Elementary Principles 301 - 8.47 Basis: 226 m 3 min 273 K...

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8- 22 8.47 Basis: 226 m K 10 mol min K 22 m STP mol humid air min 33 3 273 309 415 8908 . bg = . DA = Dry air ± ( Q kJ / min) 8908 mol 0 0 /min mol H O(v) / mol] (1- (mol DA / mol) 36 C, 1 atm, 98% rel. hum. 2 o y y [ ) ± ( [ ) n y y 1 1 1 mol / min) mol H O(v) / mol] (1- (mol DA / mol) 10 C, 1 atm, saturated 2 o ± [ n 2 mol H O(l) / min], 10 C 2 o a. Degree of freedom analysis : 5 unknowns – (1 relative humidity + 2 material balances + 1 saturation condition at outlet + 1 energy balance) = 0 degrees of freedom. b. Inlet air: yP p y w 00 098 36 098 44563 0 0575 = = B . .(. . * C mm Hg) 760 mm Hg mol H O(v) mol Table B.3 2 Outlet air: yp P 1 10 9 760 mm Hg 0 0121 == = (. o 2 C) / .209 mm Hg mol H O(v) mol b g Air balance: 1 0 0575 1 0 0121 8499 11 −= = . (8908 . ±± mol / min) mol / min nn H O balance:
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