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Unformatted text preview: 7..“... . .h. .3 ~' ..,r. ..~./»...n,  ._,.. N ... .. .~ Problem 1.7 (a) We treat the instantaneous water inventory of the container as an open system
operating unsteadily, so that the first law and the principle of mass conservation read m
57/3 % (mu) : _ + (mmin ‘ (mmout (1) g" (m) : Thin “ Ihout (2) State 1 State 2 Where =:W = 0 and min = 0.} Integrating (2) and Writing 1111 and 1112 for the initial and ﬁnal mass
inventories of the system yields t . ' 2
J; mout (it = ml" m2 The objective is to calculate xz, therefore, we focus on pinpointing state (2). We already
know one property at that state (namely, P2). To obtain the second property we use the first law (1) in integral form '
2 .
m2 112 — In1 ul : _ houtJ'1 moutdt : hout (m2 _ ml) or the dimensionless form 112 m1=ﬂt(l_ﬂt)‘  V (3) q. ‘4 ~....4 _..__ tknmu .w...‘ Mw‘M—w._—;—'J—A—A km: 7. , Next, we note that thevolume of the container is ﬁxed,
V=m1V1=m2vz (4) when v2 = viz + XZVfg’Z. What follows from eq. (4)1 is a relationship between Dal/mg and X2: V V
ELI. = 1’. = E + XZ _fg£
m2 V1 V1 V1
: 0.001154 + x 0.13064 (5)
0.03944 2 0.03944
= 0.02926 + )12 (3.312) A similar relation exists between mm; and x2, . . $ I i ‘ u 11f ,2
1E = '—‘f’2 + X2 “g = 843.16 M 1751.3 . (6)
2597.1 22597. ' _= 0.3246 + x2 (0.6743) Substituting eqs. (5, 6) into eq. (3) yields an equation for x2 alone, Whose solution is X2 = 0.805. (b) The finaléilapor/ liquid volume ratio is (Kgl inn”) = x2 (3)
sz me£2 1"‘2 sz _ 0.805 0.13177 = ' 1—0.805 (0.001154) 471'4 1—8 Chapter 2 _ THESECONDIAWIOFTHERMODYNANHCS
Problem 2.1 With reference to system A sketched below, assume that W>0 and Q2>0‘ The ﬁrst law for one cycle completed by A is
m+@:W ' (n Investigating the possible signs of Q1 and Q2 we see three options:
‘ Q1 <0 andQ2<0
(ii) Q1>OandQ2>O $9 (hCh<0 Option (i) is ruled out by the first law (1) and the assumption that W is positive. Option (ii) is a
Violation of the Kelvin—Planck statement (2.2). In order to see this violation consider system B
which executes one complete cycle while communicating with (T1) such that QB=Qi Since the. net heat transfer interaction experienced by (T1) is zero, Q + QB :: O, the (T1) reservoir
completes a cycle at the end of the cycles executed by A and B. The aggregate system [A + B +
(T1)]‘ also executes a complete cycle. This cycle is executed while making contact with (T2) only. The net heat transfer interaction of this cycle is positive Q2>O which is a clear violation of eq. (2.2). In conclusion the only option possible is (iii): QlQZ < O. in mm .L Problem 2.2 With reference to the system A shown in the preceding figure, we write the first law
for one cycle Q1+Q2=W (1) and assume this time that W is negative, W<O There are three options to consider (1) Q1<0 and Q2<O v‘ r), (ii) Q1>O and Q2>O ’
(iii) Q1Q2<0 of which only option (ii) can be ruled out, because it Violatesthe first law. Option (i) is definitely
compatible with the Sign of eq. (2.27), ‘ ’ Q1 Q2 ~—+—<0
T1 T2 (2.27) Option (iii), in which Q2 is the negative of the two heat transfer interactions, produces an analysis
identical to the segment contained between eqs. (2.11) and (2.27) in the text. The second law (2.27) is valid therefore for W < O and, as showu in the text for W > O.
In the special case of W = 0 the first law requires that Q1 = Q2. The second law (2.27) 1 reduces to Q1(T1  T2) > 0 which means that
(a) if Q1 is positive, then (T1 ~ T2) cannot be negative, or (b) if Q21 is negative, then (T1  T 2) cannot be positive In less abstract terms, (a): and (b) mean that in the absence of work transfer the heat transfer :
. interaction Q1 cannot proceed in the direction of higher temperatures. v » r W 2—2 Problem 2.3 According to the problem statement, it is being assumed that m paths (1 — 2m, and
1 — 2’rev) can be traveled in both directions (see sketch below). The two paths are reversible and
adiabatic. This assumption allows us to execute the cycle 1 — Z’m, — 2m, — 1 in two ways: (i) clockwise, in which §5Q=Q21 —U2V*‘U21 <0 rev ” 2rev re (ii) counterclockwise, in which = U21  U2 >0 IBV I'BV fistQz 2' rev _ rev Note, however, that the counterclockwise option violates the Kelvin—Planck statement of the
second law. This means that the original assumption on which options (i) and (ii) are based is
false, i.e. that two reversible and adiabatic paths cannot intersect at state I. Is state 2m, unique on the V = V2 line? Worth noting is that options (i) and (ii) are both compatible with the KelvinPlanck statement in the case where state Z’rev (or, for the matter, any
other state Z’IBV on the V = V2 line) coincides with state Zrev. In this case the reading of the cycle
goes as follows: (i) clockwise §5Q=QZ 2 z 0 rev _ rev (ii) counterclockwise i : Q2rev" 2[rev : 0 Geometrically, this second law compatible limit means that state 2w, is unique, i.e. that there is I
only one state at V = V2 that can be reached reversibly and adiabatically from state 1. ...
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 Spring '08
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