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Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 1
CHAPTER ONE
INTRODUCTION TO OPEN CHANNEL FLOW
Course objectives:- atthe endof the course studentswill be expectedto:
o Differentiate betweenopenchannel &pipe flows
o Understandfluidphenomenonapplicationtodesignvarioushydraulicstructures
o Acquaintlaminar&turbulentflows,boundarylayertheory,velocity-shearstress
relationship
? WHAT IS OPEN CHANNEL
Openchannel:- It maybe definedasapassage inwhichliquidflowswithitsuppersurface
exposedtothe atmosphere.e.g. :- curvets,spillways,andsimilarhumanmade structures
Differencesb/nthe flowinpipes&openchannel flow
Openchannel flow
 Is exposedtoatmosphericpressure.
 The cross-sectional areaof the flow is
variable.(thatdependsonmany
parametersof the flow)
 The force causingmotionisgravity.
Pipe Flow
 Is closedchannel
 The top surface iscoveredbysolid
boundary
 It isnot exposedtoatmospheric
Pressure.
Where HGL - Hydraulicgrade line (coincidewithwatersurface)
EGL - Energygrade line
Hf - headlossdue to friction
g
V
2
2
Y2
Fig 1(a) Pipe flow
Z2 Z1
Z2
Y1
Y1
Y2
Hf
HGL
EL
HGL
EL
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 2
V2
/2g- velocityhead
Types ofchannels
 Natural channels:These channelsnaturallyexistwithoutthe influence of humanbeings.
E.g. Rivers,streams,tidal estuaries,aqueducts.
 Artificial channels:Such channelsare formedbyman’sactivityforvarious
purposes.E.g.irrigationchannel,navigationchannel,sewerage channel,culverts,power
canal…… etc.
 Prismaticchannel:- channelswithconstantshape andslope.
 Non-prismaticchannels:- channelswithvaryingshape andslope.
 Open channel:-A channel withoutanycoverat the top.
 Closedchannel:-The channel havingacoverat the top.
ACTIVITY1.1
What isopenchannel?
What are the differenttypesof channel? Give example ineachcase.
1.1 Types of flow in open channel
The flow ina channel classifiedintothe followingtype,dependingonthe change inthe
depthof flowwithrespecttospace and time.
a) Steadyflow&Unsteadyflow
b) Uniformflow&Nonuniformflow
c) Steadyuniformflow&Unsteadyuniformflow
d) . Unsteadyuniformflow
Time as criteria
Steady flow & Unsteady flow
Whenthe flow characteristic(suchasdepthof flow,flow velocity andthe flow rate atany cross
section)donotchange withrespecttotime,the flow inachannel isto be steady.
Mathematically, 0

t
V


, 0

t
p


and 0

t
y


The flowissaidto be unsteadyflow whenthe flow parametervarywithtime.
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 3
Mathematically,
0

t
V


,
0

t
p


and
0

t
y


Space as a criterion
Uniformflow& Non uniformflow
Flowina channel issaidto be uniformif the depth,slope,cross-sectionandvelocityremain
constantovera givenlengthof the channel.
Mathematically,
0

s
V


, and
0

s
y


Flowinchannel issaidto be non-uniform(varied)whenthe channel depthvariescontinuously
fromone sectionto another.
Mathematically,
0

s
V


,and
0

s
y


Time and space as a criteria
Steady uniformflow: - The depthof flow doesnotchange duringtime interval andspace under
consideration.
Unsteadyuniform flow:- Thisis a flow inwhichthe depthisvaryingtime butnot withspace.
Unsteadynon uniform flow:- Is the flow inwhichthe depthisvaryingwithspace and time.
ACTIVITY 1.2
Explainbrieflythe following:
1. Steadyand Un steadyflow
2. Uniformand nonuniformflow
3. State the conditionunderwhichuniform andnonuniformflowsare produced.
1.2 Geometric elements of open channel section
Geometricelementsare propertiesof achannel sectionthatcan be definedentirelybythe
geometryof the sectionandthe depthof flow.The mostusedgeometricpropertiesinclude:
1. Depth of flow(y):itthe vertical distance fromthe lowestpointof the channel tothe free
surface.
2. Top width (T): itis the widthof channel sectionatfree surface.
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 4
3. Stage (h):is the elevationorvertical distanceof the free surface above a datum.
4. Wettedperimeter(p):itisthe lengthof the channel boundarywhichisincontact with
water.
5. Wettedarea (A):is the cross-sectional areaof the flow normal tothe directionof flow.
6. Hydraulic radius(hydraulicmean depth)(R) : itis the ratioof wettedareato itswetted
perimeter
R=
P
A
7. Hydraulic depth(D):the ratio of wettedareato the top width,
D=
T
A
8. Sectionfactor (Z):is the productof the wettedareaand the two-thirdpowerof the
hydraulicradius
Z=A D =A
T
A
=
2
1
3








T
A
=A 3
2
R
9. Conveyance (K) :
Q=VA………………………….V= 2
1
3
2
1
S
R
n
Q=A 2
1
3
2
1
S
R
n
=A 3
2
R 2
1
1
S
n
=K 2
1
S S= bedslope
K=
n
1
A 3
2
R n= Mannings constant
= CA R c= Chezy’sconstant
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 5
Where S0- bedslope of channel
Sw- Water surface slope
S- Slope of EGL
W – Weightof water
0 – Shearforce
L- Lengthof channel
Uniformflowisthe resultof exactbalance betweenthe gravityandfrictionforce
Wsin= o
 .P.L…………………………….(1)
A L sin= o
 .P.L
But sin = hf/L= S, solvingfor o
 ,
o
 = S
R
S
P
A
.
. 
  ………………………………… (2)
Where - unitweightof the water
The shear stressisassumedproportional tothe square of the meanvelocity,
or o= kV
2
…………………………………..……..(3)
Therefore,kv2
=RS
V2
= RS
k

,
L
W
Wsinө
S0
Sw
Y0
S
0
X
Hf (Z)
Fig. 1.2
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 6
Let 2
C
k


-constant(b/c&k- are constant)
.
RS
C
V  ……………………………………………….... (4)
Thisis the Chezy –formula
C= chezycoefficient(chezy’sresistancefactor)
V= Average velocityof flow
ManningFormula
V= 2
1
3
2
0
1
S
R
n
………………………………………………(5)
 The bestas well asmostwidelyusedformulaforuniformlyforuniformflow.
n- isthe roughnesscoefficient.
A relationbetweenthe Chezy’sCandManning’sn maybe obtainedbycomparingeqn(4) & (5)
n
R
C
6
1
 …………………………………………..(6)
 The value of n ranges from0.009 (forsmoothstraightsurfaces) to0.22 (for very
dense floodplainforests).
? What is hydraulic efficiency channel (most economical channel) means
 A channel sectionissaidtobe efficient(economical)if itgivesthe maximumdischarge
for the givenshape,areaandroughness.
1.3 Most economical channel section
Most economical rectangular channel section
Let B and Y be the base widthanddepthof flow respectively
A=BY………………………….(i)
P=B=2Y……………………….(ii)
From eqn.(i),B=A/Y
Substitutingin(ii) P=A/Y+2Y………….(iii)
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 7
For maximumQ,P- is minimum.
0
)
2
/
(
0 


 Y
Y
A
dY
d
dY
dp
0
2 



Y
A
Y
B
Y
A *
2 2



So,B=2Y (orY=B/2)
Thus the rectangularchannel ismostefficientandeconomical whenthe depthof waterisone
half of the widthof the channel andthe discharge flow will be maximum.
EXAMPLE -1
1 .A rectangularchannel isto be dug inthe rocky portionof a soil.Finditsmosteconomical
cross-sectionif itstoconvey12 m3
/sof waterwithan average velocityof 3 m/s.Take chezy
constantC=50
Given
Q=12 m3
/s
V=3 m/s
C=50
Solution
The geometricrelationsforoptimumdischargethrougharectangularchannel are
Then
WhenB,Y andR are base width,depthof flow andhydraulicradiusrespectively
Now
From thisequationsolvefordepthof flow
Therefore base widthof flow
Hydraulicradius,
Also chezyformula
Hence
Mosteconomicaltrapezoidalchannelsection
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 8
But for mosteconomicsection
EXAMPLE-2
An irrigationchannel of trapezoidal sectionhasside slope,m=2and carries a discharge of
15m3
/s on a longitudinalslope of 1in5000. The channel isto be linedforwhichthe value of
frictioncoefficientinManning’sformulaisn=0.012. Findthe dimensionof the mosteconomic
sectionof the channel.
GIVEN
Side slope m=2
Discharge Q=15m3
/s
Longitudinal slopeS=1:5000
Manning´scoefficientn=0.012
SOLUTION
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 9
ACTIVITY1.3
What do youmeanby mosteconomical sectionof anopenchannel?How isit determined?
What are the conditionsforthe rectangularchannel of bestsection?
Showthat the hydraulicmeandepthof a trapezoidal
1.4 Specific energy
? What is specific energy
 Specificenergyisthe energy perunitweightof flowingliquidabove the channel
bottom.
For any cross section,shape,the specificenergy( E) at a particularsectionisdefinedasthe
energyheadtothe channel bedasdatum.Thus,
g
V
Y
E
2
2


 ……………………………………………..(1)
( - iskineticenergycorrectionfactor 1)
ET1
ET2
Datum
EGL
Z1
Z2
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 10
Fig1.3 SpecificEnergyata particularsection
For a rectangularchannel,the value of flow perunitwidthisQ/B=q,andaverage velocity
Y
q
BY
qB
A
Q
V 


Therefore eqn(1) becomes:
2
2
2
2
2 gy
q
y
g
y
q
y
E 








 …………………………………… (2)
g
q
Y
y
E
2
)
(
2
2

 (Forthe case of constantq)………………………… (3)
A plotof E VsY isa hyperbolalike withasymptotes(E-Y) =0i.e.E=Y and y=0. Such a curve is
knownas specificenergydiagram.
Y
Sub critical
section
Super critical
E
Y1
Yc
Y2
Ec E0
Specific Energy diagram
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 11
For a particularq, we see there are twopossible valuesof Yfora givenvalue of E.These are
knownas Alternativedepths(fore.g.Y1 & Y2 on fig.above)
 The two alternative depthsrepresenttwototallydifferentflow regimesslow &deepon
the upperlimpof the curve (subcritical flow) &fastand shallow onthe lowerlimbof
the curve.(supercritical flow)
? What is critical depth
 Depthof flowatwhich specificenergyis minimumis calledcritical depth.
The velocityof flowatcritical depthisknownas critical velocity.
For example,arelationforcritical depthinawide rectangularchannel canbe foundby
differentiationEof eqn.2withrespecttoY to findthe value of Y for whichE isa minimum.
3
2
1
gy
q
dY
dE

 …………………………………………….. (4)
AndwhenE isa minimumY=Ycand 0

dy
dE , so that
3
2
3
2
1
0 c
c
gy
q
gY
q



 ………………………………. (5)
Substitutingq=vy= VC*Yc, gives
c
c gy
V 
2
c
c
c
y
q
gy
V 

 ……………………………………….. (6)
It may be expressedas:
3
1
2
2










g
q
g
V
y c
c ……………………………………….. (7)
From eqn(7) ,
2
2
2
c
c y
g
V
 hence,
c
c
c
c
c
c y
y
y
g
V
y
E
E
2
3
2
1
2
2
min 




 ……………… (8)
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 12
And min
3
2 E
yc  ……………………………………………………………..(9)
From eqn.(7):
3
max c
gy
q  ……………………………………….………….(10)
For nonrectangularcross sectionthe specificenergyeqn.
2
2
2gA
Q
y
E 
 …………………………………………………….. (11)
[V=Q/A]
To findthe critical depth,
dy
dA
gA
Q
dy
dE
3
2
1
 ………………………………………………….. (12)
Fromfig1.3 (b) dA = dy*T (at Yc, T= Tc)
Therefore the above equationbecomes:
1
3
2
max

c
c
gA
T
Q
…………………………………………………………….. (13)
The critical depthmustsatisfythisequation
From eqn.(13)
c
c
T
gA
Q
3
2
 and substitute in eqn.(11) then
c
c
c
c
T
A
y
E
2

 …………………………………………………………..(14)
eqn.(13) can be solvedbytrial & error forirregularsectionbyplotting 3
2
)
(
gA
T
Q
y
f  and
critical depthoccurs forthe value of y whichmakesf(y)=1
What are sub critical, critical, and super critical flow?
 Subcritical flow:-whenthe depthof flow inachannel isgreaterthanthe critical
depth(Yc) inthiscase Fr <1
 Critical flowisone inwhichspecificenergyisminimum.A few correspondingtocritical
depthalsoknownas critical flow.
 Supercritical flow:-whenthe depthof flow inachannel islessthancritical depth(Yc) in
thiscase Fr>1.
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 13
If specificenergycurve forQ- constantis redraw alongside asecondcurve of depthagainst
discharge forconstantE, will showthe variationof discharge withdepth.
Y
yc
q qmax
For a givenconstantdischarge fig
i) The specificenergycurve hasa minimumvalue Ecat pointC witha corresponding
depthYc knownas critical depth.
ii) For any othervalue of E there are two possible depthof flow knownasalternative
depthone of whichistermedsubcritical (y>Yc) and the othersupercritical (Y<Yc).
a) For a given constantspecificenergy( fig.1.5(b))
i) the depthdischarge curve showsthatdischarge isa maximumat the critical
depth
ii) For all otherdischargesthere are twopossible depthof flow ( sub- &super
critical) forany particularvalue of E,
Fromeqn.(13) above if we substitute
Q= AV (continuityequation),we get
1
3
2

gA
T
Q
1
1
2
3
2
2



gA
T
V
gA
T
V
A
butA/T = D ( Hydraulicdepth),then[ D=Y forrectangularsection)
gy
V
gy
V


 1
2
……………………………(*)

1
gy
V
Froude numberatcritical state.
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 14
gy
V
F  ……………………………………….(**)
Thus, i) F= 1critical flow
ii) F< 1 subcritical flowType equationhere.
iii)F>1Supercritical
ACTIVITY1.4
What isspecificenergy andspecificenergycurve?
What do youunderstandbycritical depthof an openchannel whenthe flow initisnotuniform?
Examples
1. For constant specificenergyof ,calculate the maximumdischargethatmay occur in
a rectangular channel 5m width.
Given
Solution
For constant specific energy discharge is maximum
2. Most efficient rectangular channel, which is laid on a bottom slope of 0.0064, is to carry
20m3
/s of water.Determine the widthof the channel whenthe flow isincritical condition. Take
n=0.015.
Given
Solution
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 15
1.5 Hydraulic jump
What is hydraulic jump?
 A flowphenomenonwhich occurswhensupercritical flow hasitsvelocityreducedtosub
critical.There issuddenrise inwaterlevel atthe pointwhere hydraulicjumpoccurs
e.g (Rapidlyvariedflow).
Hydraulicjumponhorizontal bedfollowingoveraspillway
Where V1-velocitybeforejump
V2 –velocityafterjump
Y1 –waterdepth before jump
Y2 –waterdepthafterjump
Lj –lengthof jump
Y2
V2
Y1
V1
Lj
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 16
Purposes of hydraulic jump:-
i) To increase the waterlevel onthe d/sof the hydraulicstructures
ii) To reduce the netup liftforce byincreasingthe downwardforce due tothe
increaseddepthof water,
iii) To increase the discharge froma sluice gate byincreasingthe effective headcausing
flow,
iv) For aerationof drinkingwater
v) For removingairpocketsina pipe line
vi) Reduce downstreamerosion
vii) Veryuseful &effective formixingfluids
 Analysis of hydraulic jump
Assumptions
a. The lengthof the hydraulicjumpissmall,consequently,the lossof headdue tofriction
isnegligible,
b. The channel ishorizontal asit has a verysmall longitudinal slope. The weight
componentinthe directionof flow isnegligible.
c. The portionof channel inwhichthe hydraulicjumpoccursistakenas a control volume
& it isassumedthe justbefore &afterthe control volume,the flow isuniform&
pressure distributionishydrostatic.
Let usconsidera small reachof a channel inwhichthe hydraulicjumpoccurs.
The momentumof waterpassingthroughsection(1) perunittime isgivenas:
1
1
1
QV
g
rQV
t
p


 ……………………………………….(i)
Momentumat section(2) perunittime is:
2
2
2
QV
g
rQV
t
p


 ………………………………………….(ii)
Rate of change of momentumb/nsection1& 2
)
( 1
2 V
V
Q
t
P



 ……………………………………….(iii)
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 17
The net force inthe directionof flow =F1-F2 ………………..(iv)
2
2
2
1
1
1 , Y
A
F
Y
A
F 
 

 2
1 &Y
Y are the centerof pressure atsection(1) & (2)
Therefore F1-F2 =M=Q (V2-V1)
)
( 1
2
2
2
1
1 V
V
g
Q
Y
A
Y
A 




 ……………………………………(v)
From continuityeqn.Q=A*V, V= Q/A,so
)
.........(
..........
..........
..........
..........
1
1
1
2
2
2
2
1
1
1
2
2
2
1
1
iv
A
A
g
Q
Y
A
Y
A
A
Q
A
Q
g
Q
Y
A
Y
A





 


















Rearrangingthiseqn.:
2
1
2
2
2
2
1
1
1
2
M
M
Y
A
gA
Q
Y
A
gA
Q















= Constant.…………… (vii)
M1and M2 are the specificforcesatsection(1) & (2) indicatesthatthese forcesare equal
before &afterthe jump.
Y1= initial depth
Y2 = sequentdepth
Hydraulicjumpina rectangularchannel
A1=By1 the sectionhasuniformwidth(B)
A2= By2
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 18
2
,
2
2
2
1
1
Y
Y
Y
Y 

Nowfromeqn.(Vii) above:
)
..(
..........
..........
..........
..........
..........
..........
2
2
2
*
2
2
2
2
2
2
1
1
2
2
2
2
1
1
1
2
viii
By
Bgy
Q
By
gBy
Q
y
By
gBy
Q
y
By
gBy
Q


















Flowperunitwidthof q= Q/B Q=qB, theneqn.(viii)becomes
2
2
2
2
2
2
2
2
1
1
2
2
By
Bgy
B
q
By
Bgy
B
q



2
1
1 2
1
2
2
2
1
2
y
y
y
y
g
q 







 ………………………………… (.ix)
 
 
1
2
2
1
2
2
2
1
2
2
y
y
y
y
y
y
g
q



)
.........(
..........
..........
..........
..........
..........
0
2
)
........(
..........
..........
..........
..........
..........
).........
(
2
2
2
2
1
2
1
2
2
1
2
1
2
xi
g
q
y
y
y
y
x
y
y
y
y
g
q





Thisis quadraticeqn.& the solutionisgivenas
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 19
)
........(
..........
..........
..........
..........
2
2
2
)
)(
.(
..........
..........
..........
..........
..........
2
2
2
2
2
2
1
1
2
2
2
2
2
2
1
b
gy
q
y
y
y
a
xii
gy
q
y
y
y




















)
)(
...(
..........
..........
..........
..........
..........
8
1
1
(
2
)
..(
..........
..........
..........
..........
).........
8
1
1
(
2
3
1
2
1
2
3
2
2
2
1
d
xii
gy
q
y
y
c
gy
q
y
y








The ratio of conjugate depths;
)
(
..........
..........
..........
..........
..........
8
1
1
(
2
1
)
)(
.......(
..........
..........
..........
..........
8
1
1
(
2
1
3
1
2
1
2
3
2
2
2
1
f
gy
q
y
y
e
xii
gy
q
y
y








3
2
2
2
2
2
2
1
1
1 ,
gy
q
gy
y
q
gy
V
F
gy
V
F 



)
..(
..........
..........
..........
..........
8
1
1
(
2
1
)
.....(
..........
).........
8
1
1
(
2
1
2
1
1
2
2
2
2
1
h
F
y
y
g
F
y
y
Therefore








Energy dissipation in a Hydraulic Jump
The head losshl.f causedby the jumpisthe drop inenergyfromsection(1) to (2) or:
hlf= E = E1 - E2
a
g
V
y
g
V
y )
1
.........(
..........
..........
..........
2
2
2
2
2
2
1
1 



















)
......(
..........
..........
..........
2
2 2
2
2
2
2
1
2
1 b
gy
q
y
gy
q
y 



















Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 20
From eqn.(x) substituting: )
(
2
2
1
2
1
2
y
y
y
y
g
q

 intothiseqn.& byrearranging:
  )
2
......(
..........
..........
..........
..........
4 2
1
3
1
2
y
y
y
y
E
hlf




Therefore powerlost= Q hlf (kw)…………………(3)
ACTIVITY1.5
What ismean byhydraulicjumpinopenchannel andhow itoccurs?
Types of Hydraulic jump
Hydraulicjumpsare classifiedaccordingtothe upstream Froude numberanddepthratio.
F1 Y2/y1 Classification
<1 1 Jumpimpossible
1-1.7 1-2 Undularjump(standingwave)
1.7-2.5 2-3.1 Weakjump
2.5-4.5 3.1-5.9 Oscillatingjump
4.5-9.0 5.9-12 Steadyjump(45-70% energyloss)
>9.0 >12 Strongor choppingjump(=85% energyloss)
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 21
Examples
A 3∙6m wide rectangularchannel conveys of waterwithavelocityof .
a. Is there a conditionforhydraulicjump occur?If so calculate the height,length and
strengthof the jump.
b. What islossof energy?
Given
Solution
a.
b. Loss of energyforrectangularchannel
Hydraulics II Module 2011
Hydraulic and Water Resource Engineering Dep. AMIT 22
Exercises
1. A rectangular channel which is laid on a bottom slope of 0.0064 is to carry 20m3
/s of water.
Determine the width of the channel when the flow is in critical condition. Take C=66
2.An irrigationcanal of trapezoidal sectionhavingside slope 2in3 isto carry a flow of 10m3
/s on
a longitudinal slope of 1in5000. The canal is linedforwhichthe value of frictional coefficient in
Manning’s formula is n=0.012. Find the dimension of the most economical section
3. Determine the side slopeof the mosthydraulicallyefficienttriangularsection. .Show that the
head loss in a hydraulic jump formed in a rectangular channel may be expressed as
ΔE= (V1 –V2)3
/ [2g (V1 +V2)]
4. A rectangular channel there occurs a jump corresponding to Froude number (F=2.5).
Determine the critical depth and head loss in terms of the initial depth y1.
5. A trapezoidal channel havingbottomwidth10mand side slope 2:1(H:V) carries a discharge of
100m3
/s. Findthe depthconjugate tothe initial depthof 1mbefore the jump.Also determine
the loss of energy in the jump.

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Chapter 1

  • 1. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 1 CHAPTER ONE INTRODUCTION TO OPEN CHANNEL FLOW Course objectives:- atthe endof the course studentswill be expectedto: o Differentiate betweenopenchannel &pipe flows o Understandfluidphenomenonapplicationtodesignvarioushydraulicstructures o Acquaintlaminar&turbulentflows,boundarylayertheory,velocity-shearstress relationship ? WHAT IS OPEN CHANNEL Openchannel:- It maybe definedasapassage inwhichliquidflowswithitsuppersurface exposedtothe atmosphere.e.g. :- curvets,spillways,andsimilarhumanmade structures Differencesb/nthe flowinpipes&openchannel flow Openchannel flow  Is exposedtoatmosphericpressure.  The cross-sectional areaof the flow is variable.(thatdependsonmany parametersof the flow)  The force causingmotionisgravity. Pipe Flow  Is closedchannel  The top surface iscoveredbysolid boundary  It isnot exposedtoatmospheric Pressure. Where HGL - Hydraulicgrade line (coincidewithwatersurface) EGL - Energygrade line Hf - headlossdue to friction g V 2 2 Y2 Fig 1(a) Pipe flow Z2 Z1 Z2 Y1 Y1 Y2 Hf HGL EL HGL EL
  • 2. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 2 V2 /2g- velocityhead Types ofchannels  Natural channels:These channelsnaturallyexistwithoutthe influence of humanbeings. E.g. Rivers,streams,tidal estuaries,aqueducts.  Artificial channels:Such channelsare formedbyman’sactivityforvarious purposes.E.g.irrigationchannel,navigationchannel,sewerage channel,culverts,power canal…… etc.  Prismaticchannel:- channelswithconstantshape andslope.  Non-prismaticchannels:- channelswithvaryingshape andslope.  Open channel:-A channel withoutanycoverat the top.  Closedchannel:-The channel havingacoverat the top. ACTIVITY1.1 What isopenchannel? What are the differenttypesof channel? Give example ineachcase. 1.1 Types of flow in open channel The flow ina channel classifiedintothe followingtype,dependingonthe change inthe depthof flowwithrespecttospace and time. a) Steadyflow&Unsteadyflow b) Uniformflow&Nonuniformflow c) Steadyuniformflow&Unsteadyuniformflow d) . Unsteadyuniformflow Time as criteria Steady flow & Unsteady flow Whenthe flow characteristic(suchasdepthof flow,flow velocity andthe flow rate atany cross section)donotchange withrespecttotime,the flow inachannel isto be steady. Mathematically, 0  t V   , 0  t p   and 0  t y   The flowissaidto be unsteadyflow whenthe flow parametervarywithtime.
  • 3. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 3 Mathematically, 0  t V   , 0  t p   and 0  t y   Space as a criterion Uniformflow& Non uniformflow Flowina channel issaidto be uniformif the depth,slope,cross-sectionandvelocityremain constantovera givenlengthof the channel. Mathematically, 0  s V   , and 0  s y   Flowinchannel issaidto be non-uniform(varied)whenthe channel depthvariescontinuously fromone sectionto another. Mathematically, 0  s V   ,and 0  s y   Time and space as a criteria Steady uniformflow: - The depthof flow doesnotchange duringtime interval andspace under consideration. Unsteadyuniform flow:- Thisis a flow inwhichthe depthisvaryingtime butnot withspace. Unsteadynon uniform flow:- Is the flow inwhichthe depthisvaryingwithspace and time. ACTIVITY 1.2 Explainbrieflythe following: 1. Steadyand Un steadyflow 2. Uniformand nonuniformflow 3. State the conditionunderwhichuniform andnonuniformflowsare produced. 1.2 Geometric elements of open channel section Geometricelementsare propertiesof achannel sectionthatcan be definedentirelybythe geometryof the sectionandthe depthof flow.The mostusedgeometricpropertiesinclude: 1. Depth of flow(y):itthe vertical distance fromthe lowestpointof the channel tothe free surface. 2. Top width (T): itis the widthof channel sectionatfree surface.
  • 4. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 4 3. Stage (h):is the elevationorvertical distanceof the free surface above a datum. 4. Wettedperimeter(p):itisthe lengthof the channel boundarywhichisincontact with water. 5. Wettedarea (A):is the cross-sectional areaof the flow normal tothe directionof flow. 6. Hydraulic radius(hydraulicmean depth)(R) : itis the ratioof wettedareato itswetted perimeter R= P A 7. Hydraulic depth(D):the ratio of wettedareato the top width, D= T A 8. Sectionfactor (Z):is the productof the wettedareaand the two-thirdpowerof the hydraulicradius Z=A D =A T A = 2 1 3         T A =A 3 2 R 9. Conveyance (K) : Q=VA………………………….V= 2 1 3 2 1 S R n Q=A 2 1 3 2 1 S R n =A 3 2 R 2 1 1 S n =K 2 1 S S= bedslope K= n 1 A 3 2 R n= Mannings constant = CA R c= Chezy’sconstant
  • 5. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 5 Where S0- bedslope of channel Sw- Water surface slope S- Slope of EGL W – Weightof water 0 – Shearforce L- Lengthof channel Uniformflowisthe resultof exactbalance betweenthe gravityandfrictionforce Wsin= o  .P.L…………………………….(1) A L sin= o  .P.L But sin = hf/L= S, solvingfor o  , o  = S R S P A . .    ………………………………… (2) Where - unitweightof the water The shear stressisassumedproportional tothe square of the meanvelocity, or o= kV 2 …………………………………..……..(3) Therefore,kv2 =RS V2 = RS k  , L W Wsinө S0 Sw Y0 S 0 X Hf (Z) Fig. 1.2
  • 6. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 6 Let 2 C k   -constant(b/c&k- are constant) . RS C V  ……………………………………………….... (4) Thisis the Chezy –formula C= chezycoefficient(chezy’sresistancefactor) V= Average velocityof flow ManningFormula V= 2 1 3 2 0 1 S R n ………………………………………………(5)  The bestas well asmostwidelyusedformulaforuniformlyforuniformflow. n- isthe roughnesscoefficient. A relationbetweenthe Chezy’sCandManning’sn maybe obtainedbycomparingeqn(4) & (5) n R C 6 1  …………………………………………..(6)  The value of n ranges from0.009 (forsmoothstraightsurfaces) to0.22 (for very dense floodplainforests). ? What is hydraulic efficiency channel (most economical channel) means  A channel sectionissaidtobe efficient(economical)if itgivesthe maximumdischarge for the givenshape,areaandroughness. 1.3 Most economical channel section Most economical rectangular channel section Let B and Y be the base widthanddepthof flow respectively A=BY………………………….(i) P=B=2Y……………………….(ii) From eqn.(i),B=A/Y Substitutingin(ii) P=A/Y+2Y………….(iii)
  • 7. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 7 For maximumQ,P- is minimum. 0 ) 2 / ( 0     Y Y A dY d dY dp 0 2     Y A Y B Y A * 2 2    So,B=2Y (orY=B/2) Thus the rectangularchannel ismostefficientandeconomical whenthe depthof waterisone half of the widthof the channel andthe discharge flow will be maximum. EXAMPLE -1 1 .A rectangularchannel isto be dug inthe rocky portionof a soil.Finditsmosteconomical cross-sectionif itstoconvey12 m3 /sof waterwithan average velocityof 3 m/s.Take chezy constantC=50 Given Q=12 m3 /s V=3 m/s C=50 Solution The geometricrelationsforoptimumdischargethrougharectangularchannel are Then WhenB,Y andR are base width,depthof flow andhydraulicradiusrespectively Now From thisequationsolvefordepthof flow Therefore base widthof flow Hydraulicradius, Also chezyformula Hence Mosteconomicaltrapezoidalchannelsection
  • 8. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 8 But for mosteconomicsection EXAMPLE-2 An irrigationchannel of trapezoidal sectionhasside slope,m=2and carries a discharge of 15m3 /s on a longitudinalslope of 1in5000. The channel isto be linedforwhichthe value of frictioncoefficientinManning’sformulaisn=0.012. Findthe dimensionof the mosteconomic sectionof the channel. GIVEN Side slope m=2 Discharge Q=15m3 /s Longitudinal slopeS=1:5000 Manning´scoefficientn=0.012 SOLUTION
  • 9. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 9 ACTIVITY1.3 What do youmeanby mosteconomical sectionof anopenchannel?How isit determined? What are the conditionsforthe rectangularchannel of bestsection? Showthat the hydraulicmeandepthof a trapezoidal 1.4 Specific energy ? What is specific energy  Specificenergyisthe energy perunitweightof flowingliquidabove the channel bottom. For any cross section,shape,the specificenergy( E) at a particularsectionisdefinedasthe energyheadtothe channel bedasdatum.Thus, g V Y E 2 2    ……………………………………………..(1) ( - iskineticenergycorrectionfactor 1) ET1 ET2 Datum EGL Z1 Z2
  • 10. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 10 Fig1.3 SpecificEnergyata particularsection For a rectangularchannel,the value of flow perunitwidthisQ/B=q,andaverage velocity Y q BY qB A Q V    Therefore eqn(1) becomes: 2 2 2 2 2 gy q y g y q y E           …………………………………… (2) g q Y y E 2 ) ( 2 2   (Forthe case of constantq)………………………… (3) A plotof E VsY isa hyperbolalike withasymptotes(E-Y) =0i.e.E=Y and y=0. Such a curve is knownas specificenergydiagram. Y Sub critical section Super critical E Y1 Yc Y2 Ec E0 Specific Energy diagram
  • 11. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 11 For a particularq, we see there are twopossible valuesof Yfora givenvalue of E.These are knownas Alternativedepths(fore.g.Y1 & Y2 on fig.above)  The two alternative depthsrepresenttwototallydifferentflow regimesslow &deepon the upperlimpof the curve (subcritical flow) &fastand shallow onthe lowerlimbof the curve.(supercritical flow) ? What is critical depth  Depthof flowatwhich specificenergyis minimumis calledcritical depth. The velocityof flowatcritical depthisknownas critical velocity. For example,arelationforcritical depthinawide rectangularchannel canbe foundby differentiationEof eqn.2withrespecttoY to findthe value of Y for whichE isa minimum. 3 2 1 gy q dY dE   …………………………………………….. (4) AndwhenE isa minimumY=Ycand 0  dy dE , so that 3 2 3 2 1 0 c c gy q gY q     ………………………………. (5) Substitutingq=vy= VC*Yc, gives c c gy V  2 c c c y q gy V    ……………………………………….. (6) It may be expressedas: 3 1 2 2           g q g V y c c ……………………………………….. (7) From eqn(7) , 2 2 2 c c y g V  hence, c c c c c c y y y g V y E E 2 3 2 1 2 2 min       ……………… (8)
  • 12. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 12 And min 3 2 E yc  ……………………………………………………………..(9) From eqn.(7): 3 max c gy q  ……………………………………….………….(10) For nonrectangularcross sectionthe specificenergyeqn. 2 2 2gA Q y E   …………………………………………………….. (11) [V=Q/A] To findthe critical depth, dy dA gA Q dy dE 3 2 1  ………………………………………………….. (12) Fromfig1.3 (b) dA = dy*T (at Yc, T= Tc) Therefore the above equationbecomes: 1 3 2 max  c c gA T Q …………………………………………………………….. (13) The critical depthmustsatisfythisequation From eqn.(13) c c T gA Q 3 2  and substitute in eqn.(11) then c c c c T A y E 2   …………………………………………………………..(14) eqn.(13) can be solvedbytrial & error forirregularsectionbyplotting 3 2 ) ( gA T Q y f  and critical depthoccurs forthe value of y whichmakesf(y)=1 What are sub critical, critical, and super critical flow?  Subcritical flow:-whenthe depthof flow inachannel isgreaterthanthe critical depth(Yc) inthiscase Fr <1  Critical flowisone inwhichspecificenergyisminimum.A few correspondingtocritical depthalsoknownas critical flow.  Supercritical flow:-whenthe depthof flow inachannel islessthancritical depth(Yc) in thiscase Fr>1.
  • 13. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 13 If specificenergycurve forQ- constantis redraw alongside asecondcurve of depthagainst discharge forconstantE, will showthe variationof discharge withdepth. Y yc q qmax For a givenconstantdischarge fig i) The specificenergycurve hasa minimumvalue Ecat pointC witha corresponding depthYc knownas critical depth. ii) For any othervalue of E there are two possible depthof flow knownasalternative depthone of whichistermedsubcritical (y>Yc) and the othersupercritical (Y<Yc). a) For a given constantspecificenergy( fig.1.5(b)) i) the depthdischarge curve showsthatdischarge isa maximumat the critical depth ii) For all otherdischargesthere are twopossible depthof flow ( sub- &super critical) forany particularvalue of E, Fromeqn.(13) above if we substitute Q= AV (continuityequation),we get 1 3 2  gA T Q 1 1 2 3 2 2    gA T V gA T V A butA/T = D ( Hydraulicdepth),then[ D=Y forrectangularsection) gy V gy V    1 2 ……………………………(*)  1 gy V Froude numberatcritical state.
  • 14. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 14 gy V F  ……………………………………….(**) Thus, i) F= 1critical flow ii) F< 1 subcritical flowType equationhere. iii)F>1Supercritical ACTIVITY1.4 What isspecificenergy andspecificenergycurve? What do youunderstandbycritical depthof an openchannel whenthe flow initisnotuniform? Examples 1. For constant specificenergyof ,calculate the maximumdischargethatmay occur in a rectangular channel 5m width. Given Solution For constant specific energy discharge is maximum 2. Most efficient rectangular channel, which is laid on a bottom slope of 0.0064, is to carry 20m3 /s of water.Determine the widthof the channel whenthe flow isincritical condition. Take n=0.015. Given Solution
  • 15. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 15 1.5 Hydraulic jump What is hydraulic jump?  A flowphenomenonwhich occurswhensupercritical flow hasitsvelocityreducedtosub critical.There issuddenrise inwaterlevel atthe pointwhere hydraulicjumpoccurs e.g (Rapidlyvariedflow). Hydraulicjumponhorizontal bedfollowingoveraspillway Where V1-velocitybeforejump V2 –velocityafterjump Y1 –waterdepth before jump Y2 –waterdepthafterjump Lj –lengthof jump Y2 V2 Y1 V1 Lj
  • 16. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 16 Purposes of hydraulic jump:- i) To increase the waterlevel onthe d/sof the hydraulicstructures ii) To reduce the netup liftforce byincreasingthe downwardforce due tothe increaseddepthof water, iii) To increase the discharge froma sluice gate byincreasingthe effective headcausing flow, iv) For aerationof drinkingwater v) For removingairpocketsina pipe line vi) Reduce downstreamerosion vii) Veryuseful &effective formixingfluids  Analysis of hydraulic jump Assumptions a. The lengthof the hydraulicjumpissmall,consequently,the lossof headdue tofriction isnegligible, b. The channel ishorizontal asit has a verysmall longitudinal slope. The weight componentinthe directionof flow isnegligible. c. The portionof channel inwhichthe hydraulicjumpoccursistakenas a control volume & it isassumedthe justbefore &afterthe control volume,the flow isuniform& pressure distributionishydrostatic. Let usconsidera small reachof a channel inwhichthe hydraulicjumpoccurs. The momentumof waterpassingthroughsection(1) perunittime isgivenas: 1 1 1 QV g rQV t p    ……………………………………….(i) Momentumat section(2) perunittime is: 2 2 2 QV g rQV t p    ………………………………………….(ii) Rate of change of momentumb/nsection1& 2 ) ( 1 2 V V Q t P     ……………………………………….(iii)
  • 17. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 17 The net force inthe directionof flow =F1-F2 ………………..(iv) 2 2 2 1 1 1 , Y A F Y A F      2 1 &Y Y are the centerof pressure atsection(1) & (2) Therefore F1-F2 =M=Q (V2-V1) ) ( 1 2 2 2 1 1 V V g Q Y A Y A       ……………………………………(v) From continuityeqn.Q=A*V, V= Q/A,so ) .........( .......... .......... .......... .......... 1 1 1 2 2 2 2 1 1 1 2 2 2 1 1 iv A A g Q Y A Y A A Q A Q g Q Y A Y A                          Rearrangingthiseqn.: 2 1 2 2 2 2 1 1 1 2 M M Y A gA Q Y A gA Q                = Constant.…………… (vii) M1and M2 are the specificforcesatsection(1) & (2) indicatesthatthese forcesare equal before &afterthe jump. Y1= initial depth Y2 = sequentdepth Hydraulicjumpina rectangularchannel A1=By1 the sectionhasuniformwidth(B) A2= By2
  • 18. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 18 2 , 2 2 2 1 1 Y Y Y Y   Nowfromeqn.(Vii) above: ) ..( .......... .......... .......... .......... .......... .......... 2 2 2 * 2 2 2 2 2 2 1 1 2 2 2 2 1 1 1 2 viii By Bgy Q By gBy Q y By gBy Q y By gBy Q                   Flowperunitwidthof q= Q/B Q=qB, theneqn.(viii)becomes 2 2 2 2 2 2 2 2 1 1 2 2 By Bgy B q By Bgy B q    2 1 1 2 1 2 2 2 1 2 y y y y g q          ………………………………… (.ix)     1 2 2 1 2 2 2 1 2 2 y y y y y y g q    ) .........( .......... .......... .......... .......... .......... 0 2 ) ........( .......... .......... .......... .......... .......... )......... ( 2 2 2 2 1 2 1 2 2 1 2 1 2 xi g q y y y y x y y y y g q      Thisis quadraticeqn.& the solutionisgivenas
  • 19. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 19 ) ........( .......... .......... .......... .......... 2 2 2 ) )( .( .......... .......... .......... .......... .......... 2 2 2 2 2 2 1 1 2 2 2 2 2 2 1 b gy q y y y a xii gy q y y y                     ) )( ...( .......... .......... .......... .......... .......... 8 1 1 ( 2 ) ..( .......... .......... .......... .......... )......... 8 1 1 ( 2 3 1 2 1 2 3 2 2 2 1 d xii gy q y y c gy q y y         The ratio of conjugate depths; ) ( .......... .......... .......... .......... .......... 8 1 1 ( 2 1 ) )( .......( .......... .......... .......... .......... 8 1 1 ( 2 1 3 1 2 1 2 3 2 2 2 1 f gy q y y e xii gy q y y         3 2 2 2 2 2 2 1 1 1 , gy q gy y q gy V F gy V F     ) ..( .......... .......... .......... .......... 8 1 1 ( 2 1 ) .....( .......... )......... 8 1 1 ( 2 1 2 1 1 2 2 2 2 1 h F y y g F y y Therefore         Energy dissipation in a Hydraulic Jump The head losshl.f causedby the jumpisthe drop inenergyfromsection(1) to (2) or: hlf= E = E1 - E2 a g V y g V y ) 1 .........( .......... .......... .......... 2 2 2 2 2 2 1 1                     ) ......( .......... .......... .......... 2 2 2 2 2 2 2 1 2 1 b gy q y gy q y                    
  • 20. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 20 From eqn.(x) substituting: ) ( 2 2 1 2 1 2 y y y y g q   intothiseqn.& byrearranging:   ) 2 ......( .......... .......... .......... .......... 4 2 1 3 1 2 y y y y E hlf     Therefore powerlost= Q hlf (kw)…………………(3) ACTIVITY1.5 What ismean byhydraulicjumpinopenchannel andhow itoccurs? Types of Hydraulic jump Hydraulicjumpsare classifiedaccordingtothe upstream Froude numberanddepthratio. F1 Y2/y1 Classification <1 1 Jumpimpossible 1-1.7 1-2 Undularjump(standingwave) 1.7-2.5 2-3.1 Weakjump 2.5-4.5 3.1-5.9 Oscillatingjump 4.5-9.0 5.9-12 Steadyjump(45-70% energyloss) >9.0 >12 Strongor choppingjump(=85% energyloss)
  • 21. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 21 Examples A 3∙6m wide rectangularchannel conveys of waterwithavelocityof . a. Is there a conditionforhydraulicjump occur?If so calculate the height,length and strengthof the jump. b. What islossof energy? Given Solution a. b. Loss of energyforrectangularchannel
  • 22. Hydraulics II Module 2011 Hydraulic and Water Resource Engineering Dep. AMIT 22 Exercises 1. A rectangular channel which is laid on a bottom slope of 0.0064 is to carry 20m3 /s of water. Determine the width of the channel when the flow is in critical condition. Take C=66 2.An irrigationcanal of trapezoidal sectionhavingside slope 2in3 isto carry a flow of 10m3 /s on a longitudinal slope of 1in5000. The canal is linedforwhichthe value of frictional coefficient in Manning’s formula is n=0.012. Find the dimension of the most economical section 3. Determine the side slopeof the mosthydraulicallyefficienttriangularsection. .Show that the head loss in a hydraulic jump formed in a rectangular channel may be expressed as ΔE= (V1 –V2)3 / [2g (V1 +V2)] 4. A rectangular channel there occurs a jump corresponding to Froude number (F=2.5). Determine the critical depth and head loss in terms of the initial depth y1. 5. A trapezoidal channel havingbottomwidth10mand side slope 2:1(H:V) carries a discharge of 100m3 /s. Findthe depthconjugate tothe initial depthof 1mbefore the jump.Also determine the loss of energy in the jump.