8- 208.45 (cont’d) Substituting for from Table B.1 and for from Table B.2 kJ / mol, kJ / molvΔ±±.±.HCHHpfv⇒==38 3627 30Energy balance: ²±±..QnHnH=−=−×−×∑∑outoutininkJ / s = kW116 10116 10338.46 a. Basis: 100 mol humid air fed100 mol y1(mol H2O/mol) 1-y1(mol dry air/mol) 50oC, 1 at m, 2osuperheat n2(mol), 20oC, 1 atm y22O/mol), sat’d2(mol dry air/mol) n3(mol H2O(l)) There are five unknowns (n2, n3, y1, y2, Q) and five equations (two independent material balances, 2oC superheat, saturation at outlet, energy balance). The problem can be solved. b. 2 C superheatC°⇒=∗°ypp148bgsaturation at outletypp220 Cdry air balance:100 11122bg bg−=−ynyH O balance: 210023bgbgbgyn=+c. References: Air 25°C, HO C2l,20°SubstanceAirin molHOinkJmolininoutout22nHyH
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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.