Distillation Distillation ColumnColumn

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DistillationDistillation:: “ “Process in which a liquid or vapour mixture Process in which a liquid or vapour mixture

of two or more substances is separated into of two or more substances is separated into its component fractions of desired purity, its component fractions of desired purity, by the application and removal of heat”by the application and removal of heat”

CHOICE BETWEEN PLATE AND CHOICE BETWEEN PLATE AND PACKED COLUMNPACKED COLUMN

The choice between use of tray column The choice between use of tray column or a packed column for a given mass or a packed column for a given mass transfer operation should, theoretically, be transfer operation should, theoretically, be based on a based on a detail cost analysisdetail cost analysis for the two for the two types of contactors. types of contactors. However,However, the decision the decision can be made on the basis of a can be made on the basis of a qualitative qualitative analysis of relative advantages and analysis of relative advantages and disadvantagesdisadvantages, eliminating the need for a , eliminating the need for a detailed cost comparison. detailed cost comparison.

Which are as followsWhich are as follows

Liquid dispersion difficultiesLiquid dispersion difficulties Capable of handling wide Capable of handling wide

rangesranges liquid ratesliquid rates CleaningCleaning.. Non-foaming systemsNon-foaming systems Periodic cleaningPeriodic cleaning weightweight of the columnof the column Design informationDesign information Inter stage coolingInter stage cooling Temperature changeTemperature change DiametersDiameters

As my system is As my system is non foamingnon foaming and and diameter diameter calculated is larger than calculated is larger than 0.67 m so I am going to use 0.67 m so I am going to use Tray Tray column.column.

Also as Also as average temperatureaverage temperature calculated for my distillation column calculated for my distillation column is higher that is approximately equal is higher that is approximately equal to to 9898oocc. So I prefer . So I prefer Tray columnTray column..

PLATE CONTACTORS:PLATE CONTACTORS:

Cross flow plateCross flow plate are the most commonly are the most commonly used plate contactor in distillation. In used plate contactor in distillation. In which liquid flows downward and vapours which liquid flows downward and vapours flow upward. The liquid move from plate flow upward. The liquid move from plate to plate via to plate via down comerdown comer. A certain level of . A certain level of liquid is maintained on the plates by liquid is maintained on the plates by weirweir. .

Three basic types of cross Three basic types of cross flow trays used areflow trays used are

Sieve Plate (Perforated Plate)Sieve Plate (Perforated Plate) Bubble Cap PlatesBubble Cap Plates Valve plates (floating cap Valve plates (floating cap

platesplates))

Selection of Selection of Trays:Trays:

I prefer I prefer Sieve PlateSieve Plate because: because: Pressure dropPressure drop is low as compared to is low as compared to

bubble cap trays bubble cap trays Their Their fundamentals fundamentals are well established, are well established,

entailing low risk.entailing low risk. The trays are The trays are low in costlow in cost relative to many relative to many

other types of trays.other types of trays. They can easily handleThey can easily handle wide variations in wide variations in

flowflow rates.rates. They are They are lighter in weightlighter in weight. It is easier and . It is easier and

cheaper to cheaper to install.install. Maintenance costMaintenance cost is reduced due to the is reduced due to the

ease of cleaning.ease of cleaning.

Sieve TraySieve Tray

Label Diagram Label Diagram (sieve (sieve

tray)tray)

Major BeamPlate Support

Ring

DowncomerAndWeir

CalmingZone

ManWay

FACTORS AFFECTING FACTORS AFFECTING DISTILLATION COLUMN DISTILLATION COLUMN OPERATIONOPERATION

Adverse vapour flow conditions can Adverse vapour flow conditions can cause:cause:

BlowingBlowing Coning Coning Dumping Dumping RainingRaining WeepingWeeping Flooding Flooding

FEED

REFLUX DRUM

(1) Methyl Iodide = 0.074(2) Acetic Acid = 0.65(3)Methyl Acetate = 0.215(4) Water = 0.065

(1) Methyl Iodide = 0.212(2) Acetic Acid = 0.0005(3)Methyl Acetate = 0.62(4) Water = 0.167

Condenser

Pump

Reboiler

FEED

(1) Methyl Iodide = 0.07(2) Acetic Acid = 0.65(3)MethylAcetate=0.22 (4) Water = 0.065

REFLUX DRUM

(1) Methyl Iodide = 0.21(2) Acetic Acid = 0.0005(3)Methyl Acetate = 0.62(4) Water = 0.17

(1)Acetic Acid = 0.99 (2)Water = 0.01

FLOW SHEET

From Material Balance:From Material Balance:

Heavy Key Component = Acetic AcidHeavy Key Component = Acetic Acid Light Key Component = WaterLight Key Component = Water

ComponentComponent FeedFeed

FractionFraction

xxff

BottomBottom

FractionFraction

xxbb

TopTop

FractionFraction

xxdd

(1) Methyl (1) Methyl

IodideIodide

0.070.07

00 0.210.21

(2) Acetic (2) Acetic

AcidAcid 0.650.65

0.990.99 0.00050.0005

(3) Methyl (3) Methyl

AcetateAcetate

0.220.22

00 0.620.62

(4) Water(4) Water 0.070.07 0.010.01 0.170.17

DESIGNING STEPS OF DESIGNING STEPS OF DISTILLATION COLUMNDISTILLATION COLUMN

Calculation of Minimum number of Calculation of Minimum number of stages.stages.NNminmin

Calculation of Calculation of Minimum Reflux Ratio RMinimum Reflux Ratio Rmm.. Calculation of Calculation of Actual Reflux RatioActual Reflux Ratio.. Calculation of Calculation of theoretical number of stagestheoretical number of stages.. Calculation of Calculation of actual number of stagesactual number of stages.. Calculation of Calculation of diameter diameter of the column.of the column. Calculation of Calculation of weeping point, entrainment.weeping point, entrainment. Calculation of Calculation of pressure droppressure drop.. Calculation of the Calculation of the height height of the column.of the column.

Calculation of Minimum no. of Calculation of Minimum no. of Plates:Plates:

The minimum no. of stages NThe minimum no. of stages Nminmin is is obtained obtained

from from Fenske equationFenske equation which is, which is,

NNminmin = LN[(x = LN[(xLKLK/x/xHKHK))DD(x(xHKHK /x /xLKLK))BB]] LN (αLN (αLKLK//HKHK) ) averageaverage

Average geometric relative volatility = Average geometric relative volatility =

1.531.53 So,So, NNminmin = 24 = 24

Calculation of Minimum Reflux Calculation of Minimum Reflux Ratio RRatio Rmm::

Using Underwood equationsUsing Underwood equations

As feed is entering as saturated vapors so,As feed is entering as saturated vapors so,

q = 0q = 0

By trial,By trial, = 1.68 = 1.68

Using equation of minimum reflux ratio,Using equation of minimum reflux ratio,

Putting all values we get,Putting all values we get,

RRmm = 4.154 = 4.154

1Rθα

α

θα

αm

B

DBB

A

DAA

xx

q1θα

α

θα

α

B

fBB

A

fAA

xx

Actual Reflux Actual Reflux Ratio:Ratio:

The rule of thumb is: The rule of thumb is:

R = (1.2 ------- 1.5) R R = (1.2 ------- 1.5) R minmin

R = 1.5 R R = 1.5 R minmin

R = 6.23R = 6.23

Theoretical no. of Theoretical no. of Plates:Plates: Gilliland related the number of Gilliland related the number of

equilibrium stages and the equilibrium stages and the minimum reflux ratio and the no. minimum reflux ratio and the no. of equilibrium stages with a plot of equilibrium stages with a plot that was transformed by Eduljee that was transformed by Eduljee into the relationinto the relation;;

From which the theoretical no. of From which the theoretical no. of stages to be,stages to be,

N= 39N= 39

566.0

minmin1175.01 R

RRN

NN

Calculation of actual number Calculation of actual number of stages:of stages:Overall Tray EfficiencyOverall Tray Efficiency::

avgavgoE .log5.3251

α avg =average relative volatility of light key component =1.75

μ avg = molar average liquid viscosity of feed evaluated at average temperature of column

Average temperature of column =(118+71)/2 Average temperature of column =(118+71)/2

= = 95 95 ooCC

Feed viscosity at average temperature = Feed viscosity at average temperature = avgavg

= = 0.39 mNs/m0.39 mNs/m22

So,So,

EEoo = 56.60% = 56.60%

So, So,

No. of actual traysNo. of actual trays = 39/0.566 = 39/0.566 = 68= 68

Location of feed Plate:Location of feed Plate: The Kirk bride method is used to determine the The Kirk bride method is used to determine the

ratio of trays above and below the feed point.ratio of trays above and below the feed point.

From which,From which,

Number of Plates above the feed tray = Number of Plates above the feed tray = ND = 47ND = 47

Number of Plates below the feed tray = Number of Plates below the feed tray = NB = 21NB = 21

2

log206.logDHK

BLK

LK

HK

B

Dx

xx

xD

BN

N

Determination of the Determination of the Column Diameter:Column Diameter:Flow Parameter:Flow Parameter:

FFLVLV = Liquid Vapor Factor = = Liquid Vapor Factor = 0.0560.056

0.5

L

v

n

nLV ρ

ρ

V

LF

Capacity Parameter:Capacity Parameter:

Assumed tray spacing = 18 inch (0.5 m)Assumed tray spacing = 18 inch (0.5 m) From Fig (15-5) Plant Design and Economics From Fig (15-5) Plant Design and Economics

for Chemical Engineering, sieve tray for Chemical Engineering, sieve tray flooding capacity,flooding capacity,

CCsbsb = = 0.0760 0.0760 m/Secm/Sec Surface tension of Mixture = Surface tension of Mixture = σσ == 18.35 18.35

dynes/Cmdynes/Cm

VVnfnf=1.67=1.67 m/sec m/sec Assume 90% of flooding thenAssume 90% of flooding then VVnn=0.9V=0.9Vnfnf So, actual vapor velocity, So, actual vapor velocity, VVnn==1.511.51 m/sec m/sec

5.02.0

20

v

vlCV sbnf

Net column area used in separation is Net column area used in separation is

AAnn = m = mvv/V/Vnn

Volumetric flow rate of vapors = Volumetric flow rate of vapors = mmvv

mmvv = (mass vapor flow rate /(3600) = (mass vapor flow rate /(3600) vapor density)vapor density)

mmvv = 2.1184m = 2.1184m33/sec/sec

Now, net area Now, net area AAnn = m= mvv/V/Vnn = 1.41m = 1.41m22

Assume that downcommer occupies Assume that downcommer occupies 15%15% of of cross sectional Area (Across sectional Area (Acc) of column thus:) of column thus:

AAcc = A = Ann + A + Add

WhereWhere, , AAdd = downcommer area = downcommer area

AAcc = A = Ann + 0.15(A + 0.15(Acc))

AAcc = A = Ann / 0.85 / 0.85

AAcc=1.65 m=1.65 m22

So Diameter of Column IsSo Diameter of Column Is

AAcc =(π/4)D =(π/4)D22

D = (4AD = (4Acc/π)/π) D = 1.45 meter = 5ft D = 1.45 meter = 5ft (based upon bottom conditions)(based upon bottom conditions)

Liquid flow Liquid flow arrangement:arrangement:In order to find liquid flow arrangement In order to find liquid flow arrangement

first find maximum liquid volumetric first find maximum liquid volumetric flow rateflow rate

So liquid flow rate =So liquid flow rate = ((Liquid mass rate)/ (3600) (Liquid Liquid mass rate)/ (3600) (Liquid

density)density)Max Liquid Rate Is At the bottom of Max Liquid Rate Is At the bottom of

column so using "Lcolumn so using "Lmm" values" valuesSo Maximum liquid flow rate = 0.005 So Maximum liquid flow rate = 0.005

mm33/sec/secSo from So from fig11.28fig11.28 Coulson & Richardson Coulson & Richardson

6th volume 3rd edition6th volume 3rd edition cross flow cross flow single pass platesingle pass plate is selected is selected

Provisional Plate Provisional Plate Design:Design:

Column Diameter DColumn Diameter Dcc= 1.4513 m= 1.4513 mColumn Cross-sectional Area(AColumn Cross-sectional Area(Acc)= 1.65 m)= 1.65 m22

Down comer area Down comer area AdAd = 0.15A= 0.15Acc = 0.25 m = 0.25 m22

Net Area (ANet Area (Ann) = A) = Acc - A - Add =1.41 m =1.41 m22

Active areaActive area A Aaa=A=Acc-2A-2Add = 1.16 m = 1.16 m22

Hole area Hole area AAhh take 10% take 10% AAaa = 0.1 × 1.16 = 0.1 × 1.16 =0.0462 m=0.0462 m22

Weir lengthWeir lengthAd / Ac = 0.248 / 1.654 = 0.15Ad / Ac = 0.248 / 1.654 = 0.15

From From figure 11.31figure 11.31 Coulson & Coulson & Richardson 6th volume 3rd Richardson 6th volume 3rd editionedition

LLww / dc = 0.80 / dc = 0.80 LLww = 1.452*0.80 = 1.452*0.80 == 0.733 m 0.733 m Weir length should be 60 to 85% Weir length should be 60 to 85%

of column diameter which is of column diameter which is satisfactorysatisfactory

Take weir height, hTake weir height, hww= = 50 mm50 mm Hole diameter, dHole diameter, dhh = = 5 mm5 mm Plate thickness = 5 mmPlate thickness = 5 mm

Check Weeping:Check Weeping:

where where UUminmin is the minimum design is the minimum design vapor velocity.vapor velocity.

The vapor velocity at weeping The vapor velocity at weeping point is the minimum velocity for point is the minimum velocity for the stable operation.the stable operation.

In order to have KIn order to have K22 value from value from fig11.30 fig11.30 Coulson & Richardson 6th volume Coulson & Richardson 6th volume 3rd edition3rd edition we have to 1st find we have to 1st find hhowow(depth of the crest of liquid over (depth of the crest of liquid over the weir)the weir)

where where hhowow is calculated by is calculated by following formula:following formula:

2/12

min

4.259.0

v

dKU h

hhowow=750[(L=750[(Lmm/l/lww*ρ)*ρ)2/32/3]]

Maximum liquid rate “LMaximum liquid rate “Lmm”= 4.7 kg/sec”= 4.7 kg/secMinimum Liquid Rate At Minimum Liquid Rate At 70% turn down ratio70% turn down ratio = 3.3Kg/sec= 3.3Kg/sec

At Maximum rate At Maximum rate ( h( howow)=)= 20 mm Liquid 20 mm LiquidAt Minimum rateAt Minimum rate ((hhowow)) = 16 mm Liquid = 16 mm Liquid hhww + h + howow = 50 + 16 = 66 mm Liquid = 50 + 16 = 66 mm Liquid

from from fig 11.30fig 11.30, Coulson and Richardson Vol.6, Coulson and Richardson Vol.6 KK22 = 30.50 = 30.50So, So,

U U (min)(min) = 9 m/sec = 9 m/sec

Now maximum volumetric flow rate Now maximum volumetric flow rate (vapors)(vapors)

Base = 2.12 mBase = 2.12 m33/sec/sec Top = 1.14 mTop = 1.14 m33/sec/sec

At At 70% turn down ratio70% turn down ratio

Actual minimum vapor velocity Actual minimum vapor velocity ==minimum vapor rate / minimum vapor rate /

AAhh

= = 12.81 m/sec12.81 m/sec So minimum vapor rate will be So minimum vapor rate will be

well above the weep point.well above the weep point.

Plate Pressure Drop Plate Pressure Drop (P.D):(P.D):

Consist of Consist of dry plate P.Ddry plate P.D (orifice loss), (orifice loss), P.D due to P.D due to static head of liquidstatic head of liquid and and residual P.Dresidual P.D (bubbles formation (bubbles formation result in energy loss)result in energy loss)

Dry Plate Drop:Dry Plate Drop: Max. Vapor velocity through holes Max. Vapor velocity through holes

(Uh) = (Uh) = Maximum Volumetric Flow Maximum Volumetric Flow Rate /Rate / Hole Area = 18.30 m/secHole Area = 18.30 m/sec

Perforated area Ap (active area) Perforated area Ap (active area) ==1.16 m1.16 m22

Ah/Ap = 0.100Ah/Ap = 0.100

From figFrom fig. 11.34 . 11.34 ((Coulson & Richardson Coulson & Richardson 6th volume 3rd edition) 6th volume 3rd edition) for for plate thickness/hole diameterplate thickness/hole diameter = 1.00 = 1.00

We get, We get, CCoo = 0.84 = 0.84

This equation is derived for orifice This equation is derived for orifice meter pressure drop.meter pressure drop.

hhdd= 48 mm Liquid= 48 mm Liquid

Residual Head (hResidual Head (hrr):):

hr = (12.5*10hr = (12.5*1033 / ρ / ρLL)) = = 13.3 mm Liquid13.3 mm Liquid

L

V

o

hd C

Uh

2ˆ51

So,So, Total pressure dropTotal pressure drop =48+(50+20)+13.32=48+(50+20)+13.32 hhtt = 131.35 mm liquid = 131.35 mm liquid

Total column pressure dropTotal column pressure drop Pa, (N/m Pa, (N/m22) ) = (9.81*10= (9.81*10-3-3) h) httρρLLNN = 82771.6 Pa = 82 kPa= 82771.6 Pa = 82 kPa

rowwdt hhhhh )(

Down comer Liquid Down comer Liquid Backup:Backup: Caused by Pressure Drop over the plate and Caused by Pressure Drop over the plate and

resistance to flow in the downcomer it self.resistance to flow in the downcomer it self.

hhbb = (h = (hww+ h+ howow) + h) + htt + h + hdcdc

The main resistance to flow in downcomer will be The main resistance to flow in downcomer will be caused by constriction in the downcomer outlet, and caused by constriction in the downcomer outlet, and head loss in the down comer can be estimated using head loss in the down comer can be estimated using the equation given as,the equation given as,

where Lwhere Lwdwd is the liquid flow rate in downcomer, is the liquid flow rate in downcomer, kg/seckg/sec and Aap is the clearance area under the downcomer, and Aap is the clearance area under the downcomer,

mm22

AAapap =h =hapapLLww

2

166

apL

wddc A

lh

Where Where hhapap the height of bottom edge the height of bottom edge of apron above the plate.of apron above the plate.

hhapap = h = hww – (5 to 10 mm) – (5 to 10 mm)

hhapap = 40 mm = 40 mmso, so, Area under apron “AArea under apron “Aapap” = 0.05 m” = 0.05 m22

As this is less than area of downcomer As this is less than area of downcomer AAdd so using A so using Aapap values in above values in above formula. formula.

So,So, hhdcdc = 1.95 mm = 1.95 mm

As a result, As a result, hhbb = 203.24 mm = 203.24 mm = 0.203 m= 0.203 m hhbb < ½ (Tray spacing + weir height) < ½ (Tray spacing + weir height)

0.20 < 0.250.20 < 0.25 So tray spacing is acceptableSo tray spacing is acceptable

Check Residence Check Residence Time:Time:

Sufficient residence time should be Sufficient residence time should be allowed in the downcomer for the allowed in the downcomer for the entrained vapors to disengage from entrained vapors to disengage from liquid stream to prevent aerated liquid liquid stream to prevent aerated liquid being carried under the downcomer.being carried under the downcomer.

ttrr =A =Add h hbcbc ρ ρLL/L/L(max)(max)

ttr r = 10 sec= 10 sec

It should be > 3 sec. so, result is It should be > 3 sec. so, result is satisfactorysatisfactory

Check Entrainment:Check Entrainment: (u(unn) actual velocity = (maximum volumetric ) actual velocity = (maximum volumetric

flow rate at base flow rate at base VVmm / net area / net area AAnn) )

((uunn) actual velocity = ) actual velocity = 1.51 1.51 m/sec m/sec

Velocity at flooding condition Velocity at flooding condition UUff = = 1.671.67 m/secm/sec

So Percent flooding =So Percent flooding =uunn/ u/ uff = 0.90 = = 0.90 = 90%90%

Liquid flow factor Liquid flow factor FFLVLV = 0.056 = 0.056 From From fig. 11.29fig. 11.29 Coulson & Coulson &

Richardson 6th volume 3rd editionRichardson 6th volume 3rd edition fractional entrainment fractional entrainment ψψ can be found can be found

out.out. Fractional entrainment (ψ) = Fractional entrainment (ψ) =

0.07500.0750

Well below the upper limit of (Well below the upper limit of (ψψ) which ) which is is 0.10.1. Below this the . Below this the effect of effect of entrainment on efficiency is small.entrainment on efficiency is small.

No of Holes:No of Holes:

Area of 1 Hole = (π/4) DArea of 1 Hole = (π/4) Dholehole22

= 0.00002 m= 0.00002 m22

Area of Area of NN Holes = 0.1158 m Holes = 0.1158 m22

So,So,

Number OF Holes = 5900Number OF Holes = 5900

Height of Distillation ColumnHeight of Distillation Column

Height of column HHeight of column Hcc= (N= (Nactact-1) H-1) Hss+ ∆H+ plates + ∆H+ plates

thickness thickness No. of plates = 68No. of plates = 68Tray spacingTray spacing HHss = 0.50 m = 0.50 m ∆∆H= 0.5 meter each for liquid hold up and H= 0.5 meter each for liquid hold up and vapor disengagementvapor disengagement ∆∆H=1 mH=1 mTotal thickness of trays = 0.005*68 = Total thickness of trays = 0.005*68 = 0.34 m0.34 mSo,So, Height of columnHeight of column = (68-1)*0.50+ 1+0.34 = (68-1)*0.50+ 1+0.34 = = 35 meters35 meters

1.45m height=35m

Hole diameter=5mm

No. of holes=5900

hap=40 mm

h W=50 mm

how=Weir crustPlate Specifications

Specification Sheet Of Distillation Column:

Identification:Item Distillation column

No. required 1 Tray type Sieve trayFunction: Separation of Acetic Acid from iodo methane and Reaction by products. Operation: Continuous

FeedFeed TopTop BottomBottom

AmountAmount 4755 Kg/hr4755 Kg/hr 1968 Kg/hr1968 Kg/hr 2786 Kg/hr2786 Kg/hr

CompositioComposition n

of of

Acetic AcidAcetic Acid

0.640.64 0.0050.005 0.990.99

Temp.Temp. 119119ooCC 7171ooCC 118118ooC C

Material handled:

Design data:

No. of tray= 68No. of tray= 68 Pressure = 101.325 KpaPressure = 101.325 Kpa Height of column = 35 Height of column = 35 mm Diameter of Diameter of column=1.45m column=1.45m Hole size = 5 mm Hole size = 5 mm Pressure drop per Pressure drop per

tray=1.2 Kpa tray=1.2 Kpa Tray thickness = 5 mmTray thickness = 5 mm

Active holes = 5900 Active holes = 5900 Weir height = 50 mmWeir height = 50 mm Weir length = 1 m Weir length = 1 m Reflux ratio = 6.23 Reflux ratio = 6.23 Tray spacing =0.5 m Tray spacing =0.5 m Active area = 1.16 mActive area = 1.16 m22 Percent Flooding =90% Percent Flooding =90% Entrainment = 0.075Entrainment = 0.075

ReferencesReferences Coulson & Richardson 6th volume Coulson & Richardson 6th volume

3rd edition3rd edition Plant Design and Economics for Plant Design and Economics for

Chemical EngineeringChemical Engineering Coulson & Richardson 2th volume Coulson & Richardson 2th volume

5th edition5th edition Perry’s Chemical engineer’s hand Perry’s Chemical engineer’s hand

bookbook

The EndThe End