124 LLE4

8/8/2019 124 LLE4

1/31

CHE-396 Senior Design Extraction

1

Liquid-Liquid Extraction

Senior Design CHE 396

Matrix Corporation

Zachary FijalConstantinos LoukerisZhaleh Naghibzadeh

John Walsdorf

Prof. Andreas Linninger

Akhil Jain

8/8/2019 124 LLE4

2/31


2

Table of Content

Introduction ___________________________________________________________ 3

Flowsheet _____________________________________________________________ 3

Process Operation ______________________________________________________ 4

Limitations ____________________________________________________________ 5

Applicability ___________________________________________________________ 6

Theory________________________________________________________________ 7

Theory Ternary Phase Diagram__________________________________________ 8

Theory - General Flow Sheet for Extractor Design ___________________________ 10

Properties ____________________________________________________________ 21

Costs ________________________________________________________________ 23

Case Study ___________________________________________________________ 25

Alternatives___________________________________________________________ 30

References ___________________________________________________________ 31

8/8/2019 124 LLE4

3/31


3

IntroductionExtraction is a process that separates components based upon chemical

differences rather than differences in physical properties. The basic principle behind

extraction involves the contacting of a solution with another solvent that is immiscible

with the original. The solvent is also soluble with a specific solute contained in the

solution. Two phases are formed after the addition of the solvent, due to the differences

in densities. The solvent is chosen so that the solute in the solution has more affinity

toward the added solvent. Therefore mass transfer of the solute from the solution to the

solvent occurs. Further separation of the extracted solute and the solvent will be

necessary. However, these separation costs may be desirable in contrast to distillation

and other separation processes for situations where extraction is applicable.

Flowsheet

Figure 1. Extraction Flowsheet for an Extractor Column

A general extraction column has two input stream and two output streams. The

input streams consist of a solution feed at the top containing the solute to be extracted and

a solvent feed at the bottom which extracts the solute from the solution. The solvent

containing the extracted solute leaves the top of the column and is referred to as the

extract stream. The solution exits the bottom of the column containing only small

8/8/2019 124 LLE4

4/31


4

amounts of solute and is known as the raffinate. Further separation of the output streams

may be required through other separation processes.

Process OperationThere are certain design variables that must be assigned in an extraction process.

Operating Temperature

Operating Pressure

Feed Flow Rate

Composition

Temperature of entering stream

Pressure of entering stream

As in many separation processes, the pressure and temperature conditions play a

large role in the effectiveness of the separation. In order for a good split of the feed the

pressure and temperature must be such so as to ensure that all components remain in the

liquid phase. The process will be adversely affected if one or more of the components are

allowed to become a vapor, or the extraction may not occur at all if a large enough

portion of a component is allowed to vaporize. In addition, the temperature should be

high enough that the components are all soluble with one another. If extremes in

temperature are present, finding a suitable solvent for extraction can be problematic.

This is however generally not the case since one of the biggest benefits in the extraction

process is that it can be done at ambient pressures and temperatures.

In many applications, a separation process is desired where an extreme

temperature will destroy the desired product such as the pharmaceutical industry. For

these applications, extraction is ideally suited, since the only temperature requirement is

that dictated by the solubility. At this point the biggest challenge would be finding a

suitable solvent for the extraction.

We can also use the pharmaceutical industry in another example for the benefits

of extraction and this has to do with the volumes involved for effective extraction. The

extraction process can become very expensive if the solvent needed to be used is costly

these expenses can be contained if a batch process is being used and this is often the case

in medicines. In a non-batch process the solvent would need to be constantly supplied

8/8/2019 124 LLE4

5/31

8/8/2019 124 LLE4

6/31


6

5. Separation may only occur for compositions in the region between the feed

composition and that apex of the carrier.

ApplicabilityWith all the key components in the design of an extractor system to be discussed,

the equipment selection can be evaluated. We must determine which extractor would

apply for the situation at hand. The specifications for each of these different systems are

relatively the same. The following design constraints should be placed on each system in

order to optimize the individual process -- (1) maximize surface area of mass transfer,

and (2) adjust flow feeds for maximum solute recovery. [2]

In general, there are three main types of extractors to focus on:

Mixer-settlers

Mixer-settlers are used when there will only be one equilibrium stage in the

process. For such a system, the two liquid phases are added and mixed. Due to

their density differences, one phase will settle out and the mixture will be

separated. The downfall to this type of extractor is that it requires a large-volume

vessel and a high liquid demand. [1]

Contacting columns

Contacting columns are practical for most liquid-liquid extraction systems. The

packings, trays, or sprays increase the surface area in which the two liquid phases

can intermingle. This also allows for a longer flow path that the solution can

travel through. In the selection of a packing, it is necessary to select a material

that is wetted by the continuous phase. [1] Lastly, the flow in a column should

always be counter-current.

Centrifugal contractors

Centrifugal contractors are ideal for systems in which the density difference is

less than 4%. In addition, this type of system should be utilized if process

requires many equilibrium stages. [2] In these systems, mechanical devices are

used to agitate the mixture to increase the interfacial area and decrease mass

transfer resistance. [1]

8/8/2019 124 LLE4

7/31


7

Many different types of centrifugal contractors exist, and each has its own guidelines for

operation and selection. A more detailed view can be found in. [1]

Table 1. Advantages and disadvantages of the various liquid-liquid extractor types [1].

Unit of Operation Advantages Disadvantages

Mixer-Settler Efficient Low head room Induces good contacting Can handle any number

of stages

Large floor High set-up costs High operation costs

Columns(without agitation)

Small investment costs Low operating costs

High head room Difficult to scale up

from lab

Less efficient than

mixer-settler

Columns(with agitation)

Good dispersion Low investment costs Can handle any number

of stages

Difficult to separatesmall densitydifferences

Does not tolerate highflow ratios

Centrifugal Extractors Can separate smalldensity differences

Short holding time Small liquid inventory

High set-up cost High operating and

maintenance costs

Cannot handle many

stages

TheoryWhen talking about liquid-liquid extraction, liquid-liquid equilibrium must be

considered. This is best represented by equating the chemical potential of both liquid

phases:

LII

i

LI

i = (1)

This relationship reduces to an expression, which is dependent only on the liquid molefractions and activity coefficients:

LII

i

LII

i

LI

i

LI

i xx = (2)

We can use activity coefficient models, such as UNIFAC (UNIquac Functional-group

Activity Coefficient), UNIQUAC (universal quasichemical), and NRTL (nonrandom

8/8/2019 124 LLE4

8/31


8

two-liquid) to determine the mole fractions. All three models above apply for liquid-

liquid equilibrium, it rolls down to which is easier to use and what properties we have

available. For a multi-component system, the UNIQUAC equation for the liquid-phase

activity coefficient is represented as follows: [3]

)residual(ln)ialcombinator(lnln iii += (3)

The combinatorial and residual activities are based on the statistical mechanical

theory and allowed the local compositions to result from the size and energy differences

between the molecules in the mixture. The relationships for these two activities are made

available to us through.

Theory Ternary Phase DiagramWe then are able to relate this data from the activity into a ternary phase diagram.

Ternary phase diagrams are unique in that they show all three components of a reactor

system on one plot. There are general principles that govern ternary phase diagrams, and

those are the following:

Sum of the perpendicular distances from any point within the triangle

to the three sides equals the altitude of the triangle.

Each apex of the triangle represents one of the pure components.

Any point of a side of the triangle represents a binary mixture.

Lines may be drawn parallel to the sides of the equilateral triangle forthe plotting of the compositions.

Figure 2. Phase diagram for a three component system. [2]

8/8/2019 124 LLE4

9/31


9

The ternary phase diagram may be constructed directly from experimental data.

The saturation curve (miscibility boundary), represented by JDPEK in Figure 2, can be

obtained experimentally by a cloud point titration. For example, a solution containing

components A & C with some composition is made, and then component B is added until

the onset of cloudiness due to the formation of a second phase occurs. Then the

composition is know for the mixture of the three components and can plotted onto the

ternary phase diagram. [1]

Tie lines are lines that connect points on the miscibility boundary. The tie lines

may also be presented onto the ternary phase diagram from an experiment. A mixture

may be prepared with composition that of point H (40% A, 40% C, 20% B) from Figure

2. If we allow it to equilibrate, then we can chemically analyze the final extract (E) phase

and the raffinate (R) phase. [1]

Point F is a feed composition into the extractor while point S is the solvent feed to

the extractor. Point H represents the composition of the two feeds at equilibrium. This

point is determined by summing the feed (F) and solvent (S) compositions for each

component. Points R and E are the compositions of the raffinate and extract from the

unit, respectively, and the line between them forms the tie line. The tie lines move above

and below this line based on the relationship between the raffinate and the extract. Point

P represents the plait point. At this point, only one liquid phase exists and the

compositions of the two effluents are equal. The curve represented by JRDPEK is the

equilibrium between all three components. The area under the curve is the region where

two liquid phases will exist. Above the curve, there will only be one liquid phase. If a

line is drawn from F to E or from S to R, this will represent the operating line. Although

this diagram is not the basic theory behind liquid-liquid extraction, it is helpful to review

this procedure before continuing with an in-depth discussion.

In addition to the above-mentioned considerations, equilibrium constraints must

be satisfied. This implies that

AE = AR (4)

where AE is the activity coefficient for the solute A in the extract and AR is the activity

coefficient of A in the raffinate. This condition is one of the most important aspects of

8/8/2019 124 LLE4

10/31


10

liquid-liquid extraction since it allows for calculations and assumptions that based on

equilibrium systems (e.g. the ternary phase diagram).

One consideration to be made is for the separation factor. We want this factor to

as far away from unity as possible. This leads to a better separation in the extraction

process. The separation factor is represented as follows: [4]

R

*E

x

y= (5)

One of the last essential points to the theory behind liquid-liquid extraction is

mass transfer. The driving force for this mass transfer arises from the concentration

difference of the solute in each of the solvents. In general, it is assumed that the system

is at an equilibrium state when mass transfer is occurring. Solute fluxes in the raffinate

and extract can be expressed as

N = KEA(xEi - xE) (6)

N = KRA(xRi - xR) (7)

where KE and KR are the overall mass transfer coefficients, A is the cross-sectional area,

xE and xR are the concentrations of solute in the extract and raffinate respectively, and xEi

and xRi

are the concentrations of solute in each phase at the liquid-liquid interface.

Theory - General Flow Sheet for Extractor Design

With the key components of liquid-liquid extraction discussed, the following

general flowchart can be utilized for almost any process. Figure 3 (a) illustrates a general

ternary diagram for a desired solute (C), an extracting solvent (B) and a carrier solvent

(A). In this process, depicted in Figure 3 (b), we will assume that the feed (F) contains

components A and C. A solvent (S) is introduced in such a way that it will extract C

from the feed. The raffinate composition (R) is specified with respect to the recovery of

C that is needed. Table 2 summarizes the components, flows, and unknowns of such a

system. The following steps can be utilized to determine the extract composition and the

number of stages needed for most liquid-liquid extraction problems.

8/8/2019 124 LLE4

11/31


11

(a) (b)

Figure 3. (a) A general ternary phase diagram using for designing an extractor, and (b) a generalprocess diagram relating the ternary phase diagram to physical meaning.

Table 2. Table summarizing the general extraction phase diagram and process diagram inFigure 3.

Stream Components Is the composition given? Is the flow rate given?

F A and C Yes Yes

S B Yes, usually pure orrelatively pure component

B

No, determined bycalculation

E A with large Cconcentration

No, determined fromcalculation

No, determined fromcomponent mass balance

R B with small Cconcentration

Yes, recovery amountneeded of solute C from

design specifications

No, determined fromcomponent mass balance

where:

A is the carrier solvent

B is the solvent used to extract a certaincomponent

C is the component that is to be extracted fromA

8/8/2019 124 LLE4

12/31


12

Step 1:Determine the minimum solvent-to-feed ratio (S/F)min. This calculation needs to

be completed because E1, the extract composition, needs to be found. This

procedure begins by drawing an operating line from S to R that extends beyond

the boundaries of the diagram. Next, each tie line is considered to be apinch

point, and a line drawn from each tie line to the operating line is designated a P1,

P2, ,Pn. The Pi farthest away from R is called Pmin. After Pmin has been

established, a line is drawn from Pmin, through F (the feed composition), and to

the other side of the equilibrium curve.

Figure 4. Sample Ternary Diagram used to calculate Pmin for Step 1 of the general procedure fordesigning an extractor.

This point will represent E1. Figure 4 represents a general ternary diagram for a

Pmin calculation. After E1 is known, a mass balance around the system can be

utilized to determine the mixing point (point M in Figure 5). This is completed by

saying that:

F + Smin = R + E1 = M (8)

Solving for Smin/F, we will obtain the minimum solvent-to-feed ratio as

SAMA

MAFAmin

)x()x(

)x()x(

F

S

= (9)

8/8/2019 124 LLE4

13/31


13

where (xA)is are the fractions of A (the solute) in the feed (F), the solvent (S), and

at the mixing point (M). Generally, a solvent-to-feed ratio for an extraction system

is 1.5 times Smin/F:

(S/F)actual = 1.5(S/F)min (10)

This point can be found on the diagram by saying that (S/F) = MS/FM , and the

new mixing point (M) can be determined by moving along the FS line until the

new ratio point is reached.

Figure 5. A general ternary phase diagram showing the mixing point based on the (S/F)min (M)and the actual mixing point (M).

Step 2:Determine the extract (E) composition. This calculation is done very easily after

the first step. A line should be drawn from the raffinate composition, through

the new mixing point (M) and to the other side of the equilibrium line. This

will be the extract composition of solute-rich solvent.

8/8/2019 124 LLE4

14/31


14

Step 3:Find operating point. The operating point is a graphical point that represents the

difference in the overall flow; in addition, it is merely a point for which

calculations are computed around on a graph. Draw a line connecting the solvent

(S) and raffinate (R) points on the diagram. Follow this line beyond the diagram

to the left and right this is the operating line. Draw a line connecting the extract

(E) and the feed (F). The point at which these two lines intersect (P) is the

operating point. Figure 6 depicts such a diagram for this calculation.

Figure 6. A ternary phase diagram depicting the procedure for determining the operating point(P) and number of stages for an extraction column.

Step 4:Calculate the number of stages. Following the tie line from point E to the other

side of the equilibrium curve will give the composition of an intermediate

raffinate stage. Another operating line is drawn from the operating point, through

this intermediate point, and ends at point E. This is a stage of the system. This

procedure should be repeated until stages have been constructed to R, the raffinate

composition. Figure 6 shows this procedure for a general case.

Step 5:Calculate unknown flowrates. Since the extract and raffinate flows have not been

specified up to this point, this would be an appropriate level at which to be this.

This involves an overall mass balance on individual components. For the sake of

generality, choose A and B. It follows that

xAF(F) +xAS(S) = xAR(R) + xAE(E) (11)

8/8/2019 124 LLE4

15/31


15

xBF(F) +xBS(S) = xBR(R) + xBE(E) (12)

where the xAs and xBs are the fractions of A and B for the specified streams,

and F, S, R, and E are the flow rates of the feed, solvent, raffinate, and extract. R

and E are the only unknowns, and they can be solved for by a simple system of

equations.

Step 6.Determination of Extraction Column Diameter [1]

The diameter of the column must be large enough to permit two phases to flowcounter-currently through the column without flooding.

Estimation of column diameter for liquid liquid contacting devices is far morecomplex and uncertain than liquid-vapor contactors due the larger number of

important variables.

Variables necessary for calculating extractor column diameter include: Individual phase flow rates Density differences between the two phases Interfacial tension Direction of mass transfer Viscosity and density of continuos phase Geometry of internals

Column diameter may be best determined through scale-up of laboratory test runs.The necessary experimental data are obtained by: Use laboratory or pilot plant test unit with system components of interest. Use laboratory or pilot plant test unit with a diameter of one inch or more. Measurements of superficial velocities in each phase are made. The sum of these velocities may be assumed to hold constant for larger

scaled-up commercial units.

The superficial velocity data will be used to calculate the column diameterthrough the following correlation derivation.

The following notation is utilized in the correlation derivation:

uD = Actual average velocity of the dispersed (droplet) liquid phase uC = Actual average velocity of the continuous liquid phase UD = Superficial velocity of the dispersed liquid phase UC = Superficial velocity of the continuous liquid phase D = Volume fraction of dispersed liquid phase in column ur = Average droplet rise velocity relative to the continuous phase C = Capacity Parameter

8/8/2019 124 LLE4

16/31


16

CD = Drag Coefficient M = Density (volumetric mean) D = Density of dispersed phase C = Density of continuous phase f{1-D} = Factor which accounts for hindered rising effect of other

droplets u0 = Characteristic rise velocity for a single droplet = Viscosity (subscript will determine component) = Interfacial tension (subscript will determine component) AC = Column cross sectional area DT = Column diameter g = Acceleration due to gravity MD = Mass flowrate of the dispersed phase MC = Mass flowrate of the continuous phase

Figure 7. Counter-current flows of dispersed and continuous liquid phase in a column.

Diameter Calculation ProcedureStep A Determination of Column Total Capacity

Figure 7 illustrates lower density liquid droplets rising through the denser downwardflowing continuous liquid phase. The actual average velocities of each componentrelative to the column wall are:

D

DD

Uu

= (13)

D

CC

1

Uu

= (14)

The average droplet rise velocity relative to the continuous phase is the sum of these

equations:

D

c

D

Dr

1

UUu

+

= (15)

8/8/2019 124 LLE4

17/31


17

This relative velocity may also be expressed in terms of the forces acting upon thedroplet including drag forces, gravitational forces, and buoyancy forces. Thesevariables are combined into one parameter called C:

=

D

p

C3

gd4C (16)

If the droplet diameter dp is not known C may be obtained through a correlationprovided in Seader [1] equation (6-42), which was developed through experimentaldata from operating equipment. Taking into account density and rising effects ofother components the relative velocity may be expressed as:

( ) }1{f1Cu D21

D

2

1

C

DCr

= (17)

From experimental data, Gayler et al. found that the right-hand of the equation maybe expressed as:

( )D0r 1uu = (18)Eliminating the relative velocity by combining equation (17) and (18) gives:

( )D0D

C

D

D 1u1

UU=

+

(19)

This equation is a cubic in D. A graph of UD/u0 vs. D may be generated at somevalue of UC/uo. This graph represents the holdup curve for the liquid-liquid extractioncolumn. A typical value of UC/uo may be assumed 0.1.

Figure 8. Typical holdup curve for liquid-liquid extraction

8/8/2019 124 LLE4

18/31


18

At fixed UC, an increase in UD results in a increased value of holdup D until theflooding point is reached at the maximum of Figure 8:

0U

CUD

D

=

(20)

On the other hand, with UD fixed, UC may be increased until the flooding point isachieved at:

0U

DUD

C =

(21)

Inserting these derivatives into equation (19) results in the following expression for

Dat flooding conditions. The subscript f denotes flooding:

( )

+

=

1U

U4

3U

U81

D

C

5.

D

C

f (22)

Apply derivatives of Equation (19) into Equation (22), the expression solvedsimultaneously resulting in Figure 9 for the variation of total capacity as a function ofphase flow ratio:

Figure9. Total Capacity vs. Phase flow ratio

8/8/2019 124 LLE4

19/31


19

The total capacity may be read directly from the figure for a given phase flow ratioand will be essential for calculating the column diameter. The phase flow ratio isfound by:

D

C

C

D

D

C

MM

UU

=(23)

Step B Determination of Characteristic Rise Velocity

The dimensionless quantity [(u0CC)/()] may be assumed to be approximately0.01, as found by (Olney). Therefore the characteristic rise velocity for a singledroplet may be expressed as:

( )

CC

o

01.u

= (24)

Step C Determination of the superficial velocities at 50% of flooding value

The column extractor should be operated at 50% of the flooding velocity for bestperformance. The sum of the superficial velocities is found by reading the totalcapacity from figure 9 and multiplying by the characteristic rise velocity then dividethe quantity by two:

( )

( )( )

2

uu

UU

UU

0

0

fcD

Flooding%50DC

+

=+ (25)

Step D Determination of the Total Volumetric FlowrateThe total volumetric flow rate is a function of the mass flow rates:

+

=C

C

D

Dtotal

UUQ (26)

Step E Determination of Column Cross-Sectional AreaThe cross-sectional area is the total mass flowrate divided by the sum of thesuperficial velocities at 50% of flooding:

( )Flood%50dc

TotalC

UUQA

+= (27)

Step F Determination of Column Diameter

The column diameter may be found from the cross-sectional area:

8/8/2019 124 LLE4

20/31


20

2

1

cT

A4D

= (28)

Step 7.Determining the Height of the Column [1]

HETS (Height Equivalent to a Theoretical Stage) will be considered since it can beapplied directly to determine column height from the number of equilibrium stages.

For a well designed and efficiently operated column, experimental data suggest thedominant physical properties influencing HETS are:

Interfacial tension Phase viscosities Density difference between phases

HETS is best estimated by conducting small-scale laboratory experiments with the

systems of interest to determine the diameter of the column as discussed in step 6.

These values are scaled to commercial-size column by assuming that the HETS varieswith the column diameter raised to an exponent, which may vary from .2 to .4depending on the type of system. For the general approximation in step A, theexponent is arbitrarily set at 1/3. Figure 10 plots HETS for columns and rotarycontactor

Figure 10. HETS as a function of diameter vs. interfacial tension

Height Calculation ProcedureStep A - Find Value of HETS/DT

1/3

8/8/2019 124 LLE4

21/31


21

Using Figure 10 to determine the value (x) of HETS/DT1/3 at a specified interfacial

tension for the component system.

Step B - Solve for HETSThe value of (x) is known from above as well as the column diameter:

(29)Step C - Determine the Height of the ColumnThe total height of the column is derived from the number of equilibrium stagesderived in Step 4 multiplied by the HETS:

Total Height = (HETS)(Number of Equilibrium Stages) (30)

To compare calculated results to the performance of several types of extractor column(Seader) has provided average values of HETS and the sum of superficial velocities

(see Table 3).

Table 3. Performance of Several Types of Column Extractors

Extractor Type 1/HETS, (m-1

) UD+Uc, (m/hr)

Packed Column 1.5 2.5 12 30

Sieve-Plate Column .8 1.2 27 60Rotating Disk Contactor 2.5 3.5 15 30

Karr Column 3.5 7.0 30 40

PropertiesThe following are a partial list of the needed physical properties in liquid-liquid

extraction separations. It is by no means complete, other properties will be needed for

some of the calculations, and especially those needed to size the diameter of the column.

It is however complete as it relates to the described theory.

Temperature plays a smaller role in extraction than in other separation processes. Itis only dependent upon the temperatures of the streams fed into the column. There is

not a heating requirement for the process and H of mixing is generally insignificant.

For these reasons, extraction can be regarded as an isothermal process.

( ) 3/1TDxHETS =

8/8/2019 124 LLE4

22/31

8/8/2019 124 LLE4

23/31


23

CostsSeveral economic trade-off exist for the design of an extraction process. The total

cost of the process will be directly related to the key extraction design variables and type

of extraction equipment utilized. The following is a brief analysis of several design

variables that effect the economic balance:

At a fixed solvent feed ratio, the amount of solvent extracted increases with increased

number of trays. Thus, the value of the unextracted solute may be balanced against

the cost of the extraction equipment required to recover it.

For a fixed extent of reaction, the number of stages required decreases as the solvent

rate or reflux ratio increases. The capacity of the equipment necessary for handling

the larger liquid flow must increase with the larger reflux rate. Thus, the cost of the

equipment passes through a minimum when the minimum number of stages are

utilized.

As reflux ratio and solvent rates are increased the extract solutions become more

dilute. Therefore, the cost of solvent removal is increased as well as the operating

cost for increased utilities.

As a result of these economic balances the total annualized cost (investment &operating costs) must pass through a minimum at the optimum solvent reflux rate.

Further cost must be considered for the recovery of the saturated raffinate product as well

as the extract.

Cost models have been developed for the various types of extractor design. The

following are models for a column type extractor, mixer-settler, and continuous

centrifugal extractor.

Column Type Extractor

Douglas documents cost correlations for column in general. [5] The capital cost refers

to the purchase cost plus the installation cost of the column:

( )( )cFHD +

= 1829101 8020661 .. ..280

S&M$ColumnofCostCapital

8/8/2019 124 LLE4

24/31


24

(31)

Where D = diameter (ft), H = height (ft)

Fc = cost factor = Fp (pressure cost effects) + Fm (material cost effects)

(These values may be found in Douglas)

Depending on the column extractor type, trays or packing internals may be used.

Douglas has supplied a cost model for the purchase and installation of these materials

based upon correction factors to the following model:

(32)

The correction factor is the sum of the correction factors for the spacing, internal

type, and internal material. [5]

The total capital cost is the capital cost of the column plus the capital cost of the

internals. To annualize this investment, specify a payback period n, and divide the

total capital cost over this time period:

(33)Operating costs add to the at Total Annualized Cost (TAC). The operating costs:

(34,35,36)

including the utilities, labor, and maintenance costs. Pratt has estimated these cost as

of 1977. Inflation of these values may be accounted for using the M&S index:

Therefore the Total Annualized Cost for a column type extractor is:

Total Annualized Cost (TAC) = Total Annualized Capital Cost + Operating Cost

(37)

cFHDSM

= 55174

280

..&

$InternalsofCostCapital

+=

n

InternalsofCostCapitalColumnofCostCapitalCostCapitaldAnnualizeTotal

=

=

=

Yr

CostCapital3%TotaleMaintenanc

S&M

YrWorker

20000$Labor

600

S&M

kWhr

.04$Electicity

600

8/8/2019 124 LLE4

25/31


25

Mixer-Settler Type Extractor

Woods establishes a cost model for mixer-settler extractor which includes: carbon-

steel mixer-settler, labor and maintenance, explosion-proof motor, drive, piping,

concrete, steel, instruments, electrical, insulation, and paint:

(38,39)

The desired capacity must be specified in terms of Mgal/yr. These capital cost must

be added to the operating cost as described in the column extractor section.

Continuous Centrifugal Extractor

Woods establishes a cost model for continuous centrifugal extractors based upon a

centrifugal extractor made of 316 stainless steel including flexible connectors,

explosion-proof motor, variable speed driver, instrumentation, pumps, labor, and

maintenance:

(40)The desired capacity is in units of Mgal/yr. As above, the capital cost must be added

to the operating cost as defined in the column extractor section.

Case Study

Question for Liquid-Liquid Extraction

An extractor is to be designed such that acetone will be extracted from a feed mixture of30% acetone and 70% ethyl acetate. Water will be used to extract the acetone, and the

water is assumed to be pure. The raffinate will have a composition of 7% acetone and

93% ethyl acetate (point B), while the extract will have a composition of 12% acetone,

8% ethyl acetate, and 80% water (point D). A ternary phase diagram is given for this

=

=

600

S&MCapacityDesired1014.8CostCapital

600

S&M

CapacityReference

CapacityDesiredCostReferenceCostCapital

.3

70

31010

n

=

600

S&M

102.2

CapacityDesired1051CostCapital

.

3

358

8/8/2019 124 LLE4

26/31


26

system, along with corresponding tie lines. The feed to the column has a flowrate of

20,000 kg/hr and the solvent-to-feed ratio is assumed to be 1.75.

S

D

Solution to Liquid-Liquid Extraction Case Study

The number of stages needed for this problem

The number of stages can be stepped-off in a fashion analogous to that presented

in the Theory Section of this paper. the operating line is drawn from S to B and is

extended to the left of the diagram. Another line is drawn from F to D and intersects the

operating line from S to B this is the operating point. The tie line from point D is

followed to the other side of the equilibrium curve. From this point, another operating

line is drawn back to the operating point. The point at which is line intersected the

extract side of the equilibrium curve is located, and the tie line is drawn back to the other

side of the curve to obtain another point from which to draw another operating line. The

total number of stages is four once all the equilibrium lines have been drawn.

Determine:

The number of stages needed for this problem The solvent, extract and raffinate flowrates The height and diameter of the column The cost of this trayed column

8/8/2019 124 LLE4

27/31


27

The solvent, extract and raffinate flowrates

Since the basis behind extraction processes is mass conservation -- as it is withany process a material balance is done around the process. This is analogous to

the mass balances done in the Theory Section (Feed + Solvent = Extract +Raffinate). In addition, we recall that S/F = 1.75. From the design specifications,it follows that

75.1F

S= kg/hr35,000kg/hr)000,20(75.1F75.1S ===

After determining S (the solvent feed rate), we can complete the necessarymaterial balances to solve for R (the raffinate rate) and E (the extract rate):

F + S = E + R

Balance on acetone:

0(S) + (0.30)(20,000) = (0.07)(R) + (0.12)(E)

Balance on water:

0(F) + (1.0)(35,000) = (0)(R) + (0.80)(E)

This yields:

E = 43,750 kg/hrR = 10,714 kg/hr

It is seen that this method does not give a 100% mass balance. This is accountedfor by assuming that the graphical method is not as accurate as one would like,but it practical for our design purposes.

The height and diameter of the column

The first step is to establish all necessary physical properties. Much of thisinformation is available in Langes Handbook of Chemistry. Subscript D is the

dispersed liquid phase (organic) and subscript C is the continuous liquid phase(inorganic).

Acetone = 791 kg/m3 Ethyl Acetate = 789 kg/m3

Use the organic phase composition data from above to find density of the organicphase:

8/8/2019 124 LLE4

28/31


28

Organic = (.3)(791) + (.7)(789) = 789.6 kg/m3 = .001955 lbf/ftInorganic = 1000 kg/m3 C = .000021 lbf s

Calculate the phase flow ratio from the mass flow rates and densities

72369.

m

kg6.789

mkg1000

hr

kg35000

hrkg20000

M

M

U

U

3

3

D

C

C

D

C

D =

=

=

Using Figure 9 and the phase flow ratio find the total column capacity

( )34.

u

UU

0

fDC =+

(D-less)

Determine the characteristic rise velocity

( )s

ft19988.

cm

g0.1

ft

slbf000021.

cm

g2104.

ft

lbf001995.01.

01.u

32

3

CC

o =

=

=

Calculate superficial flooding velocity

(UD + UC)f= (.34)(.19988 ft/s) = .067959 ft/s

Calculate superficial velocity at 50% of flooding velocity

(UD + UC)50% Flooding Velocity = (.067959 / 2)ft/s (3600 s/min) = 122.327 ft/hr

Determine the total volumetric flowrate:

hr

ft5.2130

m

ft3146.35

m

kg1000

hr

kg35000

m

kg6.789

hr

kg20000

UUQ

3

3

3

33C

C

D

Dtotal =

+=

+

=

Determine cross-sectional area:

( )2

3

Flood%50dc

TotalC ft4165.17

hr

ft327.122

hr

ft5.2130

UU

QA =

=+

=

Calculate the diameter of the column:

8/8/2019 124 LLE4

29/31


29

ft70.44165.174A4

D2

1

2

1

cT =

=

=

To find the height of the column use figure 10 with an interfacial tension of 29.15

dyne/cm. This results in a value of 6.4 for the y-axis.

HETS/DT1/3 = 6.4

Solve for HETS using the determined column diameter

ft728.107.44.6D4.6HETS 3/13/1

T ===

Thus, the column height is:

Total Height =(HETS)(Number of Equilibrium Stages) = (10.728)(4) = 42.9ft

Extraction Column DimensionsDiameter, ft 4.70Height, ft 42.9

The cost of this trayed column

Using Douglas [5] correlation for capital costs:

Column Cost $ = ( ) 50.965,40$9.427.49.1012801061 802.066.1 =

Tray Cost $ = ( ) 32.8411$9.427.47.4280

1061 55.1 =

Assuming a payback period of six years, n = 6

Total Annualized Capital Cost = ($40,965 + $8411) / 6 years = $8229.33 / yr.

The operating cost of electricity, labor, and maintenance are negligible in contrastto the operating cost associated with solvent recovery through further separations.

Further, introducing a new material into the process to extract a solute result innew material cost and is a function of solvent recovery.

Total Annual Cost = $8229 + operating costs + solvent & product recovery cost

8/8/2019 124 LLE4

30/31


30

AlternativesIn the case where extraction does not apply for a certain feed, we must consider

our other options. Alternatives to extraction include the following:

1.

Distillation

Extractive

Azeotropic

Reactive

2. Crystallization

3. Adsorption

In discussing distillation, we must evaluate under what conditions extraction

would be valid over distillation. Extraction is preferred to distillation for the following

reasons: [1]

Case of dissolved or complexed inorganic substance in organic or

aqueous solutions

Removal of a component present in small concentrations

A high-boiling component is present in relatively small quantities in a waste stream

Recovery of heat-sensitive materials, where extraction may be less expensive than

vacuum distillation

Case of separation of a mixture according to chemical type rather than relative

volatility

Case of the separation of close-melting or close-boiling liquids, where the difference

in solubilities can be exploited

Case of mixtures that form azeotropes

If the situation does not meet any of the above reasons, then distillation can be

considered. For example, if the boiling points of two components where not close, then

distillation would be preferred over extraction.Using crystallization over extraction, one would have to consider the difference in

the freezing points of the components and also have information for a solid-liquid phase

diagram. This diagram is necessary to determine the eutectic point, which is the point

where one component becomes fused into the other.

8/8/2019 124 LLE4

31/31


Liquid adsorption could be used for certain components by contacting a liquid

mixture with a porous solid. The solid acts as the adsorbent and must be insoluble in the

liquid mixture. There is no available theory regarding adsorption equilibrium curves;

however, experimental data at a fixed temperature is used for plotting curves

References

1. Seader, J.D. and Henley, E.J. Separation Process Principles. John Wiley and Sons.New York, 1999.

2. Strigle, R.F. Packed Tower Design and Applications. Gulf Publishing Company.Houston, 1994.

3. Sandler, S.I. Chemical and Engineering Thermodynamics. John Wiley and Sons.New York, 1998.

4. Treybal, R.E. Mass Transfer Operations. McGraw-Hill. New York, 1980.

5. Douglas, J.M. Conceptual Design of Chemical Processes. McGraw-Hill. New York,1988.

6. Hanson, C., Baird, M.H.I., Lo, T.C. Handbook of Solvent Extraction. John Wileyand Sons. New York, 1983.

7. Reid, R.C., Prausnitz, J., Poling, B. The Properties of Gases and Liquids. 4th Ed.McGraw-Hill. New York, 1987.

Date post:	10-Apr-2018
Category:	Documents
Upload:	bluestardiver
View:	219 times
Download:	0 times

124 LLE4

Documents