Split_00058 - 8.80(contd Inlet air Tdb = 37 C hr = 50 Moles...

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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 ± . b g 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 kg DA 2 0 0198 0 0106 b g = 113 . kg H O h withdrawn 2 Reference states for enthalpy calculations: H O 2 l b g , dry air at 0 o C. (C p ) H2O(l) = 4.184 kJ kg C o H O 12 C kJ / kg 2 l H C dT p , : ± . ° = = z b g 0 12 50 3 Overall system energy balance: ² ² ² ± ² ± Q H m H m H c i i i i = = Δ out in = + L N M O Q P F H G I K J F H G I K J = − 113 50 3 421 88 1 1 155 . . . . kg H O h kJ kg H O 1226 kg DA h kJ kg DA h 3600 s kW 1 kJ / s kW 2
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