Assignment_1

Assignment_1 - PRbeLEM L15 The mass o¥ Wakr (6 m = gV ;...

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Unformatted text preview: PRbeLEM L15 The mass o¥ Wakr (6 m = gV ; behave V is He volume o¥ We sphcrlcal ‘walg Thus v-. 53—11% -. (3;)cmc3om3 =l.l31xao‘~c+3 0.th «We. mm: is _ __ 3 m— SV -0024 gamma-Sm 1 moéon“lb¢______________________*EL The weigh/{'15 quv’ m9 _ momma] J? 32.2 [kW/S" =7qum‘we Fmv F’KDfiLEM L55" Pl: ‘5- IL‘F/Ilu" <— V. =3.s‘¥+’ F1: :00 IMM— The Jamssuve-volume weld—(on is Wear carrth +5.; Wow“. Thu: — F2"?! ' "’ V-V, ) F a ( VIM )( (9", 115013 in/evn daft: M (Ibo-IE) rum“ = 7'5 ———- v— .5 3 F .g '“ + (LS-2.5) “3 ( 1 “1‘ = 227.5-85V when V= 2M #3 p = 2115—9512.!) : qu Miami—4 0n f—V wordtkafes R‘- 08 P (lint/"312) w 0 t: O l w L5 1.0 2.5 V ((1?) PROBLEM 1.3 4 M) The Prone” is deter-bed b3 FV: cons-hut, TL. conrfimt “A be. evalunhd using da+A 4+ sun I P "V = £9 WS+ant = PI VI = ( lbar)(l L) = 1 bath). So, {or awn] s+u+< Luv-M3 +M Pumas: Drew +11: “(when FV: [bar'- L When P=3harJ v = ler-L : L '5 3 bar when P: 4baq _ lbar.L __ 0.1SL. 1— 4hr Phfiinj ‘Hxa [clad-fan on [unsure —v¢luwtc Coordinng we but 1 bar. L P: 'V' +0 0 b +m‘w 4 Z 3 4———— ? meav- Pia-stare — Volvmg 00M) 2 re h-Honshlp pf Fad-(b) P'Vgcqm‘aut runHthi P I o‘- pnr| (a) 1 ° V C 1- ) 0.7.5 0-5 0-75 #0 (b) For compan‘san, 1m. linear Wgrurepvoiuwa rehfimiklio is Show” on +1“. phi: above. . Tlu yaw-M; cerr-esrwneknj +0 {3: 3bm can \u ob-HM‘MJ sumth us-nfi +|~n slope HQ +6.... 3%“sz Mac bemeem l mull: m;— nbw: (3d) a, v: 0.5L. 4— .0-0.2? L- U.o-V) This vduc also cam be read from +19. plot. [slope] 1' W PROBLEM 1.43 5 : $8I mls" gn'na : KY) where K: IOJooe Er HUI PIE-Ian: M: lo K5 A: 7.8x10'3 1m?- For-Me m‘r: AV: 3.9x154m3 TM‘k‘Ur I X10 and here I'd be :prm, fur-cc ach‘nJ an '14" fl-é‘fon- AL”; ‘FVI'CHun be+wun. +|~. pain. and H»: cfijmdu- on.“ Cam 6-: I'jnored‘ AccorslinJvJ 'Hne firce excaer b3 fhc arr- leM'n 4"“ Cal-Adi» on 444.; barium of pus‘l‘an is equal +0 +11: wq'ald of“ng P-‘S'hm Plus +51: fort: exuded by‘f'kt af-w-csphcrg apt-LL 'fap mf- +l~n. pis-hm : EMA J’ 13:0 '. _—"“3 --——-— 121A: P.+.A-|-V"3 PIA 1": Pn‘huri- g A p tum-4. (IoKEX‘I-vnls") 'N Ibnr I 7.31/63m‘ tea-NM, Io‘N/u‘l fl: [.17.er e———-—n Fina”) 1N fine QYCr‘hd ‘93 +54. 0dr wH‘NH ‘hau. eulnédtr 0.4 “N 50m“ “51"”- PISJ‘DM IS 910A! +0 'fkc w-u‘ak‘} lb-Hu. PI‘S‘HWI \aluu +kn. {Lon—z eyed-:4 b3 1k- ¢+W’PkEIL an +04 +°p af—H-k pm’mn HE film. ~Furu. We.er 53 -Hu q;me an +uhpo++h yrs-m: PROBLEM 1.4 3 (Comf’d) yr" 'lgr;°lainA -0I x—- . 91A: quA + mg + Eprimj 1A 71'- Pnin—l- .”fl-l- Jaw . fl A E»- m sprde FEW-"3: Ky, wan-L x A's find. vain} +k. Mun»: In valvva .4 3 cf +5.: «N: x: flgfl 30-05”, A 7.8 Var-1M” Cotloc‘tihj resulh M (tamoN/uflomSm) lbar P1: “M”.- 7‘3 + I: (“LBXw‘J n‘) rioSNImL 1; 142.6 bar + 0.54! bar ‘ = his? bar «1—4,. FRDSLEM 155* GIVEN DATA: From H“ aim at T =32“: T (°F) R (m AfI-B‘IQZO: [2+ :(Isyzgmh A R. = 51.39 .0. ——_—“" T= = . _ T as _ T :1 32 5139 ; S T212 We 51.72 51.12 = 51.39 [1+ «(Wham =5 0( =3FUSS 22%).] quy R = swath ammo“ (T ’51)] 2=Ra[l+d(T-T.)Y ' (1:51.3294- 1.0\zzxto‘3'r‘ Lei To=321°F T=qqg,q1(k -St.32r.) 1— TLK=$|.~:1) = ‘ll-lo" F *———rimwfirir ...
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This note was uploaded on 04/15/2008 for the course MAE 210 taught by Professor Chelliah during the Spring '08 term at UVA.

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Assignment_1 - PRbeLEM L15 The mass o¥ Wakr (6 m = gV ;...

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