Elementary Principles 299 - 8.45(cont'd Substituting for H...

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

View Full Document Right Arrow Icon
8- 20 8.45 (cont’d) Substituting for from Table B.1 and for from Table B.2 kJ / mol, kJ / mol v Δ ± ± . ± . HC HH p fv ⇒= = 38 36 27 30 Energy balance : ² ±± .. Qn H n H =− = × × out out in in kJ / s = kW 116 10 116 10 33 8.46 a. Basis: 100 mol humid air fed 100 mol y 1 (mol H 2 O/mol) 1-y 1 (mol dry air/mol) 50 o C, 1 at m, 2 o superheat n 2 (mol), 20 o C, 1 atm y 2 2 O/mol), sat’d 2 (mol dry air/mol) n 3 (mol H 2 O(l)) There are five unknowns (n 2 , n 3 , y 1 , y 2 , Q) and five equations (two independent material balances, 2 o C superheat, saturation at outlet, energy balance). The problem can be solved. b. 2 C superheat C °⇒ = ∗° y p p 1 48 bg saturation at outlet y p p 2 20 C dry air balance: 100 1 1 12 2 b g b g −= yn y H O balance: 2 100 2 3 b g b g b g y n =+ c. References : Air 25 ° C , HO C 2 l ,20 ° Substance Air in mol HO i n k Jm o l in in out out 2 2 nH yH
Background image of page 1
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