Quiz-5 - / EAL.” [TE 37% Fluid Mechanics Homework Quiz 5...

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Unformatted text preview: / EAL.” [TE 37% Fluid Mechanics Homework Quiz 5 III].F [41115 Name: SHO‘W ALL L'N ITS 1_ an empirical equatiun fur the xelncit} distrihutlnn In a horizontal. ItclangLilar. upen channel is. given by n um. r 1- :1“ 3". where H — Helm-cit} at a digtance _1' meters abm e the I'lmr ut' the channel; {.I’ — depth nT't'qu — 1.2 m: 1.3...“ - maximum u - 3 til-'5; and n - empirical exponent — '«':. :1. Find The volumetric discharge pur Inctm' u idth nt'channel. h. Find the mean auluuii}. / cl , .“1 , ,rd 1' . 2.2m a “J: . Um] I . u n ‘ : u I, .m d — “AL. -. ,’ "aw *“1 1. J L” >13. “MK-i1) [ 1 3' _ a": K d '31 Z a": F? 3 ll jg l.— 'CI _ {Era ' r3 III: {‘1‘} E [ 2 ELM-fir; F 1;? I _ 4 4D if 3.: ( 3) L~ ' -' J ' 5 _. 1'3 Taf— _——- T d “1-D r JC— _ 2 C} ‘3‘— fl "“ - .5 2‘. Determine whethEr cavitation 1mill occur in this venture meter. mun? , Q T -‘ m"-'5 D = DAG m [A — I]. I 25? m‘] 53 .4 d = 0.1:; m m. — 0.00m .n-‘y ’ Pressure gage at A. = llflflflfl Nim: 1 1 f...—-—- Atmmphcric pressure -— ml .304] ':’m' I T=25°f —h----—————————__ 41's: 4.. F4- :35: 'I «171*: .1. 7.5K .1.- .1: \5 graham: 9:7“ 036%)1- T1" [11 01 E) M + FF“?— _ + 5 CHEN WE 1(EE‘15L) qTBI fig L(1BL%L] LEN—fl” L’W-fi" L'"“r—’ ELLSM 0.051m 1151M M T 2 (234m mam—11.5w fiTflIga) = HF ‘icm. $4055) 1 . “Ii-$341!!" prisgur-E 0* 7—51: [email protected]) Sim». [[91:34:- '3? 'Sl'Tfl EfiUth-fl‘h ‘1‘”;5 Lib arc-tun" t J 3. F ind the rate D'Fwater level rise nr fii". h-E’L dmmctcr 11—h ['J "333“ .-L=1I.HETIII: D: 0.2511 ~ t' H1 [15. ."L-=U.1H‘HI1' _h *— ‘L-I. 4I1-. l)7=ll.5l.l1'l ‘ A: = n" r? Tfi's i ‘— Ni :2 ._ Audi: -- [1&8 {‘3 g V FI; hf CI é d’d- '— _ _* I: F’ F[*U1Ar* U151: "113*: d5 1 __ _ 1 Fl. 3 2 [ 3 M - 09§X0.0511H) - (1r .EXoHLH‘I) + @% mom}? :43 3 23.1? W- {LEW T .314 3;} {11% fig 3 ‘ 3 m 3% _ 0.311%- -~ 1214 £43- + chm?— dg ’- 23 1? “1 = E" pf C E 51111! Fluid Mechanics Hnmen‘nrk Quiz 5 lflfl4ffl5 Name. gt: [ML '1' M1 _ SHOW ALL UNITS | . .-“tn empirical equatinn fur the x'elnuit} dintributtun in t1 hur'imntni. rectangular. open channel in git. en b}. u n..,_ .- 1- cf .- “here u aelneit} ntndistunee1'1nelerxzihuwtheflunr ntlhe channel: u" -depth ul'l'ltm 1.3 In: :t,.,. maximum n 2.5 PH 3'. and H etnpirien1 expuan 2:. Find the wltnnetrie difieharge per meter width nfehannel. h Find the mean helneit}. d _. .Irll.f {A r _, A II l_'| 1 ‘J- M II LEM :2 r s a) w 01.; 2 L - a a __ rJ : (ll-“L ] F 15L '3 j a d D ycmfjllltm) ;. 33'; ‘31.? _ :II'T. :: ( J 5,.” Ham) .. D I :___ 3'00 :4: ref": {Lara} L IL J _-, —r' 1— 4, 3 ‘2 3,00 13]- f. 2. Determine whether eat-‘itatien will Occur in this venture meter. 0 — {1.22111'1-"5' n = 0.4:] m m = (1.125? m-‘t d=0.|2 mm: &: flung-.111 Pressure gage at A ; Iliflflfl' Nr'm2 Atmospheric pressure = NH .300 Him" T=3IJ°C Ur a in _ as 11mg Hg .- —~ __ : I, D t PM a: list-m1 VI : g :: 0'11"?“33 — fl ! I 1‘.— L _—¥i Jr %| A: __I|U£l— : in + % + FEET“— it if) 3’ 1 L3 I lzergne J‘J—l _ 'L L N” +— “73mg ._ {Pl (£147???) 0:111 a: w H M 1% “LEE 5:. W71 lip, 1(‘131 f; L’fi—Y—‘F’J LF-Y————J W—h—J lilbm 0.15sm FLSILM I. + — Ht'z‘m ( TH E“ N vb .2 .23.: Lm + gnaw - ‘T “a : E‘LDH‘: gtg‘us) D I‘d Uaimr Prf’rlkk're 63‘ 30L 2 +19: 1;; 0}“) 54;” 31mg ;: +3.93“, {swim—Ham MN 1:131" occur. J" 3. Find the rat: 0f water law-cl TiSE nr fall. H-ft diamulur q—-b u- H.413 n .-1.-=n.|:5.*-1t= 1L=u1u11 ‘ 14-44115- .x; unann- nx, 4m __.. ‘— 113=l1fillli :13 1:232:11? x'gans Ii X/M : r g gfid'fi? dt :5 Cd fit : {—EAI + mat ugh] Egg}: El} 2: %)(fl.l‘2-ST§JI1) *- (IE; %XD.131T 4‘11) .1. @%Xn.5314 EFL) 'I..—-—-\{.—._.J 3 50-11541 1.1m fl; ' 1.1%: E;— 043 Q? A m ' 3 3i; _‘1‘l '— 1 "— LTD '1' + j dE Eur-.11 F41 ,I.‘+_ .. . 3TH! Fluid Mechanics Homework Quiz 5 [WHMS Name: £11 H ’l'l‘l’éfi'k anew ALL UNITS |..-1in entpirieal eqttatinn thr the telneit} distribution In at Imrimnlnl. reelangulnt‘. npen eltnnnel i5 gixen h}. H arm-u. {1' ct’ 1": 1r'ihet'e t: ' telneitt- at a tiixtaneei' meter}; above the finer with: channel: at“ -- depth ul'flmi 2.7? nt; um... — maximum tr :.''IH.-".H".fl1‘llil.if" empirieai expentent — '5.- tt. Find the t-nlnmetr-ie dieteharge per meter width ut‘eltnnnei. l'I. Find the mean t'eineit}. d .Jt ; , ‘39: Em = g urinal-j = iingt-gficig : been mitt (1%)“ “‘3 If HIM" it 1 i'_ r 3"; IE 5m " —-'Tr:' i. filth] 7- 1125 .51: it] L it at]: s i -. ' do |_ ,7 ,- l. ML f2, ' F Ir _ 5}: DH; F —;. [,mg Niki II E) .2»:an _ r '.1— my: “’13 L 53-33 t _ {gt a 33 “if; "it ' ' ‘~ =- 155:: .5 , Determine whether cavitation 1will occur in this venture meter. Q -— {1.21 “We ’11:: 1 “:x D = {145 In {A = fl.|59U m'] -"' I7 {1'} 4 d—[Hll‘nlfl‘l' Argue”: m1“ 1 Pressure gage at A = IESJJUD New 1 Atmospheric: pressure = Ifll .EUD N-“nt' T—Efluf m———m—u—-—-———-~-——-— a “if I—//rT\—_____m_ U: — = (3'14" J = [50?“ I fit 0.145110% 5 3 Ur t3; _ 0.; “‘1‘; turn at 0.0113M1F 1’ 5 at? F + Lit-i t1 4e «.1 H + Y: a,. cam E? by )1“ L Jr W; L *‘Pz— + (at W“: q ht 91 r F firm at 1(1'5' 51: 311014;? L(1.EL%;) L—___Y__.J W Li-I—W—J 2.4.rt-m D. Ilia W- 11511 m "19;. == (14-.tfirm + 9.1mm -* 333%“)(‘111‘3 2—”) 7"" "ztgam'fii GL5) Jib-Par prf'sture. Ll left; : 234-3 %L(ekt) $1.112 113m. :* 124-0) cavir‘m‘nw cine: m; occur“. ll'l'fl HIE MIC 01' W'IBVGJ NSC Ell' Hill. H-fi diam-:Icr l]. = H.411 H x. unzs‘uf 11; "Jun _ r, was .d.~,=u.u.1|4 Fr 1:1 4 It's _*. h 1}. Lawn 1 A: = ".232? n- ‘v’; firflfs- [A ' t” = H g UHJA (I; CE MP _. — _ 15% : ‘ [F ULA1+ ULAZ _ U3 A3 3. :15 if} 1 ‘ 1 n. 1 [TV (4%)] R = (3 5)(U.11W FA) (é‘giXOLEL‘H—i) + @EXDJEM :4) 90.17?” mm E; [h f}; 0.3%? " l- g: : Wig '— Ilflnfii'j *‘ m: $1.11 $41— : - (3.011! IF”: Ghilfhj) ...
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This note was uploaded on 04/03/2008 for the course CE CE 3700 taught by Professor Peyton during the Fall '05 term at Missouri (Mizzou).

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Quiz-5 - / EAL.” [TE 37% Fluid Mechanics Homework Quiz 5...

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