Homework 3 Solutions F11

Homework 3 Solutions F11 - PEoBLEM “2.93 ENGINEERING...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PEoBLEM “2.93 ENGINEERING MODEL: 1. The gas is a closed system. 2. The system undergoes a polytropic process in which pi!” = constant. 3. Kinetic and potential energy effects are negligible ANALYSIS: The heat transfer can be detennined from the energyr balance nKE+nPE+nU=Q~W Neglecting changes in kinetic energy (nKE = 0) and potential energy (nPE = 0) and solving for heat transfer give Q = nU + W internal energy change is determined floor the given maas and change in specific internal energy at! = m an = {0.2 kg){50 kJIkg) = 10 kJ To solve for Work _ a a {constarxiMV W - f. M = . Integrating and simplifying W: {comment/J” -(cansranr}PiH" 2 {psVJ'UVJ'l'} -(ithV.”}J”.""3 = pi”; -p.V1 1—1.3 1—1.3 1—1.3 The volume at state 1 is determined from PLVII'3=P2V2L3 _ 1 1.3 f m = V2[&] = (0.02 IIISIEJI 3 = 0.053 m3 p, 2 bar Substituting volume to solve for work yields N 3 3 105— W=(8har!fl.filni 1420340053111} m2 311a LM-fiu 1—1.3 lbar 10 N-m The work is negative, denoting energy transfer by work into the gas as it compresses. Heat transfer is determined by substituting internal energy change and work values in the energy equation _ Q=lflkJ+(—l4.6kJ)=—4.6kJ «In—"*— T he heat transfer is negative, denoting energy transfer by heat fi'om the gas as it compresses. PROBLEM 2.?9 5%: A 3‘15 V“d'rj°'€5 5- ‘l'htr'W-Od-JHMA’L cyqi-L flan-th‘sf‘l'nj 0‘)“ Jr‘H-u. Prfld-BSHP- FWD: Sire-ick-f‘H G19”- W‘ 4‘ P'V fl‘fir‘m' C‘I““L“L“ W‘EQLJQIEJCQE'; cud. dekrmvfi. wh‘HI-I-r‘ -Hu (yr—Lt us A. fower or“ rtfrt'jerah'm (you Scmmfl‘nc icwgu mm: L pV = fiansl'nnf l-‘l'. V: 9.013h-EJ 1:13-117 =1b.4-:J 2—3: Pv=ms+nn+J 1132111 3-H [31:45”, N1l=—lfl.5k3' ENQR-lLuOUEL: I- Tim 5a.: Lé—tw cloud S‘ffl'km. 1‘ gr+m rvrkm‘ QKE=QPE :0, 'S_ vol-Juan kaa‘ “Li,” only Wort mdql *flfli'“ “’1' wficlc = Wu +Wu + Wu . SH“... vatume g. mnmn+ “1.1. tram.“ can? us'hu Mir “Ur-‘- mh; le'I-G- Tu fun: Lu” V3 V's V v _ c ~_ 1M1 —. v in; wug‘EFavuffiaxr c: an $33 L ‘9 "1 Lwi‘fi To 9 Valud-c V3 W W31: [pdV—r 96:45.) ~27 v3: U". — Warfp V3 1 J 3 hr V3 -_ o.o13m3_ [FIOA‘SFU)1.'bM‘ “3 'u I“ : OJDE W‘ Q.4’ bar] VFW“; H-J The“ a: 1:21 II: VI”! '5 0.103 1 —_ [EREIEJ ‘34}: 2. lb”- (OJOSMJ 3““ (obi-D) logy-m Wudcle fi'ndli1J -. :-_ _ . 'Tl-u d... * amr mm”: c: + tel-“+8 -r (4" 3") Ylam‘ of“? “’“r‘ a a u {‘3 A“ """J‘y baffl'“ 1C" 1'3 fiiw @131 Mi'QtKE-Wi: {Du-W13 fibflszIJ 325 (d) To 999-1de C133: Efjan Lur‘fk fin cum-av, is». hung h9¢fhgr “(f-i... as: umPh‘n-LZ U.-TJ‘5 ‘: Chill— W3: :17 (23‘ :(‘Lfi—Ua] 4. Lynn. Sinu ‘H’il-I'i' IJ'! M OWALL by; hwfld Quin—3.3 cYu-LJT(¢U_):D LUm-U.\+C1J‘r-Ufl+ LUZ-113}: o a) kU.—Ufi: flunar —— (a) = -2J.’.-¢H-.:I 93: @Tm C9221: [-chr) + [—w-r): — 3w m WW I- A: 6. CLICK) moi-f Jrh-d Gtc~f¢Le=Lucycte 37¢ a“; at +fl haul-mu {or l—~L an”: C9n:2.t..+|c:|. SD (await 103+Lfiglgic-agqj : duct? Jriu Maul-l .12 P‘nctb)_ PRDBL E m 13'; WE: A 1‘” W'+"‘"" “ F”hn"':l"“d“" ant-minty uvuder‘jacr a. +h€fmodjmm‘tcyc1t tunru‘h‘fij 5F {rhymee qu'EIPJ. I-H Jerrer thb; Dik'fh-HHF- all} Wat; and {all _ Dc ‘I-CfMHe ¢.P in: Ll: on“ b".— FDun-ea" cuticle aw mkq. Ft‘Fy’.JflfflfiWlt’-Im. Em E. MQDEL; !- The 3415 u': 1"": dam-d :yrf‘m- 2'. Vfltume change [5 HM "HT {um-IA”- Wile. P w3i=:n.§rc.] 3. gr cad-L racer“ AKE:{1PE= Q. WtfalE-i' «3.3K? pl: Mu up“ ' V5 ‘—' 9.023 m3 V Fig. P237 ANALYSIS: (0.} To {Lma 1r.) Mu +k-n.+ "ww— IIFdV‘ F [V"V3] '3‘? V=W"w—ai .s P xkfluF-Pa. V hr ‘3 IDJSKJ‘ “33mm” iflp‘ .l 1 Chiba“: 4—‘1— Rn: 9-023‘"* W “a mm!" 'L {a} "In: In“; ML; mart WWW-.w-n (“1.3+w‘3“ S‘m'l 'mhlwa'é' ,L. ' ledWC-k'hart‘ eta... (Liar-vi. L; 6%.1.L?~. SUM-Huh T M W . w :0. *M-u due um Mir-L wt (barter: 1-?C WWW W m’h'd)’ 3'7 flu: V3.1: U’cfl‘fl“% 3‘ w.L If? “Thu: 1 i I. Luv—4.. W {wach-q Watt-e14 '75“ I c (aide ’5‘" PW" mild-A H- 1 L»: “ EM “143-; '9“!- 44m; \JJ‘fixbLa. : guf'fh" L‘th can” L. Tu Fund» @L'Ll w'r'k u :5) CQ'széf—JG—t Wu. "U235: =3 @549?” fi—fi PROBLEM ‘2- B? EMOwfl: Si-Mdr—‘Q-l-wk opernz‘l‘mj died-Ia cit-«J. cast da+¢k an: Pray-Ruled {or an rcFr::rOr-+fir. Ewan: Dei-rm'ne HM: Witt a-F clack-'q'fu in 6» man-Pk. whth «HM. A4391] Mad-m- agar-n+0: {w- 3-29 hour; San-amnth 3' awn: bflTfi: encrus-xmme Mops; \ flaw / I. TIM-II {IIIko Jhgm I'm Icl'thIfi-C --; Uudfirjur a. refrljflrqfi'm cydc I l 5-1:ka W l: 1- Energy want-hr: are perifi‘ve in c : 6‘24 4‘“ hrE‘h‘M 9“: 1“” arrows. '- t: 3- m cycle l-Pcrn'i'fll‘ M17 {up G'“ = “WM” 360 km; s... 4-“ parka-{ar- Ecfirutfl-‘lflr \ . . SPA“! W‘H-x under" tenacity-af-m. DUm+GfiI+ if fill¢kabfh15 .41 {\Ich't'fi‘L-h II! VA'V'JII-vl‘t filo-0'9 I! chat p..- haw-k p..- sW-h. G 360 Phoqu 015 opcr‘fi'm ANALYSIS-1 Main? E5.1.'f‘$_ M a... h‘y—J Am Lgni‘tl'i'} @ : @ffl 7:.) whit-It = QM : GOO = '13-}: numb 73— M “r: TMnJ ' $0.03 uw #’ =CWFWHJ<Mfi+h [ m IW _ I: na- unwrir'mr. 1w: 3.48% in «g 1.11/mon-rh .‘__.W__._. PROBLEM 3.44— Wa-l-Qr' .‘g +La Subs-fa...¢-¢_ m) P : 5 b“; V ‘05 “3&5 735fed‘3; “5“”? =0- T: 53.0% F‘ T 35a:- g: my; “5 f 3 ,. , MSW“: 0.0058- I.0132xm'3 If 'U' .1. 0.32 5 u_ = :41; + 7cm? - My) IT Results 3 56L(Eff-5251(35—q3-b*5‘l.§) = 2:05.? uzr/lag 2195 kJIkg (b) T=a£0'c '11-: 0.05 "13/123 Tablaflr-l; U>123wf 310°: $135k A-‘f; A4 320%: He: 54-911; 44115. bai-wttaq {90 «not 80 but”. ln+earpalaahhj , F=7$W bar = 735'? mph LL= 1016.0 023! L5 :TResuIts; p = 7.356 MPa, u = 2682 kJJ'kg [a] F=18mpa= 180 bar, T: 520‘6 Tabb. 1-1-2 1; = 0.01020 mlfkg 1': : 3:“i2.3 Ala/Lg {LEE-HIE v = 0.0102 maikg h = 3192 kJJ'kg (J‘: T=10”C’U=100m3fk3 TabicA-zj 15:41:41.3 deem. W5; =o_or22.§ bar =1.‘223|«:Pq f' T x = fl = “"3"? '00-‘51‘1—1mooqmo' :. G-Q‘!‘ in: H; + 2: lug = Lilo” maul-411.1) -_ 237; miles] 11‘ IT Resuits; x = 0.94. h : 233"] kakg 1.223 In?“ 2. :01: v : tau run-1&5 [El] P= q MPG. = '40 bar, T= the“: Tami—3 11': ‘flobcorofif =lsn.q‘t gimlf‘c =39 lizm'd 5M: (1 473;“ _ i. Tfiakt’tHEJ don-H12 g-nHrPdG-ILW 1: =1.o73¥xw" 1.07.55an Tsl‘w’c LL = 532 81 586.: i. 3 Tawny; 1;" infirm V {.1230 ma": "" = * 7 $‘i‘.b5 M— 11“: 1~D71Hxlf3 twat a: 58148 = :1 Ema-3 Iao‘t U 1' H The»: af ‘f-oma’a, mo“: --u = 1.1.5“wa msfkj L1 : 071‘“ mifllj IT Reaults v =1.102 x103 maxkg u = 671).? kakg ...
View Full Document

This note was uploaded on 04/03/2012 for the course ENES 232 taught by Professor Hines during the Spring '12 term at Maryland.

Page1 / 8

Homework 3 Solutions F11 - PEoBLEM “2.93 ENGINEERING...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online