Homework 3 Solutions F11

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

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

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

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