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Split_00020

# Split_00020 - 8.45(cont'd Substituting for H v from Table...

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8- 20 8.45 (cont’d) Substituting for from Table B.1 and for from Table B.2 kJ / mol, kJ / mol v Δ ± ± . ± . H C H H p f v = = 38 36 27 30 Energy balance : ² ± ± . . Q n H n H = = − × × out out in in kJ / s = kW 116 10 116 10 3 3 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 atm, 2 o superheat n 2 (mol), 20 o C, 1 atm y 2 (mol H 2 O/mol), sat’d 1-y 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 b g saturation at outlet = ° y p p 2 20 C b g dry air balance: 100 1 1 1 2 2 b gb g b g = y n y H O balance: 2 100 1 2 2 3 b gb g b gb g y n y n = + c. References : Air 25 ° C b g , H O C 2 l , 20 ° b g Substance Air in mol H O in kJ mol H O in in out out
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