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

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

This preview shows pages 1–3. Sign up to view the full content.

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: 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: ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online