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Unformatted text preview: Miriam mouth” PROBLEM 72 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 _' Qnnw ' Eatu 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 + ”lethallair + miﬂlUiee; U Rev Deccrrhcr'i, l5“)? Page I all ingest Frublurn T~2 Snluh'en
RISE: ' f. "latrines! ' ”anti "" macrtiTJTJJ +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'ltg
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 petlure, 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. $11 We?” "1' Fiﬁ1"“: P. ' 3i 3 " :3
mg: a] C =Elﬁ13xlﬂ Ea} 505mm 1: 595% kgwpﬁlr.
Mal {WEEK iEUUKi ' We can new salve fer the temperature at state 2. by ignnring the 1rulurue rat" the liquid
and by using the ideal gas law. Ill”F. = Frum Daltnn's Law nf partial pressures we knew that
P: — 0'4 NEE. 7' Pain: + {11,111.39 Page 1 pH pages Pinhleui TZ Snltitiuu Thus
in'RTI _ (sesasirassm) F MidstP“ _ e.4::1rr"'1‘‘wa a. VG where T; must be in degree K. and PM in Pa. Irerating this equation using the steam tables gives ﬁnal temperature and the partial pressure of the water vaper as:
T2 = 402.4 K rt.r1 = eesss MPa I New that we knew ﬁnal temperatures and pressures, we ehtain the felluwing values from stem tables
vr = s em rise mlrtcg vrﬁ= 0.451943 mjlkg
Ur: 54s .4 krtkg ufa = lease the;
I New the tetsl volume efthe saniraterl mixture ﬁlls the enntainrnent, therefore
Vﬁlmwp + mill‘fr + mtg]
Solving for x and substituting gives L v sreee useless:
_ _ (msﬁmJ ‘ _ {asts4t+m,) ' ' (1
"x _ vii _ at 51943 3 I if we subsittute uur derived values intu [1} we get
n1wp{uﬂ 4' mtg: ' ”it 1+ mscv'sz'Ti} “mime“mm 'Uil} ‘0
Thus
merruﬂ. r xurﬂz 464341514 Sﬂﬁiﬁfﬂi19](402.43ﬂtljtm1un'tsum +3 issxisﬁi =3 and
294594? {543.32+1995,9xlﬁﬂ4.] 5}+$.39 silt}5 + nt,{543.32+ E995.97t+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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 Fall '08
 acosta
 Physics

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