MAE 103 homework 4 sol

MAE 103 homework 4 sol - 2.139 I'll-eunknfliquidinthefigme...

Info iconThis preview shows pages 1–5. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2.139 I'll-eunknfliquidinthefigme P1139 aeeelentes re flse right srifl] the fluid figid-Isedy mefinn. {3} Compute as in ur's'. {Is} Why doesn‘t the selutien to part is} depend upon fluid density? [:2] Campfire gag-e pressure at paint EL if the fluid is Fig. Plum elf-rain at 20" C. E-nlu lion: [.1] The slupe efflse liquid us the aeeelemtien: tanfi =—‘ =—- (113. er: fi= 14° Thus a; = Ctlflg = 0213:9313 = 1.13 m-‘s1 Am. (a) {b} flea-Iris schrtient-s [a] is pmel? geemmiesnddeanet ins-eke fluid JHE. lie) me Table 11-3 fer ghreain, p = 1251] kg."m3. Them are nun}; ways tn compute PA. for example: we can 34;: shaight dawn cm the le’fl side. e111? entity: p! = para: = (1251] kg,"m3}(5|.31 m-‘sEJ-HIEE m} = 3-160 Pa {gage} Arms. {e} D: we can shut uni-he right side, ens deem 15 em'wifls g and amass lEIIIIl emwith 9:: p! = pgsz+ max =fllfifl]{9.31}[fl_1S}+{125E+]{1.23){1.flfl} = 1354+ 1150'! = 54151113: Jim. 13:]: 2.141 The same tank fi'eln P'mh. 113-9 is new accelelat'ing while telling up a 30'" inclined plane. as shewn. Assuming rigid- litltl}r metien, eempute (a) the aeeelerahen a. {h} ssheflier the aeeeletahan is up er deem, and {e} the pressure at peinrfl if he fluid 'E n:Ler-::ur'_l.r at 313°C. Fig. F1141 E-elun'en: The fi'ee surface is tilted at the angle §= -3E|"=' + T41“ = -El5§"’_ This angle must salist Eq. {1.55}: tan I9 = taut-31.59"} = 43.416 = a 1 -'I:g + at} But the 3-D" int-line eenshains the acceleratie-n su-eh that a}: = flflfifia. a1 = Ill-.5 a. Thus mS=—fl.4lfi=%, selt'efer ai—Sflflgfllmrn} sins.[a..h:l The eastesian eempenents are a: = —3.29 Isa-'s2 and a: = —l.9[|' Iii-'52. {e} The distance 5.5 nmmal finmthe surEa-ee dawn te paint-13L is {13 cash]- em. Thus p... =,::[a: + [g + a]2 f": = [13 ENE—3.29]: +£9.31—1.5|C|}3]L1I:C|.23 ens 7.41“) s 31:01} Pa [gage-j :1il1:.s. lie]- Z.l-'l-'-r The tank of meta in fig. F114”? a-eeelentes millille by milling without fiiefien damn fine 313° ineltued plane. 'What -._, is the angle Enf the fee siesta-eelI Can fen I ' ' __L.. explain this intluesting result?I - ' E-eluIi-en: Iffli-ehanlfis,El:=‘Wsinfi=ma alengflaeinifliueandfllma =gsiu3C|°=C|.5g. Fig. F231 4T Tl Ema: a = 0.5gees3CI' _ - s 1' fia' fi=5|l'! :1 ". g+aI g—ClfigstuE'U‘“ JD“ 1 IL The fee flufae-e aligns itselfexaetlf parallel with the 1-D": incline. 1.151} A cheap aeeelaemeter can he made firm the U—tuhe at right. If L = 13mandD=amm=nhatniflhheif 1 a3. = 15 Ill-'5'? Entrlulien: Ere assume that the diameter is an anal Eli-1:. L. that the flee surface is a “paint.” Then Eq. {1.55) applies: and +5.0 ma: - =—=r:+.512 5:31.5- 3” 9.31 ’ M Then h={L’3]1anfi'=[‘5|em}[fl-fi12}=5.5cm Am. Sine-e i: = {9 «inhale, the scale leadings are indeed ]inea.1‘:i.11 a1, but I defltreeemmend it as an actual accelerometer: therne are ten marl}r tnaeema-E'LE and diaadwamages. P2154 A retjr' tell Imam-diameter vase tannins 113'3 em” ef water 'When gm sl'ealfl}r to achieve figid-he-dy mum: in 4-:JJJ-diamem d1? met afiean at the hemmnfthe fine- 'Whatjs the ntetinn IEEE. 5111:]an emdiiinn? hehlinn: it I: mherahng thattte ElIlS'i'iEt' hasnething t: do withthe waterdeerem-L The value efl 17:5- eubie eeutimetem was Izhesen t4:- maketheru'dqefilaniee number: U=11T3m1= xcsmfe :sclL'e H = liflm Due wag: wuuldhe tuiutegmte and find flJevellmle affine elude-I". liquid P2154 in tum: af-r—ue radius R and dlf-spetLadius r" Thatde j-‘iEld HIE fuller-deg fimzmla: .fu =ece3—rrfje, hut :=fl:r:-"1g', hence d: = {Elf-"eke R 'I 'I 'I Thle u— Lx(e*—r53{n-r.-'Ejdr — 2 4 I 2 4 3” (Ii—R r” +1. = eeememj :malb': L|= g 4 1 4 1115' r _ --+ — EEI- I 5 311.11 Selrefer R=E|2C|im= rl__=E|-.C|Em :fl2 = 333E: fl = 513 I'he fun-mule: in HE text: centering the palahnleids Diwali": would, in the mil-E's. minim. he Ififleultte apply because efthe flee safe-2e extmding helm: the he-lte-m nfthe vase. 2.155 Supp-use the U-tube {If P1101}. 2.151 npm fl is muted about axis DC. If the flmd is ~ _ I] wata' 1t 113T and amsphann: pmsmme t " ‘nfu- r. 2115 pusfa: at what ruiafinn r1112 wfll the .n "'- - I flud begin t-n filpurLze'?‘ At what paint in The tube 1|..‘IJ1 “this happen? 1 E L "I'--:n—r E-nlulin-n: {Lt 122°]: = 50°C. flan: Tabla A—1 and A-5. far water: p= 5'33 kg.":|:|13 {In-1' 131331113311”) deW= 11.34 LE? [01 153 [:31]. When :pinnmg alum-$11313: flu! fiEE smile-e {Elma dawn fi'nm paint 1'11 tn- 3 pflsitinll bafaw point D: as shown. I'hEtEfiJrE the fluid prawn! is lama-t: upumiD Hm}. 1Lilith I] 35 Shawn in the figum: p”, = pm. = 133=pm-,LIEI1= 3115—1.91?(31.2:IIL_ h=fl=1=flgjr Balm! f-nr 11 3 30.1 ft [F] 111115 the flaming 1': mildly dlflfll'l'Eli and the dashed liI1E £31]: 1111' belnwp-nint C! {The 5011111011 E came-ct: however.) 54:11.12 far {11 = 113 2.1}[3 0.1}{1 E}: 01': fl =44 136.": = 4211 rn'.'1.1:u'.11. Arm. ...
View Full Document

This note was uploaded on 11/06/2011 for the course MAE 103 taught by Professor Kim during the Spring '08 term at UCLA.

Page1 / 5

MAE 103 homework 4 sol - 2.139 I'll-eunknfliquidinthefigme...

This preview shows document pages 1 - 5. Sign up to view the full document.

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