Design Of Distillation Column

Distillation Column A Distillation Column is used to separate a

mutilcomponent liquid mixture into distillates and bottoms due to differences in their boiling points.

They are of following two types based upon construction.

Tray Column Packed Column

Choice b/w Tray & Packed Column

Plate column are designed to handle wide range of liquid flow rates without flooding.

For large column heights, weight of the packed column is more than plate column.

Man holes will be provided for cleaning in tray Columns. In packed columns packing must be removed before cleaning.

When large temperature changes are involved as in the distillation operations tray column are often preferred.

Random-Packed Column generally not designed with the diameter larger than 1.5 m and diameters of commercial tray column is seldom less than 0.67m.

Selection Of Tray Type Sieve trays are selected due to following main

reasons. High capacity. High Efficiency . Lowest Cost per unit area than all other types with

the downcomer. Good flexibility in operation(Turndown ratio).

Designing Steps Of Distillation Column Calculation of Minimum Reflux Ratio Rm. Calculation of optimum reflux ratio. Calculation of theoretical number of stages. Calculation of actual number of stages. Calculation of diameter of the column. Calculation of weeping point. Calculation of pressure drop. Calculation of the height of the column. Calculation of thickness of the shell & Head.

Design Calculations:

1. Nature Of Feed

Pressure of feed = P=101.325psiAt, Temperature=190oCLight Key (lk) =ODCBHeavy Key (hk) =TDICheck at Temperature =194oC=381.2

oF

Comp Xf Pv(PSI) K K*Xf

COCl2 0.0005 650 44.217 0.022

ODCB 0.8451 16.6 1.13 0.954

TDI 0.1362 2.43 0.165 0.0225

Res. 0.0182 0.0085 0.00578 1.05*10-5

∑K*Xf 0.9998

As, Boiling Point (TB) is very close to feed temperature i.e. 190oC it is assumed that feed is saturated liquid at its boiling point so that q = 1

2.Determination Of Minimum Reflux Ratio

Colburn method is used to fine out the minimum reflux ratio i.e.

Rm = (1/ (αlk-1)*[(XlkD/XlkR)— αlk(XlkD/XlkR)] (1)

Where Xlkn= [rF/ (1+rF)*(1+∑αXhF)] (2)

Xhkn= (XlkR/rF) (3)

Where ∑αXhf for every component heavier than

the heavy key In Our case the component heavier than the

heavy key is residues

2…

For that α= (0.0011/2.78)=0.0004 So that ∑αXhf = 0 rF=(XlkF/XhkF) =6.22 αlk = (17.9/2.8) = 5.85XlkD=0.9987 and XhkD=0.0016Then XlkR = [6.22/ (1+6.22)] =0.861 (By 2) XhkR= (0.862/6.24)=0.138 (By 3)

Rm = 0.22 (By 1)

3.Calculation Of Optimum Reflux RatioReflux Ratio = R=1.5*Rmin (1.2----1.5) R = 1.5 * 0.22 = 0.33 Ln = R * D Ln = 69.23 Kgmole/hr Vn = Ln+D = 276.6 Kgmole/hrAs, the feed is at its boiling point, q = 1 Lw = Ln+qF = 305.41Kgmole/hrVw = Ln – B = 276.6 Kgmole/Kgs

4.Minimum Number Of Stages

Using Fenske’s equation

Nm + 1 = Log [(Xl/Xh)d (Xh/Xl)s] / Log (αlk)ave

Nm + 1 = Log [(0.9978/0.0016)d (0.879/0.115)s]

Log ((α5.29)ave)

Nm = 6

5.Number Of Ideal Stages Ideal number of Stages can be found by Lewis Matheson Method. Average Temperature = 213.61 oC = 416.5 oF Relative Volatilities are

Comp α

COCl2 144.34

ODCB 5.237

TDI 1

Res 0.00948

5…

BELOW THE FEED PLATE:

The SOL Equations are

Y m = Lw*(Xm+1 / Vw) - W*(Xw / Vw)

Ym,ODCB = 1.106Xm+1-0.00068

Ym,TDI = 1.106*Xm+1-0.119

Ym,RES = 1.106*Xm+1-0.01564

And other equations are

Yi = (ai*Xi)/∑(ai*Xi)

For every component

5….

comp Xb a*Xb Yb X1 a*X1 Y1

ODCB

0.005 0.02618

0.028 0.026 0.136 0.124

TDI 0.879 0.879 0.97 0.96 0.958 0.875

Res 0.115 0.00109

0.0012

0.0148

0.00014

0.00013

∑ 0.999 0.906 1 0.999 1.09 1

5…--- X5 a*X5 Y5 X6 a*X6 Y6 X7

--- 0.803 4.207 0.958 0.844 4.421 0.969 0.85

--- 0.182 0.182 0.041 0.141 0.141 0.030 0.13

--- 0.013 0.0003 0.0002

0.013 0.000 0.000 0.01

--- 0.999 4.389 1 0.999 1 1 0.99

The plate 7 has composition very close to the feed plate so it is considered as feed plate.

5…

ABOVE THE FEED PLATE:

The ROL equations are

Y n+1 = Ln*(Xn+1 / Vn) + D *(xD / Vn)

Y n,ODCB = (0.248 * X n+1) + 0.748

Y n,TDI = (0.248 * X n+1) + 0.0012

Y n,COCl2 = ( 0.248* X n+1) + 0.00045

5…

Comp X7 a*X7 Y7 X8 a*X8 Y9

COCl2 0.0005

0.000015

0.00045

0.00014

0.0020

0.00045

ODCB 0.853 0.9244 0.96 0.857 4.488 0.966

TDI 0.132 0.0755 0.036 0.142 0.142 0.030

Res 0.0137

0.00013

0.00013

0.00052

4.9*10^-6

1.04*0^-6

∑ 0.992 1 0.997 1 4.633 0.997

5…--- Y14 X15 a*X15 Y15 X16 a*X16 Y16

--- 0.00053

2.6*10-5

0.004

0.0005

3.1*10^-5

0.0045

0.0058

--- 0.992 0.983

5.148

0.994

0.991

5.193

0.997

--- 0.005 0.011

0.016

0.003

0.083

0.008

0.002

--- 0 0 0 0 0 0 0

--- 0.999 1 5.16 0.99 1 5.02 1.00

The Plate 16 has nearly same composition as that of the top product so it is the last plate from top to bottom.

6.Efficiency Of The Column

The efficiency of the column is given by the following empirical relation

Eo = 51 - 32.5 Log (µa * αa)

Where

μa = Average viscosity of the feed = 0.1156

αa = Average relative volatility of light to heavy

key = 5.29

Then,

Eo = 65%

7.Actual Number Of Stages Total Ideal Stages=16-1=15(Excluding Reboiler)Actual number of stages = Ideal number of stages/Eo

= 15/0.65

Actual number of stages = 23

Sieve Trays are used.

8-Provisional Plate Design:Top Condition Bottom Conditions

Ln =69.23Kgmole/hr

Lw =10178.45 Kgs/hr

Vn = 276.6 Kgmole/hr

Vw =40687.85 Kgs/hr

M aver. = 147.01 Kg/Kgmol

T = 160oC

Liq density = dL = 1306 Kg/m3

Vap density = dV = 4 Kg/m3

Lm = 305.41 Kgmol/hr

Lw = 59226.02 Kg/hr

Vm = 276.6 Kgmol/hr

Vw = 53641.04Kg/hr

M aver=193.923Kg/Kgmol

T = 252.22oC

Liquid density = dL = 1202 Kg/m3

Vapor density = dV = 4.5 Kg/cm3

8… Flooding Velocity:

Flv=(Lw/Vw)(dv/dl)^0.5

Flv = 0.0675

From figure11.27, Coulson and Richardson, 6th Ed.

At 18 inch spacing or 0.457 m

K1 = 0.08

Uc = 0.952 m/s (By above equation)

Let, flooding = 80%

Uc* = 0.8 * 0.952

= 0.762 m/s

8… Maximum volumetric flow rate of vapors :qv = Vw /dv = 3.31 m3/s Net area required: An = qv / Uc*== 4.33 m2

Column Cross sectional Area:Column area = Ac = An / 0.88 = 4.92 m2

Diameter:Diameter =Dc = (4*Ac/3.14) 0.5= 2.5mThe calculated diameter at the top of column is 2.2 m.

8… Downcomer Area:Ad = 0.12*Ac = 0.59 m2 Net Area:An = Ac – Ad= 4.33 m2

Active Area:Aa = Ac-2Ad = 3.74 m2

Hole Area:Ah = 0.11*0.579= 0.41 m2(by trial)Lets take, Weir height = hw = 50mm Plate thickness = 5mm Hole diameter = dh = 5mm

9. Weep Point Weir Length: Factor (Ad/Ac)*100 = 12 At (Ad / Ac) * 100 = 12 From Graph b/w (Ad/Ac)*100 vs. lw / Dc on page # 572

by “Coulson and Richardson’s”, 6th Ed. lw / Dc = 0.77 lw = 1.92 m Weir Liquid Crest:Maximum liquid rate = Lw = 59226.05/3600 = 16.45Kgs/secMinimum liquid rate)= Lw*=16.45*0.7(at 70% turn down) =11.5kgs/sec how =750*(Lw/dl*lw)2/3 max how =27.778mm

min how = 21.88

At minimum liquid rate, hw + how = 50 + 21.88 =71.88 mmFrom graph 11.30, page # 571,”Coulson and Richardson”

Vol. 6 At hw + how =71.88 mm K2 =30.6 mm Weep point:Uh(min) = [K2-0.9(25.4-dl)]/dv

0.5

= [30.6-0.9(25.4-5)]/4.50.5

=5.76 m/sActual Uh(min) based on active hole area is given as:Actual Uh(min) = 0.7*(Vw/dv)*A h = (0.7*53641.04)/(4.5*3600*0.41) = 5.65 m/s As, actual minimum velocity is less than Uh(min) , so we

change the hole area so that Actual Uh (min) becomes well above Uh(min) .

Another Trial For Hole Area:Aa = 3.74 m2

Ah=0.08*3.74=0.3m2

So, Actual Uh(min) = 7.72 m/sec

Since Actual Uh(min) is well above Uh (min) so our

new trial is correct

10.Plate Pressure Drop Dry Plate Pressure Drop:Maximum vapor velocity through holesUh(max) = Vm / dv*An

= 11.037 m/s(Ah / Aa) * 100 = (0.23/2.87)*100 =8.02From figure 11.34,6th Ed. “Coulson and Richardson’s”At (Ah/Aa)*100=8.02, When Plate thickness to plate diaratio is 1.Then, Co = 0.83hd = 51 (Uh / Co)2 (dv/dl) = 33.76 mm liquid

Residual Drop:hr = 12.5*1000/dl

= 10.4 mm liquid Total Plate Pressure Drop: ht= hd + hr + (hw +how) = 33.76 + 10.4+ 71.88 = 116.04 mm liquid ∆Pt = 9.81*10-3*(ht)*dl

= 9.81*10-3*116.04*1202 =1368.3Pa = 1.36 KPa = 0.1973 psi

11. Residence Time Downcomer Liquid backup/ Liquid height in

downcomer:Let, hap= hw-10 =40 mm = 0.04mArea under apron = hap*lw = 0.04*1.92 = 0.0768m2

As Aap is less than Ad = 0.59m2

hdc=166*(Lw/dl*Aap)2

=166*(59226.02/1202*3600*0.0768)2

=5.27mm

Hb = ht + hdc + (hw + how) = 116.04 + 5.27+ (71.88) =193.2 mm=0.2m Since, Hb < 0.5*(tray spacing +weir height) 0.200<0.253So, tray spacing is acceptable. Residence Time: tr = Ad * hbc * dl

Lw

tr = 0.59 * 0.20* 1202 = 8.26 sec 16.45As residence time is greater than 3 sec, so satisfactory

12. Entrainment Check

Uv = (Un / Uc)* 100WhereUn = Vw/(dv*An) = 0.764 m/s Uv = (0.764/0.952)*100 = 80.3 % (Our Assumption is correct.) Flv = 0.0675 From Graph 11.29, 6th Ed. “Coulson and Richardson” Fractional entrainment= ψ = 0.052As, entrainment is less than 0.1, process is satisfactory

13.Plate Specification Use Sectional Construction. The Plates are

supported on a ring welded around the vessel wall, and on the beams about 50mm wide. Allow 50mm wide claming zones.

lw/Dc = 1.92/3.3 = 0.77 Ө = 104o

Angle subtended at plate edge by unperforated strip = 180o – 104o

= 76o

Length of unperforated edge strips : (2.5 – 50*10-3) *76 = 3.25 m 180 Area of unperforated edge strip =Au = 50*10-3*3.25 = 0.162 m2

Mean length of Claming Zone: (2.5-50*10-3)*Sin(76o/2) = 1.508mArea of calming zone = Acz = 2*50*10-3*1.508 = 0.15m2

Total area available for perforations: Ap = Aa – (Au + Acz) = 3.42 m2

Ah/Ap = 0.3/3.42 = 0.087From Graph 11.33, 6th Ed., Coulson and Richardson

lp/dh = 3.2 (satisfactory i.e. b/w 2.5—4.0)Hole Pitch: lp/dh = 3.2, lp=16mm TriangularNumber Of Holes per plate:Number Of Holes = Total hole area = 9307 Area of one hole

14.Height Of The Column

No. of plates = 23Tray spacing = 0.457 m Distance between 23 plates = 10.5 mTop clearance = 0.5 mBottom clearance = 0.5 mTray thickness = 5 mm/plate Total thickness of trays = 0.005* 23= 0.115 mTotal height of column = 10.5+ 0.5 + 0.5 + 0.115

Ht = 11.6m

15.Material Of Construction SHELL:Diameter of the tower =Dc = 2500 mm = 2.5 mWorking/Operating Pressure = 1.01325 bar =101325 PaDesign pressure = 1.1×Operating Pressure = 1.1×101325 = 1.11*105PaWorking temperature = 525.22 ºKDesign temperature = 1.1*525.22=577.7 ºKShell material = Stainless steel,Type:317Permissible tensile stress (ft) = 540 MN/m2 Elastic Modulus (E) = 210000 MN/mm2

Insulation material = Diatomaceous earthMaximum Working Temperature=650oFInsulation thickness = 2”= 50.8 mmDensity of insulation = 288 kg/m3

HEAD - TORISPHERICAL DISHED HEAD:Material = Stainless steel,Type:317Allowable tensile stress = 540 MN/m2

SUPPORT SKIRT:Height of support = 5000 mm = 5 mMaterial – Carbon Steel

16.Minimum Shell ThicknessConsidering the vessel as an internal pressure vessel.ts = ((P×Rc)/ ((ft×J)- 0.6P)) + CWhere ts = thickness of shell, mmP = design pressure, PaRc = diameter of shell, mft = permissible/allowable tensile stress, MN/m2C = Corrosion allowance,3 mmJ = Joint factorts = 3.26 mm Taking the thickness of the shell = 6 mm (standard)

17.Head Design

Type: Torispherical head:Thickness of head = th = (P×Rc×0.885)/ (ft×J-0.1P)P =internal design pressure, PaRc = radius of shell, mth = (111457.5×1.25×0.885)/ (540E+06×1-0.1*111457.5) = 3.23mmIncluding corrosion allowance take the thickness of

head = 3 mmth = ((111457.5×1.25×0.885)/ (420*106×1.00)) + C = 3.23 mmThickness of shell=8mm (Standard)

Specification Sheet Of Distillation Column(D-310)

SPECIFICATION SHEETIdentification:

Item Distillation columnItem No. D-310No. required 1

Tray type Sieve trayFunction: Separation of ODCB from TDI and Reaction Residues. Operation: Continuous

Material handled:

Feed Top Bottom

Quantity 36453.6Kg/hr

30843.8Kg/hr

5586Kg/hr

Composition of ODCB

84.51% 99.78% 0.5%

Temp. 194oC 160oC 252.22oC

Design data:

No. of tray=27Pressure = 101.32 KPaHeight of column = 13.47mDiameter of column =2.5m Hole size = 5 mm Pressure drop per tray = 1.36 KPa Tray thickness = 5 mm

Active holes = 9307 Weir height = 50 mmWeir length = 1.92 m Reflux ratio = 0.33:1 Tray spacing =0.457m Active area = 3.74 m2 Flooding = 80.3 % Entrainment=5.2%

Condenser A condenser is a two-phase flow heat exchanger in

which heat is generated from the conversion of vapor into liquid (condensation) and the heat generated is removed from the system by a coolant.

Types of Condensers:Steam Turbine Exhaust Condensers/surface

condensers1. Plate Condensers2. Air-Cooled Condensers3. Direct Contact Condensers4. Shell & tube type

Selection Of Condenser Configuration

Four Condenser Configuration are Possible Horizontal with condensation in shell side and cooling

medium in the tubes. Horizontal with condensation in tube side cooling

medium in shell side. Vertical with condensation in the shell. Vertical with condensation in the tubes. Horizontal shell side and vertical tube side are the

most commonly used types of condensers. In this process we have used the horizontal with condensation in shell side & cooling medium in tube.

Designing Steps Of Condenser (E-312)

Heat Balance. Assumed Calculations. Calculations Of Heat Transfer Coefficients. Calculations Of Pressure Drops.

Heat Balance

Vapor: Qvap = mHv

Qvap = 11202300KJ/hr Water: Q = wCp( t2- t1) w = 88907.14 Kg/hr LMTD: LMTD = (T2 - t1) - (T1- t2) = 114.34oC Ln (T2 - t1) (T1- t2)

T2=160C

t1=30Ct2=60C

T1=160C

Assumed Calculations: Assume Design overall coefficient =UD = 850 W/m2 oC

Heat Transfer area: A=Q /(UD*LMTD)=32.017m2

Tube Lay out & size: Length = 2.4m , Passes = n=2

OD, BWG, pitch(Pt) = 19mm, 16 BWG, 24mm Triangular pitch.

Out side surface area per linear ft =a"t = 0.06m2

No. of tubes = Nt = A/ (a”t.L) = 222Shell side: From the nearest count on Table 9, “Process

Heat Transfer by Kern” ID = 0.438m , No. Of Tubes=Nt=224

Suppose Baffle spacing=B = 0.8m and Passes = 1

Calculation Of Heat Transfer Coefficients Cold Fluid: tube side (water) Flow area: at = Nt*a"t = 0.0224m2 144*n Mass velocity: Gt = w/at= 3969068.75Kg/hr.m2

Velocity : V = Gt/3600*ρw= 1.1 m/sFrom Graph 25 On Kern hi = 5678.3W/m2.oC

hio = hi*ID OD

hio = 4692 W/m2.oC

Hot fluid: shell side (Vapors) Flow area:as = ID*C*B C=Pt-do =24mm 144*Pt as = 0.073m2

Mass velocity:Gs =W/as =557367.94Kg/hr.m2 Loading: G"=W/L*Nt

2/3= 394.2 Kg/hr. m Assume ho = 2000W/m2.oC tw = ta + ho (Tv - ta) =85C (hio + ho) tf = tw + Tv = 122.5C 2 ho = 1750 W/m2.oCClean Overall Coefficient:UC = hio*ho = 1274.6 W/m2.oC hio+ho Dirt Factor: Rd = (Ud-Uc)/(Ud*Uc) = 0.00039 (Satisfactory)

Determination Of Pressure Drop Shell Side: De = 0.014 m(Table 10, Kern) Res = De*Gs = 197018.2 uf = 0.1728 (From Graph 26 Kern) s =0.004 No. Of Crosses: N + 1= (L/B) =3 ∆Ps = __f*Gs2*Ds*(N + 1) =24.3KPa 2*5.22*1010*De*s

Tube Side: water Ret = 26226.6 f = 0.037 ∆Pl = _f*Gt

2*L*n_______ = 3.2KPa 5.22*1010*D*s*t

∆Pr= 4nV2 =4.84KPa s*2g’ ∆Pt= ∆Pr+ ∆Pl=8.04KPa

Specification Sheet Of Condenser (E-312)

Identification: condenserNo. Required = 1 Function: Condense vapor mixture of (COCL2,ODCB,TDI) by removing the latent heat of vaporization Type: 1-2 Horizontal Condenser Shell side condensationHeat Duty =1.12×107 KJ/hr

Tube Side:Fluid handled: cold waterFlow rate =88907 Kg/hrTemp. = 30 oC to 60 oC

Tubes: OD:19mm,16BWG224 tubes each 2.4 m long 2 passes24 mm triangular pitchpressure drop = 8.04 KPa

Shell Side:Fluid handled :ODCB (99.78%)Flow rate= 40687.86 Kg/hr

Shell: 0.438 m diameter 1 passBaffles spacing 0.8 mPressure drop = 24.3KPaTemperature= 160oC

Reboilers

Reboilers are heat exchangers provided at the bottom of the fractionator to generate the stripping vapors stream.

Classification: 1. Forced circulation reboiler.2. Kettle reboiler.3. Fired reboiler.4. Thermosiphon reboiler.

Selection of reboiler: The best choice is the kettle reboiler due to

following main reasons. High residence time. Rate of vaporizations is very high (about

90%) of the feed. The viscosity of the system is not very high Comparatively less costly under above

conditions .

Designing Steps Of Reboiler(E-113) Heat Balance. Assumed Calculations. Calculation of heat transfer coefficients. Calculation of Pressure Drops.

Heat Balance Hot Fluid: Thermal fluid (Dimethyle Siloxane) Cold Fluid: Bottoms (87.5%TDI) Vapor load:53641.04Kg/hr Heat duty:1.12*107KJ/hr Flow rate of Dimethyle Siloxane=W=

247422.47Kg/hr LMTD:62.13oC

t2=252.2CT1=326C

t1=252.2C

T2=304C

Assumed Calculations Let UD=596 W/m2oCA=Q/(UD*LMTD)=83.8m2

Tube specification:19mm OD, 24mm Triangular Pitch, 16BWG

No.Of Tubes=Nt=(83.8/3*0.06)=465Corrected UD=(Q/a’1*Nt)=103 Btu/hr.ft2.oFFrom Table 10(Kern) By the Nearest CountNt=506 Diameter Of tube Bundle=Db=(OD)*(n/C)1/2.21

Db=0.6m Where C=0.249 Shell Dia.=1.08m (Ratio Of Bundle to Shell Dia is 1.8)

Calculation Of Heat Transfer Coefficients Tube Side:Dimethyle Siloxane Flow Area=at´=0.0002m2/tubeat = (a’t*Nt/n) = 0.0506 m2

W=Q/Cp*ΔT=247422.47Kg/hr Gt = W/at = 1358.27Kg/m2.hrV= (Gt/3600*ρ)=1.82m/sechi=1567.2 W/m2.Chio=hi*(ID/OD)=1295 W/m2.C

Shell Side:BottomsAssume ho=1703.5 W/m2.C

tw=ta + (hio/ho+hio)*(Ta-ta)

tw= 279.33C

Δtw = 27.11C From graph 11.5(Kern) ho>1703.5 W/m2.Cso use ho=1703.5 W/m2.C

Uc=ho*(hio/(ho+hio)) = 736 W/m2.C

Dirt Resistance=Rd=(Uc-Ud)/(Uc*Ud)

= 0.00032m2C/W (Satisfactory)

Calculation Of Pressure Drops Tube Side: Ret = (Gt*D/μ )

Ret= 50000f=0.0026, s=0.672 ΔPt = (f*Gt*L*n/5.22*1010 * D*s)

ΔPt = 9.16KPa ΔPr = (4/s) *( n) * (V2/g´)

ΔPr = 13.5KPa ΔPT = ΔPt + ΔPr

ΔPT = 22.6KPa Shell Side: Negligible

Specification Sheet Of Reboiler (E-113) Identification: Item name:ReboilerItem no.:E-113Type: Kettle Reboiler No. Required = 1 Function: To Vaprize The bottom Product Of Distillation Column Heat Duty = 1.12*107KJ/hr

Tube Side:Fluid handled: Dimethyle SiloxaneFlow rate = 247422.47 Kg/hrPressure = 202 KpaTemp. = 326 oC to 304 oC Tubes: OD:19mm,16BWG506 tubes each 3 m long2 passes24 mm triangular pitchpressure drop = 22.6 KPa Shell Side:Fluid handled :Bottoms Of Distillation ColumnVapor Load= 53641.04Kg/hrShell: 1.08 m diameter 1 passPressure drop = NegligibleTemperature= 252.22oC

Utilities: ThermalFluid (Dimethyle Siloxane)

UD assumed = 596 W/m2.oC Uc calculated = 736 W/m2oC Calculated dirt factor = Rd = 0.00032 m2oC/W Allowable dirt factor = Rd = 0.0006 m2oC/W