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homework_A4 - Wm M 3 HR jC S—R%’ 5(5556" New...

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Unformatted text preview: Wm. M 8222272 75 04 . _ 3 HR jC; S—R %’ 5/; (5556"?) New gimme %aQ mfflcs Y 1’3“? '1“; :42: {ma/e. baa/W gm) . Y2 [-7 (ix/222K , CF? ’32 iv 23 wafi“? w? PAV 2w112 ”€3A7’ ‘LCP22gfi . Q: __npc,,/_\T +PAV chj+aAAf2 ”gym“ ,. 5) Cami V, 3..., , ., W; O .mgmmmwmmmwm-m “QAT AUMLOE 03.. Cam-7L F (11‘ 1: nag/AH PAY , . -22... at.) C2251? V.-- ‘ .3; mc.¢../25T.. 3... CV.'?"...5/§2.‘?:-.. .. 2 _'.:.E. CF37- So 5222222227“ v 222,? A2 2.714% 5 22 :L) [(21223 62:22ch 7:) we, comm 772+ W Same, Q TED W mam out,“ erhW {ML .237 ( gm ~< cf)“ <22 ATM > A7“) ,, VL‘NIWS UAW»! whim? PU; “RT— ; (2%!“— 0212222542,)5“ “MES 22 . CL gags?» A? {xv W}MT\/ .522. ”éffiahe’... APGL , .1142. 5m __.§:..__ _ Ne: .. 1W: WI..I?Tmaw 2% 2.3.9323]? (5295’ _...._____ .. A ”Y5? .. . .. Ha“ .. .. 1.1.5. V9.5 . 1C6. . Yes. WE»? 990360 >429 fls’ V99 0 (9&3: 773’. . 5‘00 . . v.62)... Mal/3 M . b ,. 5%): h w :a. 444 4M f’l/efe»..- f!” Ems. (5942,... you W..1e9 1/666 26" 029W ”(mg mg Wmfimw ".7192. [2575 6559mm (:ch 671% & M/e, ézcé {23 def (Eye-aw . c9952, 5% am 2%.. m P2032 W = —PB (VC — VB ) = —3.00 atrn(0.400 — 0.090 0) m3 = -942 k] AIS-mt = Q + W em 5— —Em B = (100 94. 2) kj Em C— Em B- = 5.79 k} Since T is constant, Ewen — Em, c = 0 FIG. P2032 Wm z .130;A — VD) = 4.00 attn{0.200 w 1.20) m3 = +1.01 k] Eintr A 43““, D = —150 kI+(+101 k1}: ——48.7 k} Now, Eint. a — Em, A = _[(Eint, C ‘“ E'mt, B) 4' (Em, D “ Eint, c ) + (Em, A — 59m, D )1 EM 3 — Em A = —[5.79 k3+ 0 — 48.7 k1] = P2038 (a) The work done during each step of the cycle equate the negative of the area under that segment of the PV curve. L B 3 - _ _ WT~WDA +WAE +WBC +WCD W=—H(%—3‘4)+0—39(314»~V.-)+0=- (b) The initial and final values of T for the system are equal. H — _ A1 ID Therefore, AEW=0 andQ=—W:. 0 30 m) c W:w4P-V-=—4 RT-=—41.00 8.314 273 = —9.(}8k () . . n . ( )( x ) Hams P2059 W 2W2}: +WBC +ch +WDA H C D A W=—deV—jpdv:jpdv—j:d:v A V w — —nRT11nEVB] 1320/C — VB ) 11er V1n[— :2 ]— 191 (VA — VB) 1 C Now PIVA = PZVB and P2Vc = PIVD, so oniy the togarithmic terms do not cancel out VB: 13—1 and-1L =fl V1 P 2 Vc P1 P ZW=~nRT11n[P—1—]— 11R? Zin[— i—2 J: +nRT1 ln[w1:2-}- nRT21n[%]~—* —nR(T2 — T1)ln[£] A150, 2 Moreover PIT/2 2 11121"; and PIV1 = nR'F1 ZW= 4w —V1)in[lI:—2] 1 P21.14 P21 .30 (a) (b) (d) (e v nRT The piston'moves to keep pressure constant. Since V = P , then nRAT AV = for a constant pressure process. anCPAT=n(CV +R)AT so Ar=..—Q——-=—Q—2£ 11(CV + R) n{5R/2+R) 711R and AV “132] ZQ_EQV P 7nR hfifinm“ [4.40 x 103 ])(5.00 L) AV s 3 = 2.52 L 7 (1.00 mol)(8.314 j/mol-K)(300 K) Thus, VI = Vi + AV = 5.00 L + 2.52 E. z 7.52 I. See the diagram at the right. P 19ng : PCVE 3P. —~ 5 MW? = EVE Vc = 31W; = 35/714 = 2.19%. PBVB = nRTa = 3PM = 3:an '- T5 =- Pi —— A . After one whole cycle, T A = V‘ Adiabatic FIG. P2130 5 in AB, (2,.B = nCVAT = ”[3 R)(3Ti — in.) = (5.00)nRI} QEC = G as this process is abiabafie PCVC = :1ch = M21914) = 2.19nRTi so TC = 2.191",- QCA = nCPAT = :{g ij — 2.19%) = 4.1711131",- For the Whole cycle, QABCA = QAB + QBC + QCA = (5.00 "’ 4.1?)TIRT1' : 0.830711%]? (Agint)ABCA = 0 : QABCA + WABCA WABCA 3 _QABCA = —0.830?’1RT; = _0'830Pi'VI . -W‘ pip} (1.013 ><1Cl5 Pa)(2.45 x10“4 fl : —~— : m3) . RT, (8.314 I/mol -K)(300 K} n = 9.97x10—3 moi Adiabati ' ‘ c compresswn. P; = 101.3 kPa + 800 kPa = 901.3 kPa r _ . ('3) RV.- ~9fo —.__, P ‘1/7 V; : REL = 2,45x104 1113(191-3 5/”; f 901.3 w = ‘3 5 (9.9% 10 moi)§(8.314 I/mol ‘K)(560 — 300) K Now imagine this energy being shared with the inner wall as the gas is held at constant volume. The pump wall has outer diameter 25.0 mm+ 2.00 mm+ 2.00 min = 29.0 mm, and volume w... 2 2 [2r(14.5><10‘3 m) —;:(12.5 x104 m) J4.00x10"7‘ m=6.79x10’6 m3 and mass pV = [7.86 x 103 kg/m3)(6.79 x 104’ m3] = 53.3 g The overail warming process is described by 53.9 I: chAr + chT 53.9 1=(9.97 x 10—3 mol)§{8.314 I/mol-KXTE — 300 K) +(53.3 x 10*"3 kg)(448 J/kg -K)(Tfi — 300 K) 53.9 I: (0.207 E/K + 23.9 9mm. — 300 K) if, _300 K: 2.24}: P21.67 (a) (b) (C) (d) (f) _ pV ~ (1013 x 105 Pa)(5.00 210-3 m3) — — = 0.203 1 3 B " RT (8.314 J/m61.1<)(3001<) r 6 =5 .5. 566%} 900K ‘ B A P 1.00 A c I A TC = TB = 900 K 0 ‘ V0) T 900 0 5 10 15 V =V —C =5.00 14—]: 15.0 L C ‘4 T A j 300 FIG. P21.67 3 3 Em A zEnRTA = 3(0203 m61)(8.314 J/mol - K)(300 K) = 3 3 Em B = Em C = 561113 = 5(0203 m61)(8.314 I/mol-K)(900 K): 2.28 k} P (am) V(L) T(I<) Em (kl) A 1.00 5.00 300 0.760 B 3.00 5.00 900 2.28 C 1.00 15.00 900 2.28 For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. For BC, keep the sample in the oven while gradually letting the gas expand to lift a load on the piston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gas cool without touching the piston. For AB: w = E] 115m. = Em, B — 151nm = (2.28 — 0.760) k] = For BC: Afimt = E w z MRI}; 1%?) ' H w = —(0.203 m61}(8.314 J/mol -1<)(900 K)1n(3.00) 5 Q = ABM — w = For CA: 85m. = 8m” — Em C = (0.760 — 2.28) k} = w = —PAV = —nRAT = —(0.203 moi)(8.314 I/mol -K)(—6DO K) = Q = AEW .. w = -152 k] —1.01 k] z —2.53 10 We add the amounts of energy for each process to find them for the whole cycle. QABCA = +1.52 10+ 1.67 k}— 2.53 k] = 0.656 10 WAR“ = 0 —1.67 k]+1.01 k]: (AEim hm. = +1.52 k] + 0 —1.52 k] = [E *P22.5 (a) The input energy each hour is (7.39 x 103 I/revolution)(2 500 rev/min] 60 mm = 1.18 x 109 I/h 1 h 1 L im 1' fuelin ut 1.18 109 h u—__ 3 29.4 L h M P 1 x v1.9.1.7.) (b) Q; = Weng +IQc I. For a continuous—transfer process we may divide by time to have 91 = was .121 At At At W Useful power output = ig— = 9’1. _ @ At At At $1.38 x 105 w _ 7.89x1o3 ]_ 4.58x103} 2500 rev 1 min revolution revolution 1 min 60 s — 1 hp agng = 1.38 x 10D WI746 WI = 185 hp agng 1.38 >< 105 1/5 I 1 rev I C LI}? : :1) : = -————»~—-— = 527 N m ( ) ens m r a) (2 500 rev/6O s) Zarad 3 (d) Ich :M @931]: 1.91 x105 W At revoluton 50 S P2122 (a) 0?) First, consider the adiabatic process D -~> A: . r 5/3 PDV5 = PAW 50 PD =PA fl :1 400 kPa[1G'O L] = 712 kpa D 15.0L , Also 11121]; V5 z nRTA v; VD VA V H 10.0 2’3 T 2T —A =720K— 2 549K. or o M [m] D Now, consider the isothermal process C —> D:- Tc : To = 549 K . V }' swag]: p414} {Lo} Pat}; _ 1 400 kPa(10.D If” P = 445 kPa C , 14014150102!”3 N ext, consider the adiabatic process B —> C : PEI/g : PCVE . PAW; . . . VA But, PC = __1 from above. Aiso consuiermg the 1sothermai process, PB 2 PA —~ . V545 VB V P V7r . V V 10.0 LOAD L) Hence, 13417:ng =[WJV5 wh1ch reduces to V3 = g}; =—15fi— = 15.0 L . Finall P —P E w14001<13a[10'0L)— 875 kPa Y’ 5 A V5 16.0 L " ' State COUCH): A544 = nCV AT = E VB 16.0 so Q m 4w = nRTln(V—] = 2.34 moi{8.314 I/mol - K){72{) K}1n[m] = _. For the isothermal process A —> B: A For the adiabatic process B v» C : Q = E 3 7 .4 A544 2 HQ; (TC — T3) = 2.34 mol[ (8:314 I/mol -K)](549 — 720) K = W and W = —Q + ASEim 20 +(—4.98 RI) = 4.98 k] . For the isothermal process C —> D: AE-mt = nCVAT = E and Q = —w = fiRT1n[:—:] = 2.34 moi(8.314 I/moi - K)(549 K)h1[%] = $02 k} . Finally, for the adiabatic process D —> A: Q = [E Ash. = no, (TA — TD) = 2.34 moi[§—(8.314 J/mol - K)]{720 — 549) K = +4.98 k] anciW=—Q+AEW =U+4.98 14:- Process L The work done by the engine is the negative of the work input. The output work Wen‘g is . 4 given by the work column in the table with all signs reversed. W... —w 1 56 k} (c) e = —5 = ACE = '———— = 0.237 or 23.7% in <2.-. 6.5%; - r: _ 549 _ EC —1—ji:;—3—;;"O“—0.237 01' 23.7% .1 P2226 COP = O-IGOCOPCarx-iotcycle or E = [1100(fl) = 010% 1 W gamut Cycle Carnot efficiency w lth [ T J __ = 0100 h = 0-100[ 293 K _ W 293 K — 268 K "1‘17 FIG. 1322.26 Thus, 1.17 joules of energy enter the room by heat for each joule of work done. P2257 (a) For an isothermal process, Q = nRTln[—:—:‘—J Eggsirgsai §herefore, Q1 2 nR(3T1-)1n 2. ® 3T: and Q3 = nR(T,-)ln[%] @ For the constant voiurne processes, Q2 = AIS-mL 2 2 3—an — 312-) 2 and 4 = AEint 4 33711387} ‘71) I ' 2 v 2V V The net energy by heat h'ansferred is then ' ’ Q = Q1 "’ Q2 + Q3 + Q4 FIG. P2257 or Q = MRI} in 2 . (b) A positive vaiue for heat represents energy transferred into the system. Therefore, [th = Q1 + Q4 = 324127;. {1 + 1112) Since the change in temperature for the complete cycle is zero, sew = 0 and weng = Q W Therefore, the efficiency is .3G = wig—FT = IEQT 2 fl)— = h h n ...
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