m m
m
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
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.
½
½ieve 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).
m ½ m
Calculation Calculation Calculation Calculation Calculation Calculation Calculation Calculation Calculation
of Minimum Reflux Ratio Rm. of optimum reflux ratio. of theoretical number of stages. of actual number of stages. of diameter of the column. of weeping point. of pressure drop. of the height of the column. of thickness of the shell & Head.
m
3
d d !"#$ %& !#' ( ) * +
Y YY
0.0005
650
44.217
0.022
Y
0.8451 0.1362
16.6 2.43
1.13 0.165
0.954 0.0225
0.0182
0.0085
0.00578 K*Xf
1.05*10-5 0.9998
$ ,d, $!& ( -. ,,-
m
! Colburn method is used to fine out the minimum reflux ratio i.e. Rm = (1/ (Įlk-1)*[(XlkD/XlkR)² Įlk(XlkD/XlkR)] (1) Where Xlkn= [rF/ (1+rF)*(1+ĮXhF)] (2) Xhkn= (XlkR/rF) (3) Where ĮXhf for every component heavier than the heavy key In Our case the component heavier than the heavy key is residues
For that Į= (0.0011/2.78)=0.0004 ½o that ĮXhf = 0 rF=(XlkF/XhkF) =6.22 Įlk = (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)
u
! 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
è ½ *sing Fenske¶s equation Nm + 1 = Log [(Xl/Xh)d (Xh/Xl)s] / Log (Įlk)ave Nm + 1 = Log [(0.9978/0.0016)d (0.879/0.115)s] Log ((Į5.29)ave) Nm = 6
J " ½ Ideal number of ½tages can be found by Lewis Matheson Method. Average Temperature = 213.61 oC = 416.5 oF Relative Volatilities are
Y YY
144.34
Y
5.237 1
0.00948
J BELOW THE FEED PLATE: The ½OL 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,RE½ = 1.106*Xm+1-0.01564 And other equations are Yi = (ai*Xi)/(ai*Xi) For every component
J comp Xb
a*Xb
Yb
X1
a*X1
Y1
ODC B
0.005 0.026 0.028 0.026 0.136 0.124 18
TDI
0.879 0.879 0.97
Res
0.115
0.999 0.906 1
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
J ---
X5
---
0.803 4.207
0.958 0.844 4.421 0.969 0.8 5
---
0.182 0.182
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
---
a*X5
Y5
X6
a*X6
Y6
X7
The plate 7 has composition very close to the feed plate so it is considered as feed plate.
J 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
J Comp X7
a*X7
COCl2 0.000 5 ODCB 0.853 TDI Res
Y7
X8
a*X8
Y9
0.0000 0.000 0.000 15 45 14 0.9244 0.96 0.857
0.002 0 4.488
0.000 45 0.966
0.132
0.0755 0.036 0.142
0.142
0.030
0.013 7 0.992
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
J ---
Y14
---
0.0005 2.6*1 0.00 4 3 05 0.992 0.98 5.14 3 8
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
---
0.005
---
0
0.011 0.01 6 0 0
0.00 3 0
0.08 3 0
0.00 8 0
0.00 2 0
---
0.999
1
0.99
1
5.02
1.00
---
X15
a*X15 Y15
5.16
X16
a*X16 Y16
The Plate 16 has nearly same composition as that of the top product so it is the last plate from top to bottom.
x The efficiency of the column is given by the following empirical relation Eo = 51 - 32.5 Log (µa * Įa) Where ȝa = Average viscosity of the feed = 0.1156 Įa = Average relative volatility of light to heavy key = 5.29 Then, Eo = 65%
½ ' / 01 23( ,4. ! Actual number of stages = Ideal number of stages/Eo = 15/0.65 Actual number of stages = 23 ½ieve Trays are used.
ÿ # m 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
ÿ 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 *c = 0.952 m/s (By above equation) Let, flooding = 80% *c* = 0.8 * 0.952 = 0.762 m/s
ÿ Maximum volumetric flow rate of vapors : qv = Vw /dv = 3.31 m3/s Net area required: An = qv / *c*== 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 diameter at the top of column is 2.2 m.
ÿ 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
£ $
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 ³Coulson 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/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: *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 *h(min) based on active hole area is given as: Actual *h(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 *h(min) , so we change the hole area so that Actual *h (min) becomes well above *h(min) .
Another Trial For Hole Area: Aa = 3.74 m2 Ah=0.08*3.74=0.3m2
½o, Actual *h(min) = 7.72 m/sec ½ince Actual *h(min) is well above *h (min) so our new trial is correct
3 m Dry Plate Pressure Drop: Maximum vapor velocity through holes *h(max) = Vm / dv*An = 11.037 m/s (Ah / Aa) * 100 = (0.23/2.87)*100 =8.02 From figure 11.34,6th Ed. ³Coulson and Richardson¶s´ At (Ah/Aa)*100=8.02, When Plate thickness to plate dia ratio is 1. Then, Co = 0.83 hd = 51 (*h / 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
33
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 ½ince, Hb < 0.5*(tray spacing +weir height) 0.200<0.253 ½o, tray spacing is acceptable. Residence Time: tr = Ad * hbc * dl Lw tr = 0.59 * 0.20* 1202 = 8.26 sec 16.45 As residence time is greater than 3 sec, so satisfactory
3
*v = (*n / *c)* 100 Where *n = Vw/(dv*An) = 0.764 m/s *v = (0.764/0.952)*100 = 80.3 % (Our Assumption is correct.) Flv = 0.0675 From Graph 11.29, 6th Ed. ³Coulson and Richardson´ Fractional entrainment= ȥ = 0.052 As, entrainment is less than 0.1, process is satisfactory
3u ½ *se ½ectional 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 Ԧ = 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)*½in(76o/2) = 1.508m Area 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 = 0.087 From Graph 11.33, 6th Ed., Coulson and Richardson lp/dh = 3.2 (satisfactory i.e. b/w 2.5²4.0) Hole Pitch: lp/dh = 3.2, lp=16mm Triangular Number Of Holes per plate: Number Of Holes = Total hole area = 9307 Area of one hole
3è No. of plates = 23 Tray spacing = 0.457 m Distance between 23 plates = 10.5 m Top clearance = 0.5 m Bottom clearance = 0.5 m Tray thickness = 5 mm/plate Total thickness of trays = 0.005* 23= 0.115 m Total height of column = 10.5+ 0.5 + 0.5 + 0.115 Ht = 11.6m
3J ½HELL: Diameter of the tower =Dc = 2500 mm = 2.5 m Working/Operating Pressure = 1.01325 bar =101325 Pa Design pressure = 1.1×Operating Pressure = 1.1×101325 = 1.11*105Pa Working temperature = 525.22 ºK Design temperature = 1.1*525.22=577.7 ºK ½hell material = ½tainless steel,Type:317 Permissible tensile stress (ft) = 540 MN/m2 Elastic Modulus (E) = 210000 MN/mm2
Insulation material = Diatomaceous earth Maximum Working Temperature=650oF Insulation thickness = 2´= 50.8 mm Density of insulation = 288 kg/m3 HEAD - TORI½PHERICAL DI½HED HEAD: Material = ½tainless steel,Type:317 Allowable tensile stress = 540 MN/m2 ½*PPORT ½KIRT: Height of support = 5000 mm = 5 m Material ± Carbon ½teel
3x ½ Considering the vessel as an internal pressure vessel. ts = ((P×Rc)/ ((ft×J)- 0.6P)) + C Where ts = thickness of shell, mm P = design pressure, Pa Rc = diameter of shell, m ft = permissible/allowable tensile stress, MN/m2 C = Corrosion allowance,3 mm J = Joint factor ts = 3.26 mm Taking the thickness of the shell = 6 mm (standard)
3 m Type: Torispherical head: Thickness of head = th = (P×Rc×0.885)/ (ft×J-0.1P) P =internal design pressure, Pa Rc = radius of shell, m th = (111457.5×1.25×0.885)/ (540E+06×10.1*111457.5) = 3.23mm Including corrosion allowance take the thickness of head = 3 mm th = ((111457.5×1.25×0.885)/ (420*106×1.00)) + C = 3.23 mm Thickness of shell=8mm (½tandard)
½ ½
m %m u3 ½PECIFICATION ½HEET Identification: Item Distillation column Item No. D-310 No. required 1 Tray type ½ieve tray Function: ½eparation of ODCB from TDI and Reaction Residues. Operation: Continuous
Material handled:
Quantity Compositi on of ODCB Temp.
Feed 36453.6Kg/h r 84.51%
Top 30843.8Kg/ hr 99.78%
Bottom 5586Kg/hr 0.5%
194oC
160oC
252.22oC
Design data: No. of tray=27 Pressure = 101.32 KPa Height of column = 13.47m Diameter 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 mm Weir 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 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: ½team Turbine Exhaust Condensers/surface condensers 1. Plate Condensers 2. Air-Cooled Condensers 3. Direct Contact Condensers 4. ½hell & tube type
½ 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.
m ½ % u3
Heat Balance. Assumed Calculations. Calculations Of Heat Transfer Coefficients. Calculations Of Pressure Drops.
& T1=160C
t2=60C
T2=160C
t1=30C
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)
Assume Design overall coefficient =*D = 850 W/m2 oC Heat Transfer area: A=Q /(*D*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) = 222 ½hell side: From the nearest count on Table 9, ³Process Heat Transfer by Kern´ ID = 0.438m , No. Of Tubes=Nt=224 ½uppose Baffle spacing=B = 0.8m and Passes = 1
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*ȡw= 1.1 m/s From 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*Nt2/3= 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.oC Clean Overall Coefficient: *C = hio*ho = 1274.6 W/m2.oC hio+ho Dirt Factor: Rd = (*d-*c)/(*d**c) = 0.00039 (½atisfactory)
m m ½hell ½ide: De = 0.014 m(Table 10, Kern) Res = De*Gs = 197018.2 u f = 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 ½ide: water Ret = 26226.6 f = 0.037 ¨Pl = _f*Gt2*L*n_______ = 3.2KPa 5.22*1010*D*s*Ët ¨Pr= 4nV2 =4.84KPa s*2g¶ ¨Pt= ¨Pr+ ¨Pl=8.04KPa
½ ½
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Reboilers are heat exchangers provided at the bottom of the fractionator to generate the stripping vapors stream. Classification: Forced circulation reboiler. Kettle reboiler. Fired reboiler. Thermosiphon reboiler.
½
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 .
m ½ % 33u
Heat Balance. Assumed Calculations. Calculation of heat transfer coefficients. Calculation of Pressure Drops.
& Hot Fluid: Thermal fluid (Dimethyle ½iloxane) Cold Fluid: Bottoms (87.5%TDI) Vapor load:53641.04Kg/hr Heat duty:1.12*107KJ/hr Flow rate of Dimethyle ½iloxane=W= 247422.47Kg/hr LMTD:62.13oC
)
0
Let *D=596 W/m2oC A=Q/(*D*LMTD)=83.8m2 Tube specification:19mm OD, 24mm Triangular Pitch, 16BWG No.Of Tubes=Nt=(83.8/3*0.06)=465 Corrected *D=(Q/a¶1*Nt)=103 Btu/hr.ft2.oF From Table 10(Kern) By the Nearest Count Nt=506 Diameter Of tube Bundle=Db=(OD)*(n/C)1/2.21 Db=0.6m Where C=0.249 ½hell Dia.=1.08m (Ratio Of Bundle to ½hell Dia is 1.8)
!"# Flow Area=at=0.0002m2/tube at = (a¶t*Nt/n) = 0.0506 m2 W=Q/Cp*ǻT=247422.47Kg/hr Gt = W/at = 1358.27Kg/m2.hr V= (Gt/3600*ȡ)=1.82m/sec hi=1567.2 W/m2.C hio=hi*(ID/OD)=1295 W/m2.C
Assume ho=1703.5 W/m2.C tw=ta + (hio/ho+hio)*(Ta-ta) tw= 279.33C ǻtw = 27.11C From graph 11.5(Kern) ho>1703.5 W/m2.C so use ho=1703.5 W/m2.C *c=ho*(hio/(ho+hio)) = 736 W/m2.C Dirt Resistance=Rd=(*c-*d)/(*c**d) = 0.00032m2C/W (½atisfactory)
m Ret = (Gt*D/ȝ ) Ret= 50000 f=0.0026, s=0.672 ǻPt = (f*Gt*L*n/5.22*1010 * D*s) ǻPt = 9.16KPa ǻPr = (4/s) *( n) * (V2/g´) ǻPr = 13.5KPa ǻPT = ǻPt + ǻPr ǻPT = 22.6KPa Negligible
½ ½
% 33u Identification: Item name:Reboiler Item no.:E-113 Type: Kettle Reboiler No. Required = 1 Function: To Vaprize The bottom Product Of Distillation Column Heat Duty = 1.12*107KJ/hr
Fluid handled: Dimethyle ½iloxane Flow rate = 247422.47 Kg/hr Pressure = 202 Kpa Temp. = 326 oC to 304 oC Tubes: OD:19mm,16BWG 506 tubes each 3 m long 2 passes 24 mm triangular pitch pressure drop = 22.6 KPa Fluid handled :Bottoms Of Distillation Column Vapor Load= 53641.04Kg/hr ½hell: 1.08 m diameter 1 pass Pressure drop = Negligible Temperature= 252.22oC
* ThermalFluid (Dimethyle ½iloxane)
*D assumed = 596 W/m2.oC *c calculated = 736 W/m2oC Calculated dirt factor = Rd = 0.00032 m2oC/W Allowable dirt factor = Rd = 0.0006 m2oC/W
