Elementary Principles 331 - 8.73 (a) = 10.0 kg water...

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8-52 8.73 (a) 400 2 44 97 56 10 0 kg kg water kg air kg water evaporates / min . min . . = (b) h a == 10 400 0 025 kg H O min kg dry air min kg H O kg dry air 2 2 ., T db C = ° 50 Fig. 8.4-1 dew point kJ kg dry air C, C ± . HT h T wb r =− = = ° = = ° 116 11 115 33 32%, 285 bg (c) T db C 10 , saturated ⇒= = hH a 0 0077 29 5 ± . kg H O kg dry air kJ kg dry air 2 (d) 400 kg dry air 0.0250 kg H O min kg dry air kg H O min condense 2 2 = 00077 692 . . References: Dry air at 0 C, H O at 0 C 2 °° l substance ² m in ± H in ² m out ± H out Air HO 2 l 400 115 400 6.92 29.5 42 ² m air in kg dry air/min, ² m 2 in kg/min ± H air in kJ/kg dry air, ± H 2 in kJ/kg C C 22 ll ,, 02 0 °→ ° b g : ± . H = −° ⋅° = 754 10 0 42 J 1 mol C 1 kJ 10 g mol C 18 g 10 J 1 kg kJ kg 3 3 out in 34027.8 kJ 1 min 1 kW ˆˆ 565 kW min 60 s 1 kJ/s ii Q H mH =Δ = = ∑∑ ²² (e) T>50°C, because the heat required to evaporate the water would be transferred from the air, causing its temperature to drop. To calculate (T air ) in , you would need to know the
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