------------------------------------------------------------------------------

The tridiagonal matrix system can be written as

Ax = D

where

A =

0176.21000000010192.21000000010208.21000000010224.21000000010240.21000000010256.21000000010272.21000000010288.21000000010304.2

x = T 1 2 3 4 5 6 7 8 9x x x x x x x x xD = T 112.1104.0096.0088.0080.0072.0064.0056.0952.1

The tridiagonal matrix equations are solved by the Matlab program listed in Table 6.2-1.

------------ Table 6.2-1 Matlab program using the Thomas algorithm -------------

% Table 6.2-1 Matlab program using the Thomas algorithm -------------%% Example 6.2-1z=1.2:.2:2.8;h=.2;N=length(z);

6-9

B=-(2+h*h*(1-z/5));C=ones(1,N);A=C;D=h*h*z;D(1)=D(1)-2;D(N)=D(N)+1;p=C;q=C;x=C;p(1)=C(1)/B(1);q(1)=D(1)/B(1);for n=2:N p(n)=C(n)/(B(n)-A(n)*p(n-1)); q(n)=(D(n)-A(n)*q(n-1))/(B(n)-A(n)*p(n-1));endx(N)=q(N);for n=N-1:-1:1 x(n)=q(n)-p(n)*x(n+1);enddisp('x =');disp(x')

>> tridiagx = 1.3513 0.7918 0.3110 -0.0974 -0.4362 -0.7055 -0.9025 -1.0224 -1.0579

------------------------------------------------------------------------------

There are a wide variety of iterative solution procedures for solving the MESH equations since no single method could solve all types of problems3. Some of the common methods are

1. The bubble point methods (BP)2. The sum-rates methods (SR)3. The Newton methods4. The global Newton or simultaneous correction (SC) methods5. The inside-out methods

The BP methods use the tridiagonal matrix algorithm to calculate the component flow rates or compositions. These are used to calculate the temperatures by solving the bubble-point equation. An alternative is to calculate the new temperature directly, without iterating on the bubble-point equation. These new temperatures are approximate but as long as the internal compositions are properly corrected during each column trial, the temperature profile will converge toward the solution. With either alternative, the energy balances are used to find the total flow rates. The BP methods generally work well for narrow-boiling, ideal or nearly ideal systems, where composition has a greater effect on temperature than the latent heat of vaporization.

3 Friday, J.R., and B.D. Smith, AIChE J. 10, 698-707 (1964)

6-10

The SR methods can be used when the stage energy balance is much more sensitive to stage-temperatures than to component flow rates. In this case, temperatures are the dominant variables and are obtained by a Newton-Raphson solution of the stage energy balances. The component flow rates are found by the tridiagonal matrix method. These are summed to get the total rates, hence the name sum rates (SR).

In the Newton methods, the MESH equations can be regarded as a large system of interrelated, nonlinear algebraic equations. The Newton-Raphson iteration is then applied to all or part of the MESH equations. The remaining MESH equations can be solved by other methods. The solution gives the steady-state values of the column variables: temperatures, flow rates, compositions, etc.

In the global Newton methods, all of the MESH equations are solved together in a Newton-Raphson technique. If N and all zi,n, Fn, TFn, PFn, Pn, Wn, Un, and Qn are specified, the global Newton methods solve directly for xi,n, yi,n, Tn, Vn and Ln. The global Newton methods generally require initial guesses near the answer. It may be necessary to use another method, such as a BP or SR methods, to obtain initial values approaching the solution before the global Newton method can be used.

In the first four groups of methods, the MESH variables of temperatures, total flow rates, and composition flow rates are the primary solution variables and are used to generate enthalpies and vapor liquid equilibrium (VLE) K-values. These methods update the MESH variables in an outer loop with the K-values and enthalpies updated whenever the MESH variables change. The inside-out concept reverses this by using rigorous thermodynamic calculations to generate parameter for simple K-value and enthalpy models. These parameters become the variables for the outside loop, while the inside loops consists of the MESH equations.

The followings are extracted from the ProII Reference Manual: “The Inside/Out (I/O) algorithm is partitioned into an inner and outer loop. In the inner loop, the heat, material and design specifications are solved. Simple thermodynamic models for enthalpy and vapor liquid equilibrium (VLE) K-values are used in the inner loop. This is one of the factors that allow the inner loop to be solved quickly and reliably. In the outer loop, the simple thermodynamics model parameters are updated based on the new compositions and the results of rigorous thermodynamic calculations. When the rigorously computed enthalpies and K-values match those of the simple thermodynamic models, and the design specifications are met, the algorithm is solved. When more than one algorithm can be used to solve a problem, the I/O algorithm will usually have the fastest convergence. The I/O algorithm can solve most refinery problems. It also solves many chemical systems, and when possible should be the first choice for system with a single liquid phase.”

6-11

Chapter 6 Example 6.2-2 ------------------------------------------------------------------------------A butane-pentane splitter is to operate at 122 psia with the following feed composition

Mole feed (lbmol/hr)Propane,Isobutane,Normal butane,Isopentane,Normal pentane

C3iC4nC4, light keyiC5, heavy keynC5

515252035

For a specification of not more than 1 mol of the light key in the bottom product and not more than 1 mol of the heavy key in the top product, and a reflux ratio of 2.5, make a stage by stage calculation to determine the product composition and number of stages required.

Equilibrium K values =============================================================

ln K = A/T2 + B C ln(P) + D/P, where P is in psia, T is in oR Compound A B C D

=============================================================Propane 970688.6 7.15059 .76984 6.90224 i-Butane 1166846 7.72668 .92213 0 n-Butane 1280557 7.94986 .96455 0 i-Pentane 1481583 7.58071 .93159 0 n-Pentane 1524891 7.33129 .89143 0 =============================================================

Enthalpy Data( H and h are in Btu/lbmol, T is in oR)Vapor, H = AT2 + BT + C Liquid, h = aT2 + bT + c

A B C a b cC3iC4nC4 iC5nC5

0.06430.13940.04290.02680.1823

-65.15-153.1-29.10-9.231-210.8

30435.059166.020610.018043.084367.0

0.1592-0.0133

00.06630.1062

-158.061.5342.99-33.00-91.01

46265.0-22068.0-14947.07751.6029300.0

Solution ----------------------------------------------------------------------------------------------To estimate the temperatures at distillate and bottoms, we assume that nothing heavier than the heavy key appears in the distillate and nothing lighter than the light key in the bottoms. The distribution of the products are listed in Table E-1


C3iC4nC4, light key iC5, heavy keynC5

5152410

0.1111 0.3333 0.5333 0.0222

0

0011935

00

0.0182 0.3454 0.6364

45 55

6-12

We assume negligible pressure drop across the column so that the pressure is at 122 psia throughout the column.

xb

4nC = 0.0182iC = 0.3454nC = 0.6364

5

5

Partial reboiler

The bottoms temperature is calculated with a bubble point calculation since part of the liquid leaving the last stage is returned to the column and is in equilibrium with the bottoms liquid product with composition xb = [0.0182 0.3454 0.6364]. Since the vapor mole fractions are unknown, we start with the equation

= 1 (E-1)

m

iiy

1

Using the K-values: Ki = , Eq. (E-1) becomes = 1, or i

i

xy

m

iii xK

1

f(T) = = 0 (E-2)1

lnm

i ii

K x

Equation (E-2) can be solved numerically with Newton method using the following Matlab codes:

% P=122; % psiaA=[1280577 1481583 1524891]; B=[7.94986 7.58071 7.33129]; C=[0.96455 0.93159 .89143];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)));zi=[1 19 35]/55;xi=zi;Te=xi*Tsat'; % Estimate mixture temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P))';

6-13

dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=log(xi*Ki'); Tsav=T;T=T+dT; Ki=eval(fK);fdT=log(xi*Ki'); eT=fT*dT/(fdT-fT); T=Tsav-eT;endTb=T;TC=T/1.8-273.15fprintf('Bubble point temperature, T(R) = %8.2f, T(C) = %8.2f\n',T,TC)Ki=eval(fK)

>> e6d2d2_bubbleBubble point temperature, T(R) = 698.98, T(C) = 115.17Ki =

2.0039e+000 1.0756e+000 9.3028e-001

1

xd

y1 = x0= xd

x1y2

Total condenser

Table E-1Distillate y1 xd

C3iC4nC4 iC5

0.1111 0.3333 0.5333 0.0222

0.1111 0.3333 0.5333 0.0222

The top temperature is calculated with a dew point calculation with the vapor composition listed in Table E-1. Since the liquid mole fractions are unknown, we start with the equation

= 1 (E-3)

m

iix

1

6-14

Using the K-values: Ki = , Eq. (E-3) becomes f(T) = 1 = 0 (E-4)i

i

xy

m

i i

i

Ky

1


% Dew point temperature calculationP=122; % psiaA=[970688.6 1166846 1280577 1481583]; B=[7.15059 7.72668 7.94986 7.58071]; C=[0.76984 0.92213 0.96455 0.93159];D=[6.90224 0 0 0];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)+D/P));zi=[5 15 24 1]/45;yi=zi;Te=yi*Tsat'; % Estimate temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P)';dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=sum(yi./Ki)-1; Tsav=T;T=T+dT; Ki=eval(fK);fdT=sum(yi./Ki)-1; eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;TC=T/1.8-273.15;fprintf('Dew point temperature, T(R) = %8.2f, T(C) = %8.2f\n',T,TC)Ki=eval(fK)

>> e6d2d2_dewDew point temperature, T(R) = 608.80, T(C) = 65.07Ki =

2.4348e+000 1.1601e+000 8.7030e-001 4.0981e-001

We will do a stage-by-stage calculation from the top to the bottoms. We need to estimate the composition of other components in the distillate and bottoms so that they can be included in the stage calculations. The estimation of other components in the distillate and bottoms at total reflux is given by the following formula


ib

x Dx B

hd

hb

x Dx B

In this equation, xhd is the mole fraction of heavy key in the distillate, xhb is the mole fraction of heavy key in the bottoms, and Nmin is the minimum stages for total flux, which can be estimated from

6-15

Nmin = (E-6) ,

log[( / )( / )]logld hd hb lb

l ave

x x x x

In this equation

l,ave = ld lb

ld = (KnC4/KiC5)d = (0.8703/0.40981) = 2.124

lb = (KnC4/KiC5)b = (2.0039/1.0756) = 1.863

l,ave = = 1.989(2.124)(1.863)


nC4, light key iC5, heavy key

241

0.5333 0.0222

119

0.0182 0.3454

The minimum stages for total flux is then

Nmin = ,

log[( / )( / )]logld hd hb lb

l ave

x x x x

Nmin = = 9

log[(0.5333/ 0.0222)(0.3454 / 0.0182)]log 1.989

For nC5

= (nC5,ave)Nmin5,

5,

nC d

nC b

x Dx B

hd

hb

x Dx B

nC5,ave = 5 5nC d nC b

nC5,d = (KnC5/KiC5)d = (0.3446/0.4098) = 0.841

nC5,b = (KnC5/KiC5)b = (0.93028/1.0756) = 0.865

nC5,ave = = 0.853(0.841)(0.865)

= (0.853)9 = 0.012583 xnC5,dD = 0.012583 xnC5,bB5,

5,

nC d

nC b

x Dx B

119

6-16

xnC5,dD + xnC5,bB = 35 xnC5,bB = 35/1.012583 = 34.565

xnC5,dD = 0.435

We can now update Table E-1 to Table E-2 where there are all five components in the distillate.


C3iC4nC4, light key iC5, heavy keynC5

515241

0.435

0.1100 0.3301 0.5282 0.02200.0097

00119

34.565

00

0.0183 0.3482 0.6335

45.435 54.565

The liquid stream enters stage 1 is L0 = RD = (2.5)(45.435) = 113.59 lbmol/hr. The vapor stream leaving stage 1 is V1 = (R + 1)D = (3.5)(45.435) = 159.02 lbmol/hr.

Stage 1

Stage 2

L x0 0,

L x1 1,

L x2 2,

V y1 1,

V y2 2,

V y3 3,

Starting with the liquid stream entering stage 1 and vapor stream leaving stage 1 we could perform the stage by stage calculation using the following procedure:

1) Start with the following informationx0 L0 y1 V1

C3 0.1100 113.59 lbmol/hr 0.1100 159.02 lbmol/hriC4 0.3301 608.80 oR 0.3301 T = ?nC4, light key 0.5282 65.07 oC 0.5282 iC5, heavy key 0.0220 122 psia 0.0220 122 psianC5 0.0097 0.0097

2) Perform a dew point calculation with the known pressure and composition of the vapor stream leaving the stage to determine the stage temperature and the composition of the liquid stream leaving the stage.

6-17

Since the liquid mole fractions are unknown, we start with the equation

= 1 (E-3)

m

iix

1

Using the K-values: Ki = , Eq. (E-3) becomes f(T) = 1 = 0 (E-4)i

i

xy

m

i i

i

Ky

1


% Dew point temperature calculationP=122; % psiaA=[970688.6 1166846 1280577 1481583 1524891]; B=[7.15059 7.72668 7.94986 7.58071 7.33129]; C=[0.76984 0.92213 0.96455 0.93159 0.89143];D=[6.90224 0 0 0 0];% Estimate the pure species saturation temperature by setting Ki=1Tsat=sqrt(A./(B-C*log(P)+D/P));yi=[0.1100 0.3301 0.5282 0.0220 0.0097];Te=yi*Tsat'; % Estimate temperature from saturation temperature and compositionT=Te;fK='exp(-A/(T*T)+B-C*log(P)+D/P)';dT=1;eT=1;while abs(eT)>.1 Ki=eval(fK);fT=sum(yi./Ki)-1; Tsav=T;T=T+dT; Ki=eval(fK);fdT=sum(yi./Ki)-1; eT=fT*dT/(fdT-fT); T=Tsav-eT;endTd=T;TC=T/1.8-273.15;fprintf('Dew point temperature, T(R) = %8.2f, T(C) = %8.2f\n',T,TC) Ki=eval(fK);xi=yi./Ki; fprintf('Liquid composition leaving the tray, xi = %8.4f, %8.4f, %8.4f, %8.4f, %8.4f\n',xi)

>> e6d2d2_stageDew point temperature, T(R) = 610.46, T(C) = 65.99

Liquid composition leaving the tray, xi = 0.0445, 0.2797, 0.5957, 0.0525, 0.0275

Note: since = 1 we do not need to normalize xi otherwise xi = xi/5

1i

ix

5

1i

ix

3) Estimate the compositions of the vapor stream entering the stage

Assume that L0 = L1 and V1 = V2 and making material balance over stage 1 give

6-18

L0x0 + V1y2 = L0x1 + V1y1 y2 = (L0/ V1)x1 + y1

The following Matlab codes (continue from the previous codes) evaluate y2:

printf('Liquid composition leaving the tray, xi = %8.4f, %8.4f, %8.4f, %8.4f, %8.4f\n',xi) Dist=45.435;R=2.5; L0=R*Dist;V1=(R+1)*Dist; x0=yi;y1=yi;x1=xi; y2=L0*(x1-x0)/V1+y1

y2 =

6.3244e-002 2.9412e-001 5.7639e-001 4.3810e-002 2.2435e-002

4) Estimate the temperature of the vapor stream entering the stage and evaluate enthalpies of all streams entering and leaving the stage

An approximate calculation using Erbar-Maddox correlation can be performed to determine the number of equilibrium stages needed. There are about 14 ideal stages so the temperature change from stage to stage will be about (699.0 608.8)/14 6.5oR. Let T2 = 617oR, the following Matlab codes (continue from the previous codes) evaluate h0, h1, H1, and H2:

y2=L0*(x1-x0)/V1+y1 AH=[0.0643 0.1394 0.0429 0.0268 0.1823]; BH=[-65.15 -153.1 -29.10 -9.231 -210.8]; CH=[30435.0 59166.0 20610.0 18043.0 84367.0]; ah=[0.1592 -0.0133 0 0.0663 0.1062]; bh=[-158.0 61.53 42.99 -33.00 -91.01]; ch=[46265.0 -22068.0 -14947.0 7751.60 29300.0];T0=608.8;T=T0;h0i=ah*T^2+bh*T+ch;T1=610.5;T=T1h1i=ah*T^2+bh*T+ch;H1i=AH*T^2+BH*T+CH; T2=T1+6.5;T=T2; H2i=AH*T^2+BH*T+CH; h0=h0i*x0'; h1=h1i*x1'; H1=H1i*y1'; H2=H2i*y2'; fprintf('T0(R) = %8.2f, h0(Btu/lbmol) = %8.2f\n', T0,h0) fprintf('T1(R) = %8.2f, h1(Btu/lbmol) = %8.2f\n', T1,h1) fprintf('T1(R) = %8.2f, H1(Btu/lbmol) = %8.2f\n', T1,H1)fprintf('T2(R) = %8.2f, H2(Btu/lbmol) = %8.2f\n', T2,H2)

T0(R) = 608.80, h0(Btu/lbmol) = 10779.25T1(R) = 610.50, h1(Btu/lbmol) = 11099.01T1(R) = 610.50, H1(Btu/lbmol) = 18106.34T2(R) = 617.00, H2(Btu/lbmol) = 18621.14

6-19

Note: The specific enthalpies of the streams are determined from the pure enthalpies by the following expressions:

H = and h = 5

1i i

iy H

5

1i i

ix h

[x1 x2 x3 x4 x5]’ =

1

2

3

4

5

xxxxx

x1 = 0.0445, 0.2797, 0.5957, 0.0525, 0.0275

= [h1 h2 h3 h4 h5] (Matlab statement: h1=h1i*x1' )5

1i i

ix h

1

2

3

4

5

xxxxx

5) Determine the flow rates of the vapor stream (V2) entering the stage and the liquid stream (L1) leaving the stage from energy and material balances:

L0h0 + V2H2 = L1h1 + V1H1

L0 + V2 = L1 + V1

The following Matlab codes (continue from the previous codes) evaluate V2 and L1:

fprintf('T2(R) = %8.2f, H2(Btu/lbmol) = %8.2f\n', T2,H2)Amatrix=[1 -1;H2 -h1]; Bvector=[V1-L0; V1*H1-L0*h0];V2L1=Amatrix\Bvector;V2=V2L1(1);L1=V2L1(2);fprintf('V2(lbmol/hr) = %8.2f, L1(lbmol/hr) = %8.2f\n', V2,L1)

V2(lbmol/hr) = 152.97, L1(lbmol/hr) = 107.53

6) Evaluate again the compositions of the vapor stream entering the stage from the component material balance

L0x0 + V2y2 = L1x1 + V1y1 y2 = (L1/V2)x1 (L0/V2) x0 + (V1/V2)y1

The following Matlab codes (continue from the previous codes) evaluate y2:

fprintf('V2(lbmol/hr) = %8.2f, L1(lbmol/hr) = %8.2f\n', V2,L1)y2=(L1/V2)*x1-(L0/V2)*x0+(V1/V2)*y1;sumy2=sum(y2);

6-20

y2=y2/sumy2

y2 =

6.3984e-002 2.9469e-001 5.7563e-001 4.3464e-002 2.2233e-002

7) Repeat steps (4)-(7) until there is negligible change in the values of the vapor composition entering the stage.

For this example we accept the values of the vapor compositions y2 calculated above and repeat steps 1-7 for another stage with the following information

x1 L1 y2 V2C3 0.0445 107.53 lbmol/hr 0.0640 152.97 lbmol/hriC4 0.2797 610.5 oR 0.2947 T = ?nC4, light key 0.5957 0.5756 iC5, heavy key 0.0525 122 psia 0.0435 122 psianC5 0.0275 0.0222 The stage by stage calculation will continue until the ratio of the compositions of the key components in the liquid stream is less than that of the feed stream

< = = 1.25LK

HK

xx

,

,

LK Feed

HK Feed

xx

2520

The feed would be introduced at this stage. If the composition of the nonkey components on the plate does not match the feed composition it would be necessary to adjust the top composition and repeat the calculation. The stage by stage calculation is then continued to the last stage of the column.

The stage by stage calculation can also be started from the boiler and bottom stage to the top stage. The concentration of the nonkey component in the bottom product could be estimated from the formula


ib

x Dx B

hd

hb

x Dx B

This approach will require bubble point calculations along with material and energy balance. If the vapor composition at the feed point does not match with the bottom-up calculation, the assumed concentration of the nonkeys in the bottom product is adjusted and the calculation repeated.

Equilibrium stage calculations are available from a free ChemSep program that could be downloaded from the website: http://www.chemsep.com/downloads/index.html. This free version is limited to 150 stages and 10 components with the equilibrium flash/column simulator and 194 component library.

http://www.chemsep.com/downloads/index.html

6-21

Chapter 6

Algorithm for Wang-Henke BP methods for distillation1

The Bubble-Point (BP) method is frequently used for species that cover a relatively narrow range of vapor-liquid equilibrium ratios (K-values). The algorithm for Wang-Henke BP methods is listed in Table 6.2-1. In this algorithm a new set of stage temperatures is computed during each iteration k from bubble-point equations.

Table 6.2-1 Algorithm for Wang-Henke BP methods for distillationStart

Specify: all Fj, zi,n, feed conditions (TFn, PFn, or hFn), Pn, Wn, Un; and Qn except Q1 and QN; N; L1 (reflux rate), V1 (vapor distillate rate)

Step 1: Initialize tear variables Tn, VnStep 2: Compute the mole fraction of component i over all N equilibrium stages from the following tridiagonal matrix:

An xi,n-1 + Bn xi,n + Cn xi,n+1 = Dn

An = , 2 n N 1

11

n

n m m mm

V F U W V

Bn = ,1 n N 1 1 ,1

n

n m m m n n n i nm

V F U W V U V W K

Cn = Vn+1 Ki,n+1, 1 n N 1

Dn = Fnzi,n, 1 n N

In this step, composition of each component is calculated over all N equilibrium stages successively.Step 3: Normalize xi,n for each stage by

(xi,n)normalized = ,

,1

i nC

i ni

x

x

Step 4: Calculate new Tn from bubble-point equation: 1.0 = 0, ,1

C

i n i ni

K x

Calculate new yi,n from: yi,n = Ki,nxi,n

Step 5: Calculate Q1 from the enthalpy balance over the condenser:

1 J. D. Seader and E. J. Henley, Separation Process Principles, Wiley, 2006, pg. 369

6-22

Q1 = V2H2 + F1hF1 (L1 + U1)h1 (V1 + W1)H1

Calculate QN from the enthalpy balance over the column:

QN = V1H1 LNhN 1

N

n Fn n n n nn

F h U h W H

1

1

N

nn

Q

Step 6: Calculate new Vn from the enthalpy balance from each stage:

Vn = 1 1 1

1

n n n

n

V

n = (hn hn-1) + Fn(hn hFn) + Wn(Hn hn) + Qn 1

1

n

n m m mm

V F U W

n = hn-1 Hn

n = Hn+1 hn

Calculate Ln from

Ln = 1 11

n

n m m mm

V F U W V

Step 7: Calculate the convergence criterion defined by

= where k denotes the iteration number2

( ) ( 1)

1

Nk k

n nn

T T

If 0.01N then the stage temperatures are accepted (converged). If not go back to step (2).

6-23

Stage 2

Stage 3

Stage 2Stage 4

Totalcondenser

1

Partial reboiler

5

U1

V2 L1

V3 L2

V4L3

V5L4

Q1

Q5

L5

F3

Example 6.2-3 ---------------------------------------------2Calculate a new set of temperature Tn from

1.0 = 0, ,1

C

i n i ni

K x

Use one iteration of the BP method and the following K-values:

ln K = A/T2 + B C ln(P) + D/P where P is in psia, T is in oR

Propane (C3) n-Butane(nC4) n-Pentane(nC5)A 970688.6 1280557 1524891B 7.15059 7.94986 7.33129C .76984 .96455 .89143D 6.90224 0 0

Feed F3 is saturated liquid 100 lbmol/h at 100 psia. The reflux ratio (L1/U1) is 2. The bottom product L5 is 50 lbmol/h. All stages are at 100 psia.

Solution ----------------------------------------------------------------------------------------------By overall total material balance

Liquid distillate = U1 = F3 L5 = 100 50 = 50 lbmol/h

Then L1 = (L1/U1)U1 = (2)(50) = 100 lbmol/h

By total material balance around the total condenser,

V2 = L1 + U1 = 100 + 50 = 150 lbmol/h

Initial guesses of tear variables Vn and Tn are

Stage n Vn, lbmol/h Tn1 Fixed at 0 by specifications 652 Fixed at 150 by specifications 903 150 1154 150 1405 150 165


6-24

At 100 psia, the estimated K-values at the assumed stage temperatures are

Ki,n

Stage 1 2 3 4 5C3 1.1649 1.5928 2.0927 2.6592 3.2851nC4 0.32041 0.48415 0.69402 0.95198 1.2582nC5 0.099626 0.16287 0.25007 0.36434 0.50785

We now need to solve the matrix equation for the mole fraction in the liquid phase leaving each stage. For the first component C3

An xi,n-1 + Bn xi,n + Cn xi,n+1 = Dn

An = , 2 n N 1

11

n

n m m mm

V F U W V

Since V1 = 0 and there is no vapor stream leaving the column, Wm = 0, the coefficient An becomes:

An = , 2 n 5 1

1

n

n m mm

V F U

Thus A5 = V5 + (F1 U1) + (F2 U2) + (F3 U3) + (F4 U4)

A5 = 150 + (0 50) + (0 0) + (100 0) + (0 0) = 150 50 + 100 = 200 lbmol/h

A4 = V5 + (F1 U1) + (F2 U2) + (F3 U3)

A4 = 150 + (0 50) + (0 0) + (100 0) = 150 50 + 100 = 200 lbmol/h

A3 = V5 + (F1 U1) + (F2 U2) = 150 + (0 50) + (0 0) = 100 lbmol/h

A2 = V5 + (F1 U1) = 150 + (0 50) = 100 lbmol/h

Bn = ,1 n N 1 1 ,1

n

n m m m n n n i nm

V F U W V U V W K

V1 = 0 and there is no vapor stream leaving the column, Wm = 0, the coefficient Bn becomes:

Bn = , 1 n 5 1 ,11

n

n m m n n i nm

V F U U V K

B5 = [V6 + ( U1) + (F3) + V5K1,5] = [100 50 + (150)(3.2851)] = 542.77

B4 = [V5 + ( U1) + (F3) + V4K1,4] = [150 + 100 50 + (150)(2.6592)] = 598.87

6-25

B3 = [V4 + ( U1) + (F3) + V3K1,3] = [150 + 100 50 + (150)(2.0927)] = 513.9

B2 = [V3 + ( U1) + V2K1,2] = [150 50 + (150)(1.5928)] = 338.92

B1 = [V2 + ( U1) + U1 + V1K1,1] = [150 50 + 50] = 150

Cn = Vn+1 Ki,n+1, 1 n N 1

C1 = V2 K1,2 = (150)(1.5928) = 238.92

C2 = V3 K1,3 = (150)(2.0927) = 313.9

C3 = V4 K1,4 = (150)(2.6592) = 398.87

C4 = V5 K1,5 = (150)(3.2851) = 492.77

Dn = Fnzi,n

D3 = (100)(0.3) = 30, D1 = D2 = D3 = D4 = 0

Substitution of the coefficients into An xi,n-1 + Bn xi,n + Cn xi,n+1 = Dn gives

=

150 238.92 0 0 0100 338.92 313.9 0 0

100 513.9 398.87 00 200 598.87 492.770 0 200 542.77

1,1

1,2

1,3

1,4

1,5

xxxxx

003000

Solving the above triangular matrix system we obtain

x1,1 = 0.5642, x1,2 = 0.35422, x1,3 = 0.20271, x1,4 = 0.097155, and x1,5 = 0.0358

The matrix equations (An xi,n-1 + Bn xi,n + Cn xi,n+1 = Dn ) for nC4 and nC5 are solved similarly to give

xi,n Stage 1 2 3 4 5

C3 0.5642 0.35422 0.20271 0.097155 0.0358

nC4 0.18542 0.38299 0.45696 0.49485 0.41458

nC5 0.018645 0.11448 0.33004 0.49295 0.78135

0.768265 0.85169 0.98971 1.084955 1.23173i,ni

x

6-26

The compositions at each stage are normalized to the following values:

xi,n Stage 1 2 3 4 5

C3 0.734382 0.415902 0.204818 0.089547 0.029065

nC4 0.241349 0.449682 0.461711 0.456102 0.336584

nC5 0.024269 0.134415 0.333471 0.454351 0.634352

1 1 1 1 1

The bubble-point temperatures at 100 psia are then calculated. For stage 1 we have

0.734382K1,1 + 0.241349K2,1 + 0.024269K3,1 1 = 0

In this equation, temperature is the only unknown and can be solved by Newton’s method. The result is 69.9oF. The following Matlab codes can be used to solve for the new set of temperatures:

Example 6.2-3 ------- Matlab codes ----------------% BP (Bubble Point) MethodN=5;F=zeros(1,5); U=F;V=150*ones(1,5); R=2;C=F;L=F;L(5)=50; F(3)=100;U(1)=F(3)-L(5);V(1)=0;L(1)=R*U(1);F(3)=100;zin=zeros(3,5);xin=zin;zin(1,3)=0.3;zin(2,3)=0.3;zin(3,3)=0.4;a=[970688.6 1280577 1524891]; b=[7.15059 7.94986 7.33129]; c=[0.76984 0.96455 0.89142];d=[6.90224 0 0];% Assume temperature at each equilibrium stageTv=[65 90 115 140 165]+460;fK='exp(-a/(T*T)+b-c*log(P)+d/P)';P=100;% Evalue K ratios for each component for the five equilibrium stagesfor n=1:5; T=Tv(n);K=eval(fK);Km(1,n)=K(1);Km(2,n)=K(2);Km(3,n)=K(3);end

i,ni

x

6-27

% Start iteration to determine compositions at each equilibrium stagefor i=1:3;for n=2:5; sum=0; for m=1:n-1 sum=sum+F(m)-U(m); end A(n)=V(n)+sum;end for n=1:4; sum=0; for m=1:n; sum=sum+F(m)-U(m); end B(n)=-(V(n+1)+sum+U(n)+V(n)*Km(i,n));end sum=0; for m=1:5; sum=sum+F(m)-U(m); endB(5)=-(sum+V(5)*Km(i,5));for n=1:4; C(n)=V(n+1)*Km(i,n+1);end for n=1:5; D(n)=-F(n)*zin(i,n);end p=ones(1,N);q=p;x=p;p(1)=C(1)/B(1);q(1)=D(1)/B(1);for n=2:N p(n)=C(n)/(B(n)-A(n)*p(n-1)); q(n)=(D(n)-A(n)*q(n-1))/(B(n)-A(n)*p(n-1));endx(N)=q(N);for n=N-1:-1:1 x(n)=q(n)-p(n)*x(n+1);endxin(i,:)=x;endxinsave=xin;% Normalize the mole fractionsfor n=1:N sumx=0; for i=1:3 sumx=sumx+xin(i,n); end for i=1:3

6-28

xin(i,n)=xin(i,n)/sumx; endend % Recalculate temperature at each equilibrium platesxi=zeros(1,3);dT=0.1;for n=1:N for i=1:3 xi(i)=xin(i,n); end T=Tv(n); for m=1:30 Ki=eval(fK);fT=log(xi*Ki'); Tsav=T;T=T+dT; Ki=eval(fK);fdT=log(xi*Ki'); eT=fT*dT/(fdT-fT); T=Tsav-eT; if abs(eT)<0.1;break;end endTF=T-460;fprintf('T guessed (F) = %8.1f, T(F) = %8.1f\n',Tv(n)-460,TF)end

>> BPmethodT guessed (F) = 65.0, T(F) = 69.9T guessed (F) = 90.0, T(F) = 98.3T guessed (F) = 115.0, T(F) = 130.8T guessed (F) = 140.0, T(F) = 156.0T guessed (F) = 165.0, T(F) = 181.4

6-29

Chapter 6

6.3 Rate-Based Model for Distillation

yi,n+1

n+1

n+1

HTV

Vn+1

ValveV


yi,n

n

n

HT V

Vaporside streamWn

Vn

HeadHs

xhT

i,n

n

nL

xhT

i,n-1

n-1

n-1L


Ln-1

Ln

Liquidside stream

Un

Stage n

N

e

fVi,n

HnVF

fLi,n

hnLF

rnL

QnLQn

V

rnV

Figure 6.3-1 Non-equilibrium stage for rate-based method3.

Figure 6.3-1 shows a schematic diagram of a non-equilibrium stage, consisting of a tray or a segment of a packed section. This diagram indicates all the material and energy streams entering and leaving a tray. Recall that the equilibrium stage model consists of 2C + 3 MESH equations for each stage. They are C mass balances for components, C phase equilibrium relations, 2 summations of mole fractions, and 1 energy balance. In the rate-based model, the mass and energy balance around each equilibrium stage are each replaced by separate balances for each phase around a stage. These balances are written in terms of the differences between the masses (or energies) leaving and entering a phase. These differences are called the residuals which become zero when the computations are converged. For liquid-phase component balance we have

Mi,nL = (1 + rn

L)Lnxi,n Ln-1xi,n-1 fi,nL Ni,n

L = 0 (6.3-1)


6-30

In this equation Mi,nL is the residual for component i in the liquid phase of stage n, fi,n

L is the entering molar flow rate of component i, and Ni,n

L is the entering molar rate of component i across the phase boundary. For vapor-phase component balance

Mi,nV = (1 + rn

V)Vnyi,n Vn+1yi,n+1 fi,nV + Ni,n

V = 0 (6.3-2)

In this equation Mi,nV is the residual for component i in the vapor phase of stage n, fi,n

V is the entering molar flow rate of component i, and Ni,n

V is the exiting molar rate of component i across the phase boundary. For liquid-phase energy balance

EnL = (1 + rn

L)Lnhn Ln-1hn-1 + QnL = 0 (6.3-3),

1

CL

i ni

f

LF

nh Lne

In this equation EnL is the residual liquid phase energy of stage n, fi,n

L is the exit molar flow rate of component i, Qn

L is the heat leaving stage n, and is the heat entering the liquid Lne

phase. For vapor-phase energy balance

EnV = (1 + rn

V)VnHn Vn+1Hn+1 + QnV + = 0 (6.3-4),

1

CV

i ni

f

VF

nH Vne

In this equation EnV is the residual vapor phase energy of stage n, fi,n

V is the entering molar flow rate of component i, Qn

V is the heat leaving stage n, and is the heat leaving the vapor Vne

phase. At the interface, I,

EnI = = 0 (6.3-5)V

ne Lne

Phase equilibrium for each component is assumed to exist only at the phase interface:

Qi,nI = K i,n(xi,n)I (yi,n)I = 0, i = 1,2 , …, C (6.3-6)

Rate-based or non-equilibrium models use the same K-value and enthalpy correlations as equilibrium-based models. However equilibrium only exists at the interface between the vapor and liquid phases on trays or in packing. In general, the K-value correlation, whether calculated using an equation-of-state or an activity-coefficient model, is a function of phase-interface temperature and compositions, and tray pressure. Enthalpies are evaluated only at the conditions of the phases as they exit a tray. For the equilibrium-based model, the exit vapor is at the dew-point temperature and the exit liquid is at the bubble-point temperature. These two temperature are the same and at the stage temperature. For the non-equilibrium models, the stage temperature is not the same as the temperatures of the exit streams. The exit liquid is subcooled and the exit vapor is superheated. Predictions of heat and mass transfer rates are needed for non-equilibrium models. These rates depend upon transport coefficients, interfacial area, and driving forces. The mass-transfer rates across the vapor and liquid films, Ni,n

V and Ni,nL, respectively, in Eqs. (6.3-1 and 2) are needed for the calculations.

6-31

The general forms for component mass-transfer rates across the vapor and liquid films, respectively, on a tray or in a packed segment, consist of contributions from both diffusive and convective flow.

Ni,nV = Ji,n

V + yi,n NT,nV (6.3-7)I

na

Ni,nL = Ji,n

L + xi,n NT,nL (6.3-8)I

na

In these equations, is the total interfacial area for the stage, Ji,nV and Ji,n

L are the vapor and Ina

liquid diffusion fluxes, respectively, relative to the molar average velocity. For a binary mixture, these fluxes are given by

JiV = ct

VkiV(yi

V yiI)avg (6.3-9)

JiL = ct

LkiL(xi

I xiL)avg (6.3-10)

In these equations ctP is the total molar concentration where P stands for the phase (V or L),

kiP is the mass transfer coefficient for a binary mixture based on a mole fraction driving

force, and (yiV yi

I)avg [or (xiI xi

L)avg] is the mean driving force over the stage. The positive direction of mass transfer is assumed to be from the vapor phase to the liquid phase. From the definition of the molar diffusive flux:

= 0 (6.3-11)1

C

ii

J

For a binary mixture, J1 = J2.The diffusive flux for a general multi-component system is more complex than the binary system because of component-coupling effects4. For the ternary system, the fluxes for the first two components are

J1V = ct

Vk11V(y1

V y1I)avg + ct

Vk12V(y2

V y2I)avg (6.3-12)

J2V = ct

Vk21V(y1

V y1I)avg + ct

Vk22V(y2

V y2I)avg (6.3-13)

The flux for the third component is not independent of the other two, but is obtained from Eq. (6.3-11):

J3V = J1

V J2V (6.3-14)

For the general multicomponent system (1, 2, …, C), the independent fluxes for the first C 1 components are given in the matrix equation form as

JV = ctV[kV](yV yI)avg (6.3-15)

JL = ctL[kL](xI xL)avg (6.3-16)

4 Taylor, R., and R. Krishna, Multicomponent Mass Transfer, Wiley, New York, 1993

6-32

In these equations JP, (yV yI)avg, and (xI xL)avg are column vectors of length C 1 and [kP] is a (C 1)×(C 1) square matrix. The mass transfer coefficient matrix can be obtained from a reciprocal mass-transfer coefficient function, R. For an ideal-gas solution

[V] = [RV]-1 (6.3-17)

For a non-ideal liquid solution:

[L] = [RL]-1 (6.3-18)L

In these equations, the elements of RP in terms of general mole fractions, zi, are:

= + (6.3-19)PiiR i

PiC

zk 1

CkP

k i ik

zk

= zi (6.3-20)PijR 1 1

P Pij iCk k

Vapor Liquid

yi

yiI

xi

xi


AI

I

Figure 6.3-2 Concentration gradient for mass transfer of species i.

In these equations, the values of k are binary-pair mass-transfer coefficients obtained from correlations of experimental data and zi is the average concentration of species i in each phase. If a linear concentration is assumed then

ziV = 0.5(yi + yi

I) and ziL = 0.5(xi + xi

I)

For four component vapor-phase system,

JV = ctV[RV]-1(yV yI)avg (6.3-21)

The four fluxes can be obtained from

= ctV (6.3-22a)

1

2

3

V

V

V

JJJ

1

11 12 13

21 22 23

31 32 33

V V V

V V V

V V V

R R RR R RR R R

1 1

2 2

3 3

V I

avg

V I

avg

V I

avg

y y

y y

y y

6-33

= (6.3-22b)4VJ 1 2 3

V V VJ J J

The mass transfer functions are given by

= + + + (6.3-23a)11VR 1

14V

yk

2

12V

yk

3

13V

yk

4

14V

yk

= y1 (6.3-23b)12VR

12 14

1 1V Vk k

The matrix with dimension (C 1)×(C 1) contains the thermodynamic factors that L

correct for non-ideality. When an activity-coefficient model is used:

= ij + xi (6.3-24)Lij

, ,

ln i

j T P k jx

In this equation, ij is the Kronecker data, which is 1 if i = j and 0 if i j.

Example 6.3-1.5 ----------------------------------------------------------------------------------Consider tray n of a ternary distillation at 14.7 psia, involving acetone (1), methanol (2), and water (3) in a 5.5-ft-diameter column using sieve trays with a 2-inch-high weir. Vapor and liquid phases are assumed to be completely mixed. Determine the molar diffusion rates, the mass-transfer rates, and the Murphree vapor-tray efficiencies using the following data from a rate-based calculation:

Component yn yn+1 ynI Kn

I xn122

0.29710.46310.2398

0.17000.42900.4010

0.35210.46770.1802

2.7591.2250.3673

0.14590.38650.4676

1.0000 1.0000 1.0000 1.0000

The total vapor flow rate entering the tray was estimated as Vn+1 = 1,491 lbmol/h; the vapor leaving the tray was estimated as Vn = 1,545 lbmol/hr. The gas-phase binary mass-transfer coefficients were given as follows:

k12 = k21 = 1,955 lbmol/h, k13 = k31 = 2,407 lbmol/h, k23 = k32 = 2,797 lbmol/h.

Solution -----------------------------------------------------------------------------------------

(a) Compute the molar diffusion rates

Assuming linear mole-fraction gradients: ziV = 0.5(yi + yi

I)


6-34

z1 = 0.5(0.2971 + 0.3521) = 0.3246

z2 = 0.5(0.4631 + 0.4677) = 0.4654

z3 = 0.5(0.2398 + 0.1802) = 0.2100

The the elements of the reciprocal function, RV, in terms of general mole fractions, zi, are:

= + + = + + = 0.00046011VR 1

13V

zk

2

12V

zk

3

13V

zk

0.32462,407

0.46541,955

0.21002,407

= + + = + + = 0.00040822VR 2

23V

zk

1

21V

zk

3

23V

zk

0.46542,797

0.32461,955

0.21002,797

= z1 = 0.3264 = 0.000031212VR

12 13

1 1V Vk k

1 11,955 2,407

= z2 = 0.4654 = 0.000071721VR

21 23

1 1V Vk k

1 11,955 2,797

In matrix form: [RV] = 0.000460 0.00003120.0000717 0.000408

[V] = [RV]-1 = 2, 200 168.2386.6 2, 480

Since the off-diagonal terms in the matrix [kV] are much smaller than the diagonal terms, the effect of coupling in this example is small.

= 1

2

V

V

JJ

11 12

21 22

V V

V V

1 1

2 2

V I

V I

y y

y y

= ( ) + ( ) = 2,200(0.2971 0.3521) + 168.2(0.4631 0.4677)1VJ 11

V 1Vy 1

Iy 12V 2

Vy 2Iy

= 121.8 lbmol/h1VJ

= ( ) + ( ) = 386.6(0.2971 0.3521) + 2,480(0.4631 0.4677)2VJ 21

V 1Vy 1

Iy 22V 2

Vy 2Iy

= 32.7 lbmol/h2VJ

= = 154.5 lbmol/h3VJ 1

VJ 2VJ

6-35

Note: , , and are the mass transfer rates, not the mass fluxes.1VJ 2

VJ 3VJ

(b) Compute the mass-transfer rates

= + z1 = 121.8 + 0.32461VN 1

VJ VTN V

TN

The total mass-transfer rate is the difference between the vapor flow rates entering and VTN

leaving the tray. For the non-equilibrium model, the mass transfer from the vapor to the liquid is taken as positive,

= Vn+1 Vn = 1,491 1,545 = 54 lbmol/hVTN

= 121.8 + 0.3246 = 121.8 + 0.3246( 54) = 139.4 lbmol/hr1VN V

TN

Similarly

= 32.7 + 0.4654 = 32.7 + 0.4654 ( 54) = 57.8 lbmol/hr2VN V

TN

= 154.5 + 0.2100 = 154.5 + 0. 2100 ( 54) = 143.2 lbmol/hr3VN V

TN

(c) Calculate the Murphree vapor-tray efficiencies

EMVi = , , 1

, , , 1

i n i nIi n i n i n

y yK x y

EMV1 = = 0.5470.2971 0.1700(2.759)(0.1459) 0.1700

EMV2 = = 0.7670.4631 0.4290(1.225)(0.3865) 0.4290

EMV3 = = 0.7030.2398 0.4010(0.3673)(0.4676) 0.4010

6-36

6.4 Equilibrium Distillation Calculations using ChemSep program

“ChemSep is a software system for simulation of distillation, absorption, and extraction operations. ChemSep integrates flash calculations, the classic equilibrium stage column model and a nonequilibrium or rate-based column model in one easy to use program”.6 ChemSep-LITE is available for FREE. It only includes the equilibrium column model and is limited to 10 components and 150 stages. Appendix A provides instruction for the use of ChemSep. Included in the program are thermodynamic databases for commonly used chemical as well. For a license fee of about $25.00 (2009) a student version of ChemSep is available with rate-based calculations.

Example 6.4-1.7 ----------------------------------------------------------------------------------The following feed, at 353.15 K and 1,035 kPa is to be fractionated at the rate of 1 kmol/s. The following feed compositions are given:

Component Flow rate (kmol/s)MethaneEthanePropane,n-butane,n-pentane,n-hexane

C1C2C3C4C5C6

.03

.07

.15

.33

.30

.12

Specifications for the example are given in the following table:

Operation Thermodynamics Properties FeedsEquilibrium columnSimple distillationPartial condenserPartial reboilerNumber of stages: 14Feed stage: 6

K-value: EOSEquation of state: Soave-RKEnthalpy: Excess (from EOS)Leave blank for Thermodynamic Model Parameters

State: Temperature & Pressure

Specifications Column SpecificationsConstant pressureColumn heat loss: 0Default stage efficiency: 1

Bottom product flow rate = 0.6200 kmol/s

Solution -----------------------------------------------------------------------------------------

The predict separation is as follows:

Streams data:

6 http://www.chemsep.org/program/index.html7 Benitez, J. Principle and Modern Applications of Mass Transfer Operations, Wiley, 2009, p. 377

http://www.chemsep.org/downloads/index.html

6-37

Stream Feed1 V.Feed1 L.Feed1 Top Bottom

Stage 6 6 6 1 14 Pressure (N/m2) 1035000 1035000 1035000 1035000 1035000 Vapour fraction (-) 0.33185 1 0 1 0 Temperature (K) 353.15 353.15 353.15 321.311 385.388 Enthalpy (J/kmol) -8920000 657713 -10130000 Entropy (J/kmol/K) -36977.8 -5218.79 -30413.4

Mole flows (kmol/s) Methane 0.03 0.0269517 0.00304832 0.03 1.6569E-11 Ethane 0.07 0.051421 0.018579 0.0699991 9.4915E-07 Propane 0.15 0.0807888 0.0692112 0.147659 0.00234088 N-butane 0.33 0.109 0.221 0.1288 0.2012 N-pentane 0.3 0.0531321 0.246868 0.00353048 0.29647 N-hexane 0.12 0.0105562 0.109444 1.1212E-05 0.119989

Total molar flow 1 0.33185 0.66815 0.38 0.62

Enthalpies, entropies, entropy production

Stage T Hv Hl Sv Sl Q Sirr(K) (J/kmol) (J/kmol) (J/kmol/K (J/kmol/K (J/s) (J/kmol/K

1 321.311 657713 -1.7E+07 -5218.79 -59104 -9006000 647.6472 335.49 1951100 -1.6E+07 -1856.4 -54050.7 386.6813 341.466 2579800 -1.5E+07 323.764 -51242.3 216.2594 346.277 3105300 -1.5E+07 2534.47 -48545.3 228.3965 351.538 3697000 -1.5E+07 5063.34 -45536.8 2816.016 361.01 4909700 -1.4E+07 8234.7 -40870.5 8042.17 365.147 5528800 -1.3E+07 9016.61 -39308.1 139.988 367.844 5949500 -1.3E+07 9540.79 -38360.6 92.40969 369.873 6274100 -1.2E+07 9925.86 -37681.7 70.1342

10 371.552 6546700 -1.2E+07 10244.3 -37123.2 56.975811 373.169 6809200 -1.2E+07 10627.8 -36539.5 52.685612 375.187 7133600 -1.1E+07 11339.9 -35683 74.947113 378.56 7676200 -1.1E+07 12931.8 -34030.7 208.42614 385.388 8813900 -1E+07 16603.1 -30413.4 11892000 276.767

Total entropy production = 13309.4 (J/kmol/K)

From this table, the condenser duty is 900.6 kW and the reboiler duty is 11,892 kW.

6-38

The results can also be plotted:

2

4

6

8

10

12

14 320 340 360 380

Sta

ge

Temperature (K)

Temperature profile

ChemSep

T

2

4

6

8

10

12

14 0 0.2 0.4 0.6 0.8 1

Sta

ge

Vapour mole fraction

Vapour phase composition profiles

ChemSep

MethaneEthane

PropaneN-butane

N-pentaneN-hexane

7-1

Chapter 7

Heat Exchangers

7.1 Introduction

Heat exchangers are devices for transferring heat between two fluid streams. Heat exchangers can be classified as indirect contact type and direct contact type. Indirect contact type heat exchangers have no mixing between the hot and cold streams, only energy transfer is allowed as shown in Figure 7.1-1.

Hot fluid

Cold fluid

Wall separates streamsFigure 7.1-1 Indirect contact type heat exchanger.

Direct contact type heat exchangers have no wall to separate the cold from the hot streams as shown in Figure 7.1-2.

Noncondensiblebleed

Warm water

Cold waterSteam

Figure 7.1-2 Direct contact type heat exchanger.

Heat exchangers can be classified according to 1) transfer process, 2) number of fluids, 3) surface compactness, 4) construction, 5) flow arrangements, and 6) heat transfer mechanism. Table 7.1-1 shows the classification for the first three types. The complete table can be found from the Heat Transfer Handbook page 17.3.

7-2

Table 7.1-1. Classification of heat exchangers

Classification according to transfer process

Indirect contact type Direct contact type

Direct transfer type

Immiscible fluid

Gas-liquid Liquid-vaporStorage type

Fluidized bed

Single phase Multiphase

Classification according to number of fluids

Two-fluid N-fluid (N > 3)Three-fluid

Classification according to surface compactness

Gas-to-fluid Liquid to liquid or phase change

Compact(> 700 m /m )2 3

Non-compact(< 700 m /m )2 3

Compact(> 400 m /m )2 3

Non-compact(< 400 m /m )2 3

Heat exchangers can be found in automotives (radiators) or in power cycles. Figure 7.1-3b shows a schematic diagram of an air-standard gas turbine with directions for principal heat transfers indicated by arrows. Gas turbines are usually lighter and more compact than the vapor power system even though a larger portion of work developed by the gas turbine is required to drive the compressor.

Combustion chamber

TurbineCompressor

Fuel

Wnet

Heat exchanger

TurbineCompressor

Heat exchanger1

2 3

4

Qin

Qout

Wcycle

Air Product

Simple gas turbine

Air-standardBrayton cycle

Figure 7.1-3 (a) Simple gas turbine. (b) Air-standard gas turbine cycle

An open gas turbine engine is shown in Figure 7.1-3a. Air is continuously drawn into the compressor of this engine, where it is compressed to a high pressure. The air then enters a

7-3

combustor, a combustion chamber, where it is mixed with fuel and combustion occurs, resulting in combustion products at an elevated temperature. The combustion products do work by expanding through the turbine and are subsequently discharged to the surroundings. Part of the turbine work developed is used to drive the compressor. An air-standard study analysis is used to study the open gas turbine engine with the assumptions that air is the working ideal gas and the energy generated by combustion is accomplished by a heat transfer source.

With the air-standard idealization, ambient air enters the compressor at state 1 and later returns to the surrounding at state 4 with a temperature higher than the ambient temperature. The discharged air would eventually return to the same state as the air entering the compressor so we can consider the air passing through the gas turbine engine as undergoing a thermodynamic cycle. The air-standard Brayton cycle represents the states visited by the gas with an additional heat exchanger for the air to release heat to the surroundings and return to its original state 1. The air-standard Brayton cycle consists of two heat exchangers, a compressor, and a turbine.

The condenser and evaporator shown in Figure 7.1-4 are the heat exchangers. This figure depicts the most common refrigeration. In step 4 1, heat is removed at the temperature TL from the system being refrigerated by the evaporation of a liquid under the pressure PL. In step 1 2, saturated vapor at PL is compressed isentropically to PH where it becomes superheated vapor. In step 2 3, heat QH is transferred to the surrounding by condensation at TH. In step 3 4, the cycle is closed by throttling the liquid to the lower pressure PL. There is no change in enthalpy during this step.

Condenser

Evaporator

Throttlevalve Compressor

4 1

23

QL

QH

PL

PH

T

s

Liquid

VaporIsenthalp

1

2

3

4Wa b c

Figure 7.1-4 A vapor-compression refrigeration cycle and its Ts diagram.

7-4

7.2 Heat Exchanger types.

We will discuss three types of tubular heat exchangers: concentric tube, cross flow, and shell-and-tube heat exchangers. A concentric tube or double pipe heat exchanger is the simplest heat exchanger for which the hot and cold fluids move in the same or opposite directions as shown in Figure 7.2-1.

Hot fluid

Cold fluid

Hot fluid

Cold fluid

Parallel flow Counter flow

Figure 7.2-1 Concentric tube heat exchangers.

Figure 7.2-2 shows cross flow heat exchangers where fluid flows perpendicular across the tube bank rather than parallel with it. There is no mixing of the fluid outside the tube in the y-direction for the arrangement in the left side of Figure 7.2-2, while there is mixing of the fluid outside the tube in the y-direction for the arrangement in the right side of Figure 7.2-2. The fluid inside the tubes is considered unmixed since it is confined and cannot mix with other stream. Cross flow heat exchangers are usually used to heat or cool a gas such as air.

x

y

Tube flow

Tube flow

Cross flow = ( , )T f x y Cross flow

= ( )T f x

Figure 7.2-2 Cross flow heat exchangers.

Shell-and-tube heat exchanger is the most common configuration. There are many different forms of shell-and-tube heat exchangers according to the number of shell-and-tube passes. A common form with one shell pass and two tube passes is shown in Figure 7.2-3. Baffles are usually installed to increase the heat transfer coefficient of the fluid by introducing turbulence and cross-flow in the shell side.

7-5

Shell inlet

Shell outlet

Tube oulet

Tube inlet

Baffles

Figure 7.2-3a Shell-and-tube heat exchanger with one shell pass and two tube passes.

Figure 7.2-3b Details construction of a shell-and-tube heat exchanger.

The advantages of shell and tube exchangers are as follows1:

1) Large surface area in a small volume;2) Good mechanical layout: a good shape for pressure operation;3) Well-established fabrication techniques;4) Construction from a wide range of materials;5) Easy maintenance;6) Well-established design procedures.

In general, the tube side is used for fluid which is more corrosive or dirtier or at a higher pressure. The shell side is used for liquid of high viscosity or gas. It is usually easier to clean the inside of tubes than to clean the shell side. Heat exchanger shutdowns are most often caused by fouling, corrosion, and erosion. Figure 7.2-4 shows type designation for shell-and-tube heat exchangers by the American Tubular Heat Exchanger Manufacturers Association, TEMA. The TEMA standards cover three classes of exchanger: class R for exchangers in the generally severe duties of the petroleum and related industries, class C for exchangers in moderate duties of

1 Tower and Sinnott, Chemical Engineering Design, Elsevier, 2008, pg. 801

7-6

commercial and general process application, and class B for exchangers in the chemical process industries2.

Figure 7.2-4 TEMA Heat Exchanger Types

2 Tower and Sinnott, Chemical Engineering Design, Elsevier, 2008, pg. 803

7-7

The TEMA specifies the standards related to the mechanical design features, materials of construction, and testing of the shell and tube exchangers3 with a unique nomenclature. The TEMA nomenclature is a three-letter designation based on the mode of differential thermal expansion between the shell and the tubes, their degree of disassembly, and the shell-side flow arrangement. The first letter of the three-letter designation indicates the front end head type. The second letter of the three relates to the shell type. The last letter of the three indicates the rear end head type.

(a) Single -segmental baffle

(b) Double-segmental baffle

(c) Disk-and-doughnut baffle

Figure 7.2-5 Types of baffle used in shell and tube heat exchangers.

Baffles improve the rate of heat transfer by increasing the fluid velocity and directing the fluid stream across the tubes. The most commonly used type of baffle is the single baffle shown in Figure 7.2-5a; two other types are shown in Figure 7.2-5a and b.

3 Tower and Sinnott, Chemical Engineering Design, Elsevier, 2008, pg. 809.

7-8

Pt

Flow

Triangular

Pt

Pt

Square Rotated square

Figure 7.2-6 Tube patterns.

The tubes in exchangers are usually arranged in an equilateral, square, or rotated square patterns as shown in Figure 7.2-6 where Pt denotes tube pitch. The triangular and rotated square patterns have higher heat transfer rates and higher pressure drop than the square pattern. A square or rotated square pattern is used for heavily fouling fluids, where mechanical cleaning of the outside tubes is necessary. The tube pitch (distance between tube centers) is normally 1.25 times the tube outside diameter unless process requirements dictate otherwise.

7-9

Chapter 7

7.3 Analysis of Heat Exchangers

All heat exchangers analyses require energy balances between the fluids. These balances will be performed for steady state systems with the following assumptions:

a) Heat capacity cp is not a function of temperature;b) Heat transfer coefficient h is constant and does not vary along tube length. This is the

case for fully developed flow where effect of property variations is not important;c) The system is well insulated so that there is no heat loss to surroundings;d) Longitudinal heat conduction in the fluid and walls is negligible. There is no conduction

along the flow direction;e) Tube walls are smooth without scale of dirt or oxidation;f) Fluid potential and kinetic energies are negligible;g) Flow is characterized by bulk or mean velocity and mean temperature at any cross

section.

Vx

um Tm

Tw

Tw

T

The mean velocity um and mean temperature Tm is defined by the following equations:

um = and Tm = 1A xA

V dA1

mu A xAV TdA

We will analyze a simple parallel flow heat exchanger depicted in Figure 7.3-1.

Tube

Shell (insulated)

Thi Tho

Tc i

Tc o

shell (insulated)

tube

Centerline

x

ri ro

Tube wall

Insulated

Tc

Th

x

dqh

dqc

A = 2 r Lo o

A = 2 rLi i

Figure 7.3-1 Simple parallel flow heat exchanger.

7-10

Let Th and Tc be the mean fluid temperatures of the tube and shell sides respectively. The energy change of the tube side fluid along the x-direction is equal to the energy transferred through the tube wall

q0”Pdx = q0”dAs = dTh (7.3-1), ,h m h p hu c A

In this equation P is the tube perimeter, A is the tube cross sectional area, and q0” is the heat flux through the tube wall. In terms of the mass flow rate = , Eq. (7.3-1) becomeshm ,h m hu A

dqh = q0”dAs = cp,h dTh (7.3-2)hm

We have a similar equation for the energy received by the cold fluid

dqc = q0”dAs = cp,c dTc (7.3-3)cm

The energy given up by the hot fluid is absorbed by the cold fluid,

dqc = dqh = dq (7.3-4)

Equation (7.3-2) can be integrated from the inlet to the outlet of the hot stream,

= hdq ,ho

hi

T

h p h hTm c dT

qh = cp,h(Tho Thi) (7.3-5)hm

Similarly for the cold stream

qc = cp,c(Tco Tci) (7.3-6)cm

Since qc = qh = q, equations (7.3-5 and 6) can be written as

q = cp,h(Thi Tho) (7.3-7)hm

q = cp,c(Tco Tci) (7.3-8)cm

In the differential form we have

dq = cp,hdTh (7.3-9)hm

dq = cp,cdTc (7.3-10)cm

We now define the capacitance rates for the hot and cold streams Ch and Cc respectively

Ch = cp,h and Cc = cp,chm cm

7-11

Equations (7.3-7, 8, 9, and 10) become

q = Ch(Thi Tho) (7.3-11)

q = Cc(Tco Tci) (7.3-12)

dq = ChdTh (7.3-13)

dq = CcdTc (7.3-14)

Centerline

x

ri ro

Tube wallTc

Th

dx

dqh

dqcA = 2 r Lo o

A = 2 rLi i

dAo

dAi

Cold stream

Hot stream

Control volume

Figure 7.3-2 Cross stream heat transfer.

The heat transfer between the hot and cold streams shown in Figure 7.3-2 is now considered. Without fouling resistance, the heat transfer between the hot and cold streams for the chosen control volume is given by

dq = = ln( / )1 12

h c

o i

i i o o

T Tr r

h dA kdx h dA

ln( / )1 1

2 2 2

h c

o i

i i o o

T Tr r

h rdx kdx h r dx

The heat transfer between the streams can be written in terms of the overall heat transfer coefficients Ui or Uo as follows:

dq = = Ui(Th Tc)dAi

2ln( / )1 1

h c i

i o i i

i o o

T T rdxr r r r

h k h r

dq = = Uo(Th Tc)dAo

2ln( / )1 1

h c o

o o o i

i i o

T T r dxr r r r

h r k h

The overall heat transfer coefficients Ui and Uo are defined as

7-12

Ui = or =

1ln( / )1 1i o i i

i o o

r r r rh k h r

1

i iU dA1

ln( / )1 12

c

o i

i i o o

r rh dA kdx h dA

Uo = or =

1ln( / )1 1o o o i

i i o

r r r rh r k h

1

o oU dA1

ln( / )1 12

c

o i

i i o o

r rh dA kdx h dA

dq = ChdTh = Uo(Th Tc)dAo = (7.3-15)h

h c

dTT T

o o

h

U dAC

dq = CcdTc = Uo(Th Tc)dAo = (7.3-16)c

h c

dTT T

o o

c

U dAC

Subtracting Eq. (7.3-16) from Eq. (7.3-15) gives

= h

h c

dTT T

c

h c

dTT T

o o

h

U dAC

o o

c

U dAC

= UodAo (7.3-17) h c

h c

d T TT T

1 1

h cC C

Tube

Shell (insulated)

Thi Tho

Tc i

Tc o

shell (insulated)

tube

Figure 7.3-3 Parallel flow heat exchanger.

Integrating Eq. (7.3-17) over the surface area of a parallel flow heat exchanger shown in Figure 7.3-3 and assuming that Uo is independent of x, the distance along the heat exchanger, we obtain

= UodAo ,

,

ho co

hi ci

T T h c

T Th c

d T TT T

1 1oA

h cC C

ln = UoAo (7.3-18)hi ci

ho co

T TT T

1 1

h cC C

From Equations (7.3-11 and 12) we have

7-13

= and = 1

hChi hoT T

q 1

cCco ciT T

q

Therefore = + = = 1 1

h cC C hi hoT T

q co ciT T

q hi ci ho coT T T T

q 1 2T T

q

We have defined T1 = Thi Tci and T2 = Tho Tco for parallel flow. Substituting = 1 1

h cC C

into Eq. (7.3-18) gives1 2T Tq

ln = UoAo 1

2

TT

1 2T Tq

Solving for q we obtain

q = UoAoTlm where Tlm =

(7.3-19a)

1 2

1 2ln /T T

T T

We can repeat the procedure with the following expressions

dq = ChdTh = Ui(Th Tc)dAi = h

h c

dTT T

i i

h

U dAC

dq = CcdTc = Ui(Th Tc)dAi = c

h c

dTT T

i i

c

U dAC

We then obtain the heat transfer rate based on the inside surface area Ai of the tube.

q = UiAiTlm (7.3-19b)

For counter flow heat exchanger as shown in Figure 7.3-4, we will obtain a similar expression for the heat transfer rate

q = UoAoTlm = UiAiTlm (7.3-20)

Tube

Shell (insulated)

Thi Tho

Tci

Tco

shell (insulated)

tube

Figure 7.3-4 Counter flow heat exchanger.

7-14

Thi Thi

ThoThoTco

Tco

TciTci

(a) Parallel flow (b) Countercurrent flow

Figure 7.3-5 Flow arrangements in heat exchanger

For countercurrent flow, Tlm is defined by the following equation

Tlm =

= (7.3-21)

ciho

cohi

cihocohi

TTTT

TTTT

ln

)()(

1 2

1 2ln /T T

T T

Hence, for counter flow T1 = Thi Tco and T2 = Tho Tci. For other heat exchanger geometries such as cross flow and shell and tube heat exchanger, the heat transfer rate is given by

q = UoAoFTlm

In this equation, Tlm is based on counter flow and is given by Equation (7.3-21). F is the correction factor to account for the configuration in a heat exchanger for which the flow is neither parallel nor counter current. The F factor for cross-flow heat exchangers with both fluids unmixed are shown in Figure 7.3-6.

Figure 7.3-6 F factor for cross flow heat exchangers with both fluids unmixed.

7-15

The F factor can be obtained from a chart similar to the one shown in Figure 7.3-7 for shell-and-tube heat exchangers with one shell pass and two tube passes. Ti and To are the inlet and outlet temperatures of the fluid on the shell side, respectively. ti and to are the inlet and outlet temperatures of the fluid on the tube side, respectively.

The two parameters R and P required to read the chart are defined as

R = = and P = =

io

oi

ttTT

hi ho

co ci

T TT T

ii

io

tTtt

co ci

hi ci

T TT T

The F factor may also be evaluated from

F =

2

2

2

1 (1 )ln1 (1 )

2 1 1ln

2 1 1

R PR RP

P R R

P R R

Figure 7.3-7 F factor for shell-and-tube heat exchanger with one shell pass and two tube passes.

7-16

If R = 1, the above equation becomes indeterminate, which reduces to

F =

2

2

2

11

2 1 1ln

2 1 1

P RP

P R R

P R R

For exchangers with N shell passes, P is replaced by Px where

Px = for R 1 and Px = for R = 1N

N

PRPR

PRP

/1

/1

11

111

)( PNPNP

The parameter R = is equal to the cold fluid flow rate times the cold fluid mean hi ho

co ci

T TT T

specific heat, divided by the hot fluid flow rate times the hot fluid specific heat. R is simply the capacitance ratio of the cold stream to the hot stream. This definition can be obtained from the following energy balances:

q = cp,h(Thi Tho) Thi Tho = hmh

qC

q = cp,c(Tco Tci) Tco Tci = cmc

qC

Therefore R = =

= hi ho

co ci

T TT T

//

h

c

q Cq C

c

h

CC

Either definition of R can be used, whichever is more convenient.

7-17

Chapter 7

Example 7.3-1. ----------------------------------------------------------------------------------A process fluid having a specific heat of 3800 J/kgK and flowing at 4 kg/s is to be cooled from 100oC to 50oC with chilled water (specific heat of 4180 J/kgK). Cooling water is available at 15oC with the outlet temperature limited to 35oC. Assuming an overall heat transfer coefficient Uo of 2000 W/m2K, calculate the required water flow rate and the heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counter flow, and (c) shell-and-tube, one shell pass and 2 tube passes.Water flows in the tube side of length 9 ft and ¾ in. O.D. with wall thickness = 0.083 in. Determine the velocity of water.

Solution -----------------------------------------------------------------------------------------

Calculate the required water flow rate

q = cp,h(Thi Tho) = (4)(3800)(100 50) = 760,000 J/shm

The required water flow rate is

= = = 9.09 kg/scm, ( )p c co ci

qc T T

760,000(4180)(35 15)

(a) Parallel flow heat exchanger

T1 = Thi Tci = 100 15 = 85oC

T2 = Tho Tco = 50 35 = 15oC

Tlm = = = 40.36oC

1 2

1 2ln /T T

T T

85 15

ln 85 /15

Area required for parallel flow heat exchanger

Ao = = = 9.42 m2

o lm

qU T

760,000(2000)(40.36)

(b) Counter flow heat exchanger

T1 = Thi Tco = 100 35 = 65oC

T2 = Tho Tci = 50 15 = 35oC

Tlm = = = 48.46oC

1 2

1 2ln /T T

T T

65 35

ln 65 / 35

7-18

Area required for counter flow heat exchanger

Ao = = = 7.841 m2

o lm

qU T

760,000(2000)(48.46)

(c) Shell-and-tube, one shell pass and 2 tube passes, heat exchanger

Tlm = = = 48.46oC

1 2

1 2ln /T T

T T

65 35

ln 65 / 35

Area required for shell-and-tube heat exchanger

Ao = = = m2

o lm

qU T F

760,000(2000)(48.46)F

7.841F

The F factor can be evaluated from

R = = 2.5, P = = 0.2353hi ho

co ci

T TT T

co ci

hi ci

T TT T

F = = 0.9228

2

2

2

1 (1 )ln1 (1 )

2 1 1ln

2 1 1

R PR RP

P R R

P R R

Area required for shell and tube heat exchanger

Ao = = 8.497 m27.8410.9228

Outside area of one tube = π(0.75)(9)/12 = 1.7671 ft2 = 0.1642 m2 ( 1 ft2 = .0929 m2)

Number of tubes required = 8.497/0.1642 = 52 tubes

Number of tubes per pass = 52/2 = 36

Cross-sectional area of one tube = π(0.75 2×0.083)2/4/144 = 1.86×10-3 ft2 = 1.728×10-4 m2

Water velocity in tube = 9.09/(1000×36×1.728×10-4) = 1.46 m/s

7-19

Example 7.3-2. ----------------------------------------------------------------------------------Oil having a specific heat of 2350 J/kgK is to be cooled from 160oC to 100oC with chilled water (specific heat of 4181 J/kgK) which is supplied at a temperature of 15oC and a flow rate of 2.5 kg/s. Water flows on the tube side with heat transfer coefficient of 3060 W/m2K and oil on the shell side with heat transfer coefficient of 400 W/m2K. There are 10 thin wall multiple-passes tubes with diameter of 25 mm. If the water outlet temperature is 85oC, determine the heat transfer rate, the oil mass flow rate, and the tube length.

Solution -----------------------------------------------------------------------------------------

(a) The heat transfer rate

qc = cp,c(Tco Tci) = (2.5)(4181)(85 15) = 7.312×105 Wcm

(b) The oil mass flow rate

= = = 5.19 kg/shm, ( )p h hi ho

qc T T

57.312 10(2350)(160 100)

(c) The tube length

Area required for shell-and-tube heat exchanger with one shell pass and multiple of two tube passes.

A = lm

qU T F

T1 = Thi Tco = 160 85 = 75oC

T2 = Tho Tci = 100 15 = 85oC

Tlm = = = 79.9oC

1 2

1 2ln /T T

T T

75 85

ln 75 / 85


R = = 0.86, P = = 0.48hi ho

co ci

T TT T

co ci

hi ci

T TT T

F = = 0.88

2

2

2

1 (1 )ln1 (1 )

2 1 1ln

2 1 1

R PR RP

P R R

P R R

7-20

For thin wall

U = = = 354 W/m2K11 1

i oh h

11 1

3060 400

A = = = 29.4 m2

lm

qU T F

57.312 10(354)(79.9)(0.88)

Length of tube L = = = 37.4 mtube

An D

29.4(10)( )(0.025)

---------------------------------------------

Equation q = UAFTlm can easily be used to determine either q or UA if the inlet fluid temperatures are known and the outlet temperatures are specified or readily determined from the following energy balance

q = cph(Thi - Tho) = cpc(Tco - Tci) (7.2-3)hm cm

where = mass flow rate of the hot stream hm = mass flow rate of the cold stream cm

This procedure is called the log mean temperature difference (LMTD) method. If only the inlet temperatures are known, an iterative procedure is required by the LMTD method.

Example 7.3-3. ----------------------------------------------------------------------------------A process fluid at 100oC having a specific heat of 3800 J/kgK and flowing at 4 kg/s is to be cooled with chilled water (specific heat of 4180 J/kgK). Cooling water with flow rate of 9.09 kg/s is available at 15oC. Assuming an overall heat transfer coefficient Uo of 2000 W/m2K, calculate the outlet temperatures and the heat transfer rate for a counter flow heat exchanger with heat transfer area of 7.841 m2.

Solution -----------------------------------------------------------------------------------------

The heat transfer rate can be obtained from the energy balances:

q = cph(Thi - Tho) = cpc(Tco - Tci)hm cm

q = (4)(3800)(100 - Tho) = (9.09)(4180)( Tco - 15)

q = (15,200)(100 - Tho) = (38,000)( Tco - 15)

7-21

q = (15,200)(100 - Tho) = UATlm = (2000)(7.841) (100 ) ( 15)100ln

15

co ho

co

ho

T TT

T

We need to solve the following two equations for two unknowns Tho and Tco:

(15,200)(100 - Tho) = (38,000)( Tco - 15)

(15,200)(100 - Tho) = (2000)(7.841) (100 ) ( 15)100ln

15

co ho

co

ho

T TT

T

The following Matlab codes can be used to solve the above equations: (Note: The codes between the dash line (----) must be saved under the file name “LMTD” before the statement p=fminsearch('LMTD',[60 40]) can be used in the command window to find the solution).

-----------------------function y=LMTD(p)Tho=p(1);Tco=p(2);f1=(15200)*(100 - Tho) - (38000)*( Tco - 15);f2=(15200)*(100 - Tho) - 2000*7.841*((100 - Tco) - (Tho - 15))/log((100 - Tco)/(Tho - 15));y=f1*f1+f2*f2;---------------------->> p=fminsearch('LMTD',[60 40])

p =

5.0001e+001 3.5000e+001

Therefore Tho = 50oC and Tco = 35oC. The heat transfer rate is then

q = (15,200)(100 - Tho) = (15,200)(100 - 50) = 760,000 W---------------------------------------------

In Example 7.3-3 we need to solve two nonlinear algebraic equations since the outlet temperatures are unknown. There is alternative procedure called the effectiveness-NTU method that may be used to avoid an iterative procedure by the LMTD method.

7-22

7.4 The Effectiveness-NTU Method

If only the inlet temperatures are known, the effectiveness-NTU method may be used to avoid an iterative procedure by the LMTD method. The NTU method can also be applied even when this is not the case. From the parallel flow heat exchanger analysis in section 7.3 we list equation (7.3-18)

ln = UoAo (7.3-18)hi ci

ho co

T TT T

1 1

h cC C

Tube

Shell (insulated)

Thi Tho

Tc i

Tc o

shell (insulated)

tube

Figure 7.4-1 Parallel flow heat exchanger.

We now define Cmin to be the minimum of Cc and Ch and Cmax to be the maximum of Cc and Ch. We also define T1 = Thi Tci and T2 = Tho Tco. Equation (7.3-18) becomes

ln = UA (7.4-1)1

2

TT

min max

1 1C C

Factoring out Cmin and defining NTU to be we havemin

UAC

ln = = NTU2

1

TT

min

UAC

min

max

1 CC

min

max

1 CC

or

= = exp (7.4-2)2

1

TT

ho co

hi ci

T TT T

min

max

1 CNTUC

We now define the effectiveness of the heat exchanger as the ratio of the actual heat transfer rate to the maximum heat transfer rate:

= (7.4-3)max

qq

7-23

In this equation, the actual heat transfer rate is given by

q = cph(Thi Tho) = cpc(Tco Tci) (7.4-4)hm cm

The maximum heat transfer rate qmax is defined for the heat transfer in an infinite long counter flow heat exchanger as shown in Figure 7.4-2 for the cases Ch > Cc and Ch < Cc.

Thi

Tco

Tho

Tc i

Thi

Tco

Tho

Tc i

C > Ch c C < Ch c

Figure 7.4-2 Infinite long counter flow heat exchanger.

In an infinite long counter flow heat exchanger, when the hot stream has more energy that can be received by the cold stream (Ch > Cc), the outlet temperature of the cold stream Tco

will eventually reach the inlet temperature of the hot stream, i.e., Tco = Thi. Hence

qmax = cpc(Tco Tci) = Cc(Thi Tci) = Cmin(Thi Tci)cm

On the other hand, when all the energy of the hot stream can be received by the cold stream (Ch < Cc), the outlet temperature of the hot stream Tho will eventually reach the inlet temperature of the cold stream, i.e., Tho = Tci. Hence

qmax = cph(Thi Tho) = Ch(Thi Tci) = Cmin(Thi Tci)hm

The maximum heat transfer rate is therefore always given by

qmax = Cmin(Thi Tci) (7.4-5)

The heat exchanger effectiveness is then

= =

=

(7.4-6)

max

qq

min

h hi ho

hi ci

C T TC T T

min

c co ci

hi ci

C T TC T T

If Cmin = Ch then =

hi ho

hi ci

T TT T

7-24

If Cmin = Cc then =

co ci

hi ci

T TT T

We now want to express = exp in term of only. Adding and ho co

hi ci

T TT T

min

max

1 CNTUC

subtracting Thi and Tci

= = ho co

hi ci

T TT T

ho hi hi co ci ci

hi ci

T T T T T TT T

ho hi hi ci co ci

hi ci

T T T T T TT T

= + ho co

hi ci

T TT T

ho hi

hi ci

T TT T

hi ci

hi ci

T TT T

co ci

hi ci

T TT T

From Equation (7.4-6),

= = min

h hi ho

hi ci

C T TC T T

ho hi

hi ci

T TT T

min

max

CC

Let Cmin = Cc and Cmax = Ch then = . Hence

co ci

hi ci

T TT T

= + 1 = 1 = exp ho co

hi ci

T TT T

min

max

CC

min

max

1CC

min

max

1 CNTUC

Solving for , we obtain with the definition Cr = Cmin/Cmax

= = (7.4-7)

min

max

min

max

1 exp 1

1

CNTUC

CC

1 exp 11

r

r

NTU CC

Exercise: If Cmin = Ch and Cmax = Cc show that the above equation is still valid.

In general = f . If NTU, Cmin, Cmax, Thi, and Tci are known then Tho and Tco can , rNTU Cbe computed directly with the effectiveness-NTU method with no iteration as with the LMTD method.

= =

=

max

qq

min

h hi ho

hi ci

C T TC T T

c co ci

h hi ci

C T TC T T

Since is known, the above equation can be solved for Tho and Tco.

7-25

Chapter 7

Table 7.4-1 shows the expressions of effectiveness for a variety of heat exchangers. For shell-and-tube heat exchanger with n shell passes, (NTU)1 would first be calculated using the heat transfer area for one shell, 1 would then be calculated from Equation (7.4-9), and would finally be calculated from Equation (7.4-10). For Cr = 0, as in a boiler or condenser, the heat exchanger behavior is independent of flow arrangement and is given by Equation (7.4-14). Figure 7.4-3 shows the effectiveness of parallel and counter flow heat exchanger.

Table 7.4-1 Heat Exchanger Effectiveness Relations1

Concentric tubeParallel flow

= 1 exp 1

1r

r

NTU CC

(7.4-7)

Counter flow =

1 exp 11 exp 1

r

r r

NTU CC NTU C

For Cr = 1, = 1

NTUNTU

(7.4-8)

Shell-and-tubeOne shell pass

2, 4,…tube passes1 = 2

11/ 221

2 1/ 21/ 22

1

1 exp 11 (1 )

1 exp 1

r

r r

r

NTU CC C

NTU C

(7.4-9)

n Shell pass2n, 4n,…tube passes = 1

1

1 11

n

rC

1

1

1

11

n

rr

C C

(7.4-10)

Cross-flow (single pass)Both fluids unmixed

= 1 exp 0.22 0.781 ( ) exp ( ) 1rr

NTU C NTUC

(7.4-11)

Cmax(mixed),Cmin(unmixed), = (1 exp{Cr

-1[1 exp( NTU)]})1

rC

(7.4-12)

Cmin(mixed),Cmax(unmixed),

= (1 exp{Cr-1[1 exp( NTU)]}) (7.4-13)

Exchangers with Cr = 0 = 1 exp( NTU) (7.4-14)

1 Incropera, F. P. and DeWitt, D. P., Fundamentals of Heat Transfer, Wiley, 2007, pg. 689

7-26

Figure 7.4-3 Effectiveness of parallel and counter flow heat exchanger.

Example 7.4-1. ----------------------------------------------------------------------------------

Derive the relation = for a parallel-flow concentric tube heat r

r

CCNTU

1)]1(exp[1

exchanger. You can assume < and first evaluate the outlet temperature of the cpccm , hphcm ,

cold fluid Tco in terms of the known parameters: Tci, Thi, , , A, and U.cm hm

Parallel flow

x x+dx ThoThi

Hot fluid

Cold fluid

Shell volume dV = Adx

Tci Tco

Hot fluidCold fluidx

Figure E7.4-1. A schematic of a parallel-flow concentric tube heat exchanger.

Solution -----------------------------------------------------------------------------------------

7-27

Choose a control volume Acdx at a distance x from the inlets where Ac is the cross-secrional area of the tube. Appling the energy balance around this control volume gives

Tc|x Tc|x+dx + UPdx(Th Tc) = 0cpccm , cpccm ,

Let Cmin = . Rearranging the equation and dividing by dx givescpccm ,

= UP(Tc Th) min c cx dx x

C T Tdx

In the limit dx 0, we have

Cmin = UP(Tc Th) (E-1)cdTdx

In this equation, P is the perimeter of the inner tube. Before integrating the equation over the length of the heat exchanger we need to express Th as a function of Tc. Choose a control volume over the first part of the heat exchanger from 0 to x, the energy supplied to the cold stream by the hot stream is given by

qx = (Thi Th)hphcm ,

qx is also the energy received by the cold stream, therefore

qx = (Thi Th) = (Tc Tci)hphcm , cpccm ,

Cmax(Thi Th) = Cmin(Tc Tci)

x ThoThi

Hot fluid

Cold fluid

Tci Tco

Th

Tc

Th

Solving for Th gives

Th = Thi (Tc Tci) = Thi Cr(Tc Tci)min

max

CC

7-28

Substituting the expression for Th into Eq. (E-1) we obtain

Cmin = UP(Tc Th) = UP[Tc Thi + Cr(Tc Tci)]cdTdx

Cmin = UP[(1 + Cr)Tc Thi Cr Tci]cdTdx

Separating the variables and integrating over the length of the heat exchanger gives

= (1 )

c

r c hi r ci

co

ci

T

TdT

C T T C T min

UPC 0

Ldx

ln = = = NTU11 rC

(1 )(1 )

r co hi r ci

r ci hi r ci

C T T C TC T T C T

min

UPLC min

UAC

The outlet temperature Tco of the cold stream can then be evaluated from the following expression:

= exp[ NTU(1 + Cr)] (E-2)(1 )(1 )

r co hi r ci

r ci hi r ci

C T T C TC T T C T

We can rearrange equation (E-2) to solve for .

Since = = = max

qq

min

min

( )( )

co ci

hi ci

C T TC T T

co ci

hi ci

T TT T

The left hand side of Eq. (E-2) will be expressed in terms of

= exp[ NTU(1 + Cr)]( )r co ci co hi

ci hi

C T T T TT T

Cr + = exp[ NTU(1 + Cr)]co ci ci hi

ci hi

T T T TT T

Cr + 1 = (1 + Cr) + 1= exp[ NTU(1 + Cr)]

Hence

= r

r

CCNTU

1)]1(exp[1

7-29

Example 7.4-2. ----------------------------------------------------------------------------------(p. 9.271) Gray whales have counter-flow heat exchange in their tongues to preserve heat. The tip of the tongue is cooled with the cold sea water. The heat exchange is between the incoming warm bloodstream (entering with the deep-body temperature) flowing through the arteries and the outgoing cold bloodstream (leaving the tongue surface region) flowing through the veins. This is shown in Figure 7.4-1. In each heat exchanger unit, nine veins of diameter Dc completely encircle (no heat loss to the surroundings) the central artery of diameter Dh. The length of the heat exchange region is L. Determine the exit temperature of the cold bloodstream Tc,o. The inlet temperature of the cold bloodstream Tc,i is 2oC. The inlet temperature of the warm bloodstream Th,i is 36oC. L = 55 cm, Dh = 3 mm, Dc = 1 mm, uh = 1 mm/s, uc = 1 mm/s. The resistance to heat transfer by conduction through the tongues tissues is Rcond = 5oC/ W. The resistance in the bloodstreams can be obtained from the following relations

= Ah Nuh , = Ac Nuc , for Re < 10, Nu = 4.36hconvR ,

1

h

f

Dk

cconvR ,

1

c

f

Dk

Use the following properties for blood: ρf = 1,000 kg/m3, kf = 0.590 W/m-K , νf = 1.13 × 10-6 m2/s, cp,f = 4,186 J/kg-K

Figure E7.4-2. A schematic of the vascular heat exchanger in the gray whale tongue.

Solution -----------------------------------------------------------------------------------------

Since only the inlet temperatures are given, we will use the effectiveness-NTU method to solve for the exit temperature of the cold blood stream. First, we evaluate the Reynolds numbers in the artery and in the veins

Reh = = = 2.655f

hh Du )/(1013.1

)(103)/(1026

33

smmsm

1 Kaviany, Principles of Heat Transfer, Wiley, 2002, p. 350

7-30

Rec = = = 0.885f

cc Du )/(1013.1

)(10)/(1026

33

smmsm

The values of the Reynolds number are less than 10. Therefore we can use the given expression to determine the thermal resistance in the blood streams.

= Ah Nuh = DhLNuh = LNuhkfhconvR ,

1

h

f

Dk

h

f

Dk

= (0.55)(4.36)(0.59) = 4.448 W/oK Rconv,h = 0.225 K/WhconvR ,

1

= Ac Nuc = 9DcLNuc = 9LNuhkfcconvR ,

1

c

f

Dk

c

f

Dk

= 9(0.55)(4.36)(0.59) = 40.0 W/oK Rconv,c = 0.025 K/WcconvR ,

1

The total resistance is the sum of the individual resistances

Rt = Rconv,c + Rcond + Rconv,h = 0.025 + 5.0 + 0.225 = 5.25 K/W

UA = = 0.1905 W/KtR

1

The cold and hot blood flow rates are given by

= 9f uc = (9)(1000) 10-3 = 7.0710-6 kg/scm4

2cD

4)10( 23

= f uc = (1000) 10-3 = 7.0710-6 kg/shm4

2hD

4)103( 23

The hot and cold heat capacity rates are then calculated

Ch = = (7.0710-6)(4186) = 0.0296 W/Khphcm ,

Cc = = (7.0710-6)(4186) = 0.0296 W/Kcpccm ,

Therefore Cr = = 1 and the number of transfer unit NTU is: max

min

CC

7-31

NTU = = = 6.437minC

UA0296.01905.0

For counter flow heat exchanger with Cr = 1, we have

= = 0.8656 = = NTU

NTU1 )(

)(

,,min

,,

icih

icocc

TTCTTC

)()(

,,

,,

icih

icoc

TTTT

The exit temperature of the cold bloodstream, Tc,o, is finally

Tc,o = Tc,i + 0.8656(Th,i Tc,i) = 2 + 0.8656(36 2) = 31.43oC

Example 7.4-3. ----------------------------------------------------------------------------------

Oil having a specific heat of 2000 J/kgK and flowing at 1 kg/s is to be cooled from 340 K to 310 K with water. Cooling water is available at 290 K with the outlet temperature limited to 300 K.

(a) Determine UA for a shell-and-tube heat exchanger with one shell pass and two tube passes.

(b) Water enters the exchanger with UA = 2212 W/K at 290 K and a capacitance rate of 6300 W/K. If oil with a specific heat of 2100 J/kgK enters the exchanger at 370 K and 0.75 kg/s, determine the outlet temperatures of oil and water.

Solution -----------------------------------------------------------------------------------------(a) Determine UA

T1 = Thi Tco = 340 300 = 40 K

T2 = Tho Tci = 310 290 = 20 K

Tlm = = = 28.85 K

1 2

1 2ln /T T

T T

40 20

ln 40 / 20

UA required for shell-and-tube heat exchanger

AU = lm

qT F

q = cp,h(Thi Tho) = (1)(2000)(340 310) = 60,000 J/shm


R = = 3, P = = 0.2hi ho

co ci

T TT T

co ci

hi ci

T TT T

7-32

F = = 0.935

2

2

2

1 (1 )ln1 (1 )

2 1 1ln

2 1 1

R PR RP

P R R

P R R

AU = = = 2224 W/Klm

qT F

60,000(28.85)(0.935)

(b) Outlet temperatures of oil and water

We use the effectiveness-NTU method.

Ch = (2100)(0.75) = 1575 W/K < Cc = 6300 W/K

NTU = = = 1.4044, Cr = = = 0.25min

UAC

22121575 max

min

CC 1575

6300

= 2 = 0.686

11/ 22

2 1/ 21/ 22

1 exp 11 (1 )

1 exp 1

r

r r

r

NTU CC C

NTU C

= 0.686 = = min

( )( )

h hi ho

hi ci

C T TC T T

hi ho

hi ci

T TT T

Tho = Thi 0.686(Thi Tci) = 370 0.686(370 290) = 315.1 K

= 0.686 = min

( )( )

c co ci

hi ci

C T TC T T

Tco = Tci + 0.686(Thi Tci)

= 290 + 0.686(370 290)

= 303.7 Kmin

c

CC

15756300

7-33

Chapter 7

7.5 Special Operations for Liquid/Gas Heat Exchangers

The effect of the heat exchanger configuration on its performance depends on the interaction between the temperature changes of the two fluid streams. When one fluid stream has a capacity rate that is much greater than the other then one fluid stream does not change its temperature significantly and the capacity ratio approaches zero. This condition occurs when there is a phase change: either boiling or condensation. Let consider a counter flow and parallel flow exchanger where the cold fluid boils. The temperature distributions shown in Figure 7.5-1(a) and (b) are identical when Tco = Tci and therefore configuration is not important.

x

Thi

Tho

Tc iTco

(a) x

Thi

Tho

TcoTc i

(b)Figure 7.5-1 Temperature distribution in an exchanger when the cold fluid boils.

The energy received by the cold fluid is given by

q = Cc(Tco Tci)

Since Tco Tci because of phase change Cc will approach infinity: Cc ∞. The temperature distributions for the cases where the hot fluid condenses are shown in Figure 7.5-2(a) and (b). The energy supplied by the hot fluid is given by

q = Ch(Thi Tho)

For this case Thi Tho and Ch will approach infinity: Ch ∞.

x

ThiTho

Tc i

Tco

(a) x

Thi Tho

Tco

Tc i

(b)Figure 7.5-2 Temperature distribution in an exchanger when the hot fluid condenses.

7-34

Consider the counter flow heat exchanger where Cc = Ch, since

q = Cc(Tco Tci) = Ch(Thi Tho)

Then

Tco Tci = Thi Tho Thi Tco = Tho Tci.

For counter flow T1 = Thi Tco and T2 = Tho Tci, therefore T1 = T2. The log mean temperature is just an average temperature, hence

Tlm =

= = T1 = T2

ciho

cohi

cihocohi

TTTT

TTTT

ln

)()(

1 2

1 2ln /T T

T T

The above result could be proved using L’Hospital ‘s rule. When 0.33 < < 3, the geometric 1

2

TT

average Tave = 0.5(T1 + T2) is within 10% of the log mean average as shown in Figure 7.5-3.

Figure 7.5-3 Tave /Tlm at various T1/T2.

8-1

Chapter 8 Absorption and Stripping

8.1 Introduction

In absorption (also called gas absorption, gas scrubbing, or gas washing), there is a transfer of one or more species from the gas phase to a liquid solvent. The species transferred to the liquid phase are referred to as solutes or absorbate. Absorption involves no change in the chemical species present in the system. Absorption is used to separate gas mixtures, remove impurities, or recover valuable chemicals. The operation of removing the absorbed solute from the solvent is called stripping. Absorbers are normally used with strippers to permit regeneration (or recovery) and recycling of the absorbent. Since stripping is not perfect, absorbent recycled to the absorber contains species present in the vapor entering the absorber. When water is used as the absorbent, it is normally separated from the solute by distillation rather than stripping.

1

30

Exit gas25 C, 90 kPao

kmol/h O 0.009 N 0.017Water 1,926.0Acetone 10.25

2

2

Liquid absorbent25 C, 101.3 kPao

kmol/hWater 1943

Feed gas25 C, 101.3 kPao

Exit liquid 22 C, 101.3 kPao

kmol/hArgon 6.9 O 144.3 N 536Water 5.0Acetone 10.3

2

2

kmol/hArgon 6.9 O 144.291 N 535.983Water 22.0Acetone 0.05

2

2

Figure 8.1-1 Typical absorption process.

A typical industrial operation for an absorption process is shown in Figure 8.1-11. The feed, which contains air (21% O2, 78% N2, and 1% Ar), water vapor, and acetone vapor, is the gas


8-2

leaving a dryer where solid cellulose acetate fibers, wet with water and acetone, are dried. Acetone is removed by a 30-tray absorber using water as the absorbent. The percentage of acetone removed from the air stream is

×100 = 99.5%10.2510.3

Although the major component absorbed by water is acetone, there are also small amounts of oxygen and nitrogen absorbed by the water. Water is also stripped since more water appears in the exit gas than in the feed gas. The temperature of the exit liquid decreases by 3oC to supply the energy of vaporization needed to strip the water. This energy is greater than the energy of condensation liberated from the absorption of acetone.

Three approaches have generally been employed to develop equations used to predict the performance of absorbers and absorption equipment: mass transfer coefficients, graphical solution, and absorption factor. The use of mass transfer coefficient is covered in Chapter 2.2. The graphical solution is simple to use for one or two components and provides explicit graphical presentation of the interrelationships of the variables and parameters in an absorption process. However the graphical technique becomes very tedious when several solutes are present and must be considered. The absorption factor approach can be utilized either for hand or computer calculation. Absorption and stripping are conducted mainly in packed columns and plate columns (trayed tower) as shown in Figure 8.1-2.

Packed column2 Plate column3

Figure 8.1-2 Equipment for absorption and stripping.

2 www.mikropul.com/products/wscrubber/packed.htm (Aug. 25 2009)3 http://www.cgscgs.com/ga_tt.htm (Aug. 25 2009)

http://www.mikropul.com/products/wscrubber/packed.html

http://www.cgscgs.com/ga_tt.htm

8-3

8.2 Single-Component Absorption

Most absorption or stripping operations are carried out in counter current flow processes, in which the gas flow is introduced in the bottom of the column and the liquid solvent is introduced in the top of the column. The mathematical analysis for both the packed and plated columns is very similar.

Lx

t

A,t

Lx

b

A,b

Vy

t

A,t

Vy

b

A,b

L V

Figure 8.2-1 Countercurrent absorption process.

The overall material balance for a countercurrent absorption process is

Lb + Vt = Lt + Vb (8.2-1)

where V = vapor flow rateL = liquid flow ratet, b = top and bottom of tower, respectively

The component material balance for species A is

LbxA,b + Vt yA,t = LtxA,t + Vb yA,b (8.2-2)

where yA = mole fraction of A in the vapor phasexA = mole fraction of A in the liquid phase

For some problems, the use of solute-free basis can simplify the expressions. The solute-free concentrations are defined as:

= = (8.2-3a)AX1

A

A

xx

mole fraction of A in the liquidmole fraction of non-A components in the liquid

= = (8.2-3b)AY1

A

A

yy

mole fraction of A in the vapormole fraction of non-A components in the vapor

If the carrier gas is completely insoluble in the solvent and the solvent is completely nonvolatile, the carrier gas and solvent rates remain constant throughout the absorber. Using

8-4

to denote the flow rate of the nonvolatile and to denote the carrier gas flow rate, the L Vmaterial balance for solute A becomes

+ = + (8.2-4)L ,A bX V ,A tY L ,A tX V ,A bY

or = + (8.2-5),A tY LV ,A tX ,

,A b

A b

LXY

V

The material balance for solute A can be applied to any part of the column. For example, the material balance for the top part of the column is

= + (8.2-6),A tY LV ,A tX A

ALXYV

In this equation, and are the mole ratios of A in the liquid and vapor phase, AX AYrespectively, at any location in the column including at the two terminals. Equation (8.2-6) is

called the operation line and is a straight line with slope when plotted on -LV AX AY

coordinates.

The equilibrium relation is frequently given in terms of the Henry’s law constant which can be expressed in many different ways:

PA = HCA = mxA = KxA (8.2-7)

In this equation, PA is the partial pressure of species A over the solution and CA is the molar concentration with units of mole/volume. The Henry’s law constant H and m have units of pressure/molar concentration and pressure/mole fraction, respectively. K is the equilibrium constant or vapor-liquid equilibrium ratio. Table 8.2-1 list Henry’s law constant m for various gases in water.

Table 8.2-1 Henry’s Law constant for Gases in water4 (m×10-4 atm/mole fraction)T(oC) CO2 CO C2H6 C2H4 He H2 H2S CH4 N2 O2

010203040

0.07280.1040.1420.1860.233

3.524.425.366.206.96

1.261.892.633.424.23

0.5520.7681.021.27

12.912.612.512.412.1

5.796.366.837.297.51

0.02680.03670.04830.06090.0745

2.242.973.764.495.20

5.296.688.049.2410.4

2.553.274.014.755.35

Example 8.2-1. 5----------------------------------------------------------------------------------A solute A is to be recovered from an inert carrier gas B by absorption into a solvent. The gas entering into the absorber flows at a rate of 500 kmol/h with yA = 0.3 and leaving the absorber with yA = 0.01. Solvent enters the absorber at the rate of of 1500 kmol/h with xA =

4 Geankoplis, C.J., Transport Processes and Separation Process Principles, 4th edition, Prentice Hall, 2003, pg. 9885 Hines, A. L. and Maddox R. N., Mass Transfer: Fundamentals and Applications, Prentice Hall, 1985, pg. 255

8-5

0.001. The equilibrium relationship is yA = 2.8 xA. The carrier gas may be considered insoluble in the solvent and the solvent may be considered nonvolatile. Construct the x-y plots for the equilibrium and operating lines using both mole fraction and solute-free coordinates.

Solution -----------------------------------------------------------------------------------------

The flow rates of the solvent and carrier gas are given by

= Lt(1 xA,t) = 1500(1 0.001) = 1498.5 kmol/hL

= Vb(1 yA,b) = 500(1 0.3) = 350 kmol/hV

The concentration of A in the solvent stream leaving the absorber can be determined from the following expressions:

xA,b = b

b

Moles A in Moles A in +

LL L

Moles of A in Lb = Moles of A in Lt + Moles of A in Vb Moles of A in Vt

Moles of A in Lb = 1500×0.001 + 500×0.3 Moles of A in Vt

yA,t = 0.01 = t

t


VV V

t

t

Moles A in Moles A in + 350

VV

Moles of A in Vt = 350×0.01/(1 0.01) = 3.5354 kmol/h

Moles of A in Lb = 1.500 + 150 3.5354 = 147.965 kmol/h

xA,b = = = 0.0898b

b


LL L

147.965147.965 + 1498.5

For the solute free basis:

= , = AX1

A

A

xx AY

1A

A

yy

= = = 0.0010,A tX ,

,1A t

A t

xx

0.00101 0.0010

= = = 0.0987,A bX ,

,1A b

A b

xx

0.08981 0.0898

8-6

= = = 0.0101,A tY ,

,1A t

A t

yy

0.0101 0.010

= = = 0.4286,A bY ,

,1A b

A b

yy

0.301 0.30

The equilibrium curves in both mole fraction and solute-free coordinates are calculated from the following procedures:

1) Choose a value of xA between 0.001 and 0.10

2) Evaluate the corresponding = AX1

A

A

xx

3) Evaluate yA = 2.8 xA

4) Evaluate the corresponding = AY1

A

A

yy

The operating lines in both mole fraction and solute-free coordinates are calculated from the following procedures:

1) Choose a value of xA between 0.001 and 0.0898

2) Evaluate the corresponding = AX1

A

A

xx

3) Evaluate = + AY LV AX ,

,A b

A b

LXY

V

4) Evaluate the corresponding yA = 1

A

A

YY

The following Matlab codes plot the equilibrium and operating lines in both mole fraction and solute-free coordinates.

-------------------------------------------% Example 8.2-1xe=linspace(0.001,0.1);ye=2.8*xe;Xe=xe./(1-xe);Ye=ye./(1-ye);x=linspace(0.001,0.0898);Lbar=1498.5;Vbar=350;X=x./(1-x);Xb=.0898;Yb=0.4286;LoV=Lbar/Vbar;Y=LoV*X+Yb-LoV*Xb;y=Y./(1+Y);plot(xe,ye,x,y,'--')legend('Equilibrium curve','Operating line',2)xlabel('x');ylabel('y')Title('Equilibrium and Operating lines on mole fraction coordinates')figure(2)plot(Xe,Ye,X,Y,'--')legend('Equilibrium curve','Operating line',2)

8-7

xlabel('X');ylabel('Y')Title('Equilibrium and Operating lines on solute free coordinates')

-------------------------------------------

The equilibrium relation in the mole fraction coordinates is a straight line while the operating line in the solute-free coordinates is a straight line. Normally the equilibrium relation is not a straight line in the mole fraction coordinates. Therefore it is advantage to use solute-free coordinates because the operating line will always be straight.

8-8

Y

X

Yb

Yt

Xt Xb

Yb

Yb

Yt

Yt

XtXtXbXb

Yc

Xc

(A) (B) (C)

Figure 8.2-2 Limiting conditions for absorption process.

The driving force for mass transfer becomes zero whenever the operating line intersects or touches the equilibrium curve. This limiting condition represents the minimum solvent rate to recover a specified quantity of solute or the solvent rate required to remove the maximum amount of solute. In Figure 8.2-2A, the intersection of the equilibrium and operating lines occurs at the bottom of the absorber. This condition defines the minimum solvent rate to recover a specified quantity of solute. This minimum solvent rate can be calculated from the following expression:

= (8.2-8a)min

LV

b t

b t

Y YX X

In Figure 8.2-2B, the intersection of the equilibrium and operating lines occurs at the top of the absorber. This condition represents the solvent rate required to remove the maximum amount of solute. This solvent rate can be calculated from the following expression:

= (8.2-8b)max

LV

b t

b t

Y YX X

Equation (8.2-8b) is exactly the same as Eq. (8.2-8a) except in this case the bottom compositions are fixed so that the maximum slope of the operating line occurs when the operating line intersects the equilibrium curve at the top of the column.

Figure 8.2-2C shows the case when the operating line becomes tangent to the equilibrium curve. The minimum liquid-to-vapor ratio for this case can be determined from

= (8.2-9)min

LV

c t

c t

Y YX X

In this equation, and are the coordinates of the tangent point. cY cX

8-9

Chapter 8 Example 8.2-2. 6----------------------------------------------------------------------------------A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 20oC. The entering gas contains 20 mol % SO2 and that leaving 2 mol % at a total pressure of 101.3 kPa. The inert gas flow rate is 150 kg air/hm2, and the entering water flow rate is 6000 kg water/hm2. Assuming an overall tray efficiency of 25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 20oC. Equilibrium data for SO2 –water system at 20oC and 101.3 kPa are given:

x 0 .0001403 .000280 .000422 .000564 .000842 .001403 .001965 .00279y 0 .00158 .00421 .00763 .01120 .01855 .0342 .0513 .0775x .00420 .00698 .01385 .0206 .0273y .121 .212 .443 .682 .917

Solution -----------------------------------------------------------------------------------------

The vapor and liquid molar flow rates are calculated first

+ = + L ,A bX V ,A tY L ,A tX V ,A bY

= 150/29 = 5.18 kmol inert air/hm2V

= 6000/18 = 333 kmol inert water/hm2L

We have yb = 0.20, yt = 0.02, and xt = 0. For the solute-free basis

= , = AX1

A

A

xx AY

1A

A

yy

= = = 0,A tX ,

,1A t

A t

xx

01 0

= = = 0.0204,A tY ,

,1A t

A t

yy

0.0201 0.020

= = = 0.250,A bY ,

,1A b

A b

yy

0.201 0.20

can be determined from the component balance (A = SO2):,A bX

+ = + L ,A bX V ,A tY L ,A tX V ,A bY


8-10

= + ,A bX ,A tX VL ,A bY V

L ,A tY

= 0 + ×0.250 ×0.0204 = 0.00357,A bX 5.18333

5.18333

The operating line and the equilibrium curve can be plotted using the following Matlab codes:

% Example 8.2-2xe=[0 .0001403 .000280 .000422 .000564 .000842 .001403 .001965 .00279 .00420 .00698];ye=[0 .00158 .00421 .00763 .01120 .01855 .0342 .0513 .0775 .121 .212];Xe=xe./(1-xe);Ye=ye./(1-ye);X=[0 .00357];Y=[.0204 .25];plot(Xe,Ye,X,Y,'--')legend('Equilibrium curve','Operating line',2)xlabel('X');ylabel('Y')Title('Equilibrium and Operating lines on solute free coordinates')grid on

Figure E-1 Theoretical number of trays.

The number of theoretical trays is determined by simply stepping off the number of trays as shown in Figure E-1. This gives 2.4 theoretical trays. The actual number of trays is 2.4/.25 = 10 trays.

8-11

8.3 Multicomponent Absorption Using Shortcut Methods

The equilibrium relationship between the mole fractions in the vapor and the liquid phases for any equilibrium plate n in the absorber may be expressed as

yn = Knxn (8.3-1)

Equation (8.3-1) can be rewritten in terms of the molar flow rates as

= Kn (8.3-2)n

n

vV

n

n

lL

In this equation, vn is the moles of component i in the vapor stream leaving plate n, Vn is the total moles of the vapor stream leaving plate n, ln is the moles of component i in the liquid stream leaving plate n, and Ln is the total moles of the liquid stream leaving plate n. Solving for the molar flow rate of of component i in the liquid stream leaving plate n gives

ln = vn (8.3-3)n

n n

LK V

The absorption factor for any component on plate n is defined as

An = (8.3-4)n

n n

LK V

123

n

N 123

n

N

V1 L0

VN +1 LN

Vy

n

n

Lx

n-1

n-1

Vy

n+1

n+1

Lx

n

n

Lx

n+1

n+1

Vy

n

n

Lx

n

n

Vy

n-1

n-1

LN+1 VN

L1 V0

Figure 8.3-1 Schematic diagram of an absorber (left) and a stripper (right).

8-12

For the development of equations in multicomponent absorption, all vapor stream compositions are based on the feed gas to the absorber and all liquid stream compositions are based on the lean solvent entering the absorber. For plate n these concentrations are defined as

Xn’ = (8.3-5)0

n nx LL

Yn’ = (8.3-6)1

n n

N

y VV

The material balance around plate n shown in Figure 8.3-1 is

L0 Xn-1’ + VN+1 Yn+1’ = L0 Xn’ + VN+1 Yn’ (8.3-7)

Solving for Yn+1’ gives

Yn+1’ = Xn’ + Yn’ Xn-1’ (8.3-8)0

1N

LV

0

1N

LV

Equation (8.3-8) can be written for the overall column as

YN+1’ = XN’ + Y1’ X0’ (8.3-9)0

1N

LV

0

1N

LV

Equation (8.3-9) is a straight line on Y’-X’ coordinates with a slope of and a Y’ 0

1N

LV

intercept of Y1’ X0’. This is true for any absorption process and is not limited to 0

1N

LV

systems in which the carrier gas is insoluble in the solvent and the solvent is nonvolatile.

Substituting xn = Xn’ and yn = Yn’ into the equilibrium relation yn = Knxn we obtain0

n

LL

1N

n

VV

Yn’ = Kn Xn’1N

n

VV

0

n

LL

Yn’ = Xn’ = Xn’ (8.3-10a)0 1//( )

N

n n n

L VL K V

0 1/ N

n

L VA

Xn’ = Yn’ (8.3-10b)0 1/

n

N

AL V

Substituting Eq. (8.3-10b) into Eq. (8.3-8) gives

8-13

Yn+1’ = Yn’ + Yn’ Yn-1’0

1N

LV 0 1/

n

N

AL V

0

1N

LV

1

0 1/n

N

AL V

Yn+1’ = An Yn’ + Yn’ An-1 Yn-1’

Yn’ = (8.3-11)1 1 1' '1

n n n

n

Y A YA

From (10.3-10a) Y0’ = X0’, for one-plate absorber Eq. (8.3-11) is0 1/ N

n

L VA

Y1’ = = = (8.3-12)2 0 0

1

' '1

Y A YA

0 12 0 0

0

1

/' '

1

NL VY A XAA

02 0

1

1

' '

1N

LY XV

A

Eq. (8.3-11) for an absorber with two plates is

Y2’ = (8.3-13)3 1 1

2

' '1

Y AYA

Substituting Y1’ = into Eq. (8.3-13) we obtain

02 0

1

1

' '

1N

LY XV

A

Y2’ =

02 0

13 1

1

2

' ''

1

1

N

LY XVY A

A

A

(1 + A2)Y2’ = + 1 2

1

'1AY

A 1 3 1 0 0 1

1

1 ' '/1

NA Y A L X VA

(1 + A1)(1 + A2)Y2’ = A1Y2’ + (1 + A1)Y3’ + A1L0X0’/VN+1

Y2’ = (8.3-14) 1 3 1 0 0 1

1 2 2

1 ' '/1

NA Y A L X VA A A

Follow the same procedure for an absorber with three plates

Y3’ = (8.3-15) 1 2 2 4 1 2 0 0 1

1 2 3 2 3 3

1 ' '/1

NA A A Y A A L X VA A A A A A

8-14

Finally, for an absorber with N plates

YN’ = (8.3-16) 1 2 1 2 3 1 1 1 1 2 1 0 0 1

1 2 2 3 3 4

... ... ... 1 ' ... '/... ... ... ... 1

N N N N N N

N N N N

A A A A A A A Y A A A L X VA A A A A A A A A A

An overall component material balance around the column can be applied to eliminate YN’ from the above equation to obtain an expression in terms of the absorber terminals. The overall material balance is

L0(XN’ X0’) = VN+1(YN+1’ Y1’) (8.3-17)

Substituting XN’ = YN’ from Eq. (8.3-10b) into Eq. (8.3-17) gives0 1/

N

N

AL V

ANVN+1YN’ L0X0’ = VN+1(YN+1’ Y1’)

YN’ = (8.3-18)1 1 0 0 1' ' '/N N

N

Y Y L X VA

Equating Eq. (8.3-16) to Eq. (8.3-18) we obtain

+ = 1 1' 'N

N

Y YA

0 0 1'/ N

N

L X VA

1 2 1 2 3 1 1 1

1 2 2 3 3 4

... ... ... 1 '... ... ... ... 1

N N N N

N N N N

A A A A A A A YA A A A A A A A A A

+ 1 2 1 0 0 1

1 2 2 3 3 4

... '/... ... ... ... 1

N N

N N N N

A A A L X VA A A A A A A A A A

= 1 1

1

' ''

N

N

Y YY

1 2 2 3 1

1 2 2 3 3 4

... ... ...... ... ... ... 1

N N N N N

N N N N

A A A A A A A A AA A A A A A A A A A

+ 0 0

1 1

''N N

L XV Y

1 2

1 2 2 3 3 4

... 1... ... ... ... 1

N

N N N N

A A AA A A A A A A A A A

= 1 1

1

' ''

N

N

Y YY

1 2 2 3 1

1 2 2 3 3 4

... ... ...... ... ... ... 1

N N N N N

N N N N

A A A A A A A A AA A A A A A A A A A

(8.3-19)0 0

1 1

''N N

L XV Y

2 3 3 4

1 2 2 3 3 4

... ... ... 1... ... ... ... 1

N N N

N N N N

A A A A A A AA A A A A A A A A A

Equation (8.3-19) can be used to determine the terminal stream flow rates in a multicomponent absorber. However the use of this equation requires values of the liquid and vapor flow rates for each tray in the column in addition to the temperature of each tray, since these are required to determine the absorption factor for the tray. If an average value of the absorption factor is used for each tray then A = A1 = A2 = … = AN, A2 = A1A2, A3 = A1A2 A3, AN = A1A2 …AN. With this assumption, Eq. (8.3-19) becomes

8-15

= (8.3-20)1 1

1

' ''

N

N

Y YY

1 2

1 2

...... 1

N N

N N N

A A A AA A A A

0 0

1 1

''N N

L XV Y

1 2

1 2

... 1... 1

N N

N N N

A A AA A A A

Using the following identities

= AN + AN-1 + … + A2 + A1

1

NA AA

= AN + AN-1 + … + A2 + A + 11 1

1

NAA

= AY0’0 0

1

'

N

L XV

Eq. (8.3-20) becomes

= A1 1

1

' ''

N

N

Y YY

1

1 1

N

N

A AA

0

1

''N

YY

1 2

1 2

... 1... 1

N N

N N N

A A AA A A A

= 1 1

1

' ''

N

N

Y YY

1

1 1

N

N

A AA

0

1

''N

YY

2

1 2

...... 1

N N

N N N

A A A AA A A A

= = 1 1

1

' ''

N

N

Y YY

1

1 1

N

N

A AA

0

1

''N

YY

1

1 1

N

N

A AA

1

1 1

N

N

A AA

1 0

1

' ''

N

N

Y YY

= = a (fraction absorbed) (8.3-21)1 1

1 0

' '' '

N

N

Y YY Y

1

1 1

N

N

A AA

Equation (8.3-21) is known as the Kremer and Brown equation where A is the average absorption factor and Y0’ is the moles of the component in the vapor in equilibrium with the lean solvent per mole of entering wet gas. The average absorption factor is calculated as

Aavg = (8.3-22)0

1avg N

LK V

The value of the average equilibrium constant Kavg is determined for each component at the average temperature and pressure in the absorber. Equation (8.3-21) is plotted in Figure 8.3-2 and may be used for graphical solution.

8-16

Figure 8.3-2 Fraction absorbed for absorber with various trays.

Example 8.3-1. 7----------------------------------------------------------------------------------An absorber containing six theoretical plates is used to process a natural gas stream having the composition listed below:

Component C1 C2 C3 nC4 nC5 nC6Mol % 76.524 13.110 4.938 2.126 2.09 1.208

The gas is available for processing at a pressure of 500 psia. If 65% of the propane in the gas stream is to be absorbed, what will be the composition of the gas stream leaving the absorber. The average absorber temperature may be assumed to be 100oF.

Solution -----------------------------------------------------------------------------------------

Basis: 100 mol of feed gas. The equilibrium constants at 100oF (560oR) and 500 psia are calculated using the following formula:

=============================================================(1) ln K = A/T2 + B C ln(P) + D/P2

(2) ln K = A/T2 + B C ln(P) + D/P, where P is in psia, T is in oRCompound A B C D Form=============================================================Methane 292860 8.2445 .8951 59.8465 (1)

7 Hines, A. L. and Maddox R. N., Mass Transfer: Fundamentals and Applications, Prentice Hall, 1985, pg. 268

8-17

Ethane 687248.2 7.90694 .866 49.02654 (1)Propane 970688.6 7.15059 .76984 6.90224 (2)n-Butane 1280557 7.94986 .96455 0 (1)n-Pentane 1524891 7.33129 .89143 0 (1)n-Hexane 1778901 6.96783 .84634 0 (1)=============================================================

Component C1 C2 C3 nC4 nC5 nC6K 5.744 1.396 0.4891 0.1191 0.0464 0.0190

The absorption factor A for propane is calculated from the Kremer and Brown equation

= a = 0.651

1 1

N

N

A AA

7

7 1A AA

>> fun=inline('(A^7-A)-.65*(A^7-1)');>> A=fsolve(fun,0.1,optimset('Display','off'))

A =

0.6716

From Eq. (8.3-22) Aavg = = Aavg Kavg = (0.6716)(0.4891) = 0.32850

1avg N

LK V

0

1N

LV

The absorption factor for the other components is determined from

Aavg = = 0

1avg N

LK V

0.3285

avgK

Component C1 C2 C3 nC4 nC5 nC6Aavg 0.0572 0.2353 0.6716 2.7582 7.0797 17.2895

The factional absorption, a, is then determined from

a = = 1

1 1

N

N

A AA

7

7 1A AA

Component C1 C2 C3 nC4 nC5 nC6a 0.0572 0.2353 0.6500 0.9986 1.0000 1.0000

Moles of each component absorbed = a×(moles in feed)

8-18

Component C1 C2 C3 nC4 nC5 nC6Moles in feed 76.524 13.110 4.938 2.126 2.09 1.208Moles absorbed 4.3764 3.0846 3.2100 2.1229 2.0900 1.2080

The composition of the gas stream leaving the absorber is given in the following table

Component C1 C2 C3 nC4 nC5 nC6Moles in feed 76.524 13.110 4.938 2.126 2.09 1.208Moles exit stream 72.1476 10.0254 1.7280 0.0031 0 0Mole % in exit stream 85.9882 11.9487 2.0595 0.0037 0 0

A-1

Appendix A A.1 Distillation Calculation using ChemSep Software

“ChemSep is a software system for simulation of distillation, absorption, and extraction operations. ChemSep integrates flash calculations, the classic equilibrium stage column model and a nonequilibrium or rate-based column model in one easy to use program”.1 ChemSep-LITE is available for FREE. It only includes the equilibrium column model and is limited to 10 components and 150 stages.

Click on the cs.bat and the following screen will appear

You can click on the menu: Components, Operation, Properties, Feeds, and Specifications on the left column to specify your operation and conditions. When all the necessary specifications are entered into the program, the menu will have a green check next to it. Let consider the following example:

A butane-pentane splitter is to operate at 8.3 bar with the following feed compositionSaturated liquid feed (kmol/s)

Propane,Isobutane,Normal butane,Isopentane,Normal pentane

C3iC4nC4, light keyiC5, heavy keynC5

.05

.15

.25

.20

.35For a specification of a reflux ratio of 2.5 and split between two components in the bottom iso-pentane/n-pentane = 0.45, determine the compositions of the product streams.

1 http://www.chemsep.org/program/index.html

http://www.chemsep.org/downloads/index.html

A-2

You can enter the data given in the example and the following information into the program:

Operation Thermodynamics Properties FeedsEquilibrium columnSimple distillationTotal condenserPartial reboilerNumber of stages: 14Feed stage: 6

K-value: EOSEquation of state: Soave-RKEnthalpy: Soave-RKLeave blank for Thermodynamic Model Parameters

State: Pressure & Vapor fraction

Specifications Column SpecificationsConstant pressureColumn heat loss: 0Default stage efficiency: 1

Split between two componentsiso-pentane/n-pentane = 0.45

You will need to explore this program to be familiar with the menu and sub-menu. You should read the ChemSep Help and ChemSep Book to know more about the program and its functions. After you enter all the necessary information into the program you will arrive at the following screen:

Choose Quick Solve in the menu and you will get the following results:

A-3

Stream Feed1 Top Bottom

Stage 6 1 14 Pressure (N/m2) 830000 830000 830000 Vapour fraction (-) 0.000000 0.000000 0.000000 Temperature (K) 355.242 338.339 384.767 Enthalpy (J/kmol) -1.428E+07 -1.548E+07 -1.032E+07 Entropy (J/kmol/K) -41313.7 -48735.5 -33575.1

Mole flows (kmol/s) Propane 0.0500000 0.0499994 5.5389E-07 Isobutane 0.150000 0.149712 2.8757E-04 N-butane 0.250000 0.246904 0.00309628 Isopentane 0.200000 0.0630532 0.136947 N-pentane 0.350000 0.0456737 0.304326

Total molar flow 1.00000 0.555342 0.444658

Mole fractions (-) Propane 0.0500000 0.0900335 1.2456E-06 Isobutane 0.150000 0.269586 6.4674E-04 N-butane 0.250000 0.444597 0.00696328 Isopentane 0.200000 0.113539 0.307983 N-pentane 0.350000 0.0822443 0.684406

B-1

Appendix B Previous Exams (2009)

CHE 425 (Fall 2009) __________________ LAST NAME, FIRST

Quiz #1Note: Your answers must be correct to 3 significant figures and have the appropriate units.

I. In a hot combustion chamber, oxygen diffuses through a stagnant film of air with thickness L to the carbon surface where it reacts to make CO and CO2. The mole fraction of oxygen at z = L is 0.21. The reaction may be assumed to be instantaneous. No reaction occurs in the gas film. The chamber is at 1.5 atm, 1000oK, and L = 0.1 m. The diffusivity of oxygen at these conditions is 0.35 cm2/sec. The following reaction occurs at the carbon surface (Gas constant = 0.08205 atmm3/kmoloK):

4C + 3O2 2CO + 2CO2

(1) a) NO2 = NCO2 b) NO2 = 1.5 NCO2 c) NO2 = NCO2 d) None of the above23

23

(2) The concentration of O2 at z = L in kmol/m3 is ____________

II.

1 ft

z

3 ft

2 ft

A tank containing water has its top open to air. The tank is cylindrical with a diameter of 2 ft. The liquid surface is maintained at a level 3 ft below the top of the tank as shown in the sketch. The surrounding air is at 310oK, 1bar, and 30 percent relative humidity. The vapor pressure of water at 310oK is 6.28 kPa. Note: 1 bar = 100 kPa.

(3) If R is the radius of the tank in the range 0 z 3 ft, then an expression for R as a function of

z is _______________

(4) The mole fraction of water vapor at the top of the tank is _________

B-2

III. (5) The flux of oxygen across a 75-mil-thick polypropylene film at 30oC is 3510-9 mol/m2.s per atmosphere of oxygen partial pressure difference. Determine the flux of oxygen across a 30-mil-thick film where the left side is open to air at 1 atm with 21 mol % oxygen and the right side is in vacuum with no oxygen.

__________

IV. Nitric oxide (NO) emissions from automobile exhaust can be reduced by using a catalytic converter, and the following reaction occurs at the catalytic surface:

NO + CO → 0.5 N2 + CO2

The concentration of NO is reduced by passing the exhaust gases over the surface, and the rate of reduction at the catalyst is governed by a first order reaction of the form

r"A(kmol/m2∙s) = k”1CA where k”1 = 0.10 m/s and A denotes NO

z=0yA0

z=L

Catalytic surface

Exhaust gas at T and P

As a first order approximation it may be assume that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface. Consider a situation for which the exhaust gas is at 750oK and 1.2 bars and the mole fraction of NO is yA0 = 0.25 if DA,mix = 10-4 m2/s and the film thickness is L = 1 mm.

(6) The molar flux of A (NO) can be obtained by integrating the following expression

(A) NA = (B) NA = (1 0.5 )

AB

A

cDy

Adydz (1 0.5 )

AB

A

cDy

Adydz

(C) NA = cDAB (D) None of the aboveAdydz

B-3

(7) If the molar of A is given by NA = cDAB(yA0 yAL)/L determine yAL

V. A homogeneous reaction A P takes place inside a porous cylindrical catalyst with radius R. The two ends of the catalyst particle are impermeable to A so diffusion is only in the r direction. The length of the catalyst particle is L = 1 cm. The effective diffusivity, De, of A within the catalyst is 1.210-7 m2/s. The concentration of A within the catalyst is given by

CA = CAo , where CAo = 0.15 kmol/m3 and R = 1.00 mm

Rrsinh20.0

(8) The concentration of A at the surface of the catalyst is __________

(9) The molar flux of A at r = R/2 is __________

(10) The moles of A within the cylindrical catalyst can be determined from the expression

(a) R2LCA (b) 0

R

AC rLdr(c) (d) None of the above

02

R

AC RLdr

B-4


I. A mixture contains 55 mole % isobutane and 45 mole % isopentane is at 30 psia. The K values for these compounds can be obtained from

ln K = A/T2 + B + C ln P where T is in oR and P is in psiaCompound A B CIsobutane -1,166,846 7.72668 -.92213Isopentane -1,481,583 7.58071 -.93159

1) The vapor mole fraction of isobutane at the bubble point of 527.6oR is ___________

2) The liquid mole fraction of isobutane at the dew point of 549.7oR is ___________

3) The mixture is flash at 536.0 oR, 30 psia where V/F = 0.4, then the mole fraction of isobutane (iC4) in the liquid phase is __________

4) If the mixture is flash at 526.0 oR, 30 psia thenA) V/F = 0 B) V/F = 1 C) .2 < V/F < .8 D) None of the above

II. A system contains 20 mol % toluene, 30 mol % ethylbenzene, and 50 mol % water. Mixtures of ethylbenzene and toluene obey Raoult's law and the hydrocarbons are completely immiscible in water. The system pressure is 1 atm. (Note T = toluene, EB = ethylbenzene, W = water, Po = vapor pressure)

5) The bubble point temperature can be determined by solving the following equation:

A) .5 + .2 + .3 = 1 atm B) + .2 + .3 = 1 atmowP o

TP oEBP o

wP oTP o

EBPC) + .4 + .6 = 1 atm D) None of the aboveo

wP oTP o

EBP

B-5

III. A liquid containing 50 mol % benzene and 50 mol % toluene is subjected to a flash vaporization process at a pressure of 760 mmHg where the temperature inside the flash drum is 95oC. At 95oC,

= 1176.8 mmHg, = 476.87 mmHgobenzeneP o

tolueneP

(6) The mole fraction of benzene in the exit liquid is ____________

IV. A homogeneous reaction A P takes place inside a porous cylindrical catalyst with radius R. The two ends of the catalyst particle are impermeable to A so diffusion is only in the r direction. The length of the catalyst particle is L = 1 cm. The effective diffusivity, De, of A within the catalyst is 1.210-7 m2/s. The concentration of A within the catalyst is given by

CA = CAo , where CAo = 0.10 kmol/m3 and R = 1.00 mm

Rrsinh20.0

(7) The molar flux of A at r = R/2 is __________

(8) The consumption rate of A within the catalyst is __________

B-6

V. (9) A process that uses p-xylene to perform stereoschemically specific reactions is supplied with a stock chemical which is received as a mixture of three isomers: meta, para, and ortho. The composition is roughly half m-xylene and half p-xylene with trace amounts of the ortho isomer. In order for the reaction to proceed correctly, a feed stream with a purity of at least 85% p-xylene is required. Consider the following processes:a) Absorption; b) Chemical Reaction; c) Condensation; d) Crystallization ; e) Distillation f) Adsorption; g) Membrane; h) Extraction (Liquid-Liquid).

The following processes can separate the p-xelene from the other isomers.A) a, b, c, dB) a, b, c, eC) b, c, d, gD) b, d, f, h E) e, f, g, h

VII. (10) A process that mixes different solid crystal components into a uniform blend by adding a liquid solvent (isopropanol), emits an unsatisfactory amount of isopropanol (IPA) when the crystals are dried. Currently the EPA standards for isopropanol is 90% control of total daily emissions.Consider the following processes:a) Absorption; b) Chemical Reaction; c) Condensation; d) Crystallization ; e) Distillation; f) Adsorption; g) Membrane; h) Extraction (Liquid-Liquid); i) Stripping; j) Evaporation

The following processes can separate IPA from air.A) a, b, c, dB) a, b, c, eC) a, b, c, f D) a, b, c, hE) a, b, c, j

B-7


I. An ethanol-water separation is achieved in a distillation column at 1 atm. Feed flow rate is 5,000 lb/hr. The distillate contains 90 wt% ethanol, and the bottoms contains 5 wt% ethanol. Assume theoretical trays, a total condenser, a partial reboiler, and saturated reflux.1) The minimum number of equilibrium stages (including the reboiler is)

A) 3 B) 4 C) 5 D) 6 E) 7

2) If the reflux ratio R is 1.8, the enthalpy H’ of the upper operating point D’ in Btu/lb is

A) 1120 B) 1170 C) 1220 D) 1270 E) 1320

3) If the feed contains 30 wt% ethanol and H’ = 1,200 Btu/lb, the condenser duty is___________

4) If the feed is saturated liquid with 30 wt% ethanol and the enthalpy of W’ = 100 Btu/lb, then the enthalpy of the upper operating point D’ in Btu/lb is

A) 750 B) 700 C) 650 D) 600 E) 550

5) If the feed is saturated vapor at 190oF, then the minimum reflux ratio is

A) 1.16 B) 1.06 C) 0.96 D) 1.26 E) 1.36

6) If the feed is saturated liquid, the enthalpy H’ of the upper operating point D’ is 1100 Btu/lb, and the enthalpy of W’ is 100 Btu/lb then temperature of the feed in oF is

A) 186 B) 188 C) 190 D) 192 E) 194

B-8

II. A saturated vapor mixture of benzene and toluene containing 40 mole percent benzene is to be separated into an overhead product containing 98 mole percent benzene and a bottom product containing 2 mole percent benzene. Benzene and toluene mixture may be considered as ideal. Use optimum feed plate.

ln Psat = A B/(T + C) P where T is in K and Psat is in kPa

Compound A B Cbenzene 14.1603 2948.78 44.5633toluene 14.2515 3242.38 47.1806

7) If the feed is at 420 K, the feed pressure is __________

8) Consider an equilibrium tray at 415 K and the liquid mole fraction of benzene leaving the tray is 0.3, the vapor mole fraction of benzene leaving the tray is

____________

9) If the reflux ratio is 3.0 and the liquid mole fraction of benzene leaving an equilibrium tray is 0.3, the vapor mole fraction of benzene entering the tray is

____________

10) If the reflux ratio is 3.0 and the vapor mole fraction of benzene entering an equilibrium tray is 0.35, the liquid mole fraction of benzene leaving the tray is

B-10


I. A feed with q = 0.4, consisting of 100 mol/s phenol and cresols, is separated in a plate column with a total condenser and a partial reboiler. Assume CMO with a reflux ratio of 3. The following compositions and relative volatilities are given

Compound Feed, mol% Distillate, mol% Bottoms, mol% Relative volatility1) Phenol2) o-cresol3) m-cresol

402535

94.24.51.3

5.2 1.301.000.70

1) The mole fraction of o-cresol leaving the column in the bottoms is ____________

2) The mole fraction of phenol in the liquid stream leaving the top (first) tray is ____________

3) The mole fraction of o-cresol in the liquid stream leaving the top tray is ____________

B-11

4) The liquid flow rate in the stripping section of the column is ____________

5) The vapor flow rate in the stripping section of the column is ____________

The mole fraction of phenol and o-cresol in the liquid stream leaving the second tray are 0.8945 and 0.0680, respectively.

6) The mole fraction of phenol in the vapor stream entering the second tray is ____________

7) The mole fraction of o-cresol in the vapor stream entering the second tray is ____________

II. 10,000 lb/hr of saturated liquid at 1 atmosphere is fed into an adiabatic flash drum operating at 0.1 atmosphere. The feed stream is a binary mixture of 29 wt% ethanol and 71 wt% water.

8) The temperature in the flash drum is ____________

9) The mole fraction of ethanol in the vapor phase is ____________

III. The degree of freedom for an equilibrium stage with feed and heat transfer for 10 components is____________

B-13


I. 1000 kg/hr of a saturated liquid with 30 wt% ethanol in water is fed into a distillation column. We desire a distillate composition xD = 0.90 mass fraction ethanol and a bottoms composition xB = 0.02 mass fraction ethanol. Reflux is return as a saturated liquid with a reflux ration of 1.5. The column is adiabatic with a partial reboiler and a total condenser. The enthalpies of ethanol-water system are given below

h(kJ/kg) = 114x2 279x + 393 = enthalpy of saturated liquid mixtureH(kJ/kg) = 1614y + 2676 = enthalpy of saturated vapor mixturex, y = mass fraction of ethanol in the liquid and vapor mixture, respectively

1) The condenser duty in kJ/hr, QC, is ____________

2) If QC = 2.0×105 kJ/hr, determine the reboiler duty, QR, in kJhhr ____________

II A. The MESH equations can be applied to all of the equilibrium stages in the column, including the

reboiler and condenser.B. When the equilibrium model is applied to multicomponent mixtures the Murphree tray efficiency

could become negative

a. A and B are true b. Only A is true c. Only B is true d. A and B are false

III. The degree of freedom for an adiabatic equilibrium stage with feed 6 components is____________

B-14

IV. Consider an air cooler where the flow rate of water is large compared to the air, and the effectiveness, , of the cooler is known to depend on the flow rate of air according to the relation, ~ . The air inlet temperature is 25oC. 0.3

airm

5) The water inlet temperature is 100oC. If the air flow rate increases 50%, the percentage increase in the heat removal rate is

__________

6) At a given air flow rate, the outlet air temperature is 50oC. When the air flow rate increase 50%, the heat transfer rate increase 20%, the new air outlet temperature is

__________

V. Component A in a water stream is to be stripped out using an air stream in a counter current staged stripper. Inlet air is pure with a flow rate of 600 lbmol/hr. Inlet liquid stream has 25 mol % A and a flow rate of pure water of 500 lbmol/hr. The desired outlet mole fraction of A is x = 0.05. Assume that water is nonvolatile and air is insoluble. Equilibrium data can be represented by y = 1.3x for x < 0.5

.

7) At a location in the stripper where mole ratio X = 0.12 then Y = __________

8) At a location in the stripper where mole ratio Y = 0.12 then X = __________

B-15

VI. A cross flow heat exchanger consists of 120 tubes, arranged in four rows with two passes. Process water in the tube at 100oC, 20 kg/s is to be cooled to 80oC by cross-flow of air at 30oC, 1 atm. The tube has an inner diameter of 20 mm, outer diameter of 24 mm, and length of 12.0 m. The mass flow rate of air is 40 kg/s. The following data are given: water = 995 kg/m3, air = 1.16 kg/m3, cp,water = 4178 J/(kgK), and cp,air = 1007 J/(kgK).

9) The total inside area for heat transfer is ___________

10) The velocity of water is ___________

B-16

Answer to 2009 Quizzes

Quiz #11) (d)2) 3.8410-3 kmol/m3

3) R = 0.5 + z/64) 0.01885) 18.37510-9 mol/m2.s6) (A)7) 0.1258) 0.2063 kmol/m3

9) 2.0310-5 kmol/m2s10) (d)

Quiz #21) 0.819272) 0.265373) 0.430094) (A)5) (C)6) 0.404497) 1.3510-5 kmol/m2s8) 1.1610-9 kmol/s9) (D)10) (C)

Quiz #31) (C)2) (E)3) 1.63×106 Btu/hr4) (B)5) (A)6) (B)7) 327.7 kPa8) 0.47919) 0.4710) 0.1821

B-17

Quiz #41) 0.38162) 0.91933) 0.05714) 157.3 mol/s5) 96.4 mol/s6) 0.90647) 0.06238) 97oF9) 0.6810) 38

Quiz #51) 7.87×105 kJ/hr2) 2.19×105 kJ/hr3) (a)4) 25 5) 32.82% 6) 45oC7) 0.05618) 0.19669) 90.5 m2

10) 1.0664 m/s

C-1

Appendix C Previous Exams (2010)



I. Helium and nitrogen gas are contained in a conduit 5 mm in diameter and 0.2 m long at 298 K and a uniform constant pressure of 1.0 atm abs. The partial pressure of He at the left end of the tube is 0.060 atm and 0.020 atm at the other end. The diffusivity of Helium in nitrogen at 298 K is 6.87×10-5 m2/s. (Gas constant = 0.08206 atmm3/kmolK):

(1) Flux of He in kmol/s∙m2 is ____________

(2) Partial pressure of He at 0.05 m from the left end is ____________

II. (3) The flux of oxygen across a 75-mil-thick polypropylene film at 30oC is 3510-9 mol/m2.s per atmosphere of oxygen partial pressure difference. Determine the flux of oxygen across a 60-mil-thick film where the left side is open to air at 1 atm with 21 mol % oxygen and the right side is pure oxygen at 1 atm.

__________

III. (4) A long glass capillary tube, of diameter 0.01 cm, is in contact with water at one end and dry air at the other. If the molar flux of water through the capillary is 3.9010-8 mol/scm2. Determine the required time in seconds for one gram of water to evaporate through this system.

____________

C-2

IV. Nitric oxide (NO) emissions from automobile exhaust can be reduced by using a catalytic converter, and the following reaction occurs at the catalytic surface:

NO + CO → 0.5 N2 + CO2

The concentration of NO is reduced by passing the exhaust gases over the surface, and the rate of reduction at the catalyst is governed by a first order reaction of the form

r"A(kmol/m2∙s) = k”1CA where k”1 = 0.05 m/s and A denotes NO

z=0yA0

z=L

Catalytic surface


As a first order approximation it may be assume that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface. Consider a situation for which the exhaust gas is at 750oK and 1.2 bars and the mole fraction of NO is yA0 = 0.25 if DA,mix = 10-4 m2/s and the film thickness is L = 1 mm.

(5) If the molar of A(NO) is given by NA = cDAB(yA0 yAL)/L determine yAL

(6) If k1” = 10,000 m/s, what are the two boundary conditions (numerical values) required to solve for the molar flux of NO.

_______________________ _______________________

V. (7) A vessel initially containing propane at 20oC is connected to a nitrogen cylinder, and the pressure is increased to 300 psia. Assuming that the nitrogen is insoluble in liquid propane, what is the mol fraction of propane in the vapor phase? Vapor pressure of propane at 20oC is 8.2 atm.

____________

C-3

VI. Consider a glass tube with length L where the end plates and fittings of the tube impermeable to hydrogen. The molar solubility ratio S of H2 (species A) in glass is 0.2 (mol/cm3) H2 in glass/(mol/cm3) H2 in gas. The inside and outside radius of the tube are ri and ro, respectively. DAB is the diffusivity of hydrogen in glass. cgi and cgo are the molar concentrations of H2 inside and outside the tube, respectively. (Gas constant = 82.06 atmcm3/molK). (8) The leak rate, WA, of hydrogen can be estimated from the following equation:

(A) WA = (B) WA = AB gi go

o i

2 ( )ln( / )

LD S c cr r

AB gi go

o i

2 ( )ln( / )

LD c cr r

(C) WA = (D) None of the aboveAB gi go

o i

2 ( )LD S c cr r

(9) If H2 inside the tube is at 2 atm and 300 K, determine the concentration (mol/cm3) of H2 in the glass at ri

____________

(10) The glass tube is surrounded by atmospheric air at 300 K, and the partial pressure of hydrogen in air is 3.810-4 torr (1 atm = 760 torr). Determine the concentration (mol/cm3) of H2 in the glass at ro

____________

C-4



1). Solve f(x) = 3x3 + 2x - 1 = 0 using Newton method with the initial guess x = 1. Only 1 iteration is required.

___________

2) f(x) = (ln x)2 + 2x - 1 = 0 using Newton method with the initial guess x = 1 and numerical derivative with dx = 0.1. Only 1 iteration is required. ___________

3) Consider the following Matlab codesP = 1;% Use the Wilson equation with parametersG12 = 1.2042; G21 = 0.39799;% Vapor pressure data: P(atm), T(K)p1sat = 'exp(9.33984-2696.79/(T-46.14))';p2sat = 'exp(9.20324-2697.55/(T-48.78))';Tb1 = 2696.79/(9.33984-log(P))+46.14;Tb2 = 2697.55/(9.20324-log(P))+48.78;x1=[0 .002 .004 .006 .008 .01 .015];x2=linspace(.02,.92,46);x3=[.93 .94 .95 .96 .97 .98 .985 .990 .995 1];xp=[x1 x2 x3];np=length(xp);yp=xp;Tp=xp;Tp(1)=Tb2;Tp(np)=Tb1;dT = .01; T=Tb2;for i=2:np x1=xp(i);x2=1-x1;% Evaluate activity coefficients tem1 = x1 + x2*G12; tem2 = x2 + x1*G21; gam1 = exp(-log(tem1)+x2*(G12/tem1-G21/tem2)); gam2 = exp(-log(tem2)+x1*(G21/tem2-G12/tem1)); for k=1:20 fT=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; T=T+dT; fT2=x1*gam1*eval(p1sat)+x2*gam2*eval(p2sat)-P; dfT=(fT2-fT)/dT; eT=fT/dfT; T=T-dT-eT; if abs(eT)<.001, break ,end

C-5

endTp(i)=T;yp(i)=x1*gam1*eval(p1sat)/P;

A) These codes construct a Pxy diagram.B) Ideal gas is assumed in the equilibrium calculation.


IV.) A mixture contains 35 mol% isobutane, 35 mol% isopentane, and 30 mol% n-hexane is at 30 psia. This mixture is flashed at 582.74oR where 60 mol% of the feed is evaporated and at these conditions KiC4 = 3.1718, KiC5 = 1.051, and KnC6 = 0.3169.

4) The mole fraction of isobutane in the liquid phase is ___________

5) The mole fraction of n-hexane in the vapor phase is ___________

V.) A mixture contains 35 mole % isobutane and 65 mole % isopentane is at 30 psia. The K values for these compounds can be obtained from

ln K = A/T2 + B + C ln P where T is in oR and P is in psiaCompound A B CIsobutane -1,166,846 7.72668 -.92213Isopentane -1,481,583 7.58071 -.93159

6) The vapor mole fraction of isobutane at the bubble point of 542.3oR is ___________

7) The liquid mole fraction of isobutane at the dew point of 562.11oR is ___________

C-6

8) The mixture is flash at 552.1 oR, 30 psia where V/F = 0.4, then the mole fraction of isobutane (iC4) in the liquid phase is __________

VI.) (4 pts) Consider a glass tube with length L = 2 m where the end plates and fittings of the tube impermeable to hydrogen. The molar solubility ratio S of H2 (species A) in glass is 0.2 (mol/cm3) H2 in glass/(mol/cm3) H2 in gas. The inside and outside radius of the tube are ri and ro, respectively. DAB is the diffusivity of hydrogen in glass. cgi and cgo are the molar concentrations of H2 inside and outside the tube, respectively. (Gas constant = 82.06 atmcm3/molK). The diffusivity of H2 in glass at 300 K, is 0.4×10-8 cm2/s. The hydrogen pressure inside the tube is 2 atm at 300 K and the partial pressure outside the tube is zero. If ri = 1 cm and ro

= 1.5 cm, determine the hydrogen leak rate in gmol/s. Show all your work!

C-7



1) f(x) = (ln x)2 + 2x - 1 = 0 using Newton method with the initial guess x = 2 and numerical derivative with dx = 0.1. Only 1 iteration is required. ___________

II) A distillation column with 100 kmol/h feed of 60% A and 40% B produces a distillate product with xD = 0.95 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and the equilibrium data is given by

y = 21

xx

2) For total reflux, determine (numerically) the composition (y) of the vapor stream entering the second equilibrium plate from the top.

__________

3) For total reflux, determine (numerically) the composition (y) of the vapor stream leaving the bottom equilibrium plate (above the reboiler).

__________

C-8

4) The equilibrium data is plotted in Figure Q3-4. The minimum number of equilibrium stages (including the reboiler) isA) 7 B) 8 C) 9 D) 10 E) 11

Figure Q3-4 Equilibrium data5) If the reflux ratio R is 2, determine (numerically) the composition (y) of the vapor stream entering the top equilibrium plate.

__________

6) If R = 2 and q = 0.8, determine (numerically) the composition (x) of the liquid stream entering the reboiler).

__________

C-9

7) If q = 0.8, the minimum reflux ratio is___________

III.) A liquid mixture containing 20 mol % n-heptane and 80 mol % n-octane is fed at its boiling point to the top of a stripping tower where the feed stream is the saturated liquid and the distillate stream is the saturated vapor. There is no reboiler or condenser in a stripping tower. The bottoms are to contain 98 mol % n-octane. For every 3 mol of feed, 2 mol of vapor is withdrawn as product. 8) The mole fraction of n-heptane in the distillate stream is __________

IV) A liquid feed at the boiling point contains 3.3 mol % ethanol and 96.7 mol % water and enters the top tray of a stripping tower shown in Figure Q3-9.

F

B, xB

V, yD

S (Steam)

xF

Lx

Vy

Figure Q3-9 Stripping tower and direct steam injection.

C-10

Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 95% of the alcohol in the feed. Assume equimolar overflow for this problem. Equilibrium data for mole fraction of alcohol are as follows at 101.32 kPa abs pressure.

x 0 0.0080 0.020 0.0296 0.033y 0 0.0750 0.175 0.250 0.270

(9) For an infinite number of theoretical steps, calculate the minimum moles of steam needed per mole of feed.

___________

(10) Using 24 moles of steam per 100 moles of feed, calculate the mole fraction of alcohol in the vapor

___________

C-11



1) f(x) = sinh(2x) + x2 - 8 = 0 using Newton method with the initial guess x = 1 and analytical derivative. Only 1 iteration is required.

___________

2) Solve f(x) = exp(x2) + 2x - 10 = 0 using Newton method with the initial guess x = 2 and numerical derivative with dx = 0.1. Only 1 iteration is required. ___________

II) A distillation column with 100 kmol/h feed of 60% A and 40% B produces a distillate product with xD = 0.95 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and

the equilibrium data is given by y = 2.51 1.5

xx

3) For total reflux, determine (numerically) the composition (y) of the vapor stream entering the second equilibrium plate from the top.

__________

4) For a reflux ratio of 2, and q = 0.5, determine the liquid composition (x) of the feed point (the intersection of the q-line and the operating lines).

__________

C-12

III) Consider a distillation column with the McCabe-Thiele diagram given in Figure Q4-III. The column has a total condenser and a partial reboiler. Note: the points A, C, E, a, c, and e are on the equilibrium curve and the points B, D, F, b, d, and f are on the operating lines.

Figure Q4-III McCabe-Thiele diagram for binary distillation.

5) A) The x coordinate of point D gives the mole fraction of the volatile species in the liquid stream entering equilibrium tray 4 from the top (with the top tray as tray 1).

B) The y coordinate of point C gives the mole fraction of the volatile species in the vapor stream entering equilibrium tray 3 from the top.

a. A and B are true. b. Only A is true c. Only B is true d. A and B are false

6) A) The y coordinate of point d gives the mole fraction of the volatile species in the vapor stream entering equilibrium tray N-2 (with the bottom tray before the reboiler as tray N).

B) The x coordinate of point d gives the mole fraction of the volatile species in the liquid stream leaving tray N-2.


7) A) The feed is introduced into tray 6 from the top.B) The feed is introduced at the optimum location in the column.


C-13

F

B, xB

V, yD

S (Steam)

xF

Lx

Vy

IV) A liquid feed at the boiling point contains 5 mol % of species A and 95 mol % water and enters the top tray of a stripping tower shown in Figure Q4-IV. Saturated steam is injected directly into liquid in the bottom of the tower. The overhead vapor which is withdrawn contains 98% of A in the feed. Assume equimolar overflow for this problem. Equilibrium data for mole fraction of A is given by y = 10x for x 0.08.

(8) For an infinite number of theoretical steps, calculate the minimum moles of steam needed per mole of feed.

___________Figure Q4-IV Stripping tower

(9) Using 15 moles of steam per 100 moles of feed, calculate the mole fraction of A in the vapor

___________

C-14

V. (10) A distillation column operated at 4 kg/cm2 is used to separate ammonia from water. The saturated liquid feed contains 30 wt. % ammonia. The distillate is a saturated liquid at 95 wt. % ammonia.

Determine the minimum reflux ratio for this operation. ___________

C-15



I) Consider a distillation column with the McCabe-Thiele diagram given in Figure Q4-III. The column has a total condenser and a partial reboiler. Note: the points A, C, E, a, c, and e are on the equilibrium curve and the points B, D, F, b, d, and f are on the operating lines.

Figure Q4-III McCabe-Thiele diagram for binary distillation.

1) A) The x coordinate of point D gives the mole fraction of the volatile species in the liquid stream leaving equilibrium tray 3 from the top (with the top tray as tray 1).

B) The y coordinate of point C gives the mole fraction of the volatile species in the vapor stream leaving equilibrium tray 3 from the top.


2) A) The y coordinate of point d gives the mole fraction of the volatile species in the vapor stream leaving equilibrium tray N-2 (with the bottom tray before the reboiler as tray N).

B) The x coordinate of point e gives the mole fraction of the volatile species in the liquid stream leaving tray N-2.


C-16

II. Component A in a water stream is to be stripped out using an air stream in a counter current staged stripper. Inlet air is pure and flow rate is 480 lbmol/hr. Inlet liquid stream has 25 mol % A and a flow rate of pure water of 500 lbmol/hr. The desired outlet mole fraction of A is x = 0.05. Assume that water is nonvolatile and air is insoluble Equilibrium data can be represented by y = 1.3x. 3) At a location in the stripper where mole ratio X = 0.12 then Y = ___________

4) At a location where x = 0.15, kx = 3.810-2 kmol/sm2, and ky = 6.210-2 kmol/sm2 then xi = ___________

5) At a location where y = 0.15, kx = 3.810-2 kmol/sm2, and ky = 6.210-2 kmol/sm2 then yi = ___________

6) If the inlet air flow rate is not 480 lbmol/hr, determine the

minimum air flow rate required for the stripper ___________

C-17

III. A feed with q = 0.6, consisting of 100 mol/s phenol and cresols, is separated in a plate column with a total condenser and a partial reboiler. Assume CMO with a reflux ratio of 3. The following compositions and relative volatilities are given

Compound Feed, mol% Distillate, mol% Bottoms, mol% Relative volatility1) Phenol2) o-cresol3) m-cresol

403030

94.04.51.5

5.0 1.401.000.70

7) The mole fraction of o-cresol leaving the column in the bottoms is ____________

8) The mole fraction of phenol in the liquid stream leaving the top (first) tray is ____________

9) The mole fraction of o-cresol in the vapor stream entering the top tray is ___________

C-18

IV. Consider a very long, concentric tube heat exchanger having hot and cold oil inlet temperature of 100 and 40oC, respectively. The flow rate of the hot oil is four times that of the cold oil. Assume equivalent hot and cold oil specific heats.

10) For counter flow the hot oil outlet temperature is ___________

Answer to 2010 Quizzes

Quiz #11) 5.6210-7 kmol/sm2

2) 0.05 atm3) 3.45610-8 mol/m2.s4) 1.81371010 s5) 0.16676) z = 0, yA0 = 0.25 z = L, yAL = 07) 0.40158) (A)9) 1.6210-5 mol/cm3

10) 4.0610-12 mol/cm3

Quiz #21) 0.63642) 0.52173) (c)4) 0.15205) 0.16116) 0.65247) 0.14278) 0.24029) 2.008310-10 mol/s

C-19

Quiz #31) 0.71102) 0.82613) 0.14294) 95) 0.91996) 0.06997) 1.448) 0.299) Minimum 0.116 mol steam/mol feed10) 0.1304

Quiz #41) 1.35422) 1.82563) 0.75254) 0.535) b6) a7) c8) Minimum 0.098 mol steam/mol feed9) 0.326710) 0.1

Quiz #51) a2) c3) 0.07024) 0.1078 5) 0.1776 6) 291.5 lbmol/hr7) 0.46538) 0.91009) 0.057010) 85oC

D-1

Appendix D Previous Exams (2012)



I. A gas mixture at a total pressure of 124 kPa and 313 K contains 45% H2 and 55% O2 by volume. The absolute velocity of each species are -2 and 10 m/s, respectively, all in the direction of the z-axis. Gas constant = 8.314 m3kPa/kmolK. Determine:

1) the mass diffusion flux of O2 __________

Mass diffusion flux of O2 = 0.48958 kg/m2s

2) the molar diffusion flux of O2 __________

Molar diffusion flux of O2 = 0.14152 kmol/m2s

II. (3) Consider a radial flow fixed bed reactor where the reaction rate per unit volume rA = k1CA

2 occurs between the radial locations R3 and R4 (the shaded area). The length of the bed is L, the diffusivity of A in the bed is DA. If the concentration of A depends on r only, CA(r) can be obtained from the following differential equation ________

A) + CA = 0 B)

CA = 0 A1 dCd r

r dr dr

1

A

kD

A1 dCd rr dr dr

1

A

kD

C) + CA

2 = 0 D) CA

2 = 0 (Ans.)A1 dCd rr dr dr

1

A

kD

A1 dCd rr dr dr

1

A

kD

E) None of the above.

D-2

III. Species A diffuses into a cylindrical pore where it reacts at the cylindrical surface to produce B (A B) according to a zero order reaction r”A(mol/cm2s) = k1 where k1 = 1.2×10-6 mol/cm2s. There is no reaction at the end of the pore and the end is impermeable to A. The inside diameter of the pore is 0.1 cm and its length L = 1.0 cm. You can assume CA(z) is a function of z where z is the distance along the pore with z = 0 at the entrance to the pore. The diffusivity of A in the pore is DAB = 0.06 cm2/s and CA(z = 0) = CA0 = 2.5×10-3 mol/cm3. The inside pore diameter is D.

4) Determine the consumption rate of A within the pore in mol/s. ___________

3.77×10-7 mol/s

5) Determine the molar flux of A at z = 0.2 cm in mol/scm2 ___________

3.84×10-5mol/scm2

6) Determine the molar flux of B at z = 0.5 cm in mol/scm2

2.4×10-5 mol/scm2

The concentration of A inside the pore is give by CA(z) = 2 z2 + B1z + B21

AB

kDD

7) B2 = 2.5×10-3 mol/cm3

8) B1 = 8×10-4 mol/cm4

D-3

IV. A 3-ft depth of stagnant water at 25oC lies on top of a 0.10-in. thickness of NaCl. At time t < 0, the water is pure. At time t = 0, the salt begins to dissolve into the water. The concentration of salt in the water at the solid-liquid interface is maintained at saturation (0.00544 mol NaCl/cm3) and the diffusivity of NaCl is 1.2×10-5 cm2/s, independent of concentration. The semi-infinite medium solution can be used with

NA0 = − DAB = 0

A

x

cx

0.5, ,( )AB A s A iD c c

t

9) Determine the molar flux of salt at the interface in mol/scm2 at t = 100 s. __________

1.06×10-6 mol/scm2

10) After 1 hr, determine the mole of salt dissolved into the water for an interfacial area of 100 cm2 in mol.__________

0.128 mol.

D-4



I). The following mixture is fed to a flash drum at 100 psia and 600oRPropane: 30 mol%, KC3 = 2.66 n-pentane: 70 mol%, KC5 = 0.364

F =

1)1()1(

i

iFi

KfxK

1) Determine at f = 0.4 ___________dfdF

0.807891

2) Solve F = = 0 using Newton method with the initial guess f = 0.5 and =

1)1()1(

i

iFi

KfxK

dfdF

0.8556. Only 1 iteration is required.___________

f = 0.0551

II. 3) Water is evaporating into initially dry air in the closed vessel. The vessel is isothermal at 25oC, so the water’s vapor pressure is 23.8 mmHg. This vessel has 0.8 liter of water with 200 cm2 of surface area in a total volume of 20 liters. After 60 minutes, the air is sixty percent saturated. What is the mass transfer coefficient?

__________

kc = 1.466 cm/minute.

D-5

III. Consider an adiabatic vapor liquid equilibrium stage with 5 components, 3 input streams, and three output streams. The streams are either saturated liquid or saturated vapor. The exit streams are one vapor and two liquid streams.4) Determine the variables required to describe the system ___________

With 3 streams in and 3 streams out, and no heat transfer,

NV = 48

5) Determine the number of equations relating the variables ___________

NE = 26

IV) A) The tie lines on a Txy are parallel.B) The tie lines on an enthalpy concentration diagram are parallel

a. A and B are true b. Only A is true(A) c. Only B is true d. A and B are false

V. Nitric oxide (NO) emissions from automobile exhaust can be reduced by using a catalytic converter, with the reaction NO + CO → 0.5 N2 + CO2 occurs at the catalytic surface. The concentration of NO is reduced by passing the exhaust gases over the surface, and the rate of reduction at the catalyst is governed by a first order reaction of the form: r"A(kmol/m2∙s) = k”1CA where k”1 = 0.08 m/s and A denotes NO

z=0yA0

z=L

Catalytic surface


As a first order approximation it may be assume that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface. Consider a situation for which the exhaust gas is at 750oK and 1.2 bars and the mole fraction of NO is yA0 = 0.35 if DA,mix = 10-4 m2/s and the film thickness is L = 1 mm.(7) If the molar of A is given by NA = cDAB(yA0 yAL)/L determine yAL _________________

0.1944

D-6

VI. A mixture contains 35 mol% isobutane, 35 mol% isopentane, and 30 mol% n-hexane is at 30 psia. This mixture is flashed at 582.74oR where 60 mol% of the feed is evaporated and at these conditions KiC4 = 3.1718, KiC5 = 1.051, and KnC6 = 0.3169.8) The mole fraction of isobutane in the liquid phase is ___________

xiC4 = 0.1520

9) The mole fraction of n-hexane in the vapor phase is ___________

ynC6 = 0.1611

VII. Saturated-liquid feed of F = 40 mol/h, containing 50 mol% A (more volatile species) and B, is supplied to the apparatus in the following figure.

Feed F

Still pot

Heat Bottoms W

Vapor V

Condenser

Reflux R

Distillate D

(10) The condensate is split so that reflux/condensate = 0.5. (a) If heat is supplied such that W = 30 mol/h and relative volatility = 2.5, determine the mole fraction of A in the bottoms.

__________

xA = 0.444446

D-7

Liquid feed

h , V, T , P , yV V V i

h , L, T , P , xL L L i

h , F, T , P , zF F F i

Flash drum



I. A feed liquid mixture of acetone (A) and water is available at 125oC and 687 kPa and contains 57 mol% A. Before entering the distillation tower, the feed passes through an expansion valve and is partially vaporized at 61oC and 101.3 kPa. From the data below, determine the vapor fraction in the feed.Equilibrium data for acetone-H2O at 101.3 kPaT(oC) 56.7 57.1 60.0 61.0 63.0 71.7 100Mol% A in liquid 100 92.0 50.0 33.0 17.6 6.8 0Mol% A in vapor 100 94.4 85.0 83.7 80.5 69.2 0

________f = 0.473

II. Consider an adiabatic equilibrium flash shown on the right with 6 components. (Numerical values required)

2) Determine the variables required to describe the system

__________27

3) Determine the number of equations relating the variables: 18 __________

III. A system contains 2 moles benzene, 1 mole toluene, and 2 moles water in the liquid phase. Mixtures of benzene and toluene obey Raoult's law and the hydrocarbons are completely immiscible in water. At 75oC, the vapor pressures of benzene, toluene, and water are 85.4 kPa, 32.4 kPa, and 38.3 kPa. Determine the total pressure of the system at equilibrium.

_________

P = 106 kPa

D-8

IV. A distillation column with 100 kmol/h feed of 50% A and 50% B produces a distillate product with xD = 0.95 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and


xx

5) If the feed is saturated vapor, determine the minimum reflux ratio __________

Rmin = 2.1857

For questions 6, 7, and 8 the reflux ratio is 1.5 and the feed is a saturated mixture with q = 0.6, determine:

6) = __________ 7) = __________V L

= 135.82 kmol/hL

= 86.374 kmol/hV

8) The mole fraction of A in the vapor stream entering the top equilibrium stage __________

y2 = 0.91271

D-9

V. A liquid containing 60 mol% toluene and 40 mol% benzene (the more volatile species) is continuously distilled in a single equilibrium stage at 1 atm. Assume a relativity volatility of 2.2. The vapor fraction is 0.3.

9) Determine the percent of benzene in the feed leaves as vapor. ________

40 %

10) Determine the percent of toluene in the feed leaves as liquid. _________

76.7%

D-10

Liquid feed

V, T , P , yV V i

F, T , P , zF F i

Flash drum

L , T , P , xI I I IL L i

L , T , P , xII II II IIL L i

Vapor

Liquid I

Liquid II



I. A feed liquid mixture of acetone (A) and water is available at 125oC and 687 kPa and contains 40 mol% A. Before entering the distillation tower, the feed passes through an expansion valve and is partially vaporized at 63oC and 101.3 kPa. From the data below, determine the molar ratio of liquid to vapor in the feed.Equilibrium data for acetone-H2O at 101.3 kPaT(oC) 56.7 57.1 60.0 61.0 63.0 71.7 100Mol% A in liquid 100 92.0 50.0 33.0 17.6 6.8 0Mol% A in vapor 100 94.4 85.0 83.7 80.5 69.2 0

________1.808

II. Consider a non-adiabatic equilibrium flash shown on the right with 6 components. (Numerical values required)

2) Determine the variables required to describe the system

__________37

3) Determine the number of equations relating the variables: 27 __________

III. (4) A maximum boiling azeotrope has two components with equal saturation pressures. At the azeotropic composition, the activity coefficient for component M is ____________ the activity coefficient for component N.

A) greater than B) less than C) equal to ✓ D) not enough information

D-11

IV. Consider the following data for methanol-water system at 1 atm where enthalpies (H or h) are in Btu/lbmol:

x or y 0 5 10 15 20 30 40 50 60 70 80 90 100H 20720 20520 20340 20160 20000 19640 19310 18970 18650 18310 17980 17680 17390

h 3240 3070 2950 2850 2760 2620 2540 2470 2410 2370 2330 2290 2250

Vapor liquid equilibrium data (x and y are mole percentage of methanol) :x 0 2 4 6 8 10 15 20 30 40 50 60 70 80 90 95 100y 0 13.4 23.0 30.4 36.5 41.8 51.7 57.9 66.5 72.9 77.9 82.5 87.0 91.5 95.8 97.9 100

T,oC 100 96.4 93.5 91.2 89.3 87.7 84.4 81.7 78.0 75.3 73.1 71.2 69.3 67.6 66.0 65.0 64.5

One hundred lbmol/h of a saturated liquid mixture of 40 mol% methanol in water at 1 atm is to be separated by distillation at the same pressure into a liquid distillate containing 90 mol% methanol and a bottoms liquid product containing 94 mol% water. (Note: the answers for 5, 6, and 7 still need to be accurate to 3 significant figures using linear interpolation)

5) Determine the slope (Btu/lbmolmol%) of the operating line on the enthalpy concentration diagram corresponding to the condition of minimum reflux ratio:

____________

Slope = 476.42 Btu/lbmolmol%

6) If the slope of the operating line is 600 Btu/lbmolmol%, determine the enthalpy of the upper operating point.

____________

H’ = 32540 Btu/lbmol

7) Determine the mol% of methanol in the vapor stream enter the last (bottom) tray in the distillation column.

y = 30.4 %

D-12

V. A distillation column with 100 kmol/h feed of 40% A and 60% B produces a distillate product with xD = 0.98 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and the

equilibrium data is given by y = 2.41 1.4

xx

8) If the feed is saturated liquid, determine the minimum reflux ratio __________

Rmin = 1.6929

9) With a reflux ratio of 1.5Rmin, given mole fraction x = 0.35 of the liquid entering an equilibrium stage n, we can determine the mole fraction of the liquid stream leaving stage n by

A) Use the upper operating line then use the bubble point calculation.

B) Use the lower operating line then use the bubble point calculation.

C) Use the upper operating line then use the dew point calculation.

D) Use the lower operating line then use the dew point calculation. (Ans.)

E) Use the bubble point calculation then use the lower operating line.

VI. A liquid containing 50 mol% toluene and 50 mol% benzene (the more volatile species) is continuously distilled in a single equilibrium stage at 1 atm. Assume a relativity volatility of 2.4. The vapor fraction is 0.4.

10) Determine the percent of toluene in the feed leaves as vapor. ________

29.68 %

D-13



I. A feed liquid mixture of acetone (A) and water is available at 125oC and 687 kPa and contains 60 mol% A. Before entering the distillation tower, the feed passes through an expansion valve and is partially vaporized at 61oC and 101.3 kPa. From the data below, determine the molar ratio of liquid to vapor in the feed.Equilibrium data for acetone-H2O at 101.3 kPaT(oC) 56.7 57.1 60.0 61.0 63.0 71.7 100Mol% A in liquid 100 92.0 50.0 33.0 17.6 6.8 0Mol% A in vapor 100 94.4 85.0 83.7 80.5 69.2 0

________0.878

II. Component A in a water stream is to be stripped out using an air stream in a counter current staged stripper. Inlet air is pure and flow rate is 600 lbmol/hr. Inlet liquid stream has 20 mol % A and a flow rate of pure water of 500 lbmol/hr. The desired outlet mole fraction of A is x = 0.05. Assume that water is nonvolatile and air is insoluble. Equilibrium data can be represented by y = 1.4x.

2) If the mole fraction of A in the liquid stream entering an equilibrium stage is 0.10, determine the mole fraction of A in the liquid stream leaving this stage:

_________

x = 0.033191

3) If the mole fraction of A in the vapor stream entering an equilibrium stage is 0.10, determine the mole fraction of A in the vapor stream leaving this stage:

_________y = 0.21953

D-14

III. Consider the following data for methanol-water system at 1 atm where enthalpies (H or h) are in Btu/lbmol:

x or y 0 5 10 15 20 30 40 50 60 70 80 90 100H 20720 20520 20340 20160 20000 19640 19310 18970 18650 18310 17980 17680 17390

h 3240 3070 2950 2850 2760 2620 2540 2470 2410 2370 2330 2290 2250

Vapor liquid equilibrium data (x and y are mole percentage of methanol) :x 0 2 4 6 8 10 15 20 30 40 50 60 70 80 90 95 100y 0 13.4 23.0 30.4 36.5 41.8 51.7 57.9 66.5 72.9 77.9 82.5 87.0 91.5 95.8 97.9 100

T,oC 100 96.4 93.5 91.2 89.3 87.7 84.4 81.7 78.0 75.3 73.1 71.2 69.3 67.6 66.0 65.0 64.5

One hundred lbmol/h of a saturated vapor mixture of 82.5 mol% methanol in water at 1 atm is to be separated by distillation at the same pressure into a liquid distillate containing 90 mol% methanol and a bottoms liquid product containing 94 mol% water. (Note: the answers for 4, 5, and 8 still need to be accurate to 3 significant figures using linear interpolation)

4) Determine the slope (Btu/lbmolmol%) of the operating line on the enthalpy concentration diagram corresponding to the condition of minimum reflux ratio:

____________

Slope = 688.67 Btu/lbmolmol%

5) If the slope of the operating line is 800 Btu/lbmolmol%, determine the enthalpy of the upper operating point.

____________

H’ = 23905 Btu/lbmol

6) If the slope of the operating line is 800 Btu/lbmolmol%, determine the enthalpy of the lower operating point.

____________hw’ = -43295 Btu/lbmol

7) If the enthalpy of the upper operating point is 24,000 Btu/lbmol, determine the reflux ratio

____________R = 0.4107

D-15

8) If the enthalpy of the upper operating point is 24,000 Btu/lbmol and the distillate flow rate is 1500 lbmol/hr, determine the condenser duty

QC = 3.26×108 Btu/hr

IV. (9) Determine the degree of freedom for an equilibrium feed stage with 5 components, heat transfer and feed.

____________F = NV NE = 41 18 = 23

V. (10) A distillation column with 100 kmol/h feed of 50% A and 50% B produces a distillate product with xD = 0.95 and a bottom stream with xbot = 0.04 of the more volatile species A. CMO is valid and


xx

If the feed is saturated liquid, determine the minimum reflux ratio __________

Rmin = 1.1857

unit operationstknguyen/che425/lecture/che-425.pdfin chemical potential is due to the variation in...

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