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prob.vii-2

# prob.vii-2 - Miriam mouth” PROBLEM 7-2 SOLUTION Figure I...

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Unformatted text preview: Miriam mouth” PROBLEM 7-2 SOLUTION Figure I a We will approach this as a control mass problem, including the primary in our analysis .I'hut ignoring the secondary (Figure I]. _-I hastening that no water vapor is initially present in the air. I Our starting energ},r equation then becomes: I U: ' UL _' Qn-nw ' Eat-u -W 7. Here assume that the containment is adiabatic. i.e. no heat transfer to the structure - I Assume complete hlowdown .. 3 3' There is no shaft work and no boundary work in this problem as the containment volume is ﬁsetl l Thus our equation becomes ' U1 ' U] ‘0 'l The internal energyr at state i is that of the ice. the air in the containment, and the tiriniary coolant. . i All of me iee will melt, thus the interns] energy at state 2 is that of a saturated mixture and the air in the containment. ' Assume air is an ideal gas ' The mass in the containment is the sum of the mass of the air, ice, and primary coolant. sites.» we can break the above equation into the sum ol'tlte intents] energ3,I changes of the different mass components: mFriinWﬁuFrhnmy + ”lethal-lair + miﬂlU-iee; U Rev Deccrrhcr'i, l5“)? Page I all ingest Frublurn T~2 Snluh'en RISE: -' f. "latrines! ' ”anti "" macrtiTJ-TJJ +l'ﬂr'ilJis‘Ui1) = '5" [1} o The internal energj,r of the primer}.r eenlaut at states 1 and 2 is given by um = u {g 15.5 We. samrated liquid = 16434 15 kag llwps = Uta + 3'5 Urge Since state 2 is a saturated mixture. - At state 1, the mass ef lee is l'rneen while at state 2 it is a saturated mixture in Equilibrium with primary.r ennlant. The internal energy can he appruximated with the felluwiug equatinn uc *- irtterutﬂ ettetgj,r ofiiquid water at 213 K and 1 atm = {ll-:Ir'l-tg up a heat effusiuu emf water = 3.33 x ll]j Jilrg ei a speeifie heat nfiee = 4.23 x it)1 Jiltg K T,“ —' melting point efiee -' 213 K Thus ui. = ass x In” k1ﬂrg—423 r m3 1.:th [2?3 K ease} a” = -3 ass e 1e5 Mcg=-3,'F53x 1e1 kii’kg A: state 2, in a saturated mixture “1'2 _ Ur: + XUrg I lDur uukuewns are new the mass ni' the air, primary. and ice. the ﬁnal tern pet-lure, and the final quality. 4' Fer the primary at state i all” = uuutess mﬂtzg ' and the volume of the primary lenp is given as SCH.) cubic meters. Thus VF St‘lﬂmj = ———— = star: ms kg va: use! 6351113 In; I We arrive fer the mass ut‘ the air lag neglecting the aniume of the iee and by using the ideal gasiaw statute | I. \$11- We?” "1' Fiﬁ-1"“: P. ' 3i 3 " :3 mg: a] C =Elﬁ13xlﬂ Ea} 505mm 1: 595% kgwpﬁlr. Mal {WEEK iEUUKi ' We can new salve fer the temperature at state 2. by ignnring the 1rulurue rat" the liquid and by using the ideal gas law. Ill-”F. = Frum Daltnn's Law nf partial pressures we knew that P: — 0'4 NEE. 7' Pain: + {11,111.39 Page 1 pH pages Pinhleui T-Z Snltitiuu Thus in'RTI _ (sesasirassm) F Midst-P“ _ e.4::1rr"'-1‘-‘wa a.- VG where T; must be in degree K. and PM in Pa. Ire-rating this equation using the steam tables gives ﬁnal temperature and the partial pressure of the water vaper as: T2 = 402.4 K rt.r1 = eesss MPa I New that we knew ﬁnal temperatures and pressures, we ehtain the felluwing values from stem tables vr = s em rise mlrtcg vrﬁ= 0.451943 mjlkg Ur: 54s .4 krtkg ufa = lease the; I New the tetsl volume efthe saniraterl mixture ﬁlls the enntainrnent, therefore Vﬁlmwp + mill‘fr + mtg] Solving for x and substituting gives L- v sreee useless: _ _ (msﬁmJ ‘ _ {asts4t+m,) ' ' (1 "x _ vii _ at 51943 3 I if we subsittute uur derived values intu [1} we get n1wp{uﬂ 4' mtg: ' ”it 1+ mscv'sz'Ti} “mime-“mm 'Uil} ‘0 Thus merruﬂ. r- xurﬂz 464341514 Sﬂﬁiﬁfﬂi19](402.4-3ﬂtlj-t-m1un't-sum +3 issxisﬁi =3 and 294594? {543.32+1995,9x-lﬁﬂ4.] 5}+-\$.39 silt}5 + nt,{543.32+ E995.97-t+3 T53x105)=ﬂ ,- anus; seat x1U3x+3153543ml+1995ﬂ 3|: mi=3.lﬂﬁ:~: is“ rat I lt'tve set up an iteratien where we substitute fer values nf X in equatittn [3), selve for m, and then substitute this rn inte equation (2}, we can. then compare the x from equatiurt {2} with eur assumed value. iterating until diﬁ‘erenee is sere- yields m,=t??xlﬂ5 xi=ﬂ.156 Page 3 el'3 pages ...
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