Answers Modell

8/8/2019 Answers Modell

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Answers to Selected Problems

3.1 (a) 268.8 K and 4.625 105 Pa; (b) at 90o (horizontal) 282.6 K and 4.863 105 Pa; (c) at 180o(vertical downward) 296.4 K and 5.1 105 Pa.

3.2 (a) dQ/ dt = 3.51 108 J/hr [contractor pays](b)dQ/ dt = 3.23108 J/hr [contractor collects fee]3.5(a) LA = 0.4352 m, LB = 0.0218 m, TA = TB = 311 K, PA = PB = 7 bar(b)

Same as (a)(c) There is no real solution given the constraints, we must specify more aboutthe expansion(or compression) of A or B. For example, if A expands adiabatically, PA = PB = 6.68 bar,TA = 263.6 K, TB = 881.3 K, LA = 0.3917 m, and LB = 0.0655 m.3.7Many variations are possible. If we assume gas in the cylinder is ideal and thatit expandsadiabatically, then (a) total time t = 0.9 s, (b) height h(max) = 4.78 m, by reducing tubelength to 1.64 m.3.12 (a) After 6 s, T = 349.5 K, P = 1.139

105 Pa in the bulge.(b) After 3 s, T = 454.6 K, P = 6.26103 Pa in the large tank.4.1 For the minimum condition TA(final) = 132 K and for the maximum condition TA(final) =617 K, these extrema require using the object with the lowest mass x heat capacity product.4.2W = 9.807y in J/kg evaporated, for typical values of ambient dry bulb and wet bulbtemperatures of 300 K and 280 K respectively, y = 16.8 km!4.3 (a) n

B = 0.166 mol/s, nC = 0.833 mol/s(b) TD = 299.1K(c) W = !26.4 J/s(d) .S= 13.3 J/K(e) Same as (d).(f) W = !4000 J/smax

4.4W = -1.11 104 J, Tf = 288.7 K, Pf = 0.54 bar4.6 (a) 21.6 s

(b) 391.8 K and 2.2 bar at 10 s(c) .Sgas= 4.93 103 J/K = .Suniverse


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(d) .S(gas in tank) = !813.6 J/K.S(gas vented) = +2020 J/K.S(surroundings) = +1203 J/K(e)Wmin = +2.83105 JUpdated: 1/14/04


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4.7 (a) Q..= 2400 W, T = 327.7 K, and Q = 22,200 J after 10 s(b) T = 1750 K at t = 57.7 s with all stored work consumed.4.9 Wnet = -3.17 106 J4.12 (a) T2 = 227.9 K, T1 = 392.9 K (2) after venting (1) before venting(b) T2 = 211.1 K, T1= 363.9 K

(c) W = 2547 J/mol(d) Wmax = 1.38105 J/kg4.15 (a) .S = -1.38 103 J/K hr; (b) .S = 2 J/K hr4.16 (a) Case (1) 1.9810-2 kWhr; (2) 2.78 10-2 kWhr; (3) 2.1610-2 kWhr;(4) 2.78 10-2 kWhr(b) Case (1) 420 K, 42 J/K; (2) 300 K, 0 J/K; (3) 300 K, 73.9 J/K; (4) 300 K, 0J/K4.18 W = 5.23 103 J = !Q4.22 (a) P = 6.35104 Pa(b) T = 272.8 K(c) P = 3.68104 Pa with 2 pumps(d) P = 1.91104 Pa and T .8

(e) Power (minimum) = 20.9 W4.24 (a) W = !198.76 kJ/kg (Carnot + expansion work)maxWp = ideal pump work (estimated to include PE for lifting water and for thecompression of gas space in storage tanks)

4.30 (a) W= 6.38106 J/smax

(b) .u= 0.90 = (actual power)/(maximum power)(c) no, the maximum power outputs are the samen

(n+1)()0 (n+1)

5.2 (a) y =y -TS -SjNj dy =-SdT -SN jdj -PdVj=1n

()1 ()

1

(b) y =S -(TU dy =-(T +/ )-S/ )1/ )Ud 1/ )(P TdV (T dN

jjj=1


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5.4 (a) (.P / .T )V ,N i(b) (.P / .T )T,N =GP +T (-GPT )=V -T (.V / .T )PN,

(c) U(d) .H / .V T,N = V / G -TG / Gbg

PP PTPP

Updated: 1/14/04


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5.7(a) (.S / .T )G,N =-S(.V / .T )P /V +CP / T(b)(.A / .G)T,N =-P(.V / .T )T /V= -(.lnV / .lnP)T5.10Tf = 1087 K

()21

5.12y22 =-NCv / T =-NC p / T +Vap / .T5.21Vc = .RT = (Cp/Cv) RT 330 m/s (use mass units for R with molecular weight)distance at 2 s 660 m5.22(a) Cp -Cv =Va2p / .T

5.23(b) No, the ratio as shown in Problem 5.17 is expressed in terms of PVT propertiesand an isentropic derivative which in turn requires non-PVT property informationspecifically related to the temperature dependence of energy stored in U.()

5.24(a) dy 0 =-SdT -PdV +dN +dN +F dZ11 221

()

(b)y 0 =-PV +1N1 +2 N2 +F1 Z1()(c)dy 5 =+Td S +VdP -Nd-Nd+Z dF .011 2211

(d)(.F / .)=-(.N / .Z )

11T ,V ,N ,Z 11 ,T ,V ,N

2 12()** ** 3

12

(e)y =[T (.C / .T )-T ]/(C )

SSSVV


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()5()

5.25(a) y =0 dy 5 =-SdT +VdP -N1d1 -N 2 d2 -VBdH(b)(.1/ .B)T ,V ,H ,N =V (.H/ .N1)T ,V ,2,N2

2

(c)Yes, using the Gibbs-Duhem relationshipx

d=-d=f ()

2 -2

=.2 .11 1

(1 -x )

1

VLV VLV

6.2dA =-P -P dV +

-dN =0()()

VLVL

thus, P=P and=

at equilibrium

6.3T =T /(1 +a)and a=f (T ) from chemical equilibrium criteria, intersections definei

equilibrium states.

Updated: 1/14/04


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()I ()III

II ()

6.4

(a) T =T =T()II ()III

P =P()I ()

II

=

HH

22

Svjj =0

6.5 (a) V =V =RT /10512

=

A,1A,2

(b) T = 366.5 K, PA,1 = PA,2 = 2 bar, PB,2 = 1 barNB,1 = 0 NB,2 = 1 mole NA,1 + NA,2 = 4 moles(c) P1 = 2 bar, N1 = 5 moles; V1 = 2.5 RT/105 = 0.0752 m3 with T = 367 K and nogas onside 2.6.7Fraction evaporated e=0.227Q .=136 e(

)=U in J.S =e[0.581 -0.462 (6 10-2 ) ln (4.40 e)].A =-5.63 e[1 -ln (4.40 e)]

7.1if ANN = 0 then AVV = 0 = ANN /V 27.3T = 55.1 K7.6for a= 2.5

xA xB

0.1 0.286, 0.614


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0.2 0.4 (critical pt.)0.3 no solution0.4 0.2, 0.4 (critical pt.)0.5 0.175, 0.3250.6 0.085, 0.287.11 .G =221.3J / molat10Cor =-217.8 J/mol at -10CUpdated: 1/14/04


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8.4At 365.8 K, 16.5 bar-12-52

(.S / .P)=[C / T +(.P / .T )(.V / .P)]/(.P / .T )=3.33 10 J/mol K (N/m )

Vp VTV

8.5

At T = 319.4 KPr 6H/6S Z = PV/RT0.1 0.1000.9670.2 0.1940.9320.4 0.3690.8621.0 0.7980.5984.0 0.4260.54710.0 -0.296

1.1088.9Using an isenthalpic expansion across an insulated value to 1 bar, 0.236 kg of dry icecould be produced per kg of CO2 drawn from the cylinder. For an isentropic expansion,0.41 kg dry ice/kg of CO2 is possible.8.10(a) Tf = 379.3 K at 48.3 bar(b) W = -9.1 107 J8.14

TB,f256 to 257 K, PB,f

21.5 bar, xg = 0.63 (fraction vapor)

8.15(a) Power = 508 kW(b) Power = 516 kW8.30W..= 90.9 kW and Q..= 68.3 kW8.31Yes, the vdW EOS gives at the Zeno condition (Z = 1) a straight line:Tr = T/Tc = 27/8 - (9/8) .r

9.1(a) -31.36 kJ/mole water added to acid at the start; !61.1 kJ/mole and added towater


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9.7(dS / dt)universe =[(w,in -w )n..w +(s,in -s )n..s ]/ Tw = water and s = salt (NaCl)

9.8(a) Wmin =-1069 J (PR EOS)(b) Wmin =-1812 J (ideal gas mixture )..

9.9W= 3MW9.11x salt (wt fraction) .(bar) .ID (bar),(completely dissociated)0.01 7.28.5

0.0542.2 43.90.1090.5 91.10.15156.6 142.00.20242.0 197.00.25355.0 257.09.13.=Q =

1624 JH mix.Smix =12 34 J/K

.

9.14(a) .=1.18 .=4.17 at 0.8 =x

EtOH MeOHEtOH

(b) At !85oC (188 K), there are two phases in equilibrium at xEtOH = 0.07 and 0.35.9.20dQ / dt = 675 WdWmin / dt =108 W9.21(a) W= 1.32 105 W

(b) See Section 14.4 for details, information provided in part (a) is not sufficient, idealgas state heat capacities and a mixture PVTN EOS are needed.


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9.27 .Wmax / .t =676 kW while Gyro claims -400 kW which gives .u =0.623 not 0.95H!Ho

10.2 = PV !

RT !a/VD

10.3r =2A Fij,max / kT =20.43r =3 Fij,max / kT =

6.05r =5 Fij,max / kT =1.31r =10 Fij,max / kT =0.1610.4=0i, j

Updated: 1/14/04


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D10.6 , Ar()H OH O 22 -F(CO CO)-F(10 21-J) (10 21-

J)2 288.42-2.52-10 0.0185-0.000164-

10.7 b = 4RT /( 30P ) =4V / 30c cca = 1+ 4 RT V 30

cc

2 1/2 2 1/2

10.9 (a) []/ .=(1 -b.)/V [.(1 -2a.(1 -b.) )](b) .=1/(3b) and =27b /(8a)cc

(c) for T/Tc = 1.00001 .I / I 9.29 10-5o

10.10 f=0.245CO2

11.4 (a) .12 =0.0961, .21 =

1.0273,=-0.8381(b) .8=-10.13, .8=2.4112

(c) for only positive values of .ij , a large number of values [.ij , .ji ] exist as afunction of xi that yield extrema, e.g.

for x1 = 0.2 .12 .21

0 3.2360.500 1.699


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1.000 1.0001.500 0.5862.000 0.2092.118 011.5 for x=x =0.5 a=

0.42761 2 maxfor x =0.1, x =-0.9 a=0.6177

1 2 max

11.7 (c) fraction of solution precipitated = 0.006711.8 for 2-suffix Margules Tc = w/2 k

11.9 for quasi-chemical model Tc = w/2.23 kUpdated: 1/14/04


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12.3 (a) selected values of .are given below for LiClmolality .(Meissner) .(Pitzer)

0.1 0.9651 0.96530.01 0.9032 0.90440.1 0.7850 0.79271.0 0.7636 0.77585.0 1.9586 2.022210.0 7.9948 11.0827** ..

(b) s (s +10).=exp[-.G rx / RT ] =

f (T only)given that s = 9 mol/kg for pure LiCl in H2O. Using the Meissner model for .predictions for both pure and mixed electrolyte.Ksp (9)2 (~6)2 = 2916 for pure LiCl at 25oC2 **

.(HCl(10m) +LiCl(s )(s +10) =2916Iterating s *=2.4 and .=

10 3H for mixed LiCl and HCl at 25oC

(c) Difficult to say without more information on how much Na2SO4 influences . forLiCl.For example is the effect on . large enough to compensate for the Cl! common ioneffect?

o

12.8 Some representative values for .

follow for T=25 C, D =78.5s

.-1 .-1

molality , nm (NaCl or LiCl) , nm (CuCl2)

0.1 0.96 0.551.0 0.30 0.1810.0 0.096 0.055At T = 300 oC Ds

22.3 so multiply above values by (22.3 573)/(78.5)(298) = 0.74

12.10 I = 0.546 or m = I/3 = 0.182 molal


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13.2 Using Jobacks method with Tb = 225.5K (exp. value)Tc = 364 K, Pc = 46.7 bar, and .= 0.1606Cpo = !2.83 + 0.269 T + 3.15 10!4 T 2 + 4.2 10!8 T 313.3 Tc = 663.5 K, Pc = 32.21 bar, Vc = 429.5 cm3/mol, and Tb = 450.38 KUsing Eq. (13-18) for vapor pressure:3.4723)663.57.3106(1ln --=

TPvpUpdated: 1/14/04


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5370 kg/s

mR-115 / mgf =2.87

14.6(a) Wmax =Cp (T2-T1) +ToCp ln (T2/ T1) =.B(b) .net (at T2 =600 K) =0.52(c) Using the RKEOS and a suitable mixing rule, amix and bmix parameters can becalculated and used to calculate all PVTN properties. With an ideal-gas state Cpall

derived properties needed for the cycle calculation can be obtained using a departurefunction approach. Then,wt sat sat

Wnet .t mwf (HB -HC ) -mVl (PA -Pvp (To ))

.

p

sat

.1 -....

pW .cycle ..Vg ..

14.8(a) a* =.0.428 ........cycle ..Hvap ..-3

(b) a*( NH 3 )/a * ( H2O ) =6.4 10 (about150 times smaller!)


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14.12(a) and (b) W=-1.26 106 J =0.35 kWhr so CTIs claim of 1 kWhr productionmax

violates the 2nd Law limit

(c) All heat transfer and work production and utilization steps have some irreversibility inpractical systems.14.14(a) W=-9471 kWmax

(b) no fundamental laws or concepts are violated..

(c) a =Qtotal / .T....oRTmUpdated: 1/14/04


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about 17% more area for heat exchange is required for the GH process

(d) W(GH )/W(Rankine) =1.15 which is less than 22%net net

NB: If an electric motor drive is used for the feed pump in the conventional Rankinesystem, then the net output would be reduced possibly accounting for the discrepancy.

7

15.1(.P/.T)(S-L)=1.35 10 Pa/K15.3x (Naphthalene) = 0.196

15.6(c) q (critical quality) = .TinCv .Hv(e) O2: q = 0.54 and H2: q = 0.3615.12xKCl = 0.413 and T = 624 KVL15.13 H EtOH -H EtOH =.H vap=4.0 104 J/mol15.15Anesthetic pressure of CCl4 4.9

10-3 bar.1/ .15.19 (a) ln .A =f(P, xi ) /(T )(b).A (400K) =1.17EX

(c).SA=

R ln .A ((1 -.)/ .)EX(i)if .SA =0 , solution is regularEX(ii) if .SA >0 , less structural ordering in mixture versus pure state(iii) if .EX For F

ij

iijj

16.9(a) yA = 0.764 yB = 0.236(b) not in equilibrium, estimate yB assuming equilibrium16.12W=4.868 105J/smax

5

16.14W =9.15 10 Jmax

16.16.T =7.95K using expansion valve.T =2.39 K using expansion turbineConsider net energy, entropy, and work flows using 1st and 2nd Law concepts to s

howUpdated: 1/14/04


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process is not feasible as described.

16.20(a) P = 21.9 bar(b) T = 51.5 K(c) .S universe =

2.96 104 J/K16.22T, K Cp (effective), J/gK293 4.69313 7.20333 8.29353 6.36373 3.6417.3(a) F =4 -pwith n =

3and r =1; at A: L1-L2-H-G and at B: L1-H-I-G(b) L1 = 0 and M1 = 0 use P-explicit EOS to determine requiredAijk and Aij derivativeso(H )(L )(G)(c) d ln P / d(1/ T ) =.H / R =(H -17H 1 -H )/ Rrx

(d) for xsmall, T -

T (RT 2/ .H 0)x =5.6 xhexane 00 rx hexane hexane0(H )(L )(L )

.H =H-17 H 2 -H 2 in this case

rx

(V )(V )(L1) (L2)17.6d ln P/d(1/T) = -.Hvap / R where .Hvap H -y1 H -y2 Ha2 aa

18.1P -P e


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(D -1) E / 2, T =T , and =o s18.4W =

9 106J/kg mol18.5Ti = 269 K = !4oC (winter!)18.6Tbath = 317 K, efficiency = 0.06518.8z = depth = 764 m18.9y (tube bottom) = 0.501, if L .8y.4/7He4

He4

18.11a=1.075, P(rim) =4600 bar18.14ytritium = 0.065 and ydeuterium = 0.538 at rim18.16h = 252 m, for an equilibrium ocean, fresh water cannot rise above a level 252 mbelowsea surface. For the well-mixed case, process becomes feasible if Z = depth > 9850 m(32,000 ft.).

Updated: 1/14/04


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19.1 a=2.17EtOH toH O

26 -9

19.2 yethane =0.995, P(bubble) =1.97 10 Pa, r =4.2 10 m419.5 ln Pvp (drop) =-3.93 10 / T +constant19.7 xbutanol = 0.43619.8 (a) 0.046 J/m2(b) 296 K

Updated: 1/14/04

2/22/99JWT


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