20/09/2011
1
Dr. Muhammad Ali [email protected]; Internal 652
Solid understanding of open channel flow
An introduction to hydraulic modelling
Sediment transport in open channels
Various types of Dams and forces acting onVarious types of Dams and forces acting onthem
20/09/2011
2
By the end of course the students will be abletoEvaluate geometric features of open channelsBe able to solve uniform flow problemsWork out critical flow conditionsAnalyse hydraulic jump and hydraulicstructuresUnderstanding of classification of varyingflow profiles
Understand the unsteady flow equations andtheir applicationsppUnderstand the basics of hydraulic modellingand similitudeAnalyse the forces acting on dams and designSite selection of water power projects
20/09/2011
3
Steady flow in open channels
◦ Specific energy and critical depth◦ Surface profiles and backwater curves in channels
of uniform cross sections◦ Hydraulic jump and its practical applications◦ Flow over humps and contractions◦ Critical depth meters broad crested weirs andCritical depth meters broad crested weirs and
venturiflumes.
Unsteady flow
◦ Discharge through orifices and weirs under varyinghead◦ Unsteady flow through pipe◦ Water hammer◦ Instantaneous and slow closure of valves◦ Surges in open channelsSurges in open channels
20/09/2011
4
Hydraulic similitude
◦ Similitude in hydraulic model testing◦ Similitude requirement: geometric, kinematic and
dynamic similarity◦ Various dimensionless numbers and their
significance◦ Model techniques and analysisq y
Dams
◦ Brief description of various types of storage dams◦ Forces on dams◦ Design of gravity dams
20/09/2011
5
Water power engineering
◦ Selection of water power sites◦ Components and layout of water power scheme
Sediment transport in open channels
P i f i di id l i l◦ Properties of individual particles◦ Fall velocity◦ Collection and analysis of field data
Branch of scientific and engineering disciplinethat deals with mechanical properties ofp pfluids basically water.
Applied in many civil engineering systemsthat include◦ Water resource management
Fl d d f◦ Flood defence◦ Harbour and port◦ Bridge
20/09/2011
6
◦ Building◦ Environmental protection◦ Hydropower◦ Irrigation◦ Ecosystems etc.
The stream is not completely enclosed bysolid boundaries.
The upper surface of the liquid is in contactwith the atmosphere
Flow normally takes place under the action ofgravity along the slope of the channel.
20/09/2011
7
Flow is not caused by some external head,rather it takes place under the action ofpgravity.
So it is known as free surface flow or gravityflow.
A channel with constant bed slope and samecross section along its length is known as aprismatic channel.
20/09/2011
8
Principle types of open channels include◦ Natural rivers and streams◦ Artificial canals◦ Sewers◦ Tunnels◦ Pipelines not completely filled
Accurate solutions of problems associatedAccurate solutions of problems associatedwith open channels are much morecomplicated than pressurised pipe flow.
Cross sectional shape range from circular toirregular forms of natural streams asgcompared to pipes which are normallycircular.
In pipes degree of roughness ranges fromnew smooth metal to old corroded iron or
l isteel pipes
20/09/2011
9
20/09/2011
10
20/09/2011
11
20/09/2011
12
20/09/2011
13
Smooth timber or concrete to rough irregularbeds in open channels.p
Choice of frictional coefficients is of greateruncertainty.
Some common classifications of openchannels flows are
Classification
Rapidly Varied Gradually Varied
Steady flow Unsteady flow Uniform flow Varied flow
By time By space
p yflow
yflow
20/09/2011
14
Uniform flow refers to the flow whose waterdepth, width, flow area and velocity do notp , , ychange with distance
If there is a change then the flow will becalled varied (non uniform flow).
20/09/2011
15
20/09/2011
16
20/09/2011
17
For the case of open channel flow
◦ S0-slope of channel bed◦ Sw- slope of water surface◦ S-slope of energy line
◦ Relations for all these slopes can be calculated fromthe open channel flow diagramthe open channel flow diagram
20/09/2011
18
The angle Ѳ between the channel bed and thehorizontal is small hence L the distance along thegchannel bed between the two sections is almost equalto Δx, the horizontal distance between the twosections.
Water depth, width, flow area does notchange with distance.g
Flow rate or discharge is an important term inhydraulics…..define?????
Similar to the pipe flow energy equation,replacing pipe diameter D with hydraulicradius R
20/09/2011
19
VRLfh f 2
2
=
By rearranginggRf 2
Lh
RfgV f2
=
RSCV 0RSCV =
fgCl
hS f 2;0 ==
0RSCV =
Is the Chezy’s formula (1775).
Robert Manning (1890) found that value of C varied approximately as Rh
1/6 while others observed that the proportionality was close to 1/n.
20/09/2011
20
Most widely used formula for uniform flow in open channels.p
Irish Engineer (Robert Manning; 1816-1897)
This led to the development of Manning’s formula given byg y
n = Manning’s roughness coefficient.nSR
V 03/2
=
As Q=AV then
20/09/2011
21
3/20
3/5
nPSA
Q =
Researchers have carried out a large numberof experiments to estimate the value ofManning’s n for various channels and can befound in various Hydraulics books.
nP
(in fps)nSR
V 03/2486.1
=
20/09/2011
22
similar to pipe flow, flow in open channelscan either be laminar or turbulent.
Reynold’s number is used to determine thetype of flow
For pipe flow it is well known that
μρVD
=Re
20/09/2011
23
20/09/2011
24
The limits for each type of flow arelaminar Re<2000a a e 000turbulent Re>4000
If R (hydraulic radius) is the characteristiclength and R=D/4 (for ??????)
4Re
4Re pipe
channelVDVR
===μρ
μρ
20/09/2011
25
So for open channel
laminar Re channel <500turbulent Re channel >1000
In practice, limit for turbulent flow is not well defined so normally it is taken as 2000. y
Two types of problems
Determine the discharge given the depthDetermine the depth given the discharge
In steady uniform flow, the flow depth isknown as the normal depth.p
20/09/2011
26
The normal depth of flow in a trapezoidalconcrete lined channel is 2m. The channelhas a base width of 5m and side slopes of 1Vto 2H. Manning’s n is 0.015 and the bedslope is 0.001. Determine the discharge Qand mean velocity V.
b = 5m, y = 2m, x=2, n = 0.015, So = 0.01.
To find Q, V.
Since we want to use Manning’s formula
SR 3/2
nSR
V 0=
20/09/2011
27
So determine hydraulic radius R first.
A and P needed.
A = 18 sq.mP = 13.94 m.
R = A/P = 1.29 mV = 2.3 m/s and Q = 45 cumecs.
In the previous problem, if the discharg in the channel is 30 cumecs. Find the normal depth pof flow.
b = 5m, Q = 30 cumecs,x = 2, n = 0.015, So = 0.001.
y = ?
20/09/2011
28
Get A in terms of yGet P in terms of yGet te s o yThen get the relation for Q (Manning’s eq) in terms of y
3/2
3/5
)525(])25[(23.14yyy
++
=
Use trial and error to find a value of y for which RHS is equal to 14.23. ans = 1.65.
)( y
Channel bed slope◦ 0.001◦ 0.1 percent◦ 1:1000◦ 1V:1000H◦ 1m drop over 1 km lengthChannel side slope◦ 1:2◦ 1:2◦ 1V:2H◦ X = 2 (x is the inverse slope)
20/09/2011
29
Measured velocity across a channel crosssection varies because of the presence ofpfriction along the boundary.
Due to existence of free surface the velocitydistribution is not symmetric too.
The maximum velocity is expected to befound at the free surface where the shearforce is zero, but it is not so.
Rather the maximum velocity is found justbeneath the surface.
This is due to the presence of secondarycurrents which circulate from the boundariestowards the centre.
Also due to the resistance at the air/waterinterface.
20/09/2011
30
Typical velocity distributions are shown below
20/09/2011
31
20/09/2011
32
Specific energy at a particular section isdefined with respect to channel bottom i.epz=0.
For a relatively wide channel with uniformg
VyE2
2
+=
depth, flow at the centre is unaffected by theside boundaries and flow per unit width canbe expressed as
=///
/bbAQV
bQq
E varies with y if Q remains constant .
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
===
2
2
21
///
yq
gyE
yqbyqbAQV
E varies with y if Q remains constant .Occurs if slope of a wide rectangular channelis changed with flow remaining constant
20/09/2011
33
For constant ‘q’ the specific energy equationcan be written as
constant
A plot of ‘E’ vs. ‘y’ is a parabola withasymptotes
( ) ==−gqyyE2
22
;0)( =− yE
0
;0)(
=
=
yand
yEor
yE
20/09/2011
34
Two flow regimes
‘C’ represents the dividing point between thetwo flow regimes
At ‘C’ for a given ‘q’ E is minimum and thepoint at this point is known as critical flow.
Depth of flow is critical depth yc
20/09/2011
35
To determine the critical depth differentiate E with respect to y and equate to zero.p y q
32
3
2
01
gyqgyq
dydE
=
=−=
Putting
Vyq =
cc
V
gyV =2
Where ‘c’ indicates critical flow conditions. Also
cc gyV =
3/122
⎟⎞
⎜⎛ qV
⎟⎟⎠
⎞⎜⎜⎝
⎛==
gq
gV
y cc
20/09/2011
36
c yV 12
=
Hence
cyg 22=
2
min 23
2y
gV
yEE cc
cc =+==
min32
32
22
EEy
g
cc ==
Also since2q
Differentiating w r t ‘y’ and equating to zero
)(22 yEgyq
−=
1 ⎟⎞
⎜⎛ ydq 0
212 =⎟
⎟⎠
⎞⎜⎜⎝
⎛
−−−=
yEyyEg
dydq
Eyc 32
=
20/09/2011
37
It can be said that maximum discharge occurswhen specific energy is minimum i.e thep gydepth is critical.
( ) 3max 2 cccc gyyEgyq =−=
Upper portion of the specific energy diagramrepresents subcritical (tranquil or upperp q ppstage) flow.
Lower portion represents supercritical (rapidor lower stage) flow
Flow is critical at a point between upper andlower portions of the curve.
20/09/2011
38
The slope required to give uniform flow atcritical depth is known as critical slope:p p
So>Sc-steep slope; depth of flow<critical
3/1
2
cc y
gnS =
o c p p ; pdepth; supercritical flow.
So<Sc-mild slope; depth of flow>criticaldepth; subcritical flow.p ;
A steep slope for a channel with smoothlining could be a mild slope for the same flowwith rough lining.
Slope may be mild for a lower discharge andsteep for a higher one.
20/09/2011
39
Froude number
Hydraulic depth
hgyVF =
A is the x-sectional area.B is the width
BAyh /=
Irregular shaped section of area A and flow Q.
Specific energy eq. will be
Differentiating w.r.t ‘y’ and equate to zero for
2
2
2gAQyE +=
Differentiating w.r.t y and equate to zero for ‘y’ against minimum E
022
1 3
2
=⎜⎜⎝
⎛⎟⎟⎠
⎞−=
dydA
AgQ
dydE
20/09/2011
40
If dA=Bdy or dA/dy=B thenAQ ⎟
⎞⎜⎛ 32
substituting in above eq.cyyB
AgQ
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
ccVAQ =
cc AV= ;
2
( )chc
cc
c
ygBgA
V
Bg
==
= ;
For a rectangular channel
yBA
Result in the same relations as already developed.
ccc yBA =
20/09/2011
41
20/09/2011
42
