Elementary Principles 337 - 8.80 (cont'd) Inlet air: Tdb =...

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8-58 8.80 (cont’d) Inlet air: T h h H r a db Fig. 8.4-1 2 C 0.0198 kg H O kg DA 88.5- 0.5 kJ kg DA kJ kg DA = = == R S T 37 50% 88 0 1 1 ± . bg Moles dry air: ² m a 1250 kg 1 kg DA h 1.0198 kg kg DA h 1226 Outlet air: T h H a db Fig. 8.4-1 2 C, sat'd kg H O kg DA kJ kg DA = = R S T 15 0 0106 421 2 . ± . Overall water balance ⇒= ² .. m c 1226 kg DA kg H O h k g D A 2 0 0198 0 0106 = 113 . kg H O h withdrawn 2 Reference states for enthalpy calculations: HO 2 l , dry air at 0 o C. (C p ) H2O(l) = 4.184 kJ kg C o HO 1 2C k J/k g 2 lH C d T p ,: ± . °= = z 0 12 50 3 Overall system energy balance: ²² ² ± ² ± Q H mH ci i i i Δ out in =+ L N M O Q P F H G I K J F H G I K J =− 50 3 421 88 11 155 . . . . kg H O h kJ kg H O 1226 kg DA h kJ kg DA h 3600 s kW 1 kJ / s
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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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