sm4_114 - 4.114 Awell-insulated rigid tank of volume 7 ft3...

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Unformatted text preview: 4.114 Awell-insulated rigid tank of volume 7 ft3 initially contains helium at 160°F and 30 lbf/in.2 A valve connected to the tank is opened, and helium is withdrawn slowly until the pressure within the tank drops 41> PL. 2. An electrical resistor inside the tank maintains the temperature at 160°F (a) Determine the mass of helium withdrawn, in lb, and the energy input to the resistor, in Btu, when Pm: \3 (b) Plot the quantities of part (a) versus p ranging from 15 to 30 lbf/in.'2 PROBLEM 4-134“ gnaw”: Heiwm is wi‘ékdrowv‘ siovly rfirom « Qw~mgalnl~ld %»k. 3W wle Wm «km pasture with.» What»; 3», "PM fiwirakrf w.%~.~a {hawkmk {3 hair}- Cent‘kni b3 «51 chva-wt N31473:)”. 5,3313: it] {whys ‘3‘: (315%)“), dgkrwum M MAS: h; hill’M dmdm (may alan “519 «flu rumor, (1:) P1911 wqwnfi’fm ofpuu‘a; W’rms 9 («9‘3th from IFS-{p 30 itrflia} 5: gm “a & 6’95“ BMW: Ewéexnovfl: l- W cm 4m voamg, a: O . is slaw“ .‘mwfiu scywmwht. <2. firm /// m cah‘hml volthn Qch-0 J, and i *e’ fitnefi‘tfpoi‘enhm emergu sj—jrcflw ®é£t¢5+ r (’1’ 3“. zfgl'wtex Z . Ndh'vw y; WMQé' new}? 1““186"F I 62: a.» 4 Kim-J? Sax. 4.‘ M mg: w} {NF/‘0 3230‘th 4M Antiviral“ .‘t that emak’w in gamed, m3“ FMA m,a W MA war“ 0.5 l—‘onc ap Pa. A»mvu:: TM mg: a; “(1mm warm—em 'gwa. “hwy: :nurm.qwo rm _ . 4~Htiam bahu¢cn W m;fo «Mod 7 m}; w; +9... We. M M Hausa—w 4M 4‘ WW P7. — \Kfs «wk 046% M...- ‘iwmc: @muafi' w«’+hdarm)n my» Ma. 3...“ 7;: 14.144». am 3» “may”? 3% W1“ W umwfivk M M {airmgwma pay—.19; AW : a? * gr h _ I :1;- Caf-irhévm W £73: '- (E! P )RT. ( 3) (.I) - m m »» "‘6’ - L "‘ ‘ MW). e @a-PH42M53) lb {7.) Lara ———> mafiosow O ' v ’szL ” maul : 1'7/ ’ WV} i? ‘ML '— fikadme MWPW’Wm CafsW-l’ smocT rs WSW'F so “7.: “i , (MZ'MD (1.0+ wow} = — a,“ I (wk-m7) |A (oer ludF) we\a 3 (miymzwfiw- Wt lQO°F : Me WD’Q/ka-F P5129] VW 7.0m :: me (—- ‘9an / =-w\C RE a = —- .0506 Xbmflm mvolwir‘llab) R lB‘l'W = "l’Z,O=aL"Rl>'¥lr < 713N245 : S‘15.53> BM! 4.114 Awell-insulated rigid tank of volume 7 ft3 initially contains helium at 160°F and 30 lbf/in.2 A valve connected to the tank is opened, and helium is withdrawn slowly until the pressure within the tank drops to p. An electrical resistor inside the tank maintains the temperature at 160°F. (a) Determine the mass of helium withdrawn, in lb, and the energy input to the resistor, in Btu, when p=18psi (b) Plot the quantities of part (a) versus p ranging from 15 to 30 lbf/in.2 PROBLEM 4. “4* Known: Helm-m is wi‘lkdrow». slowly “Ham ax ww-rncufakd th03 CW volum urn-kl -Hu. [Manure WIH-aH Wknhq ’0. TM Run-penal“: chum-o1 'fkq‘l‘nk 1'5 Ktp'i- Cow-shat b: an clack-Int Msl‘SfDr‘. Fmbt “l twink ftrfil‘h—F/m’, dekruum HM. M4,“ a4 lule corn-de annual-HA. {aura} Input) +9 «flu Ntu'sfbr, Us} Pflo‘f +k'fiwuh'f'"f “{Pfli‘“) VP’FMJ P mnjuns frown f5“ #0 30 Ibfh'mi w GIVEN Dam: ENQQJ’HDDEL.’ I- Thu ecu-\th volume. IIS 5:th IlW‘I'N :c-Pwma—i‘p" '2' RV“ Ctran votoM'J chzo J “*4 Y—meh‘q‘fwfenho-fi eneffij ‘5'?th kw. ignore. 2. Haul/mu; Md»! (:1 5,, A” 4 «A Sac. 4—. TM ms: of {*4 Mpfigr is that enoka b-l namedfl ANALYSIS: TM ms: 01‘ helium wad-fnm awa— +‘Mfi (Ah/U11 HUAL 1““ d-Huenu. b4: haunt -{~u. "uh-A «mad 7 M51 withhflt H ‘1'“- M‘L‘“ M {div-k F5444. My 'hV-tel @M‘Ud tun-+11%”): Wt-Mz. Sun-4.- 7:: I’LJ ‘Hp'h ' JM j“ "[IJa-fimvs Shetl- wle-L LAM-J. DNWfl-fi M fittmjwma PlgP' 3m: L-Y._v__lr.~_(e.—ml u) RT’ In W _1 ( 3) : Lawpfl'i’viffl 244:: J strum m it :51 ur'fi- (mm) = (éo-P)(4.2m53) lb 17-3 Th! “‘1 N49 htanu dlM:rlld,f-: aha , Th4 cur}; Am has-«LA...th mm f—o and” = 37:: '6‘“ ““J‘ib‘c- Tnh‘flfinj 44'“ M15“ mzt. ‘91.“;an and mh"*§ {'14-‘11 5‘ rem»? cath blah-14.. hA-pnraN—dJ-o Lamina-d") $3" ‘ ‘ w‘” + he ‘22:" a“? AU‘V = —U-’cv +3“? 0% g 0ch): Mil-l; -h.u’—LW;-W,]lfle I Elm “want-Pure L; Quihu'f/ ulzutTJ,u,-.u{1‘)) L19 2 MC?) +Qv)e_ Tu... © Q—UU‘V) =1 {m,-w;1pt.9e' ~37 LuHLfk- {Eb-«(544 31‘).th SMJfi'w‘V):£mf'mL)R—[ 3155.0] LM.-|ML\ =(P.-P}V!RTJ ‘J-WMQ ('W‘V) =(Pl-PYV- Tax-“3; (WW): {ho-pjflgfgl 'fjfl‘fllw‘)|%b¥ 2 1.236(30-9) an (3) (a) (ulna Flown-.3, 5T“) 5mm (M.w1fk¢rm):O-OSOSK315?‘”) 5”" ‘3’“— (-wcu‘): 155$ RM' (no) PLOTS Pa u—k mkdr‘wn) lb an (“U3”)! EM l5 Pugh-$1) 3O 15' Pabflhg) 30 1. To msnht“ “u. +fw-Pcrg41-er wifluum hat cons-ha} {Wu myth..- Murt farm/rd... can” +0 in. lam-hum w: thk him-l 1'0 +94 4W3? mnl‘rd ad by Hutu- war-r; out {1. ...
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