Elementary Principles 217 - 6.71 Basis: 1000 kg/h product...

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6- 51 6.71 Basis : 1000 kg/h product n A1 (mol A/h) n E1 (mol E/h) 280°C reactor n A2 (mol A/h) n E2 (mol E/h) condenser n H2 (mol H /h) 2 n C (mol/h) still 0.550 A 0.450 E liquid, –40°C n 0 (mol E/h) Fresh feed n 3 (mol/h) vapor, –40°C y A3 (mol A/mol), sat'd y E3 (mol E/mol), sat'd y H3 (mol H /h) 2 scrubber n H4 (mol H /h) 2 Product 1000 kg/h n p (mol/h) 0.97 A 0.03 E Scrubbed Hydrocarbons n A4 (mol A/h) n E4 (mol E/h) E = C H OH ( 2 5 M = 46.05) A = CH CHO ( 3 M = 44.05) P = 760 mm Hg n r (mol/h) 0.05 A 0.95 E Strategy Calculate molar flow rate of product ± n p di from mass flow rate and composition Calculate y A3 and y E3 from Raoult’s law: yy y H3 A3 E3 = 1 . Balances about the still involve fewest unknowns ( ±± ) nn cr and Total mole balance about still A balance about still U V W ± , ± A, E and H 2 balances about scrubber ± , ± A4 E4 , and ± n H4 in terms of ± n 3 . Overall atomic balances on C, H, and O now involve only 2 unknowns ( ± , ± 03 ) Overall C balance Overall H balance U V W ± , ± A balance about fresh feed-recycle mixing point
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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