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1
DESIGN OF KARR
EXTRACTOR
PRESENTED BY: GROUP # 04
WALEED SHAHID (UW-13-Ch.E-BSC-10)
MOHSIN RAZA (UW-13-Ch.E-BSC-18)
HASEEB IQBAL (UW-13-Ch.E-BSC-21)
SAEED (UW-13-Ch.E-BSC-23)
ALI HASSAN (UW-13-Ch.E-BSC-31)
M. FAHAD (UW-13-Ch.E-BSC-52) 2
MUHAMMAD FAHAD
(UW-13-CH.E-BSC-052)
3
DESIGN PROBLEM
The extractor is designed to separate a solution of methylene chloride and
methanol with the input mass flow rates of 2185 lb/h (991 kg/h) and 33 lb/h
(15.0 kg/h) respectively, using water as selective solvent with a mass flow rate
of 2218 lb/h (995 kg/h) to recover 95% methanol. Calculate the diameter and
height of the given system.
4
LIQUID-LIQUID EXTRACTION
Liquid-liquid extraction is a mass transfer operation in which a
solution (feed: mixture of solute and carrier liquid) is brought into
intimate contact with an immiscible or slightly miscible liquid
(selective solvent) in order to achieve transfer of the solute from the
feed to the solvent. The two liquid phases that have different
densities are then separated.
5
STEPS USED DURING LIQUID-LIQUID
EXTRACTION
• Bringing the feed and the solvent into intimate contact by
dispersing one phase into the other as droplets.
• Separation of the extract and the Raffinate phases that have
different densities.
• Removal and recovery of the solute from the extract phase in
relatively pure form (by evaporation, crystallization, etc.).
• Removal and recovery of the solvent from each phase, usually by
distillation. 6
LLE Vs. DISTILLATION
DISTILLATION LLE
Phase creation is achieved through Heat. No heat required.
Vapor and liquid are chemically similar. Phases are chemically different.
Used for liquids having large volatility difference.
Separation on the basis of boiling point.
Used for liquids having low or negligible volatility
difference.
It is an expensive method of separation. Relatively cheaper.
7
M.WALEED SHAHID
(UW-13-CH.E-BSC-010)
8
ADVANTAGES AND DISADVANTAGES
ADVANTAGES DISADVANTAGES
Liquid-liquid extraction can be used in the
separation of azeotropes.
Can be time consuming, especially if attainment of
equilibrium is slow
It can be operated at low to moderate temperature
for recovery of heat sensitive products
Can be affected by small impurities in the
solvent(s).
LLE is cheaper and can be used instead of using
the chemical methods.
Cumbersome for a large number of samples or for
large samples.
It has rapid and very selective separations that are
usually highly efficient.
Formation of emulsions can interfere with the
phase-separation process
Also used for components that have close boiling
points
Counter-current process can be complicated and
can require complicated equipment
9
APPLICATIONS
• Extraction of Fermentation Broth by using Karr Reciprocating-Plate Extractor.
• Extraction of Penicillin in Karr Reciprocating-Plate Extractor.
• Use of Ionic Liquid in a Karr Reciprocating-Plate Extractor.
• Recovery of tightly hydrogen-bonded organics from water; such as formaldehyde,
formic acid and acetic acid.
• Removal of high boiling organics from wastewater; such as phenol, aniline and
nitrated aromatics
• Essential oil extraction; such as pharmaceuticals, flavors, fragrances and food
products.
• Such as acrylates, nitrated organics and chloro-benzene compounds.
10
LIQUID-LIQUID EXTRACTION
EQUIPMENT
The equipment used to carry out the LLE are called Liquid-Liquid Extractors.
Common extractors are divided into following classes:
11
FLUID MOVEMENT MECHANICALLY AGITATED
Sieve column KARR column
Packed column SCHEIBEL Column
Spray column Pulsed column
Mixer settler Rotating disk contractor
ALI HASSAN
(UW-13-CH.E-BSC-031)
12
SIMPLE GUIDLINE FOR EXTRACTOR
SELECTION
Process
Minimum residence time
Emulsifying tendency
Flow rate over 400 gal/min , minimum residence
Small number of stages required
No
No
No
No
Centrifugal contractor
Reciprocating plate column , centrifugal
contractor
Mixer settler
Mixer settler (with out mechanical agitation)
Mechanically agitated column
Yes
Yes
Yes
Yes
13
SELECTION OF EQUIPMENT
We selected Karr Reciprocating-Plate Column for the given system, because it
is suitable for the systems having intermediate to low interfacial tension and
emulsifying tendency.
14
KARR RECIPROCATING-PLATE
COLUMN
The Karr Reciprocating-Plate column is a descendent of the pulsed sieve-tray
column. It has a reciprocating shaft with perforated plates mounted on it. In
this column plates move up and down approximately 2–7 times per second.
Also, the close spacing of the plates (25–50 mm) promotes high turbulence and
minimizes axial mixing, thus giving high mass-transfer rates and low HETS.
The central shaft, which supports sets of plates, is reciprocated by a drive at the
top of the column.
15
KARR EXTRACTOR
16
WORKING OF KARR EXTRACTOR
17
SAEED AKHTAR
(UW-13-CH.E-BSC-023)
18
CASE
The extractor is designed to separate a solution of methylene chloride and
methanol with the input mass flow rates of 2185 lb/h (991 kg/h) and 33 lb/h
(15.0 kg/h) respectively, using water as selective solvent with a mass flow rate
of 2218 lb/h (995 kg/h) to recover 95% methanol. Calculate the diameter and
height of the given system.
19
GIVEN DATA
Feed:
• Methylene chloride = 2185 lb/hr
• Methanol = 33 lb/hr
• Total = mF =2218 lb/hr
• Recovery coefficient = K = 2
• Methanol recovery % = =0.95
• Concentration mass unit= C = 0.5

20
CONTINUED...
Densities:
• Methylene chloride =82.41
• Methanol = 48.7
• Water =62.4
• x1 = mass fraction = 33/2218 = 0.0148
.
3
lb ft
3
lb ft
3
lb ft
21
DESIGNING STEPS FOR THE KARR
EXTRACTOR
Step 1: Mass Fraction in Raffinate
Step 2: Minimum Solvent Flow Rate
Step 3: Minimum Solvent Rate and Operating Rate
Step 4: Mass Fraction of Solute in Extract
Step 5: Calculation of Extraction Factor AE
Step 6: Calculate the Number of Equilibrium Stages
Step 7: Extractor Diameter
22
SOLUTION
Step 1: Mass Fraction in Raffinate
For Finding the mass fraction in Raffinate we use following equation.
2 1x = ( 1- ε ) x
2x 0.000744
23
CONTINUED...
Step 2: Minimum Solvent Flow Rate
F 1 2
SM 1 2
m Kx - y
=
m (x - x )
2(0.01488)-0
= = 2.105
(0.01488-0.000744)
24
CONTINUED...
Step 3: Minimum Solvent Rate and Operating Rate
F F
S SM
m m
= C
m m
 
 
 
=0.5×2.105
F
S
m
= 1.053
m
Fm = 2218 lb hr
s
2218
m = = 2106 lb hr
1.063
,
F
SM
m
= 2.105
m
25
CONTINUED...
Step 4: Mass Fraction of Solute in Extract
 F
1 2 1 2
S
m
y = y + x -x
m
 
 
 
=0+(1.053)(0.01488-0.00074)
1 0.014885y 
26
CONTINUED...
Step 5: Calculation of Extraction Factor AE
 F S
E
m /m
A =
K
E
1.063
=
2
A = 0.5265
27
Step 6: Calculate the Number of Equilibrium Stages
CONTINUED...
 
 
   1 2
E E E E
2 2
x - y K
N = log 1-A + A log (1 A )
x - y K
 
 
 
   E
0.014878-0
N =log 1-0.5265 +0.5265 log(1 0.5265)
0.000744
 
  
EN =3.587
EN 4
28
HASEEB IQBAL
(UW-13-CH.E-BSC-021)
DESIGN OF KARR EXTRACTOR 29
CONTINUED…
• Step 7: Extractor Diameter and Height
For calculation of extractor Height we require HETS
As
For HETS calculation, we require Diameter of extractor.
As
 E EZ =N HETS
 
0.38
1
1
D
HETS= HETS
D
 
 
 
30
31
CONTINUED…
• Now we have to calculate Diameter of our system for which we calculate
area, which is
 
1
1
2
D = 12 inches
Min HETS = 5.6
Volumetric Throughput = 1193 gal h.ft
SF F Sm ρ +m ρ
A =
J
32
CONTINUED…
Volumetric Flow Rates of both Feed and Selective solvent.
Feed =
Selective Solvent
Area Volumetric Throughput
3F
F
m 2185 33
= + = 27.19 ft hr
ρ 82.41 48.7
3S
S
m 2106
= = = 33.7514 ft hr = 252.4942 gal hr
ρ 62.43
SF F Sm ρ +m ρ
=
J
2203.418+262.4942
= = 0.3812 ft
1193
2
gal
J= 1193
hr.ft
33
CONTINUED…
So for our system
Diameter
So this calculated Diameter is less than 30 inch and we have standard pipe size of diameter
10.42 inch, so we use 10.42 inch pipe.
HETS
0.5 0.5
A 0.38067
= 4× = 4× = 0.6975ft
π 3.14
   
      
0.38 0.38
1
1
D 10.42
=(HETS) =5.6 =5.307in
D 12
   
     
34
CONTINUED…
For our design we increase value of HETS by 20% to avoid flooding.
Rounding off ZE to nearest 3
On both top and bottom we have installed settlers which have the diameter 50% greater than
the extractor diameter and also height of each settler is equal to settler diameter.
E E
Corrected HETS=5.307×1.20=6.368 in
Z =N (HETS)=4(6.368)=25.47
EZ =27
35
CONTINUED…
Diameter of settler =
Height of both settler =
To join Extractor with settler we require reducers which are a foot long.
Now reduces height =
1.5×10.42=15.63 inch
2×15.63=31.26 inch
2 12 24 
E SZ=Z +Z +Reducers height
=27+31.26+24=82.26inch
Z=6.86ft
36
CONTINUED…
Approximated height is:
Z=7ft
37
MOHSIN RAZA
(UW-13-CH.E-BSC-018)
38
EFFECT OF CONCENTRATION ON
HEIGHT
7
8 8
10
12
5
6
7
8
9
10
11
12
13
0.4 0.5 0.6 0.7 0.8 0.9 1
Height(ft)
Concentration (lb/ft3)
39
EFFECT OF RECOVERY % ON
HEIGHT
40
6.772
6.272
5.6885.6885.688
5.1885.1885.1885.1885.188
5
5.5
6
6.5
7
0 10 20 30 40 50 60 70 80 90 100
Height(ft)
Recovery (%)
EFFECT OF CONCENTRATION ON
HEIGHT
41
8.4 8.5
8.8
9
9.2
9.4
9.7
9.9
10.1
10.3
10.5
10.7
10.9 11
11.2
11.4
11.6
11.8
12 12.12
8
8.5
9
9.5
10
10.5
11
11.5
12
12.5
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Diameter(in)
Solvent Rate in Feed (lb/hr)
42
43

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Final design of-karr extractor

  • 1. 1
  • 2. DESIGN OF KARR EXTRACTOR PRESENTED BY: GROUP # 04 WALEED SHAHID (UW-13-Ch.E-BSC-10) MOHSIN RAZA (UW-13-Ch.E-BSC-18) HASEEB IQBAL (UW-13-Ch.E-BSC-21) SAEED (UW-13-Ch.E-BSC-23) ALI HASSAN (UW-13-Ch.E-BSC-31) M. FAHAD (UW-13-Ch.E-BSC-52) 2
  • 4. DESIGN PROBLEM The extractor is designed to separate a solution of methylene chloride and methanol with the input mass flow rates of 2185 lb/h (991 kg/h) and 33 lb/h (15.0 kg/h) respectively, using water as selective solvent with a mass flow rate of 2218 lb/h (995 kg/h) to recover 95% methanol. Calculate the diameter and height of the given system. 4
  • 5. LIQUID-LIQUID EXTRACTION Liquid-liquid extraction is a mass transfer operation in which a solution (feed: mixture of solute and carrier liquid) is brought into intimate contact with an immiscible or slightly miscible liquid (selective solvent) in order to achieve transfer of the solute from the feed to the solvent. The two liquid phases that have different densities are then separated. 5
  • 6. STEPS USED DURING LIQUID-LIQUID EXTRACTION • Bringing the feed and the solvent into intimate contact by dispersing one phase into the other as droplets. • Separation of the extract and the Raffinate phases that have different densities. • Removal and recovery of the solute from the extract phase in relatively pure form (by evaporation, crystallization, etc.). • Removal and recovery of the solvent from each phase, usually by distillation. 6
  • 7. LLE Vs. DISTILLATION DISTILLATION LLE Phase creation is achieved through Heat. No heat required. Vapor and liquid are chemically similar. Phases are chemically different. Used for liquids having large volatility difference. Separation on the basis of boiling point. Used for liquids having low or negligible volatility difference. It is an expensive method of separation. Relatively cheaper. 7
  • 9. ADVANTAGES AND DISADVANTAGES ADVANTAGES DISADVANTAGES Liquid-liquid extraction can be used in the separation of azeotropes. Can be time consuming, especially if attainment of equilibrium is slow It can be operated at low to moderate temperature for recovery of heat sensitive products Can be affected by small impurities in the solvent(s). LLE is cheaper and can be used instead of using the chemical methods. Cumbersome for a large number of samples or for large samples. It has rapid and very selective separations that are usually highly efficient. Formation of emulsions can interfere with the phase-separation process Also used for components that have close boiling points Counter-current process can be complicated and can require complicated equipment 9
  • 10. APPLICATIONS • Extraction of Fermentation Broth by using Karr Reciprocating-Plate Extractor. • Extraction of Penicillin in Karr Reciprocating-Plate Extractor. • Use of Ionic Liquid in a Karr Reciprocating-Plate Extractor. • Recovery of tightly hydrogen-bonded organics from water; such as formaldehyde, formic acid and acetic acid. • Removal of high boiling organics from wastewater; such as phenol, aniline and nitrated aromatics • Essential oil extraction; such as pharmaceuticals, flavors, fragrances and food products. • Such as acrylates, nitrated organics and chloro-benzene compounds. 10
  • 11. LIQUID-LIQUID EXTRACTION EQUIPMENT The equipment used to carry out the LLE are called Liquid-Liquid Extractors. Common extractors are divided into following classes: 11 FLUID MOVEMENT MECHANICALLY AGITATED Sieve column KARR column Packed column SCHEIBEL Column Spray column Pulsed column Mixer settler Rotating disk contractor
  • 13. SIMPLE GUIDLINE FOR EXTRACTOR SELECTION Process Minimum residence time Emulsifying tendency Flow rate over 400 gal/min , minimum residence Small number of stages required No No No No Centrifugal contractor Reciprocating plate column , centrifugal contractor Mixer settler Mixer settler (with out mechanical agitation) Mechanically agitated column Yes Yes Yes Yes 13
  • 14. SELECTION OF EQUIPMENT We selected Karr Reciprocating-Plate Column for the given system, because it is suitable for the systems having intermediate to low interfacial tension and emulsifying tendency. 14
  • 15. KARR RECIPROCATING-PLATE COLUMN The Karr Reciprocating-Plate column is a descendent of the pulsed sieve-tray column. It has a reciprocating shaft with perforated plates mounted on it. In this column plates move up and down approximately 2–7 times per second. Also, the close spacing of the plates (25–50 mm) promotes high turbulence and minimizes axial mixing, thus giving high mass-transfer rates and low HETS. The central shaft, which supports sets of plates, is reciprocated by a drive at the top of the column. 15
  • 17. WORKING OF KARR EXTRACTOR 17
  • 19. CASE The extractor is designed to separate a solution of methylene chloride and methanol with the input mass flow rates of 2185 lb/h (991 kg/h) and 33 lb/h (15.0 kg/h) respectively, using water as selective solvent with a mass flow rate of 2218 lb/h (995 kg/h) to recover 95% methanol. Calculate the diameter and height of the given system. 19
  • 20. GIVEN DATA Feed: • Methylene chloride = 2185 lb/hr • Methanol = 33 lb/hr • Total = mF =2218 lb/hr • Recovery coefficient = K = 2 • Methanol recovery % = =0.95 • Concentration mass unit= C = 0.5  20
  • 21. CONTINUED... Densities: • Methylene chloride =82.41 • Methanol = 48.7 • Water =62.4 • x1 = mass fraction = 33/2218 = 0.0148 . 3 lb ft 3 lb ft 3 lb ft 21
  • 22. DESIGNING STEPS FOR THE KARR EXTRACTOR Step 1: Mass Fraction in Raffinate Step 2: Minimum Solvent Flow Rate Step 3: Minimum Solvent Rate and Operating Rate Step 4: Mass Fraction of Solute in Extract Step 5: Calculation of Extraction Factor AE Step 6: Calculate the Number of Equilibrium Stages Step 7: Extractor Diameter 22
  • 23. SOLUTION Step 1: Mass Fraction in Raffinate For Finding the mass fraction in Raffinate we use following equation. 2 1x = ( 1- ε ) x 2x 0.000744 23
  • 24. CONTINUED... Step 2: Minimum Solvent Flow Rate F 1 2 SM 1 2 m Kx - y = m (x - x ) 2(0.01488)-0 = = 2.105 (0.01488-0.000744) 24
  • 25. CONTINUED... Step 3: Minimum Solvent Rate and Operating Rate F F S SM m m = C m m       =0.5×2.105 F S m = 1.053 m Fm = 2218 lb hr s 2218 m = = 2106 lb hr 1.063 , F SM m = 2.105 m 25
  • 26. CONTINUED... Step 4: Mass Fraction of Solute in Extract  F 1 2 1 2 S m y = y + x -x m       =0+(1.053)(0.01488-0.00074) 1 0.014885y  26
  • 27. CONTINUED... Step 5: Calculation of Extraction Factor AE  F S E m /m A = K E 1.063 = 2 A = 0.5265 27
  • 28. Step 6: Calculate the Number of Equilibrium Stages CONTINUED...        1 2 E E E E 2 2 x - y K N = log 1-A + A log (1 A ) x - y K          E 0.014878-0 N =log 1-0.5265 +0.5265 log(1 0.5265) 0.000744      EN =3.587 EN 4 28
  • 30. CONTINUED… • Step 7: Extractor Diameter and Height For calculation of extractor Height we require HETS As For HETS calculation, we require Diameter of extractor. As  E EZ =N HETS   0.38 1 1 D HETS= HETS D       30
  • 31. 31
  • 32. CONTINUED… • Now we have to calculate Diameter of our system for which we calculate area, which is   1 1 2 D = 12 inches Min HETS = 5.6 Volumetric Throughput = 1193 gal h.ft SF F Sm ρ +m ρ A = J 32
  • 33. CONTINUED… Volumetric Flow Rates of both Feed and Selective solvent. Feed = Selective Solvent Area Volumetric Throughput 3F F m 2185 33 = + = 27.19 ft hr ρ 82.41 48.7 3S S m 2106 = = = 33.7514 ft hr = 252.4942 gal hr ρ 62.43 SF F Sm ρ +m ρ = J 2203.418+262.4942 = = 0.3812 ft 1193 2 gal J= 1193 hr.ft 33
  • 34. CONTINUED… So for our system Diameter So this calculated Diameter is less than 30 inch and we have standard pipe size of diameter 10.42 inch, so we use 10.42 inch pipe. HETS 0.5 0.5 A 0.38067 = 4× = 4× = 0.6975ft π 3.14            0.38 0.38 1 1 D 10.42 =(HETS) =5.6 =5.307in D 12           34
  • 35. CONTINUED… For our design we increase value of HETS by 20% to avoid flooding. Rounding off ZE to nearest 3 On both top and bottom we have installed settlers which have the diameter 50% greater than the extractor diameter and also height of each settler is equal to settler diameter. E E Corrected HETS=5.307×1.20=6.368 in Z =N (HETS)=4(6.368)=25.47 EZ =27 35
  • 36. CONTINUED… Diameter of settler = Height of both settler = To join Extractor with settler we require reducers which are a foot long. Now reduces height = 1.5×10.42=15.63 inch 2×15.63=31.26 inch 2 12 24  E SZ=Z +Z +Reducers height =27+31.26+24=82.26inch Z=6.86ft 36
  • 39. EFFECT OF CONCENTRATION ON HEIGHT 7 8 8 10 12 5 6 7 8 9 10 11 12 13 0.4 0.5 0.6 0.7 0.8 0.9 1 Height(ft) Concentration (lb/ft3) 39
  • 40. EFFECT OF RECOVERY % ON HEIGHT 40 6.772 6.272 5.6885.6885.688 5.1885.1885.1885.1885.188 5 5.5 6 6.5 7 0 10 20 30 40 50 60 70 80 90 100 Height(ft) Recovery (%)
  • 41. EFFECT OF CONCENTRATION ON HEIGHT 41 8.4 8.5 8.8 9 9.2 9.4 9.7 9.9 10.1 10.3 10.5 10.7 10.9 11 11.2 11.4 11.6 11.8 12 12.12 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Diameter(in) Solvent Rate in Feed (lb/hr)
  • 42. 42
  • 43. 43

Editor's Notes

  1. Principle of heat and mass transfer by binay k dutta
  2. they can handle systems that tend to emulsify and feeds that contain particulates. Surface tension of methylene chloride is 26.50mN/m…. Water 72.80 mN/m methanol 22.70
  3. Height equivalent of theoretical stage.
  4. Interfacial Tension data Not available for our system so we switch to another method
  5. Volumetric throughput= Volumetric flow through a unit area as you are seeing from the units.
  6. For Feed we have firstly divided flow rate of solute with its density and then flow rate of solvent with its density and also done this for selective solvent
  7. Now, our diameter is less than 30 inches to be precise (8.37 inches) so we can select a standard size pipe rather than design one of exact measurements