Elementary Principles 304

Elementary Principles 304 - 8.49(cont'd c(1 n0 = 5.32 mol...

This preview shows page 1. Sign up to view the full content.

8- 25 8.49 (cont’d) c. 3 01 0 12 0 (1) 5.32 mol feed gas/s (2) 6.58 10 mol A(v)/mol outlet gas (3) 0.147 mol A(v)/mol feed gas (4) 4.57 mol outlet gas/s (5) 0.75 mol A(l)/s ˆ (6) 48.1 kJ/ A ny y nn H ⇒= ⇒= × ⇒ = ± ±± 1 ˆ mol, 34.0 kJ/mol ˆˆ (7) 3.666 kJ/mol, 1.245 kJ/mol (8) 84.1 kW A aa H HH Q = = ± 8.50 a. Feed : 22 2 3 42 3 3 m (35) cm 1 m 273 K 850 torr 1 kmol 10 mol mol 50.3 s 10 cm (273+40)K 760 torr 22.4 m (STP) s π = Assume outlet gas is at 850 mm Hg. Degree-of-freedom analysis 6 unknowns 0123 (,,,,,) ynnnTQ ± ±±± – 2 independent material balances – 2 Raoult’s law (for feed and outlet gases) – 1 60% recovery equation – 1 energy balance 0 degrees of freedom All unknowns can be calculated.
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online