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EXTRACTION
1. General ............................................................................................................................................. 2
1.1 Symbols and notes ..................................................................................................................... 2
2. Theory .............................................................................................................................................. 2
2.1 Mass Fraction coordinates ......................................................................................................... 2
2.2 Phase Equilibrium ...................................................................................................................... 3
2.3 Calculation Models .................................................................................................................... 4
2.4 Mass transfer Ratio .................................................................................................................... 6
2.5 Calculating the operating line .................................................................................................... 6
2.6 Overall Efficiency ...................................................................................................................... 7
3. Equipment ........................................................................................................................................ 7
3.1 Column ....................................................................................................................................... 7
3.2 Measuring Equipment ................................................................................................................ 8
4. Operating the colunm ....................................................................................................................... 8
4.1 Arrangements ............................................................................................................................. 8
4.2 Starting the Work ....................................................................................................................... 8
4.3 Measurements ............................................................................................................................ 9
4.4 Finishing the Work..................................................................................................................... 9
5. Specification..................................................................................................................................... 9
6. Nomenclature ................................................................................................................................. 11
7. References ...................................................................................................................................... 11
8. Appendices ..................................................................................................................................... 11
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1. GENERAL
In this work only continuous extraction is considered. Batch distillation, which is time dependent, is
not handled. Extraction as a continuous and industrial unit operation is usually done in one unit,
which is called an extraction column. After the extraction the solvent has to be separated from the
products (usually by distillation), so the whole process includes several steps. Therefore distillation
is often economical choice if it can be used.
Extraction as a unit operation means separating the components of a mixture. Separation is based on
the difference of solubility of components when a certain solvent is used. Notice the analogy
between distillation and extraction especially when recycle flows are included.
1.1 SYMBOLS AND NOTES
No recycle flow is used in this work, so the extraction apparatus looks like the schematic
representation in figure 1.
Va = extract
water phase
heavy product
Vb = solvent
water
heavy feed
La = feed
Hac + MIK
light feed
Lb = raffinate
MIK phase
light product
Figure 1. Extraction apparatus
Notice that extract phase is V phase and its concentration is marked with y and raffinate phase is L
phase and its concentration is marked with x (Treyball, 1980, p 477). In this special case the
markings are a contrary to the general manner where the heavier phase is L and the lighter is V
phase.
If HAc or acetone was to be extracted from water with MIK as in McCabe et al. (1993) page 633
figure 20.10, MIK phase would be the extract phase and water phase the raffinate phase. Notice in
figure 20.10 that equilibrium data does not define which phase is the extract phase but the direction
where component A transfers. By definition, component A transfers from raffinate to extract
(Treyball, 1980, p 477).
2. THEORY
The following notions are assumed to be familiar:
 Ideal stage, ideal stage model, calculation of the amount of ideal stages
 Mass transfer ratio, equimolar, equimass
 Phase equilibrium
 Saturated flow
2.1 MASS FRACTION COORDINATES
In this case the mass transfer is neither equimolar nor equimass so the operating line will not be
straight in mole or mass fraction coordinates. The equilibrium data of the system under study is
given in mass fractions (appendixes 1 and 2). Therefore mass fraction coordinates are used as base
coordinates in this work.
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The mass transfer ratio is defined analogically with mole fraction coordinates:
z
d (Vy )
dV
z
d ( Lx)
,
dL
(1)
where concentrations are in mass fractions and flow rates are total mass flows despite the markings.
If only equations for the equilibrium line and operating line are used in calculations, as in this case,
instead of molar studies relating to mass transfer and diffusion between components (actual
multicomponent calculation), the use of mass fractions instead of mole fractions gives the same
number of ideal steps. In general case this method is incorrect due to facts mentioned above.
2.2 PHASE EQUILIBRIUM
Only three-component systems are discussed here. Equilibrium data of such a system can always be
represented in triangular coordinates as in figure 20.10 in McCabe et al. (1993). Though, a more
useful method to plot three-component data is to use rectangular coordinates as in figure 2. Tie lines
in figure 2 connect extract and raffinate compositions, which are in equilibrium state together. A
diagram based on these points gives the equilibrium line in the x,y-coordinates.
Pure
component
A
xA, yA
Solubility line
plait
point
Tie line
pure
solvent
pure
component
B
xS, yS
Figure 2. Rectangular coordinates of three-component system.
If all the flows in an extraction process are saturated and the solvent concentration of saturated
extract and saturated raffinate is a linear function of concentration of component A, the mass
transfer ratio is a constant in extraction.
In order to use a constant mass transfer ratio in the model, the solubility line must be linearized. The
easiest way to do this is shown in figure 3.
xA, yA
Solubility line
linearization
xS, yS
Figure 3. Linearization of the solubility line.
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Hence, the following equations are obtained for the solubility lines:
yS  M y y A  By
(2a)
x S  M x x A  Bx
(2b)
2.3 CALCULATION MODELS
No recycle flow is used in the laboratory equipment so figure 1 represents the process and a normal
ideal stage model shown in figure 4 can be used to calculate the process.
ya
Va
V1
V2
Vn-1
L1
Ln-2
1
La
xa
L0
Vn
n-1
Vn+1
n
Ln-1
Vn+2
VN
Ln+1
LN-1
n+1
Ln
VN+1
yb
Vb
N
LN
Lb
xb
Figure 4. Ideal stage model
The inner flows Ln and Vn+1 and the exiting flows Lb and Va are assumed to be saturated, that is they
are on the solubility line. This is a very accurate assumption in this case and in mass transfer
processes in general. The aim of mass and heat transfer between the phases is to make it as efficient
as possible.
If the operating and equilibrium line can be assumed to be straight lines, the amount of the ideal
stages can be calculated from equation (3).
ya  ya*
xa  xa*
ln
ln
yb  yb*
xb  xb*
N

,
ln S
ln S
(3)
where the stripping factor S is the ratio of slopes of the equilibrium line and operating line. With the
assumptions mentioned above, the stripping factor S is a constant and is defined as follows:
S
m
m

yb  ya
L /V
xb  xa
(4)
Since the solvent feed Vb is pure water and since in the feed La there is no water at all or only a
slight amount of it, the feed and solvent feed are unsaturated.
However, all the flows in the model are assumed to be saturated so that the mass transfer ratio could
be assumed to be constant in the process. This means that two internal flows Li and Vi must be
added to the model in order to make the flows L0 and VN+1 saturated. The concentration of an
internal flow is the same as that flow, from which it separates. So the state of flows Li and LN is the
same and the state of flows Vi and V1 is the same. This specified ideal stage model is shown in
figure 5 and the states of the flows are shown in figure 6.
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ya
Va
V1
Vi yi
La
xa
V2
Vn-1
1
L0
Vn
n-1
L1
Ln-2
Vn+1
n
Ln-1
Vn+2
VN
n+1
Ln
VN+1
N
Ln+1
yb
Vb
LN-1
Li xi
LN
Lb
xb
Figure 5. Specified ideal stage model of the process.
xA, yA
Va, Vi
L0
La
Vn+1
Ln
Lb, Li
Vb
VN+1
xS, yS
Figure 6. The states of the flows in the process.
The flow rates of the internal flows are obtained from the balances over them. For example:
Overall balance: L0  La  Vi
Component A: L0 x0, A  La xa, A  Vi ya, A
(5a)
(5b)
L0 x0, S  La xa, S  Vi ya, S
(5c)
Solvent S:
Now, the flow Va = V1 is known to be saturated, so the concentrations ya are known and equal to the
concentrations yi of the internal flow Vi. Since L0 is wanted to be saturated and the solubility line is
linearized, by substituting xs from equation (2b) to equation (5c), equation (6) is obtained:
L0 (M x x0, A  Bx )  La xa, S  Vi ya, S
(6)
By substituting equations (5a) and (5b) into equation (6) and solving Vi, we get
Vi  La
xa , S  ( M x xa , A  Bx )
( M x ya , A  Bx )  ya , S
(7)
Analogically, for Li:
Li  Vb
yb , S  ( M y yb , A  By )
( M y xb, A  By )  xb, S
(8)
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2.4 MASS TRANSFER RATIO
According to chapter 2.2 the mass transfer ratio is constant with certain assumptions and the mean
mass transfer ratio is obtained as follows, if the solubility lines are linearized as in equation (2) and
the flows are saturated (notify the signs):
z
B y  Bx
Bx  B y
d (Vy ) d ( Lx)



dV
dL
My Mx My Mx
(9)
2.5 CALCULATING THE OPERATING LINE
If the mass transfer ratio z is constant or the mean mass transfer ratio can be assumed to be constant
in the whole process, it is possible to calculate the operating line of the process, when
feeds,
products and
constant mass transfer ratio
of the process are known.
The algorithm for the calculation is following:
1.
Some values for concentration x or y are chosen:
x
2.
L or V is calculated from either end of the process with the help of z:
L
3.
(10)
y
z  xa
z  xb
La 
Lb
zx
zx
z  ya
z  yb
Va 
Vb
zy
zy
(11)
The unknown V or L is calculated from the overall balance:
V  L  D
4.
V 
L  V  D
(12)
The concentration y or x corresponding to the chosen concentration x or y is calculated from
the component balance:
y
L
C
x
V
V
x
V
C
y
L
L
(13)
D and C are defined as:
D
 Va  La
 Vb  Lb
C
 Va ya  La xa
 Vb yb  Lb xb
(14)
(15)
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2.6 OVERALL EFFICIENCY
If the real process is a stage process, the change from the ideal stage model to the actual stage
process is done with the efficiency ε of an actual stage. There are many different definitions for
efficiencies but the simplest one is the overall efficiency εTOT, which is defined as:
 TOT 
N IDEAL
N ACTUAL
(16)
3. EQUIPMENT
3.1 COLUMN
The extraction column used to this work is a York-Scheibel column. It is a countercurrent
multistage extraction column developed by E. G. Scheibel in the late 40’s and made by Otto H.
York Inc. in 1952. One mixing zone and one settling zone form one actual stage. According to the
data sheet of the device the efficiency of an actual stage for MIK-water-HAc system was measured
to be 0.5 – 0.9 when flow rate was between 0 – 0.06 m3 / sm2 and the diameter of the column was
12 in. The mixing speed was not told. According to Scheibel (1948) the mixing speeds of other
substances have varied between 1000 and 1600 rpm.
According to the data sheet the nominal capacity of a 1-inch (25 mm) column in MIK-water-HAc
system is a gallon per hour or 1*10-6 m3 / s, which gives a liquid load of 0.02 m3/m2/s.
Figure 7. Extraction equipment
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3.2 MEASURING EQUIPMENT
Feed rotameters are attached to the laboratory equipment. When the rotameter reading is higher
than 6, the flow rate is:
Water:
Flow rate (g/s) = 0,0429reading - 0,1921
MIK+HAc: Flow rate (g/s) = 0,0301reading - 0,1323
Also






A stopwatch
Tachometer for defining mixing speeds
NaOH-burette for defining concentrations + NaOH solution and an indicator.
Magnetic mixer and a magnet
3 x 100 ml Erlenmeyer-bottles for titration
2 x 500 ml Erlenmeyer-bottles for defining the flow rates
4. OPERATING THE COLUNM
4.1 ARRANGEMENTS
Water is used to extract acetic acid (HAc) from methyl isobutyl ketone (MIK). So, the extracted
component is HAc, the diluent is MIK and water is the solvent:
component A
component B
solvent S
= HAc
= MIK
= water
The continuous phase is the water phase into which the MIK phase is dispersed by mixing. Water is
led to the column from top and MIK-HAc mixture from the bottom. Flow from the feed tanks to the
column is achieved with compressed air. The feed rates are controlled with valves and measured
with the rotameters.
The phase boundary is adjusted to the upper part of column, below the exit point of the lighter
phase. Adjusting is done with an overflow pipe.
The solvent fed to the system is pure water, so yS,b = 1.0.
The feed is a mixture of pure MIK and pure HAc.
CAUTION!
MIK-HAc mixture is corrosive! Watch out for skin contact. Use safety gloves if needed!
4.2 STARTING THE WORK





Define the concentration of NaOH solution with 0,1 M HCl solution.
Define the concentration of the HAc-MIK feed twice.
Open the pressure air valve and adjust the pressure to 0,3 bar.
Remove the air from the rotameters by opening the feed valves almost completely open for a
moment.
Fill up the column with water so that the water level is about 10 cm below the water feeding
point.
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


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Adjust both the feed flow rates to their desired values.
Switch the mixer on.
Keep the phase boundary between the uppermost settling zone and the exit point of the lighter
phase.
4.3 MEASUREMENTS









Calculate the amount of ideal stages.
Assistant gives the flow rates and the mixing speed.
Adjust the feed rates and the mixing speed to their desired values.
Take a sample every 10 minutes from the extract phase and define its HAc concentration.
When the HAc concentration is not changing remarkably, take the other measurements:
Take samples from the products and titrate them.
Define the product rates by collecting a sample at least for 10 minutes and by weighting it.
Check and note the mixing speed.
Adjust the feed rates and the mixing speed to their desired values and repeat the measurements.
4.4 FINISHING THE WORK





Show your results to the assistant and after permission:
Close the pressure air valve and the feed valves.
Empty the column to the canister in the fume cupboard.
Empty the collector bottles of the extract and raffinate to the same canister.
Clean up the surroundings.
5. SPECIFICATION
Specification is done as told in general instructions. The both members of work pair take care of
one stabilized state measurements of the column.
1.
2.
3.
4.
Show the stabilization of column graphically.
Match the measured balances with workbook tasmays.xls.
Summarize the results.
Use mass fraction coordinates.
Ignore following facts:
 Mass transfer is not equimass
 Equilibrium is not linear
 Feed and solvent are not saturated
Assume that the operating line is a straight line between points (xa, ya) and (xb, yb) in mass fraction
coordinates and assume a straight equilibrium line in the same coordinates.
4.1. Draw the equilibrium line points and linearize the equilibrium line. Define the constants m
and b of the linearized equilibrium line. Draw the end points of the operating line (xa, ya) and
(xb, yb) and draw a straight line between them. Define the number of ideal stages NG1
graphically by stepping off the column.
4.2. Define numerically (and in this case analytically) the number of ideal stages NN1.
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5. Use mass fraction coordinates.
Notify the facts:
 Mass transfer is not equimass
 Equilibrium is not linear
Ignore the facts:
 Feed and solvent are not saturated.
So, assume that all flows in the process are saturated including feed flow and assume that the mass
transfer ratio is a constant.
5.1. Linearize the solubility line.
5.2. Calculate the mean mass transfer ratio.
5.3. Draw the equilibrium line.
Calculate and draw the operating line.
Define the number of ideal stages NG2by stepping off the column.
5.4. Calculate numerically the number of ideal stages NN2 between points (xa, ya) and (xb, yb). This
can be done with a spreadsheet (Excel).
6. Use mass fraction coordinates.
Notify the facts:
 Mass transfer is not equimass
 Equilibrium is not linear
 Feed and solvent are not saturated.
Assume that only the flows inside the process are saturated and the mass transfer ratio is constant.
6.1. Calculate the internal flows Vi and Li and the flow rates and concentrations of VN+1 and L0.
6.2. Calculate numerically the number of ideal stages NN3 between points (x0, y1) and (xN, yN+1) for
example with a spreadsheet program.
7. Compare the numbers of ideal stages (NG1 with NG2, NN1 with NN2, NN3 with others, graphically
defined with numerically defined, and actual number of stages with all others).
8. Calculate the overall efficiency.
9. Incorrect estimate
 Estimate the errors in balances quantitatively.
 Estimate gross and systematical errors qualitatively.
10. Conclusions
 What can you say about the results? Why the results are like they are? Do the results fit the
theory? If not, why?
 Errors of the results are not discussed here.
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6. NOMENCLATURE
b
B
C
D
f
g
L
m
M
V
x
y
z
Constant of the linearized equilibrium line (y*=mx+b)
Constant of linearized solubility line, dimensionless
Net flow of component into the direction of V phase, usually for component A, kg/s
total net flow into the direction of V phase, kg/s
function of the operating line
function of the equilibrium line, y*=g(x) and x*=g-1(y)
total mass flow in the raffinate, kg/s
slope of the linearized equilibrium line (y*=mx+b)
Constant of linearized solubility line, dimensionless
total mass flow in the extract, kg/s
mass fraction in raffinate, usually for component A, dimensionless
mass fraction in extract, usually for component A, dimensionless
mass transfer ratio of component A, dimensionless
Subindexes
a
the end of the device where L phase is fed
b
the end of the device where V phase is fed
A
component A
B
component B
i
internal flow
n
tray n
S
solvent
x
raffinate
y
extract
Superscripts
x*
Equilibrium value of x
7. REFERENCES
McCabe, W.L., Smith, J.C. and Harriot, P., Unit Operations of Chemical Engineering, 5th ed.,
McGraw-Hill, 1993.
Treyball, R.E., Mass-Transfer Operations, 3rd ed., McGraw-Hill, 1980
8. APPENDIXES
1. The solubility and equilibrium data of the HAc-MIK-H2O –system.
2. Matching the balances of extraction
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APPENDIX 1
1
The solubility and equilibrium data for HAc – MIK – water system.
A
HAc
C2H4O2
60.05
B
MIK
C6H12O
100.16
SOLUBILITY
w
A
B
0.000
0.020
0.242
0.053
0.286
0.073
0.311
0.094
0.329
0.125
0.342
0.172
0.341
0.254
0.333
0.322
0.322
0.402
0.312
0.432
0.295
0.492
0.279
0.535
0.260
0.580
0.237
0.624
0.217
0.661
0.193
0.698
0.167
0.741
0.136
0.790
0.000
0.977
S
water
H2O
18
S
0.980
0.705
0.641
0.595
0.546
0.486
0.405
0.345
0.276
0.256
0.213
0.186
0.160
0.139
0.122
0.109
0.093
0.074
0.023
w = weight fraction
wy= weight fraction in water phase
wx= weight fraction in MIK phase
Linearization of the solubility line
Relative % relative
S(calc)
difference difference difference
0.98
0.00
0.00
0.11
0.70
0.01
0.01
0.97
0.65
0.01
0.01
0.90
m= -1.17
b= 0.98
0.17
0.15
0.14
0.13
0.12
0.10
0.09
0.01
m= 0.56
b= 0.01
Relative % relative
difference difference difference
0.02
0.11
11.15
0.01
0.03
3.37
0.00
0.02
1.96
0.01
0.07
6.98
0.01
0.07
7.41
0.01
0.11
10.52
0.01
0.15
15.05
0.01
0.61
61.12
EQUILIBRIUM
wx
wy
0
0
0.069
0.091
0.126
0.153
0.229
0.256
0.276
0.301
0.308
0.324
0.324
0.338
Source:
Othmer, D.F., White, R.E., Trueger, E., Liquid-liquid Extraction data,
Ind. Eng. Chem. 33(1941) 1240-8
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APPENDIX 1
2
Solubility of HAc-MIK-H2O -system
1.000
0.900
0.800
0.700
w(HAc)
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
1.000
w(water)
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APPENDIX 1
3
Equilibrium of HAc-MIK-H2O -system
1
0.9
0.8
w(HAc) water-phase
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
w(HAc) MIK-phase
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APPENDIX 2
MATCHING THE BALANCES
This workbook contains the following sheets:
MENU
This sheet.
BASIC CASE
Protected therefore Solver does not work.
EXCTRACTION
Protected therefore Solver does not work.
BINARY DISTILLATION Protected therefore Solver does not work.
MACRO
Contains the macros.
1
8.4.2005
Color codes
GIVE
MATCHED
CONSTRAINTS
TARGET FUNCTION
GENERAL PRINCIPLE OF MATCHING THE BALANCES
The sum S = SUM( Pi * ((xi-Xi) / xi)^2 ) is minimized
xi = measured value of quantity i
Xi = matched value of quantity i
Pi = weighting coefficient of quantity i
so that total and component balances hold true and the sum of mole fractions is one.
The task at hand is a constrained optimization task, which constraints are the balances and the sum of mole fractions
and the minimized function is the weighted sum of relative differences of
matched and measured values.
There is no universal algorithm to solve a problem like this, which would always lead to a solution.
Therefore matching should be done with few different initial values.
INSTRUCTIONS
0 Go to sheet and use the macro CopyPage to make an unprotected sheet.
1 Give the measured values to cells:
MEASURED
2 If needed give the weigthing coefficients to cells:
WEIGHT
The greater the weighting coefficient the smaller is the difference between matched and measured values.
Weighting coefficient of an accurate measument should be large.
Weighting coefficient of an inaccurate measument should be small.
Weighting coefficient can be in the range (0, positive infinity). Here the coefficients should be >= 1.
If there is no special need for weighting, use 1.
3 Give initial values (e.g. measured values) to cells:
INITIAL VALUES
Leave all other cells UNTOUCHED!
4 Choose Tools / Solver / Solve: Solver starts the optimization.
The target cell of the optimization becomes active.
The constraints are:
Balances hold true, in other words the cells IN-OUT = 0
Sum of mole fractions equals to 1, cells SUMx = 1
All matched values are positive, MATCHED >= 0
5 If the result is not reasonable, try other initial values or change the weighting coefficients.
Notice that the best result is obtained only by trial and error.
6 If a reasonable result is not found, save the result and restart solver.
If a reasonable result is not found after a few times, return to 5 and try other initial values.
7 Extraction:
Substance A (component to be extracted) = HAc
Substance B (diluter) = MIK
Substance S (solvent) = vesi
if HAc, of which the feed is made, contains water p -weight %,
feed contains xB=1.0-(1.0+p/(100-p))xA,
since xS/xA is always p/(100-p) in feed, which is made by adding substance B
In the solvent yS=1.0 because it is pure water.
Raffinate and extract are saturated, so xS is obtained from the saturation curve.
In raffinate xS=0.56*xA+0.01 and in extract xS=-1.17*xA+0.98.
In both xB=1.0-xA-xS
Extra constraint: in raffinate and extract xA >= 0.001
8 Binary distillation
Notice that xB=1.0-xA
MENU
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APPENDIX 2
2
MATCHING THE BALANCES
EXTRACTION (A: HAc - B: MIK - S: water)
Extract
Solvent
PROCESS
Feed
Water in HAc=
Raffinate
20.000 weight%
WEIGHTING COEFFICIENTS
IN
Feed
TOTAL
xA
xB
xS
COLOR CODES
GIVE
MATCHED
CONSTRAINTS
TARGET FUNCTION
OUT
Solvent
1.0
Raffinate
1.0
Extract
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
1.0
MEASURED
IN
Feed
TOTAL
xA
xB
xS
SUMx
0.200
0.750
0.050
1.000
Solvent
2.000
0.400
1.500
0.100
0.000
0.000
1.000
1.000
2.000
0.000
0.000
2.000
OUT
Raffinate
3.000
0.300
0.900
0.522
1.566
0.178
0.534
1.000
Extract
0.300
0.071
0.629
1.000
3.000
0.900
0.213
1.887
IN-OUT
-2.000
-1.400
-0.279
-0.321
4.400
1.936
0.417
2.047
IN-OUT
-4.400
-2.904
-1.037
-0.459
MATCHED
IN
Feed
TOTAL
xA
xB
xS
SUMx
0.110
0.863
0.028
1.000
Solvent
1.100
0.121
0.949
0.030
0.000
0.000
1.000
1.000
2.200
0.000
0.000
2.200
OUT
Raffinate
3.300
0.330
1.089
0.475
1.568
0.195
0.643
1.000
Extract
0.440
0.095
0.465
1.000
RELATIVE DIFFERENCE
IN
Feed
TOTAL
xA
xB
xS
TOTAL
xA
xB
xS
0.450
0.450
-0.150
0.450
Solvent
-0.100
OUT
Raffinate
Extract
-0.100
-0.467
-0.100
-0.467
0.090
-0.335
-0.094
0.260
WEIGHTING COEFFICIENT * SQUARE OF RELATIVE DIFFERENCE
IN
OUT
Feed
Solvent
Raffinate
Extract
0.203
0.010
0.010
0.218
0.203
0.010
0.218
0.023
0.008
0.112
0.203
0.009
0.068
SUM
1.293
15.9.2015