Design Of Distillation Column

Distillation Column

A Distillation Column is used to separate a mutilcomponent liquid mixture into d istillates and bottoms due to differences in their boiling points. They are of f ollowing 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 floo ding. For large column heights, weight of the packed column is more than plate c olumn. Man holes will be provided for cleaning in tray Columns. In packed column s packing must be removed before cleaning. When large temperature changes are in volved 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 di ameters 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 Effi ciency . Lowest Cost per unit area than all other types with the downcomer. Good flexibility in operation(Turndown ratio).

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

Design Calculations:

1. Nature Of Feed Pressure of feed = P=101.325psi At, Temperature=190oC Light Key (lk) =ODCB Heavy Key (hk) =TDI Check at Temperature =194oC=381.2 oF

Comp COCl2 ODCB TDI Res. Xf 0.0005 0.8451 0.1362 0.0182 Pv(PSI) 650 16.6 2.43 0.0085 K 44.217 1.13 0.165 0.00578 K*Xf

K*Xf 0.022 0.954 0.0225 1.05*10-5 0.9998 As, Boiling Point (TB) is very close to feed temperature i.e. 190oC it is assume d 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/ ( lk-1 )*[(XlkD/XlkR) lk(XlkD/XlkR)] (1) Where Xlkn= [rF/ (1+rF)*(1+ XhF)] (2) Xhkn= (Xlk R/rF) (3) Where Xhf for every component heavier than the heavy key In Our case t he component heavier than the heavy key is residues

2 For that = (0.0011/2.78)=0.0004 So that Xhf = 0 rF=(XlkF/XhkF) =6.22 lk = (17.9/ 2.8) = 5.85 XlkD=0.9987 and XhkD=0.0016 Then XlkR = [6.22/ (1+6.22)] =0.861 XhkR = (0.862/6.24)=0.138 Rm = 0.22 (By 2) (By 3) (By 1)

3.Calculation Of Optimum Reflux Ratio Reflux 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/hr As, the feed is at its boiling point, q = 1 Lw = Ln+qF = 305.41Kgmole/hr Vw = Ln B = 276.6 Kgmole/Kgs

4.Minimum Number Of Stages Using Fenskes equation Nm + 1 = Log [(Xl/Xh)d (Xh/Xl)s] / Log ( lk)ave Nm + 1 = L og [(0.9978/0.0016)d (0.879/0.115)s] Log (( 5.29)ave) Nm = 6

5.Number Of Ideal Stages Ideal number of Stages can be found by Lewis Matheson Method. Average Temperatur e = 213.61 oC = 416.5 oF Relative Volatilities are Comp COCl2 ODCB TDI Res 144.34 5.237 1 0.00948

5 BELOW THE FEED PLATE: The SOL Equations are Y m = Lw*(Xm+1 / Vw) - W*(Xw / Vw) Y m,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 ODC B TDI Res

a*Xb Yb X1 a*X1 Y1

0.005 0.026 0.028 0.026 0.136 0.124 18 0.879 0.879 0.97 0.115 0.96 0.958 0.875 0.001 0.001 0.014 0.000 0.000 09 2 8 14 13 0.999 1.09 1 0.999 0.906 1

5 ----------X5 a*X5 Y5 X6 a*X6 Y6 X7 0.803 4.207 0.182 0.182 0.958 0.844 4.421 0.9 69 0.8 5 0.041 0.141 0.141 0.030 0.1 3 0.013 0.0003 0.000 0.013 0.000 0.000 0.0 2 1 0.999 4.389 1 0.999 1 1 0.9 9 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,COCl 2 = ( 0.248* X n+1) + 0.00045

5 Comp X7 COCl2 0.000 5 ODCB 0.853 TDI Res 0.132 0.013 7 0.992 a*X7 Y7 X8 a*X8 0.0 02 0 4.488 0.142 Y9 0.000 45 0.966 0.030 0.0000 0.000 0.000 15 45 14 0.9244 0.96 0.857 0.0755 0.036 0.142 0.0001 0.000 0.000 3 13 52 1 0.997 1 4.9*10 1.04*0 ^-6 ^-6 4.633 0.997

5 ----Y14 X15 a*X15 Y15 X16 a*X16 Y16 0.0005 2.6*1 0.00 4 3 05 0.992 0.98 5.14 3 8 0.005 0 0.999 0.011 0.01 6 0 0 1 5.16 0.00 3.1*1 0.00 0.00 45 58 05 0^ -5 0.99 0.99 5.19 0.99 4 1 3 7 0.00 3 0 0.99 0.08 3 0 1 0.00 8 0 5.02 0.00 2 0 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 = 5 1 - 32.5 Log (a * a) Where a = Average viscosity of the feed = 0.1156 a = 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 Ln =69.23Kgmole/hr Lw =10178.45 Kgs/hr Vn = 276.6 Kgmole/hr Vw =40 687.85 Kgs/hr M aver. = 147.01 Kg/Kgmol T = 160oC Liq density = dL = 1306 Kg/m3 Vap density = dV = 4 Kg/m3 Bottom Conditions 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 requir ed: 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= 2.5m The calculated di ameter 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 Richardsons, 6th Ed. lw / Dc = 0.7 7 lw = 1.92 m Weir Liquid Crest: Maximum liquid rate = Lw = 59226.05/3600 = 16.4 5Kgs/sec Minimum 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 mm From graph 11.30, page # 571,Coulson and Richardson Vol. 6 At hw + how =71.88 mm K2 =30.6 mm Weep point: U h(min) = [K2-0.9(25.4-dl)]/dv0.5 = [30.6-0.9(25.4-5)]/4.50.5 =5.76 m/s Actual 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 th an Uh(min) , so we change the hole area so that Actual Uh (min) becomes well abo ve 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 o ur new trial is correct

10.Plate Pressure Drop Dry Plate Pressure Drop: Maximum vapor velocity through holes Uh(max) = Vm / dv* An = 11.037 m/s (Ah / Aa) * 100 = (0.23/2.87)*100 =8.02 From figure 11.34,6th Ed . Coulson and Richardsons At (Ah/Aa)*100=8.02, When Plate thickness to plate dia ra tio is 1. Then, Co = 0.83 hd = 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. 04m Area 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