Practice Problem Solutions_Exam 2

Practice Problem Solutions_Exam 2 - W = 3545 kg/hr and n 0...

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1. 100 kmol/h V kmol /h 0.20 kmol n-H/kmol 0.80 kmol HC/kmol 0.50 kmol n-H/kmol 0.50 kmol HC/kmol L kmol /h x kmol n-H/kmol (1-x) kmol HC/kmol 1 atm T o C (100)(0.50)(0.90) = Lx or Lx = 45 kmol/h Balances: 100 = L+V 50 = V(0.20) + Lx So: V = 25 kmol/h L = 75 kmol/h x = 0.60 moles/mole Raoult’s law: (P)(0.20) = p* n-H (0.60) p* n-H = 253.3 mmHg (P)(0.80) = p* HC (0.40) p* HC = 1520 mmHg Antoine equation parameters for n-H: A = 6.90253, B = 1267.828, C = 216.823 C 65.0 T T C B - A (253.3) log o 10 = + = More volatile species: HC
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2. Evaporator Centrifuge Balance on overall process: n 0 x S = 0.22 x W = 0.78 n 1 x 1 Slurry = Solution n R, x R + Crystals 1000 kg/hr Crystals 1000 kg/hr Solution n R, x R n W 0.78 n 0 = n W 0.22 n 0 = 1000 Ö n
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Unformatted text preview: W = 3545 kg/hr and n 0 = 4545 kg/hr Solution leaving evaporator is saturated with 0.2 kg salt/kg water Ö x R = 0.2/(1+0.2) = 0.167 Ratio of crystals to solution = 0.67 = 1000/n R => n R = 1500 kg/hr n 1 = n +n R = 6045 kg/hr Mass balance on salt at the mixing point: x 1 . n 1 = 0.22 n + 0.167 n R => x 1 = 0.207 3. 4. Use the ternary phase diagram: Initial point is 0.2 W, 0.33 A, 0.47 M. This is in the 2-phase region. So it will slpit (using the tie line) into two points on the equilibrium curve: Water rich phase: 0.7 W, 0.25 A, 0.05 M MIBK-rich phase: 0.57 M, 0.35 A, 0.08 W The total mass is not required in the problem....
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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 Tech.

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Practice Problem Solutions_Exam 2 - W = 3545 kg/hr and n 0...

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