Elementary Principles 184

# Elementary Principles 184 - 6.30 a. Room air - T = 22 C , P...

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6- 18 6.30 a. Room air C −=° T 22 , P = 1atm, h r = 40% : () ( ) 2 1H O 1 2 0.40 19.827 mm Hg 0.40 22 C 0.01044 mol H O mol 760 mm Hg yP p y =∗° = = Second sample C T 50 , P = 839 mm Hg , saturated: 2 2H O 2 2 92.51 mm Hg 50 C 0.1103 mol H O mol 839 mm Hg yP p y =∗ ° ⇒ = = ln ln yb H a ya e bH =+ = , yH 11 001044 5 == ., , 22 01103 48 b yy HH = = = ln ln . . . 21 01103 0 01044 48 5 0 054827 bg b g ln ln ln . . . exp . . ay b H a =− = = = = × 3 001044 0054827 5 48362 48362 7 937 10 b g b g 7 937 10 0 054827 3 .e x p . b. Basis: 1 m delivered air 273K 1 k mol 10 mol 22 273 K 22.4m STP 1 kmol mol air delivered 33 3 + = b g 4131 . Saturation condition prior to reheat stage : ( ) ( )( ) * HO 0.01044 760 mm Hg 7.93 mm Hg 7.8 C (from Table B.3) yPp T T =⇒ = ⇒= ° Humidity of outside air: mol H O mol Part a 2 Hy =⇒= 30 0 0411 0 ( ) ( )( ) 00 0 41.31 0.9896 Overall dry air balance: 1 41.31 0.9896 42.63 mol 10 . 0 4 1 1 ny n −=
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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.

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