Elementary Principles 95 - 4.62 a. i - C 4 H 10 + C 4 H 8 =...

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4- 54 4.62 a. i -C H C H 41 0 48 81 8 + = Basis : 1-hour operation reactor (n-C H ) 4 n 51 0 (i-C H ) 4 n 61 0 (C H ) 8 n 71 8 (91% H SO ) 2 m 84 (n-C H ) 4 n 21 0 (i-C H ) 4 n 31 0 (C H ) 8 n 11 8 (91% H SO ) 2 m 44 decanter still (C H ) 8 n 8 (n-C H ) 4 n 0 (i-C H ) 4 n 0 (C H ) 8 n 8 (n-C H ) 4 n 0 F P D E C B A (kg 91% H SO ) 2 m (i-C H ) 4 n 0 40000 kg kmol n 0 0.25 0.50 0.25 i-C H 4 10 n-C H 4 10 C H Units of n : kmol Units of m : kg Calculate moles of feed MM M M =++ = + = −− 0 25 050 0 25 0 75 5812 0 25 5610 57 6 ... . . . . . LCH nCH CH 0 0 kg kmol bg b g b g n 0 40000 1 694 == kg kmol 57.6 kg kmol b g Overall n - C H balance: 0 n 2 050 694 347 . kmol n - C H in product 0 C H balance: 8 n 1 694 1735 0.25 kmol C H react 1 mol C H 1 mol C H kmol C H in product 8 88 b g . At (A) , 5 mol - C H 1 mole C H n mol - C H kmol 0 0 A moles C H at A=173.5 -C H at
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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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