# 5 689 ml 1124581 2236722 21758625535

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Unformatted text preview: Also given, V = 446.5 lbmol/h and L = 553.5 lbmol/h In Fig. 6.24, FLV LM L V = VMV 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), - V Uf =C L V From Eq. (6-44), 4VMV DT = fU f V 1/ 2 1/ 2 33.2 - 113 . = 0.27 113 . 1/ 2 = 144 ft/s . 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 = L 33.2 4V 4(208) Now compute the height, H, from Eq. (7-45). H = V2 = = 33.6 ft DT 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 1/ 3 208 = 3.14 1/ 3 = 4.05 ft From Eq. (7-46), H = 4DT = 4(4.05) = 16.2 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. Analysi...
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## This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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