Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Given column top and bottom conditions in fig 750

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Unformatted text preview: . The value of FLV is obtained first, using the conditions in Fig. 7.48. MV = [187.6(58.1) + 176.4(72.2) + 82.5(86.2)]/446.5 = 68.9 ML = [112.4(58.1) + 223.6(72.2) + 217.5(86.2)]/553.5 = 74.8 From a simulation program, ρV = 1.13 lb/ft3, compared to 0.965 from the ideal gas law and ρL = 33.2 lb/ft3. Also given, V = 446.5 lbmol/h and L = 553.5 lbmol/h In Fig. 6.24, FLV LM L ρV = VMV ρ L 1/ 2 = 553.6(74.8) 113 . 446.5(68.9) 33.2 1/ 2 = 0.248 From Fig. 6.24, CF = 0.27 ft/s. From Eq. (6.42), C = (1)(1)(1)0.27 = 0.27 ft/s. From Eq. (6-40), ρ − ρV Uf =C L ρV 1/ 2 33.2 − 113 . = 0.27 113 . 1/ 2 = 144 ft/s . From Eq. (6-44), 4VMV DT = fU f πρV 1/ 2 4(446.5 / 3600)(68.9) = 0.85(1.44)(314)(113) . . 1/ 2 = 2.81 ft Assume a liquid residence time, t, of 5 minutes (0.0833 h), half full . 2 LM L t 2(553.5)(74.8)(0.0833) = = 208 ft3 From Eq. (7-44), vessel volume = VV = ρL 33.2 4V 4(208) Now compute the height, H, from Eq. (7-45). H = V2 = = 33.6 ft πDT 314(2.81) 2 . Thus, H/D = 33.6/2.81 = 12.0. This is too large, so re-dimension the volume to give H/D = 4. From Eq. (7-46), V DT = V π 1/ 3 208 = 3.14 From Eq. (7-46), H = 4DT = 4(4.05) = 16.2 ft 1/ 3 = 4.05 ft Exercise 7.46 Subject: Sizing of a horizontal reflux drum. Given: Flow conditions leaving reflux drum in Fig. 7.49. Find: Drum length and diameter. Analysis: Liquid rate leaving drum = (3 + 1)120 = 480 lbmol/h Assume properties of pure n-hexane. From Fig. 2.4, boiling point of nC6 at 1 atm = 150oF. From Fig. 2.3, ρL = 0.63 g/cm3 = 39 lb/ft3 Assume a liquid residence time in the reflux drum of 5 minutes (0.0833 h), half full. From Eq. (7-44), 2 LM L t 2(480)(86.2)(0.0833) VV = = = 177 ft3 ρL 39 From Eq. (7-46), V DT = V π 1/ 3 177 = 3.14 1/ 3 = 3.83 ft From Eq. (7-46), H = 4DT = 4(3.83) = 15.3 ft Might use a drum of 4 ft diameter and 16 ft long. Exercise 7.47 Subject: Sizing of a sieve-tray distillation column separating methanol and water. Given: Column top and bottom conditions in Fig. 7.50. Assumptions: Column feed is a bubble-point liquid. Constant molar overflow. Find: (a) Column diameters at top and bottom. Should column be swaged? (b) Length and diameter of horizontal reflux drum. Analysis: First, compute the liquid and vapor flow rates at the top and bottom of the column. Molecular weight of distillate = 0.9905(32.04) + 0.0095(18.02) = 31.91 Distillate flow rate = D = 462,385/31.91 = 14,490 lbmol/h Molecular weight of bottoms = 0.0101(32.04) + 0.9899(18.02) = 18.16 Bottoms flow rate = B = 188,975/18.16 = 10,410 lbmol/h From these values and the given compositions, the overall material balance is: Lbmol/h: Component Feed Distillate Bottoms Methanol 14,457 14,352 105 Water 10,443 138 10,305 Total: 24,900 14,490 10,410 From the reboiler duty, assuming the bottoms is pure water, can compute the boilup rate. Heat of vaporization of water at 262.5oF = 937 Btu/lb = 16,885 Btu/lbmol Boilup rate = V = 442,900,000/16,885 = 26,230 lbmol/h Assume constant molar overflow with saturated liquid feed. T...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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