HW07soln

HW07soln - ?RoBLEM 3.45 wad-er as +He. Subs-fanf-‘C. (b)...

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Unformatted text preview: ?RoBLEM 3.45 wad-er as +He. Subs-fanf-‘C. (b) p: 2:: metal”: us +13“; 12H: A'3E V; (V (V: =§ T: 2.21% °F p X: y» Vf- ‘ = lb-0.0lb$’3 29.0‘l—O-fllb8’3 =- 0-?‘No a". U‘: OLf-exCus—LLF) v v ': IQGJ‘i-l- 0-?96U0514q5‘4q) _ q0l_z_q {sh/lb ERBJJH—S; T = 228 °F, x = 0.7962, u = 901.3 Btu/lb (L) T: 9oo°Fa P: How-HM?- p I7OIU‘HI'12' Table A-H-E J in+erpolnk qwoF avi- QDO°F P v: 4.7-“ #3115 _- 1418.0; BNIb IT Results v = 4.718 ft3IIb h = 1478 Btu/lb (a) T: (90006 v:o‘e.PIJ//b Tweak-2.6, V‘s-V3 at «00°F. g Table A-‘l-E . Ai- 6°°°F «a Sim-Ir Falls he’ll-'99" Boo and. Q00 lb'f'ln'hl. TM-i-OI-PIA‘HV‘} ’ p: 395. a, “Win1 u: “633” Bh/lb 0 I- 3 La) T=4o I5 V-t‘15‘0% Tame A915 V4v1-V our 40°F. 'I'kusfp': N i? “"6 x, \f-Vf: used-0'5 p _ Vfi-Vg 2w)-—a.OIb :0. 738 h: h; +1: VLQ 3 39:2 4 o~ ?(‘°"‘"“) = gulp 3W”? V ITVRVBSUHS; p = 0.1217 Ibf/in.2,_ X = 0.7975, 1'] = 861.9 Btu/lb PRoBLEM 3.4-3" ( Confi'nued’) (h) In: boo [Id-fin}; T2310°F Table A86 4‘ 6°” “'“ML TgAi-a ‘4'“: .33”: a; Iquia S'Hd'e- 73H: v-domb'c. Ini-or‘ula Hon =soolb‘fliul' T: 3009!: v: -o‘ol?‘ilb 0.0!?37‘7 u: lbyfql 168v 1-: 400 ‘1:ngst 0.08350 ‘4: 3?3-(=8” 311.5? _ d: goo Mac 3009‘: V: DIWI-O‘, U“ lei-1‘8 : . ‘- H.000? v 0 0| L1: 373-7'3‘ Then, ox Gaoibflinl, 310°F‘ V: o.ol7e%F3/“' ur- LSQ-‘TI 5h!“ I T Results v = 0.01766 its/lb u =_ 288.3 Btu/lb PROBLEM 3.525" Cumin-ined (#1 Q P‘s-fur"- ;'\lVDfVEJ' +hr¢e~ Purifi- FIN D: I De‘i'Or‘m'he %“Hd3fi_’v pm” each o-F-Hde ‘Hnr‘ee. qu'f‘soghowmrA—cw‘j‘g SCHEMATlC t_' a; UHF” DATA: ___________________‘ Emma. MODEL: —___________ _ __”_1 I. TMW¢+¢FLV§WPISPUM‘¢IJL1‘ndLA_ :Mhrl ‘ . anemblj Q14“ aloud 515th.“ . L... _. _._j 2' TIN“ (“fl-2‘3? Bccurr 9.41" WEhhT Preasu’re‘ Quasi-ant Wcmre=z°_£‘b 3’ {ML Volum than: g-Hu mfg, war-k Pvt-ode, [Add W=Sho°F+D=aiquv£d KIWC‘HC and eféQ'f'Sd—tg b"* c. '. S‘tu c...» cl: Sufi-Vapor +1 TZ=SOD°F heauj;bl<. - AMALTSIS: Far-each [amass we oust wzfpaxm pmr=v%= pau— cm) and 1-h- fullDMJIHS “day-9:310»; clam-"9Q! «Ewan Encrsgj bowl—amm: C9 ‘WT:AU+1::’: = 4u+P41r (a) Duh fimTfibjel‘ 14-2) 3-3) A -4., “04¢;At I) “I‘VWCTTJ _________________I__ (D |'-"> CL: 3%: ptVa—UT) 4.3 .44.} 18%. = Goffj‘o'mmw'mw") TJMIF‘ mm- : O .003 Bi“: LHEJHJ‘Ib‘GJ 4- Ala“ lb 2 _ 8+!»- Qm 1 aqa.1q—.1?.DQ-t 0.003 1'}? [3 T5. 0 {H b" ma .. ? 3H3!»- wab : 20(Zo.oq—o.olci'3)l 3,615,- 4' T “.10 M 4—-— - ‘ m... (Saab : @3314: -l‘3b.l°i) + 7-4.31— 960 I2. Tl: ab‘z‘. 5-1 m ' ‘0 13M (El—2': _ Q|03‘1—~10'EL.0) 4,- ?.4‘0 = 35 lb \44. PROBLEM 3.66 (Hum-UN: A Plsfinnczlnudcr anemia]? corn-Fmth Qumm‘a, wlu'cku'a Sinai} hea‘l'ml wLLJLI. werecrure v'nm'e: lihegf’? Ludo“- SPc mqea‘c. [Io/UMC. $4141 dab-M 1;: Pr owdaj. FIND: Pratt's—r a; {-kn. gonna-mfg. on min—V— d—A'Q-‘Jr'qm_- fir "PM. amuswa :Vaiuctn. -H~e. Luna—k anal keg-f {Yams-Fng each»: h?/_K:j~_ SCH‘EMRTIC t' GIUSM Dav-4,- ——————_- -___-_———.I Q Enema-u:th Mona..- " The “Minna-{A v3 +k4 CHOSch Syskm. 2. Ear fkanmnvm'fi- («unsure Var-'9: “HM”? “WH‘ sfec—tfi‘c VOIU‘Mf- 3' V°‘”“"e Chou“ at. -'-s +he only warm Mode." ‘ 4:- Kb‘ehr. «ma. pa+eh+hue fauna), cf—ch'fs Can ’24 J'ahbrE-cl ANALYSIS: “Sum-5 5:7 . 3-1-7- w.fk acu'umpb‘m'z) Xi: I Pau- = anew-£41) =(w‘2w147) =Arap.(I—z—a..b—I) M I 1 Dad-I. 'FIM 3 V= U- L‘" 'V =1£2§§ 0.: 01.23343» 2? :o.‘511‘t-"=- (Ewen-:3 ' V’ g H m3 "' ( 733;) Pa Rstblc A-II‘, \rz : 0.353% 0 v.3: 1:5 I LN; s (_(.bar+ moan; a“)! to‘NZu‘Kggggqo- 03,2491: 1:3- : 20.3551 4— m '2. lbar‘ "5 ‘° W" ‘37 Pm Chtrsn bdance Niece; M “Fallows: Air-t-gwé-i—QPE‘: @"W é g --' —%- "i‘ (“I‘d-4:) ha I I ¥ ~ 215—47.. LTechA-dli) Lu: u4+ntu5-u;) : 325401-051 uzqqggfgacw) (,qsg c i ‘41-; nut-*1 .22. (Rb-'rA-lr) co _. zen-E! + ape-1.1L—ms.fz_)5r Tu" ' =5 F; = Ci?“S‘-?5'El .4__.____._ ‘5 PROBLEM 343 Known: Ammonia in a piston~cylinder assembly undergoes two processes in series; the first at constant pressure and the second at constant volume. Data are known at each of the end states. Find: Determine (a) the heat transfer and work for Process 1-2 and (b) the heat transfer for Process 2-3, all in Btu. Schematic and Given Data: State I m = [.2 lb Ammonia p, = 100 lbfiin.2 XI = l Process 1*» 2 Constantp process T (01:) Cooling process State 2 P2 =P1 x2 = 0.75 Process 2—+ 3 Constant v process ‘00 lbmnz ________ _, Heating process State 3 v3 2 W V (l't3/lb) I3 = 2.2187 2.9497 Engineering Model: (1) The ammonia is a closed system. (2) The processes occur at constant pressure and constant volume, respectively. (3) Volume change is the only work mode. (4) Kinetic and potential energy effects are negligible. Analysis: The energy transfers are found using energy balances. First, fix each of the end states. State 1: p; = 100 [bf/in.2 and x1 = 1, using Table A-14E: PROBLE M 333 (Confi‘wued) ——..___,__,_____—__. u, = u, = 57121131111115 v, = v, = 2.9497 ft3/lb State 2: p2 : p1 = 100 ibf/in.l and x2 = 0.75, using Table A—14E: u2 = uf + x; (u, — 11f): 103.87 Btu/lb + 075(57121 — 103.87) Btu/lb = 454.38 Btu/lb v; = w + x; (123 — v1) : 0.02584 113/11; + 075(29497 — 0.02584) a3/1b = 2.2187 ft3/lb State 3: V3 = v; and x3 = 1, using Table A—14E, the value of V3 = vg = 2.2187 ft3/lb exists between 130 and 140 lbf/in.2 Using linear interpolation, the corresponding :43: 573.61 Btu/lb. (a) The energy balance for Process 1 to 2 reduces to: AKE+APE+m(u, —u,)=Ql2 —W,, Q12=m(”2_u1)+W12 (1) The work for Process 1 to 2 can be determined using Eq. 2.17: Wu =mjpdv=mp(v2 —v.)= 3 i 2 W12 =1.21b[100£](2.2187—2.9497)§L 144‘? fl = —16.24 Btu 1n. lb 1ft 77‘8ft - lbf The heat transfer for Process 1 to 2 can now be determined using Eq. (1). Btu Q,, = m(u2 — u, )4 W =1.21b(454.38 — 571.21)F + — 16.24 Btu) = @ Q12 2 —140.2 Btu + (— 16.24)Btu = —156.44 Btu (b) The energy balance for Process 2 to 3 reduces to: @ Q23 = m(u3 — 11,) = 1.2 lb(573.61—454.38)B;—;u = 143.08 Btu 1. The negative sign for W12 denotes energy transfer into the system, as expected during compression. 2. The negative sign for Q; denotes energy rejected from the system by heat transfer, as expected during cooling. 3. The positive sign for Q23 denotes energy added to the system through heat transfer, as expected during heating. ...
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HW07soln - ?RoBLEM 3.45 wad-er as +He. Subs-fanf-‘C. (b)...

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