Practice Problem Solutions_Exam 2

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

This preview shows pages 1–11. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## 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.

### Page1 / 11

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

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online