CALCULATIONS: D =91 666.667 kg hr-1 Lo F= 155 000 kg hr-1 Vd Ln 60% Methanol Vd Lm 99.9% Methanol 40% Water W=63 333.333kg hr-1 2.25% Methanol DIAGRAM OF MASS BALANCE OF DISTILLATION COLUMN The feed flow rate F is estimated to be 155 000 kg hr -1 Overall Balance: F = D+W 155 000=91 666.667 + W W= 63 333.333 kg hr-1 FXf=DDw + WWf 155 000(0.6) = 91 666.667(0.999) + 63 333.333(Wf) Wf=0.0225 Converting the weight percentage fractions to mole fractions: Xf = XD= Xw= 0.6/32____ (0.6/32) + (0.4/18) =0.458 0.999/32_____ (0.999/32) + (0.01/18) =0.983 0.0225/32______ (0.0225/32) + (0.9775/18) =0.013 1 TABLE SHOWING EQUILIBRIUM DATA FOR METHANOL AND WATER AT CONSTANT PRESSURE TEMPERATURE (C) 100.0 96.4 93.5 91.2 89.3 87.7 84.4 81.7 78.0 75.3 73.1 71.2 69.3 67.5 66.0 65.0 64.5 MOLE FRACTION OF LIQUID 0.00 0.02 0.04 0.06 0.08 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 1.00 MOLE FRACTION OF VAPOUR 0.00 0.134 0.230 0.304 0.365 0.418 0.517 0.579 0.665 0.729 0.779 0.825 0.870 0.915 0.958 0.979 1.00 Perry, R.H., Green, D.W., Perry’s chemical Engineers’ Handbook seventh edition, Section 13 pg 13, McGraw-hill. 1997 1 Assuming that the feed thermal condition is at saturated liquid feed so q=1. Calculating the minimum reflux ratio, Rmin Gradient of line l = 0.98-0.76 0.98-0.458 = 0.421 Now Gradient of l =____Rmin______ Rmin + 1 0.421 = ____Rmin______ Rmin + 1 Rmin = 0.727 The optimum reflux ratio will lie between 1.2 to 1.5 times the minimum reflux ratio. Assuming that the optimum value is 1.30. The optimum reflux ratio R = 1.30 x 0.727 = 0.945 The equation of the Upper Operating Line: y= X R x+ D R+1 R+ 1 = 0.945 0.983 x+ 1.945 1.945 = 0.486 x +0.505 From the graph, the number of theoretical stages was found to be 14. The liquid and vapour amounts in the rectifying section; R=flow returned as reflux flow of top product R= LO D 0.945= Lo 91 666.667 LO = 86 625.000 mol hr-1 = 86.625 kmol hr -1 In accordance with equimolar overflow, the molar vapour and liquid flows from each stage are constant. Ln=LO Ln= 86.625 kmol hr-1 Material balance for the rectifying section: V n=Lo + D V n=86 625.000+ 91666.667 = 178 239.667 mol hr -1 =178. 292 kmol hr -1 Now Lm=F + Ln Vm=Vn Lm = 155 000 +86 625 Vm= 178.292 kmol hr-1 = 241 625 mol hr-1 =241.625 kmol hr-1 . Table showing temperature and density of methanol and water at the feed, distillate and residue positions. Molar compositio n of liquid methanol FEED DISTILLATE RESIDUE Molar compositio n of vapour methanol 0.458 0.983 0.013 Temperature (C) 0.760 1.00 0.02 Density of water (liquid) Kg m-3 Density of water (vapour) Kg m-3 74.0 62.2 98.0 Density of liquid mixture at 74C = 975(0.542) + 746(0.458) = 870.118 kg m-3 Density of liquid mixture at 62.2C=982(0.017) + 754(0.983) =757.876 kg m-3 Density of methanol (liquid) Kg m-3 975 0.25 982 0.13 960 0.58 Density of methanol (vapour) Kg m-3 746 754 712 1.50 1.10 4.30 Density of liquid mixture at 98C= 960(0.987) + 712(0.013) =956.776 kg m-3 Density of vapour mixture at 74C=0.25(0.24) + 1.50(0.760) =1.20 kg m -3 Density of vapour mixture at 62.2C=0.13(0) + (1.10)(1.00) =1.10 kg m-3 Density of vapour mixture at 98C=0.58(0.98) + 4.30(0.02) =0.654 kg m -3 ρl =(870.118+757.876) /2 Density of the liquid mixture in the rectifying section =813.997 kg m-3 Density of the vapour mixture in the rectifying section Density of liquid mixture in the stripping section Density of vapour mixture in stripping section ρv =1.15 kg m-3 ρl = 857.326 kg m-3 ρv = 0.877 kg m-3 Calculation for the diameter of the column, 1m is the typical size for a water-methanol system. Liquid vapour factor, FL = v L V √ ρv ρl 1.15❑ 813.997 ¿ For the stripping section, 86.625 √¿ FL = 178.295 v =0.0182 U f =K Flooding velocity, 1 √ ρl −ρ K 1 = 0.084 (constant given in Figure 11.27 v ρv R.K. Sinnott. 2005.Coulson & Richardson`s Chemical Engineering Design. Volume 6. 4th ed, 568. Oxford: Butterworth-Heinemann. ) U f =0.084 Net area An = A c –AD √ 831.997−1.15 1.15 = 2.26 m s-1 Ac = cross sectional area of the column AD = down comer area (taken to be 12% of Ac) An = 0.88Ac Let flooding condition = 80% Uf Actual vapour velocity, Ua = 0.8 (2.26) = 1.81 m s-1 Assuming that inside the column methanol has a composition of 50% so it will have a molecular weight of 25 kg kmol-1. Now Vn =(178.292 x 25)/(1 x 3600) = 1.24 m3 s-1 V n=¿ U An a ¿ A n = 0.69 m2 Now AC = 0.784m2 But Ac= Dc2 /4 Dc = 0.9991 m = 1.00 m (as expected) The same diameter is used for the stripping section. Murphree plate efficiency = 80% Weir height = 15% of plate spacing = 0.15 x 0.5 =0.075m Weir length, lw= 50mm R.K. Sinnott. 2005.Coulson & Richardson`s Chemical Engineering Design. Volume 6. 4th ed, 598. Oxford: Butterworth-Heinemann. 2 The tray spacing is estimated to be 0.5m since the diameter of the column is 1m. Plate hole diameter is taken to be 5mm as the size must be less than 6.5mm because entrainment is greater with larger hole sizes and a thickness of 5mm. To calculate the number of holes in a plate in the rectifying section: A a =A c −2 A d A a =A c −( 0.12 ) (2 A c ) A a = 0.596 m2 Assuming that the holes present are 10% of Entire hole area of a tray = 0.1 x 0.596 =0.0596 m2 Area of a hole = 1.96 x 10-5 m2 Aa Total number of holes in a plate= 0.0596/ 1.95 x 10-5 = 3 041 Similarly, the number of holes in a plate from the stripping section = 2 024 Column Efficiency calculation: Using Van Winkle`s correlation: mV =¿ 0.07 Dg 0.14 Sc0.25 ℜ0.08 E¿ σl μl μv Dg = surface tension number = Where σ l =¿ liquid surface te-nsion μl = liquid viscosity μv = superficial vapour velocity Sc = liquid Schmidt number = μl ρl D LK D LK = liquid diffusivity ℜ=¿ Reynolds number = hw uv ρv / μl (FA ) hw = weir height FA = fractional area= area of holes∨risers total column cross sectional area Assumptions for calculation: the temperature of the liquid is the same as the average column temperature and the properties of water and the liquid mixture are the same. Average column temperature = (74 + 98)/2 = 86C σl = 0.06 N/m Uv=Ua= 1.81 m s-1 Average densities: μl = 355 x 10-6 N s/m2 ρl ρv = 844.662 kg m-3 D LK = 5 x 10-9 m2 s-1 =1.014 kg m-3 hw =0.075m FA = 0.0596/ 0.785 = 0.076 Dg = 0.06 355 x 10−6 x 1.81 = 93.378 −6 355 x 10 Sc = 844.662 x 5 x 10−9 ℜ = = 84 0.075 X 1.81 X 1.014 355 x 10−6 X 0.076 = 5101.95 m V =¿ 0.14 0.25 0.08 E¿ 0.07(93.378 ) (84 ) (5101.95 ) = 0.79 = 79% Actual number of stages= Number of ideal stages/ = Em V 13 0.79 = 17 Height of the column Hc= (actual number of stages-1)(tray spacing) + H + plate thickness = (17-1)(0.5) + 1 + (17x0.005) = 9.1 m Feed tray location: From Graph 1, the theoretical location is the feed tray is at the 10th plate. Actual location of feed plate= 10/0.79 = 13th plate. http://www.engineeringtoolbox.com/surface-tension-d_962.html http://www.suppliersonline.com/propertypages/317.asp http://guide.rockwool-rti.com/products/industry/prorox-sl-930-(rockwoolmultiboard).aspx