Elementary Principles 96 - 4.62 (cont'd) i - C 4 H 10...

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4- 55 4.62 (cont’d) in n - C H balance around second mixing point l C H in recycle E 41 0 0 + = ⇒= 867 5 34 700 33,800 kmo 6 6 ., Recycle E : Since Streams (D) and (E) have the same composition, n n ni n 5 2 6 3 5 moles n - C H moles n - C H moles - C H moles - C H 16,900 kmol n - C H 0 E 0 D 0 E 0 D 0 bg =⇒ = n n n n n 7 1 6 3 7 moles C H moles C H 8460 kmol C H 81 8 E 8 D 8 =⇒= Hydrocarbons entering reactor: 347 16900 5812 + F H G I K J b g kmol n - C H kg kmol 0 . ++ F H G I K J + F H G I K J 867 5 33800 1735 5610 . ... b g kmol - C H kg kmol kmol C H kg kmol 0 48 i + F H G I K J 8460 kmol 114.22 4 00 10 6 CH kg kmol kg 8 . . H SO solution entering reactor and leaving reactor 4.00 10 kg HC 2 kg H SO aq 1 kg HC kg H SO aq 24 6 = × 800 10 6 . mn nn m 8 6 5 25 8 6 784 10 H SO in recycle H SO leaving reactor n - C H in recycle n - C H leaving reactor kg H SO aq in recycle E 0 0 . . × = + × m 4 5 16 10 = H SO entering reactor H SO in E kg H SO aq recycled from decanter . 091 1480 5 .. ×= di
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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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