CEE 366 - Joseph Wartman - Spring 2011 - Homework 4 Solutions

CEE 366 - Joseph Wartman - Spring 2011 - Homework 4 Solutions

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Unformatted text preview: 7-1. AdeensWhewyepenneebMyoHJx 10"arv§endevwmdo.45mplecedme haumlpemveewyeppemmgesshovmmflg. 7.2. Wemmmwmryendm eeepegevefoaersrheheedAngoesflomObBOcm. Themse—eecrbndereedmmm 91991395001“. endme misunplelso.65m long. SOLUTION: Eq 7.5: v = In . ) (800m 1 M890 11m . V = (4.5)(10) — =(4.5)(10) (1.231) = 0.0055 an]: \656'0 v: 0.56m]: v e 0.45 WM: =—; =—=—=O.31 —v' n n 1+0 1+O.45 0.55 = — =1." v. —W8 1-2. AwnpleofmedlummmnteuedMememneedpemeemelen WW3 diemel'eflsmmendnruengmls 130m.LMdereneppnedheedd60cm 119an’flows mmmesenmh5nnn. mu,olmem1p1m4109.cwwm(a)moamymmmd permeabllly. (mmemecnelyemlomxandmnheeeepege velocity. {Mom Casegrande.) SOLUTION: 0L g119an’313.0unw , E 7.9:k=—= =0.003 =3 10 (a) Q M1 (60m)::(6.0cm)’(5 mnxeo 16..) am " we ’6011111‘= \1acm, (0) “charge velocity: v = Id = (0.003 We) (c) seepage velocity: v, = 1; determine n. assume p. = 2.709/6111’ n M 4109 , V.;—‘-: p. 2.70 cm’ v, = 35Lfo 13: 3615666111”; v, = v, - v. :367.566- 151.65 = 215.716cm’ _ 215.116 - = 0.537 V| $7.555 V 1-10. Ammheadpommmmmpedmmdmammnzanxzanaqmmzs Milena Thenceddlfemnceapphodmrhgvnmals 18mm65an’mmdowram (#100090. (a) mmmmmbmammummwm MAW hmrestlstobomanhemmspemwnmmsmmm-tfi 1003)andlhe mndpipodlamol'eflso.8cm. Hanoverogehoadmamnamddbo 18cmwha¢mm and")? SOLUTION: . _$_ 5405m’g175ang41 _ _ 3 (3) Eq 7.9. k - m - (38 mm 75 an)“, a) - 000052 We - 8.52x 10 an]: A=2x2=4cm’ 0!. Sem’ 2.50m , k=—=—LL—L. =0.0017360ml =1.n 10 w M: (1BmX4an)‘(1003) ' " ’ (b) Falnghoad m uerq. 7.100: nuimwfi M! n, Adam"; a=nM¥X=0503afi z gasoaxzsun; 2L 0.001736 an]: 2.3 a an. m m 3) no.0 n: log", %- : 0240 I '-“-=12714 3 h. = 1211a, h. +h, =18(2) n. 4127100641,) h. = 20.1561“. 11, = 15.85am 7-11. The coefldenlofpenmaamyoladeanssndwas 389x firmware mmdflja 5mm mapwnmryolmmmmmIauots 0.61. SOLUTION: Assunpflmssnm. v, =10000m’ = V‘ ¢ V“ lg,” ;389(10) ‘ an]: e,=%;o.3a:$’t —. V‘ :72434 cm’ (assumev, 00930010113090) 9, = 0.61: -. v, = «2.03 an‘ 724.64 car m; V‘ , ; “2.03 + 724.6! :1166.67an‘ kispmpommaltomevoumeotmasamw, —o Rwak 1166.67 kn: ‘ 369(10) "‘ 100° Eu. nu v1” ; 0.0‘5 cm]: 7-30. Form WMMMIOIFIO. P130. mavens mmm dam peme minnow”):de 4.2x 10‘ ants. 511.720: q=m #- Frommofiownu: N,=3. N.=93. l\_=6.3m 3 q: (42:10)‘ "you (5.3 m): E= 0.53.. 10‘ my.me we: ...
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This note was uploaded on 03/31/2012 for the course CEE 366 taught by Professor Arduino,p during the Spring '08 term at University of Washington.

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CEE 366 - Joseph Wartman - Spring 2011 - Homework 4 Solutions

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