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Split_00026 - 8.50(cont'd Air balance n3 = 50.3(1 0.178 =...

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8- 26 8.50 (cont’d) Air balance: ( )( ) 3 50.3 1 0.178 41.3 mol air s n = = ± Mole fraction of hexane in outlet gas: ( ) ( ) ( ) 2 2 3 3.58 67.8 mm Hg 3.58 41.3 850 mm Hg H H p T n p T n n = = = + + ± ± ± Saturation at outlet : Table B.4 * H ( ) ( ) 67.8 mm Hg 7.8 C H p T p T T = = ⎯⎯⎯⎯→ = ° Reference states: C H C 6 14 l , . 7 8 ° b g , air (25 ° C) Substance ± n in ² H in ± n out ² H out ( ) 6 14 C H v ( ) 6 14 C H l Air 8.95 41.3 37.5 0.435 3.58 5.37 41.3 32.7 0 –0.499 ± n in mol/s ² H in kJ/mol ( ) 6 14 C H v : ² ² . , ² . . . H C dT H C dT C
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