Practice Exam 1Solution - EXAM 1 Solution 1. A....

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EXAM 1 Solution 1. A. Accumulation Rate = Input Rate – Output Rate (no Generation or Consumption) Acc. Rate = 5 – 3 = 2 kg/sec B. K . m ) s / J ( 07 . 0 1 x kg g 1000 x K 1 C 1 x C . g J 4 . 1 x s 3600 hr 1 x m 1 ft 281 . 3 x lb 1 kg 454 . 0 x hr . ft lb 400 k C Pr p = μ = => Pr = 3310 C. Let required mass flow rate of oxygen be X Consider 1 mol of mixture. It contains 0.14 mol butane and 0.86 mol oxygen Mass fraction of butane = 23 . 0 32 x 86 . 0 58 x 14 . 0 58 x 14 . 0 = + So 500 kg/hr contains 115 kg/hr butane and 385 kg/hr oxygen After adding X kg/hr of oxygen, the mol % of butane is reduced to 10% => 1 . 0 32 / ) X 385 ( 58 / 115 58 / 115 = + + X = 186 kg/hr
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Balances around distillation column Ethanol: 4 3 n 02 . 0 5000 x 97 . 0 n 53 . 0 + = Total: 4 3 n 5000 n + = Solving, we get n 4 = 4314 kg/hr and n 3 = 9314 kg/hr Balances around the mixing tank Ethanol: 2 1 n 8 . 0 n 2 . 0 9314 x 53 . 0 + = Total: 9314 n n 2 1 = + Solving, we get n 1 = 4191 kg/hr and n 2 = 5123 kg/hr Distillation Unit Overhead
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This note was uploaded on 05/01/2011 for the course CHBE 2100 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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Practice Exam 1Solution - EXAM 1 Solution 1. A....

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