Problem 4.2 Solution

1 CEE345 Spring2002 Problemset# concretesewerpipe4 ft indiameteris laidsoit hasa sewage(assumethepropertiesarethesameasth oseofwater) owsata depthof2 ft inthepipe,whatwillbethedischarge?Solutio n:Rh=AP=18D2 12D =D4=1:0 ftAssumingn=0:013,wethenhaveQ=1:49nAR23h S120=1:490:013 4(4ft)2(1:0 ft)23 1:00ft1000ft!12=22:8 widewitha slopeof10ft in8000ftisdesignedfora watertemperatureof40 F, estimatethedepthofthe :Assumingn=0:015,Q=1:49nAR23hS120=1:49nb y byb+2y!23S120where,bis thechannelwidthandyis the ow depth600cfs=1:490:015 (12ft)53 y53(12ft+y)23 10ft8000ft!12 Solvingthisforywegetthesolutionsof 4:11ft and5:60ft. Obviously, onlythelatteroneis possible,soy=5 # andsideslopesof1verticalto2 horizontalis designedtocarrya theslopeofthechannelis ,whatwouldbethedepthof ow inthechannel? Solution :Assumingn=0:012AR2 3hb83=Qn1:49S120b83=3000cfs 0:0121:49 0:00112 (10ft)83=1 ,y=b=0:90ory=9 ,Idaho,whenthedepthis 4 :Q=1:49nAR23hS120 FirstcalculateAandRhfromthegiven foundtobeA=300ft2.

2 Likewise,byapproximationit isfoundthatP=125ft. ThusRh=AP=300ft2125ft=2:40ft:Assumen=0 :49nAR23hS120=1:490:038 300ft (2:40ft)23 5 ft5280ft!12=1200cfs2 CEE345 Spring2002 Problemset# dischargeof200cfs,andit willhave it istobedesignedsothaterosionofthecanalwil lnotoccur?Choosea :FromTable4-3themaximumpermissibleveloci tyforcoarsegravel is givenas4:00ft=sandn=0 :V=1:49nR23hS120orRh=0 BBBBBBB@V n1:49S1201 CCCCCCCA32=2:17ft:AlsoA=QV=200cfs4:00ft= s=50ft2 Assumesideslopeswillbe1 verticalto2 horizontalP=ARh=50ft22:17ft=23:0 ftAlsoP=b+2yp1+4=b+2p5y=23:0 ftA=by+2y2=50ft2 Solvingtheabove two equationsforthebottomwidthbanddepthyyiel dsb=7:59ft andy=3 # owsata depthof10cmwitha velocityof6 m=s ina ow subcriticalorsupercritical?Whatis thealternatedepth? Solution :CheckFroudenu mberFr=Vpgy=6 m=sp9:81=s2 0:1 m=6:06>1sothe ow is +V22g=0:1 m+(6m=s)22 9:81m=s2=1:935mSolvingforthealternatedep thforanE=1:935myieldsyalt=1 a formulaforcriticaldepth, :A3cTc=Q2gwhereforthischannelAc=d2candTc =2dcso(d2c)32dc=12d5c=Q2gordc= 10-ftwiderectangularchannelis verysmoothexceptfora smallreachthatis owsinthechannelata rateof200cfsandata depthof1:00ft.

3 Assumefrictionless owexceptovertheroughenedpartwherethetota ldragofalltheroughness(alltheangleirons) is assumedtobe2000lb. :Usethemomentumequationwrittenfromthesec tionupstreamoftheangleirons(callit 1)toa sectiondownstreamofthem(section2).Writei t perfootofwidthofchannel. y212 y222 Fab= V21y1+ V22y2ory21 y22 2 Fab = 2gV21y1+2g Qby2!2y2whereFais theforceontheangleironsandbis (1ft)2 y22 2 2000lb10ft 62:4 lb=ft3= 2 (20ft=s)2 1 ft32:2 ft=s2+2 (200cfs)232:2 ft=s2 (10ft)21y2 Solvingthisequationfory2wegetthreesoluti ons, 4:95ft, 1:43ft and3:52ft. The rstoneisofcourseimpossible,buttheothertw o owswitha velocityof2 m=s andata depthof3 mina thechangeindepthandinwatersurfaceelevati onproducedbya gradualupwardchangeinbottomelevation(ups tep)of60cm?Whatwouldbethedepthandelevati onchangesif therewerea gradualdownstepof15cm? Whatis themaximumsizeofupstepthatcouldexistbefo reupstreamdepthchangeswouldresult?

4 #3 SolutionsSolution:E1=y1+V212g=3 m+(2m=s)22 9:81m=s2=3:20m:E2=E1 z=3:20m 0:60m=2:60m:AlsoE2=y2+q22gy22=y2+(6m3=s= m)22 9:81m=s2 y22=2:60msoy2=2:24m. y=y2 y1= 0:76msowatersurfacedrops0 downwardstepof15cmwehaveE2=E1 z=3:20m ( 0:15m)=3:35m:givingy2=3:17mand y=y2 y1=0:17msowatersurfacerises0 ectingupstreamwatersurfacelevelsisfory2= ycyc=3sq2g=3s(6m3=s=m)29:81m=s2=1:54 dischargeof1:2 m3=s :y0+q22gy20=y1+q22gy215 m+(1:2 m3=s=m)22 9:81m=s2 (5m)2=y1+(1:2 m3=s=m)22 9:81m=s2 y216 CEE345 Spring2002 Problemset#3 Solutionssolvingfory1wegety1=0 (1:2 m3=s=m)2p9:81m=s2 (0:123m)3=8:88:y2=y12 q1+8Fr21 1!=0:123m2 p1+8 8:882 1 =1:48 owingasshownunderthesluicegateina horizontalrectangularchannelthatis6 ft and1 ft respectively. Whatwillbethehorsepowerlostinthehydrauli cjump? Solution :Assumenegligibleenergylos sfor ow sectionupstreamofthesluicegatetoa +V202g=y1+V212g65ft+0=1 ft+V212 32:2 ft=s2V1=p(65ft 1 ft) 2 32:2 ft=s2=64:2 ft=sFr1=V1pgy1=64:2 ft=sp32:2 ft=s2 1 ft=11:3 Now solve forthedepthafterthejumpy2=y12 q1+8Fr21 1!

5 =1 ft2 p1+8 11:32 1 =15:5 fthL=(y2 y1)24y1y2=(15:5 ft 1 ft)24 1 ft 15:5 ft=49:2 ft7 CEE345 Spring2002 Problemset#3 SolutionsP=Q hL=V by hL=64:2 ft=s 6 ft 1 ft 62:4 lb=ft3 49:2 ft550ft lbs=s=HP=2150 leshownis fora rectangularchannelthatis 3 m wideandhaswater owinginat a rateof5 m3=s. Sketchinthemissingpartofwatersurfacepro leandidentifythetype(s). Solution :Flow overweir:Q=(0:40+0:05HP)Lp2gH32=(0:40+0: 05H1:6 m) 3 mp2 9:81m=s2H32=5 m3=sgivingH=0:917msothedepthupstreamofth eweiris 0:917m+1:60m=2 m3=s3 mp9:81m=s2 (0:3 m)3=3:24>1supercriticalFr1=qpgy3=1:67m3= s=mp9:81m=s2 (0:52m)3=0:133< :3 m2 p1+8 3:242 1 =1:23mThewatersurfacepro ledownstreamofthehydraulicjumpandabove theslopeis S1andabove thehorizontalbottomis mwideandcarriesa dischargeofwaterof12m3= #3 SolutionsSolution:q=Qb=12m3=s4 m=3 m3=s=myc=3sq2g=3s(3m3=s=m)29:81m=s2=0:97 2msostartatx=4yc=3:89mRe=V 4Rh =3:09m=s 4 0:654m10 6m2=s=8 106usingvaluesfromthe rstsection,givingf=0 stepsolutiongivesthebelow then1 xxmm/sm/sm2 # ,whatkindofwatersurfacepro le(classi cation)is upstreamofthejump?

6 Whatkindofwatersurfacepro leis downstreamofthejump?If ba e :Upstreamofjumpthepro e blockswillcausethedepthupstreamofAtoincr ease,thereforethejumpwillmove cationindicatethatthemeanvelocityalonga verticallineina widestreamis closelyapproximatedbythevelocityat rivercrosssectionaremeasured,whatis thedischargeintheriver? Solution :Q=XViAi1 0 CEE345 Spring2002 Problemset#3 SolutionsVAV Am/sm2m3 A=549m3= oodcausedwaterto owovera highwayasshownbelow. Thewatersurfaceele-vationupstreamofthehi ghway(atA) wasmeasuredtobe101:00ft. Theelevationatthetopofthecrownofthepavem entofthehighwayis 100:10ft. Estimatethedischargeovera stretchofhighwaywiththiselevation,whichi s 100ft owat thecrownofthehighway? Solution :The ow overthehighwayis asif ow wereoccurringovera broadcrestedweirQ=0:385 CLp2gH32 AssumeC=1Q=0:385 1 100ftp2 32:2 ft=s2(101:0 ft 100:1 ft)32=264cfsCriticaldepthwilloccurat q2g!

7 13= (2:64ft2=s)232:2 ft=s2!13=0:60ft:11