Unformatted text preview: Prob. 12.17 Pwvn : 0522 xﬂ3—
PF’Wl.1 101.324.5013 Similarly for (02. Pwu‘ g = glzPsAT (@ T2) where
.352 = 1, Psm (@ 20°C) =2.3388 kPa, so Pwv_1 = 1 x 2.3388 = 2.3388 kPa PW #1622 2.3388 .wa ><101.322.3388 at = 0.522 = 0.0289 kg t kg Dry Air we = 0.622 = 0.014? kg t‘ kg Dry Air The energy balance equation can be rearranged to solve for Gem/mm : ‘  l'h m . (mww ' mwvz)
QOUT : m0n:(hoA.1 ' hDA.2] + [ﬂ] hww ‘[_r;1wf]hwv,a ' Tm QDUT m = (hem ’ hon: ) + “hhww ' m2'r\t\t.:e _ (a): ' ‘02)hL DA
We can obtain the following enthalpies values from the property tables:
hW1 = hV (T = 35°C] = 2564.4 thkg (Table 105) hm2 = hV (T = 20°C) 2 2537.2 thkg (Table'lOs)
hL = hL (T = 22°C) = 92.20 thkg (TablelOs)
hum = 309.09 thkg (Table 55)
hm = 294.01 thkg (Table 55) Finally, the heat transfer rate is QM (3090929401)+(0.0289x2554.4)(0.014?x253?.2)[ U ]
(0.02890.0147)(92.20) ngA E = 50.585 [k—J]
A DISCUSSION OF RESULTS: Using psychrometric chart, the liquid enthalpy is neglected. so the energy
balance equation becomes Q kJ
QOUT _ momma h1J “‘3" f = {“2 ' hdkgﬁ
DA
From Figure AJS h1z 110 kgklgA and h2 :: 58 kgkﬁ
00.“ kJ kJ
— = ”HO58 — = 52 —
mm [ ] kg DA kg DA The psychrometric chart results agree with the detailed propertybased calculations
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 Fall '09
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