Split_00009 - 8.26 (cont'd) (5) Energy balance: . 10(0169)...

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8- 9 8.26 (cont’d) (5) Energy balance: = 10 0169 0002 1002 1000 18 0847 1000 1002 1000 29 0727 91.84 0020 0169 0 779 1000 18 10 6 7 2 1000 29 33 44 4 4 (. ) . . ) . . ) ) ) ) ( ) ) + F H G I K J F H G I K J + F H G I K J F H G I K J + F H G I K J +− F H G I K J mm my m y Solve Eqs. (3)–(5) simultaneously m 3 = 2.55 kg/s, m 4 = 2.70 kg/s, y 4 = 0.0564 kg H 2 O/kg 2 00255 .55 kg humid air / s 100 mol gas/ s kg humid air mol gas = . Mole fraction of water : kg H O (1-.0564) kg dry air kg DA kmol DA kmol H O .0963 kmol H O kmol humid air 22 2 2 0 0564 29 1 18 0963 0 1 0 0963 0 0878 . . (.) . = + = Relative humidity: C mm Hg mm Hg HO o 2 2 p p * ) ( ) . . 48 0 0878 760 8371 100% 79 7% ej = c. The membrane must be permeable to water, impermeable to CO, CO 2 , O 2 , and N 2 , and both durable and leakproof at temperatures up to 50 o C. 8.27 a. y p P 2 2 C mm Hg 760 mm Hg mol H O mol = ° == * . . 57 129 82 0171 bg 28.5 m STP 1 mol h 0.0224 m STP mol h mol H O h 3 3 2 =⇒ 1270 217 2 . 391 . k g HOh 2 1270 217 2 1053 =
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This note was uploaded on 01/07/2012 for the course CH E 210 at Penn State.

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