Download - BFC21103 Chapter2
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BFC21103Hydraulics
Chapter2.UniformFlowinOpenChannel
TanLaiWai,WanAfnizan&[email protected]
Updated:February2015
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LearningOutcomes
Attheendofthischapter,studentsshouldbeableto:
i. Understandtheconceptofuniformflow
ii. CalculatenormalflowdepthinvariablechannelsectionsusingChezyandManningequations
iii. Determinethebesthydraulic/effectivesectionofopenchannel
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OpenChannelFlow
ClassificationbasedonTime
ClassificationbasedonSpace
Steady Unsteady Uniform NonUniform
GVF RVF
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Uniformflowisconsideredtobesteadyonly,sinceunsteadyuniformflowispracticallydoesnotexist.
Steadyuniformflowisrareinnaturalstreams,onlyhappensinprismaticchannels.
Weadopt/assumeuniformflowformostflowcomputationsbecauseuniformflowcalculationissimple,practicalandprovidesatisfactorysolution.
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The132kmlongAllAmericanCanallinksCalifornia'sImperialValleytotheColoradoRiver.Thisnewconcretelinedsectionsavesabout3.8millionofwaterayearoveritsleakyearthenforerunner
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TheconcretechannelofLosAngelesRiver(NGM,2010)
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TheKlangRiver,KualaLumpur&Selangor
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Inuniformflow,thenormaldepthyo occurswhendepthofwateristhesamealongthechannel.
Normaldepthyo impliesthatthewaterdepth,flowarea,wettedperimeter,velocityanddischargeateverysectionofthechannelareconstantwithinaprismaticchannel.
Thus,inuniformflow,theenergyline,watersurfaceandchannelbottomareparallel,i.e.theslopesareequalSf =Sw =So =S.
gV2
2
EnergylinegradientSf
yo
WatersurfaceSw
BottomslopeSo
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2.1VelocityDistribution
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Naturalchannel
0.840.820.800.760.700.620.48
Vmax
Vmax
0.52
0.500.450.400.35
yo
0.53
Rectangularchannel
Vaverage
0.6yo
Vmax0.2yo
V
y
yo
Velocitydistribution
Dependsonthegeometryofthechannelandwettedboundaryroughness
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2.2ChezyandManningEquations
Twomostcommonequationsusedintheuniformflowcomputations:
1. Chezyformula
2. Manningformula
21
21
oSCRV =
21
321
oSRnV =
xo
xSRV constant=Thus,thegeneraluniformflowequation:
C =Chezyroughnesscoefficient
n =Manningroughnesscoefficient
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611
Rn
C =DifferencebetweenChezyandManningformulae
Factorsdeterminingtheroughnessaresurfaceroughness,vegetation,channelirregularity,channelalignment,siltingandscouring,obstruction,sizeandshapeofchannel,stageanddischarge,seasonalchange,andsuspendedmaterialandbedload.
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TheChezytwoassumptionsare:
1. Theforceresistingtheflowperunitareaofthechannelbedisproportionaltothesquareofthevelocity:
2. Theeffectivecomponentofthegravityforcecausingtheflowmustbeequaltothetotalforceofresistance.Thisisalsothebasicprincipleofuniformflowwhereuniformflowwillbedevelopediftheresistanceisbalancebythegravityforces:
DerivationofChezyequation
PLVkFf2=
sinALFg =
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gV2
2
EnergylinegradientSf
yo
L
W
W sinW cos
Datum
WatersurfaceSw
BottomslopeSokV2PL
A
P
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sin2 ALPLVk =oALSPLVk =2
oSPA
kV
=2
21
212
1
oSRkV
=
21
21
oSRCV = whereC =Chezycoefficient
1221 sin MMpFWp f =+ Sinceforuniformflow, 2121 and MMpp ==
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Arectangularchannel2.0mwidecarrieswateratadepthof0.5 m.Thechannelislaidonaslopeof0.0004.TheChezycoefficientis73.6.Computethedischargeofthechannel.
GivenB =2.0m,y =0.5m,So =0.0004andC =73.6
A =By =1m2,P =B +2y =3m,R =1/3m
B
y
oRSACQ =
0004.031
6.731 =Q
/sm850.0 3=Q
Activity2.1
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Water flows in a triangular channelwith side slope 1.5(H) : 1(V),bottom slope 0.0002 andChezy coefficient of 67.4. The depth offlow is 2.0m. Find the flow rate and average velocity. Based onFroudenumber,determinethestateofflow.
zy1
Giveny =2.0m,z =1.5,So =0.0002andC =67.4
A =zy2 =6m2,P =2y =7.211m,R =A/P =0.832m,D =A/T =6/2zy =1m
Activity2.2
oRSCV =0002.0832.04.67 =V
m/s869.0=V
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AVQ =869.06=Q
/sm217.5 3=Q
gDV=Fr
181.9869.0
Fr =flowlsubcritica277.0Fr =
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ChezyresistancefactorC
ThefollowingtwoequationscanbeusedtodetermineChezycoefficient:
1. GanguilletKutter
2. Bazin
Rn
S
nSC
o
o
++
++=
00155.0231
100155.023
Rm
C+
=1
87
n =Kuttercoefficient
m =Bazincoefficient
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Table2.1aValuesofManningroughnesscoefficientn
Surfacecharacteristics Rangeofn
(a)Linedchannelswithstraightalignment
Concrete
i.formed,nofinish 0.013 0.017
ii.trowelfinish 0.011 0.015
iii.floatfinish 0.013 0.015
iv.gunite,goodsection 0.016 0.019
v.gunite,wavysection 0.018 0.022
Concretebottom,floatfinish,sidesasindicated
i.dressedstoneinmortar 0.015 0.017
ii.randomstoneinmortar 0.017 0.020
iii.cementrubblemasonry 0.020 0.025
iv.cementrubblemasonry,plastered 0.016 0.020
v.dryrubble(riprap) 0.020 0.030
Tile 0.016 0.018
Brick 0.014 0.017
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Table2.1bValuesofManningroughnesscoefficientn
Surfacecharacteristics Rangeofn
Sewers(concrete,asbestoscement,vitrifiedclaypipes)
0.012 0.015
Asphalt
i.smooth 0.013
ii.rough 0.016
Concretelined,excavatedrock
i.goodsection 0.017 0.020
ii.irregularsection 0.022 0.027
Laboratoryflumessmoothmetalbed,glassorperspexsides
0.009 0.010
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Manningroughnesscoefficientn=0.020 0.022
Manningroughnesscoefficientn=0.020 0.022
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Manningroughnesscoefficientn=0.022 0.024
Manningroughnesscoefficientn=0.020
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Surfacecharacteristics Rangeofn
(b)Unlined,nonerodiblechannels
Earth,straightanduniform
i.clean,recentlycompleted 0.016 0.020
ii.clean,afterweathering 0.018 0.025
iii.gravel,uniformsection,clean 0.022 0.030
iv.withshortgrass,fewweeds 0.022 0.033
Channelswithweedsandbrush,uncut
i.denseweeds,highasflowdepth 0.050 0.120
ii.cleanbottom,brushonsides 0.040 0.080
iii.denseweedsoraquaticplantsindeepchannels
0.030 0.035
iv.grass,someweeds 0.025 0.033
Rock 0.025 0.045
Table2.1cValuesofManningroughnesscoefficientn
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Surfacecharacteristics Rangeofn
(c)Naturalchannels
Smoothnaturalearthchannels,freefromgrowth,littlecurvature
0.020
Earthchannels,considerablycoveredwithsmallgrowth
0.035
Mountainstreamsincleanloosecobbles,riverswithvariablesectionwithsomevegetationonthebanks
0.040 0.050
Riverswithfairlystraightalignment,obstructedbysmalltrees,verylittleunderbrush
0.060 0.075
Riverswithirregularalignmentandcrosssection,coveredwithgrowthofvirgintimberandoccasionalpatchesofbushesandsmalltrees
0.125
Table2.1dValuesofManningroughnesscoefficientn
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Manningroughnesscoefficientn=0.11
Manningroughnesscoefficientn=0.20
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Grassedswale
Table2.2ValuesofManningroughnesscoefficientforgrassedswale
Surfacecover
Manningn
Shortgrass 0.030 0.035
Tallgrass 0.035 0.050
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Table2.3ProposedvaluesofBazincoefficientm
Descriptionofchannel Bazincoefficientm
Verysmoothcementofplanedwood 0.11
Unplanedwood,concrete,orbrick 0.21
Ashlar,rubblemasonry,orpoorbrickwork 0.83
Earthchannelsinperfectcondition 1.54
Earthchannelsinordinarycondition 2.36
Earthchannelsinroughcondition 3.17
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Calculatethevelocityanddischarge inatrapezoidalchannelhavingabottomwidthof20m,sideslopes1(H):2(V),andadepthofwater6m.GivenKutter'sn =0.015andSo =0.005.
zy1
B
Activity2.3
GivenB =20m,y =6.0m,z =0.5,So =0.005andn =0.015
A =By +zy2 =138m2,
P =B +2y =33.42m,
R =A/P =4.13m
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Rn
S
nSC
o
o
++
++=
00155.0231
100155.023
GanguilletKutter
13.4015.0
005.000155.0
231
015.01
005.000155.0
23
++++
=C
769.76=C
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oRSCV =Chezyvelocity005.013.4769.76 =V
m/s03.11=V
AVQ =Discharge03.11138=Q
/sm14.1522 3=Q
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FindtheequivalentBazincoefficientm forthequestion inActivity2.3andcomparetheChezycoefficientsobtainedfromKuttern &Bazinm.
AssumethatforconcretewithKuttern =0.015,Bazinm =0.21
Rm
C+
=1
87Bazin
KnownA =138m2,P =33.42m,R =4.13m
13.421.0
1
87
+=C
Kutter)Ganguillet(from76.769Bazin)(from852.78 =C
Activity2.4
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A trapezoidal channel is 10.0mwide and has a side slope of1.5(H):1(V).Thebedslope is0.0003.Thechannel is linedwithsmooth concrete n = 0.012. Compute the mean velocity anddischargeforadepthofflowof3.0m.
zy1
B
GivenB =10m,y =3.0m,z =1.5,So =0.0003andn =0.012
A =By +zy2 =43.5m2,
P =B +2y =20.817m,
R =A/P =2.090m
21 z+
Activity2.5
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21
321
oSRnV =Manningvelocity
21
32
0003.0090.2012.01 =V
m/s359.2=V
AVQ =Discharge359.25.43 =
/sm625.102 3=
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InthechannelofExample2.5,findthebottomslopenecessarytocarryonly50m3/softhedischargeatadepthof3.0m.
Activity2.6
GivenB =10m,y =3.0m,z =1.5andn =0.012
andA =43.5m2,P =20.817m,R =2.090m
21
321
oSARnQ =Manningdischarge
21
32
09.25.43012.01
50 oS=
0000712.0=oS51012.7 =oS
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A triangularchannelwithanapexangleof75 carriesa flowof1.2m3/satadepthof0.80m. Ifthebedslope is0.009,findtheroughnesscoefficientC andn ofthechannel.
Activity2.7
Giveny =0.80m,So =0.009, =75,andQ =1.2m3/s
andA =zy2 =0.491m2,P =2y =2.017m,
R =A/P =0.2435m
21 z+z
y1 75
=2
tan
z
=275
tano
0.767=
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21
321
oSARnQ =UsingManningequation
21
32
009.02435.0491.01
2.1 =n
0151.0=n
21
21
oSCARQ =UsingChezyequation21
21
009.02435.0491.02.1 =C197.52=C
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A trapezoidal channel of bottom width 25 m and side slope2.5(H):1(V)carriesadischargeof450m3/swithanormaldepthof3.5m.Theelevationsatthebeginningandendofthechannelare685 m and 650 m, respectively. Determine the length of thechannelifn =0.02.
GivenB =25m,z =2.5,yo =3.5,n =0.02,andQ =450m3/s
zy1
B
A =By +zy2 =118.125m2
P =B +2y =43.848m21 z+R =A/P=2.694m
Activity2.8
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21
321
oSARnQ =Manningequation,
21
32
694.2125.11802.01
450 oS=
00155.0=oS
Ho L
zS
=
( )HL650685
00155.0=
m13.22601=HL
Manningequation,
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2.3Conveyance
ConveyanceK ofachannelsectionisameasureofthecarryingcapacityofthechannelsectionperunitlongitudinalslope.ItisdirectlyproportionaltodischargeQ.
1. Chezyformula
2. Manningformula
21
21
oSCARQ =
21
321
oSARnQ =
21
CARK =
321
ARn
K =
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SectionfactorZ intheManningformulaisAR2/3,whichisafunctionofthedepthofflow.
InManningformula 21
321
oSARnQ =
Therefore,21
32
oS
QnAR =
SectionfactorAR2/3 isnormallyusedtocomputethenormaldepthyowhenthedischargeQ,bottomslopeSo andManningroughnesscoefficientn areprovided.
Computationofyo couldbethrougheitherdirecttrialanderrorcomputation,basedongraph,orthroughprovideddesignchart.
2.4SectionFactor
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Activity2.9
A trapezoidalchannel5.0mwideandhavinga side slopeof1.5(H) :1(V)islaidonaslopeof0.00035.Theroughnesscoefficientn =0.015.Findthenormaldepthforadischargeof20m3/sthroughthischannel.
GivenB =5.0m,z =1.5,So =0.00035,n =0.015,andQ =20m3/s
zy1
B
A =By +zy2 =5y +1.5y2
P =B +2y =5+2y21 z+ 25.3( )( )yyyPAR 25.325 5.15
2
++==
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ArrangingManningequationasafunctionofsectionfactor,
21
32
oS
QnAR =
21
321
oSARnQ =Manningequation,
( )21
32
22
00035.0
015.02025.3255.15
5.15=
+++
o
oooo y
yyyy
( )( ) 036.1625.325
5.15
32
35
2
=+
+o
oo
y
yy
Therefore,yo =1.820m
yo (m)( )( )32
35
2
25.325
5.15
o
oo
y
yy
++
Bytrialanderror:
1
2
1.8
1.820
5.391
19.159
15.706
16.035
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Graphically,
( )( ) 036.1625.325
5.15
32
35
2
=+
+o
oo
y
yy
yo (m)( )( )32
35
2
25.325
5.15
o
oo
y
yy
++
1
2
1.5
1.7
5.391
19.159
11.198
14.115
1.8
1.9
15.706
17.387 0
0.5
1
1.5
2
2.5
0 5 10 15 20 25
AR 2/3
y
o
(
m
)
yo =1.82m
16.036
Therefore,yo =1.820m
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DesignChartisavailable,
Circular
Rectangular(z =0)
B
38
32
38
32
and
od
AR
B
AR
ody
By
and
0.2194
0.37
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036.1632
=AR
38
38
32
5
036.16=B
AR2194.0=
37.0=By
537.0 =yTherefore,yo =1.85m
Atthexaxis,
Intersectingatz =1.5oftrapezoidalchannelgives
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DesignchartforlinedopendrainfromUrbanStormwaterManagementManualforMalaysia(DepartmentofIrrigationandDrainage,2000)
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Activity2.10
A concretelined trapezoidal channelwithn = 0.015 is tohave asideslopeof1(H):1(V).Thebottomslopeistobe0.0004.Findthebottom width of the channel necessary to carry 100 m3/s ofdischargeatanormaldepthof2.50m.
zy1
B
Givenyo =2.5m,z =1,So =0.0004,n =0.015,andQ =100m3/s
A =By +zy2 =2.5B +6.25
P =B +2y =B +7.07121 z+( )( )071.7
25.65.2++==
BB
PA
R
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Manningequationasafunctionofsectionfactor,
21
32
oS
QnAR =
( )21
32
0004.0
015.0100071.725.65.2
25.65.2=
+++
BB
B
( )( ) 75071.7
25.65.2
32
35
=++
B
B
Bytrialanderror,B =16.33m
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Activity2.11
Waterflowsuniformlyat10m3/sinarectangularchannelwithabasewidthof6.0m,channelslopeof0.0001andManning'scoefficientn =0.013.Usingtrialanderrormethod,findthenormaldepth.
B
y
GivenQ=10m3/s,B =6.0m,So =0.0001andn =0.013
A =By =6y
P =B +2y =6+2y
( )yy
R += 33
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21
32
oS
QnAR =
21
32
0001.0
013.01033
6=
+ oo
o yy
y
167.233 3
2
=
+ oo
o yy
y
Bytrialanderror,yo =1.942m
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Asewerpipeof2.0mdiameterislaidonaslopeof0.0004withn =0.014.Findthedepthofflowwhenthedischargeis2m3/s.
2rD
yo
( ) sin228
2
DAreaA =
PerimeterP = D
Activity2.12
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21
32
oS
QnAR =Manningequation:
21
32
0004.0
014.02=AR
38
38
32
2
4.1=D
AR
2205.038
32
=D
AR
Fordesignchart:
6.0=Dyo
26.0 =oy =1.20mIntersectingatcircularsectiongives
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DesignChart:
Circular
Rectangular(z =0)
B
38
32
38
32
and
od
AR
B
AR
ody
By
and
0.2205
0.6
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SimplificationforWide RectangularChannel
Widechannel: 02.0