MAE 103 homework 4 sol

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

This preview shows pages 1–5. Sign up to view the full content.

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

View Full Document

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

View Full Document
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
Ask a homework question - tutors are online