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Chap12sol

# Chap12sol - Prob 12.17 Pwvn 0522 xﬂ3— P-F’Wl.1...

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Unformatted text preview: Prob. 12.17 Pwvn : 0522 xﬂ3— P-F’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 (30909-29401)+(0.0289x2554.4)-(0.014?x253?.2)[ U ] -(0.0289-0.0147)(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 — = ”HO-58 — = 52 —- mm [ ] kg DA kg DA The psychrometric chart results agree with the detailed property-based calculations within approximately 2% due to neglect of the water enthalpy term. Excerpts from this work may be reproduced by instruclors for dislrlbullon on a nol‘for-proi‘it basis for testing or lnslrucllonal purposes only to students enrolled in courses for which the textbook has boon adopted. My other mpmductton or translation oftttts went beyond that permttted by Sections 10? or £08 0! the 1.9% United States Cﬂpyrt'gt'l'f Act without the pentttsst'on ot'ttre coyyrtghr owrrer is nuts wtltt ...
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